Solution Manual For Estimating in Building Construction, 10th Edition by Steven J. Peterson Frank R.

Solution Manual For To Accompany: Estimating In Building Construction Tenth Edition

Steven J. Peterson Professor Emeritus Weber State University

Frank R. Dagostino (deceased)

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Table of Contents Welcome

iv

New to the 10th Edition

v

Outcomes

vii

Instructor Resources

ix

Chapter 1—Introduction to Estimating

1

Chapter 2—Contracts, Bonds, and Insurance

6

Chapter 3—Project Manual

11

Chapter 4—The Estimate

15

Chapter 5—Computers in Estimating

19

Chapter 6—Overhead and Contingencies

22

Chapter 7—Labor

25

Chapter 8—Equipment

39

Chapter 9—Specialty Contractors

46

Chapter 10—Excavation

48

Chapter 11—Concrete

78

Chapter 12—Masonry

106

Chapter 13—Metals

118

Chapter 14—Wood

124

Chapter 15—Thermal and Moisture Protection

151

Chapter 16—Doors and Windows

166

Chapter 17—Finishes

172

Chapter 18—Electrical

196

Chapter 19—Plumbing

201

Chapter 20—Heating, Ventilating, and Air Conditioning

207

Chapter 21—Profit

212

Chapter 22—Other Estimating Methods

215

Appendix F—BidSim

222

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New to the 10th Edition This edition has been extensively edited to improve its readability. The general outline of the book remains the same, and much of the information remains unchanged; however, most of the paragraphs have been rewritten to improve the flow and make it a faster, easier read for students. Hopefully, this will make reading this book a more enjoyable experience. The following is a summary of the fundamental changes that have been made to this edition: General 

Learning outcomes have been added to each chapter.

Chapter 4 

Rewrote Section 4-14 Errors and Omissions.

Chapter 5 

Updated Section 5-6 Building Information Modeling to Autodesk Revit® 2023.



Added Section 5-7 Software Selection.

Chapter 6 

Updated Example 6-1 so that the allocation of home office costs is now based on construction costs rather than revenues, allowing it to be used as a markup to the bid. In the previous edition, the allocation of the home office costs was calculated as a percentage of revenues, which required iteration to determine the home office costs to allocate a project. Every time the home office costs were added, the revenues increased, decreasing the home office costs as a percentage of the bid. By using the construction costs, this percentage can be added to the construction costs without iteration. The relationship between these two is as follows:

 Percentage of revenues  Markup     1  Percentage of revenues  

Rewrote Section 6-4 Scheduling.

Chapter 7 

Updated the payroll taxes in Section 7-6 Labor Burden.

Chapter 10 

Rewrote the cross-section method in Section 10-6 Earthwork—New Site Grades and Rough Grade.

Chapter 12 

Updated the standard sizes of bricks in Figure 12.11. Figures 12.6 and 12.13, Example 12-2, and Problems 17 and 18 have been updated to reflect this change. iv Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14 

Updated the stud length to 92 5 / 8 inches in Examples 14-11 to 14-14 to reflect current practices.



Added an example, Example 14-24, covering the estimating of lookouts.



Updated Figures 14.9 and 14.29 to be consistent with Figure 14.16.

Chapter 15 

Updated Figure 15.8 to be consistent with Figure 14.16.

Chapter 17 

Updated Examples 17-3, 17-4, 17-7, and 17-8 to include the revisions from Figure 14.9.



Updated Figure 17.12 to be consistent with Figure 14.16.

Chapter 18 

Added two figures to Chapter 18.

Chapter 19 

Added two figures to Chapter 19.

Appendices 

Redrew the drawings in Appendices A, B, C, and D and the associated figures throughout the book to improve clarity and correct drawing errors. This change affects the solutions to the related examples and practice problems in Chapters 11 through 17.



Revised many of these files in Appendix J to enhance their functionality and correct errors.

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Outcomes Many higher education institutions use outcome-based learning, which requires them to measure their students’ ability to meet the required outcomes. These outcomes are often based on accreditation standards. Currently, in the United States, there are four accreditation standards for construction management and construction engineering programs, which are as follows: American Council for Construction Education (A CCE); ABET—Engineering Accreditation Commission, for construction engineering; ABET— Engineering Technology, for construction engineering technology; and ABET—Applied Science, for construction management. Although each of these standards is different, they all focus on three general outcomes, which can be summarized as follows.

Construction management/engineering students should be able to: 

Prepare construction cost estimates. This book includes multiple sets of plans that may be used to prepare an estimate. Each chapter, where applicable, provides homework problems related to these plans. By completing the following problems, the students will complete an estimate for the Real Estate Office: o Chapter 6 Problems 12 and 13 o Chapter 7 Problems 31 and 32 o Chapter 9 Problem 7 o Chapter 10 Problem 48 o Chapter 11 Problem 64 o Chapter 12 Problem 22 o Chapter 14 Problems 45 and 46 o Chapter 15 Problems 29 through 31 o Chapter 16 Problems 18 and 19 o Chapter 17 Problems 33 through 37 o Chapter 18 Problem 9 o Chapter 19 Problem 9 o Chapter 20 Problem 6



Effectively communicate in writing. Chapter 9 Problem 7 requires the student to write scopes of work for the Real Estate Office in Appendix E that clearly communicate the work to be performed by the specialty contractor. The following problems required the student to prepare reflective written responses: o Chapter 1, Problems 11 and 12 o Chapter 2 Problem 14 o Chapter 3 Problem 13 o Chapter 4 Problems 9 and 10 vi Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

o Chapter 5 Problems 8 and 9 o Chapter 8 Problem 10 o Chapter 9 Problem 7 

Understand ethics as it relates to estimating: o Chapter 1, Problem 14

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Instructor Resources The instructor’s website contains the following resources: Chapter 1 – Introduction to Estimating The resources for this chapter include: 1. Problem 01-13.xlsx, a spreadsheet solution for Problem 13.

Chapter 10 – Excavation The resources for this chapter include: 1. Problem 10-09.xlsx, a spreadsheet solution for Problems 18 and 19.

Appendix A – Small Commercial Building The resources for this appendix include: 1. Appendix A - Small Commercial Building v10-1 BW.pdf, a black-and-white set of PDF drawings that can be used to prepare an estimate for the project. The drawings are also found on the Student Resource website.

2. Appendix A - Small Commercial Building v10-1 Color.pdf, a color set of P DF drawings that can be used to prepare an estimate for the project. The drawings are also found on the Student Resource website.

3. Appendix A - Small Commercial Building v10 Estimate.xlsx, a spreadsheet solution for the project. This spreadsheet contains solutions for many of the example problems.

Appendix B – Residential Building Project The resources for this appendix include: 1. Appendix B - Residential Building Project v10 B W.pdf, a black-and-white set of PDF drawings that can be used to prepare an estimate for the project. The drawings are also found on the Student Resource website.

2. Appendix B - Residential Building Project v10 Color.p df, a color set of PDF drawings that can be used to prepare an estimate for the project. The drawings are also found on the Student Resource website.

3. Appendix B - Residential Building Project v10 Estimate.xlsx, a spreadsheet solution for the project. This spreadsheet contains solutions for the following problems: 1. Chapter 12 Problem 19 (see the Masonry tab) 2. Chapter 14 Problem 41 (see the Floor tab) viii Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

3. Chapter 14 Problem 42 (see the Walls 1, Walls 2, Walls 3, Walls 4, and Walls 5 tabs) 4. Chapter 14 Problem 43 (see the Ceiling tab) 5. Chapter 14 Problem 44 (see the Roof 1, Roof 2, Roof 3, and Roof 4 tabs) Appendix C – Commercial Building Project The resources for this appendix include: 1. Appendix C - Commercial Building Project v10 BW.p df, a black-and-white set of PDF drawings that can be used to prepare an estimate for the project. The drawings are also found on the Student Resource website.

2. Appendix C - Commercial Building Project v10 Color.p df, a color set of PDF drawings that can be used to prepare an estimate for the project. The drawings are also found on the Student Resource website.

3. Appendix C - Commercial Building Project v10 Estimate.xlsx, a spreadsheet solution for the project. This spreadsheet contains solutions for the following problems: 1. Chapter 10 Problem 45 (see the Excavation tab) 2. Chapter 10 Problem 46 (see the Topsoil tab) 3. Chapter 10 Problem 47 (see the Pavement tab) 4. Chapter 11 Problem 61 (see the Concrete 1 and Concrete 2 tabs) 5. Chapter 12 Problem 20 (see the Masonry 1 and Masonry 2 tabs) 6. Chapter 13 Problem 14 (see the Steel tab) 7. Chapter 15 Problem 25 (see the Damp proofing section on the Thermal tab) 8. Chapter 15 Problem 26 (see the Insulation section on the Thermal tab) 9. Chapter 15 Problem 27 (see the Roofing section on the Thermal tab) 10. Chapter 15 Problem 28 (see the Sheet Metal section on the Thermal tab) 11. Chapter 16 Problem 16 (see the Windows and Curtain Walls section on the Openings tab) 12. Chapter 16 Problem 17 (see the Metal Doors, Wood Doors, and Hardware sections on the Openings tab) 13. Chapter 17 Problem 30 (see Drywall 1 and Drywall 2 tabs) 14. Chapter 17 Problem 31 (see the Ceramic Tile and V CT sections of the Finishes 1 tab and the Carpet section of the Finishes 2 tab) 15. Chapter 17 Problem 32 (see the Paint section of the Finishes 2 tab) Appendix D – Billy’s C-Store ix Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

The resources for this appendix include: 1. Appendix D - Billys C-Store v10 BW.pdf, a black-and-white set of PDF drawings that can be used to prepare an estimate for the project. The drawings are also found on the Student Resource website.

2. Appendix D - Billys C-Store v10 Color.pdf, a color set of PDF drawings that can be used to prepare an estimate for the project. The drawings are also found on the Student Resource website.

3. Appendix D - Billys C-Store v10 Estimate.xlsx, a spreadsheet solution for the project. This spreadsheet contains solutions for the following problems: 1. Chapter 11 Problem 62 (see the Footings tab) 2. Chapter 11 Problem 63 (see the S OG 1 and SOG 2 tabs) 3. Chapter 12 Problem 21 (see the Masonry 1 and Masonry 2 tabs) 4. Chapter 13 Problem 15 (see the Steel tab) Appendix E – Real Estate Office The resources for this appendix include: 1. Appendix E - Real Estate Office 2012.r vt, a Revit drawing file that can be used to prepare an estimate for the project. This file is also found on the Student Resource website. The students will need Autodesk Revit 2012 or later to open the file.

2. Appendix E - Real Estate Office v10 BW.pdf, a black-and-white set of PDF drawings that can be used to prepare an estimate for the project. The drawings are also found on the Student Resource website.

3. Appendix E - Real Estate Office v10 Estimate.xlsm, a spreadsheet solution for the project. This spreadsheet contains solutions for the following problems: o Chapter 6 Problem 13 (see Division 1 – Overhead on the Detailed Cost Breakdown tab) o Chapter 7 Problem 31 (see the Labor Cost tab) o Chapter 7 Problem 32 (see the Crews tab) o Chapter 10 Problem 48 (see Division 31 – Grading and Excavation on the Detailed Cost Breakdown tab) o Chapter 11 Problem 64 (see Division 3 – Concrete on the Detailed Cost Breakdown tab) o Chapter 12 Problem 22 (see Division 4 – Masonry, the Detailed Cost Breakdown tab) o Chapter 14 Problem 45 (see Division 6 – Wall Framing on the Detailed Cost Breakdown tab)

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o Chapter 14 Problem 46 (see Division 6 – Roof Framing on the Detailed Cost Breakdown tab) o Chapter 15 Problem 29 (see Division 7 – Insulation on the Detailed Cost Breakdown tab) o Chapter 15 Problem 30 (see Division 7 – Roofing the Detailed Cost Breakdown tab) o Chapter 15 Problem 31 (see Division 7 – Soffit, Fascia, and Rain Gutters on the Detailed Cost Breakdown tab) o Chapter 16 Problem 18 (see Division 8 – Windows and Storefronts on the Detailed Cost Breakdown tab) o Chapter 16 Problem 19 (see Division 8 – Doors and Hardware on the Detailed Cost Breakdown tab) o Chapter 17 Problem 33 (see Division 9 – Drywall on the Detailed Cost Breakdown tab) o Chapter 17 Problem 34 (see Division 9 – Flooring on the Detailed Cost Breakdown tab) o Chapter 17 Problem 35 (see Division 9 – Paint on the Detailed Cost Breakdown tab) o Chapter 17 Problem 36 (see Division 9 – Ceramic Tile on the Detailed Cost Breakdown tab) o Chapter 17 Problem 37 (see Division 9 – Acoustical Ceilings on the Detailed Cost Breakdown tab) o Chapter 18 Problem 9 (see Division 26 – Electrical on the Detailed Cost Breakdown tab) o Chapter 19 Problem 9 (see Division 22 – Plumbing on the Detailed Cost Breakdown tab) o Chapter 20 Problem 6 (see Division 23 – HVAC on the Detailed Cost Breakdown tab) Appendix F – Bid Sim The resources for this appendix include: 1. BidSim Bid Form.docx, a bid form that the students use to submit their bid.

2. BidSim All Bids.pdf, which includes the supplier and subcontractor bids used in the bid simulation game.

3. Web Files folder that contains the web code, the individual supplier and subcontractor bids, and the background picture for the web-based simulation.

The instructions for the BidSim are found at the end of this guide. xi Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Appendix J – Spreadsheets The resources for this appendix include the following spreadsheets: 1. Bid-Tab.xls x 2. Cut-Fill.xls x 3. Est-Summary.xls x 4. Rebar.xls x 5. Recap.xlsx 6. Slab-Rebar.xls x 7. Steel.xlsx 8. Warehouse.xlsx 9. Work-Sheet.xlsx PowerPoints Slides This folder contains the PowerPoint Slides for the individual chapters.

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Chapter 1—Introduction to Estimating Learning Outcomes This chapter provides an overview of estimating and the estimator’s role in preparing the estimate. After reading this chapter, the reader should be able to: Learning Outcome 1.1: Define estimating and its role in the construction industry.

Learning Outcome 1.2: Describe the types of estimates and when they might be used.

Learning Outcome 1.3: Describe the job opportunities for estimators.

Learning Outcome 1.4: Identify the skills needed to be an estimator.

Learning Outcome 1.5: Define a quantity survey and explain its role in estimating.

Learning Outcome 1.6: Describe the different types of bids used in the construction industry.

Learning Outcome 1.7: Describe the components of the contract documents and their functions.

Learning Outcome 1.8: Describe the sources of estimating information.

Learning Outcome 1.9: Define ethics as it relates to estimating.

Activities 1. Invite an estimator from the industry to discuss their job with the students. Have them answer the following questions: 

How do they go about estimating a project?



What are the different types of estimating methods they use?



What role does estimating play in the success of the company?



What does it take to be a successful estimator?

Please encourage them to tell estimating-related stories.

2. Discuss Problems 11, 12, and 14 from the chapter.

Instructional Resources PowerPoint Slides: Chapter 01.ppt Solutions to the Review Questions 1. What information is contained in the working drawings?

The plans (drawings, illustrations) from which the project will be built. They contain the dimensions and locations of building elements and materials required, delineating how they fit together.

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2. What information is contained in the technical specifications?

The technical specifications are written instructions describing the material’s quality and performance. The technical specifications supplement the information on the working drawings.

3. What is the relationship between the working drawings and the technical specifications?

The working drawings contain the dimensions and locations of the building elements and materials required, delineating how they fit together. While the technical specifications are written instructions describing the material quality and performance, the technical specifications supplement the information on the working drawings.

4. How does the work involved in being an estimator for a general contractor differ from that of an estimator who works for a specialty contractor?

The estimator for the general contractor is responsible for a detailed estimate for the whole project. They must compile costs on every project component and assemble a comprehensive bid. An estimator for a subcontractor will prepare an estimate only for the work they will perform. For example, the masonry specialty contractor will prepare a bid for the masonry on the project.

5. What is the difference between doing a quantity survey and a complete detailed estimate?

A quantity survey or quantity takeoff (QTO) estimates the amount of in-place materials required to construct a project. A complete detailed estimate covers everything needed for the project’s construction and includes costs and quantities for materials, labor, equipment, and subcontractor costs.

6. What additional skills must the estimator have to be able to take a quantity survey and turn it into a detailed estimate? a. By looking at the drawings, visualize the project through its various construction phases. b. Have enough construction experience to possess a good knowledge of job conditions, including methods of handling materials on the job, the most economical methods of construction, and labor productivity. c. Have sufficient knowledge of labor operations and productivity to convert them into costs on a project. d. Be able to keep a database of information on costs of all kinds, including labor, material, project overhead, and equipment, and a knowledge of the availability of all the required items. e. Be computer literate, know how to manipulate and build various databases, and use spreadsheet programs and other estimating software. f. Be able to meet bid deadlines and remain calm. g. Have good writing and presentation skills.

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h. Be able, from looking at the drawings, to visualize the project through its various phases of construction.

7. What is the difference between competitive and negotiated bidding?

Competitive bidding involves each contractor submitting a lump-sum bid or a proposal in competition with other contractors to build the project. The project may be awarded based on the price or the proposal that provides the best value to the owner.

In negotiated work, the parties arrive at a mutually agreed-upon price, terms and conditions, and contractual relationship. This arrangement often entails negotiations back and forth on virtually all aspects of the project, such as materials used, sizes, finishes, and other items that affect the project’s price.

8. What is the difference between a detailed estimate and a square-foot estimate?

The detailed estimate includes determining the quantities and costs of everything required to complete the project. This includes materials, labor, equipment, insurance, bonds, and overhead, as well as an estimate of the profit.

Square-foot estimates are prepared by multiplying the square footage of a building by a cost per square foot and then adjusting the price to compensate for differences in the building heights, length of the building perimeters, and other building components.

Square-foot estimates require less information to prepare and are less accurate.

9. What are the contract documents, and why are they so important?

The contract documents include the invitation to bid, instructions to bidders, bid form, owner-contractor agreement, general conditions, supplementary general conditions, technical specifications, and the working drawings, including all addenda incorporated in the documents before the contract’s execution. All these documents become part of the contract.

10. Why is it important to bid only from a complete set of contract documents?

The contractor must have a complete set of contract documents to prepare a detailed estimate. If part of the documents is missing, that portion of the project would be left out of the bid.

11. For this assignment, you will explore the role estimating plays in the construction industry by interviewing a person whose job duties include estimating. Begin by setting up an interview with an estimator, project manager, project engineer, superintendent, foreperson, architect, engineer, construction material salesperson, or freelance estimator. During the interview, ask the person the following questions and follow-up questions as necessary. Be respectful of their time, and limit your interview to 20 minutes unless the person offers to extend the interview. Be sure to thank the person before you leave, and send them a thank-you note within 48 hours of the interview. After the interview, prepare written responses to the following questions and be prepared to discuss your findings in class if your instructor chooses to do so: 3 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

a. What are the estimates used for (ordering materials, preliminary budgets, etc.)? b. At what stage of the construction process (early-design, late-design, bidding, construction, etc.) does the estimate occur? c. What are the consequences if the estimate is slightly wrong? If it is very wrong? d. How do they prepare an estimate? After the interview, decide which estimating method (detailed, assembly, square-foot, parametric, model, or project comparison) best describes the type of estimates they prepare. e. How long does it take to prepare an estimate? f. What skills are required to become a good estimator? g. What experience is required to get a job like theirs?

The answers to these questions will vary from interviewee to interviewee.

12. Review a copy of a construction project's contract documents (drawings and project manual). Contract documents may be reviewed at a contractor’s, specialty contractor’s, architect’s, or engineer’s office or downloaded from the Internet. Write a summary of how the contract documents are organized. Be sure to discuss both the project manual and the drawings. Be prepared to discuss your findings in class if your instructor chooses to do so.

The answers to this question will vary from project to project.

13. Using the Warehouse.xlsx Excel file found on the companion website for this text, determine the estimated cost of a warehouse with the following parameters: Building length—210 feet Number of bays on the length side of the building—7 each Building width—120 feet Number of bays on the width of the building—4 each Wall height above grade—22 feet Depth to top of footing—12 inches Floor slab—6 inches thick with wire mesh Number of roof hatches—2 each Number of personnel doors—4 each Number of 14-foot-wide by 14-foot-high overhead doors—14 each Number of 4-foot by 4-foot skylights—28 each Fire sprinklers are not required.

Two bathrooms are required.

The estimated cost is $1,137,495. See Problem 01-13.xls x, which is available on the instructor’s website.

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14. Using the American Society of Professional Estimators (A SPE) Code of Ethics (www.aspenational.org/resource/resmgr/bylaws/2018/2018_code_of_ethics.pdf), describe under what conditions the following situations would be considered ethical or unethical: a. Sharing a supplier’s or subcontractor’s bid with their competitors b. Going to lunch with a supplier c. Attending a sporting event with a subcontractor d. Receiving free travel for purchasing a certain amount from a supplier e. Choosing a subcontractor other than the low bidder f. Asking a subcontractor to lower their bid g. Using company equipment to complete an improvement project on your home h. Using company equipment to do side jobs on the weekend The responses will be different for each student.

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Chapter 2—Contracts, Bonds, and Insurance Learning Outcomes This chapter examines the role of contracts, bonds, and insurance in estimating. After reading this chapter, the reader should be able to: Learning Outcome 2.1: Describe the types of construction agreements.

Learning Outcome 2.2: Describe the components of a construction agreement.

Learning Outcome 2.3: Describe the types of bonds used in the construction industry and how to obtain a bond.

Learning Outcome 2.4: Describe the types of insurance used in the construction industry.

Activities 1. Invite an insurance agent to the class to discuss the types of insurance required on a construction project. Have them answer the following questions: 

What types of insurance are needed for a construction project, and what kinds of losses do they cover?



How do you go about getting a project insured?



What steps can you take to reduce insurance costs?

2. Invite a bonding agent to the class to discuss construction bonds. Have them answer the following questions: 

How does one go about establishing a bonding line?



What factors does the bonding company consider when establishing the maximum project size and the total work on hand for a contractor?



How can a contractor increase the bonding limits for the company?

3. Obtain two different subcontracts. They may be obtained from local contractors or downloaded from the internet. Divide the students into groups and have them compare the two contracts and answer the following questions: 

Which contract is better? Why is it better?



Are there clauses included in the contract you did not choose that you would like to have added to the contract you chose? What are they? Why would you like to add them?



Are there clauses in the contract you chose that you would like to delete? What are they? Why would you like to delete them?

Have the groups present their responses or discuss their answers in class. You may want the groups to take different perspectives, such as the general contractor’s, subcontractor’s, architect’s/engineer’s, or owner’s perspective.

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4. Discuss Problem 14 from the chapter. You may have the students obtain a set of construction documents from a contractor, or you may provide them with a set of construction documents. Ideally, the students should all have different sets.

Instructional Resources PowerPoint Slides: Chapter 02.p pt The surety information office (www.sio.org) has educational resources on bonding.

Solutions to the Review Questions 1. What is a single contract, and what are its principal advantages and disadvantages for the owner?

The single contract comprises all work required to complete a project and is the responsibility of a single prime contractor (often called the general contractor). Its benefits are: 

The general contractor’s responsibility is to plan, direct, and coordinate all parties involved in completing the project.



The specialty contractors and material suppliers involved in the project are responsible directly to the general contractor, who is responsible directly to the owner.



The general contractor must ensure that all work is completed in accordance with the contract documents, that the work is completed on time, and that all subcontractors and vendors have been paid.

2. What are separate contracts, and what are the principal advantages and disadvantages for the owner?

Under separate contracts, the owner signs separate agreements to construct various portions of a project. Its benefits are: 

The owner retains the opportunity to select the contractors for the various essential phases of the project.



The general contractor does not get a markup on the separate contracts.



The responsibility for the separate contracts is directly between the owner and contractors rather than through the general contractor.

Its disadvantages are: 

If one of the subcontractors cannot perform, the owner (rather than the general contractor) absorbs the added cost of finding a replacement and any associated delays.



The owner assumes responsibility for managing and coordinating the project and its associated risks.

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

The owner must hire someone if they do not have the talents or personnel to accomplish these tasks.

3. With separate contracts, describe three options available to the owner for managing the contractor’s work on the project.

The owner may pay the general contractor to do the coordination, pay the architect/engineer or a construction management company, or handle it themselves.

4. List and briefly define the types of agreements that may be used for the owner’s payment to the contractor.

Lump-Sum Agreement: The contractor agrees to construct the project in accordance with the contract documents for a set price arrived at through competitive bidding or negotiation.

Unit-Price Agreement: The contractor bases the bid on estimated quantities of work and the completion of the work in accordance with the contract documents.

Payments are made based on the contractor’s price for each unit of work and field measurements of the completed work.

Cost-Plus-Fee Agreement: The contractor is reimbursed for the construction costs as defined in the contract. However, the contractor is not reimbursed for all items.

5. What is the ―time of completion,‖ and why must it be clearly stated in the contract provisions?

The ―time of completion‖ determines the amount of time the contractor has to complete the project. The agreement should specify the starting and completion times. The completion date is expressed as the number of calendar or working days or a specific date. Often, there are bonuses for early completion and costs (liquidated damages) for late completion.

6. What are progress payments and why are they important to the contractor?

Progress payments are the periodic payments made by the owner to the contractor based on the completed work and stored materials. They are important to the contractor because they use this money to pay all expenses associated with the project. The timing and payment procedures should be detailed in the contract documents.

7. What is retainage, where is the amount specified, and why is it used?

Retention is a percentage of the payments from the owner to the contractor that is withheld as a protection for the owner. It ensures the contract’s completion and payment of the contractor’s financial obligations. The agreement (a contract between the owner and contractor) specifies the retainage requirements.

8. What is a bid bond, and how does it protect the owner?

The bid bond ensures that if a contractor is awarded the bid within the time specified, the contractor will enter into the contract and provide all other specified bonds. If the contractor fails to do so without justification, the bond will be forfeited to the owner.

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9. Where would information be found on whether a bid bond was required and, if so, its amount?

Bid bond requirements are stated in the advertisement of bidders and reiterated in the project manual.

10. What are performance bonds? Are they required on all proposals?

The performance bond guarantees to the owner that the contractor will perform all work in accordance with the contract documents and that the owner will receive the project built in substantial agreement with the contract documents. It protects the owner against the contractor's default up to the bond amount. Performance bonds are only required on some projects.

11. How are the various surety bonds required on a specific project obtained?

Contractors most commonly use a corporate surety that specializes in construction bonds. The contractor will have to request the bond from the surety company.

12. How does insurance differ from a surety bond?

With an insurance policy, the responsibility for specified losses is shouldered by the insurance company. In contrast, with a bond, the bonding companies will fulfill the bond’s obligations and turn to the contractor to reimburse them for all the money they expended on their behalf.

13. What types of insurance should a contractor maintain, and what do they cover?

Workers’ Compensation: A workers’ compensation insurance policy provides benefits to employees or their families if the employee is killed or injured on the job.

Builder’s Risk: Builder’s risk insurance protects projects under construction against direct losses due to fire and lightning. The policy may be extended to additional risks, including windstorms, hail, explosions, riots, civil commotion, vandalism, malicious mischief, earthquakes, and sprinkler leakage.

Commercial General Liability: Commercial general liability insurance covers liability arising from negligent actions by the contractor and its employees. It includes bodily injury, property damage or loss, and other personal injury such as slander or damage to reputation.

Automotive. Automotive insurance covers vehicles used on public roads, including cars, trucks, portable job site office trailers, and construction equipment that drives on public roads, such as dump trucks.

Marine. Marine insurance covers equipment used on waterways, such as barges and boats.

Inland Marine. Inland marine insurance covers off-road construction equipment, such as backhoes, scrapers, and dump trucks not licensed for use on public roads.

Property. Property insurance covers real property (real estate) the contractor owns, such as office buildings, shops, and warehouses.

Business Personal Property. Business personal property insurance covers the contents of a building, such as computers and furniture.

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Errors and Omissions. Errors and omissions insurance covers liability arising from errors or omissions by project designers.

Umbrella. An umbrella insurance policy goes on top of a company’s other insurance policies, increasing the coverage limits.

Life. Life insurance policy on key personnel (such as the owner) where the company is the beneficiary to protect the company against losses incurred due to the employee’s death.

14. Review a copy of the contract documents (drawings and project manual) for a construction project. Contract documents may be reviewed at a contractor’s, specialty contractor’s, architect’s, or engineer’s office or downloaded from the Internet. Answer the following questions and be prepared to discuss your findings in class if your instructor chooses to do so. a. What type of agreement (lump sum, unit price, or cost-plus-fee) is used for the project? If it is a cost plus fee agreement, how is the fee determined, and is there a guaranteed maximum price? b. What is the scope of the work for the project? c. What provisions are included in the contract documents regarding the time of completion? What penalties are there for failing to meet the completion date? Is there a bonus for completing the project ahead of schedule? d. How are progress payments handled? When are they due? How quickly will they be paid? e. Will retention be withheld? If so, how much? What are the requirements for the release of retention? f. How is final acceptance handled? What inspections are required? What forms, documents, maintenance and operation manuals, certifications, red-line drawings, and so on must be submitted before final acceptance? g. What bonds are needed for the job? h. What are the insurance requirements for the project?

The answers to this exercise will vary from project to project.

10 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 3: Project Manual

Chapter 3—Project Manual Learning Outcomes This chapter examines the role of the project manual in estimating. After reading this chapter, the reader should be able to: Learning Outcome 3.1: Describe the contents of the project manual and their uses.

Learning Outcome 3.2: Describe the CSI MasterFormat and how it is used.

Learning Outcome 3.3: Explain what an addendum is and how it is used.

Learning Outcome 3.4: Describe how to handle errors in the specifications.

Activities 1. Invite a specification writer or architect to the class to discuss how the project manual is prepared. Have them answer the following questions: 

How are the specifications for the different types of materials chosen?



Why is it important to read and understand the project manual?



What happens when a contractor does not follow the requirements in the Instructions to the Bidders?

2. Discuss Problem 13 from the chapter. You may have the students obtain a set of construction documents from a contractor, or you may provide them with a set of construction documents. Ideally, the students should all have different sets.

Instructional Resources PowerPoint Slides: Chapter 03.ppt Solutions to the Review Questions 1. What types of information are found in the project manual?

1. Invitation to bid (advertisement for bidders) 2. Instructions to bidders 3. Bid (or proposal) forms 4. Form of owner-contractor agreement 5. Form of bid bond 6. Forms of performance and payment bonds 7. General conditions of the contract 8. Supplementary general conditions 9. Specifications (technical specifications) 11 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 3: Project Manual

2. Why is it important for the estimator to carefully review the entire project manual?

The contractor is responsible for everything in the project manual and the drawings, including anything stated in the proposal (or bid form), the information to bidders, the general conditions, and the supplementary general conditions. Many essential items are stated only in the project manual and not on the drawings, ranging from the types of bonds and insurance required to the type, quality, and color of materials used on the job.

3. Describe the CSI MasterFormat and how it is used.

The CSI MasterFormat is a numbering system that divides the major areas involved in the construction project into 50 divisions (0 through 49), with many of the divisions reserved for future use. Each division is subdivided into specific areas. The CSI MasterFormat ties in easily with computer programs and cost accounting systems.

4. What types of information are provided in the invitation to bid?

Generally, the advertisement (invitation to bid) describes the work’s location, extent, and nature. It designates the authority under which the project is being built. It identifies how to obtain the bidding documents and lists the time, manner, and place where bids will be received. It also lists the bond requirements and the start and completion dates for the work.

5. Why is it important that the bids be delivered at the proper time and place?

The bid must be submitted according to the requirements in the instructions to bidders. Any deviation from these requirements may result in the bid being rejected.

Contractors must submit the bid at the right place at the right time. Countless bids have never been opened because they were delivered a few minutes late.

6. How do the supplementary general conditions differ from the general conditions?

The supplementary general conditions amend or supplement portions of the general conditions. It is through these supplemental conditions that the general conditions are tailored to the local requirements and individual project needs. Part of the supplemental conditions cancels or amends the articles in the general conditions, while the remaining portion adds provisions.

7. What information is contained in the project manual’s specifications (technical specifications)?

The specifications (sometimes called technical specifications) include the type of materials required, their required performance, and the method used to obtain the specified result. The material portion of the specifications usually mentions the material’s physical properties, performance requirements, handling, and storage.

8. Explain what an alternate is and how it is handled during the bidding process.

An alternate is a price for alternate methods or materials of construction requested by the owner that adds or deducts the price to the base bid should the alternative be chosen. When an owner has a limited budget, the alternates allow them to choose how to best spend the available money.

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Chapter 3: Project Manual

9. Why is it important to prepare the alternate pricing carefully and thoroughly?

The alternates are added to and subtracted from the base bid to determine the contract amount. The price for any alternates must be complete and include all taxes, overhead, and profit. Alternates deserve the same estimating care and consideration as the rest of the project.

10. Explain what an addendum is and when it is used.

Any amendments, modifications, revisions, corrections, and explanations issued by the architect or engineer during the bidding period are effected by issuing an addendum. The statements, and any drawings included, serve to revise the contract documents. They notify the bidder of document corrections, interpretations, additional requirements, and other similar matters.

11. Why is it important that estimators make sure that they have received all addenda before submitting a bid?

The addenda become part of the contract documents, making amendments, modifications, revisions, corrections, and explanations to the original documents.

Without the addenda, the bid may be incomplete. The addenda may affect specialty contractors’ and material suppliers’ bids, so the contractors should verify that the addenda have been received and incorporated into their bids.

12. How should the estimator handle any errors or omissions found in the contract documents?

If the error is discovered early in the bidding period and no immediate answer is needed, the error should be kept on a list of errors and omissions. The architect or engineer should not be contacted about each issue separately, but the problems should be addressed in groups. Most specifications require that all requests for interpretations be made in writing and state how many days before the bid date they must be received.

13. Review a copy of a construction project's contract documents (drawings and project manual). Contact documents may be reviewed at a contractor’s, specialty contractor’s, architect’s, or engineer’s office or downloaded from the Internet.

Answer the following questions and be prepared to discuss your findings in class if your instructor chooses to do so. a. How can a contractor obtain a set of contract documents? b. Are the contract documents on file at plan rooms, trade associations, or government agencies? If so, where are they on file? c. When and where is the bid due? d. What needs to be submitted for the bid to be complete? e. Do the contract documents use a standard contract prepared by the American Institute of Architects, a trade association, or a government agency? If so, who prepared the contract?

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Chapter 3: Project Manual

f. Do the contract documents use a standard set of general conditions prepared by the American Institute of Architects, a trade association, or a government agency? If so, who prepared the general conditions? g. Do the contract documents contain supplementary general conditions? If so, how do they modify the general conditions? h. What alternatives need to be included in the bid? i. Have any addenda been issued for the project? If so, how do they modify or clarify the contract documents? j. What allowances, if any, must be included in the bid? k. Select one section from the technical specifications and answer the following questions. Be sure to select a section that deals with specific components of the building (e.g., doors, concrete, and masonry). i.

What scope of work does this section cover? What is specifically included?

What is expressly excluded?

ii.

What submittals, if any, are required under this section?

iii.

How does this section address quality?

iv.

What materials are required to complete the work under this section?

v.

What installation procedures are required by this section?

vi.

What warranties are required for this section?

vii.

What else is contained in this section?

viii.

What did you find interesting about this section?

The answers to this exercise will vary from project to project.

14 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 4: The Estimate

Chapter 4—The Estimate Learning Outcomes This chapter examines the role of the estimate in the construction industry and the process used to prepare an estimate. After reading this chapter, the reader should be able to: Learning Outcome 4.1: Describe the estimator’s role in a construction company.

Learning Outcome 4.2: Describe the process and keys to preparing an estimate.

Learning Outcome 4.3: Describe the factors to consider when deciding whether to bid on a project.

Learning Outcome 4.4: Describe the use of workup and summary sheets.

Learning Outcome 4.5: Describe the common errors made in an estimate and how to avoid them.

Activities 1. Invite a representative of the local plan room to class to discuss what services the plan room offers. Have them answer the following questions: 

How do plan rooms benefit the construction community?



In addition to providing access to contract documents, what services do they provide?



How do plan rooms help the owners of buildings?



What are the benefits of being a member of the plan room?



What does it cost to be a member?

2. Review Figure 4-1: This figure is a visual guide that students can use to trace their way through the estimating process. In addition, it shows the relationships between the estimating data. For example, the number of labor hours affects the duration of the task and the project’s duration, which in turn affects the indirect field costs.

3. Review the spreadsheets from the companion website. This website contains several spreadsheets for specific aspects of the quantity takeoff and pricing. These spreadsheets were designed for this text and may not perform as desired on other projects.

4. Discuss Problems 9 and 10 from the chapter.

Instructional Resources PowerPoint Slides: Chapter 04.ppt Solutions to the Review Questions 15 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 4: The Estimate

1. Why should the estimate be documented? What documentation should be included?

The estimator should document each estimate so they have a record of what was done, what numbers were used, and all the history of the estimate in one place.

2. Why might a contractor decide not to bid on a particular project?

The reasons for not bidding on a project include the following: 

We are unlikely to win the job.



We cannot make a profit on this job.



We do not have the cash to fund the job.



We do not have the bonding capacity.



We do not have the expertise to complete this job successfully.



We lack the needed specialty contractors and material suppliers.



The architect or engineer is hard to work with.



The construction documents are incomplete.



The owners do not have the funding to pay for the project.



The owners do not pay on time.



The owners are hard to work with.



We do not have the needed trade and management personnel.



We do not have the needed equipment.

3. Why should estimators check carefully to be sure that they have all the contract documents before bidding on a project?

The estimator must clearly understand the scope of work that needs to be performed to prepare a complete estimate for the construction project. This understanding comes from the construction documents (plans and project manual). It is impossible to accurately estimate the project cost if the estimator does not have a complete set of contract documents.

4. The contract documents often require the contractor to visit the site. Why is the site investigation important?

From the site visit, estimators learn important aspects of the project. The importance of the visit and the items to be checked vary depending on the type of project and its location. Examples of the kinds of information that should be collected during the site visit include the following: 1. Site access 2. Availability of utilities (electric, water, internet) 3. Site drainage 4. Transportation facilities 16 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 4: The Estimate

5. Any required protection or underpinning of adjacent property 6. A rough layout of the site, locating the proposed office trailer, storage trailers, and equipment locations 7. Subsurface soil conditions 5. Why is it important for a contractor to become familiar with the surrounding area when bidding on a project in a new and unfamiliar area?

To identify the following information: 1. Local ordinances and regulations, noting any special requirements (permits, licenses, barricades, fences) 2. The local labor situation and union rules 3. Availability of construction equipment rentals, including the type, condition, and cost 4. Prices and delivery information from local material suppliers 5. The availability of local specialty contractors (names, addresses, and type of work) 6. The conditions of the roads leading to the project (low bridges and load limits on roads or bridges) 7. Housing and feeding facilities, if workers must be imported 8. Banking facilities 6. Explain what a specialty contractor is and their relationship with the owner and the general contractor.

A specialty contractor is a separate contractor most commonly hired by the general contractor to perform specific work packages and becomes a subcontractor to the general contractor. As such, their relationship with the owner is through the general contractor.

7. Why should material quotes be in writing? What items must be checked in these quotes?

It is best to obtain written quotations that spell out the exact terms of the freight, taxes, delivery time, materials included in the price, how long the quote is good for, and the payment terms. The written quotations should be checked against the specifications to ensure that the specified materials were bid.

8. What is the difference between workup sheets and summary sheets?

Workup sheets include a variety of forms customized to what is being quantified and may be used for material, labor, equipment, or a combination of these. Summary sheets are condensed versions of the workup sheets, where all costs contained on the workup sheets are condensed and totaled.

9. Review a copy of the contract documents (drawings and project manual) for a construction project. Contract documents may be reviewed at a contractor’s, 17 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 4: The Estimate

specialty contractor’s, architect’s, or engineer’s office or downloaded from the Internet. Discuss the pros and cons of bidding on the project. Be prepared to present your findings in class if your instructor chooses to do so.

The pros and cons will vary from project to project.

10. You are working as an estimator for a small general contractor. The company has its own finish carpentry, framing, and concrete crews. The finish carpentry crew can perform all types of finish carpentry. The framing and concrete crews can perform typical framing (both wood and metal studs) and typical concrete work, respectively.

The framing and concrete crews cannot perform complex or specialized framing or concrete work. Review a copy of the contract documents (drawings and project manual) for a construction project and answer the following questions. Contract documents may be reviewed at a contractor’s, specialty contractor’s, architect’s, or engineer’s office or downloaded from the Internet. Be prepared to present your findings in class if your instructor chooses to do so. a. What part of the work would your company’s crews perform? How many hours do you think it will take you to bid on this work? b. From what material suppliers will you need prices to complete the bid? c. From what specialty contractors will you need prices to complete the bid? d. What else do you need to complete the bid? e. How many days will it take to prepare the bid, including getting material quotes and specialty contractor bids?

Each response will be different. Emphasis should be placed on preparing an organized approach to the estimate and anticipating any problems.

11. Identify one potential supplier for each supplier identified in Problem 10b.

The list will be different for each student.

12. Identify one potential specialty contractor for each of the specialty contractors identified in Problem 10c.

The list will be different for each student.

18 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 5: Computers in Estimating

Chapter 5—Computers in Estimating Learning Outcomes This chapter looks at the use of computers in the estimating process. After reading this chapter, the reader should be able to: Learning Outcome 5.1: Describe why computers are essential to the estimating process.

Learning Outcome 5.2: Describe the advantages and disadvantages of using computers to prepare an estimate.

Learning Outcome 5.3: Describe how spreadsheets are used in estimating.

Learning Outcome 5.4: Describe the role of specialized estimating software in estimating.

Learning Outcome 5.5: Describe the role of takeoff software in estimating.

Learning Outcome 5.6: Describe how BIM can be integrated into estimating.

Learning Outcome 5.7: Determine the appropriate estimating software for a company.

Activities 1. Invite a representative of an estimating software package to class to demonstrate the company’s software package. Have them answer the following questions: 

What are the advantages of the software?



What are the disadvantages of the software?



What are the costs of the software?



How much effort is needed to set up and maintain the software?

2. Teach the students to use WinEst, On-Screen Takeoffs, Revit, or another software program used in estimating. This is a substantial time commitment and should not replace teaching the students the fundamentals of estimating.

3. Discuss Problems 8 and 9 from the chapter.

Instructional Resources PowerPoint Slides: Chapter 05.ppt Solutions to the Review Questions 1. What are the benefits of computerized estimating?

1. It can reduce calculation errors, providing a more accurate estimate.

2. It increases the speed at which the estimate is prepared by performing the math for the estimator.

19 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 5: Computers in Estimating

3. Many computerized estimating packages allow the estimator to track where the quantities were found on the drawings.

4. It allows the estimator to quickly change a price or a productivity rate and get an instantaneous change in the project’s cost.

2. What two dangers must be avoided when using computerized estimating?

1. Turn the thinking over to the computer, making the estimator nothing more than a data-entry person.

2. Using the software for a project that it was not designed to estimate.

3. What are the four categories of estimating software? How are they used in estimating?

1. Spreadsheet. They are used to prepare entire estimates and augment specialized software packages.

2. Specialized software packages. They are used to prepare entire estimates.

3. Takeoff packages. They are used to determine the lengths, areas, and volumes of the different components in a building project.

4. Building information models (B IM). They are used to determine the lengths, areas, and volumes of the different components in a building project.

4. What steps are used to create a set of construction drawings from a building information model? What happens in each step?

1. Preparing a 3D Model: The user creates and interacts with a 3D model using several 2D views of the model.

2. Adding Details to the Model: The designer adds this information to the details by drawing them on the views.

3. Creating the Construction Drawings: The 2D views of the model and the details are placed on the drawing sheets, creating a set of traditional construction drawings.

5. What part of the building information model can quantities be extracted from?

Only the components created in the 3D model. Detailed information added in the drafting views cannot be incorporated into the quantities generated by the model.

6. What are the limitations and potential problems of using a building information model to prepare an estimate?

1. Incomplete Models: Only components that have been accurately modeled can be used to extract accurate quantities.

2. Conflicts between the Model and the Construction Drawings: Although the model is used to create the contract document, information may be added, deleted, or changed as the model is used to create the construction drawings. In many cases, the model itself does not become part of the contract documents, and as such, if the contractor chooses to use the model to prepare the estimate, they do so at their own risk.

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Chapter 5: Computers in Estimating

3. Models Do Not Include Means and Methods: The model does not address the means and methods used to construct the building. Additionally, the model often does not consider the difficulty or problems with constructing a component.

4. Models Do Not Address Waste: The model does not consider waste inherent in the design. The estimator should consider inherent waste when preparing the quantities.

7. How might a building information model be used in estimating?

1. Conceptual and Preliminary Estimates: The model may be used to prepare conceptual or preliminary estimates because they are based on designs with very little detail, making it easy to completely model the conceptual design.

2. Checking the Estimate: The quantities from the model may be used to check the estimate prepared from the construction drawings, allowing the estimator to check their quantities and look for missing items.

3. Preparing Estimates: All or part of the model may be used to prepare the estimate. The model may provide complete information for some building components, partial information for other components, and little or no information for other components.

4. Preparing Estimates from Shop Drawings: By their very nature, shop drawings accurately represent components needed for construction.

8. Select two specialized estimating software packages. How do the features of the packages compare? Which package would you choose, and why? Be prepared to present your findings in class if your instructor chooses to do so.

Each response will be different.

9. Select a takeoff software package. Discuss the advantages and disadvantages of using the selected software package compared to performing takeoffs from a set of paper plans. Be prepared to present your findings in class if your instructor chooses to do so.

Each response will be different.

21 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 6: Overhead and Contingencies

Chapter 6—Overhead and Contingencies Learning Outcomes This chapter looks at the overhead and contingencies that are added to the construction costs. After reading this chapter, the reader should be able to: Learning Outcome 6.1: Explain the difference between home office overhead and job overhead.

Learning Outcome 6.2: Describe what is included in the home office overhead and how it is allocated.

Learning Outcome 6.3: Describe what items estimators should include in the job overhead.

Learning Outcome 6.4: Explain the role of the schedule in determining project overhead.

Learning Outcome 6.5: Identify the appropriate uses of contingencies.

Activities 1. Divide the class into teams. Using one of the projects provided with the textbook, prepare a bar chart scheduled for the project. Have the teams present their schedules to the class.

2. Divide the class into teams. Using one of the projects provided with the textbook, have each team prepare a project overhead budget for the project. Have the teams present their budgets to the class.

3. Discuss Problems 12 and 13 from the chapter.

Instructional Resources PowerPoint Slides: Chapter 06.ppt The pdf drawings from Appendices A through E. They can be found at the Pearson Careers Resources for Students website (http://www.pearsonhighered.com/careersresources/) by searching for the book by title or going to Browse by Discipline and choosing Civil and Environmental Engineering.

The drawings include the following: 

Appendix A – Small Commercial Building



Appendix B – Residential Building Project



Appendix C – Commercial Building Project



Appendix D – Billy’s C-Store



Appendix E – Real Estate Office

Solutions to the Review Questions 22 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 6: Overhead and Contingencies

1. What is overhead? Why must it be included in the cost of the project?

Overhead costs are divided into home office overhead costs and job overhead costs.

The home office overhead costs include all items that cannot be readily charged to any one project but represent the cost of operating the construction company. The job overhead costs include all overhead expenses that will be incurred for a specific project. Home office overhead and project costs are part of the company’s business costs and must be recouped by including them in their bids.

2. What is the difference between the general (home office) and job overhead costs?

Home office overhead costs are costs that are not readily chargeable to one project.

These costs are expenses that the contractor must pay and are the cost of staying in business. Conversely, job overhead comprises all costs that can be readily charged to a specific project but not to a particular item of work on that project. The primary difference between the two is that the home office overhead costs are incurred independently of any project.

3. How are the items the estimator will include in each type of overhead determined?

The home office overhead costs are determined by identifying all the overhead costs the company incurs that cannot be attributed to a specific job. These costs must be allocated to the job. A percentage applied to the bid is developed using the estimated construction costs for the year.

Job overhead costs must be estimated with the same diligence and precision as the direct costs. Estimators must make a complete list of all required items and estimate the cost for each item.

4. Why is the project’s duration important in determining the overhead costs?

A primary determinant of project overhead costs is the project’s duration. This duration affects the cost of supervisory wages, the rental of trailers and toilet facilities, security guards, traffic control, and barricades. It also affects the cost of support equipment provided by the general contractor.

5. What competitive advantage is there for a company that has kept its general (home office) overhead low?

Keeping home office costs low reduces the home office cost allocation applied to each project, which reduces the project’s bid and helps keep the company competitive with other contractors in the area.

6. How may the weather, climate, and season during the project’s construction affect the overhead costs?

Cold or ice water must be provided during hot weather, and special electrolyte liquids may be required to maintain a healthy labor force if the construction is in a hot and dry area. When construction runs into or through the winter, several extra costs must be included in the overhead, including the cost of temporary enclosures, heating the enclosure, heating concrete and materials, and protecting equipment from the elements.

7. Where would the cost of temporary utilities be in the estimate?

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Chapter 6: Overhead and Contingencies

Temporary utilities would be placed under job overhead.

8. How would the estimator determine who is responsible for what portion of the temporary utility costs?

This information is often included in the project manual and must be checked for each project.

9. How can a preliminary work schedule for the project help the estimator?

The estimator can use a preliminary work schedule to estimate how long it will take to complete the project, which is particularly useful when estimating job overhead costs.

10. Define contingency amounts and how some estimators use them.

A contingency is a sum of money (or percentage) added to the bid for missed items, escalation, or unexpected conditions. Using a contingency to cover missed items is often an excuse for poor estimating practices. One appropriate use for a contingency is to cover anticipated price escalation, which occurs when the contractor cannot lock in prices for the project’s duration. A contingency also covers unexpected conditions that cannot reasonably be determined during the design and other project risks.

11. Prepare a bar chart schedule for the project’s construction using the plans and outline specifications in Appendix A. Several schedules would be correct for this exercise. The best approach is to have the students develop a schedule and then have them present it to the class.

Presenting the schedules allows students to see that there are several equally right ways to construct a project.

12. Using the outline specifications and drawings for the Real Estate Office found in Appendix E, prepare a preliminary schedule for the project, assuming the project is to begin on April 1. The schedule should be in sufficient detail to determine the duration of the project and the overhead budget. Be prepared to present your schedule in class if your instructor chooses to do so.

Each response will be different.

13. Using the outline specifications and drawings for the Real Estate Office found in Appendix E, prepare a list of overhead items needed to complete the project. Identify when and for how long each item will be needed. For example, the mobilization and demobilization of the temporary office will be required at the beginning and end of the project, and the temporary office will be required during the entire project. Use your schedule from Problem 12 to help you determine when overhead items are needed. Determine the project overhead by determining the cost of each of these items. Be prepared to present your list in class if your instructor chooses to do so.

Each response will be different. Sample overhead costs are found in Appendix E Real Estate Office v10 Estimate.xlsm on the instructor’s website.

24 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

Chapter 7—Labor Learning Outcomes This chapter looks at determining labor costs. After reading this chapter, the reader should be able to: Learning Outcome 7.1: Calculate labor productivity and labor hours.

Learning Outcome 7.2: Describe the requirements of the Fair Labor Standards Act. Learning Outcome 7.3: Describe the requirements of the Davis-Bacon Act. Learning Outcome 7.4: Compare union labor with open-shop labor.

Learning Outcome 7.5: Describe the components of the labor burden.

Learning Outcome 7.6: Calculate the average wage rate with labor burden.

Activities 1. Bring a construction component (such as a light switch) to the class that needs to be assembled. Have three or four students assemble the component and measure the time it takes them to complete the task. Using these cycle times, estimate the labor hours required to complete the installation of the construction component.

2. Obtain labor burden rates for different trades from a local contractor. Discuss the difference in burden rates for the different trades.

Instructional Resources PowerPoint Slides: Chapter 07.ppt Solutions to the Review Questions 1. What unit of time is used to measure labor? What does it represent?

The unit of time used in estimating is the labor hour. A labor hour is defined as one worker working for one hour.

2. How do climatic conditions influence the amount of work completed in an hour?

Cold, heat, winds, rain, snow, and combinations of these affect productivity.

Typically, any weather extremes will slow the work and may require additional precautions that add labor hours to the task.

3. Where would you use cycle time to estimate productivity?

A cycle time analysis is used when the work is performed in a repeatable cycle, such as a truck hauling earthen materials from the borrow pit to the job site and returning to the borrow pit to make an additional trip.

4. Where would you use the rate of progress to estimate productivity?

25 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

The rate of progress is used when the work is performed linearly, such as paving or striping a road, placing a concrete curb using a slip-forming machine, or grading a road.

5. What federal laws govern labor employment, and what do they cover?

The Fair Labor Standards Act (F LSA) requires that employers pay nonexempt employees at least the current minimum wage established by the federal government and at least one and a half times the regular pay for hours worked in excess of 40 hours per week. The Davis-Bacon Act requires that prevailing wages be paid on jobs with federal funding.

6. What effect can upcoming labor union negotiations have on a bid?

The estimator must consider how any potential changes in the agreement will affect the project’s costs. If wages are expected to be increased, the estimator will have to include enough in the estimate to cover the higher cost of the work done after the expiration of the agreement.

7. What effect could an extreme shortage of skilled workers have on the cost of a project?

Less-trained workers are used when there is a labor shortage. These less-trained workers will require more time or labor hours to complete the required task, which tends to increase construction costs unless a decrease in the hourly wage offsets the decrease in productivity.

8. How can working conditions on the job site affect worker productivity?

Job site working conditions can affect worker productivity if they cannot quickly and efficiently get the materials needed to do their job. Such factors as little working space, limited storage space, difficult delivery conditions, waiting for the crane to deliver materials, difficulty moving from floor to floor, and long distances from the time clock to the work area all adversely affect productivity.

9. What are some things an open-shop contractor can do to retain workers?

They can provide an attractive work environment, including a living wage, insurance, reasonable work schedules that include vacation and sick leave, rewards for outstanding performance, and open communication between management and the employees.

10. What costs should be included in the labor burden?

The costs should include the following: 

Cash equivalents and allowances



Payroll taxes (Social Security and Medicare taxes)



Unemployment insurance (FUTA and SUTA)



Workers’ compensation insurance



General liability insurance 26 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor



Insurance benefits



Retirement contributions



Union payments



Vacation, holidays, and sick leave

11. How do cash allowances differ from reimbursements?

Cash allowances are not reimbursements to the employee for actual expenses but an allowance given to the employee to defer the cost of providing their tools and vehicle. In contrast, reimbursements are based on the tools’ actual cost or the mileage the employee uses their vehicle for business. Cash allowances are taxable income to the employee, whereas reimbursed expenses are not.

12. How are vacation, holidays, and sick leave included in the labor burden?

They are incorporated into the labor burden by including the wages paid for the time off in the employee’s wages before other burden costs are calculated. These hours are not included in the billable hours.

13. How are crews included in the labor estimate? How does this compare with using individual workers?

When using crews comprised of different crafts or workers of varying skill levels being paid different wage rates, a weighted average burdened wage rate is used.

This wage rate is determined by calculating the total labor cost for the crew for an hour and by dividing that cost by the number of persons on the crew. This step is not needed when using individual workers.

14. During the construction of a recent residence, your company’s finish carpentry crew installed 14 interior doors in 18 labor hours. Determine the productivity rate in labor hours per door.

Using Formula 7-1, we get the following: Productivityrate 

18Ihr.  1.29 Ihr. door 14doors

15. Using the productivity from Problem 14, how many hours would it take to install 27 interior doors?

Using Formula 7-2, we get the following:

Labor hours = 27 doors × 1.29 lhr./door = 34.8 lhr.

16. During the construction of a recent office building, your company’s framing crew installed 515 linear feet of 8-foot-high, metal-stud wall in two days. The crew consisted of two carpenters and one laborer and worked 8 hours per day. Determine the productivity rate in labor hours per foot of wall.

Determine the number of labor hours:

Labor hours = 3 ea. × 8 hr./day × 2 days = 48 lhr.

27 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

Using Formula 7-1, we get the following: Productivityrate 

48Ihr.  0.093 Ihr. ft.

515 ft.

17. Using the productivity from Problem 16, determine the labor hours required to install 810 linear feet of 8-foot-high metal-stud wall. If the crew consisted of two carpenters and a laborer, how many days would it take to frame the wall? They work 8 hours per day.

Using Formula 7-2, we get the following:

Labor hours = 810 ft. × 0.093 lhr./ft. = 75.3 lhr.

Days 

75.3Ihr.  3.1 days 3ea. 8Ihr. day

18. A commercial project requires the installation of 1,856 square feet of acoustical ceiling. Using a productivity rate of 0.013 labor hours per square foot and a productivity factor of 0.95, determine the labor hours required to install the ceiling.

Using Formula 7-2, we get the following: Labor hours = 1,856 s.f. × 0.013 lhr./s.f. = 24.13 lhr.

Using Formula 7-3, we get the following:

Adjusted labor hours = 24.13 lhr. × 0.95 = 22.9 lhr.

Use 23 lhr.

19. A commercial project requires the installation of 127 square yards of carpet. Using a productivity rate of 0.108 labor hours per square yard and a productivity factor of 1.05, determine the labor hours required to install the carpet.

Using Formula 7-2, we get the following:

Labor hours = 127 s.y. × 0.108 lhr./s.y. = 13.72 lhr.

Using Formula 7-3, we get the following:

Adjusted labor hours = 13.72 lhr. × 1.05 = 14.4 lhr.

Use 15 lhr.

20. Several dump trucks haul asphalt from the plant to a road construction project. The estimated cycle time for the trucks is 55 minutes. The trucks carry 15 tons of asphalt per trip. Using a system efficiency of 50 minutes per hour and a productivity factor of 1.1, determine the labor hours required to haul 1 ton of asphalt. Each truck requires one operator.

Determine the productivity rate as follows: Productivity rate 

55min.1.1  1 operator  0.081Ihr. ton 50min. hr. ×15 tons 28

Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

21. Scrapers are used to haul dirt from a borrow pit to the cap of a landfill. The estimated cycle time for the scrapers is 9.5 minutes. The scrapers carry 10 cubic yards of soil per trip. Using a system efficiency of 45 minutes per hour and a productivity factor of 1.0, determine the labor hours required to haul 1 cubic yard of dirt. Each scraper requires one operator.

Determine the productivity rate as follows: Productivity rate 

95min.1.0  1 operator  0.021Ihr. c.y.

45min. hr. ×10c.y.

22. An asphalt paving crew is used to pave a road 8,000 feet long. The road is 48 feet wide, and the paving machine can pave a 12-foot-wide strip of the road on each pass. The machine paves 13 linear feet of 12-foot-wide road per minute. The road requires two lifts of asphalt. After completing a pass, it takes the crew 15 minutes to turn the paving machine around and prepare for the next pass. The crew consists of a foreperson, seven operators, and a laborer. Using a system efficiency of 45 minutes per hour, determine the labor hours required to place 1 linear foot of road 12 feet wide.

Determine the productivity rate as follows: Number of passes 

48 ft.  2lifts=8passes 12ft. pass

Quantity = 8 passes × 8,000 ft. = 64,000 ft.

There will be one less turn than passes.

Travel time = 7 turns × 15 min./turn = 105 min.  64,000   13 ft. min.  105min.   9persons  Productivity rate    0.0157Ihr. ft.

45min. hr. × 64,000 ft.

23. An asphalt paving crew is used to pave a 500-foot by 300-foot parking lot. The parking lot will be paved in 12-foot-wide strips running in the long (500-foot) direction. The machine paves 11 linear feet of 12-foot-wide parking lot per minute.

The parking lot requires a single lift of asphalt. After completing a pass, it takes the crew 10 minutes to turn the paving machine around and prepare for the next pass.

The crew consists of a foreperson, six operators, and a laborer. Using a system efficiency of 50 minutes per hour, determine the labor hours required to place 1 linear foot of parking lot 12 feet wide.

Determine the productivity rate as follows: Number of passes =

300 ft.  1 lifts  25passes 12ft. pass

Quantity = 25 passes × 500 ft./pass = 12,500 ft.

There will be one less turn than passes.

29 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

Travel time = 24 turns × 10 min./turn = 240 min.  12,500 ft.   11ft. min.  240min.   8persons  Productivity rate =   0.0176Ihr ft.

50min. hr. × 12,500 ft.

24. Given the following information, determine a carpenter’s burdened hourly wage rate.

Assume the carpenters take full advantage of the retirement benefit.

Wage rate—$23.00 per hour Hours worked—45 hours per week for 30 weeks and 40 hours per week for 20 weeks Vacation, holidays, and sick leave—two weeks at 40 hours per week Overtime—time-and-a-half for any hours over 40 per week Gas allowance—$150 per month Annual bonus—$300 Social Security—6.2 percent on the first $160,200 of wages Medicare—1.45 percent of all wages FUTA—0.6 percent on the first $7,000 of wages SUTA—5.8 percent on the first $10,000 of wages Workers’ compensation insurance—$12.25 per $100 of wages General liability insurance—0.55 percent of wages Health insurance (company’s portion)—$450 per month per employee Retirement—$0.50 per $1.00 contributed by the employee on up to 6 percent of the employee’s wages.

Determine the wages and burden for a year as follows:

Wages =  40 hr./wk. × 52 wk. × $23.00/hr. +  5 hr./wk. × 30 wk. × $23.00/hr. × 1.5  + $150/mo. × 12 mo. + $300

Wages = $55,115 Social Security = $55,115 × 0.062 = $3,417

Medicare = $55,115 × 0.0145 = $799

FUTA = $7,000 × 0.006 = $42 SUTA = $10,000 × 0.058 = $580 Workers'compensationinsurance  $55.115 

$12.25  $6,752 $100

30 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

General liability insurance = $55,115 × 0.0055 = $303

Health insurance = $450/mo. × 12 mo. = $5,400

The company contributes up to three percent (50 percent of six percent) of the employee’s wages to their retirement account.

Retirement = $55,115 × 0.03 = $1,653

Benefits = $3,417 + $799 + $42 + $580 + $6,752 + $303 + $5,400 + $1,653 Benefits = $18,946 Billable hours = 45 hr./wk. × 30 wk. + 40 hr./wk. × 20 wk. = 2,150 hr.

Burdened hourly wage rate =

$55.115  $18,946  $34.45 hr.

2,150hr.

25. Given the following information, determine an equipment operator's burdened hourly wage rate. Assume the equipment operators take full advantage of the retirement benefit.

Wage rate—$36.25 per hour Hours worked—50 hours per week for 40 weeks and 40 hours per week for 9 weeks Vacation, holidays, and sick leave—3 weeks at 40 hours per week Overtime—time-and-a-half for any hours over 40 per week Gas allowance—$100 per month Annual bonus—$750 Social Security—6.2 percent on the first $160,200 of wages Medicare—1.45 percent of all wages FUTA—0.6 percent on the first $7,000 of wages SUTA—2.5 percent on the first $22,000 of wages Workers’ compensation insurance—$9.27 per $100 of wages General liability insurance—0.65 percent of wages Health insurance (company’s portion)—$385 per month per employee Retirement—$0.75 per $1.00 contributed by the employee on up to 6 percent of the employee’s wages Determine the wages and burden for a year as follows: Wages =  40 hr./wk. × 52 wk. × $36.25/hr. + 10 hr./wk. × 40 wk. × $36.25/hr. × 1.5  + $100/mo. × 12 mo. + $750 31 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

Wages = $99,100

Social Security = $99,100 × 0.062 = $6,144 Medicare = $99,100 × 0.0145 = $1,437 FUTA = $7,000 × 0.006 = $42 SUTA = $22,000 × 0.025 = $550

Workers's compensation insurance  $99,100 

$9.27  $9,187 $100

General liability insurance = $99,100 × 0.0065 = $644 Health insurance = $385/mo. × 12 mo. = $4,620

The company contributes up to 4.5 percent (75 percent of six percent) of the employee’s wages to the employee’s retirement account.

Retirement = $99,100 × 0.045 = $4,460

Benefits = $6,144 + $1,437 + $42 + $550 + $9,187 + $644 + $4,620 + $4,460 Benefits = $27,084

Billable hours = 50 hr./wk. × 40 wk. + 40 hr./wk. × 9 wk.

Billable hours = 2,360 hr.

Burdened hourly wage rate 

$99,100  $27,084  $53.47 hr.

2,360hr.

26. Given the following information, determine a project manager's burdened monthly wage rate. Assume the project manager takes full advantage of the retirement benefit.

Wage rate—$13,000 per month Gas allowance—$200 per month Annual bonus—$5,000 Social Security—6.2 percent on the first $160,200 of wages Medicare—1.45 percent of all wages FUTA—0.6 percent on the first $7,000 of wages SUTA—3.1 percent on the first $18,000 of wages Workers’ compensation insurance—$6.60 per $100 of wages General liability insurance—0.8 percent of wages Health insurance (company’s portion)—$400 per month per employee Retirement—$0.50 per $1.00 contributed by the employee on up to 6 percent of the employee’s wages 32 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

Determine the wages and burden for a year as follows:

Wages = $13,000/mo. × 12 mo. + $200/mo. × 12 mo. + $5,000 Wages = $163,400 Social Security is only paid on the first $160,200 of wages.

Social Security =$160,200 × 0.062 = $9,932 Medicare =$163,400× 0.0145 = $2,369 FUTA = $7,000 × 0.006 = $42 SUTA = $18,000 × 0.031 = $558

Workers' comp.insurance  $163,400 

$6.60  $10,784 $100

General liability insurance =$163,400 × 0.008 = $1,307 Health insurance = $400/mo. × 12 mo. = $4,800

The company contributes up to 3.0 percent (50 percent of six percent) of the employee’s wages to the employee’s retirement account.

Retirement =$163,400× 0.03 = $4,902

Benefits = $9,932 + $2,369 + $42 + $558 + $10,784 + $1,307 + $4,800 + $4,902 Benefits = $34,694 Burdened monthly wage rate 

$163,400  $34,694  $16,508 mo.

12mo.

27. A masonry crew consists of six masons and four laborers. The bare hourly wage rate and labor burden for each class of employee is found in Table 1. What is the weighted average burdened wage rate for the crew?

The labor costs are as follows: Masons = 6 ea. ×  $32.50/ea. + $9.13/ea. = $249.78 Laborers = 4 ea. ×  $17.90/ea. + $6.12/ea. = $96.08

Total = $249.78 + $96.08 = $345.86 The weighted average burdened wage rate for the crew is as follows: Weighted average burdened wage rate 

$345.86  $34.59 hr.

6ea.  4ea.

28. A carpentry crew consists of one foreperson, five carpenters, and three laborers.

The bare hourly wage rate and labor burden for each class of employee is found in Table 2. What is the weighted average burdened wage rate for the crew?

The labor costs are as follows: 33 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

Foreman = 1 ea. ×  $18.75/ea. + $6.29/ea. = $25.04 Carpenters = 5 ea. ×  $16.25/ea. + $5.78/ea. = $110.15 Laborers = 3 ea. ×  $13.80/ea. + $5.27/ea. = $57.21 Total = $25.04 + $110.15 + $57.21 =$192.40

The weighted average burdened wage rate for the crew is as follows: Weighted average burdened wage rate 

$192.40  $21.38 hr.

1ea.  5ea.  3ea.

29. It is estimated that constructing a block wall will take 217 labor hours. Using an average burdened hourly rate of $38.75, determine the labor cost to construct the wall.

Determine the labor cost as follows: Labor cost = 217 lhr. × $38.75/lhr. = $8,409

30. It is estimated that hanging drywall in a commercial building will take 538 labor hours. Using an average burdened hourly rate of $27.49, determine the labor cost to hang the drywall.

Determine the labor costs as follows: Labor cost = 538 lhr. × $27.49/lhr. = $14,790

31. The worker classes in Table 3 will be used to construct the Real Estate Office in Appendix E. Given the following information, determine the burdened hourly wage rate for each worker class. Assume the workers take full advantage of the retirement benefit.

Wage rate—as per Table 3 Hours worked—40 hours per week for 50 weeks Vacation, holidays, and sick leave—80 hours per year Social Security—6.2 percent on the first $160,200 of wages Medicare—1.45 percent of all wages FUTA—0.6 percent on the first $7,000 of wages SUTA—1.25 percent on the first $33,600 of wages Workers’ compensation insurance—as per Table 3 General liability insurance—1.05 percent of wages Health insurance (company’s portion)—$625 per month per employee Retirement—$0.50 per $1.00 contributed by the employee on up to 6 percent of the employee’s wages Labor rate calculations are as follows: 34 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

Superintendent

Class

General Cement Laborer Mason Mason

Rough Finish Carpenter Carpenter

Wage rate ($/hr.)

65.00

16.55

21.65

27.20

24.17

28.24

Hours per week

40

40

40

40

40

40

Weeks per year

50

50

50

50

50

50

Vacation & Sick Leave (hr.)

80

80

80

80

80

80

6.25

8.42

11.54

9.20

10.66

6.95

135,200

34,424

45,032

56,576

50,274

58,739

Billable Hours

2,000

2,000

2,000

2,000

2,000

2,000

Social Security

7,886

2,134

2,792

3,508

3,117

3,642

Medicare

1,960

499

653

820

729

852

SUT A

420

420

420

420

420

420

FUT A

42

42

42

42

42

42

Workers' Compensation

8,450

2,899

5,197

5,205

5,359

4,082

General Liability

1,420

361

473

594

528

617

Employer Paid Health Ins.

7,500

7,500

7,500

7,500

7,500

7,500

Retirement

4,056

1,033

1,351

1,697

1,508

1,762

166,934

49,312

63,459

76,362

69,477

77,656

83.47

24.66

31.73

38.18

34.74

38.83

blank

blank

blank

blank

blank

Workers' Compensation Wages

Annual Cost of Employee Average Hourly Wage Rate Monthly Wage Rate

13,911.20

35 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

HVAC Class

Roofer

Plumber

Equipment

Installer

Electrician

Operator

Wage rate ($/hr.)

19.35

33.25

43.50

31.75

21.25

Hours per week

40

40

40

40

40

Weeks per year

50

50

50

50

50

Vacation & Sick Leave (hr.)

80

80

80

80

80

Workers' Compensation

12.21

8.25

8.45

8.60

6.25

Wages

40,248

69,160

90,480

66,040

44,200

Billable Hours

2,000

2,000

2,000

2,000

2,000

Social Security

2,495

4,288

5,610

4,094

2,740

Medicare

584

1,003

1,312

958

641

SUT A

420

420

420

420

420

FUT A

42

42

42

42

42

Workers' Compensation

4,914

5,706

7,646

5,679

2,763

General Liability

423

726

950

693

464

Employer Paid Health Ins.

7,500

7,500

7,500

7,500

7,500

Retirement

1,207

2,075

2,714

1,981

1,326

Annual Cost of Employee

57,833

90,919

116,674

87,408

60,096

Average Hourly Wage Rate

28.92

45.46

58.34

43.70

30.05

The solution can also be found in Appendix E - Real Estate Office v10 Estimate.xlsm.

32. The crews in Table 4 will be used to construct the Real Estate Office in Appendix E. Using the burdened hourly wage rates from Problem 31, determine each crew’s average burdened hourly wage rate.

Crew rate calculations are as follows: Concrete Cement Mason

Count

Cost 3

31.73 36

Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

Concrete

Count

General Laborer Average Plumbing

Cost 2

24.66

blank

28.90

Count

Cost

Plumber

1

45.46

General Laborer

1

24.66

blank

35.06

Average Masonry

Count

Cost

Mason

3

38.18

General Laborer

2

24.66

blank

32.77

Average HVAC

Count

Cost

HVAC Installer

1

58.34

General Laborer

1

24.66

blank

41.50

Average Rough Carpentry

Count

Cost

Rough Carpenter

5

34.74

General Laborer

2

24.66

blank

31.86

Average Electrical

Count

Cost

Electrician

2

43.70

General Laborer

1

24.66

blank

37.35

Average Finish

Count

Finish Carpenter

Cost 2

38.83 37

Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 7: Labor

Finish

Count

General Laborer Average Earthwork

Cost 1

24.66

blank

34.10

Count

Cost

Equipment Operator

2

30.05

General Laborer

1

24.66

blank

28.25

Average Roofing

Count

Cost

Roofer

2

28.92

General Laborer

1

24.66

blank

27.50

Average

The solution can also be found in Appendix E - Real Estate Office v10 Estimate.xls m.

38 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 8: Equipment

Chapter 8—Equipment Learning Outcomes This chapter looks at determining equipment costs. After reading this chapter, the reader should be able to: Learning Outcome 8.1: Identify factors that affect equipment selection.

Learning Outcome 8.2: Describe how equipment productivity is determined.

Learning Outcome 8.3: Determine the costs for contractor-owned equipment.

Learning Outcome 8.4: Describe how costs are determined for rented and leased equipment.

Learning Outcome 8.5: Describe how the cost of miscellaneous tools is handled.

Activities 1. Download a copy of the Caterpillar Equipment Handbook (https://wheelercat.com/company/resources/cat-performance-handbook/). Divide the students into groups, and using Section 22—Estimating Owning and Operating Costs, determine the ownership and operating cost of a piece of equipment. This section includes a worksheet and instructions on using the data in the handbook to calculate equipment costs. Have the students present their calculations to the class.

Instructional Resources PowerPoint Slides: Chapter 08.p pt Solutions to the Review Questions 1. What are the advantages to a small contractor of renting equipment instead of owning it?

Usually, the rental company assumes responsibility for scheduled maintenance and repairs for normal wear and tear on the equipment. It may be desirable to rent equipment for a short period and try it out on a job before committing to purchasing or leasing it, thus reducing the risk of making a long-term commitment to the wrong piece of equipment.

2. What is depreciation on equipment?

Depreciation is the loss in the value of the equipment. As the equipment is used, it begins to wear out, and over time, it will become completely worn out or obsolete.

3. What operating costs must be considered?

Operating costs are those costs that are most closely related to the equipment’s operation. These costs increase as the hours of operation increase and include the following: 39 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 8: Equipment



Fuel



Lubrication (oil, grease) and fluids (hydraulic, transmission)



Filters



Tires



Tire repair



Repairs or repair reserves

4. Why should interest be included in the equipment costs if the contractor paid cash for the equipment?

The company is forgoing the interest it could have earned on the money used to buy the equipment.

5. Why must mobilization be included in the cost of the equipment?

Mobilization is the cost of transporting all equipment to and from the project. It includes the cost of erecting some equipment (such as hoists, scaffolding, or cranes), loading the equipment at the contractor’s yard, and unloading it at the job site. These costs are directly attributable to the project and should be charged to the project.

6. Why is it important that field reports be kept for the equipment?

Field reports documenting the equipment time must include only the time when the equipment is in use. When excessive idle time occurs, estimators must check to see if it can be attributed to bad weather, poor working conditions, or management problems on the project.

7. What factors may contribute to excessive idle time for equipment on the job?

Factors contributing to excessive idle time for equipment on a project include bad weather, poor working conditions, or management problems.

8. Determine the scraper’s ownership and operating costs using the following information: Actual cost (delivered)—$162,000, including tires Horsepower rating—200 hp Cost of tires—$6,000 Salvage value—$32,400 Useful life—6 years at 1,400 hours per year Total interest—7 percent per year Taxes—$3,645 per year Insurance—$3,240 per year Storage—$2,025 per year Fuel cost—$3.90 per gallon 40 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 8: Equipment

Fuel consumption rate—0.06 gallons per hp per hour Power utilization—85 percent Use factor—40 minutes per hour (66.67 percent) Lubrication—8 quarts of oil at $4.50 per quart Oiler labor—2.2 labor hours at $27.65 per labor hour Lubrication schedule—every 100 hours Oil filter—$8.70 each Grease—1 tube per 8-hour shift at $7.00 per tube Air filter—$17.00 each, changed every 200 hours Life of tires—3,500 hours Repair to tires—12 percent of deprecation Repairs to equipment—120 percent over the useful life The ownership costs are calculated as follows: Deduct the cost of the tires from the purchase price: P  $162,000  $6,000  $156,000

Find the hourly depreciation and interest cost:

$156,000 1  0.07   0.07 6

Depreciation & interest cost 

1 0.07   1 6



$32,400  0.07

1 0.07   1 6

Depreciation & interest cost  $28,199 / yr.

Depreciation & interest cost 

$28,199 yr.  $20.14 hr.

1,400hr. yr.

Find the hourly cost for taxes, insurance, and storage: Taxes cost 

$3,645 yr.  $2.60 hr.

1,400hr. yr.

Insurance cost  Storage cost 

$3,240 yr.  $2.31 hr.

1,400hr. yr.

$2,025 yr.  $1.45 hr.

1,400hr. yr.

Find the total ownership costs:

Total ownership costs = $20.14/hr. + $2.60/hr. + $2.31/hr. + $1.45/hr.

Total ownership costs = $26.50/hr.

The operating costs are calculated as follows: 41 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 8: Equipment

Find the hourly fuel cost:

Fuel cost = 200 hp × 0.6667 × 0.85 × 0.06 gal./hp  hr. × $3.90/gal.

Fuel cost = $26.52/hr.

Find the hourly lubrication cost:

Lubrication cost = 8 qt. of oil × $4.50/qt. = $36.00 Labor cost = 2.2 lhr. × $27.65/lhr. = $60.83

Oil filter = $8.70 Total cost for oil change = $36.00 + $60.83 + $8.70 = $105.53 Cost per hour 

$105.53  $1.06 hr.

100hr.

Grease cost  $7.00 tube  Air filter cost 

1tube shift  $0.88 hr.

8hr. shift

$17.00  $0.09 hr.

200hr.

Lubrication cost = $1.06/hr. + $0.88/hr. + $0.09/hr. = $2.03/hr.

Find the hourly cost for tires: Tire depreciation 

$6,000  $1.71 hr.

3,500hr.

Tire repair = 0.12 × $1.71/hr. = $0.21/hr.

Tire cost = $1.71/hr. + $0.21/hr. = $1.92/hr.

Find the hourly repair cost: Repair cost 

$156,000  1.20 =$22.29 hr.

6yr.  1,400hr.

Find the total operating costs:

Total operating costs = $26.52/hr. + $2.03/hr. + $1.92/hr. + $22.29/hr.

Total operating cost = $52.76/hr.

Find the total costs: Total costs = $26.50/hr. + $52.76/hr. = $79.26/hr.

9. Determine the ownership and operating costs of a hydraulic excavator using the following information: Actual cost (delivered)—$172,600 Horsepower rating—245 hp 42 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 8: Equipment

Salvage value—$17,200 Useful life—8 years at 1,600 hours per year Total interest—6.5 percent per year Taxes—$3,020 per year Insurance—$2,589 per year Storage—$2,157 per year Fuel cost—$3.65 per gallon Fuel consumption rate—0.04 gallons per hp per hour Power utilization—60 percent Use factor—50 minutes per hour (83.33 percent) Lubrication—7 quarts of oil at $4.65 per quart Oiler labor—2.9 labor hours at $31.27 per labor hour Lubrication schedule—every 150 hours Oil filter—$10.95 each Grease—1 tube per 8-hour shift at $8.50 per tube Air filter—$22.50 changed every 450 hours Repairs to equipment—75 percent over the useful life The ownership costs are calculated as follows: Find the hourly depreciation and interest cost:

$172,600 1  0.065   0.065 8

Depreciation & interest cost 

1 0.065   1 8



$17,200  0.065

1  0.065   1

Depreciation & interest cost = $26,640/yr.

Depreciation & interest cost 

$26,640 yr.  $16.65 hr.

1,600hr. yr.

Find the hourly cost per hour for taxes, insurance, and storage:

Total cost 

$3,020 yr.  $1.88 hr. $1,600hr. yr.

Insurance cost  Storage cost 

$2,589 yr.  $1.62 hr. $1,600hr. yr.

$2,157 yr.  $1.35 hr. $1,600hr. yr.

43

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8

Chapter 8: Equipment

Find the total ownership costs:

Total ownership costs = $16.65/hr. + $1.88/hr. + $1.62/hr. + $1.35/hr.

Total ownership costs = $21.50/hr.

The operating costs are calculated as follows: Find the hourly fuel cost:

Fuel cost = 245 hp × 0.8333 × 0.60 × 0.04 gal./hp-hr. × $3.65/gal.

Fuel cost = $17.88/hr.

Find the hourly lubrication cost:

Lubrication cost = 7 quarts of oil × $4.65/quart = $32.55

Labor cost = 2.9 hr. × $31.27/hr. = $90.68 Oil filter = $10.95

Total cost for oil change = $32.55 + $90.68 + $10.95 = $134.18

Cost per hour 

$134.18  $0.89 hr.

150hr.

Great cost  $8.50 tube  Air filter cost 

1tube shift  $1.06 hr.

8hr. shift

$22.50  $0.05 hr.

450hr.

Lubrication cost = $0.89/hr. + $1.06/hr. + $0.05/hr. = $2.00/hr.

Find the hourly repair cost:

Repair cost 

$172,600  0.75  $10.11 hr.

8 yr.  1,600hr.

Find the total operating costs:

Total operating costs = $17.88/hr. + $2.00/hr. + $10.11/hr.

Total operating costs = $29.99/hr.

Find the total costs:

Total costs = $21.50/hr. + $29.99/hr. = $51.49/hr.

10. Visit a local equipment supplier. Find out what information the supplier can provide about the ownership and operation costs of the equipment. Prepare a written summary of your findings. Be prepared to present your findings in class if your instructor chooses to do so.

44 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 8: Equipment

The response will vary from dealer to dealer.

11. Call a local equipment rental company and get the monthly rental rate for a telescoping forklift. Find out what costs are included in the monthly rental and what costs the renter must cover in addition to the monthly rent. Using the procedures in this chapter or the data supplied by the rental company, prepare a monthly cost for the forklift that includes all costs.

The response will be different for each student.

45 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 9: Specialty Contractors

Chapter 9—Specialty Contractors Learning Outcomes This chapter looks at determining equipment costs. After reading this chapter, the reader should be able to: Learning Outcome 9.1: Describe the role of specialty contractors in the construction industry.

Learning Outcome 9.2: Describe the purpose of a request for quote.

Learning Outcome 9.3: Describe how scopes of work are used.

Learning Outcome 9.4: Describe what should be considered when reviewing and selecting specialty contractors.

Activities 1. Invite a specialty contractor to your class to discuss the role they play in the local construction industry.

Instructional Resources PowerPoint Slides: Chapter 10.ppt Solutions to the Review Questions 1. What is the primary advantage of using specialty contractors?

The general contractor can transfer some of the project’s risk to the specialty contractors.

2. Why is it important to use requests for quotes when seeking bids from specialty contractors?

Requests for quotes are used to clearly communicate to the prospective specialty contractors the scope of work to be performed, when the bid is due, and any other information needed to submit a price.

3. What are the two ways of writing a scope of work?

By referencing all or part of a set of contract documents (plans and project manuals) and spelling out what is included and what is excluded from the scope of work.

4. What should estimators do if they suspect a specialty contractor’s bid is too high or too low?

They should contact the specialty contractor to discuss their concerns.

5. A bid of $7,000 has been received from an acoustical ceiling specialty contractor for the Real Estate Office in Appendix E. On a recent project for a 4,200-square-foot, wood-framed office building, the acoustical ceiling contractor’s bid was $9,600. Is the 46 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 9: Specialty Contractors

acoustical ceiling bid on the Real Estate Office too high, within an expected range, or too low?

Prepare an estimate for the ceiling using the square footage:

AreaNew building = 72 ft.  50 ft. = 3,600 s.f.  $9,600  CostBid     3,600 s.f.  $8,229  4,200 s.f. 

Bidas a percentage of estimated cost 

$7,000  0.85or 85% $8,229

The bid appears to be too low. A call to the specialty contractor is in order.

6. What should an estimator consider when selecting a specialty contractor?

The factors include price and a good track record of meeting the schedule and quality standards.

7. Prepare separate scopes of work for the following items for the Real Estate Office in Appendix E. a. Concrete b. Masonry c. Framing d. Insulation e. Roofing f. Soffit, fascia, and rain gutters g. Windows and storefronts h. Doors and hardware i. Drywall j. Flooring k. Paint l. Ceramic tile m. Acoustical ceilings n. Plumbing o. HVAC p. Electrical q. Grading and excavation The scope of work will vary from student to student.

47 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 10: Excavation

Chapter 10—Excavation Learning Outcomes This chapter looks at determining the materials and labor associated with the excavation included in Division 31 Earthwork, Division 32 Exterior Improvements, and Division 33 Utilities of the MasterFormat. After reading this chapter, the reader should be able to: Learning Outcome 10.1: Describe the three major types of excavation.

Learning Outcome 10.2: Describe the factors to consider when estimating excavation.

Learning Outcome 10.3: Describe how shrink and swell affect excavation quantities.

Learning Outcome 10.4: Describe the different types of excavation equipment and their uses.

Learning Outcome 10.5: Calculate excavation quantities for rough grading, general excavation, and special excavation.

Learning Outcome 10.6: Calculate topsoil quantities and costs.

Learning Outcome 10.7: Calculate backfill quantities and costs.

Learning Outcome 10.8: Describe how the landscaping is estimated.

Learning Outcome 10.9: Describe the role of specialty contractors in excavation.

Learning Outcome 10.10: Describe how piles are estimated.

Learning Outcome 10.11: Determine the tons of asphalt and base for paving.

Activities 1. Give the students a simple foundation plan and have them determine the quantity of excavation and backfill needed for the job and the cost to complete the excavation and backfill. Be sure to specify the working distance, the required embankment slope, productivity rates, equipment costs, and other factors needed to complete the calculations.

2. Using a plan for a simple basement or a bridge pier, determine the quantity of excavation using the cross-sectional method, the average-end area method, and the average width and length. For the average-end method, use the area at the top and bottom of the excavation. Each of these methods will likely give you a different answer. Discuss why this is the case and which method is most accurate.

Instructional Resources PowerPoint Slides: Chapter 10.ppt The pdf drawings from Appendices A through E. They can be found at the Pearson Careers Resources for Students website (http://www.pearsonhighered.com/careersresources/) by searching for the book by title 48 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 10: Excavation

or going to Browse by Discipline and choosing Civil and Environmental Engineering.

The drawings include the following: 

Appendix A – Small Commercial Building



Appendix B – Residential Building Project



Appendix C – Commercial Building Project



Appendix D – Billy’s C-Store



Appendix E – Real Estate Office

Solutions to the Review Questions 1. What type of excavation information can an estimator learn from the specifications?

The information includes the following: 

What is the extent of work?



What happens to the excess material?



Who does the clearing and grubbing?



Who removes the trees?



Must the topsoil be stockpiled for future use, and where will it be stored?



Who is responsible for any trenching required for the electrical and mechanical trades?

2. How does soil type affect the excavation estimate?

The effects include the following: 

The swell and shrinkage percentages vary by soil type, affecting the backfill, export, and import volumes and costs.



It is one of the factors affecting the amount of slope required for excavation.



It is one of the factors affecting excavation equipment selection.

3. What is the unit of measure for excavation?

Excavation is measured by the cubic yard for the quantity takeoff.

4. How do the type of soil, the shape of the excavation, and the amount of work to be done affect the equipment selection?

Each type of excavation needs to be analyzed, and the best equipment for the work needs to be selected. Certain pieces or groups of equipment are designed to handle specific types of excavation. Using the wrong equipment may take more time and add work to the project.

5. What is the difference between a bank cubic yard, a loose cubic yard, and a compacted cubic yard? Why are they important in estimating excavation?

49 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 10: Excavation

The soil in its natural state is measured in bank cubic yards (b.c.y.). Cuts and excavationsare measured in bank cubic yards.

Uncompacted excavated materials are measured in loose cubic yards (l.c.y.). The density of loose soil is less than the same soil in its bank condition. Soils are transported in their loose state; therefore, the volume of the soil transported is larger than the volume excavated.

Materials that have been placed and compacted are measured in compacted cubic yards (c.c.y.). Fills are measured in compacted cubic yards.

6. What is meant by cut and fill?

Cutting consists of removing earth to lower the grade, and filling consists of adding soil to raise the grade.

7. Why must the cuts be kept separate from the fills?

The cut and fill quantities are kept separate so the estimator can determine if the available cut material can be used for fill. Additionally, these quantities are needed to estimate the effort required to convert the cut material into fill. In addition, because of shrinkage, one cubic yard of cut is not equal to one cubic yard of fill.

8. How can one estimate the topsoil depth on the project?

The depth of the topsoil to be removed can be determined by reviewing the soil borings, which are often included in the project manual, or by visiting the site and digging a small hole with a shovel or posthole digger and measuring its depth. If the depth is measured on-site, several holes should be dug to get an accurate depth.

9. What type of excavation is considered to be general excavation?

General (mass) excavation is the removal of all soil that can be handled in large quantities, such as excavations required for a basement, mat footing, or a cut for a highway or parking area.

10. How does general excavation differ from special excavation?

General or mass excavation is the removal of all soil that can be handled in large quantities. In contrast, special excavation includes work that requires hand excavation or excavation that requires special equipment.

11. What are excess and borrow, and how are each considered in the estimate?

Excess is the soil to be removed from the site when all the excavation is complete.

When there are excess materials, the estimator must determine the quantity to be removed and find a place to haul them.

Borrow is the soil that must be brought to the site to complete the required excavation. When borrow is required, the estimator must determine the quantity to be imported and find a supply of material as close to the job site as is practical.

12. How will the possibility of a high water table affect the bid?

The possibility of a high water table affects the bid because it may require dewatering during the construction.

50 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 10: Excavation

13. When would sheet piles be required on a project?

Sheet piling is used when the excavation adjoins a property line or when the soil is not self-supporting.

14. For a project, your company must haul away 1,450 bank cubic yards of soil. If the soil has a swell of 25 percent, how many loose cubic yards of soil will need to be hauled? How many truckloads are required to haul the soil offsite if each truck can haul 9 loose cubic yards?

Determine the volume of the haul and the number of loads as follows:

Cubic yards of haul = 1,450 b.c.y. × 1 + 0.25  = 1,812 l.c.y.

Loads 

1,812 I.c.y.  202loads 9 I.c.y. load

15. For a project, your company must haul away 350 bank cubic yards of sand. If the sand has a swell of 12 percent, how many loose cubic yards of sand will need to be hauled? How many truckloads are required to haul the sand offsite if each truck can haul 12 loose cubic yards?

Determine the volume of the haul and the number of loads as follows:

Cubic yards of haul = 350 b.c.y. × 1 + 0.12  = 392 l.c.y.

Loads 

392 I.c.y.  33loads 12 I.c.y. load

16. If 750 compacted cubic yards of in-place soil are required for a project, how many loads of import will be required? The imported material has a swell of 30 percent and a shrinkage of 90 percent. The trucks can haul 10 loose cubic yards.

Determine the volume of the haul and the number of loads as follows:

Bank cubic yards 

750 c.c.y.  833b.c.y.

0.90

Loose cubic yards = 833 b.c.y.× 1 + 0.30  = 1,083 l.c.y.

Loads 

1.083 I.c.y.  109 loads 10 I.c.y. load

17. If 1,490 compacted cubic yards of in-place soil are required for a project, how many loads of import will be required? The imported material has a swell of 14 percent and a shrinkage of 95 percent. The trucks can haul 12 loose cubic yards.

Determine the volume of the haul and the number of loads as follows:

Bank cubic yards 

1,490 c.c.y.  1.568b.c.y.

0.95

51 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 10: Excavation

Loose cubic yards = 1,568 b.c.y. 1 + 0.14  = 1,788 l.c.y.

Loads 

1,788 I.c.y.  149 loads 12 I.c.y. load

18. Using the cross-sectional method, determine the cuts and fill for the project shown in Figure 10.47. The grids are 50 feet apart in both directions. The existing grade appears above the proposed grade. Express your cuts in bank cubic yards and your fills in compacted cubic yards.

The cuts and fills are shown in Figure 10-1.

Figure 10-1 The dividing line between the cuts and fills and the distances from the grid intersections to this line are shown in Figure 10-2.

52 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 10: Excavation

Figure 10-2 Unless the division line between the cuts and fills passes through a cell, the cut or fill area is 2,500 square feet (50 feet by 50 feet). The cuts and fills are calculated as follows: Cell 1:

cut1 

0 ft.+0 ft.+0 ft.+0.1ft.  2,500s.f.  62b.c.f.

4 Fill1  0c.c.f.

Cell 2:

cut 2 

0 ft.+0 ft.+0.1ft.+0.3 ft.  2,500s.f.  250b.c.f.

4 Fill2  0c.c.f.

Cell 3:

cut3 

0 ft.+0 ft.+0.3 ft.+0.9 ft.  2,500s.f.  750b.c.f.

4 53

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Chapter 10: Excavation

Fill3  0c.c.f.

Cell 4:

Cut 4 

0 ft.  0 ft.  0.9 ft.  1.1ft.  2,500s.f. 1,250b.c.f.

4 Fill4  0c.c.f.

Cell 5:

Cut 5 

0 ft.  0 ft.  1.1ft.  1.5 ft.  2,500s.f. 1,650b.c.f.

4 Fill5  0c.c.f.

Cell 6:

Cut 6 

0 ft.  0 ft.  1.5 ft.  0 ft.  2,500s.f.  938b.c.f.

4 Fill6  0c.c.f.

Cell 7:

Cut Area 7  Cut7 

0 ft.  0 ft.  0.1ft.  312s.f. 10b.c.f.

3

Fill Area 7  Fill 7 

12.5 ft.  50 ft.  312s.f.

2

37.5 ft.  50 ft.  938s.f.

2

0 ft.  0 ft.  0.3 ft.  938s.f.  94c.c.f.

2

Cell 8: Cut Area 8  50 ft.  50 ft. 

Cut 8 

 37.5 ft.  30 ft.  1,938 s.f.

2

0.1ft.  0.3 ft.  0 ft.  0 ft.  0.2ft.  1,938s.f.  233b.c.f.

5 Fill Area 8  Fill8 

37.5 ft. 30 ft.  562s.f.

2

0 ft.  0.3 ft.  0 ft.  562s.f.  56c.c.f.

3

Cell 9:

54 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 10: Excavation

0.3 ft.  0.9 ft.  0.2ft.  0.4 ft.  2,500s.f. 1,125b.c.f.

4

Cut9 

Fill9  0c.c.f.

Cell 10:

Cut10 

0.9 ft.  1.1ft.  0.4 ft.  0.7 ft.  2,500s.f. 1,937b.c.f.

4 Fill10  0c.c.f.

Cell 11:

Cut11 

1.1ft.  1.5 ft.  0.7 ft.  0.9 ft.  2,500s.f.  2,625b.c.f.

4 Fill11  0c.c.f.

Cell 12:

Cut12 

1.5 ft.  0 ft.  0.9 ft.  0 ft.  2,500s.f. 1,500b.c.f.

4 Fill12  0c.c.f.

Cell 13: Cut13  0b.c.f.

Fill13 

0 ft.  0.3 ft.  0 ft.  0.7 ft.  2,500s.f.  625c.c.f.

4

Cell 14:

Cut Area14  Cut14 

20 ft.  33.3 ft.  333 s.f.

2

0 ft.  0.2ft.  0 ft.  333 s.f.  22b.c.f.

3

Fill Area14  50 ft. 50 ft.  333 s.f.  2,167 s.f.

Fill14 

0.3 ft.  0 ft.  0 ft.  0.7 ft.  0.1ft.  2,167s.f.  477c.c.f.

4

Cell 15:

Cut Area15  50 ft. 50 ft.  Cut15 

16.7 ft.  12.5 ft.  2,396s.f.

2

0.2ft.  0.4 ft.  0 ft.  0 ft.  0.3 ft.  2,396s.f.  431b.c.f.

5 55

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Chapter 10: Excavation

Fill Area15  Fill15 

16.7 ft.  12.5 ft.  104 s.f.

2

0 ft.  0.1ft.  0 ft.  104s.f.  3c.c.f.

3

Cell 16:

Cut16 

0.4 ft.  0.7 ft.  0.3 ft.  0.6 ft.  2,500s.f. 1,250b.c.f.

4 Fill16  0c.c.f.

Cell 17:

Cut17 

0.7 ft.  0.9 ft.  0.6 ft.  0.6 ft.  2,500s.f. 1,750b.c.f.

4 Fill17  0c.c.f.

Cell 18:

Cut18 

0.9 ft.  0 ft.  0.6 ft.  0 ft.  2,500s.f.  938b.c.f.

4 Fill18  0c.c.f.

Cell 19: Cut19  0b.c.f.

Fill19 

0 ft.  0.7 ft.  0 ft.  1.2ft.  2,500s.f.  1,188c.c.f.

4

Cell 20:

Cut 20  0b.c.f.

Fill20 

0.7 ft.  0.1ft.  1.2ft.  0.9 ft.  2,500s.f.  1,812c.c.f.

4

Cell 21:

Cut Area21  Cut 21 

37.5 ft.  18.75 ft.  352s.f.

2

0 ft.  0.3 ft.  0 ft.  352s.f.  35b.c.f.

3

Fill Area21  50 ft.  50 ft.  352s.f.  2,148 s.f.

Fill21 

0.1ft.  0 ft.  0 ft.  0.9 ft.  0.5 ft.  2,148s.f.  644c.c.f.

5 56

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Chapter 10: Excavation

Cell 22:

Cut Area22  50 ft. 50 ft.  Cut 22 

31.25 ft.  41.7 ft. 1,848s.f.

2

0.3 ft.  0.6 ft.  0 ft.  0 ft.  0.1ft.  1,848s.f.  370b.c.f.

5 Fill Area22  Fil22 

31.25 ft.  41.7 ft.  652 s.f.

2

0 ft.  0.5 ft.  0 ft.  652 s.f.  109c.c.f.

3

Cell 23:

Cut 23 

0.6 ft.  0.6 ft.  0.1ft.  0.2ft.  2,500s.f.  938b.c.f.

4 Fill23  0c.c.f.

Cell 24:

Cut 24 

0.6 ft.  0 ft.  0.2ft.  0 ft.  2,500s.f.  500b.c.f.

4 Fill24  0c.c.f.

Cell 25: Cut 25  0b.c.f.

Fill25 

0 ft.  1.2ft.  0 ft.  1.3 ft.  2,500s.f.  1,562c.c.f.

4

Cell 26: Cut 26  0b.c.f.

Fill26 

1.2ft.  0.9 ft.  1.3 ft.  0.9 ft.  2,500s.f.  2,688c.c.f.

4

Cell 27: Cut 27  0b.c.f.

Fill27 

0.9 ft.  0.5 ft.  0.9 ft.  0.6 ft.  2,500s.f.  1,812c.c.f.

4

Cell 28:

Cut Area28 

8.3 ft.  8.3 ft.  34s.f.

2 57

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Chapter 10: Excavation

Cut 28 

0 ft.  0.1 ft.  0 ft.  34s.f.  1b.c.f.

3

Fill Area28  50 ft.  50 ft.  34 s.f. = 2,466 s.f.

Fill28 

0.5 ft.  0 ft.  0 ft.  0.6 ft.  0.5 ft.  2,466s.f.  789c.c.f.

4

Cell 29:

Cut Area29  50 ft.  Cut 29 

0.1ft.  0.2ft.  0 ft.  0 ft.  458s.f.  34b.c.f.

4

Fill Area29  50 ft. Fill29 

8.3 ft.  10 ft.  458s.f.

2

41.7 ft.  40 ft.  2,042s.f.

2

0.ft.  0 ft.  0.5 ft.  0.8 ft.  2,042s.f.  664c.c.f.

4

Cell 30:

Cut Area30  Cut30 

0 ft.  0 ft.  0.2ft.  250s.f.  17 b.c.f.

3

Fill Area30  Fil30 

10 ft.  50 ft.  250 s.f.

2

40 ft.  50 ft.  1,000 s.f.

2

0 ft.  0.8 ft.  0 ft.  1,000s.f.  267c.c.f.

3

Cell 31:

Cut 31  0b.c.f.

Fill31 

0.ft.  1.3 ft.  0 ft.  0 ft.  2,500s.f.  812 c.c.f.

4

Cell 32: Cut 32  0b.c.f.

Fill32 

1.3 ft.  0.9 ft.  0 ft.  0 ft.  2,500s.f. 1,375 c.c.f.

4

Cell 33: 58 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 10: Excavation

Cut 33  0b.c.f.

Fill33 

0.9 ft.  0.6 ft.  0 ft.  0 ft.  2,500s.f.  938 c.c.f.

4

Cell 34: Cut 34  0b.c.f.

Fill34 

0.6 ft.  0.5 ft.  0 ft.  0 ft.  2,500s.f.  688 c.c.f.

4

Cell 35: Cut 35  0b.c.f.

Fill35 

0.5 ft.  0.8 ft.  0 ft.  0 ft.  2,500s.f.  812 c.c.f.

4

Cell 36: Cut 36  0b.c.f.

Fill36 

0.8 ft.  0 ft.  0 ft.  0 ft.  2,500s.f.  500 c.c.f.

4

Total Cuts: Total cuts = 62 b.c.f. + 250 b.c.f. + 750 b.c.f. + 1,250 b.c.f. + 1,625 b.c.f. + 938 b.c.f. + 10 b.c.f. + 233 b.c.f. + 1,125 b.c.f. + 1,937 b.c.f. + 2,625 b.c.f. + 1,500 b.c.f. + 0 b.c.f. + 22 b.c.f. + 431 b.c.f. + 1,250 b.c.f. + 1,750 b.c.f. + 938 b.c.f. + 0 b.c.f. + 0 b.c.f. + 35 b.c.f. + 370 b.c.f. + 938 b.c.f. + 500 b.c.f. + 0 b.c.f. + 0 b.c.f. + 0 b.c.f. + 1 b.c.f. + 34 b.c.f. + 17 b.c.f. + 0 b.c.f. + 0 b.c.f. + 0 b.c.f. + 0 b.c.f. + 0 b.c.f. + 0 b.c.f.

Total Cuts 

18,591b.c.f.  689c.y.

27b.c.f. b.c.y.

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Chapter 10: Excavation

Total Fills: Total Fills = 0 c.c.f. + 0 c.c.f. + 0 c.c.f. + 0 c.c.f. + 0 c.c.f. + 0 c.c.f. + 94 c.c.f. + 56 c.c.f. + 0 c.c.f. + 0 c.c.f. + 0 c.c.f. + 0 c.c.f. + 625 c.c.f. + 477 c.c.f. + 3 c.c.f. + 0 c.c.f. + 0 c.c.f. + 0 c.c.f. + 1,188 c.c.f. + 1,812 c.c.f. + 644 c.c.f. + 109 c.c.f. + 0 c.c.f. + 0 c.c.f. + 1,562 c.c.f. + 2,688 c.c.f. + 1,812 c.c.f. + 789 c.c.f. + 664 c.c.f. + 267 c.c.f. + 812 c.c.f. + 1,375 c.c.f. + 938 c.c.f. + 688 c.c.f. + 812 c.c.f. + 500 c.c.f.

Total Fills 

17,915c.c.f.  664c.c.y.

27c.c.f. c.c.y.

19. Set up Problem 18 in a spreadsheet.

Each student’s spreadsheet layout will be unique. See Problem 10-19.xlsx for a sample solution.

20. The material in Problem 18 has a shrinkage of 95 percent and a swell of 30 percent.

Determine the net import or export required for the project and the number of 9 cubic yard truckloads needed to haul the material to or from the project.

Convert the fill from compacted cubic yards to bank cubic yards:

Required b.c.y. 

664c.c.f.  699b.c.y.

0.95

Find the net import:

Import  699 b.c.y.  689 b.c.y.  10 b.c.y.

Import  10 b.c.y.  1 + 0.30   13 l.c.y.

Trucks 

13 I.c.y.  2 load 9 I.c.y. load

21. Using the average end area method, determine the required cut volume and fill volume for a new road. Table 10.1 shows the cuts and fills for each station. Express your answer in b.c.y. and c.c.y.

The cuts are calculated as follows:

Cut10  00 to10 +50 

0s.f.  257s.f.  50 ft.  6,425b.c.f.

2

6,425b.c.f.  238 b.c.y.

27b.c.f. b.c.y.

275s.f.  175s.f.

Cut10  50 to11 +00   50 ft. 10,800b.c.f.

2 Cut10  00 to10 +50 

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Chapter 10: Excavation

10,800b.c.f.  400 b.c.y.

27b.c.f. b.c.y.

175s.f.  20s.f.

Cut11 00 to11 +00   50 ft.  4,875 b.c.f.

2 4,875b.c.f.

Cut11 00 to11 +50  181 b.c.y.

27b.c.f. b.c.y.

20s.f.  0s.f.

Cut11 50 to12 +00   50 ft.  500 b.c.f.

2 Cut10  50 to11 +00 

Cut11 50 to12 +00 

Cut12 00 to12 +20 

500b.c.f. 19 b.c.y.

27b.c.f. b.c.y.

0s.f.  0s.f.  20 ft.  0 b.c.f.

2

Cut12  00 to12 +20 

0b.c.f.  0 b.c.y.

27b.c.f. b.c.y.

Total cuts = 238 b.c.y. + 400 b.c.y. + 181 b.c.y. + 19 b.c.y. + 0 b.c.y. = 838 b.c.y.

The fills are calculated as follows:

Fill10  00 to10 +50 

0s.f.+0s.f.  50 ft.  0c.c.f.

2

Fill10  00 to10 +50 

Fill10  50 to11 +00 

0c.c.f.  0c.c.y.

27c.c.f. c.c.y.

0s.f.  50s.f.  50 ft.  1,250c.c.f.

2

1,2500c.c.f.  46c.c.y.

27c.c.f. c.c.y.

50s.f.  285s.f.

Fill11 00 to11 +50   50 ft.  8,375c.c.f.

2 8,375c.c.f.

Fill11 00 to11 +50   310c.c.y.

27c.c.f. c.c.y.

285s.f.  233s.f.

Fill11 50 to12 +00   50 ft.  12,950c.c.f.

2 12,950c.c.f.

Fill11 50 to12 +00   480c.c.y.

27c.c.f. c.c.y.

233s.f.  0s.f.

Fill12  00 to12 +20   20 ft.  2,330c.c.f.

2 2,330c.c.f.

Fill12  00 to12 +20   86c.c.y.

27c.c.f. c.c.y.

Fill10  50 to11 +00 

Total fills = 0 c.c.y. + 46 c.c.y. + 310 c.c.y. + 480 c.c.y. + 86 c.c.y. = 922 c.c.y.

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Chapter 10: Excavation

22. The material in Problem 21 has a shrinkage of 97 percent and a swell of 18 percent.

Determine the net import or export required for the project and the number of 11cubic-yard truckloads needed to haul the material to or from the project.

Convert the fill from compacted cubic yards to bank cubic yards:

Required fill 

922c.c.y.  951 b.c.y.

0.97

Find the net import: Import  951 b.c.y.  838 b.c.y. = 113 b.c.y.  1 + 0.18  = 133 l.c.y.

Find the number of trucks: Trucks 

133 l.c.y.  13 loads 11I.c.y. loads

23. Using the average end area method, determine the required cut volume and fill volume for a new road. Table 10.2 shows the cuts and fills for each station. Express your cuts in bank cubic yards and your fills in compacted cubic yards.

The cuts are calculated as follows:

Cut 8  30 to 8 +50 

0s.f.  125s.f.  20 ft.  1,250 b.c.f.

2

Cut 8  30 to 8 +50 

Cut 8 50 to 9 +00 

125s.f.  168s.f.  50 ft.  7,325b.c.f.

2

Cut 8  50 to 9 +00 

Cut 9 00 to 9 +50 

7,325b.c.f.  271b.c.y.

27b.c.f. b.c.y.

168s.f.  210s.f.  50 ft.  9,450b.c.f.

2

Cut 9  00 to 9 +50 

Cut 9 50 to 10 +00 

1,250b.c.f.  46 b.c.y.

27b.c.f. b.c.y.

9,450b.c.f.  350b.c.y.

27b.c.f. b.c.y.

210s.f.  144s.f.  50 ft.  8,850b.c.f.

2

Cut 9  50 to 10 +00 

Cut 10 00 to 10 +50 

8,850b.c.f.  328b.c.y.

27b.c.f. b.c.y.

144s.f.  75s.f.  50 ft.  5,475b.c.f.

2

Cut 10  00 to 10 +50 

5,475b.c.f.  203b.c.y.

27b.c.f. b.c.y.

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Chapter 10: Excavation

75s.f.  0s.f.  25 ft.  938b.c.f.

2

Cut 10 50 to 10 +75 

Cut 10  50 to 10 +75 

938b.c.f.  35b.c.y.

27b.c.f. b.c.y.

Total cuts = 46 b.c.y. + 271 b.c.y. + 350 b.c.y. + 328 b.c.y. + 203 b.c.y. + 35 b.c.y.

Total cuts = 1,233 b.c.y.

The fills are calculated as follows:

Fill 8 30 to 8 +50 

0s.f.  25s.f.  20 ft.  250c.c.f.

2

Fill 8  30 to 8 +50 

25s.f.  0s.f.  50 ft.  625c.c.f.

2

Fill 8 50 to 9 +00 

Fill 8  50 to 9 +00 

Fill 9 00 to 9 +50 

0c.c.f.  0c.c.y.

27c.c.f. c.c.y.

0s.f.  95s.f.  50 ft.  2,375c.c.f.

2

Fill 9  50 to 10 +00 

Fill 10 00 to 10 +50 

625c.c.f.  23c.c.y.

27c.c.f. c.c.y.

0s.f.  0s.f.  50 ft.  0c.c.f.

2

Fill 9  00 to 9 +50 

Fill 9 50 to 10 +00 

250b.c.f.  9c.c.y.

27c.c.f. c.c.y.

2,375c.c.f.  88c.c.y.

27c.c.f. c.c.y.

95s.f.  65s.f.  50 ft.  4,000c.c.f.

2

Fill 10  00 to 10 +50 

Fill 10 50 to 10 +75 

4,000c.c.f.  148c.c.y.

27c.c.f. c.c.y.

65s.f.  0s.f.  25 ft.  812c.c.f.

2

Fill 10  50 to 10 +75 

812c.c.f.  30c.c.y.

27c.c.f. c.c.y.

Total fills = 9 c.c.y. + 23 c.c.y. + 0 c.c.y. + 88 c.c.y. + 148 c.c.y. + 30 c.c.y.

Total fills = 298 c.c.y.

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Chapter 10: Excavation

24. The material in Problem 23 has a shrinkage of 94 percent and a swell of 25 percent.

Determine the net import or export required for the project and the number of 10cubic-yard truckloads needed to haul the material to or from the project.

Convert the fill from compacted cubic yards to bank cubic yards:

Required fill 

298c.c.y.  317 b.c.y.

0.94

Find the net export: Export  1,233 b.c.y.  317 b.c.y.  916 b.c.y.  1 + 0.25   1,145 l.c.y.

1,145 l.c.y.  115loads 10 I.c.y. loads

Trucks 

25. Find the perimeter of the building in Figure 10.48.

Starting at the bottom left corner and proceeding clockwise, we get: Perimeter = 30 f t. + 27 f t. + 12 f t. + 26 f t. + 42 f t. + 53 f t. = 190 f t. ee

ee

ee

ee

ee

ee

ee

26. Find the area of the building in Figure 10.48.

Two solutions are provided below: Area  27 ft.  30 ft.  26 ft.  42 ft.  1,902 s.f.

Area  53 ft.  42 ft.  27 ft.  12 ft.  1,902 s.f.

27. Six inches of topsoil is to be removed and stockpiled for the building in Figure 10.48.

The topsoil is to be removed 8 feet back from the perimeter of the building to allow for working around the building and sloping the sides of the excavation. How many bank cubic yards of topsoil need to be removed and stockpiled?

The limits of the topsoil excavation are shown in Figure 10-3.

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Chapter 10: Excavation

Figure 10-3 Two solutions are provided below: Area  27 ft.  46 ft.  42 ft.  58 ft.  3,678 s.f.

Area  69 ft.  58 ft.  27 ft.  12 ft.  3,678 s.f.

Depth = 6 in. = 0.5 f t. ee

Volume  3,678 s.f.  0.5 ft.  1,839 b.c.f.

Volume 

1,839b.c.f.  68b.c.y.

27b.c.f. b.c.y.

28. The topsoil in Problem 27 is to be removed with a bulldozer. The estimated productivity is 22 bank cubic yards per hour, and it will take 2 hours to mobilize the bulldozer. The bulldozer’s total ownership and operation cost is $42.56 per hour, and the cost of an operator is $22.14 per hour. Determine the number of hours and the cost to strip the topsoil.

Determine the number of hours and cost as follows: Hours 

68b.c.y.  2hr.  5.1 h.r.

27b.c.y. hr.

Equipment cost  5.1 hr.  $42.56/hr.  $217 Labor cost  5.1 hr.  $22.14/hr.  $113

Total cost = $217 + $113 = $330 29. Eight inches of topsoil is to be removed and stockpiled for the building in Figure 10.49. The topsoil is to be removed 10 feet back from the perimeter of the building to allow for working around the building and sloping the sides of the excavation. How many bank cubic yards of topsoil need to be removed and stockpiled?

The limits of the topsoil excavation are shown in Figure 10-4.

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Chapter 10: Excavation

Figure 10-4 Two solutions are provided below: Area  14 ft.  48 ft.  52 ft.  12 ft.  48 ft.  14 ft.  48 ft.  4,464 s.f.

Area  80 ft.  12 ft.  48 ft.  2 ea.  14 ft.  12 ft.  4,464 s.f.

Depth = 0.8 in. = 0.67 f t. ee

Volume  4,464 s.f.  0.67 ft.  2,991 b.c.f.

Volume 

2,991b.c.f.  111 b.c.y.

27b.c.f. b.c.y.

30. The topsoil in Problem 29 is to be removed with a bulldozer. The estimated productivity is 30 bank cubic yards per hour, and it will take 2.5 hours to mobilize the bulldozer. The bulldozer’s total ownership and operation cost is $47.12 per hour, and the cost of an operator is $26.88 per hour. Determine the number of hours and the cost to strip the topsoil.

Determine the number of hours and cost as follows: Hours 

111 b.c.y.  2.5hr.  6.2 hr.

30b.c.y. hr.

Equipment cost = 6.2 hr.  $47.12/hr. = $292 Labor cost = 6.2 hr.  $26.88/hr. = $167

Total cost = $292 + $167 = $459 31. Determine the quantity of excavation needed for the basement in Figures 10.50 and 10.51. A 2-foot working distance is required, and the excavation will be sloped 1.5:1 (1.5 vertical feet for every horizontal foot).

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Chapter 10: Excavation

Determine the depth as follows:

Depth  10 ft.  1 ft. + 1 ft. = 10 ft.

The distance from the toe to the top of the slope is calculated as follows: DistanceSlope 10 ft. 

1ft.  6 ft.8in.

1.5 ft.

Calculate the average width: DistanceTop of excavation  6 ft.8 in.  2ft. 11 in.  22ft.  11 in.  2ft.  6ft. 8in.

DistanceTop of excavation  41ft. 2 in.

DistanceBottom of excavation  2ft.  11 in.  22ft.  11 in.  2ft.  27ft. 10in.

DistanceAverage 

41ft. 2 in.  27 ft. 10 in.  34 ft. 6in.  34.5 ft.

2

Calculate the average length: DistanceAverage  6 ft. 8 in.  2 ft.  11 in.  34 ft.  11 in.  2 ft.  6 ft. 8 in.

DistanceTop of excavation  53 ft. 2 in.

DistanceBottom of excavation  2ft.  11 in.  34 ft.  11 in.  2ft.  39ft. 10in.

DistanceAverage 

53 ft. 2 in.  39 ft. 10 in.  46 ft. 6in.  46.5 ft.

2

General excavation  34.5 ft.  46.5 ft.  10 ft.  16,042 b.c.f.

General excavation 

16,042 b.c.f.  594b.c.y.

27b.c.f. b.c.y.

32. Determine the quantity of backfill needed for the basement in Problem 31. Ignore any backfill inside of the walls.

From Problem 31, the excavation depth was 10 feet, and the quantity of excavation was 594 b.c.y. Find the volume of the basement: Builingvolume  22ft.  34 ft.  10 ft. 

7,480 c.f.  277c.y.

27c.f. c.y.

Find the volume of the footings that are outside the foundation wall. The corners of the footings will be added to the top and bottom sides.

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Chapter 10: Excavation

Footing length = 22 ft.  35 ft. 10 in.  22 ft.  35 ft. 10 in. = 115 ft. 8 in. =115.67 ft.

Footing width = 11 in. = 0.92 f t. ee

Footing depth = 12 in. = 1 f t. ee

Footing volume  115.67 ft.  1 ft.  0.92 ft.  106 c.f.

Footing volume 

106c.f.  4c.y.

27c.f. c.y.

Find the volume of the backfill. The backfill will be in compacted cubic yards.

Backfill  594 c.y.  277 c.y.  4 c.y.  313 c.c.y.

33. Determine the quantity of excavation needed for the basement in Figures 10.52 and 10.53. A 1-foot working distance is required, and the excavation will be sloped at 2:1 (2 vertical feet for every horizontal foot). Hint: Calculate the excavation for each area separately and add them together.

Determine the depth as follows:

Depth  8 ft.  6 in.  10 in.  8 ft. 4 in.  8.33 ft.

The distance from the toe to the top of the slope is calculated as follows: DistanceSlope  8 ft. 4 in. 

1 ft.  4 ft. 2 in.

2ft.

Calculate the average width for Area A: DistanceTop of excavation  4 ft. 2 in.  8in.  25 ft.  8in.  1ft.  4 ft. 2in.

DistanceTop of excavation  36 ft. 8 in.

DistanceBottom of excavation  1 ft.  8 in.  25 ft.  8 in.  1 ft.  28 ft. 4 in.

DistanceAverage 

36 ft. 8 in.  28 ft. 4 in.  32ft. 6in.  32.5 ft.

2

Calculate the average length for Area A. The slope, footing, and working space will only occur on one side of the area: DistanceTop of excavation  10 ft.  8 in.  1 ft.  4 ft. 2 in.  15 ft. 10 in.

DistanceBottom of excavation  10 ft.  8 in.  1 ft.  11 ft. 8 in.

DistanceAverage 

15 ft. 10 in.  11 ft. 8 in.  13 ft. 9in. 13.75 ft.

2

Calculate the average width for Area B. The slope, footing, and working space will only occur on one side of the area: DistanceTop of excavation  4 ft. 2 in.  1 ft.  8 in.  25 ft.  30 ft. 10 in.

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Chapter 10: Excavation

DistanceBottom of excavation  25 ft.  8 in.  1 ft.  26 ft. 8 in.

DistanceAverage 

30 ft. 10 in.  26 ft. 8 in.  28 ft. 9in.  28.75 ft.

2

Calculate the average length for Area B. The slope, footing, and working space will only occur on one side of the area: DistanceTop of excavation  20 ft.  8 in.  1 ft.  4 ft. 2 in.  25 ft. 10 in.

DistanceBottom of excavation  20 ft.  8 in.  1 ft.  21 ft. 8 in.

DistanceAverage 

25 ft. 10 in.  21 ft. 8 in.  23 ft. 9in.  23.75 ft.

2

Calculate the average width for Area C. The slope, footing, and working space will only occur on one side of the area: DistanceTop of excavation  15 ft.  8 in.  1 ft.  4 ft. 2 in.  20 ft. 10 in.

DistanceBottom of excavation  15 ft.  8 in.  1 ft.  16 ft. 8 in.

DistanceAverage 

20 ft. 10 in.  16 ft. 8 in.  18 ft. 9in. 18.75 ft.

2

Calculate the average length for Area C: DistanceTop of excavation  4 ft. 2 in.  1 ft.  8 in.  20 ft.  8 in.  1 ft.  4 ft. 2 in.

DistanceTop of excavation  31 ft. 8 in.

DistanceBottom of excavation  1 ft.  8 in.  20 ft.  8 in.  1 ft.  23 ft. 4 in.

DistanceAverage 

31 ft. 8 in.  23 ft. 4 in.  27 ft. 6in.  27.75 ft.

2

The general excavation is calculated by determining the volume of each area and then adding them together.

General excavationA  32.5 ft.  13.75 ft.  8.33 ft.  3,722 b.c.f.

General excavationA 

3,722 b.c.f.  138b.c.y.

27b.c.f. b.c.y.

General excavationB  28.75 ft.  23.75 ft.  8.33 ft.  5,688 b.c.f.

General excavationB 

5,688 b.c.f.  211 b.c.y.

27b.c.f. b.c.y.

General excavationC  18.75 ft.  27.5 ft.  8.33 ft.  4,295 b.c.f.

General excavationC 

4,295 b.c.f.  159 b.c.y.

27b.c.f. b.c.y.

69

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Chapter 10: Excavation

General excavation  138 b.c.y.  211 b.c.y.  159 b.c.y.  508 b.c.y.

34. Determine the quantity of backfill needed for the basement in Problem 33. Ignore any backfill inside of the walls.

From Problem 33, the excavation depth was 8.33 feet, and the quantity of excavation was 508 b.c.y. Find the volume of the basement:

Building area  40 ft.  30 ft.  15 ft.  10 ft.  1,050 s.f.

Building volume 1,050s.f.  8.33 ft. 

8,746 c.f.  324 c.y.

27c.f. c.y.

The corners of the footings will be added to the top and bottom sides.

Footing length  30 ft.  26 ft. 4 in.  10 ft.  15 ft.  20 ft.  41 ft. 4 in.  142 ft. 8 in.

Footing volume  142 ft. 8 in.  10 in.  8 in.

Footing volume  142.67 ft.  0.83 ft.  0.67 ft.  79 c.f.

Footing volume 

79 c.f.  3c.y.

27c.f. c.y.

Find the volume of the backfill. The backfill will be in compacted cubic yards.

Backfill  508 c.y.  324 c.y.  3 c.y.  181 c.c.y.

35. Determine the quantity of excavation needed for the continuous footings in Figures 10.54 and 10.55. A 1-foot working distance is required, and the excavation will be sloped at 1:1 (1 vertical foot for every horizontal foot).

Determine the depth and perimeter as follows:

Depth  2 ft.  1 ft.  1 ft.  2 ft.

Perimeter = 25 f t. + 21 f t. + 5 f t. + 15 f t. + 20 f t. + 36 f t. = 122 f t. ee

ee

ee

ee

ee

ee

ee

The distance from the toe to the top of the slope is calculated as follows:

DistanceSlope  2ft.

1 ft.  2ft.

1 ft.

Calculate the average width: DistanceTop of excavation  2 ft.  1 ft.  2 ft.  1 ft.  2 ft.  8 ft.

DistanceBottom of excavation  1 ft.  2 ft.  1 ft.  4 ft.

DistanceAverage 

8 f.t.  4 f.t.  6 f.t.

2

Find the general excavation as follows:

General excavation  122 ft.  6 ft.  2 ft.  1,464 b.c.f.

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Chapter 10: Excavation

1,464 b.c.f.  54.2 b.c.y.

27b.c.f. b.c.y.

General excavation 

36. Determine the quantity of backfill needed for the continuous footing and foundation in Problem 35.

From Problem 35, the excavation depth was 2 feet, the length of the wall was 122 feet, and the quantity of excavation was 54.2 b.c.y. Find the volume of the footing and foundation: Footing volume 122ft. 2ft.  1 ft. 

244 c.f.  90c.y.

27c.f. c.y.

Foundation volume  122 ft.  1 ft.  8 in.  122 ft.  1 ft.  0.67 ft.  82 c.f.

Foundation volume 

82 c.f.  3.0c.y.

27c.f. c.y.

Find the volume of the backfill. The backfill will be in compacted cubic yards.

Backfill  54.2 c.y.  9.0 c.y.  3.0 c.y.  42.2 c.c.y.

37. Determine the quantity of excavation needed for the spread footings in Figures 10.54 and 10.56. A 1-foot working distance is required, and the excavation will be sloped at 1:1 (1 vertical foot for every horizontal foot).

Determine the depth as follows:

Depth  2 ft.  1 ft.  1 ft.  2 ft.

The distance from the toe to the top of the slope is calculated as follows:

DistanceSlope  2ft.

1 ft.  2ft.

1 ft.

The average width and length are the same because the footings are square.

Calculate the average width and length: DistanceTop of excavation  2 ft.  1 ft.  2 ft. 6 in.  1 ft.  2 ft.  8 ft. 6 in.  8.5 ft.

DistanceBottom of excavation  1 ft.  2 ft. 6 in.  1 ft.  4 ft. 6 in.  4.5 ft.

DistanceAverage 

8.5 ft.  4.5 ft.  6.5 ft.

2

General excavation  2 ea.  6.5 ft.  6.5 ft.  2 ft.  169 b.c.f.

General excavation 

169 b.c.f.  6.3 b.c.y.

27b.c.f. b.c.y.

38. Determine the quantity of backfill needed for the spread footings and columns in Problem 37.

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Chapter 10: Excavation

From Problem 37, the excavation depth was 2 feet, and the quantity of excavation was 6.3 b.c.y. Find the volume of the footings and the columns:

Footing volume  2 ea.  2 ft. 6 in.  2 ft. 6 in.  1 ft.

Footing volume  2 ea.  2.50 ft.  2.50 ft.  1 ft.  12.5 c.f.

Footing volume 

12.5 c.f.  0.5 c.y.

27c.f. c.y.

Column volume  2 ea.  1 ft.  1 ft.  1 ft. 

2 c.f.  0.1 c.y.

27c.f. c.y.

Find the volume of the backfill. The backfill will be in compacted cubic yards:

Backfill  6.3 c.y.  0.5 c.y.  0.1 c.y.  5.7 c.c.y.

39. Determine the quantity of excavation needed for the continuous footings in Figures 10.57 and 10.58. A 2-foot working distance is required, and the excavation will be sloped at 1.75:1 (1.75 vertical feet for every horizontal foot).

Determine the depth and length of the wall as follows:

Depth  2 ft.  8 in.  1 ft. 2 in.  2 ft. 6 ft.  2.5 ft.

Length of wall  40 ft.  30 ft.  40 ft.  30 ft.  15 ft.  19 ft.  174 ft.

The distance from the toe to the top of the slope is calculated as follows:

DistanceSlope  2.5 ft.

1 ft.  1.4 ft.

1.75 ft.

Calculate the average width: DistanceTop of excavation  1.4 ft.  2 ft.  2.5 ft.  2 ft.  1.4 ft.  9.3 ft.

DistanceBottom of excavation  2 ft.  2.5 ft.  2 ft.  6.5 ft.

DistanceAverage 

9.3 ft.  6.5 ft.  7.9 ft.

2.

Find the general excavation as follows:

General excavation  174 ft.  7.9 ft.  2.5 ft.  3,436 b.c.f.

General excavation 

3,436 b.c.f.  127 b.c.y.

27b.c.f. b.c.y.

40. Determine the quantity of backfill needed for the continuous footing and foundation in Problem 39.

From Problem 39, the excavation depth was 2.5 feet, the wall length was 174 feet, and the quantity of excavation was 127 b.c.y. Find the volume of the footing and foundation: 72 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 10: Excavation

Footing volume  174 ft.  2.5 ft.  1 ft. 2 in.

Footing volume  174 ft.  2.5 ft.  1.17 ft.  509 c.f.

Footing volume 

509c.f.  19 c.y.

27c.f. c.y.

Foundation volume  174 ft.  1 ft. 4 in.  10 in.

Foundation volume  174 ft.  1.33 ft.  0.83 ft.  192 c.f.

Footing volume 

192c.f.  7c.y.

27c.f. c.y.

Find the volume of the backfill. The backfill will be in compacted cubic yards:

Backfill  127 b.c.y.  19 c.y.  7 c.y.  101 c.c.y.

41. Determine the quantity of excavation needed for the spread footing in Figure 10.59.

A 3-foot working distance is required, and the excavation will be sloped at a 1:1 (1 vertical foot for every horizontal foot).

Determine the depth as follows:

Depth  7 ft.  2 ft. 6 in.  7 ft.  2.5 ft.  9.5 ft.

The distance from the toe of the slope to the top of the slope is calculated as follows:

DistanceSlope  9.5 ft.

1 ft.  9.5 ft.

1 ft.

Calculate the average width: DistanceTop of excavation  9.5 ft.  3 ft.  8 ft.  3 ft.  9.5 ft.  33 ft.

DistanceBottom of excavation  3 ft.  8 ft.  3 ft.  14 ft.

DistanceAverage 

33 ft.  14 ft.  23.5 f.t.

2

Calculate the average length: DistanceTop of excavation  9.5 ft.  3 ft.  17 ft.  3 ft.  9.5 ft.  42 ft.

DistanceBottom of excavation  3 ft.  17 ft.  3 ft.  23 ft.

DistanceAverage 

42 ft.  23 ft.  32.5 ft.

2.

Find the general excavation as follows: General excavation  23.5 ft.  32.5 ft.  9.5 ft.  7,256 b.c.f.

General excavation 

7,256 b.c.f.  269 b.c.y.

27b.c.f. b.c.y.

73

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Chapter 10: Excavation

42. Determine the quantity of backfill needed for the spread footing and column in Problem 41.

From Problem 41, the excavation depth was 9.5 feet, and the quantity of excavation was 269 b.c.y. Find the volume of the footing and the column:

Footing volume  8 ft.  17 ft.  2 ft. 6 in.  8 ft.  17 ft.  2.5 ft.  340 c.f.

Footing volume 

340c.f.  13c.y.

27c.f. c.y.

Column volume  3 ft.  12 ft.  7 ft. 

252 c.f.  9 c.y.

27c.f. c.y.

Find the volume of the backfill. The backfill will be in compacted cubic yards:

Backfill  269 c.y.  13 c.y.  9 c.y.  247 c.c.y.

43. Determine the quantity of asphalt and granular material required for the parking lot shown in Figure 10.60. The asphalt is 4 inches thick, and the granular material is 8 inches thick.

Ignoring the small radius, the area of the parking lot is calculated as follows: Main area  200 ft.  99 ft.  19,800 s.f.

Drive area  23 ft.  24 ft.  552 s.f.

Total area  19,800 s.f.  552 s.f.  20,352 s.f.

Figure 10-44 of the textbook shows that the parking lot will require 26.0 tons of asphalt and 42.0 tons of granular material per thousand square feet.

Asphalt  26.0 tons/thousand s.f.  20.352 thousand s.f.  529 tons

Granular material  42.0 tons/thousand s.f.  20.352 thousand s.f.  855 tons 44. Determine the quantity of asphalt and granular material required for the parking lot shown in Figure 10.61. The asphalt is 3 inches thick, and the granular material is 6 inches thick.

Ignoring the small radius, the area of the parking lot is calculated as follows: Main area  227 ft.  146 ft.  33,142 s.f.

Drive area  12 ft.  23 ft.  276 s.f.

Total area  33,142 s.f.  276 s.f.  33,418 s.f.

Figure 10-44 of the textbook shows that the parking lot will require 19.5 tons of asphalt and 31.5 tons of granular material per thousand square feet.

Asphalt  19.5 tons/thousand s.f.  33.418 thousand s.f.  652 tons Granular material  31.5 tons/thousand s.f.  33.418 thousand s.f.  1,053 tons

45. Calculate the general excavation for the building shown in Appendix C. 74 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 10: Excavation

The quantity of excavation is 139 b.c.y. The following workup sheet contains detailed calculations.

The solution can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

46. Calculate the topsoil to be stockpiled, put back around the building, and hauled away for the building shown in Appendix C. The topsoil to be removed is 240 b.c.y., and the total topsoil to be replaced around the site is 60 c.c.y.

These numbers can easily vary based on the order and how the plot plan was divided. However, the total should be relatively close. The following sketch and workup sheet contain detailed calculations.

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Chapter 10: Excavation

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Chapter 10: Excavation

The solution can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

47. Calculate the crushed gravel and asphalt required for the parking lot for the building shown in Appendix C. The quantity of gravel is 226 tons, and the quantity of asphalt is 175 tons. The following workup sheet contains detailed calculations.

The solution can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

48. Calculate the excavation needed for the footing and foundation for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the earthwork crew from Problem 7-32, determine the cost of the excavation.

The pricing is shown in the following workup sheet:

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

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Chapter 11: Concrete

Chapter 11—Concrete Learning Outcomes This chapter looks at determining the materials and labor associated with Division 3 Concrete of the MasterFormat. After reading this chapter, the reader should be able to: Learning Outcome 11.1: Describe the advantages and disadvantages of plant-mixed and job-mixed concrete.

Learning Outcome 11.2: Calculate concrete quantities and costs.

Learning Outcome 11.3: Calculate rebar quantities and costs.

Learning Outcome 11.4: Calculate vapor barrier quantities and costs.

Learning Outcome 11.5: Describe other components that must be included in the concrete bid.

Learning Outcome 11.6: Calculate formwork quantities and costs.

Learning Outcome 11.7: Describe how precast concrete is bid.

Activities 1. Invite a form supplier or manufacturer representative to the class to discuss how to estimate concrete formwork. Have them answer the following questions: 

What components are needed to complete the forms?



How long does it take to assemble the forms?



What is the cost of the forming system?



How many times may the forms be reused?

2. Discuss with the students that forming costs are a function of the number of times the forms are used on a project, particularly when the forms may be assembled in panels and moved as an entire panel. Divide the students into groups and provide each group with plans for a building or retaining wall with several footings and foundation walls. Have the students prepare a plan to form the walls using multiple pours. They should try to balance reusing the forms with minimizing the number of pours. Have the students present their solutions to the class.

Instructional Resources PowerPoint Slides: Chapter 11.ppt The pdf drawings from Appendices A through E. They can be found at the Pearson Careers Resources for Students website (http://www.pearsonhighered.com/careersresources/) by searching for the book by title or going to Browse by Discipline and choosing Civil and Environmental Engineering.

The drawings include the following: 78 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 11: Concrete



Appendix A – Small Commercial Building



Appendix B – Residential Building Project



Appendix C – Commercial Building Project



Appendix D – Billy’s C-Store



Appendix E – Real Estate Office

Solutions to the Review Questions 1. What is the difference between plant-mixed and job-mixed concrete?

Plant-mixed concrete is mixed at a plant and delivered to the job by the concrete company. Job-mixed concrete is mixed at the job site using a field batch plant.

2. Under what circumstances might it be desirable to have a field batch plant for jobmixed concrete?

Job-mixed concrete is advantageous when the volume of concrete is large enough to justify the cost of setting up a batch plant, when existing plants are not close enough to supply the concrete, or where nearby plants lack the needed capacity.

3. What is the unit of measure for concrete?

The unit of measure for concrete is the cubic yard (c.y.).

4. Why should the estimator keep the different places (such as walls, sidewalks, and floor slabs) where concrete will be used separate in the estimate?

They will have different requirements, such as strength, admixtures, forming, curing, and finishing.

5. Why should the different strengths of concrete be kept separate?

Different strengths require different amounts of cement. More cement is required to increase the concrete’s strength, which increases the cost.

6. Where would the estimator most likely find the strength of the concrete required?

The estimator should review the specifications to determine the strength of the concrete.

7. How is rebar taken off? In what unit of measure are large quantities ordered?

The rebar is usually taken off in linear feet of each size of bar and then converted into weight. Most rebar is sold by hundredweight (cwt) or ton.

8. How does lap affect the rebar quantities?

Lapping adds 5 to 18 percent, depending on the size of the bar and the required lap.

9. How is W W F taken off? How is it ordered?

WW F is taken off by the square feet required and converted into the number of rolls or sheets that need to be ordered.

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Chapter 11: Concrete

10. How is the vapor barrier taken off? How does the estimator determine the number of rolls required?

The vapor barrier is taken off by the square foot required and converted into the size of rolls needed. The most accurate method of determining the vapor retarder is sketching its placement layout. Often, several sketches must be made, trying rolls of various sizes and orientations before the most economical layout is found.

11. What unit of measure is used when taking off expansion joint fillers?

The expansion joint filler is taken off by linear feet.

12. What unit of measure is used when taking off concrete finishes?

Concrete finishes are estimated by the area (square feet or square yards) of the surface to be finished.

13. Why should each finish be listed separately on the estimate?

Each finish will require different amounts of material and labor.

14. Where would the estimator look to determine the concrete curing requirement?

The curing requirements are spelled out in the specifications.

15. Why must the estimator consider how the concrete will be transported to the job site?

The concrete transportation method determines the equipment cost and production rate for the given crew of workers.

16. What unit of measure is used when taking off concrete forms? How can the reuse of forms affect the estimate?

The unit of measurement used for forms is the square foot of contact area (s.f.c.a.) of the concrete against the forms. Reusing forms reduces the number of forms that must be purchased or rented.

17. What unit of measure is used for form liners?

Liners are estimated by the square footage of the surface requiring a particular type of liner and may be left in square feet or converted to the number of liner pieces.

18. What two pricing methods might a specialty contractor use for precast concrete?

Most precast concrete is priced by the unit price (such as a square foot or linear foot) or as a lump sum.

19. Many suppliers take responsibility for installing the precast units. Why might this be desirable?

Contractors often require the precast concrete supplier to install the units because the contractor lacks personnel experienced in the erection of precast concrete and because the supplier assumes responsibility for damage to the precast concrete during erection.

20. Under what conditions might it be desirable for a contractor to precast the concrete on the job site?

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Chapter 11: Concrete

A contractor might decide to precast the concrete on the job site if no precast plant is located close enough to serve the project, the pieces are too large to transport, or it is more economical to precast the concrete on the job site.

21. Using a waste factor of seven percent, determine the number of cubic yards of concrete needed to pour the continuous footings shown in Figures 11.36 and 11.37.

Assume that the wall is centered over the footing.

The footings extend 11 inches beyond the wall on both sides. The lengths of the footings are shown in Figure 11-1. Other lengths may be used, depending on how the corners are handled.

Figure 11-1

Footing length  22 ft. 10 in.  5 ft.  15 ft.  16 ft. 10 in.  37 ft. 10 in.  21 ft. 10 in.

Footing length = 119 f t. 4 in. = 119.33 f t. ee

ee

Cross-sectional area  2 ft. 6 in.  1 ft.  2.50 ft.  1 ft.  2.5 s.f.

Volume of concrete  2.5 s.f.  119.33 ft.  298 c.f.

Volume of concrete 

298c.f.  11.04c.y.

27c.f. c.y.

Add seven percent for waste and round off: Volume of concrete  11.04 c.y.  1 + 0.07   11.81 c.y.

Order 12 c.y.

22. Determine the amount of rebar needed for the continuous footings shown in Figures 11.36 and 11.37. Allow for 2 inches of cover and extend the dowels 15 inches into the wall. Add 10 percent for lap and waste to the continuous bars.

From Problem 21, the length of the footing is 119.33 feet. The dowels are calculated as follows, with the number of spaces rounded up to the next whole number: 81 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 11: Concrete

Number of spaces 

119.33 ft.  120 spaces 1 ft.

Add one to get the number of dowels: Order 121 dowels.

The dowels extend 15 inches into the wall, 10 inches 12in.  2in.  into the footing, and then turn and continue for 13 inches  30in./2  2in.  .

Length per bar  15 in.  10 in.  13 in.  38 in.  3 ft. 2 in.  3.17 ft.

Total bar length  121 dowel  3.17 ft./dowel  384 ft.

Total bar weight  384 ft.  0.668 lb./ft.  257 lb.

The continuous bars are calculated as follows:

Total length of long bars  119.33 ft./bar  4 bars  477 ft.

Total weight of long bars  477 ft.  0.668 lb./ft.  319 lb.

Add ten percent for waste and lap: Total weight of long bars  319 lb.  1 + 0.10   351 lb.

Total weight of rebar  257 lb.  351 lb.  608 lb.

23. Determine the number of linear feet of forms needed for the continuous footings shown in Figures 11.36 and 11.37.

From Problem 21, the length of the footings is 119.33 feet.

Length of forms  2 sides  119.33 ft./side  238.66 ft.

24. Using a waste factor of 8 percent, determine the number of cubic yards of concrete needed to pour the continuous footings shown in Figures 11.38 and 11.39. Assume that the wall is centered over the footing.

The footings extend one foot beyond the wall on both sides. The lengths of the footings are shown in Figure 11-2. Other lengths may be used, depending on how the corners are handled.

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Chapter 11: Concrete

Figure 11-2

Length of footing  26 ft.  42 ft.  26 ft.  42 ft.  12 ft.  15 ft.  163 ft.

Cross-sectional area  3 ft.  1 ft. 3 in.  3 ft.  1.25 ft.  3.75 s.f.

Volumeof concrete  3.75 s.f.  163 ft. 

611 c.f.  22.63 c.y.

27c.f. c.y.

Add eight percent for waste and round off: Volume of concrete  22.63 c.y.  1 + 0.08   24.44 c.y.

Order 24.5 c.y.

25. Determine the amount of rebar needed for the continuous footings shown in Figures 11.38 and 11.39. Allow for 2 inches of cover and extend the dowels 15 inches into the wall. Add 10 percent for lap and waste to the continuous bars.

From Problem 24, the length of the footing is 163 feet. The number of short bars and dowels is calculated as follows: Number of spaces 

163 ft.  163 spaces 1 ft.

Add one to get the number of dowels or short bars: Order 164 dowels and short bars.

Length of short bars  36 in.  2  2 in.  32 in.  2 ft. 8 in.  2.67 ft.

Total bar length  164 bars  2.67 ft./bar  438 ft.

Total bar weight  438 ft.  0.668 lb./ft.  293 lb.

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Chapter 11: Concrete

There are two sets of dowels, one on each side of the wall. The dowels extend 15 inches into the wall, 13 inches 15in.  2in. into the footing, and then turn and continue for 12 inches 12 in.  2 in.  2 in. .

Dowel length  15 in.  13 in.  12 in.  40 in.  3 ft. 4 in.  3.33 ft.

Total bar length  2 ea.  164 dowels  3.33 ft./dowel  1,092 ft.

Total bar weight  1,092 ft.  1.043 lb./ft.  1,139 lb.

The continuous bars are calculated as follows:

Total length of long bars  5 bars  163 ft./bar  815 ft.

Total weight of long bars  815 ft.  0.668 lb./ft.  544 lb.

Add ten percent for waste and lap: Total weight of long bars  544 lb.  1 + 0.10   598 lb.

Total weight of rebar  293 lb.  1,139 lb.  598 lb.

Total weight of rebar = 2,030 lb.

26. Determine the number of linear feet of forms needed for the continuous footings shown in Figures 11.38 and 11.39.

From Problem 24, the length of the footings is 163 feet.

Length of forms  2 sides  163 ft./side  326 ft.

27. Using a waste factor of 8 percent, determine the number of cubic yards of concrete needed to pour two of the spread footings in Figure 11.40.

Find the volume of concrete as follows:

Volume of concrete  2 ea.  2 ft. 6 in.  2 ft. 6 in.  10 in.

Volume of concrete  2 ea.  2.50 ft.  2.50 ft.  0.83 ft.  10.38 c.f.

Volume of concrete 

10.38c.f.  0.38c.y.

27c.f. c.y.

Add eight percent for waste and round off: Volume of concrete  0.38 c.y.  1 + 0.08   0.41 c.y.

Order 0.5 c.y.

28. Determine the amount of rebar needed for two of the spread footings shown in Figure 11.40. Allow for 2 inches of cover.

The dowels extend 34 inches (36in.  2in.) into the columns, 8 inches (10in.  2in.) into the footing, and turn and continue for 9 inches (30in./2  12in./2).

Dowel length  34 in.  8 in.  9 in.  51 in.  4 ft. 3 in.  4.25 ft.

84 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 11: Concrete

Total bar length  2 footings  4 bars/footing  4.25 ft./bar  34 ft.

Total bar weight  34 ft.  1.043 lb./ft.  35.5 lb.

There are eight horizontal bars per footing, four in each direction.

Length horizontal bars  30 in.  2  2 in.  26 in.  2 ft. 2 in.  2.17 ft.

Total length of horizontal bars  2 footings  8 bars/footing  2.17 ft./bar Total length of horizontal bars = 34.7 f t. ee

Total weight of horizontal bars  34.7 ft.  0.668 lb./ft.  23.2 lb.

Total weight = 35.5 lb. + 23.2 lb. = 58.7 lb.

29. Using a waste factor of 10 percent, determine the number of cubic yards of concrete needed to pour the spread footing in Figure 11.41.

Find the volume of concrete as follows: Volumeof concrete  8 ft.  17 ft.  2.5 ft. 

340 c.f.  12.59 c.y.

27c.f. c.y.

Add ten percent for waste and round off: Volume of concrete  12.59 c.y.  1  0.10   13.8 c.y.

Order 14 c.y.

30. Determine the amount of rebar needed for the spread footing Figure 11.41. The dowels extend 24 inches into the column. Allow for 3 inches of cover.

The dowels extend 24 inches into the column, 27 inches (2ft. 6in.  3 in.) into the footing, and turn and continue for 30 inches (8 ft./2  3 ft./2).

Dowel length  24 in.  27 in.  30 in.  81 in.  6 ft. 9 in.  6.75 ft.

Total bar length  28 dowels  6.75 ft./dowel  189 ft.

Total bar weight  189 ft.  4.303 lb./ft.  813 lb.

Short horizontal bars  8 ft.  2  3 in.  7 ft. 6 in.  7.5 ft./bar Spaces 

17 ft.  12in. ft.  2  3in.  22spaces 9 in.

Add one to get the number of short horizontal bars: Order 23 bars per mat.

Total length of short bars  2 mats  23 bars/mat  7.5 ft./bar  345 ft.

Total weight of short bars  345 ft.  2.67 lb./ft.  921 lb.

Long horizontal bars  17 ft.  2  3 in.  16 ft. 6 in  16.50 ft./bar 85 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 11: Concrete

Spaces 

8 ft.  12in. ft.  2  3in.  10spaces 9 in.

Add one to get the number of long horizontal bars: Order 11 bars per mat.

Total length of short bars  2 mats  11 bars/mat  16.5 ft./bar  363 ft.

Total weight of short bars  363 ft.  2.67 lb./ft.  969 lb.

Total weight = 813 lb. + 921 lb. + 969 lb. = 2,703 lb.

31. Using a waste factor of 5 percent, determine the number of cubic yards of concrete needed to pour two of the piers in Figure 11.40.

Find the cross-sectional area:

Cross-sectional area  1 ft.  1 ft.  1 s.f.

Volumeof piers  2ea.  1 s.f. ea.  3 ft. 

6 c.f.  0.22 c.y.

27c.f. c.y.

Add five percent for waste and round off: Volume in piers  0.22 c.y.  1  0.05   0.23 c.y.

Order 0.25 c.y.

32. Determine the amount of rebar needed for two of the piers shown in Figure 11.40.

Allow for 2 inches of cover.

The dowels are part of the footing rebar. The ties are 8 inches  2 in.  2  2 in. square.

Tie length  4 sides  8 in./side  32 in.  2 ft. 8 in.  2.67 ft.

Number of spaces 

3 ft.  12in. ft.  12spaces 3 in.

Add one to get the number of ties.

Order 13 ties per column.

Total length of ties  2 columns  26 ties/column  2.67 ft./tie  69.4 ft.

Total weight of ties  69.4 ft.  0.376 lb./ft.  26.1 lb.

33. Using a waste factor of 4 percent, determine the number of cubic yards of concrete needed to pour the column in Figure 11.41.

Find the cross-sectional area:

Cross-sectional area  3 ft.  12 ft.  36 s.f.

Volumeof piers  36s.f. 10 ft. 

360 c.f.  13.33 c.y.

27c.f. c.y.

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Chapter 11: Concrete

Add four percent for waste and round off: Volume in piers  13.33 c.y.  1  0.04   13.86 c.y.

Order 14 c.y.

34. Determine the amount of rebar needed for the column shown in Figure 11.41. Allow for 3 inches of cover.

The dowels are part of the footing rebar. The ties are 2 feet 6 inches  3 ft.  2  3 in. by 11 feet 6 inches 12 ft.  2  3 in. .

Length  2 ft. 6 in  11 ft. 6 in.  2 ft. 6 in  11 ft. 6 in.  28 ft.

Number of spaces 

10 ft.  12in. ft.  20spaces 6 in.

Add one to get the number of ties: Order 21 ties.

Total length of ties  21 ties  28 ft./tie  588 ft.

Total weight of ties  588 ft.  1.502 lb./ft.  883 lb.

Vertical bars length  10 ft.  3 in.  9 ft. 9 in.  9.75 ft.

Total bar length  28 bars  9.75 ft./bar  273 ft.

Total bar weight  273 ft.  4.303 lb./ft.  1,175 lb.

Total weight  883 lb.  1,175 lb.  2,058 lb.

35. A building requires four drilled piers with a shaft diameter of 24 inches, a bell diameter of 48 inches, and a bell angle of 50 degrees. The height from the bottom of the bell to the top of the pier is 10 feet. Using a waste factor of 12 percent, determine the number of cubic yards of concrete needed to pour the drilled piers.

Shaft diameter = 24 in. = 2 f t. ee

Bell diameter = 48 in. = 4 f t. ee

Find the height of the bell using Equation 11-2 or Figure 11-7 from the textbook: BS  4 ft.  2 ft.  H  tan  A     tan  50       1,192 ft.

2  2   

Find the bell volume using Equation 11-4 or Figure 11-7:

Volumeof bell  Volumeof bell 

  1.192ft.

12

H

S  B  SB 12 2

2

2ft.   4 ft.  2ft.  4ft. 2

2

Volume of bell = 8.74 c.f.

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Chapter 11: Concrete

Length of shaft  10 ft.  1.192 ft.  8.808 ft.

Find the shaft volume using Equation 11-5:

  S2  Shaft volume     Length of Shaft   4     2 ft.2    8,808 ft.  27.67c.f.

Shaft volume     4   Volume of concrete in piers  Count   Volume in bell  Volume in shaft  Volume of concrete in piers  4 ea.   8.74 c.f./ea.  27.67 c.f./ea.  146 c.f.

Volumeof concrete in piers 

146 c.f.  5.4 c.y.

27c.f. c.y.

Add 12 percent for waste and round off: Volume of concrete  5.4 c.y.  1  0.12   6.0 c.y.

Order 6 c.y.

36. A building requires ten drilled piers with a shaft diameter of 30 inches, a bell diameter of 54 inches, and a bell angle of 60 degrees. The height from the bottom of the bell to the top of the pier is 15 feet. Using a waste factor of 15 percent, determine the number of cubic yards of concrete needed to pour the drilled piers.

Shaft diameter = 30 in. = 2 f t. 6 in. = 2.5 f t. ee

ee

Bell diameter = 54 in. = 4 f t. 6 in. = 4.5 f t. ee

ee

Find the height of the bell using Equation 11-2 or Figure 11-7 from the textbook: BS  4.5 ft.  2.5 ft.  H  tan  A     tan 600      1,732 ft.

2  2   





Find the bell volume using Equation 11-4 or Figure 11-7:

Volumeof bell  Volumeof bell 

  1.732ft.

12

H

S  B  SB 12 2

2

2.5 ft.   4.5 ft.  2.5 ft.   4.5 ft. 2

2

Volume of bell = 17.12 c.f.

Length of shaft  15 ft.  1.732 ft.  13.268 ft.

Find the shaft volume using Equation 11-5: 88 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 11: Concrete

  S2  Shaft volume     Length of Shaft   4     2.5 ft.2    13,268 ft.  65.13c.f.

Shaft volume     4   Volume of concrete in piers  Count   Volume in bell  Volume in shaft  Volume of concrete in piers  10 ea.  17.12 c.f./ea.  65.13 c.f./ea.  822 c.f.

Volumeof concrete in piers 

822 c.f.  30.4 c.y.

27c.f. c.y.

Add 15 percent for waste and round off: Volume of concrete  30.4 c.y.  1  0.15   35.0 c.y.

Order 35 c.y.

37. Using a waste factor of 5 percent, determine the number of cubic yards of concrete needed to pour the foundation walls shown in Figures 11.36 and 11.37.

The wall lengths are shown in Figure 11-3. Other lengths may be used, depending on how the corners are handled.

Figure 11-3

Length of wall  21 ft.  4 ft. 4 in.  15 ft. 8 in.  18 ft. 8 in.  36 ft.  23 ft. 8 in.

Length of wall = 119 f t. 4 in. = 119.33 f t. ee

ee

Foundation wall concrete  119.33 ft.  10 ft.  8 in.

Foundation wall concrete  119.33 ft. 10 ft.  0.67 ft.  800 c.f.

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Chapter 11: Concrete

Foundation wall concrete 

800 c.f.  29.63 c.y.

27c.f. c.y.

Add five percent for waste and round off:

Foundation wall concrete  29.63 c.y.  1  0.05   31.1 c.y.

Order 31 c.y.

38. Determine the amount of rebar needed for the foundation walls shown in Figures 11.36 and 11.37. Allow for 2 inches of cover. Add 10 percent for lap and waste to the continuous bars.

From Problem 37, the length of the wall is 119.33 feet. The vertical bars are calculated as follows, with the number of spaces rounded up to the next whole number: Number of spaces 

119.33 ft.  120spaces 1 ft.

Add one to get the number of vertical bars  Order 121 bars: Vertical bars length  10 ft.  2 in.  9 ft. 10 in.  9.83 ft.

Total length of vertical bars  121 bars  9.83 ft./bar  1,189 ft.

Total weight of vertical bars  1,189 ft.  0.668 lb./ft.  794 lb.

The horizontal bars are calculated as follows: Number of spaces 

10 ft.  10spaces 1 ft.

Add one to get the number of horizontal bars needed to meet the spacing requirements, plus an additional top and bottom bar (10 + 1 + 1 + 1). Order 13 bars:

Total length of horizontal bars  13 bars  119.33 ft./bar  1,551 ft.

Total weight of horizontal bars  1,551 ft.  0.668 lb./ft.  1,036 lb.

Add ten percent for waste and lap: Total weight of horizontal bars  1,036 lb.  1  0.10   1,140 lb.

Total weight of rebar = 794 lb. + 1,140 lb. = 1,934 lb.

39. Determine the cost to install the rebar for the foundation walls in Figures 10.36 and 10.37 using a productivity rate of 10.75 labor hours per ton and an average labor rate of $25.62 per labor hour.

From Problem 38, the weight of the rebar is 1,934 pounds.

Labor hour 

1,935 Ib.  10.75Ihr. ton  10.4 Ihr. 2,000Ib. ton 90

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Chapter 11: Concrete

Labor cost  10.4 lhr.  $25.62/lhr.  $266.45

40. Determine the square foot contact area (s.f.c.a.) of the forms needed for the foundation walls shown in Figures 11.36 and 11.37.

From Problem 37, the length of the foundation wall is 119.33 feet.

Area of forms  2 sides  119.33 ft.  10 ft.  2,387 s.f.c.a.

41. Determine the cost to form the foundation walls in Figures 11.36 and 11.37 using a productivity rate of 0.125 labor hours per square foot contract area and an average labor rate of $25.62 per labor hour.

From Problem 40, the area of the forms is 2,387 s.f.c.a.

Labor hour  2,387 s.f.c.a.  0.125 lhr./s.f.c.a.  298 lhr.

Labor cost  298 lhr.  $25.62/lhr.  $7,634.76

42. Using a waste factor of 6 percent, determine the number of cubic yards of concrete needed to pour the foundation walls shown in Figures 11.38 and 11.39.

The wall lengths are shown in Figure 11-4. Other lengths may be used, depending on how the corners are handled.

Figure 11-4

Length of wall  28 ft.  40 ft.  28 ft.  40 ft.  12 ft.  17 ft.  165 ft.

Foundation wall concrete  165 ft.  12 ft.  1 ft.  1,980 c.f.

Foundation wall concrete 

1,980 c.f.  73.33 c.y.

27c.f. c.y.

Add six percent for waste and round off: Foundation wall concrete  73.33 c.y.  1  0.06   77.73 c.y.

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Chapter 11: Concrete

Order 77.75 c.y.

43. Determine the amount of rebar needed for the foundation walls shown in Figures 11.38 and 11.39. Allow for 2 inches of cover. Add 10 percent for lap and waste to the continuous bars.

From Problem 42, the length of the wall is 165 feet. The vertical bars are calculated as follows: Number of spaces 

165 ft.  165spaces 1 ft.

Add one to get the number of vertical bars  Order 166 bars per side: Vertical bars length  12 ft.  2 in.  11 ft. 10 in.  11.83 ft.

Total length of vertical bars  2 sides  166 bars/side  11.83 ft./bar Total length of vertical bars = 3,928 f t. ee

Total weight of vertical bars  3,928 ft.  0.668 lb./ft.  2,624 lb.

The horizontal bars are calculated as follows: Number of spaces 

12 ft.  12spaces 1 ft.

Add one to get the number of horizontal bars needed to meet the spacing requirements for one side. Order 13 bars per side:

Total length of horizontal bars = 2 sides  13 bars/side  165 ft./bar  4,290 ft.

Total weight of horizontal bars  4,290 ft.  0.668 lb./ft.  2,866 lb.

Add ten percent for waste and lap: Total weight of horizontal bars  2,866 lb.  1  0.10   3,153 lb.

Total weight of rebar  2,624 lb.  3,153 lb.  5,777 lb.

44. Determine the cost to install the rebar for the foundation walls in Figures 11.38 and 11.39 using a productivity rate of 10.3 labor hours per ton and an average labor rate of $31.26 per labor hour.

From Problem 43, the weight of the rebar is 5,777 pounds.

Labor hour 

5,777Ib.  10.3Ihr. ton  29.8Ihr. 2,000 Ib. ton

Labor cost  29.8 lhr.  $31.26 per lhr.  $931.55 45. Determine the square foot contact area of the forms needed for the foundation walls shown in Figures 11.38 and 11.39.

From Problem 42, the length of the foundation wall is 165 feet.

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Chapter 11: Concrete

Area of forms  2 sides  165 ft.  12 ft.  3,960 s.f.c.a.

46. Determine the cost to form the foundation walls in Figures 11.38 and 11.39 using a productivity rate of 0.11 labor hours per square foot contract area and an average labor rate of $32.17 per labor hour.

From Problem 45, the area of the forms is 3,960 s.f.c.a.

Labor hour  3,960 s.f.c.a.  0.11 lhr./s.f.c.a.  436 lhr.

Labor cost  436 lhr.  $32.17/lhr.  $14,026.12

47. A building requires 20 feet of the grade beam, whose cross-section is shown in Figure 11.42. Using a waste factor of 8 percent, determine the number of cubic yards of concrete needed to pour the grade beam.

Find the cross-sectional area:

Cross-sectional area  12 in.  14 in.  1 ft.  1.17 ft.  1.17 s.f.

Volumein grade beam  20 ft.  1.17s.f. 

23.4 c.f.  0.87 c.y.

27c.f. c.y.

Add eight percent for waste and round off: Volume in piers  0.87 c.y.  1  0.08   0.94 c.y.

Order 1 c.y.

48. Determine the amount of rebar needed for the grade beam in Problem 47. Allow for 2 inches of cover and ignore waste.

The ties are 8 inches 12 in.  2  2 in. by 10 inches 14 in.  2  2 in. .

Tie length  8 in.  10 in.  8 in.  10 in.  36 in.  3 ft.

The number of ties is calculated as follows: Number of spaces 

20 ft.  12in. ft.  30spaces 8 ft.

Add one to get the number of ties  Order 31 ties: Total length of ties  31 ties  3 ft./tie  93 ft.

Total weight of ties  93 ft.  0.376 lb./ft.  35 lb.

The horizontal bars are 20 feet long and calculated as follows:

Total length of horizontal bars  4 ea.  20 ft./ea.  80 ft.

Total weight of horizontal bars  80 ft.  1.043 lb./ft.  83 lb.

Total weight of rebar = 35 lb. + 83 lb. = 118 lb.

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Chapter 11: Concrete

49. A building requires 35 feet of the grade beam, whose cross-section is shown in Figure 11.43. Using a waste factor of 10 percent, determine the number of cubic yards of concrete needed to pour the grade beam.

Find the cross-sectional area:

Cross  sectional area  12 in.  16 in.  1 ft.  1.33 ft.  1.33 s.f.

Volumein grade beam  35 ft.  1.33s.f. 

46.6 c.f.  1.73 c.y.

27c.f. c.y.

Add ten percent for waste and round off: Volume in piers  1.73 c.y.  1  0.10   1.90 c.y.

Order 2 c.y.

50. Determine the amount of rebar needed for the grade beam in Problem 49. Allow for 3 inches of cover. Add 10 percent for lap and waste to the continuous bars.

The ties are 6 inches 12 in.  2  3 in. by 10 inches 16 in.  2  3 in. .

Tie length  6 in.  10 in.  6 in.  10 in.  32 in.  2 ft. 8 in.  2.67 ft.

The number of ties is calculated as follows: Number of spaces 

35 ft.  12in. ft.  70spaces 6 ft.

Add one to get the number of ties  Order 71 ties: Total length of ties  71 ties  2.67 ft./tie  190 ft.

Total weight of ties  190 ft.  0.376 lb./ft.  71 lb.

The continuous bars are calculated as follows:

Total length of continuous bars  4 ea.  35 ft./ea.  140 ft.

Total weight of continuous bars  140 ft.  0.668 lb./ft.  93.5 lb.

Add ten percent for waste and lap: Total weight of horizontal bars  93.5 lb.  1  0.10   103 lb.

Total weight of rebar = 71 lb. + 103 lb. = 174 lb.

51. A building requires a 100-foot-long by 83-foot-wide by 5-inch-thick concrete slab.

Using a waste factor of 10 percent, determine the number of cubic yards of concrete needed to pour the slab.

Find the quantity of concrete as follows:

Quantity of concrete  100 ft.  83 ft.  5 in.  100 ft.  83 ft.  0.42 ft.  3,486 c.f.

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Chapter 11: Concrete

Quantity of concrete 

3,486 c.f.  129 c.y.

27c.f. c.y.

Add ten percent for waste: Quantity of concrete  129 c.y.  1  0.10   142 c.y.

52. Determine the number of rolls of W W F needed for the slab in Problem 51. There are 750 square feet of W W F per roll. Add 12 percent for lap and waste.

Find the area of the slab as follows: Area of slab  100 ft.  83 ft.  8,300 s.f.

Rolls of mesh required 

8,300s.f.  11.1 rolls 750s.f. roll

Add 12 percent for waste and lap: Rolls of mesh required  11.1 rolls  1  0.12   12.43 rolls

Order 13 rolls.

53. Determine the amount of rebar needed for the slab in Problem 51. The slab is reinforced with #4 rebar at 12 inches on center both ways. Assume 2 inches of cover and add 10 percent for lap and waste.

Determine the number of long bars:

Long bars length  100 ft.  2  2 in.  99 ft. 8 in.  99.67 ft.

Long bar spaces 

83 ft.  83spaces 1 ft.

Add one to get the number of long bars. Order 84 bars:

Total length of long bars  84 bars  99.67 ft./bar  8,372 ft.

Total weight of long bars  8,372 ft.  0.668 lb./ft.  5,592 lb.

Determine the number of short bars:

Short bars length  83 ft. – 2  2 in.  82 ft. 8 in.  82.67 ft.

Short bar spaces 

100 ft.  100spaces 1 ft.

Add one to get the number of short bars. Order 101 bars:

Total length of short bars  101 bars  82.67 ft./bar  8,350 ft.

Total weight of short bars  8,350 ft.  0.668 lb./ft.  5,578 lb.

Total weight of rebar = 5,592 lb. + 5,578 lb. = 11,170 lb.

Add ten percent for waste and lap: 95 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 11: Concrete

Total weight of rebar  11,170 lb.  1  0.10   12,287 lb.

54. Determine the cost to finish the slab in Problem 51 using a productivity rate of 0.011 labor hours per square foot and an average labor rate of $29.92 per labor hour.

From Problem 52, the area of the slab is 8,300 square feet:

Labor hour  8,300 s.f.  0.011 lhr./s.f.  91.3 lhr.

Labor cost  91.3 lhr.  $29.92/lhr.  $2,731.70

55. A vapor barrier is to be placed below the slab in Problem 51. Determine the number of 10-foot by 100-foot rolls of vapor barrier needed. The vapor barrier needs to be lapped 1 foot. Hint: Subtract the lap from the width of the roll.

Run the rolls in the 100 f t. direction; therefore, one roll will cover the length of the slab. ee

Roll width  10 ft.  1 ft.  9 ft.

Number of rolls wide 

83 ft.  9.2rolls 9 ft. roll

Order 10 rolls 100 f t. long. ee

56. Using a waste factor of 12 percent, determine the amount of concrete needed to pour the slab shown in Figure 11.44. The slab is 4 inches thick.

Find the area of the slab as follows: Area of slab  25 ft.  30 ft.  15 ft.  20 ft.  1,050 s.f.

or Area of slab  40 ft.  30 ft.  15 ft.  10 ft.  1,050 s.f.

Slab thickness = 4 in. = 0.33 f t. ee

Quantity of concrete  1,050s.f.  0.33 ft. 

346 c.f.  12.8 c.y.

27c.f. c.y.

Add 12 percent for waste: Quantity of concrete  12.8 c.y.  1  0.12   14.3 c.y.

Order 14.5 c.y.

57. The slab in Problem 56 is reinforced with W W F. Determine the number of rolls of W W F needed for the slab. There are 750 square feet of W W F per roll. Add 11 percent for lap and waste.

From Problem 56, the area of the slab is 1,050 square feet: Rolls of mesh required 

1,050s.f.  1.4rolls 750s.f. roll

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Chapter 11: Concrete

Rolls of mesh required  1.4 rolls  1  0.11  1.6 rolls

Order 2 rolls.

58. Determine the amount of rebar needed for the slab in Problem 56. The slab is reinforced with #4 rebar at 18 inches on center both ways. Assume 2 inches of cover and add 10 percent for lap and waste.

Start with the 25-foot by 30-foot area. Determine the number of long bars:

Long bar length  30 ft.  2  2 in.  29 ft. 8 in.  29.67 ft.

Long bar spaces 

25 ft.  17spaces 1.5 ft.

Add one to get the number of long bars. Order 18 bars:

Total length of long bars  18 bars  29.67 ft./bar  534 ft.

Determine the number of short bars:

Short bar length  25 ft.  2  2 in.  24 ft. 8 in.  24.67 ft.

Short bar spaces 

30 ft.  20spaces 1.5 ft.

Add one to get the number of short bars. Order 21 bars:

Total length of short bars  21 bars  24.67 ft./bar  518 ft.

Calculate the rebar for the 15-foot by 20-foot area. Determine the number of long bars:

Long bar length  20 ft.  2  2 in.  19 ft. 8 in.  19.67 ft.

Long bar spaces 

15 ft.  10spaces 1.5 ft.

Add one to get the number of long bars. Order 11 bars:

Total length of long bars  11 bars  19.67 ft./bar  216 ft.

Determine the number of short bars:

Short bar length  15 ft.  2  2 in.  14 ft. 8 in.  14.67 ft.

Short bar spaces 

20 ft.  14spaces 1.5 ft.

Add one to get the number of short bars. Order 15 bars:

Total length of short bars  15 bars  14.67 ft./bar  220 ft.

Total length of rebar  534 ft.  518 ft.  216 ft.  220 ft.  1,488 ft.

Total weight of rebar  1,488 ft.  0.668 lb./ft.  994 lb.

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Chapter 11: Concrete

Add ten percent for waste and lap:

Total weight of bars  994 lb.  1  0.10   1,093 lb.

59. Determine the cost to finish the slab in Problem 56 using a productivity rate of 0.013 labor hours per square foot and an average labor rate of $31.17 per labor hour.

From Problem 56, the area of the slab is 1,050 square feet:

Labor hour  1,050 s.f.  0.013 lhr./s.f.  13.65 lhr.

Labor cost  13.65 lhr.  $31.17/lhr.  $425.47

60. A vapor barrier is to be placed below the slab in Problem 56. Determine the number of 10-foot by 50-foot rolls of vapor barrier needed. The vapor barrier needs to be lapped 1 foot.

Run the rolls in the 30-foot direction. The 25-foot by 30-foot area will require three 10-foot by 30-foot pieces, and the 15-foot by 20-foot area will require two 10-foot by 20-foot pieces. The length of the plastic is calculated as follows:

Length  3 ea.  30 ft./ea.  2 ea.  20 ft./ea.  130 ft.

Number of rolls 

130 ft.  2.6rolls 50 ft. roll

Order 3 rolls.

61. Determine the amount of concrete, reinforcing, forms, and accessories required for the Commercial Building Project in Appendix C. The following workup sheets contain detailed concrete, reinforcing, and form calculations.

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Chapter 11: Concrete

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

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Chapter 11: Concrete

62. Using the drawings for Billy’s Convenience Store found in Appendix D, determine the quantity of concrete and reinforcing required for the spread footings.

The following workup sheet contains detailed calculations for the spread footings.

The solutions can also be found in Appendix D - Billys C-Store v10 Estimate.xlsx.

63. Using the drawings for Billy’s Convenience Store found in Appendix D, determine the quantity of concrete, reinforcing, and formwork in the slab on grade.

The following workup sheets contain detailed concrete, reinforcing, and formwork calculations.

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Chapter 11: Concrete

The solution can also be found in Appendix D - Billys C-Store v10 Estimate.xlsx.

64. Determine the amount of concrete, reinforcing, forms, and accessories required for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the concrete crew from Problem 7-32, determine the cost of the concrete.

The foundations are found under walls. Wood-framed walls must be filtered out, and the walls must be sorted by width. The eight-inch-wide walls represent the foundation wall, and the two-foot-wide walls represent the footings. The following quantities can be extracted from the model.

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The footings must be taken off as follows:

Length of perimeter footing  73 ft. 4 in.  47 ft. 4 in.  73 ft. 4 in.  47 ft. 4 in.

Length of perimeter footing = 241 f t. 4 f t. = 241.33 f t. ee

ee

ee

Find the volume of concrete:

Volume of perimeter footing  241.33 ft.  1 ft.  2 ft.  483 c.f.

Volume of perimeter footing 

483 c.f.  17.9 c.y.

27c.f. c.y.

Length of interior footings  70 ft. 8 in.  70 ft. 8 in.  141 ft. 4 in.  141.33 ft.

Volume of interior footings  141.33 ft.  8 in.  2 ft.

Volume of interior footings  141.33 ft.  0.67 ft.  2 ft.  189 c.f.

Volume of interior footing 

189 c.f.  7.0 c.y.

27c.f. c.y.

Concrete volume = 17.9 c.y. + 7.0 c.y. = 24.9 c.y.

This quantity matches the quantity (24.86 c.y.) extracted from the model.

Add ten percent for waste: Concrete volume  24.9 c.y.  1  0.10   27.4 c.y.

Order 27.5 c.y.

Find the footing forms:

Forms 12 in. high   2 sides  241.33 ft./side  483 ft.

Forms  8 in. high   2 sides  141.33 ft./side  283 ft.

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Find the rebar in the footings: Horizontal #4 rebar  3 ea.   241.33 ft./ea.  141.33 ft./ea.  1,148 ft

Add ten percent for waste and lap: Horizontal #4 rebar  1,148 ft.  1  0.10   1,263 ft.

Assume a two-inch cover on the dowels in the perimeter footings:

Dowels length  10 in.  10 in.  18 in.  38 in.  3 ft. 2 in.  3.17 ft.

Number of #4dowels 

241.33 ft. 12 in. ft.  181 spaces 16in.

Add one to get the number of dowels. Order 182 dowels:

#4 rebar for dowels  182 ea.  3.17 ft./ea.  577 ft.

Determine the number of anchor bolts: Anchor bolts 

141.33 ft. 12 in. ft.  106 spaces 16in.

Add one to get the number of anchor bolts. Order 107 anchor bolts.

The foundation is taken off as follows:

Length of foundation  72 ft.  48 ft. 8 in.  72 ft.  48 ft. 8 in.

Length of foundation = 241 f t. 4 in. = 241.33 f t. ee

ee

Find the volume of concrete:

Concrete volume  241.33 ft.  8 in.  1 ft. 8 in.

Concrete volume  241.33 ft.  0.67 ft.  1.67 ft.  270 c.f.

Concrete volume 

270 c.f.  10.0 c.y.

27c.f. c.y.

This quantity matches the quantity (9.93 c.y.) extracted from the model. The minor difference is due to rounding.

Add ten percent for waste: Concrete volume  10.0 c.y.  1  0.10   11.0 c.y.

Order 11.0 c.y.

Find the foundation forms:

Forms  2 sides  241.33 ft.  1 ft. 8 in.

Forms  2 sides  241.33 ft.  1.67 ft.  806 s.f.c.a.

Find the rebar in the footings: 103 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 11: Concrete

Horizontal #4 rebar  3 bars  241.33 ft./bar  724 ft.

Add ten percent for waste and lap: Horizontal #4 rebar  724 ft.  1  0.10   796 ft.

Slabdowels 

241.33 ft. 12 in. ft.  181 spaces 16 in.

Add one to get the number of dowels. Order 182 dowels: Dowel length = 18 in. + 18 in. = 36 in = 3 f t. ee

#4 rebar for slab dowels  182 ea.  3 ft./ea.  546 ft.

Determine the number of anchor bolts: Anchor bolts 

241.33 ft. 12 in. ft.  181 spaces 16in.

Add one to get the number of anchor bolts. Order 182 anchor bolts: Find the volume of concrete in the slab:

Concrete slab  72 ft.  50 ft.  4 in.

Concrete slab  72 ft.  50 ft.  0.33 ft.  1,188 c.f.

Concrete slab 

1,188 c.f.  44.0 c.y.

27c.f. c.y.

This quantity matches the quantity (44.44 c.y.) extracted from the model. The minor difference is due to rounding.

Add ten percent for waste: Concrete volume = 44.0 c.y.  1  0.10   48.4 c.y.

Order 48.5 c.y.

Find the total rebar:

Total length of #4 rebar = 1,263 ft.  577 ft.  796 ft.  546 ft.  3,182 ft.

Weight of #4 rebar  3,182 ft.  0.668 lb./ft.  2,126 lb.

The pricing is shown in the following workup sheet:

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Chapter 11: Concrete

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

We have used new template in the page ii. Pleae check and confirm.

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Chapter 12: Masonry

Chapter 12—Masonry Learning Outcomes This chapter looks at determining the materials and labor associated with Division 4 Masonry of the Master-Format. After reading this chapter, the reader should be able to: Learning Outcome 12.1: Describe the factors to consider when bidding on masonry.

Learning Outcome 12.2: Identify the different bond patterns.

Learning Outcome 12.3: Identify the different types of masonry.

Learning Outcome 12.4: Determine the materials and labor needed to install masonry.

Activities 1. Give the students a residence that uses brick facing and have them determine the quantity of brick needed for the residence.

Instructional Resources PowerPoint Slides: Chapter 12.ppt The pdf drawings from Appendices A through E. They can be found at the Pearson Careers Resources for Students website (http://www.pearsonhighered.com/careersresources/) by searching for the book by title or going to Browse by Discipline and choosing Civil and Environmental Engineering.

The drawings include the following: 

Appendix A – Small Commercial Building



Appendix B – Residential Building Project



Appendix C – Commercial Building Project



Appendix D – Billy’s C-Store



Appendix E – Real Estate Office

Solutions to the Review Questions 1. What factors affect the costs of labor when estimating masonry?

The amount of time required for masons and their helpers to lay a masonry unit varies with the: Size, weight, and shape of the unit.

Bond or masonry pattern.

Number and size of openings.

Whether the walls are straight or have jogs in them.

The distance the units must be moved both horizontally and vertically.

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Chapter 12: Masonry

The shape and color of the mortar joint.

Labor hours are directly related to labor costs.

2. How may special patterns affect the quantity of materials required?

Special patterns require additional cutting, increasing the material waste and labor costs.

3. Why are accurate estimates required for masonry?

Accurate estimates are needed to ensure the correct materials are ordered and the masonry is priced correctly.

4. Why should local suppliers be contacted early in the bidding process when special shapes or colors are required?

If special shapes or colors are required, local suppliers should be contacted early in the bidding process to give them time to review the contract documents and determine if they will manufacture the needed masonry or ship them in from another plant.

5. Why must the estimator separate the various sizes of masonry units in the estimate?

The estimator must separate masonry by the units’ size, shape, and color to help make the estimate as accurate as possible. Different masonry units, bonds, and mortars have different costs and require different amounts of labor.

6. What is a cash allowance, and how does it work?

A cash allowance is a designated amount for the purchase of face brick, allowing the owner and architect to select the type of brick later. The estimator includes the cash allowance in the estimate in place of the brick’s cost.

7. How are stone quantities estimated?

Stone is usually estimated by the area in square feet of a specified thickness, which is converted to cubic feet and cubic yards. Stone trim is usually estimated by the linear foot.

8. How may cold weather affect the cost of a building?

Cold-weather construction is more expensive than warm-weather construction. The increased costs come from constructing temporary enclosures to protect the masonry, the higher frequency of equipment repair, thawing materials, and the need for temporary heat.

9. Determine the number of 8-inch-high by 8-inch-wide by 16-inch-long concrete (CM U) blocks required to complete 250 feet of the fence, whose cross-section is shown in Figure 12.22. If lintel blocks are needed wherever the #4 horizontal bars are located, how many plain blocks and lintel blocks are needed for the wall?

Find the area of the wall as follows: Wall area  250 ft.  6 ft. 8 in.  250 ft.  6.67 ft.  1,668 s.f.

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Chapter 12: Masonry

Figure 12.6 shows that there are 1.125 blocks per square foot. The number of blocks is calculated as follows: Number of blocks  1,668 s.f.  1.125 blocks/s.f.  1,876 blocks

There are two rows of lintel blocks, one at 4 feet and one at 6 feet 8 inches. The lintel blocks are calculated as follows: Lintel bolts  2rows 

250 ft. 12 in. ft.  375 lintel blocks 16 in.

Plain blocks  1,876 blocks  375 blocks  1,501 plain blocks 10. Determine the amount of rebar needed for the CMU fence in Problem 9. Allow for 2 inches of cover, and add 10 percent for lap and waste to the horizontal bars.

The vertical bars are calculated as follows: Number of spaces 

250 ft. 12 in. ft.  94 spaces 32in.

Add one to get the number of vertical bars. Order 95 bars. The weight is calculated as follows:

Vertical bars length  6 ft. 8 in.  2 in.  6 ft. 6 in.  6.50 ft.

Total length of vertical bars  95 bars  6.50 ft./bar  618 ft.

Total weight of vertical bars  618 ft.  1.043 lb./ft.  645 lb.

Four horizontal bars are needed: two at 4 feet and two at 6 feet 8 inches. The horizontal bars are calculated as follows:

Total length of horizontal bars  4 bars  250 ft./bar  1,000 ft.

Weight of horizontal bars  1,000 ft.  0.668 lb./ft.  668 lb.

Add ten percent for waste and lap: Total weight of horizontal bars  668 lb.  1  0.10   735 lb.

Total weight of rebar = 645 lb. + 735 lb. = 1,380 lb.

11. Using 3.5 cubic feet of mortar per 100 square feet of wall, determine the amount of mortar needed for the C MU fence in Problem 9.

From Problem 9, the area of the fence is 1,668 square feet, or 16.68 squares. The mortar is calculated as follows:

Mortar  16.68 sq.  3.5 c.f./sq.  58.4 c.f.

12. Determine the cost to lay the CMU block in Problem 9. It takes 5.5 mason labor hours and 7.5 laborer labor hours to construct 100 square feet of wall. The average burdened wage rate is $39.74 per labor hour for the masons and $22.35 per labor hour for the laborers.

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Chapter 12: Masonry

From Problem 9, the area of the fence is 1,668 square feet or 16.68 squares. The labor is calculated as follows:

Mason labor hours  16.68 sq.  5.5 lhr./sq.  91.7 lhr.

Mason labor cost  91.7 lhr.  $39.74/lhr.  $3,644

Laborer labor hours  16.68 sq.  7.5 lhr./sq.  125.1 lhr.

Laborer labor cost  125.1 lhr.  $22.35/lhr.  $2,796

Total labor cost = $3,644 + $2,796 = $6,440 13. Determine the number of 8-inch-high by 8-inch-wide by 16-inch-long concrete blocks required to complete the wall in Figures 12.23 and 12.24. The overhead doors are 10 feet wide by 12 feet high. If lintel blocks are needed wherever the #4 horizontal bars are located and above the doors, how many plain blocks and lintel blocks are required for the wall?

Find the net area of the wall as follows: Gross wall area  80 ft.  17 ft. 4 in.  80 ft.  17.33 ft.  1,386 s.f.

Area of doors  4 ea.  10 ft.  12 ft.  480 s.f.

Net area  1,386 s.f.  480 s.f.  906 s.f.

Figure 12.6 shows that there are 1.125 blocks per square foot. The number of blocks is calculated as follows: Number of blocks  906 s.f.  1.125 blocks/s.f.  1,019 blocks

There are five rows of lintel blocks located at 4 feet, 8 feet, 12 feet, 16 feet, and 17 feet. 4 inches. The bottom three rows pass through the doors and are only 40 feet long ( 80 ft. – 4  10 ft. ). The top two rows are 80 feet long. There are 40 feet ( 4  10 ft. ) of lintel block above the doors. The blocks are calculated as follows:

Lintel blocks  3 rows  40 ft./row  2 rows  80 ft./row  40 ft.  320 ft.

Lintel blocks 

320 ft. 12 in. ft.  240 blocks 16in.

Plain blocks  1,019 blocks  240 blocks  779 blocks

14. Determine the amount of rebar needed for the CMU wall in Problem 13. In addition to the rebar shown in Figure 12.24, there is one #6 rebar on either side of the overhead doors, extending from the bottom of the wall to 16 inches above the top of the door, and two #4 rebar above the doors. Allow for 2 inches of cover on the vertical bars. Add 8 percent for lap and waste to the horizontal bars and 30 percent for lap and waste to the vertical bars.

The waste may seem high for the vertical bars. Thirty percent waste is used because the bars will be ordered in 5 f t. to 5.5 f t. lengths and lapped 1 f t. to 1.5 f t. to facilitate their installation. Using shorter bars eliminates the need to lift the blocks ee

ee

ee

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ee

Chapter 12: Masonry

over a full-height vertical bar or install the vertical bars after the block has been placed. The vertical #5 bars are calculated as follows: Number of spaces 

80 ft. 12 in. ft.  30spaces 32in.

Add one to get the number of vertical bars. Order 31 bars.

Half of these bars will be 17 feet 2 inches long (17 ft. 4 in.  2 in.), and the other half will be 5 feet 2 inches long (17 ft. 4 in.  2 ft.  2 in.) because they are above the overhead doors. The weight of the vertical bars is calculated as follows:

Total length of vertical bars  16 bars  17 ft. 2 in./bar  15 bars  5 ft. 2 in./bar Total length of vertical bars  16 bars  17.17 ft./bar  15 bars  5.17 ft./bar Total length of vertical bars = 352 f t. ee

Total weight of vertical bars  352 ft.  1.043 lb./ft.  367 lb.

Add 30 percent for lap and waste: Total weight of vertical bars  367 lb.  1  0.30   477 lb.

There are eight vertical #6 bars 13 feet 4 inches long (12 f t. + 16 in.) around the doors. The weight of these bars is calculated as follows: ee

Total length of vertical bars  8 bars  13 ft. 4 in./ bar Total length of vertical bars  8 bars  13.33 ft./bar  107 ft.

Total weight of vertical bars  107 ft.  1.502 lb./ft.  161 lb.

Add 30 percent for lap and waste: Total weight of vertical bars  161 lb.  1  0.30   209 lb.

From Problem 13, there are 320 feet of lintel blocks with two #4 in them, which includes the blocks above the doors. The weight of these bars is calculated as follows:

Total length of horizontal bars  2 bars  320 ft./bar  640 ft.

Total weight of horizontal bars  640 ft.  0.668 lb./ft.  428 lb.

Add eight percent for waste and lap: Total weight of horizontal bars  428 lb.  1  0.08   462 lb.

Total rebar weight = 477 lb. + 209 lb. + 462 lb. = 1,148 lb.

15. Using 3.0 cubic feet of mortar per 100 square feet of wall, determine the amount of mortar needed for the CMU wall in Problem 13.

From Problem 13, the area of the wall is 906 square feet, or 9.06 squares. The volume of mortar is calculated as follows: 110 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 12: Masonry

Mortar  9.06 sq.  3.0 c.f./sq.  27.2 c.f.

16. Determine the cost to lay the CMU block in Problem 13. It takes 6.0 mason labor hours and 8.5 laborer labor hours to construct 100 square feet of wall. The average burdened wage rate is $42.33 per labor hour for the masons and $21.03 per labor hour for the laborers.

From Problem 13, the area of the wall is 906 square feet, or 9.06 squares. The labor is calculated as follows:

Mason labor hours  9.06 sq.  6.0 lhr./sq.  54.4 lhr.

Mason labor cost  54.4 lhr.  $42.33/lhr.  $2,303

Laborer labor hours  9.06 sq.  8.5 lhr./sq.  77.0 lhr.

Laborer labor cost  77.0 lhr.  $21.03/lhr.  $1,619

Total labor cost = $2,303 + $1,619 = $3,922 17. Determine the number of modular bricks needed for the wall in Figure 12.25. Add 4 percent for waste.

Find the net area of the wall and the number of bricks as follows:

Gross wall area  8 ft.  30 ft.  10 ft.  24 ft.  0.5  4 ft.  24 ft.  528 s.f.

Areas of openings  4 ft.  4 ft.  6 ft. 8 in.  3 ft.  7 ft.  16 ft.

Areas of openings  4 ft.  4 ft.  6.67 ft.  3 ft.  7 ft.  16 ft.  148 s.f.

Net area of wall  528 s.f.  148 s.f.  380 s.f.  3.8 sq.

Figure 12.6 shows that 686 bricks are needed per 100 square feet (square). The bricks are calculated as follows:

Number of bricks  3.8 sq.  686 bricks/sq.  2,607 bricks Add four percent for waste: Number of bricks  2,607 bricks  1  0.04   2,711 bricks

18. Determine the number of modular bricks needed for the wall in Figure 12.26. The windows are 4 feet high by 4 feet wide, and the door is 6 feet wide by 7 feet high.

Add 5 percent for waste.

Find the net area of the wall as follows:

Gross wall area  10 ft.  48 ft.  480 s.f.

Areas of openings  4 ea.  4 ft.  4 ft.  6 ft.  7 ft.  106 s.f.

Net area of wall  480 s.f.  106 s.f.  374 s.f.  3.74 sq.

Figure 12.6 shows that 686 bricks are needed per 100 square feet (square). The bricks are calculated as follows:

Number of bricks  3.74 sq.  686 bricks/sq.  2,566 bricks 111 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 12: Masonry

Add five percent for waste: Number of bricks  2,566 bricks  1  0.05   2,694 bricks

19. Determine the masonry, mortar, accessories, and any other related items required for the building in Appendix B. The following workup sheet contains detailed masonry, mortar, and accessories calculations.

The solution can also be found in Appendix B - Residential Building Project v10 Estimate.xls x.

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Chapter 12: Masonry

20. Determine the masonry, mortar, and accessories required for the building in Appendix C. The following workup sheets contain detailed masonry, mortar, and accessories calculations.

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Chapter 12: Masonry

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

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Chapter 12: Masonry

21. Determine the quantity of brick veneer and split-faced block needed for Billy’s Convenience Store, found in Appendix D. The following workup sheets contain detailed brick veneer and split-faced block calculations.

The solution can also be found in Appendix D - Billys C-Store v10 Estimate.xlsx.

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Chapter 12: Masonry

22. Determine the amount of brick, mortar, accessories, and any other related items required for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the masonry crew from Problem 7-32, determine the cost of the masonry.

The following quantities can be extracted from the model. The brick is found by doing a material takeoff of the walls and filtering out the brick.

All the 24-inch by 48-inch windows are in the exterior walls. Only two of the 36-inch by 72-inch windows are in the exterior walls. There is one 3-foot by 7-foot door and one 6-foot by 7-foot door in the exterior walls. The remaining twenty 3-foot by 7-foot doors are in the interior walls. The area of the brick is calculated as follows: Length of wall = 70 f t. + 50 f t. + 70 f t. + 50 f t. = 240 f t. ee

ee

ee

ee

ee

Area of wall  240 ft.  9 ft. 1 in.  240 ft.  9.08 ft.  2,179 s.f.

Area of openings  34 ea.  24 in.  48 in.  2 ea.  36 in.  72 in.  3 ft.  7 ft.  6 ft.  7 ft.

Area of openings  34 ea.  2 ft.  4 ft.  2 ea.  3 ft.  6 ft.  3 ft.  7 ft.  6 ft.  7 ft.

Area of openings = 371 s.f.

Area of brick  2,179 s.f. – 371 s.f.  1,808 s.f.  18.08 sq.

This quantity is similar to the 1,818 square feet extracted from the model, with the difference due to rounding.

Figure 12.6 shows that there are 686 bricks per square. The number of bricks is calculated as follows:

Number of bricks  18.08 sq.  686 bricks / sq.  12,403 bricks Add five percent for waste: Number of bricks  12,403 bricks  1  0.05   13,023 bricks

Figure 12.6 shows that there are 7.2 c.f. of mortar per square. The quantity of mortar is calculated as follows: 116 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 12: Masonry

Mortar  18.08 sq.  7.2 c.f./sq.  130 c.f.

Using one tie for every 2.66 square feet, we get: Ties 

1,808s.f.  680 ties 2.66s.f. tie

Extend the steel lintels four inches on each side. The following lintels will be needed:

34 ea. @ 32 in. for the 24 in.  48 in. windows 2 ea. @ 44 in. for the 36 in.  72 in. windows 1 ea. @ 44 in. for the 3 ft.  7 ft. door 1 ea. @ 80 in. for the 6 ft.  7 ft. door The pricing is shown in the following workup sheet:

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

117 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 13: Metals

Chapter 13—Metals Learning Outcomes This chapter looks at determining the materials and labor associated with Division 5 Metals of the Master-Format. After reading this chapter, the reader should be able to: Learning Outcome 13.1: Describe the types of metals covered by Division 5 metals of the MasterFormat.

Learning Outcome 13.2: Describe the factors to consider when bidding on metals.

Learning Outcome 13.3: Determine the quantity and weight of structural steel.

Learning Outcome 13.4: Determine the quantity and weight of metal joists.

Learning Outcome 13.5: Determine the materials and labor needed for metal decking.

Activities 1. Divide the students into groups and give each group a copy of a steel framing plan for a small commercial building and load tables for a hydraulic crane. Have the groups estimate the weight of each of the steel members, including a ten percent allowance for connections. Have the groups determine where they will place the crane to erect the steel to minimize the number of crane locations and the cost of moving the crane. Have the students present their plans for placing the steel in the class.

Instructional Resources PowerPoint Slides: Chapter 13.ppt The pdf drawings from Appendices A through E. They can be found at the Pearson Careers Resources for Students website (http://www.pearsonhighered.com/careersresources/) by searching for the book by title or going to Browse by Discipline and choosing Civil and Environmental Engineering.

The drawings include the following: 

Appendix A – Small Commercial Building



Appendix B – Residential Building Project



Appendix C – Commercial Building Project



Appendix D – Billy’s C-Store



Appendix E – Real Estate Office

Solutions to the Review Questions 1. What two materials are commonly used for structural framing metals, and how are they priced?

118 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 13: Metals

The structural products are commonly made of steel and aluminum. Both aluminum and steel are sold by weight, so the takeoff is made in pounds and converted into tons.

2. Under what conditions might it be desirable for the contractor to use a specialty contractor to erect the structural metal frame of the project?

Often, specialty contractors can erect structural steel at considerable savings compared with the general contractor’s costs. The specialty contractors have specialized, well-organized workers and specialized equipment that includes cranes, air tools, welders, and impact wrenches.

3. Why should the estimator list the different shapes separately?

The estimator must list each of the different shapes separately because the cost per ton varies depending on the type and shape of the metal. Labor productivity and costs are different for each type and shape.

4. What is the unit of measure for metal decks? What type of information needs to be noted?

Metal decking is priced by the square (100 square feet). Information to be noted includes the type of decking, depth, gauge of metal, finish, method of attachment, required accessories, and the approved manufacturers.

5. How are metal fabrications usually priced?

Estimators should contact possible suppliers to discuss pricing and installation.

6. Prepare a structural steel materials list for the roof-framing plan shown in Figure 13.16. The columns are 16 feet high. How many pounds of steel need to be purchased for the roof?

The following materials are needed: Quantity

Size

Length

Lb./ea.

Lb.

1 ea.

W18×130

41 f t. 2 in.

5,352

5,352

2 ea.

W16×89

41 f t. 2 in.

3,664

7,328

4 ea.

W14×74

34 f t. 7 in.

2,559

10,236

4 ea.

W14×74

33 f t. 3 in.

2,460

9,840

1 ea.

C15×40

14 f t. 0 in.

560

560

1 ea.

C15×40

10 f t. 0 in.

400

400

6 ea.

W14×68

16 f t. 0 in.

1,088

6,528

ee

ee

ee

ee

ee

ee

ee

blank

blank

Total

blank

40,244

119 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 13: Metals

Weight 

40,244Ib.  20.122 tons 2,000Ib. ton

7. Prepare a structural steel materials list for the roof-framing plan shown in Figure 13.17. The columns are 18 feet high and weigh 76 pounds per foot. How many pounds of steel need to be purchased for the roof?

The following materials are needed: Quantity

Size

Length

Lb./ea.

Lb.

2 ea.

W16×77

43 f t. 0 in.

3,311

6,622

2 ea.

W21×147

43 f t. 0 in.

6,321

12,642

4 ea.

W16×50

28 f t. 6 in.

1,425

5,700

4 ea.

W16×50

29 f t. 7 in.

1,479

5,916

2 ea.

W16×50

29 f t. 0 in.

1,450

2,900

2 ea.

W16×50

30 f t. 0 in.

1,500

3,000

1 ea.

C12×30

11 f t. 0 in.

330

330

1 ea.

C12×30

15 f t. 0 in.

450

450

8 ea.

TS12×12×1/2

ee

ee

ee

ee

ee

ee

ee

ee

18 f t. 0 in.

1,368

ee

blank

blank

10,944 blank

Total

48,504 Weight 

48,504Ib.  24,252 tons 2,000 Ib. ton

8. Prepare a steel and steel joist materials list for the roof-framing plan shown in Figure 13.18 and determine the number of pounds of joists, beams, and columns that need to be purchased for the roof. The columns are 17 feet high and weigh 76 pounds per foot.

Figure 13-10 shows that the joist weighs 12.0 pounds per foot.

Quantity

Size

Length

Lb./ea.

Lb.

2 ea.

W18×76

38 f t. 0 in.

2,888

5,776

1 ea.

W18×130

38 f t. 0 in.

4,940

4,940

14 ea.

24K9

39 f t. 6 in.

474

6,636

4 ea.

24K9

38 f t. 6 in.

462

1,848

6 ea.

TS12×12×1/2

ee

ee

ee

ee

17 f t. 0 in.

1,292

ee

blank

blank

Total

7,752 blank

26,952 Weight 

26,952 Ib.  13,476 tons 2,000 Ib. ton 120

Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 13: Metals

9. Prepare a steel and steel joist materials list for the roof-framing plan shown in Figure 13.19 and determine the number of pounds of joists, beams, and columns that need to be purchased for the roof. The columns are 15 feet high and weigh 58 pounds per foot.

Figure 13-10 shows that the joist weighs 9.0 pounds per foot.

Quantity

Size

Length

Lb./ea.

Lb.

2 ea.

W18×71

38 f t. 0 in.

2,698

5,396

2 ea.

W18×67

38 f t. 0 in.

2,546

5,092

4 ea.

18K7

23 f t. 6 in.

212

848

18 ea.

18K7

24 f t. 6 in.

220

3,960

2 ea.

18K7

24 f t. 0 in.

216

432

9 ea.

18K7

25 f t. 0 in.

225

2,025

8 ea.

TS12×12×3/8

ee

ee

ee

ee

ee

ee

15 f t. 0 in.

870

ee

blank

blank

6,960 blank

Total

24,713 Weight 

24,713 Ib.  12,36 tons 2,000 Ib. ton

10. Determine the amount of metal deck that needs to be purchased to construct the floor in Figure 13.16. Add 12 percent for waste and lapping. Ignore the opening.

Find the area as follows: Area  42ft.  70 ft. 

2,940 s.f.  29.4 sq.

100 s.f. sq.

Add 12 percent for waste and lap: Area  29.4 sq.  1  0.12   32.9 sq.

11. Determine the cost to install the metal deck for the floor shown in Problem 10 using a productivity of 3.1 labor hours per square (100 square feet) and an average burdened labor rate of $37.58 per labor hour.

From Problem 10, the area of the deck is 29.4 squares.

Labor hour  29.4 sq.  3.1 lhr./sq.  91.1 lhr.

Labor cost  91.1 lhr.  $37.58/lhr.  $3,424

12. Determine the amount of metal deck that needs to be purchased to construct the floor in Figure 13.17. Add 15 percent for waste and lapping. Ignore the opening.

Find the area as follows:

121 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 13: Metals

Area  42 ft.  90 ft. 

4,050 s.f.  40.5 sq.

100 s.f. sq.

Add 15 percent waste and lap: Area  40.5 sq.  1  0.15   46.6 sq.

13. Determine the cost to install the metal deck for the floor shown in Problem 12 using a productivity of 2.8 labor hours per square (100 square feet) and an average burdened labor rate of $35.11 per labor hour.

From Problem 12, the area of the deck is 40.5 squares.

Labor hour  40.5 sq.  2.8 lhr./sq.  113.4 lhr.

Labor cost  113.4 lhr.  $35.11/ lhr.  $3,981

14. Estimate the metal decking requirements for the commercial building project in Appendix C. The decking is inside the parapet walls and is 78 ft.  38 ft.

Area  78 ft.  38 ft. 

2,964 s.f.  29.64 sq.

100 s.f. sq.

Add 15 percent waste and lap: Area  29.64 sq.  1  0.15   34.1 sq.

The following workup sheet contains the metal decking calculations.

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

15. Using the drawings for Billy’s Convenience Store (Appendix D), prepare an estimate for the structural steel found in the roof structure.

The following workup sheet contains detailed structural steel calculations for the roof.

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Chapter 13: Metals

The solution can also be found in Appendix D - Billys C-Store v10 Estimate.xlsx.

123 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Chapter 14—Wood Learning Outcomes This chapter looks at determining the materials and labor associated with Division 06 Wood, Plastics, and Composites of the MasterFormat. After reading this chapter, the reader should be able to: Learning Outcome 14.1: Identify the difference between nominal and actual dimensions.

Learning Outcome 14.2: Calculate the number of board feet in a standard piece of lumber.

Learning Outcome 14.3: Determine the materials needed to frame wood floors.

Learning Outcome 14.4: Determine the materials needed to frame wood walls.

Learning Outcome 14.5: Determine the materials needed to frame wood ceilings.

Learning Outcome 14.6: Determine the materials needed to frame wood roofs.

Learning Outcome 14.7: Determine the materials needed for trim.

Learning Outcome 14.8: Determine framing labor costs.

Learning Outcome 14.9: Describe the factors to consider when bidding on wood framing.

Activities 1. Assign the students to visit a house being framed. Be sure to have the students obtain permission to visit the house. Each student should select one exterior and two interior walls and prepare a materials list for each of them based on the actual quantity of materials used in their construction. Before leaving the site, the students should measure the walls’ lengths and the openings’ locations and sizes. After the site visit, the students should prepare an estimate for the walls based on their measurements using the textbook methods and compare their estimates to the actual quantity of materials used. Have the students explain why their estimate differs from the actual amount of materials used and what they learned from this experience.

Instructional Resources PowerPoint Slides: Chapter 14.ppt The pdf drawings from Appendices A through E. They can be found at the Pearson Careers Resources for Students website (http://www.pearsonhighered.com/careersresources/) by searching for the book by title or going to Browse by Discipline and choosing Civil and Environmental Engineering.

The drawings include the following: 

Appendix A – Small Commercial Building 124 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood



Appendix B – Residential Building Project



Appendix C – Commercial Building Project



Appendix D – Billy’s C-Store



Appendix E – Real Estate Office

Solutions to the Review Questions 1. What unit of measure is used for lumber?

Lumber is priced and sold by the thousand-foot board measure. One board foot equals the volume of a piece of wood one inch thick and one foot square.

2. Determine the number of board feet for the following order quantities:

190 — 2  4 @ 8'0" Board Feet  190ea. 

2in.  4in.  8 ft.  1,013b.f.

12in. ft.

1,120 — 2  4 @ 12'0" Board Feet  1,120ea. 

2in.  4in.  12ft.  8,960b.f.

12in. ft.

475 — 2  6 @ 16'0" Board Feet  475ea. 

2in.  4in.  16 ft.  7,600b.f.

12in. ft.

475 — 2  12 @ 16'0" Board Feet  475ea. 

2in.  12in.  16 ft.  15,200b.f.

12in. ft.

18 — 2  8 @ 18'0" Board Feet  18ea. 

2in.  8in.  18 ft.  432b.f.

12in. ft.

3. How would you determine the number of studs required for a project?

The number of studs required for a project is determined by the following: 1. Review the specifications and drawings for the thickness and spacing of studs and lumber grade.

2. Exterior studs will be required around the perimeter of the building, the same distance used for the plates.

3. Divide the perimeter (length of the wall) by the stud spacing to determine the number of spaces. Then add one to close off the last space.

125 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

4. Add extra studs for corners, wall intersections (where two walls join), and openings.

4. How do you determine the number of joists required?

The floor length is divided by the joist spacing, and one joist is added for the extra joist required at the end. If the joists are to be doubled under partitions or at the side of headers, one extra joist should be added for each occurrence. Two additional joists are needed for walls not located over a joist.

5. What is the difference between ―pitch‖ and ―slope,‖ as the terms pertain to roofing?

The pitch is the rise of the roof divided by the span. The slope is the rise of the roof divided by the run (horizontal distance to the ridge), typically one-half of the span.

6. Determine the length of rafter required for each of the following conditions if the run is 16 feet:

1/12 pitch

3 in 12 slope

1/6 pitch

4 in 12 slope

Figure 14.50 shows that for a 1/12 pitch, the run is multiplied by 1.015. The length is as follows:

Length  16 ft.  1.015  16.24 ft.

Figure 14.50 shows that for a 1/6 pitch, the run is multiplied by 1.055. The length is as follows:

Length  16 ft.  1.055  16.88 ft.

Figure 14.50 shows that for a 3 in 12 slope, the run is multiplied by 1.030. The length is as follows:

Length  16 ft.  1.030  16.48 ft.

Figure 14.50 shows that for a 4 in 12 slope, the run is multiplied by 1.055. The length is as follows:

Length  16 ft.  1.055  16.88 ft.

7. What unit of measure is used for plywood when it is used for sheathing? How is plywood waste kept to a minimum?

The unit of measure for plywood used as sheathing is the number of sheets required. The sheathing is most accurately estimated by sketching a planned layout for the area to be covered.

8. What unit of measure will most likely be used for laminated beams?

The takeoff should list the linear footage required for each type and the size or style of the beam or arch.

9. What unit of measure will most likely be used for wood decking? How is it determined?

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Chapter 14: Wood

The unit of measure for wood decking is the board measure. Decking is taken off by the square footage of the area to be covered, which is multiplied by the appropriate factor shown in Figure 14.61 to get the number of board feet.

10. What unit of measure is used for wood trim? If it requires a finish, where is this information found?

The trim is taken off by the linear foot. Any finishes required are usually found in the specifications.

11. Determine the purchase quantity and number of board feet of 2  12 needed to construct the girder in Figure 14.62. The foundation walls are eight inches thick, and the bearing distance for the girder is six inches.

The number of board feet is found as follows:

Inside dimension  40 ft.  8 in.  8 in.  38 ft. 8 in.

Girder length  38 ft. 8 in.  6 in.  6 in.  39 ft. 8 in.  39.67 ft.

Total length of girder  3  39.67 ft.  119 ft.

Order nine pieces of 2  12 @ 14 ft. so the joints break over columns.

The number of board feet is calculated as follows: Board Feet  9ea. 

2in.  12in.  14 ft.  252b.f.

12in. ft.

12. A 2×4 sill plate runs around the floor in Figure 14.62. Determine the purchase quantity and number of board feet of 2  4 needed for the sill.

Order eight pieces of 10 f t. for the top and bottom walls and four pieces of 14 f t. for the side walls. ee

ee

Sill length  8 ea.  10 ft./ea.  4 ea.  14 ft./ea.  136 ft.

The number of board feet is calculated as follows: Board Feet 

2in.  4in.  136 ft.  91 b.f.

12in. ft.

13. Determine the purchase quantity and number of board feet of 2  12 joists needed for the floor in Figure 14.62. The joists are spaced 16 inches on center. Be sure to include the headers.

Figure 14.8 shows that there are 0.75 spaces per foot. The number of joists is found as follows:

Number of joist spaces  40 ft.  0.75 spaces/ft.  30 spaces Add one to get the number of joists.

Order 31 joists per row.

127 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Order 31 pieces at 14 feet for the top of the drawing, 25 pieces (31  6) at 14 feet for the bottom of the drawing, and six at 8 feet for between the opening and the center of the building. Order four additional at 14 feet for the double joists at the sides of the building, three additional joists at 14 feet for under the walls (one wall aligns with an existing joist), and four additional at 14 feet for the double joists at the sides of the opening. Order eight pieces at 10 feet for the headers on the top and bottom walls and two pieces at 8 feet for the header at the opening. The total counts are as follows: 

67 pieces 2  12 @ 14 ft.



8 pieces 2  12 @ 10 ft.



8 pieces 2  12 @ 8 ft.

Joist length  67 ea.  14 ft./ea.  8 ea.  10 ft./ea.  8 ea.  8 ft./ea.

Joist length = 1,082 f t. ee

The number of board feet is calculated as follows: Board Feet 

2in.  12in.  1,082ft.  2,164 b.f.

12in. ft.

14. Metal bridging is placed at the center of spans over 10 feet and where the joists pass over the girder. Determine the number of pieces of metal bridging needed for the floor in Figure 14.62.

Metal bridging is not needed in the joist between the center of the building and the header at the opening. Thirty pairs are needed in the top half of the drawing, 30 pairs over the girder, and 24 pairs (30  6) in the bottom half.

Number  2 ea./pair   30 pairs  30 pairs  24 pairs   168 ea.

15. Determine the number of 4-foot by 8-foot sheets of plywood needed for the floor in Figure 14.62. Ignore the opening.

The number of sheets is found as follows: 40 ft.  28 ft.  35sheets 32s.f. sheet

Sheets of plywood 

16. Determine the purchase quantity and number of board feet of 2  12 needed to construct the girders in Figure 14.63. The foundation walls are 8 inches thick, and the bearing distance for the girder is 6 inches.

The top girder is calculated as follows: Inside dimension = 13 f t. 4 in. – 8 in. – 8 in. = 12 f t. ee

ee

Girder length = 12 f t. + 6 in. + 6 in. = 13 f t. ee

ee

Order 2 at 14 f t. ee

The bottom girder is calculated as follows: 128 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Inside dimension  36 ft.  4 ft.  8 in.  31 ft. 4 in.

Girder length = 31 f t. 4 in. + 6 in. = 31 f t. 10 in. ee

ee

Order 6 at 12 feet to match the post spacing.

Total length of girder  2 ea.  14 ft./ea.  6 ea.  12 ft./ea.  100 ft.

The number of board feet is calculated as follows: Board Feet 

2in.  12in.  100 ft.  200b.f.

12in. ft.

17. A 2×4 sill plate runs around the floor in Figure 14.63. Determine the purchase quantity and number of board feet of 2  4 needed for the sill.

Starting at the top left corner and proceeding clockwise around the building, the following plates are needed: 

Top-left wall: 2 ea. 2  4 @ 12 ft.



Left-top wall: 1 ea. 2  4 @ 10 ft.



Top-right wall: 1 ea. 2  4 @ 14 ft.



Right wall: 1 ea. 2  4 @ 10 ft. and 2 ea. 2  4 @ 12 ft.



Bottom wall: 3 ea. 2  4 @ 12 ft.

 Left-bottom wall: 2 ea. 2  4 @ 12 ft.

The total counts are as follows: 

1 ea. 2  4 @ 14 ft.



9 ea. 2  4 @ 12 ft.



2 ea. 2  4 @ 10 ft.

Sill length  1 ea.  14 ft./ea.  9 ea.  12 ft./ea.  2 ea.  10 ft./ea.

Sill length = 142 f t. ee

The number of board feet is calculated as follows: Board Feet 

2in.  4in.  142ft.  95b.f.

12in. ft.

18. Determine the purchase quantity and number of board feet of 2  12 joists needed for the floor in Figure 14.63. The joists are spaced 16 inches on center. Be sure to include the headers.

Figure 14.8 shows that there are 0.75 spaces per foot. Determine the number of joist spaces for the bottom and middle row of joists:

Number of joist spaces = 36 ft.  0.75 spaces/ft.  27 spaces 129 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Add one to get the number of joists.

Order 28 joists.

Determine the number of joist spaces for the top row of joists:

Number of joist spaces  13 ft. 4 in.  0.75 spaces/ft.

Number of joist spaces  13.33 ft.  0.75 spaces/ft.  10 spaces Add one to get the number of joists.

Order 11 joists.

The joists needed are as follows: 

Bottomrow of joists: 2 ea.2  12@ 10 ft., 26 ea.2  12@ 12 ft.



Middle row of joists: 1 ea.2  12@ 12 ft.  cut in half for joist at opening , 26 ea.2  12@ 12 ft.



Top row of joists: 11 ea. 2  12@ 10 ft.



Doubled joists: 2 ea. 2  12@ 10 ft., 8 ea. 2  12@ 12 ft.



Joist headers at the top: 2 ea. 2  12@ 12 ft., 1 ea. 2  12@ 14 ft.

 

Joist headers at the bottom: 3 ea. 2  12@ 12 ft.

Joist headers at the opening: 1 @ 8 ft. (cut in half )

The total counts are as follows: 

1 ea. 2  12 @ 14 ft.



66 ea. 2  12 @ 12 ft.



15 ea. 2  12 @ 10 ft.



1 ea. 2  12 @ 8 ft.

Joist length  1 ea.  14 ft./ea.  66 ea.  12 ft./ea. + 15 ea.  10 ft./ea.  1 ea.  8 ft./ea.

Joist length = 964 f t. ee

The number of board feet is calculated as follows: Board Feet 

2in.  12in.  964 ft.  1,928b.f.

12in. ft.

19. Metal bridging is placed at the center of spans over 10 feet and where the joists pass over the girder. Determine the number of pieces of metal bridging needed for the floor in Figure 14.63.

The number of pairs of metal bridging is as follows: 130 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood



Bottom row of joists  27 spaces  3 spaces  24 spaces



Bottom girder  27 spaces  3 spaces  24 spaces



Middle row of joists  27 spaces  3 spaces  24 spaces



Top girder = 10 spaces



Top row of joist: Not required Number  2 ea./pair   24 pairs  24 pairs  24 pairs  10 pairs 

Number = 164 ea.

20. Determine the number of 4-foot by 8-foot sheets of plywood needed for the floor in Figure 14.63. Ignore the opening.

The number of sheets is found as follows:

Area  36 ft.  24 ft.  13 ft. 4 in.  10 ft.

Area  36 ft.  24 ft.  13.33 ft.  10 ft.  997 s.f.

The number of sheets is calculated as follows: Sheets of plywood 

997s.f.  31.2sheets 32s.f. sheet

Order 32 sheets.

21. The garage shown in Figure 14.64 is constructed of 2×4 lumber and includes a single treated sole plate and two top plates. Determine the purchase quantity and number of board feet of treated and untreated lumber needed for the wall plates.

Order the following for the bottom plates. Deduct the overhead door but not the personnel door. The following materials will be ordered: 

North wall: 1 ea. 2  4 @ 14 ft., 1 ea. 2  4 @ 16 ft.



East wall: 1 ea. 2  4 @ 8 ft.



South wall: 1 ea. 2  4 @ 14 ft., 1 ea. 2  4 @ 16 ft.



West wall: 2 ea. 2  4 @ 12 ft.

The total order quantities are as follows: 

2 ea. 2  4 @ 16 ft.



2 ea. 2  4 @ 14 ft.



2 ea. 2  4 @ 12 ft.



1 ea. 2  4 @ 8 ft.

Length of treated lumber  2 ea.  16 ft./ea.  2 ea.  14 ft./ea.

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Chapter 14: Wood

 2 ea.  12 ft./ea.  1 ea.  8 ft./ea.

Length of treated lumber = 92 f t. ee

The number of board feet is calculated as follows: Board Feet 

2 in.  4in.  92ft.  62b.f.

12 in. ft.

Be sure to include top plates over the opening for the overhead door. Order the following for the top plates: 

North wall: 2 ea. 2  4 @ 14 ft., 2 ea. 2  4 @ 16 ft.



East wall: 4 ea. 2  4 @ 12 ft.



South wall: 2 ea. 2  4 @ 14 ft., 2 ea. 2  4 @ 16 ft.



West wall: 4 ea. 2  4 @ 12 ft.

The total order quantities are as follows: 

4 ea. 2  4 @ 16 ft.



4 ea. 2  4 @ 14 ft.



8 ea. 2  4 @ 12 ft.

Length of untreated lumber  4 ea.  16 ft./ea.  4 ea.  14 ft./ea.  8 ea.  12 ft./ea.

Length of untreated lumber = 216 f t. ee

The number of board feet is calculated as follows: Board Feet 

2 in.  4in.  216 ft.  144b.f.

12 in. ft.

22. Determine the number of 92 5/8-inch-long 2  4 studs needed for the garage in Figures 14.64 and 14.65. The studs are spaced 16 inches on center. Add two studs for each corner. Ignore the gable ends and studs over the overhead door.

Figure 14.8 shows that there are 0.75 spaces per foot. The studs are calculated as follows:

Wall length  30 ft.  4 ft.  4 ft.  30 ft.  24 ft.  92 ft.

Wall studs  92 ft.  0.75 spaces/ft.  69 spaces Order 70 studs.

Figure 14.64 shows that one stud is needed on each side of the personnel door, and two studs are needed on each side of the overhead door.

Studs for personnel door  2 sides  1 stud/side  2 studs Studs for overhead door  2 sides  2 stud/side  4 studs 132 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Studs for corners  4 corners  2 studs/corner  8 studs Total studs = 70 studs + 2 studs + 4 studs + 8 studs = 84 studs Studs are not needed above the openings.

23. Determine the materials needed for the headers of the garage shown in Figure 14.64. The personnel door is three feet wide.

Figure 14.64 shows that the header for the personnel door bears on one stud on each side. Allow three inches for the door frame and clearance.

Length of header  3 ft.  2 sides  1.5 in./side  3 in.  3 ft. 6 in.

Order 1 ea. 2  10 @ 8 ft.  cut in half  .

Additionally, about three s.f. of O SB will be needed as a spacer in the header.

Figure 14.64 shows that the header for the garage door bears on two studs on each side.

Length of header  16 ft.  2 sides  2 studs/side  1.5 in./stud Length of header = 16 f t. 6 in. ee

Order 2 pieces 1 ¾  9 ½ LVL at 17 ft. long.

24. Determine the number of 8-foot-long studs needed for the gable ends of the garage shown in Figures 14.64 and 14.65. The gable end on the west elevation is the same as the gable end on the east.

Figure 14.8 shows that there are 0.75 spaces per foot. The number of studs is found as follows:

Number of gable-end studs  24 ft.  0.75 spaces/ft.  18 spaces/end Only 17 studs are needed because studs are not required to close off the ends of the gable.

Average height  4 ft. 

2  2ft.8in.

3

On average, three gable-end studs can be cut from one 8-foot stud. The quantity must be doubled to account for both gable ends.

Number of studs  2ends 

17ea. end  12studs 3ea. stud

25. Determine the number of 4-foot by 8-foot sheets of plywood needed to cover the garage’s exterior in Figures 14.64 and 14.65.

Ignore the personnel door opening because this material will be waste.

Area of north wall  8 ft. 1 in.  30 ft.  8.08 ft.  30 ft.  242 s.f.

Area of east wall  8 ft. 1 in.  24 ft.  0.5  4 ft.  24 ft.  7 ft.  16 ft.

133 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Area of east wall  8.08 ft.  24 ft.  0.5  4 ft.  24 ft.  7 ft.  16 ft.  130 s.f.

Area of south wall  8 ft. 1 in.  30 ft.  8.08 ft.  30 ft.  242 s.f.

Area of west wall  8 ft. 1 in.  24 ft.  0.5  4 ft.  24 ft.

Area of west wall  8.08 ft.  24 ft.  0.5  4 ft.  24 ft.  242 s.f.

Total area  242 s.f.  130 s.f.  242 s.f.  242 s.f.  856 s.f.

The number of sheets is calculated as follows: Sheets of plywood 

856s.f.  26.8sheets 32s.f. sheet

Order 27 sheets.

26. The residence shown in Figure 14.66 is constructed of 2×4 lumber and includes a single treated sole plate and two top plates. Determine the purchase quantity and number of board feet of treated and untreated lumber needed for the wall plates.

Run the plates through the doors for ease of construction. The order quantities are shown in Figure 14-1. Add one 8 foot for the 2 foot 1 inch walls at the ends of the closets. Other solutions are possible.

Figure 14-1 The total order quantities are as follows: 

2 ea. 2  4 @ 16 ft.



4 ea. 2  4 @ 14 ft.



5 ea. 2  4 @ 12 ft.



7 ea. 2  4 @ 10 ft.



3 ea. 2  4 @ 8 ft.

134 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Length of treated lumber  2 ea.  16 ft./ea.  4 ea.  14 ft./ea.  5 ea.  12 ft./ea.  7 ea.  10 ft./ea.  3 ea.  8 ft./ea.

Length of treated lumber = 242 f t. ee

The number of board feet is calculated as follows: Board Feet 

2in.  4in.  242ft.  162b.f.

12in. ft.

The top plates will require twice the above quantities, or 484 f t. The number of board feet is calculated as follows: ee

Board Feet 

2in.  4in.  484 ft.  323b.f.

12in. ft.

27. Determine the number of 92 5/8-inch-long 2  4 studs needed for the residence in Figures 14.66 and 14.67. The studs are spaced 16 inches on center. Add two studs for each door, window, intersection, and corner. Ignore the gable ends.

The studs are calculated as follows:

Exterior wall length  32 ft.  26 ft.  36 ft.  13 ft. 6 in.  4 ft.  12 ft. 6 in.

Exterior wall length = 124 f t. ee

Interior wall length  12 ft. 10 in.  19 ft. 4 in.  16 ft. 4 in.  3.5 in.  8 ft. 9 in.

 11 ft. 9 in.  3 ea.  2 ft. 1 in./ea.  3 ea.  8 ft. 4 in./ea.

Interior wall length = 100 f t. 6.5 in. ee

Order 101 f t. ee

Total wall length = 124 f t. + 101 f t. = 225 f t. ee

ee

ee

Figure 14.8 shows that there are 0.75 spaces per foot:

Wall studs  225 ft.  0.75 spaces/ft.  168.8 Add one stud and round up: Order 170 studs.

Studs for doors and windows  14 ea.  2 studs/ea.  28 studs Studs for corners  9 ea.  2 studs/ea.  18 studs Studs for intersections  14 ea.  2 studs/ea.  28 studs Total studs = 170 studs + 28 studs + 18 studs + 28 studs = 244 studs 28. Determine the materials needed for the headers for the residence shown in Figure 14.66. Two 2  10s are used for the headers in the exterior walls and the wall running east-west through the house. The remaining headers are two 2  4s . The swinging doors are 3 feet wide, and the closet doors are 6 feet wide.

135 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

The header materials are as follows: #

Opening

Length

Material

Order

1

3 f t. window

3 f t. 3 in.

2 ea. 2  10

1 ea. 2  10 @ 8 ft.

4

4 f t. window

4 f t. 3 in.

2 ea. 2  10

4 ea. 2  10 @ 10 ft.

1

5 f t. window

5 f t. 3 in.

2 ea. 2  10

1 ea. 2  10 @ 12 ft.

3

3 f t. door

3 f t. 6 in.

2 ea. 2  10

3 ea. 2  10 @ 8 ft.

2

3 f t. door

3 f t. 6 in.

2 ea. 2  4

2 ea. 2  4 @ 8 ft.

3

6 f t. door

6 f t. 6 in.

2 ea. 2  4

6 ea. 2  4 @ 92 5/8 in.

ee

ee

ee

ee

ee

ee

ee

ee

ee

ee

ee

ee

Additionally, an OSB spacer will be needed for the header. The length of the headers are as follows: Length  2  10   3 ft. 3 in.  4 ea.  4 ft. 3 in./ea.  5 ft. 3 in.  3 ea.  3 ft. 6 in./ea.

Length  2  10  = 3.25 ft.  4 ea.  4.25 ft./ea.  5.25 ft.  3 ea.  3.5 ft./ea.

Length  2  10   36 ft.

Length  2  4   2 ea.  3 ft. 6 in./ea.  3 ea.  6 ft. 6 in./ea.

Length  2  4   2 ea.  3.5 ft./ea.  3 ea.  6.5 ft./ea.  26.5 ft

Spacer  36 ft.  10 in.  26.5 ft.  4 in.

Spacer  36 ft  0.83 ft.  26.5 ft.  0.33 ft.  38.6 s.f.

Sheets of plywood 

38.6s.f.  1.2sheets 32s.f. sheet

Order 2 sheets.

29. Determine the number of 8-foot-long studs needed for the gable ends of the residence shown in Figures 14.66 and 14.67. The gable end on the east elevation is the same as on the west.

Figure 14.8 shows that there are 0.75 spaces per foot. The number of studs is found as follows:

Number of gable-end studs  26 ft.  0.75 spaces/ft.  20 spaces Only 19 studs are needed because studs are not required to close off the ends of the gable.

Average height  4 ft. 4in. 

2  2ft.11in.

3

136 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

On average, two gable-end studs can be cut from one 8 f t. stud or one 92-5/8 in. stud. The quantity must be doubled to account for both gable ends. ee

Number of boards  2ends 

19studs end  19boards 2stud board

30. Determine the number of 4-foot by 8-foot sheets of plywood needed to cover the residence’s exterior in Figures 14.66 and 14.67.

Ignore the door and window openings because this material will be waste. From Problem 27, the length of the exterior wall is 124 feet.

Area of exterior walls  8 ft. 1 in.  124 ft.  8.08 ft.  124 ft.  1,002 s.f.

Area of the gable ends  2 ends  0.5  4 ft. 4 in.  26 ft.

Area of the gable ends  2 ends  0.5  4.33 ft.  26 ft.  113 s.f.

Total area = 1,002 s.f. + 113 s.f. = 1,115 s.f.

The number of sheets is calculated as follows: Sheets of plywood 

1,115s.f.  34.8sheets 32s.f. sheet

Order 35 sheets.

31. Determine the purchase quantity of 2 6 rafters needed for the roof in Figure 14.68.

The slope of the roof is 4:12.

Figure 14.50 shows that the slope length factor is 1.055.

Run 

28 ft.  14 ft.

2

Length  14 ft.  1.055  14.77 ft.

Order 16 f t. rafters. ee

Figure 14.8 shows that there are 0.50 spaces per foot:

Number of spaces  34 ft.  0.5 spaces/ft.  17 spaces Add one to close off the end space: Order 18 for each half of the roof for a total of 36.

32. Collar ties are used to prevent the rafters in Figure 14.68 from spreading. The collar ties are 1 6s spaced 4 feet on center and are located one-third of the distance from the ridge of the roof to the top of the walls. Determine the purchase quantity of 1 6 needed for the collar ties.

The collar tie’s length is 1/3 the distance between the outsides of the walls.

Length 

24 ft.  8 ft.

3

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Chapter 14: Wood

Number of spaces 

30 ft.  8 spaces 4 ft. space

Add one to close off the end space: Order nine pieces 8 feet long.

33. Determine the purchase quantity of 1 8s needed for the ridge of the roof in Figure 14.68.

The length needs to add up to 34 f t. All the following are correct answers: ee



2 ea. 1 8 @ 10 ft. and 1 ea. 1 8 @ 14 ft.



1 ea. 1 8 @ 10 ft. and 2 ea. 1 8 @ 12 ft.

34. Determine the number of 4-foot by 8-foot sheets of plywood needed to cover the roof in Figure 14.68.

From Problem 31, the rafter length is 14.77 feet.

Area  2 sides  Rafter length  Ridge length Area  2 sides  14.77 ft.  34 ft.  1,004 s.f.

The number of sheets is calculated as follows: 1,004s.f.  31.4sheets 32s.f. sheet

Sheets of plywood 

Order 32 sheets.

35. How many linear feet of 1 8 fascia and 1/ 2 -inch A-C plywood soffit are needed for the roof in Figure 14.68?

From Problem 31, the rafter length is 14.77 feet.

Length of fascia  2 sides  34 ft./side  4 rafters  14.77 ft./rafter  128 ft.

The soffit is 2 feet wide. Be sure not to double-count the corners.

Length of soffit  2 sides  30 ft./side  4 rafters  14.77 ft./rafter  119.1 ft.

Area of soffit  119.1 ft.  2 ft.  239 s.f.

The number of sheets is calculated as follows: Sheets of plywood 

239s.f.  7.5sheets 32s.f. sheet

Order 8 sheets.

36. Determine the purchase quantity of 2  8 rafters needed for the roof in Figure 14.69.

The slope of the roof is 4:12.

Figure 14.50 shows that the slope length factor is 1.055.

138 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Run 

28 ft. 8in.  14 ft. 4in.

2

Length  14 ft. 4 in.  1.055  14.33 ft.  1.055  15.12 ft.

Order 16 f t. rafters. ee

Figure 14.8 shows that there are 0.75 spaces per foot:

Number of spaces  38 ft. 8 in.  0.75 spaces/ft.

Number of spaces  38.67 ft.  0.75 spaces/ft.  29 spaces Add one to close off the end space: Order 30 for each half of the roof for a total of 60.

37. Collar ties are used to prevent the rafters in Figure 14.69 from spreading. The collar ties are 1 6s spaced 4 feet on center and are located one-third of the distance from the ridge of the roof to the top of the walls. Determine the purchase quantity of 1×6s needed for the collar ties.

The collar tie’s length is 1/3 the distance between the outsides of the walls.

Length 

26 ft.  8.67 ft.

3

Order 10 f t. long 1 6s. ee

Number of spaces 

36 ft.  9 spaces 4 ft. space

Add one to close off the end space.

Order 10 pieces 10 f t. long. ee

38. Determine the purchase quantity of 1 10s needed for the ridge of the roof in Figure 14.69.

The length needs to add up to 38 f t. 8 in. All the following are correct answers: ee



5 ea. 1 10 @ 8 ft.



4 ea. 1 10 @ 10 ft .



1 ea. 1 10 @ 12 ft. and 2 ea. 1 10 @ 14 ft.



1 ea. 1 10 @ 8 ft. and 2 ea. 1 10 @ 16 ft.

39. Determine the number of 4-foot by 8-foot sheets of plywood needed to cover the roof in Figure 14.69.

From Problem 36, the rafter length is 15.12 feet.

Area  2 sides  Rafter length  Ridge length 139 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Area  2 sides  15.12 ft.  38 ft. 8 in.

Area  2 sides  15.12 ft.  38.67 ft.  1,169 s.f.

The number of sheets is calculated as follows:

Sheets of plywood 

1,169 s.f.  36.5 sheets 32s.f. sheet

Order 37 sheets.

40. How many linear feet of 1 10 fascia and 1/ 2 -inch A-C plywood soffit are needed for the roof in Figure 14.69?

From Problem 36, the rafter length is 15.12 feet.

Length of fascia  2 sides  38 ft. 8 in./side  4 rafters  15.12 ft./rafter Length of fascia  2 sides  38.67 ft./side  4 rafters  15.12 ft./rafter  138 ft.

The soffit is 16 in. wide. Be sure not to double-count the corners.

Length of soffit  2 sides  36 ft./side  4 rafters  15.12 ft./rafter Length of soffit = 132.5 f t. ee

Area of soffit  132.5 ft.  16 in.  132.5 ft. 1.33 ft.  176 s.f.

The number of sheets is calculated as follows: Sheets of plywood 

176s.f.  5.5sheets 32s.f. sheet

Order 6 sheets.

41. Determine the materials required to frame the floor for the residence in Appendix B. The following workup sheet contains detailed floor framing calculations.

140 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

The solution can also be found in Appendix B - Residential Building Project v10 Estimate.xls x.

42. Determine the materials required to frame the walls for the residence in Appendix B. The following workup sheets contain detailed wall framing calculations.

141 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

142 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

The solution can also be found in Appendix B - Residential Building Project v10 Estimate.xls x.

43. Determine the materials required to frame the ceiling for the residence in Appendix B. The following workup sheet contains detailed ceiling framing calculations.

The solution can also be found in Appendix B - Residential Building Project v10 Estimate.xls x.

44. Determine the materials required to frame the roof for the residence in Appendix B. 143 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

The following workup sheets contain detailed roof framing calculations.

144 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

The solution can also be found in Appendix B - Residential Building Project v10 Estimate.xls x.

45. Determine the materials required to frame the walls for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the rough carpentry crew from Problem 7-32, determine the cost to frame the walls.

The following data can be extracted from the model:

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Chapter 14: Wood

All the 24-inch by 48-inch windows are in the exterior walls. Only two of the 36-inch by 72-inch windows are in the exterior walls. There is one 3-foot by 7-foot door and one 6-foot by 7-foot in the exterior walls. The remaining twenty 3-foot by 7-foot doors are in the interior walls.

Determine the materials needed for the exterior walls. There are 241 f t. of exterior walls. Run the sill plate through the doors for ease of construction; therefore, 241 feet of treated plate is needed for the sill. ee

Board Feet 

2 in.  4in.  241 ft.  161 b.f.

12 in. ft.

The top plates will require twice the above quantities; therefore, 482 f t. or 322 b.f. of plate is needed. ee

Determine the amount of Weathermate needed between the foundation wall and the wood plate.

Weathermate 

241 ft.  5rolls 50 ft. roll

Figure 14.8 shows that there are 0.75 spaces per foot. Determine the number of studs needed:

Wall studs  241 ft.  0.75 spaces/ft.  181 Add one to close off the space. Order 182 studs.

Studs for doors/windows   34 ea.  2 ea.  1 ea.  1 ea.  2 studs/ea.

Studs for doors and windows = 76 studs

Studs for corners  4 ea.  2 studs/ea.  8 studs Studs for intersections  15 ea.  2 studs/ea.  30 studs Total studs = 182 studs + 76 studs + 8 studs + 30 studs = 296 studs The header materials are as follows: #

Opening

Length

Material

Order

34

24 in. window

2 f t. 3 in.

2 ea. 2  12

17 ea. 2  12 @ 10 ft.

2

36 in. window

3 f t. 3 in.

2 ea. 2  12

2 ea. 2  12 @ 8 ft.

1

3 f t. door

3 f t. 5 in.

2 ea. 2  12

1 ea. 2  12 @ 8 ft.

1

6 f t. door

6 f t. 5 in.

2 ea. 2  12

2 ea. 2  12 @ 8 ft.

ee

ee

ee

ee

ee

ee

146 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Additionally, a plywood spacer would be needed between the headers. The spacer is calculated as follows:

Length  34 ea.  2 ft. 3 in./ea.  2 ea.  3 ft. 3 in./ea.  3 ft. 5 in.  6 ft. 5 in.

Length  34 ea.  2.25 ft./ea.  2 ea.  3.25 ft./ea.  3.42 ft.  6.42 ft.

Length = 93 f t. ee

Area  93 ft.  1 ft.  93 s.f.

The number of sheets is calculated as follows: Sheets of plywood 

93s.f.  2.9sheets 32s.f. sheet

Order 3 sheets.

The sheathing is calculated as follows. Ignore the door and window openings because this material will be waste.

Area of exterior walls  9 ft. 1 in.  241 ft.  9.08 ft.  241 ft.  2,188 s.f.

The number of sheets is calculated as follows: Sheets of plywood 

2,188s.f.  68.4sheets 32s.f. sheet

Order 69 sheets.

Tyvek comes in 9-foot by 150-foot rolls. The number of rolls is calculated as follows: Tyvek 

241 f.t.  1.6rolls 150 ft. roll

Order 2 rolls.

Determine the materials needed for the interior walls. There are 463 f t. of interior walls. Run the sill plate through the doors for ease of construction; therefore, 463 feet of treated plate is needed for the sill. ee

Board Feet 

2 in.  4in.  463 ft.  309b.f.

12 in. ft.

The top plates will require twice the above quantities; therefore, 926 f t. or 618 b.f. of plate is required. ee

Figure 14.8 shows that there are 0.75 spaces per foot. Determine the number of studs needed:

Wall studs  463 ft.  0.75 spaces/ft.  348 spaces Add one to close off the end. Order 349 studs.

Studs for doors/windows  12 ea.  20 ea.  2 studs/ea.  64 studs 147 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Studs for corners  10 ea.  2 studs/ea.  20 studs Studs for intersections  19 ea.  2 studs/ea.  38 studs Total studs  349 studs  64 studs  20 studs  38 studs  471 studs The header materials are as follows: #

Opening

Length

Material

Order

12

3 f t. window

3 f t. 3 in.

2 ea. 2  12

8 ea. 2  12 @ 10 ft.

20

3 f t. door

3 f t. 5 in.

2 ea. 2  12

20 ea. 2  12 @ 8 ft.

ee

ee

ee

ee

Scrap from sheathing the exterior walls will be used as the spacer in the headers.

The pricing is shown in the following workup sheet:

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

46. Determine the materials required to frame the roof for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the rough carpentry crew from Problem 7-32, determine the cost to frame the roof.

The following data can be extracted from the model:

From page S1.2, the following trusses can be counted: 148 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 14: Wood

Count

Truss

11

53 f t. standard truss

24

53 f t. hip truss

48

3 f t. 6 in. single-pitch truss

ee

ee

ee

Two Simpson H1 ties are required for each standard and hip truss (one at each end where they bear on the exterior walls). One Simpson H1 is required for the singlepitch trusses (one end bears on the exterior wall, and the other attaches to a hip truss).

H1  11 trusses + 24 trusses   2 ea./truss  48 trusses  1 ea./truss H1 = 118 ea.

Order two 2  6 @ 8 ft. for the hips in the corners and two 2  4 @ 8 ft. for the lookouts in the corners.

Determine the sheathing. From the model, the area of the sheathing is 4492 s.f. The number of sheets is calculated as follows: Sheets of plywood 

4,492s.f.  140.4sheets 32s.f. sheet

Order 141 sheets.

Manually determine the sheathing:

Length  1ft.8in.  72ft.  1 ft. 8 in.  75 ft. 4 in.  75.33 ft.

Length  1ft.8in.  50ft.  1 ft. 8 in.  53 ft. 4 in.  53.33 ft.

Figure 14.50 shows that the slope factor for a 6 in 12 roof is 1.12.

Area of sheathing  75.33 ft.  53.33 ft.  1.12  4,499 s.f.

The difference is due to the rounding of 1.12. The number of sheets is calculated as follows: Sheets of plywood 

4,499s.f.  140.6sheets 32s.f. sheet

Order 141 sheets.

Determine the quantity of 2  6 fascia:

Length  75.33 ft.  53.33 ft.  75.33 ft.  53.33 ft.  257.32 ft.

Order 258 f t. ee

Determine the quantity of 2  4 sill:

Length  72 ft.  50 ft.  72 ft.  50 ft.  244 ft.

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Chapter 14: Wood

A 1 f t. 5 in. lookout (backing) is needed for every single-pitch truss, and two are needed for the standard and hip trusses: ee

Lookout  11 trusses  24 trusses   2 ea./truss  48 trusses  1 ea./truss

Lookout = 118 ea.

Five lookouts can be cut from an 8-foot-long board.

Boards 

118ea.  23.6boards 5ea. board

Order 24 ea. 2  4 @ 8 ft.

The pricing is shown in the following workup sheet:

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

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Chapter 15: Thermal and Moisture Protection

Chapter 15—Thermal and Moisture Protection Learning Outcomes This chapter looks at determining the materials and labor associated with Division 07 Thermal and Moisture Protection of the MasterFormat. After reading this chapter, the reader should be able to: Learning Outcome 15.1: Describe the different types of waterproofing and dampproofing.

Learning Outcome 15.2: Determine the materials and labor required for waterproofing and dampproofing.

Learning Outcome 15.3: Determine the materials required for insulation.

Learning Outcome 15.4: Describe the factors to consider when bidding on thermal and moisture protection.

Learning Outcome 15.5: Determine the materials required for siding, soffit, and fascia.

Learning Outcome 15.6: Determine the materials required for roofing.

Activities 1. Assign each student a different material in Division 7 of the MasterFormat. Have the students research their assigned materials and answer the following questions: 

How is the product packaged (roll, bucket, bundle, etc.)?



How much will one package of the material cover?



What additional items are needed to install the material?



Does the manufacturer have any recommended installation procedures for the product?

Have the students present their research in class.

Instructional Resources PowerPoint Slides: Chapter 15.ppt The pdf drawings from Appendices A through E. They can be found at the Pearson Careers Resources for Students website (http://www.pearsonhighered.com/careersresources/) by searching for the book by title or going to Browse by Discipline and choosing Civil and Environmental Engineering.

The drawings include the following: 

Appendix A – Small Commercial Building



Appendix B – Residential Building Project



Appendix C – Commercial Building Project 151 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 15: Thermal and Moisture Protection



Appendix D – Billy’s C-Store



Appendix E – Real Estate Office

Solutions to the Review Questions 1. What is the difference between waterproofing and dampproofing?

Waterproofing is designed to resist the passage of water and the hydrostatic pressure that occurs when the surrounding soils become saturated. Dampproofing is designed to withstand dampness and is not intended to resist the hydrostatic pressure caused by saturated soils.

2. What is membrane waterproofing, and how is it estimated?

Membrane waterproofing consists of the following: 

Built up layers of tar or asphalt and membranes (plies) into a strong, impermeable blanket or



A layer (e.g., polyethylene film) and an adhesive layer (e.g., self-adhesive rubber bitumen compound) or



A waterproof sheet (such as P VC)

that is adhered to the foundation wall. It is estimated by the square (100 square feet).

3. What is the unit of measure for rolls of batt insulation? What type of information should be noted on the estimate?

Blanket and batt insulation is ordered by the roll or the bundle, where a bundle consists of several sheets of insulation cut to a specific length to fit between standard framing members. When estimating blanket or batt insulation, the estimator should note the required thickness, type, and facing material on the workup sheets.

4. What is the unit of measure for shingle roofing? What type of information should be noted on the estimate?

Asphalt shingles are estimated by the square (sq.), which equals 100 square feet.

The estimator should note the color, style, weight, and exposure on the workup sheets.

5. What is the unit of measure for built-up roofing? What type of information should be noted on the estimate?

Built-up roofing is estimated by the square (100 square feet). The estimator should note the number of squares, base sheets, number and weight of piles, the type and number of coats of bituminous material, aggregate surface, type and thickness of insulation, flashing, trim, and miscellaneous items.

6. Determine the square feet of waterproofing needed for a rectangular basement measuring 26 feet by 40 feet. The waterproofing is 8 feet high. How many gallons of waterproofing are needed if 1 gallon of waterproofing will cover 60 square feet?

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Chapter 15: Thermal and Moisture Protection

Find the amount of waterproofing as follows:

Length of wall  26 ft.  40 ft.  26 ft.  40 ft.  132 ft.

Area to be waterproofed  132 ft.  8 ft.  1,056 s.f.

Gallons of waterproofing 

1,056s.f.  17.6gallons 60 s.f. gallon

7. Determine the cost to waterproof the basement in Problem 6 using a productivity of 1.00 labor hours per square (100 square feet) and an average burdened labor rate of $23.17 per labor hour.

From Problem 6, the area of waterproofing is 1,056 square feet, or 10.56 squares.

Labor hours  10.56 sq.  1.00 lhr./sq.  10.56 lhr.

Labor cost  10.56 lhr.  $23.17/lhr.  $244.68 8. Determine the square feet of waterproofing needed for the basement shown in Figures 15.12 and 15.13. The waterproofing starts 6 inches from the top of the wall and continues to the top outside corner of the footing. How many gallons of waterproofing are needed if 1 gallon of waterproofing will cover 65 square feet?

Find the amount of waterproofing as follows:

Length of wall  21 ft.  5 ft.  15 ft.  20 ft.  36 ft.  25 ft.  122 ft.

Height of the waterproofing  8 ft.  6 in.  11 in.  8 ft. 5 in.  8.42 ft.

Area to be waterproofed  122 ft.  8.42 ft.  1,027 s.f.

Gallons of waterproofing 

1,027 s.f.  15.8 gallons 65 s.f. gallon

9. Determine the cost to waterproof the basement in Problem 8 using a productivity of 0.95 labor hours per square (100 square feet) and an average burdened labor rate of $24.42 per labor hour.

From Problem 8, the area of waterproofing is 1,027 square feet, or 10.27 squares.

Labor hours  10.27 sq.  0.95 lhr./sq.  9.76 lhr.

Labor cost  9.76 lhr.  $24.42/lhr.  $238.34

10. How many rolls of R-19 insulation are required to insulate the floor shown in Figure 15.14? The insulation comes in rolls 15 inches wide by 56 feet long.

Find the area of the floor as follows: Area  40 ft.  28 ft.  8 ft.  6 ft. 6 in.  40 ft.  28 ft.  8 ft.  6.50 ft.  1,068 s.f.

Effective coverage  56 ft.  16 in.  56 ft.  1.33 ft.  74 s.f./roll

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Chapter 15: Thermal and Moisture Protection

Rolls of insulation 

1,068s.f.  14.4rolls 74 s.f. roll

Order 15 rolls.

11. How many rolls of R-19 insulation are required to insulate the floor shown in Figure 15.15? The insulation comes in rolls 15 inches wide by 56 feet long.

Find the area of the floor as follows:

Area  24 ft.  36 ft.  10 ft.  13 ft. 4 in.  8 ft.  4 ft.

Area  24 ft.  36 ft.  10 ft.  13.33 ft.  8 ft.  4 ft.  965 s.f.

Effective coverage per roll  56 ft.  16 in.  56 ft.  1.33 ft.  74 s.f./roll Rolls of insulation 

965s.f.  13.0rolls 74 s.f. roll

Order 13 rolls.

12. How many rolls of R-13 insulation are required to insulate the garage walls shown in Figures 15.16 and 15.17? The insulation comes in rolls 15 inches wide by 56 feet long. On the gable ends of the building (east and west sides), the insulation is run to the underside of the rafters. Be sure to deduct the area of the garage door, but do not deduct the area of the personnel door. The west side of the building is similar to the east side without the door.

Find the area of the walls as follows:

Area of the north wall  8 ft. 1 in.  30 ft.  8.08 ft.  30 ft.  242 s.f.

Area of the east wall  8 ft. 1 in.  24 ft.  0.5  4 ft.  24 ft.  7 ft.  16 ft.

Area of the east wall  8.08 ft.  24 ft.  0.5  4 ft.  24 ft.  7 ft.  16 ft.  130 s.f.

Area of the south wall  8 ft. 1 in.  30 ft.  8.08 ft.  30 ft.  244 s.f.

Area of west wall  8 ft. 1 in.  24 ft.  0.5  4 ft.  24 ft.

Area of west wall  8.08 ft.  24 ft.  0.5  4 ft.  24 ft.  242 s.f.

Total wall area = 242 s.f. + 130 s.f. + 242 s.f. + 242 s.f. = 856 s.f.

Effective coverage per roll  56 ft.  16 in.  56 ft.  1.33 ft.  74 s.f.

Rolls of insulation 

856s.f.  11.6rolls 74 s.f. roll

Order 12 rolls.

13. How many rolls of R-19 insulation are required to insulate the garage roof shown in Figures 15.16 and 15.17? The insulation comes in rolls 23 inches wide by 56 feet long. The insulation is installed between the rafters, which are spaced 24 inches on center.

154 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 15: Thermal and Moisture Protection

Figure 14.50 shows that the slope length factor is 1.055.

Run 

24 ft.  12 ft.

2

Length of rafter  12 ft.  1.055  12.66 ft.

Area of roof  2 sides  12.66 ft./side  30 ft.  760 s.f.

Effective coverage per roll  56 ft.  24 in.  56 ft.  2 ft.  112 s.f.

Rolls of insulation 

760s.f.  6.8rolls 112 s.f. roll

Order 7 rolls.

14. How many rolls of R-13 insulation are required to insulate the exterior walls of the residence shown in Figure 15.18? The wall height is 8 feet, 1 inch high. The insulation comes in rolls 15 inches wide by 56 feet long. Do not deduct for the doors or windows.

Find the length of the wall as follows: Exterior wall length = 32 f t. + 26 f t. + 36 f t. + 13 f t. 6 in. + 4 f t. + 12 f t. 6 in. = 124 f t. ee

ee

ee

ee

ee

ee

ee

Area of exterior walls  124 ft.  8 ft. 1 in.  124 ft.  8.08 ft.  1,002 s.f.

Effective coverage per roll  56 ft.  16 in.  56 ft.  1.33 ft.  74 s.f.

Rolls of insulation 

1,002s.f.  13.5rolls 74 s.f. roll

Order 14 rolls.

15. How many rolls of R-19 insulation are required to insulate the roof of the residence shown in Figure 15.18? The insulation comes in rolls 15 inches wide by 56 feet long.

The insulation is installed between the ceiling joists, which are run horizontally and are spaced 16 inches on center.

Find the area of the ceiling as follows:

Area of ceiling  36 ft.  26 ft.  4 ft.  12 ft. 6 in.

Area of ceiling  36 ft.  26 ft.  4 ft.  12.50 ft.  886 s.f.

Effective coverage per roll  56 ft.  16 in.  56 ft.  1.33 ft.  74 s.f.

Rolls of insulation 

886s.f.  12.0 rolls 74 s.f. roll

Order 12 rolls.

16. How many rolls of air barrier are required for the garage shown in Figures 15.16 and 15.17? The air barrier comes in rolls 150 feet long by 9 feet wide.

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Chapter 15: Thermal and Moisture Protection

Starting with the side of the garage door opening, 92 feet (4 f t. + 30 f t. + 24 f t. + 30 f t. + 4 f t.) of barrier is required. A 5-foot by 24-foot piece is required on each gable end (allowing for a 1-foot overlap). ee

ee

ee

ee

ee

Total length = 92 f t. + 24 f t. + 24 f t. = 140 f t. ee

Rolls 

ee

ee

ee

140 ft.  0.9rolls 150 ft. roll

Order 1 roll.

17. Determine the siding, soffit, and fascia required for the garage shown in Figures 15.16 and 15.17. The roof overhang is 16 inches.

Length of wall = 24 f t. + 30 f t. + 24 f t. + 30 f t. = 108 f t. ee

ee

ee

ee

ee

Ignore the personnel door opening, deduct the garage opening, and add the gable ends.

Area of wall  108 ft.  8 ft. 1 in.  16 ft.  7 ft.  2 ea.  0.5  24 ft.  4 ft.

Area of wall  108 ft.  8.08 ft.  16 ft.  7 ft.  2 ea.  0.5  24 ft.  4 ft.  857 s.f.

Order 9 sq. of siding.

Order 92 ft. 108 ft.  16 ft. of starter strip

.

Order four outside corners 9 f t. long. ee

Figure 14.50 shows that the slope length factor is 1.055. The length of the rafters (from the ridge to the outside of the wall) is as follows:

Length of rafter 

24 ft.  1.055  12.7 ft.

2

J-molding is required around the openings and at the top of the wall.

J-molding for top of wall  2 ea.  30 ft./ea.  4 ea.  12.7 ft./ea.  111 ft.

J-molding around doors = 7 f t. + 16 f t. + 7 f t. + 7 f t. + 3 f t. + 7 f t. = 47 f t. ee

ee

ee

ee

ee

ee

ee

Total J-molding = 111 f t. + 47 f t. = 158 f t. ee

ee

ee

Determine the area of the soffit. Be sure not to double-count the corners: Area of soffit  2 ea.  1.33 ft.  30 ft.  1.33 ft.  1.33 ft.  4 ea.  12.7 ft.  1.33 ft.

Area of soffit = 155 s.f.

Order 2 squares.

J-molding for the soffit occurs at the edge of the building. The soffit is tucked under the fascia on the other side.

J-molding for soffit  2 ea.  30 ft.  4 ea.  12.7 ft.  111 ft.

Order 111 feet of J-molding.

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Chapter 15: Thermal and Moisture Protection

Figure 14.50 shows that the slope length factor is 1.055. The length of the rafter (from the ridge to the fascia) is as follows: Rise  12 ft.  1.33 ft.  1.055  14.1 ft Length of fascia  2 ea.  1.33 ft.  30 ft.  1.33 ft.  4 ea.  14.1 ft.  122 ft.

Order 122 feet of fascia.

18. Determine the siding, soffit, and fascia required for the residence shown in Figure 15.18. The wall height is 8 feet 1 inch high, and the residence has a hip roof with an overhang of 18 inches.

Length of wall = 26 f t. + 36 f t. + 13 f t. 6 in. + 4 f t. + 12 f t. 6 in. + 32 f t. ee

ee

ee

ee

ee

ee

Length of wall = 26 f t. + 36 f t. + 13.5 f t. + 4 f t. + 12.5 f t. + 32 f t. = 124 f t ee

ee

ee

ee

ee

ee

ee

Ignore the openings: Area of wall  124 ft.  8 ft. 1 in.  124 ft.  8.08 ft.  1,002 s.f.

Order 10 squares of siding.

Order 124 feet of starter strip.

Order four outside corners 9 feet long.

J-molding is required around the openings and at the top of the wall. Allow 20 feet of J-molding per opening:

J-molding for siding  124 ft.  7 openings  20 ft./opening  264 ft.

Order 264 feet of J-molding.

The area of the soffit equals the perimeter of the building times the overhang plus the corners and the soffit by the front door.

Area of soffit  124 ft.  1.5 ft.  4 corners  1.5 ft.  1.5 ft.  12.5 ft.×4 ft.

Area of soffit = 245 s.f.

Order 3 squares of soffit.

Order 124 feet of J-molding.

Length of fascia  2 sides  1.5 ft.  36 ft.  1.5 ft. /side  2 sides  1.5 ft.  26 ft.  1.5 ft. /side

Length of fascia = 136 f t. ee

Order 136 f t. of fascia. ee

19. How many rolls of felt are needed for the roof shown in Figure 15.19? The slope of the roof is 4 in 12. The felt comes in rolls 36 inches wide by 67 feet long and is lapped 19 inches.

Figure 14.50 shows that the slope length factor is 1.055.

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Chapter 15: Thermal and Moisture Protection

Run 

28 ft.  14 ft.

2

Length of rafter  14 ft.  1.055  14.77 ft.

Area of roof  2 sides  14.77 ft.  34 ft.  1,004 s.f.

Roll  67 ft.  36 in.  67 ft.  3 ft.  201 s.f.

Exposure  36 in.  19 in.  17 in Roll coverage 

201 s.f.  17in.  94.9s.f.

36in.

Roll of felt required 

1,004s.f.  10.6rolls 94.9s.f. roll

Order 11 rolls.

20. How many squares of roofing and pieces of ridge shingles are needed for the roof in Figure 15.19? The slope of the roof is 4 in 12, and the exposure is 5 inches.

From Problem 19, the roof area is 1,004 square feet. or 10.04 squares.

Starter course  2 sides  34 ft./side  5 in.

Starter course  2 sides  34 ft./side  0.42 ft.  29 s.f.  0.29 sq.

Squares of roofing = 10.04 sq. + 0.29 sq. = 10.33 sq.

Ridge shingles 

34 ft.

408in.   82shingles 5in. shingle 5in. shingle

21. How many rolls of felt are needed for the roof shown in Figure 15.20? The slope of the roof is 5 in 12. The felt comes in rolls 36 inches wide by 134 feet long and is lapped 6 inches.

Figure 14.50 shows that the slope length factor is 1.083:

Run 

28 ft.8in.  14 ft. 4in.  14.33 ft.

2

Length of rafter  14.33 ft.  1.083  15.52 ft.

Area of roof  2 sides  15.52 ft.  38 ft. 8 in.

Area of roof  2 sides  15.52 ft.  38.67 ft.  1,200 s.f.

Roll  134 ft.  36 in.  134 ft.  3 ft.  402 s.f.

Exposure  36 in.  6 in.  30 in.

Roll coverage 

402 s.f.  30in.  335s.f.

36in.

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Chapter 15: Thermal and Moisture Protection

Rolls of felt required 

1,200 s.f.  3.6rolls 335s.f. roll

Order 4 rolls.

22. How many squares of roofing and pieces of ridge shingles are needed for the roof in Figure 15.20? The slope of the roof is 5 in 12, and the exposure is 5 inches wide.

From Problem 21, the roof area is 1,200 square feet, or 12.00 squares.

Starter course  2 sides  38 ft. 8 in./side  5 in.

Starter course  2 sides  38.67 ft./side  0.42 ft.  32 s.f.  0.32 sq.

Squares of roofing = 12.00 sq. + 0.32 sq. = 12.32 sq.

Ridge shingles 

38 ft. 8in.

464in.   93shingles 5in. shingle 5in. shingle

23. Determine the gutters and downspouts required for the garage shown in Figures 15.16 and 15.17.

The gutter is only needed on the sides of the roof perpendicular to the roof’s slope.

Gutters  2 sides  1.33 ft.  30 ft.  1.33 ft. /side  66 ft.

Locate downspouts on each corner and allow 1.33 feet of downspout to move the downspout from the eave to the wall.

Downspout  4 spouts  1.33 ft.  8 ft. / spout  38 ft.

An elbow is required at the gutter, one at the top of the wall to turn the downspout down, and one at ground level to direct the water away from the building; therefore, three elbows will be required for each downspout.

Order 12 elbows.

24. Determine the gutters and downspouts required for the residence shown in Figure 15.18. The residence has a hip roof with an overhang of 18 inches.

The gutters are needed around the entire perimeter of the roof. Be sure to include the overhang.

Gutters  2 sides  1.5 ft.  36 ft.  1.5 ft. /side  2 sides  1.5 ft.  26 ft.  1.5 ft. /side

Gutters = 136 f t. ee

Locate downspouts on each of the corners and allow for 1.5 f t. of downspout to move the downspout from the eave to the wall. ee

Downspout  4 spouts  1.5 ft.  8 ft  /spout  38 ft.

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Chapter 15: Thermal and Moisture Protection

An elbow is required at the gutter, one at the top of the wall to turn the downspout down, and one at ground level to direct the water away from the building; therefore, three elbows will be required for each downspout.

Order 12 elbows.

25. Determine the amount of dampproofing required for the building in Appendix C. The following workup sheet contains the dampproofing calculations.

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

26. Determine the amount of insulation required for the walls and ceiling of the building in Appendix C. The following workup sheet contains the insulation calculations.

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

27. Determine the amount of roofing required for the building in Appendix C. The following workup sheet contains the roofing calculations.

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

28. Determine the amount of flashing required for the building in Appendix C. 160 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 15: Thermal and Moisture Protection

The following workup sheet contains the flashing calculations.

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

29. Determine the materials required to complete the insulation for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the rough carpentry crew from Problem 7-32, determine the cost to insulate the building.

The following data can be extracted from the model:

All the 24-inch by 48-inch windows are in the exterior walls. Only two of the 36-inch by 72-inch windows are in the exterior walls. There is one 3-foot by 7-foot door and one 6-foot by 7-foot door in the exterior walls. The remaining twenty 3-foot by 7-foot doors are in the interior walls.

Determine the materials needed for the exterior walls. From the model, we see that we need 1,818 square feet of exterior wall, which requires 3.5-inch batt insulation. It may be manually calculated as follows: Area of wall  241 ft.  9 ft. 1 in.  241 ft.  9.08 ft.  2,188 s.f.

Area of openings  34 ea.  2 ft.  4 ft.  2 ea.  3 ft.  6 ft.  3 ft.  7 ft.  6 ft.  7 ft.

Area of openings = 371 s.f.

Area of wall to be insulation  2,188 s.f.  371 s.f.  1,817 s.f.

This area is similar to the 1,818 square feet extracted from the model, with the difference due to rounding.

Assume 56-foot-long rolls:

Effective coverage per roll  56 ft.  16 in.  56 ft.  1.33 ft.  74 s.f.

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Chapter 15: Thermal and Moisture Protection

Rolls of insulation 

1,817s.f.  24.6rolls 74 s.f. roll

Order 25 rolls.

Determine the materials needed for the interior walls. From the model, we see that we need 3,483 square feet of interior walls, which require 3.5-inch batt insulation.

It may be manually calculated as follows: Area of wall  463 ft.  9 ft. 1 in.  463 ft.  9.08 ft.  4,204 s.f.

Area of openings  12 windows  36 in.  72 in.  20 doors  3 ft.  7 ft.

Area of openings  12 windows  3 ft.  6 ft.  20 doors  3 ft.  7 ft.  636 s.f.

Area of wall to be insulation  4,204 s.f.  636 s.f.  3,568 s.f.

This number is close to the number calculated by the model but is off by two percent. This is a good time to discuss the potential pitfalls of using the model for takeoffs. Clearly, there is a difference in how the wall area is calculated.

Assume 56-foot-long rolls.

Effective coverage per roll  56 ft.  16 in.  56 ft.  1.33 ft.  74 s.f.

Rolls of insulation 

3,483s.f.  47.1 rolls 74 s.f. roll

Order 48 rolls.

The area of the ceiling equals the area of the floor slab. The following quantities can be extracted from the model.

The area of the ceiling from the model is 3,600 s.f. Alternatively, it may be calculated as follows: Area  72 ft.  50 ft.  3,600 s.f.

Assume 56-foot-long rolls.

Effective coverage per roll  56 ft.  24 in.  56 ft.  2 ft.  112 s.f.

Rolls of insulation 

3,600s.f.  32.1 rolls 112 s.f. roll

Order 33 rolls.

The pricing is shown in the following workup sheet. The sheet uses the quantities extracted from the model:

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Chapter 15: Thermal and Moisture Protection

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

30. Determine the materials required to complete the roofing for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the roofing crew from Problem 7-32, determine the cost to roof the building.

The following data can be extracted from the model:

Determine the area of the roofing manually: Length = 1 f t. 8 in. + 72 f t. + 1 f t. 8 in. = 75 f t. 4 in. = 75.33 f t. ee

ee

ee

ee

ee

Width = 1 f t. 8 in. + 50 f t. + 1 f t. 8 in. = 53 f t. 4 in. = 53.33 f t. ee

ee

ee

ee

ee

Figure 14.50 shows that for a 6 in 12 slope, the run is multiplied by 1.12.

Area of sheathing  75.33 ft.  53.33 ft.  1.12  4,499 s.f.

This number is similar to the 4,492 square feet extracted from the model, with the difference due to rounding.

Felt in a roll  134 ft.  36 in.  134 ft.  3 ft.  402 s.f.

Assume a 6-inch lap.

Exposure  36 in.  6 in.  30 in.

Roll coverage 

402 s.f.  30in.  335s.f.

36in.

Rolls of felt required 

4,492 s.f.  13.4rolls 335s.f. roll

Order 14 rolls.

Starter course = 75 f t. 4 in. + 53 f t. 4 in. + 75 f t. 4 in. + 53 f t. 4 in. = 257 f t. 4 in. ee

ee

ee

ee

ee

Order 258 f t. ee

Starter course  258 ft.  5 in.  258 ft.  0.42 ft.  108 s.f.  1.08 sq.

Squares of roofing  44.92 sq.  1.08 sq.  46.0 sq.

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Chapter 15: Thermal and Moisture Protection

Rise 

53 ft.4in. 6 53.33 ft. 6     13.33 ft.

2 12 2 12

Run 

53 ft.4in. 53.33 ft.   26.67 ft.

2 2

2 2 2 Length of hips  Run   Run   Rise    

0.5

2 2 2 Length of hips   26.67 ft.   26.67 ft.  13.33 ft.   

0.5

 40ft.

Length of ridge  75 ft. 4 in.  53 ft. 4 in.  22ft.

Ridge shingles  4 hips  40ft./hip  22ft.  182 ft.

Ridge shingles 

182ft.

2184in.   437 shingles 5in. shingle 5in. shingle

Drip edge  75 ft. 4 in.  53 ft. 4 in.  75 ft. 4 in.  53 ft. 4 in.  257 ft. 4 in.

Order 258 f t. ee

The roof requires three pipe flashings.

Assume a ratio of one square foot of vent per 300 square feet of roof measured in plan view. Determine the required venting area: Area of roof  75.33 ft.  53.33 ft.  4,017 s.f.

Area of vents 

4,017 s.f.  13.4s.f.

300s.f. s.f.

Use Master Flow static roof vents bronze aluminum square roof louver, which provides 50 square inches (0.35 s.f.) per vent.

Number of vents 

13.4 s.f.  38.3 vents 0.35s.f. vent

Order 39 vents.

Also, three pipe flashings will be needed for the plumbing vents.

The pricing is shown in the following workup sheet:

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Chapter 15: Thermal and Moisture Protection

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

31. Determine the materials required to complete the soffit, fascia, gutters, and downspouts for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the finish crew from Problem 7-32, determine the cost to install the soffit, fascia, gutters, and downspouts.

32. Fascia = 75 f t. 4 in. + 53 f t. 4 in. + 75 f t. 4 in. + 53 f t. 4 in. ee

ee

ee

ee

33. Fascia = 75.33 f t. + 53.33 f t. + 75.33 f t. + 53.33 f t. = 257.32 f t. ee

ee

ee

ee

ee

Order 258 f t. ee

Do not double-count corners for the soffit: Soffit = 75 f t. 4 in. + 50 f t. + 75 f t. 4 in. + 50 f t. ee

ee

ee

ee

Soffit = 75.33 f t. + 50 f t. + 75.33 f t. + 50 f t. = 250.66 f t. ee

ee

ee

ee

ee

Soffit  250.66 ft.  1ft. 8 in.  250.66 ft.  1.67ft.  419 s.f.

The gutters are needed around the entire perimeter of the roof: Gutters = 75 f t. 4 in. + 53 f t. 4 in. + 75 f t. 4 in. + 53 f t. 4 in. ee

ee

ee

ee

Gutters = 75.33 f t. + 53.33 f t. + 75.33 f t. + 53.33 f t. = 257.32 f t. ee

ee

ee

ee

ee

Order 258 f t. ee

Locate downspouts on each of the corners and allow for 2 feet of downspout to move the downspout from the eave to the wall: Downspout  4 ea.   2 ft.  9 ft. /ea.  44 ft

An elbow is required at the gutter, one at the top of the wall to turn the downspout down, and one at ground level to direct the water away from the building; therefore, three elbows will be required for each downspout.

Order 12 elbows.

The pricing is shown in the following workup sheet:

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

165 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Chapter 16—Doors and Windows Learning Outcomes This chapter looks at determining the materials and labor associated with Division 08 Openings of the MasterFormat. After reading this chapter, the reader should be able to: Learning Outcome 16.1: Determine the materials and labor needed to install windows.

Learning Outcome 16.2: Determine the materials needed for curtain walls, storefronts, and glazing.

Learning Outcome 16.3: Describe the factors to consider when bidding on openings.

Learning Outcome 16.4: Determine the materials and labor needed to install doors.

Activities 1. Visit a building on campus that has storefront entrances. Have the students measure the storefronts and prepare a material list for the storefront.

Instructional Resources PowerPoint Slides: Chapter 16.ppt The pdf drawings from Appendices A through E. They can be found at the Pearson Careers Resources for Students website (http://www.pearsonhighered.com/careersresources/) by searching for the book by title or going to Browse by Discipline and choosing Civil and Environmental Engineering.

The drawings include the following: 

Appendix A – Small Commercial Building



Appendix B – Residential Building Project



Appendix C – Commercial Building Project



Appendix D – Billy’s C-Store



Appendix E – Real Estate Office

Solutions to the Review Questions 1. What accessories should be checked for when taking off windows and curtain walls?

The accessories include glass, glazing, screens, hardware, weather stripping, millions, sills, stools, flashings, and lintels.

2. What is the definition of glazing? Why must the estimator determine who will perform the required glazing?

Glazing is the setting of whatever material (such as glass, porcelain, metal panels, plastic laminates, brick and stone veneer, and precast concrete) in the frames. The 166 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

estimator should verify who will install the glazing to ensure that it is included in the bid and not included more than once.

3. What information is required to price a door?

The information required to price a door includes the size, core type, face veneer, swing, frame type, prefinished or job finished, pre-hung or job hung, hardware, firerating, sound-rating, and accessories (such as weather stripping and thresholds).

4. What are the advantages of using premachined and prefinished doors?

Premachining is popular as it reduces labor costs. Prefinishing can save considerable finishing time and generally yields a better result than job finishing.

5. Why should the required type of finish on the door and frames be noted on the workup sheet?

The estimator should note the door and frame finish on the workup sheet and make sure that the prefinished doors are included in the door supplier’s bid and that the finishing costs for the job-finished doors and frames are included under finishes.

6. How may hardware be handled on a project?

The hardware required for a project is divided into rough and finished. Rough hardware (such as bolts, screws, nails, small anchors, and miscellaneous fasteners) is not included in the hardware schedule, but it is often required for installation.

Finished hardware (such as hinges, door-closing devices, locks, latches, deadbolts, and kickplates) is either identified on the finish hardware schedule or handled using a cash allowance.

7. What precautions must an estimator take when using an allowance from the specifications in the estimate?

They must be sure the allowance is included in the estimate, add applicable sales tax and delivery costs, and include the cost to install the hardware.

8. What is the unit of measure for glass, and why should the various types and sizes be listed separately?

Glass is estimated by the square foot, with sizes taken from the drawings. Care must be taken to keep each type separate because their costs will differ.

9. Prepare a materials list for the residential windows in Figure 16.15.

The following is a list of the required windows: Quantity

Type

1

3 ft. 0 in.  1 ft. 0 in. vinyl slider

4

4 ft. 0 in.  4 ft. 0 in. vinyl slider

1

5 ft. 0 in.  4 ft. 0 in. vinyl slider

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Chapter 17: Finishes

10. Determine the cost to install the windows in Figure 16.15 using a productivity rate of 3.0 labor hours per window and a burdened labor rate of $29.50 per labor hour.

From Problem 9, there are 6 windows.

Labor hours  6 ea.  3.0 lhr./ea.  18.0 lhr.

Labor cost  18.0 lhr.  $29.50/lhr.  $531 11. Prepare a materials list for the residential doors in Figure 16.15. Be sure to specify swing.

The following is a list of the required doors: Quantity

Type

1

3 ft. 0 in.  6 ft. 8 in. exterior solid, right hand

2

3 ft. 0 in.  6 ft. 8 in. wood, hollow core, right hand

2

3 ft. 0 in.  6 ft. 8 in. wood, hollow core, left hand

3

6 ft. 0 in.  6 ft. 8 in. bifold, hollow core

12. Determine the cost to install the doors in Figure 16.15 using the productivity rates in Figure 16.16 and a burdened labor rate of $29.50 per labor hour.

The labor cost is determined as follows:

Labor hours  1 ea.  2.5 lhr./ea.  4 ea.  2.0 lhr./ea.  3 ea.  3.0 lhr./ea.

Labor hours = 19.5 lhr.

Labor cost  19.5 lhr.  $29.50/lhr.  $576

13. Determine the linear feet of aluminum tubing needed for the storefront in Figure 16.17.

Be sure not to include tubing below the door.

Lineal feet  6 ea.  9 ft. 6 in./ea.  18 ea.  3 ft./ea.

Lineal feet  6 ea.  9.5 ft./ea.  18 ea.  3 ft./ea.  111 ft.

14. Prepare a materials list for the glass required for the storefront in Figure 16.17. How many square feet of glass is required for the storefront? Use nominal dimensions.

The glass sizes and square footages are as follows: Size

Pieces

s.f. each

Total s.f.

2 ft. 6 in.  3 ft. 0 in.

5

7.5

37.5

4 ft. 0 in.  3 ft. 0 in.

4

12.0

48.0

3 ft. 0 in.  3 ft. 0 in.

4

9.0

36.0

7 ft. 0 in.  3 ft. 0 in.

1

21.0

21.0

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Chapter 17: Finishes

Size

Pieces

s.f. each

Total s.f.

Total

142.5

15. Determine the cost to install the storefront in Figure 16.17 using a productivity rate of 0.08 labor hours per square foot and a burdened labor rate of $32.96 per labor hour.

The labor cost is determined as follows:

Area  15 ft.  9 ft. 6 in.  15 ft.  9.5 ft.  142.5 s.f.

Labor hours  142.5 s.f.  0.08 lhr./s.f.  11.4 lhr.

Labor cost  11.4 lhr.  $32.96/lhr.  $375.74 16. Determine the windows required for the Commercial Building in Appendix C. The following workup sheet contains the window calculations.

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

17. Determine the door and hardware required for the Commercial Building in Appendix C. The specifications require a $2,000 allowance for door hardware.

The following workup sheet contains a list of the doors.

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

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Chapter 17: Finishes

18. Determine the windows and storefronts needed for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the finish crew from Problem 7-32, determine the cost for the windows and storefronts.

The following data can be extracted from the model:

Two 36-inch by 72-inch windows are in the exterior walls, and the other 12 are in the interior walls.

The building requires two aluminum storefronts: one 36 inch by 84 inch and one 72 inch by 84 inch.

The pricing is shown in the following workup sheet:

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

19. Determine the doors and hardware needed for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the finish crew from Problem 7-32, determine the cost for the doors and hardware.

The following data can be extracted from the model:

The first and last doors in the above table are the storefront doors included in Problem 18. There are twenty 36-inch by 84-inch interior wood doors.

Five 8-foot pieces of trim are needed for each interior door: two for each of the jambs and one for the head. The storefront doors do not require trim.

Trim  20 doors  5 ea./door  100 ea.

Order the interior door pre-hung and pre-drilled. The hinges will be included with the pre-hung door.

The hardware includes 20 locksets (for interior doors), two cylinders (for the storefront doors), and 23 door stops; the 72-inch by 84-inch door requires two stops.

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Chapter 17: Finishes

The pricing is shown in the following workup sheet:

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

171 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Chapter 17—Finishes Learning Outcomes This chapter looks at determining the materials and labor associated with Division 09 Finishes of the MasterFormat. After reading this chapter, the reader should be able to: Learning Outcome 17.1: Describe the common types of finishes and their components.

Learning Outcome 17.2: Estimate the materials needed for drywall and metal studs.

Learning Outcome 17.3: Estimate the materials needed for flooring.

Learning Outcome 17.4: Estimate the materials needed for painting.

Activities 1. Provide the students with a residence or commercial building floor plan. Using a carpet width of 12 feet, have the students prepare a carpet layout for the building and estimate the quantity (square yards) of carpet. The carpet can only be run in one direction (up and down or left and right), and the number of seams should be minimized. Consider dividing the students into two groups and assigning each group a different direction to run the carpet. Have the students present their solutions to the class and compare the different amounts of carpet needed based on the direction of the layout and the reuse of scrap materials. Discuss how careful planning of the carpet’s layout can reduce the cost.

Instructional Resources PowerPoint Slides: Chapter 17.ppt The pdf drawings from Appendices A through E. They can be found at the Pearson Careers Resources for Students website (http://www.pearsonhighered.com/careersresources/) by searching for the book by title or going to Browse by Discipline and choosing Civil and Environmental Engineering.

The drawings include the following: 

Appendix A – Small Commercial Building



Appendix B – Residential Building Project



Appendix C – Commercial Building Project



Appendix D – Billy’s C-Store



Appendix E – Real Estate Office

Solutions to the Review Questions 1. What is the difference between drywall and wet wall?

Drywall construction uses gypsum wallboard, and wet wall uses plaster and stucco.

172 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

2. Why should walls of various heights, thicknesses, and finishes be listed separately?

The estimator should list any walls with different thicknesses, heights, types of drywall, number of layers, types of fasteners, levels of finish, or any other variations on the workup sheet because the different types of walls will have different costs.

3. What procedure is used to estimate metal studs and runners?

The walls must be separated according to thickness, stud type, stud spacing, wall height, finish, and application technique. The takeoff of metal studs is similar to wood studs, but the metal studs run the full height of the wall because the studs sit inside the metal track. The number of studs is calculated by dividing the length of walls by the stud spacing and adding one to close off the end. Additional studs are added for any openings, corners, and wall intersections, and the waste is added.

The quantity of runners equals two times the wall length because a runner is required at the top and bottom of the wall. To this quantity, waste is added.

4. What are the types of lath used for wet-wall construction? What are their units of measure?

The three types of lath are gypsum, metal, and wood. Gypsum lath is sold by the sheet. The metal and wood lath are taken off by the square yard.

5. What unit of measure is used for solid wood flooring, and what information should be noted on the workup sheets?

Wood flooring is often sold by the square foot. Estimators should note the quantity of flooring, the size and thickness of the flooring, the pattern, the finish, and the type of installation on the workup sheets.

6. What unit of measure is used for resilient tile, and what information should be noted on the workup sheets?

Resilient tile is estimated by the square foot. The estimator should note the thickness of the flooring, size and shape, color, design layout, type of subfloor, and underlayment on the workup sheet.

7. What unit of measure is used to estimate carpet, and what can be done to minimize waste?

Carpeting is estimated by the square yard. The estimator should prepare a rough layout showing where the carpet will be seamed and how the scraps will be used.

Several layouts may be needed to find the best design for keeping waste and seams to a minimum.

8. How many steel studs are needed for the office space shown in Figure 17.16? The exterior walls are existing. The studs are spaced 16 inches on center. Add two studs for each door and each intersection.

The wall length and number of studs are determined as follows:

Wall length  40 ft. 1 in.  3 ea.  11 ft. 9 in./ea.  75 ft. 4 in.  75.33 ft.

Figure 14.8 shows that there are 0.75 spaces per foot: 173 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Wall studs  75.33 ft.  0.75 spaces/ft.  57 spaces Add one to close off the end: Order 58 studs.

Studs for doors  4 ea.  2 studs/ea.  8 studs

Studs for intersections  8 ea.  2 studs/ea.  16 studs Total studs = 58 studs + 8 studs + 16 studs = 82 studs 9. How many 20-foot-long pieces of runner track are needed for the office space in Figure 17.16?

Both a top and bottom track are needed. From Problem 8, the length of the interior walls is 75.33 feet.

Track 

2 ea.  75.33 ft. ea.  8pieces 20 ft. piece

10. How many 4-foot-wide by 8-foot-high sheets of drywall are needed for the office space in Figure 17.16? The walls are 8 feet high. The exterior walls need drywall on the interior side only. The interior walls need drywall on both sides. Add 5 percent waste.

From Problem 8, the length of the interior walls is 75.33 feet.

Exterior wall length  2 ea.  40 ft. 1 in./ea.  2 ea.  23 ft. 9 in./ea.

Exterior wall length = 127 f t. 8 in. = 127.67 f t. ee

ee

Drywall  127.67 ft.  2 sides  75.33 ft./side   8 ft. = 2,227 s.f.

Drywall 

2,227s.f.  69.6sheets 32s.f. sheet

Add five percent for waste: Drywall  69.6 sheets  1  0.05   73.1 sheets

Order 74 sheets.

11. How many steel studs are needed for the office space shown in Figure 17.17? The exterior walls are existing. The studs are spaced 16 inches on center. Add two studs for each door and each intersection.

The wall length and number of studs are determined as follows:

Wall length  27 ft. 4 in.  33 ft.  11 ft. 10 in.  2 ea.  11 ft. 5 in./ea.

Wall length = 95 f t. ee

Figure 14.8 shows that there are 0.75 spaces per foot:

Wall studs  95 ft.  0.75 spaces/ft.  72 spaces 174 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Add one to close off the end: Order 73 studs.

Studs for doors  5 ea.  2 studs/ea.  10 studs Studs for intersections  10 ea.  2 studs/ea.  20 studs Total studs = 73 studs + 10 studs + 20 studs = 103 studs 12. How many 20-foot-long pieces of runner track are needed for the office space in Figure 17.17?

Both a top and bottom track are needed. From Problem 11, the length of the interior walls is 95 feet.

Track 

2 ea.  95 ft. ea.  10pieces 20 ft. piece

13. How many 4-foot-wide by 8-foot-high sheets of drywall are needed for the office space in Figure 17.17? The walls are 8 feet high. The exterior walls need drywall on the interior side only. The interior walls need drywall on both sides. Add 5 percent waste.

From Problem 11, the length of the interior walls is 95 feet.

Exterior wall length  2 ea.  45 ft. 1 in./ea.  2 ea.  27 ft. 4 in./ea.

Exterior wall length  2 ea.  45.08 ft./ea.  2 ea.  27.33 ft./ea.

Exterior wall length = 144.82 f t. ee

Drywall  144.82 ft.  2 sides  95 ft./side   8 ft.  2,679 s.f.

Drywall 

2,679 s.f.  83.7 sheets 32s.f. sheet

Add five percent for waste: Drywall  83.7 sheets  1  0.05   87.9 sheets

Order 88 sheets.

14. How many 4-foot-wide by 8-foot-high sheets of drywall are needed for the garage walls in Figure 17.18? The walls are 8 feet high, and the overhead door is 7 feet high by 16 feet wide. Only the inside of the garage walls needs drywall.

Determine the wall length without the overhead door.

Wall length = 30 f t. + 4 f t. + 4 f t. + 30 f t. + 24 f t. = 92 f t. ee

ee

ee

ee

ee

ee

Sheets of drywall  92 ft.  8 ft.  736 s.f.

Sheets of drywall 

736s.f.  23sheets 32s.f. sheet

Add one sheet to be used above the overhead door: 175 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Order 24 sheets.

15. How many 4-foot-wide by 8-foot-high sheets of drywall are needed for the ceiling of the garage in Figure 17.18? The ceiling is flat. Add 5 percent waste.

The ceiling area and the number of sheets of drywall are determined as follows:

Ceiling area  30 ft.  24 ft.  720 s.f.

Area per sheet  4 ft.  8 ft.  32 s.f.

Drywall 

720s.f.  22.5sheets 32s.f. sheet

Add five percent for waste:

Drywall  22.5 sheets  (1 0.05)  23.6 sheets Order 24 sheets.

16. How many 4-foot-wide by 8-foot-high sheets of drywall are needed for the walls of the residence in Figure 17.19? The walls are 8 feet high. The exterior walls get drywall only on the inside, and the inside walls get drywall on both sides. Add 5 percent waste.

The wall area and the number of sheets of drywall are determined as follows: Exterior wall length = 32 f t. + 26 f t. + 36 f t. + 13 f t. 6 in. + 4 f t. + 12 f t. 6 in. ee

ee

ee

ee

ee

ee

Exterior wall length = 124 f t. ee

Interior wall length  12 ft. 10 in.  19 ft. 4 in.  16 ft. 4 in.  3.5 in.  8 ft. 9 in.  11 ft. 9 in.  3 ea.  2 ft. 1 in./ea.  3 ea.  8 ft. 4 in./ea.

Interior wall length  12.83 ft.  19.33 ft.  16.33 ft.  0.29 ft.  8.75 ft.

11.75 ft.  3 ea.  2.08 ft./ea.  3 ea.  8.33 ft./ea.

Interior wall length = 100.51 f t. ee

Drywall  124 ft.  2 sides  100.51 ft./side   8 ft.  2,600 s.f.

Drywall 

2,600 s.f.  81.3 sheets 32s.f. sheet

Add five percent for waste: Sheets of drywall  81.3 sheets  1  0.05  = 85.4 sheets

Order 86 sheets.

17. How many 4-foot-wide by 8-foot-high sheets of drywall are needed for the ceiling of the residence in Figure 17.19? The ceiling is flat. Add 5 percent waste.

The ceiling area and the number of sheets of drywall are determined as follows:

Ceiling area  36 ft.  26 ft. – 4 ft.  12 ft. 6 in.

176 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Ceiling area  36 ft.  26 ft. – 4 ft.  12.50 ft.  886 s.f.

Area per sheet  4 ft.  8 ft.  32 s.f.

Sheets of drywall 

886s.f.  27.7sheets 32s.f. sheet

Add five percent for waste: Sheets of drywall = 27.7 sheets × 1 + 0.05  = 29.1

Order 29 sheets.

18. The walls of the residence shown in Figure 17.19 are plastered with a 1/4 inch of plaster over metal lath. The exterior walls get plaster only on the inside, and the inside walls get plaster on both sides. The walls are 8 feet high. How many cubic feet of plaster are needed for the walls?

From Problem 16, the length of the exterior walls is 124 feet., and the length of the interior walls is 100.5 feet.

Area of walls  124 ft.  2 sides  100.5 ft./side   8 ft.  2,600 s.f.

Area of walls 

2,600s.f.  2.89 hundreds.y.

9s.f. s.y.

Figure 16.7 shows that 5.2 cubic feet of plaster are needed for 100 square yards for each 1/16 inch of thickness plus 8.0 cubic feet for the metal lath. The plaster is 4/16 of an inch thick.

Plaster  4 ea.  5.2 c.f./ea.  8.0 c.f.  28.8 c.f./hundred s.y.

Plaster  2.89 hundred s.y.  28.8 c.f./hundred s.y.  83.2 c.f.

19. The ceiling of the residence shown in Figure 17.19 is plastered with 3/8 of an inch of plaster over metal lath. How many cubic feet of plaster are needed for the ceiling?

From Problem 17, the area of the ceiling is 886 square feet.

Area of ceilling 

886s.f.  99 s.y.

9s.f. s.y.

Figure 16.7 shows that 5.2 cubic feet of plaster are needed for 100 square yards for each 1/16 inch of thickness plus 8.0 cubic feet for the metal lath. The plaster is 6/16 of an inch thick.

Plaster  6 ea.  5.2 c.f./ea.  8.0 c.f.  39.2 c.f./hundred s.y.

Plaster  0.99 hundred s.y.  39.2 c.f./ hundred s.y.  38.8 c.f.

20. How many square feet of resilient flooring are needed for the residence in Figure 17.19? Add 5 percent for waste. How many gallons of adhesive are required using a coverage of 1 gallon per 120 square feet of floor?

177 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

The quantity of flooring and adhesive is calculated as follows:

Flooring  6 ft.  8 ft. 9 in.  6 ft.  8.75 ft.  52.5 s.f.

Add five percent for waste: Flooring  52.5 s.f.  1  0.05   55 s.f.

Adhesive 

52.5 s.f.  0.4gallons 120s.f. gallon

Order 1 gallon.

21. How many square yards of carpet are needed for the residence in Figure 17.19?

The carpet is 12 feet wide.

Run the carpet up and down. Two pieces will be needed for the large room: Length of carpet = 12 f t. 3 in. + 12 f t. 3 in. + 12 f t. 10 in. + 12 f t. 10 in. + 16 f t. 4 in. ee

ee

ee

ee

ee

Length of carpet = 12.25 f t. + 12.25 f t. + 12.83 f t. + 12.83 f t. + 16.33 f t. ee

ee

ee

ee

ee

Length of carpet = 66.49 f t. ee

Carpet  66.49 ft.  12ft. 

798 s.f.  88.7s.y.

9s.f. s.y.

22. How many gallons of primer and paint are needed to paint the walls of the residence shown in Figure 17.19? The walls will receive one coat of primer and two coats of paint. The exterior walls only need paint on the interior sides. The walls are 8 feet high. One gallon of primer will cover 275 square feet, and 1 gallon of paint will cover 350 square feet with a single coat. The swinging doors are 3 feet wide by 6 feet 8 inches high, the closet doors are 6 feet wide by 6 feet 8 inches high, and the windows are 4 feet high.

The quantity of paint and primer is calculated as follows: Exterior wall length = 32 f t. + 26 f t. + 36 f t. + 13 f t. 6 in. + 4 f t. + 12 f t. 6 in. ee

ee

ee

ee

ee

ee

Exterior wall length = 124 f t. ee

Interior wall length  12 ft. 10 in.  19 ft. 4 in.  16 ft. 4 in.  3.5 in.  8 ft. 9 in.

 11 ft. 9 in.  3 ea.  2 ft. 1 in./ea.  3 ea.  8 ft. 4 in./ea.

Interior wall length  12.83 ft.  19.33 ft.  16.33 ft.  0.29 ft.  8.75 ft.

11.75 ft.  3 ea.  2.08 ft./ea.  3 ea.  8.33 ft./ea.

Interior wall length = 100.51 f t. ee

Gross area of wall  124 ft.  2 sides  100.51 ft./side   8 ft.  2,600 s.f.

There is one 3-foot by 6 foot 8-inch exterior door, four 3-foot by 6 foot 8-inch interior doors, and three 6-foot by 6 foot 8-inch interior doors. Their area is calculated as follows: 178 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Area of doors  1 ea.  1 sides  3 ft.  6 ft. 8 in.  4 ea.  2 sides  3 ft.  6 ft. 8 in.

3 ea.  2 sides  6 ft.  6 ft. 8 in Area of doors  1 ea.  1 sides  3 ft.  6.67 ft.  4 ea.  2 sides  3 ft.  6.67 ft.  3 ea.  2 sides  6 ft.  6.67 ft.

Area of doors = 420 s.f.

Area of windows  5 ft.  4 ft.  3 ft.  4 ft.  4 ea.  4 ft.  4 ft.

Area of windows = 96 s.f.

Net area of wall  2,600 s.f.  420 s.f.  96 s.f.  2,084 s.f.

Primer 

2,084 s.f.  7.6gallons 275s.f. gallon

Order 8 gallons.

Paint  2coats 

2,084 s.f.  11.9gallons 350s.f. galloncoat

Order 12 gallons.

23. How many gallons of primer and paint are needed to paint the ceiling of the residence shown in Figure 17.19? The ceiling is to receive one coat of primer and two coats of paint. One gallon of primer will cover 275 square feet, and 1 gallon of paint will cover 350 square feet with a single coat.

The quantity of paint and primer is calculated as follows:

Ceiling area  36 ft.  26 ft.  4 ft.  12 ft. 6 in.

Ceiling area  36 ft.  26 ft.  4 ft.  12.5 ft.  886 s.f.

Primer 

886 s.f.  3.2gallons 275s.f. gallon

Order 4 gallons.

Paint  2coats 

886 s.f.  5.1gallons 350s.f. galloncoat

Order 6 gallons.

24. How many square feet of tile are needed for the floors of the residence in Figure 17.19?

The tile is calculated as follows:

Tile  6 ft. 3 in.  5 ft. 9 in.  6.33 ft.  5.67 ft.  35.9 s.f.

179 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Order 36 s.f.

25. How many square feet of resilient flooring are needed for the office space in Figure 17.16? Add 5 percent for waste. How many gallons of adhesive are required using a coverage of 1 gallon per 150 square feet of floor?

The quantity of flooring and adhesive is calculated as follows:

Flooring  11 ft. 9 in.  28 ft. 4 in.  11 ft. 9 in.  11 ft. 6 in. 11 ft. 9 in.  11 ft. 10 in.  11 ft. 9 in.  11 ft. 10 in.

11 ft. 9 in.  15 ft. 10 in.

Flooring  11.67 ft.  28.33 ft.  11.67 ft.  11.5 ft. 11.67 ft.  11.83 ft.  11.67 ft.  11.83 ft.

11.67 ft.  15.83 ft.

Flooring = 926 s.f.

Add five percent for waste: Flooring  926 s.f.  1  0.05   972 s.f.

Order 942 s.f.

Adhesive 

926 s.f.  6.2 gallons 150s.f. gallon

Order 7 gallons.

26. How many linear feet of 4-inch-wide rubber base are needed for the office space in Figure 17.16? The doors are 3 feet wide, and base is not required on the outside of the exterior walls. Add 5 percent for waste. How many gallons of adhesive are needed using a coverage of 1 gallon per 150 square feet of base?

The amount of base is calculated as follows: Gross length of base  2 ea.  11 ft. 9 in.  28 ft. 4 in.  2 ea.  11 ft. 9 in.  11 ft. 6 in.  2 ea.  11 ft. 9 in.  11 ft. 10 in.  2 ea.  11 ft. 9 in.  11 ft. 10 in. 2 ea.  11 ft. 9 in.  15 ft. 10 in. Gross length of base  2 ea.  11.75 ft.  28.33 ft.  2 ea.  11.75 ft.  11.50 ft.  2 ea.  11.75 ft.  11.83 ft.  2 ea.  11.75 ft.  11.83 ft. 2 ea.  11.75 ft.  15.83 ft.

Gross length of base = 276.1 f t. ee

Deduct for doors  1 ea.  1 side  3 ft./side  4 ea.  2 sides  3 ft./side 180 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Deduct for doors = 27 f t. ee

Net length of base  276.1 ft.  27 ft.  249.1 ft.

Add five percent for waste: Net length of base  249.1 ft.  1  0.05   262 ft

Order 262 f t. ee

Area of base  249.1 ft.  4 in.  249.1 ft.  0.33 ft.  82.2 s.f.

Adhesive 

82.2 s.f.  0.55gallons 150s.f. gallon

Order 1 gallon.

27. How many square yards of carpet are needed for the office space in Figure 17.16?

The carpet is 12 feet wide.

Run the carpet left and right to minimize the number of seams.

Length of carpet = 28 f t. 4 in. + 11 f t. 6 in. + 11 f t. 10 in. + 11 f t. 10 in. + 15 f t. 10 in. ee

ee

ee

ee

ee

Length of carpet = 28.33 f t. + 11.50 f t. + 11.83 f t. + 11.83 f t. + 15.83 f t. ee

ee

ee

ee

ee

Length of carpet = 79.3 f t. ee

Carpet 

79.3 ft.  12 ft.  105s.y.

9s.f. s.y.

Order 106 s.y.

28. How many gallons of primer and paint are needed to paint the walls of the office space shown in Figure 17.16? The walls are to receive one coat of primer and two coats of paint. The exterior walls only need paint on the interior sides. The walls are 8 feet high. One gallon of primer will cover 300 square feet, and 1 gallon of paint will cover 400 square feet with a single coat. The doors are 3 feet wide by 6 feet 8 inches high.

The quantity of paint and primer is calculated as follows: Length of wall  2 ea.  11 ft. 9 in.  28 ft. 4 in.  2 ea.  11 ft. 9 in.  11 ft. 6 in.  2 ea.  11 ft. 9 in.  11 ft. 10 in.  2 ea.  11 ft. 9 in.  11 ft. 10 in.  2 ea.  11 ft. 9 in.  15 ft. 10 in. Length of wall  2 ea.  11.75 ft.  28.33 ft.  2 ea.  11.75 ft.  11.50 ft.  2 ea.  11.75 ft.  11.83 ft.  2 ea.  11.75 ft.  11.83 ft. 2 ea.  11.75 ft.  15.83 ft.

Length of wall = 276.1 f t. ee

181 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Gross area of wall  276.1 ft.  8 ft.  2,209 s.f.

Deduct for doors  1 ea.  1 side  3 ft.  6 ft. 8 in.  4 ea.  2 sides  3 ft.  6 ft. 8 in.

Deduct for doors  1 ea.  1 side  3 ft.  6.67 ft.  4 ea.  2 sides  3 ft.  6.67 ft.

Deduct for doors = 180 s.f.

Net area of wall  2,209 s.f.  180 s.f.  2,029 s.f.

Primer 

2,029 s.f.  6.8gallons 300s.f. gallon

Order 7 gallons.

Paint  2coats 

2,036 s.f.  10.2gallons 400s.f. gallon coat

Order 11 gallons.

29. How many square feet of tile are needed for the walls and floor of the bathroom in Figure 17.20? The wall tile is four feet high, and the doors are three feet wide.

The area of the tile is calculated as follows:

Floor tile  2 ea.  8 ft.  24 ft. 1 in.  2 ea.  8 ft.  24.08 ft.  385 s.f.

Length of wall  4 ea.  24 ft. 1 in./ea.  4 ea.  8 ft./ea.

 2 doors  3 ft./door Length of wall  4 ea.  24.08 ft./ea.  4 ea.  8 ft./ea.  2 doors  3 ft./door Length of wall = 122.32 f t. ee

Wall tile  122.32 ft.  4 ft.  489 s.f.

30. Determine the quantity of materials required for the metal studs and drywall for the Commercial Building Project in Appendix C. The following workup sheets contain the metal stud and drywall calculations.

182 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

31. Determine the amount of flooring required for the Commercial Building Project in Appendix C. The following workup sheets contain the flooring calculations.

183 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

32. Determine the interior and exterior areas to be painted for the Commercial Building Project in Appendix C. The following workup sheet contains the painting calculations.

The solutions can also be found in Appendix C - Commercial Building Project v10 Estimate.xls x.

33. Determine the amounts of materials required for the drywall for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the finish crew from Problem 7-32, determine the cost to install the drywall.

The following data can be extracted from the model for the walls:

184 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

The length of the exterior walls is the same as the length of the foundation. The total length of the drywall is calculated as follows:

Length of drywall  241 ft.  2 sides  463 ft./side  1,167 ft.

The walls are 9 feet, 1 inch high. Order 4-foot by 9-foot sheets. Install the ceiling first, which leaves a 3/8 inch gap at the bottom of the wall, which the base will cover.

Area  1,167 ft.  8 ft.  9,336 s.f.

Number of sheets 

9,336 s.f.  292sheets 32s.f. gallon

Add five percent for waste: Number of sheets  292  1  0.05   307 sheets

The area of the ceiling equals the area of the floor slab. The following quantities can be extracted from the model.

The area of the ceiling from the model is 3,600 square feet. Alternatively, it may be calculated as follows: Area  72 ft.  50 ft.  3,600 s.f.

Order 4 ft.  9 ft. sheets: Area per sheet  4 ft.  9 ft.  36 s.f.

Sheets of drywall 

3,600 s.f.  100sheets 36s.f. sheet

Add five percent for waste: Sheets of drywall  100 sheets  1  0.05   105 sheets

The pricing is shown in the following workup sheet.

185 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

34. Determine the amount of flooring required for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the finish crew from Problem 7-32, determine the cost to install the flooring.

Run the carpet from the top to the bottom of the drawing. Order the following pieces of carpet: Room

Carpet

Notes

1

1 @ 12 f t.

2

1.33 @ 12 f t.

3

1 @ 12 f t.

4

1 @ 12 f t.

5

1.33 @ 12 f t.

6

1 @ 12 f t.

7

1 @ 12 f t.

8

1 @ 12 f t.

9

1.33 @ 12 f t.

10

1 @ 12 f t.

11

1 @ 12 f t.

12

1.33 @ 12 f t.

13

1 @ 12 f t.

14

1 @ 16 f t.

15

1 @ 16 f t.

16 / 17 w / hall

1.33 @ 27 f t.

Order 2 and use scrap in halls

Halls

6 @ 4 f t.

Order 6 @ 4 feet long

blank

ee

ee

Order 2 and use scrap for 5 and 9 blank

ee

blank

ee

ee

Use scrap from Office 2 blank

ee

blank

ee

blank

ee

ee

Use scrap from Office 2 blank

ee

blank

ee

ee

Use scrap from halls blank

ee

blank

ee

blank

ee

ee

ee

0.67 @ 24 f t. ee

Use scrape from 16 / 17 w/hall 186

Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Room

Carpet

Notes

0.33 @ 13 f t.

Order 1 and use scrap in office 12

ee

Carpet  14 ea.  12 ft./ea.  2 ea.  16 ft./ea.  2 ea.  27 ft./ea.

 6 ea.  4 ft./ea.  13 ft.

Carpet = 291 f t. ee

Order 291 lineal feet of 12-foot-wide carpet: Carpet  291 ft.  12 ft.=

3,492 s.f.  388s.y.

9s.f. s.y.

Run the carpet from the left to the right of the drawing. Order the following pieces of carpet: Room

Carpet

Notes

1

1 @ 11.75 f t.

2

1 @ 15.25 f t.

3

1 @ 11.25 f t.

4

1 @ 11.25 f t.

5

1 @ 15.25 f t.

6

1 @ 11.75 f t.

7

1 @ 11.75 f t.

8

1 @ 11.75 f t.

9

1 @ 15.25 f t.

10

1 @ 11.25 f t.

11

1 @ 11.25 f t.

12

1 @ 15.25 f t.

13

1 @ 11.75 f t.

14

1.33 @ 12 f t.

15

1.33 @ 12 f t.

16

1 @ 16 f t.

17

1 @ 16 f t.

Halls

2 @ 0.33 @ 48 f t.

blank

ee

blank

ee

blank

ee

blank

ee

blank

ee

blank

ee

blank

ee

blank

ee

blank

ee

blank

ee

blank

ee

blank

ee

blank

ee

ee

Order 2 and use scrap in room 15 blank

ee

blank

ee

blank

ee

ee

Order 1 @ 48 f t. ee

Order 3 @ 4 f t. ee

187 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Room

Carpet

Notes

2 @ 1.33 @ 4 f t.

Order 1 @ 4 f t. ee

ee

1@4

Carpet  5 ea.  11.75 ft./ea.  4 ea.  15.25 ft./ea.  4 ea.  11.25 ft./ea.

 3 ea.  12 ft./ea.  2 ea. 16 ft./ea.  48 ft.  4 ea.  4 ft./ea.

Carpet = 296.75 f t. ee

Order 297 feet of 12-foot-wide carpet: Carpet  297 ft.  12 ft.=

3,564 s.f.  396s.y.

9s.f. s.y.

Although this layout requires more carpet, it has fewer seams. Use the 388 square yards.

The following data can be extracted from the model for the rubber base:

The gross length of the rubber base is 1,019 f t. Alternatively, it may be manually calculated as follows: ee

Room

North (f t.)

East (f t.)

South (f t.)

West (f t.)

Total (f t.)

1

11.60

11.89

11.60

11.89

46.98

ee

ee

ee

ee

ee

188 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Room

North (f t.)

East (f t.)

South (f t.)

West (f t.)

Total (f t.)

2

15.05

11.89

15.05

11.89

53.88

3

11.05

11.77

11.05

11.77

45.65

4

11.05

11.77

11.05

11.77

45.65

5

15.05

11.89

15.05

11.89

53.88

6

11.60

11.89

11.60

11.89

46.98

7

11.60

11.89

11.60

11.89

46.98

8

11.60

11.89

11.60

11.89

46.98

9

15.05

11.89

15.05

11.89

53.88

10

11.05

11.77

11.05

11.77

45.65

11

11.05

11.77

11.05

11.77

45.65

12

15.05

11.89

15.05

11.89

53.88

13

11.60

11.89

11.60

11.89

46.98

14

11.60

15.94

11.60

15.94

55.08

15

11.60

15.94

11.60

15.94

55.08

16

15.60

10.83

0.00

10.83

37.27

17

0.00

12.28

15.60

12.28

40.17

Hall 1

12.40

0.00

16.00

0.00

28.40

Hall 2

0.00

16.73

0.00

23.94

40.67

Hall 3

47.60

12.28

40.40

12.28

112.56

Hall 4

0.00

23.94

0.00

16.73

40.67

Hall 5

12.40

0.00

16.00

0.00

28.40

blank

blank

blank

blank

ee

ee

ee

ee

Total

ee

1071.27

Where: 

Hall 1 is the southwest hallway (between Rooms 1 and 14).



Hall 2 is the west hallway.



Hall 3 is the north hallway, including the hallway to the north exit.



Hall 4 is the east hallway.



Hall 5 is the southeast hallway (between Rooms 13 and 15).

The doors must be subtracted from this number. The following data can be extracted from the model: 189 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

The exterior doors and the interior doors for the men’s room, janitor’s closet, and women’s room will only have rubber base on one side.

Doors  1 ea.  3 ft./ea.  17 ea.  2 sides/ea.  3 ft./side  3 ea.  3 ft./ea.  1 ea.  6 ft./ea.

Doors = 120 f t. ee

Base = 1,072 ft.  120 ft. = 952 ft.

Clearly, there is a problem with how the model calculates the distance. The problem is that the model includes base between the reception area (Room 16) and the hallway and between the waiting area (Room 17) and the hallway. If there were walls between these rooms, there would be two walls about 16 feet long with base on two sides. Deduct the base for these walls as follows: Base  1,019 ft.  4 ea.  16 ft./ea.  955 ft.

This quantity is similar to the manually calculated quantity. We will use the quantity manually calculated at 952 feet.

The pricing is shown in the following workup sheet.

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

35. Determine the interior areas to be painted for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the finish crew from Problem 7-32, determine the cost to paint the office space.

The following data can be extracted from the model:

190 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

All the 24-inch by 48-inch windows are in the exterior walls. Only two of the 36-inch by 72-inch windows are in the exterior walls. There is one 3-foot by 7-foot door and one 6-foot by 7-foot door in the exterior walls. The remaining twenty 3-foot by 7-foot doors are in the interior walls.

Length of walls  241 ft.  2 sides  463 ft./side  1,167 ft.

The walls are 9 f t. 1 in. high, but the acoustical ceiling is at 8 f t. Paint to 8.5 f t. ee

ee

ee

Gross area of wall  1,167 ft.  8.5 ft.  9,920 s.f.

Doors  1 ea.  3 ft.  7 ft.  20 ea.  2 sides  3 ft.  7 ft.  1 ea.  6 ft.  7 ft.

Doors = 903 s.f.

Windows  34 ea.  2 ft.  4 ft.  12 ea.  2 sides  3 ft.  6 ft.  2 ea.  3 ft.  6 ft.

Windows = 740 s.f.

Deduct the area of ceramic tile. From Problem 34, the perimeter for the ceramic tile is 60 feet 3 inches, or 60.25 feet. Alternately, this distance may be determined as follows: Room

North (f t.)

East (f t.)

South (f t.)

Men’s

5.60

5.10

5.60

5.10

21.40

Janitor’s

3.60

5.10

3.60

5.10

17.40

Women’s

5.60

5.10

5.60

5.10

21.40

Blank

Blank

Blank

Blank

ee

ee

ee

West (f t.)

Perimeter (f t.)

ee

Total

ee

60.20

Deduct the doors:

Length  60.25 ft.  3 ea.  3 ft./ea.  51.25 ft.

Area of tile  51.25 ft.  4 ft.  205 s.f.

Net area of wall  9,920 s.f.  903 s.f.  740 s.f.  205 s.f.  8,072 s.f.

The ceiling does not need to be painted because it is hidden by a dropped acoustical ceiling, except in the men’s room, janitor’s closet, and women’s room. From Problem 34, the area of these ceilings is 76 square feet. Alternately, this area may be determined as follows: Room

Width Length (f t.) (f t.) ee

ee

Area (s.f.)

191 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

Room

Width Length (f t.) (f t.)

Area (s.f.)

Men’s

5.60

5.10

28.6

Janitor’s

3.60

5.10

18.4

Women’s

5.60

5.10

28.6

Blank

Blank

ee

ee

Total

75.5

Twenty 3-foot by 7-foot doors need to be painted. Each door has 34 feet of trim, for a total of 680 feet.

The pricing is shown in the following workup sheet.

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

36. Determine the ceramic tile needed for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the finish crew from Problem 7-32, determine the cost to install the ceramic tile.

From Problem 34, the floor area is 76 square feet, and the perimeter of the rooms with tile is 60 feet 3 inches. Deduct the doors from the wall:

Length  60 ft. 3 in.  3 ea.  3 ft./ea.  51 ft. 3 in.  51.25 ft.

Allow for a 4-inch base and a 2-inch top trim piece.

Height  48 in.  2 in.  4 in.  42 in.  3.5 ft.

Area of tile  52 ft.  3.5 ft.  182 s.f.

Base = 52 f t. ee

Trim = 52 f t. ee

The pricing is shown in the following workup sheet.

192 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

37. Determine the acoustical ceiling needed for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the finish crew from Problem 7-32, determine the cost to install the acoustical ceiling.

The area of the ceiling from the model is 3,600 square feet.

Alternatively, the individual components could be taken off from the reflected ceiling plan. The components will vary based on the assumptions about the main and cross tees’ placement.

The quantities for the ceiling grid are as follows: Light 4 f t. 2 f t. Wall s& Cros Cros Angl Openin HVA Tile s s e gs C s 15 18 24 36 6 30 20 24 28 48 6 42 15 18 24 36 6 30 15 18 24 36 6 30 20 24 28 48 6 42 15 18 24 36 6 30 15 18 24 36 6 30 15 18 24 36 6 30 20 24 28 48 6 42 15 18 24 36 6 30 15 18 24 36 6 30 20 24 28 48 6 42 15 18 24 36 6 30 20 24 28 48 7 41 20 24 28 48 7 41 114 150 17 205 369 456 401 725 ee

Roo Lengt Lengt Widt Widt Mai m h ft h in h ft h in n 2 1 11 8 11 11 2 15 1 11 11 3 3 11 1 11 10 2 4 11 1 11 10 2 5 15 1 11 11 3 6 11 8 11 11 2 7 11 8 11 11 2 8 11 8 11 11 2 9 15 1 11 11 3 10 11 1 11 10 2 11 11 1 11 10 2 12 15 1 11 11 3 13 11 8 11 11 2 14 12 0 16 0 3 15 12 0 16 0 3 Halls 15 51 ee

blank

ee

blank

blank

blank

blank

blank

blank

blank

blank

ee

blank

blank

blank

blank

The layout for the grid in the hallways is as shown in the following figure, where the dark blue lines are the main runners, the squares are 4-foot tees, and the triangles are 2-foot tees:

193 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

The tiles for the hallways are counted as shown in the following figure. Tiles that are not marked are cut from straps.

The pricing is shown in the following workup sheet.

194 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 17: Finishes

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

195 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 18: Electrical

Chapter 18—Electrical Learning Outcomes This chapter looks at determining the materials and labor associated with Division 26 Electrical and Division 27 Communications of the MasterFormat. After reading this chapter, the reader should be able to: Learning Outcome 18.1: Describe the factors to consider when bidding on electrical work.

Learning Outcome 18.2: Describe the difference between single and separate contracts.

Learning Outcome 18.3: Determine electrical costs using assemblies.

Learning Outcome 18.4: Determine the electrical materials required using a detailed estimate.

Learning Outcome 18.5: Determine materials required for electrical conduit and wiring.

Activities 1. Review the plans and project manual for a project. Identify any items that need coordination between the electrical, plumbing, and HVAC contractors.

Instructional Resources PowerPoint Slides: Chapter 18.ppt Solutions to the Review Questions 1. What is the difference between single contracts and separate contracts?

Under single contracts, the general contractor is responsible for the electrical work.

The general contractor will include the selected electrical contractor’s price plus overhead and profit in the project’s bid.

With separate contracts, the electrical contractor will bid the electrical work to the owner, and the owner will contract directly with the selected electrical contractor.

2. Why are the electrical portions of the work put out under separate contracts?

Owners may use separate contracts for the electrical because they want to select the electrical contractor or to avoid the general contractor marking up the electrical contractor’s bid.

3. Why should the estimator review the electrical portions of the job, whether it involves single or separate contracts?

The estimator should review the electrical portions of the job, whether the project uses single or separate contracts, because, under either arrangement, there are coordination requirements that they must include in their bid.

196 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 18: Electrical

4. Why are cooperation and coordination necessary among the various contractors on a project?

Coordination of work among the electrical, HVAC, and plumbing contractors is also crucial since the electrical, HVAC, and plumbing contractors may all have work to do on a particular piece of equipment. There are many instances where multiple trades are required to connect to one item, and it is not always clear where one trade’s work stops and another begins. The estimator must ensure that the electrical, HVA C, and plumbing contracts do not leave any gaps between the trades and do not overlap.

5. How do the various types of construction affect the cost of the electrical work?

Different types of construction affect the cost and difficulty of the installation of the rough and finish work. While steel joists offer excellent flexibility for running conduits and conductors, precast concrete will require considerably more planning.

6. Determine the wiring and conduit needed for the drill presses in Figure 18.6. The conduit is to be run along the ceiling. The ceiling height is 20 feet.

Place a single J-box between the two drill presses. Allow five feet of flexible conduit for each drill press. Mount the J-box at 4 feet.

The length of the conduit is calculated as follows: Length = 15 f t. + 28 f t. + 16 f t. = 59 f t. ee

ee

ee

ee

Order the following conduit: 

6 ea. 3/4 in.  10 ft. EMT conduits



5 ea. 3/4 in. couplings  joint EMT conduit 



2 ea. 3/4 in. connectors  connect conduit to panel and J-box 



1 ea. J-box



1 ea. J-box cover



10 ft. of 1/2 in. flexible conduit



4 ea. 1/2 in. flexible connectors  connect to J-box and drill presses 



5 ea. hangers

Because the drill presses are single-phase, the neutral wires cannot be combined, but the grounds can. Five wires (2 hots, 2 neutrals, and 1 ground) are needed. The length of the wire, allowing for one foot to connect the ends, is calculated as follows: Length per wire = 3 f t. + 59 f t. + 5 f t. + 1 f t. = 68 f t. ee

ee

ee

ee

ee

Length  5 ea.  68 ft./ea.  340 ft.

7. What is the estimated cost of the electrical system for the residence shown in Figure 18.8 using the assembly costs in Figure 18.9? The light fixture cost includes a $30 197 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 18: Electrical

fixture allowance. The associated wiring is included in the cost of each assembly.

What is the fixture allowance for the residence?

The cost is calculated as follows: Switches  6 ea.  $50/ea.  $300 Outlets  16 ea.  $50/ea.  $800 GFCI outlets  3 ea.  $60/ea.  $180

Light fixtures  7 ea.  $80/ea.  $560 Main service  1 ea.  $1,000/ea.  $1,000

Total cost = $300 + $800 + $180 + $560 + $1,000 = $2,840

Fixture allowance  7ea.  $30/ ea.  $210 8. Write a scope of work for the electrical of the residence from Problem 7.

The scope of work may be written to cover the entire electrical package or by specifically calling out the needed items. The following is a sample scope of work that covers the entire electrical package: Furnish all labor, materials, equipment, tools, supervision, and permits to complete the electrical work for the residence shown in Figure 18.7 in accordance with the contract documents.

The following is a sample scope of work that specifically calls out the needed items: Furnish all labor, materials (including wiring), equipment, tools, supervision, and permits to install. 

6 ea. single-pole switches



16 ea. duplex outlets



3 ea. GFI outlets



7 ea. light fixtures



1 ea. main service

in the residence shown in Figure 18.7. The work includes a $210 allowance for light fixtures. All work is to be done in accordance with the contract documents.

9. Determine the electrical materials needed for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the electrical crew from Problem 7-32, determine the cost for the electrical.

The following can be extracted from the model:

198 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 18: Electrical

The following electrical components are needed: 

19 ea. single-pole switches



10 ea. three-way switches



15 ea. phone jacks



2 ea. phone jacks, floor mount



79 ea. duplex outlets



2 ea. duplex outlets, floor mount



3 ea. GFI duplex outlets



2 ea. GFI duplex outlets with weatherproof covers for use outside



83 ea. 2-foot by 2-foot fluorescent light fixtures



6 ea. can light fixtures



3 ea. surface-mount light fixtures

There are a total of 178 outlets and light fixtures. Allowing eight lights or outlets per circuit, 23 each 20-amp circuits will be needed. Additionally, circuits for the heat and cooling system (allow a 40-amp circuit) and the water heater (allow a 30-amp circuit) are needed. Allow for one 30-space 225-amp panel, one 225-amp main breaker, and one meter base.

The following approximate wiring quantities are needed. The quantities were measured in OnScreen Takeoff and will vary based upon which outlets are placed on which circuit: 

322 f t. of 3-12 Romex (2 ea. 250 f t. rolls)



2,655 f t. of 2-12 Romex (3 ea. 1,000 f t. rolls)



756 f t. of phone wire (2 ea. 500 f t. rolls)

ee

ee

ee

ee

ee

ee

199 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 18: Electrical



64 f t. of wire and 32 f t. #8 wire (connect panels)



50 f t. of #4 wire (water heater)



100 f t. of #6 wire (heating)

ee

ee

ee

ee

The pricing is shown in the following workup sheet.

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

200 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 19: Plumbing

Chapter 19—Plumbing Learning Outcomes This chapter looks at determining the materials and labor associated with Division 21 Fire Suppression and Division 22 Plumbing of the MasterFormat. After reading this chapter, the reader should be able to: Learning Outcome 19.1: Describe the factors to consider when bidding on fire suppression and plumbing.

Learning Outcome 19.2: Describe the difference between single and separate contracts.

Learning Outcome 19.3: Determine plumbing costs using assemblies.

Learning Outcome 19.4: Determine the plumbing materials required using a detailed estimate.

Learning Outcome 19.5: Determine the materials required for plumbing piping.

Learning Outcome 19.6: Determine the materials required for a simple fire sprinkler system.

Activities 1. Review the plans and project manual for a project. Identify any items that need coordination between the electrical, plumbing, and HVAC contractors.

Instructional Resources PowerPoint Slides: Chapter 19.ppt Solutions to the Review Questions 1. How do the building codes affect the installation of the plumbing?

The plumbing work must be installed in accordance with all applicable building codes, which contain information on water supply systems, sewage drainage systems, storm-water drainage, pipe sizing, vents and venting, and various design issues.

2. What type of work is most generally included under plumbing?

Plumbing usually includes water supply and sewage and storm-water drainage.

3. Why would an estimator contact a specialty contractor if it is suspected that their bid is too low?

If the estimator suspects the bid is too low and the plumbing contractor cannot make a reasonable profit, they should call it to the subcontractor’s attention, as the performance will probably suffer if the bid is too low.

201 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 19: Plumbing

4. Why should the estimator review the plumbing portions of the project, whether it involves single or separate contracts?

The estimator should review the plumbing portions of the project because there are coordination requirements that they must include in their bid.

5. How do the various types of construction affect the cost of the plumbing work?

Estimators must consider the type of construction (such as steel, concrete, and wood frame) as it affects the cost of installation of the plumbing work by making it easy or more challenging to install the piping. While steel joists offer excellent flexibility for running pipping, precast concrete will require considerably more planning.

6. What plumbing materials are needed for the fire sprinkler systems shown in Figure 19.8?

The following materials are needed: 

20 ea. sprinkler heads



20 ea. escutcheons

 

20 ea. 1-in.  24-in. flexible pipes 4 ea. 1-in. 90-degree elbows



4 ea. 1-in.  9-ft. pipes



4 ea. 1 1/4-in.  1-in.  1-in. tees



4 ea. 1 1/4-in.  9-ft. pipes



4 ea. 1 1/2-in.  1 1/4-in.  1-in. tees



4 ea. 1 1/2-in.  1 1/2-in.  1-in. tees



8 ea. 1 1/2-in.  9-ft. pipes



4 ea. 2-in.  1 1/2-in.  1-in. tees



4 ea. 2-in.  4-ft. pipes



1 ea. 2-in. 90-degree elbow



1 ea. 2-in.  10-ft. pipe



1 ea. 2 1/2-in.  2-in.  2-in. tee



1 ea. 2 1/2-in.  2 1/2-in.  2-in. tee



2 ea. 2-in.  10-ft. pipe



1 ea. 3-in.  2-in.  2 1/2-in. tee



1 ea. 3-in.  6-ft. pipe 202 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 19: Plumbing

7. What is the estimated cost to replace the plumbing fixtures in the bathroom shown in Figure 19.9 using the unit prices in Figure 19.10?

The cost is calculated as follows: Lavatory  sink   6 ea.  $695/ea.  $4,170 Water closet  toilet   6 ea.  $610/ea.  $3,660 Urinal  2 ea.  $820/ea.  $1,640 Total cost  $4,170  $3,660  $1,640  $9,470

8. Write a scope of work for the plumbing of the building shown in Figure 19.9.

The scope of work may be written to cover the entire plumbing package or by specifically calling out the needed items. The following is a sample scope of work that covers the entire plumbing package: Furnish all labor, materials (including piping), equipment, tools, supervision, and permits to replace the plumbing fixtures in the bathroom shown in Figure 19.8 in accordance with the contract documents.

The following is a sample scope of work that specifically calls out the needed items: Furnish all labor, materials, equipment, tools, supervision, and permits to replace 

6 ea. lavatories (sinks)



6 ea. water closets (toilets)



2 ea. urinals

in the bathroom, shown in Figure 19.8. All work is to be done in accordance with the contract documents.

9. Determine the plumbing needed for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the plumbing crew from Problem 7-32, determine the cost for the plumbing.

The following can be extracted from the model:

The following fixtures are needed: 

2 ea. water closets



2 ea. toilet seats 203 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 19: Plumbing



2 ea. lavatories (sinks), wall mounted



2 ea. faucets (with drain)



1 ea. mop/floor sink



1 ea. faucet (mop/floor sink)



1 ea. electric water heater



1 ea. expansion tank

The following materials are needed for the waste lines. Stop the waste ten feet outside the building: 

2 ea. 1 1/4-in. P-traps



2 ea. 1 1/4-in. male adapters



2 ea. 1 1/4-in.  6-in. pipes



2 ea. 1 1/4-in. 90-degree elbows



2 ea. 1 1/4-in.  3-ft. pipes



2 ea. 1 1/4-in. sanitary tees



2 ea. 1 1/4-in.  19-ft. pipes  vents 



2 ea. 1 1/4-in. roof flashings



2 ea. 1 1/4-in.  3-ft. pipes



2 ea. 4-in.  4-in.  1 1/4-in. sanitary tees



2 ea. 4-in. toilet flanges



2 ea. wax gaskets



1 ea. 4-in. 90-degree elbow



1 ea. 4-in. sanitary tee



1 ea. sink drain



1 ea. 1 1/4-in. P-trap



1 ea. 1 1/4-in.  1-ft. pipe



1 ea. 1 1/4-in. 45-degree elbow



1 ea. 1 1/4-in.  6-in. pipe

 

1 ea. 4-in.  4-in.  1 1/4-in. sanitary tee

1 ea. 4-in. male plug (cleanout)



1 ea. 4-in. female adapter (cleanout) 204 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 19: Plumbing

1 ea. 4-in.  3-ft. pipe

 

1 ea. 4-in. sanitary tee



50 -f t. 4-in. pipe ee

The following is needed for the water supply. Stop the water line ten feet outside the building: 

45 ft. of 1/2-in. copper soft  flexible  pipe



1 ea. 1/2-in. valve



1 ea. 1/2-in. pressure reducing valve



1 ea. 1/2-in. valve  water heater 



2 ea. 1/2-in.  24-in. braided stainless steel supply lines  water heater 



2 ea. seismic straps (water heater)



4 ea. 1/2-in.  3/8-in. shut off valves  lavatories 



4 ea. 3/8-in.  18-in. braided stainless steel supply lines  lavatories 



2 ea. 1/2-in.  3/8-in. shut off valves  water closets 



2 ea. 3/8-in.  12-in. braided stainless steel supply lines  water closets 



1 ea. 1/2-in. coupling  connect flexible pipe to ridged pipe 



4 ea. 1/2-in. male adapters  2 for water heater, 2 for floor-sink faucet 



15 ea. 1/2-in. 90-degree elbows



7 ea. 1/2-in. Tees



90 ft. of 1/2-in. copper hard  ridged pipe

Other solutions are possible depending on how you run the pipe.

The pricing is shown in the following workup sheets.

205 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 19: Plumbing

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

206 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 20: Heating, Ventilating, and Air Conditioning

Chapter 20—Heating, Ventilating, and Air Conditioning Learning Outcomes This chapter looks at determining the materials and labor associated with Division 23 Heating, Ventilating, and Air Conditioning of the MasterFormat. After reading this chapter, the reader should be able to: Learning Outcome 20.1: Describe the factors to consider when bidding on heating, ventilating, and air conditioning.

Learning Outcome 20.2: Describe the difference between single and separate contracts.

Learning Outcome 20.3: Determine heating, ventilating, and air conditioning costs for a residence using assemblies.

Learning Outcome 20.4: Determine the materials needed for a residential heating, ventilating, and air conditioning system using a detailed estimate.

Activities 1. Review the plans and project manual for a project. Identify any items that need coordination between the electrical, plumbing, and HVAC contractors.

Instructional Resources PowerPoint Slides: Chapter 20.ppt Solutions to the Review Questions 1. Would electric heat most likely be placed under electrical, plumbing, or HVAC? The HVAC contractor typically installs a forced-air electrical furnace and its associated ductwork, but it still requires a licensed electrician to provide power to the furnace.

2. How do the various types of construction affect the cost of heating work?

Estimators must consider the type of construction (such as steel, concrete, and wood frame) as it affects the installation cost of the HVAC system work by making it easy or more challenging to install the ductwork. While steel joists offer excellent flexibility for running ductwork, precast concrete will require considerably more planning.

3. Why are specialty contractors hired under single contracts to perform the HVAC work?

HVAC is another area often subcontracted to a contractor specializing in this type of work because installation of the HVAC system requires expertise that the general contractor’s crews most likely do not have. In addition, some contract documents and building codes require licensed personnel to perform the work.

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Chapter 20: Heating, Ventilating, and Air Conditioning

4. What is the estimated cost for the heating system of the residence shown in Figure 20.5? An HVAC contractor has given you a cost of $1,900 for the heating unit and a cost of $195 per register, which includes the ductwork from the heating unit to the register. A register is required in each room. Rooms over 150 square feet require one register per 150 square feet, or a fraction thereof.

The cost is calculated as follows: Area Bathroom  = 8 ft. 9 in.  5 ft. 9 in.

Area Bathroom   8.75 ft.  5.75 ft.  50 s.f.

Use one register.

Area Bedroom 1  11 ft. 6 in.  12 ft. 10 in.

Area Bedroom 1  11.50 ft.  12.83 ft.  148 s.f.

Use one register.

Area Bedroom 2   11 ft. 6 in.  12 ft. 10 in.

Area Bedroom 2   11.50 ft.  12.83 ft.  148 s.f.

Use one register.

Area Bedroom 3   11 ft. 9 in.  16 ft. 4 in.

Area Bedroom 3   11.75 ft.  16.33 ft.  192 s.f.

Registers =

192 s.f. = 1.3 ea.

150 s.f.

Use two registers.

Area  Great Room   19 ft. 4 in.  12 ft. 3 in.  6 ft.  8 ft. 9 in.

Area  Great Room   19.33 ft.  12.25 ft.  6.00 ft.  8.33 ft.  287 s.f.

Regisers 

287 s.f.  1.9 ea.

150 s.f.

Use two registers.

Cost  $1,900  7 resisters  $195/register  $3,265 5. Write a scope of work for the HVAC of the residence shown in Figure 20.5.

The scope of work may be written to cover the entire HVAC package or by specifically calling out the needed items. The following is a sample scope of work that covers the entire HVAC package:

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Chapter 20: Heating, Ventilating, and Air Conditioning

Furnish all labor, materials (including ductwork), equipment, tools, supervision, and permits to complete the heating system for the residence shown in Figure 20-5 in accordance with the contract documents.

The following is a sample scope of work that specifically calls out the needed items: Furnish all labor, materials, equipment, tools, supervision, and permits to install 

1 ea. heating unit



7 ea. resisters with ductwork

in the residence shown in Figure 20-5. The heating unit should be adequately sized for the residence. Provide two registers in the large bedroom and great room and one register in all the remaining rooms. All work is to be done in accordance with the contract documents.

6. Determine the HVAC materials needed for the Real Estate Office in Appendix E. Using the labor rates from Problem 7-31 and the crew rate for the HVAC crew from Problem 7-32, determine the cost for the HVAC. The following can be extracted from the model:

The following equipment is needed: 

1 ea. heating/cooling unit



1 ea. thermostat with low-voltage wiring



1 ea. vent pipe with roof flashing



1 ea. electrical connection



1 ea. fuel connection

The following are needed for the supply ductwork. No ductwork has been deducted for the fittings; therefore, the lengths will vary depending on how the students account for fittings: 

27 ea. 24 in. × 24 in. diffusers



31 ea. 6-in. elbows



54 f t. of 6-in. flexible duct



195 f t. of 6-in. duct



4 ea. 8-in.  8-in.  6-in. tees



43 f t. of 8-in. duct

ee

ee

ee

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Chapter 20: Heating, Ventilating, and Air Conditioning



4 ea. 10-in.  8-in.  6-in. tees



24 f t. of 10-in. duct



4 ea. 12-in.  10-in.  6-in. tees



10 ea. 12-in.  12-in.  6-in. tees



59 f t. of 12-in. duct



2 ea. 12-in.  12-in.  16-in. tees



22 f t. of 16-in. duct



1 ea. 16-in.  20-in. duct – the length will vary depending on how it is

ee

ee

ee

connected to the furnace.

The following are needed for the return ductwork. 

20 ea. 24-in.  24-in. registers



20 ea. 6-in. elbows



40 f t. of 6-in. flexible duct



120 f t. of 6-in. duct



4 ea. 6-in.  12-in. elbows



16 ea. 12-in.  12-in.  6-in. tees



108 f t. of 12-in. duct



2 ea. 12-in.  12-in.  16-in. tees



26 f t. of 16-in. duct



1 ea. 16-in.  20-in. duct – the length will vary depending on how it is

ee

ee

ee

ee

connected to the furnace.

The pricing is shown in the following workup sheet.

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Chapter 20: Heating, Ventilating, and Air Conditioning

This pricing is also found in Appendix E - Real Estate Office v10 Estimate.xlsm on the instructor’s website and will vary based on the assumptions made by the estimator.

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Chapter 21: Profit

Chapter 21—Profit Learning Outcomes This chapter looks at adding profit to the bid. After reading this chapter, the reader should be able to: Learning Outcome 21.1: Describe the factors to consider when adding profit to the bid.

Learning Outcome 21.2: Identify the sources of risk on a project.

Learning Outcome 21.3: Explain how to assess the competition when selecting the profit markup.

Activities 1. Divide the students into groups and provide each group with the plans and project manual for a construction project. Have each group assess the risks associated with the project and identify what adjustments they would make to the profit to account for these risks. Have each group present their responses to the class.

Instructional Resources PowerPoint Slides: Chapter 21.ppt Solutions to the Review Questions 1. Where should home-office overhead be included?

The allocated portion of home-office overhead should be included in the estimated cost before adding profit.

2. How should project risk affect the profit on a job?

The riskier the job, the higher the profit markup. There is no reason to take on highrisk construction projects for a low-profit markup when they can take on low-risk construction projects for the same profit markup.

3. Why should you track your competitors’ bids?

Suppose the estimator finds that only a few bidders are left and that the remaining competitors’ profit markups are usually higher than theirs. This may be an opportunity to bid and win the job at a higher profit markup. However, you may need to bid your minimum profit markup if several competitors are bidding on the job or the competition has often beaten you.

4. What is the problem with needing or wanting a job?

There is sometimes a tendency to need or want a job so badly that risks are ignored, and too little profit to cover the risks is included in the bid.

5. What risks should an estimator consider when setting the profit margin for a job?

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Chapter 21: Profit

Estimators should consider the following risks when setting the profit margin for a job: 1. Owner-related risks 2. Designer-related risks 3. Construction-document-related risks 4. Labor-related risks 5. Production-rate-related risks 6. Material-related risks 7. Regulatory-related risks 8. Weather-related risks 9. Specialty-contractor-related 6. Your company submitted a bid of $1,263,000 for the West Side Retirement Home on May 15, 20XX. Your estimated total construction costs were $1,205,068. The bid results were as follows: ABC Contractor $1,269,999 XYZ Contractor $1,267,425 What data would you record in your database for each of these contractors?

The estimated profit markups are as follows:

ABC 

$1,269,999  1  0.0539 or 5.39% $1,205,068

XYZ 

$1,267,425  1  0.0517 or 5.17% $1,205,068 Estimated Total Low

Project

Construction

Markup

Bid Date

Bid

Bid

Costs

(%)

blank

blank

blank

blank

blank

Mt. Ogden Day Care 5/10/20XX

Y

889,999

814,331

9.29

West Side Retirement Home

5/15/20XX

N

1,269,999

1,205,068

5.39

blank

blank

blank

blank

blank

Mt. Ogden Day Care 5/10/20XX

N

897,256

814,331

10.18

West Side Retirement Home

N

1,267,425

1,205,068

5.17

ABC Contractor

XYZ Contractor

5/15/20XX

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Chapter 21: Profit

7. Your company is bidding on a religious building. You estimated that its construction costs are $1,865,400, which includes project overhead. Your company marks up all construction costs 5 percent to cover main-office overhead. Your boss wants to make a 3 percent profit on the construction costs (excluding main-office overhead).

What is your company’s bid for the project?

The bid is calculated as follows: Main-office overhead = $1,865,400  0.05 = $93,270 Profit = $1,865,400  0.03 = $55,962

Bid = $1,865,400 + $93,270 + $55,962 = $2,014,632 8. Your company is bidding on an office building. The estimated construction costs are as follows: Labor = $985,466 Materials = $1,091,256 Equipment = $100,754 Subcontractor f t. s bids = $6,254,800 ee

Your company marks up labor and materials at 15 percent, equipment at 20 percent, and subcontractor’s bids at 4 percent to cover main-office overhead and profit. What is your company’s bid for the project?

The bid is calculated as follows: Labor  $985,466  1  0.15   $1,133,286 Materials  $1,091,256  1  0.15   $1,254,944 Equipment  $100,754  1  0.20   $120,905 Subcontractor  $6,254,800  1  0.04   $6,504,992

Total = $1,133,286 + $1,254,944 + $120,905 + $6,504,992 = $9,014,127

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Chapter 22: Other Estimating Methods

Chapter 22—Other Estimating Methods Learning Outcomes This chapter examines three estimating methods often used during the preconstruction stage of a construction project and value engineering. After reading this chapter, the reader should be able to: Learning Outcome 22.1: Describe different types of estimating methods.

Learning Outcome 22.2: Prepare an estimate using the project comparison method.

Learning Outcome 22.3: Prepare an estimate using the square-foot method.

Learning Outcome 22.4: Prepare an estimate using the assembly method.

Learning Outcome 22.5: Describe how value engineering is used.

Activities 1. Obtain a copy of Means Square Foot Cost Data. Using the square foot cost portion of the book, prepare an estimate for a simple building, such as a warehouse, the core and shell of an office building, or a residence. Then, using the assembly costs in the back half of the book, prepare a cost estimate for the same building. Discuss the difference between the two costs estimates and the following questions: 

What assumptions do you need to make for each cost estimate?



Which cost estimate is most accurate? Why?



Where would you use each of these estimating methods?

Instructional Resources PowerPoint Slides: Chapter 22.ppt Solutions to the Review Questions 1. How would you determine the costs for a project using the project comparison method?

The estimator starts with the costs for a similar project and adjusts these costs for any differences between the comparison project and the proposed project, including project location, inflation, and changes in market conditions.

2. How would you determine the costs for a project using the square-foot method and in-house cost data?

The following steps are used to determine the cost: 1. Determine the historical cost per square foot.

2. Determine the cost of the proposed project by multiplying its square footage by the historical cost per square foot.

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Chapter 22: Other Estimating Methods

3. Adjust for inflation and any other differences.

3. What are some of the ratios that you would use to prepare a line-by-line square-foot estimate?

You would include the following ratios: 

Area ratio



Perimeter ratio



Fixed cost (a factor of 1)

4. Determine the following using Figure 22.2: a. Base cost per square foot for a 40,000-square-foot, tilt-up concrete warehouse $76.55/s.f.

b. Base cost per square foot for a 57,000-square-foot, tilt-up concrete warehouse  57,000 s.f.  50,000 s.f.   $73.00 $75.00  $75.00 Cost      s.f.  s.f.  60,000 s.f.  50,000 s.f.   s.f.

Cost  $73.60 s.f. c. Base perimeter for a 50,000-square-foot warehouse 966 f t. ee

d. Perimeter adjustment for a 25,000-square-foot warehouse $2.60 per s.f. per 100 feet of perimeter e. Story height adjustment for a 15,000-square-foot warehouse $0.60 per s.f. per foot of height f. Added cost for a 70-foot-high aluminum flagpole $9,725 5. What components would you include in an exterior wall assembly for a residence?

The most common components would include the following: 

Bottom (sill) plate



Studs



Top plate(s)



Headers



Wood sheathing



Insulation



Drywall



Paint 216 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 22: Other Estimating Methods



Exterior finishes (siding, brick, stucco, etc.)

6. Using the project comparison method, prepare an estimate for a two-story home in your local area. The home is to have three bedrooms, two bathrooms, a living room, kitchen and dining facilities, and a two-car garage. The square footage should be between 1,800 and 2,200 square feet. Exclude the lot, all site work, and all permits from your estimate. To prepare this estimate, you will need to find two or three projects to use as a comparison. These may be obtained via the Internet or by talking to local builders. Be sure to adjust the prices as necessary to make the projects comparable, including adjustments for age and quality of materials. Be prepared to share the projects you used as a comparison and your calculations with the class if your instructor chooses to do so.

The solution will vary based on the projects selected by the students. Emphasis should be placed on making the projects comparable.

7. Last year, your construction company built a 2,400-square-foot home with a two-car garage for $255,658. Another client wants a similar home built, except the client wants a three-car garage. It is estimated that the garage will cost an additional $9,500, and those costs have risen 5 percent during the last year. Using this information, prepare a preliminary estimate for the house.

The cost is as follows: Cost  $255,658  1  0.05   $9,500  $277,941

Use $278,000 8. Your company has been asked to prepare a preliminary cost estimate for a 10,000square-foot automotive repair garage with four hydraulic lifts. From past projects, you have determined that the average cost per square foot to construct an automotive repair garage is $107.60, excluding the cost of the hydraulic lifts. The lifts cost $16,000 each. Ignoring inflation, prepare an estimate for the repair garage.

The cost is as follows: Cost  10,000 s.f.  $107.60/s.f.  4 lifts  $16,000/lift

Cost = $1,140,000 9. Your company has been asked to prepare a preliminary cost estimate for a threestory, 22,000-square-foot office building. From projects you have constructed during the last year, you have determined that the average cost per square foot to build an office building is $158.59. Prepare an estimate for the office building using an inflation rate of 5 percent.

The cost is as follows: Cost  22,000 s.f.  $158.59/s.f.  1  0.05   $3,663,429

Use $3,660,000 10. List the items you need to include in a concrete footing and foundation wall assembly.

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Chapter 22: Other Estimating Methods

The list should include the following: 

Footing concrete



Continuous footing rebar



Crosswise footing rebar



Dowels (connect footing to foundation wall)



Foundation wall concrete



Horizontal foundation wall rebar



Vertical foundation wall rebar



Anchor bolts

Other items that may be included are: 1. Waterstop 2. Footing forms 3. Foundation forms 4. Keyway 5. Waterstop 11. Using Figures 22.2 and 22.3, determine the cost for a 250-foot by 200-foot warehouse. The exterior walls will be constructed of concrete blocks and are used as bearing walls. The warehouse is 20 feet high and requires ten 7-foot by 8-foot dock levelers.

The cost is determined as follows: Area  250 ft.  200 ft.  50,000 s.f.

Perimeter = 250 f t. + 200 f t. + 250 f t. + 200 f t. = 900 f t. ee

ee

ee

ee

ee

Figure 22-2 shows that the cost per square foot for a concrete block warehouse with bearing walls and an area of 50,000 square feet is $71.05 per square foot. It also shows that the base perimeter is 966 feet, and the adjustment for perimeter length is $1.30 per square foot per 100 feet.

Deduct for perimeter  

900 ft.  966 ft.  $1.30 s.f.100 ft.

100 ft.

Deduct for perimeter  $0.86/s.f.

Figure 22-3 shows that the base story height is 24 feet, and the adjustment for story height is $0.40 per square foot per foot of height.

Deduct for story height   20 ft.  24 ft.  $0.40 per s.f. per ft.

Deduct for story height  $1.60/s.f.

Base cost  $71.05/s.f.  $0.86/s.f.  $1.60/s.f.

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Chapter 22: Other Estimating Methods

Base cost  $68.59/s.f.

Base cost  50,000 s.f.  $68.59/s.f.  $3,429,500

Figure 22-2 shows that a 7-foot by 8-foot dock leveler costs $6,225.

Dock levelers  10 ea.  $6,225/ea.  $62,250

Total cost = $3,429,500 + $62,250 = $3,491,750 Use $3,490,000 12. Using Figures 22.2 and 22.3, determine the cost for a 125-foot by 200-foot warehouse. The exterior walls will be galvanized steel siding over a steel frame. The warehouse is 26 feet high. Include 50 feet of asphalt around the building in your estimate.

The cost is determined as follows: Area  125 ft.  200 ft.  25,000 s.f.

Perimeter = 125 f t. + 200 f t. + 125 f t. + 200 f t. = 650 f t. ee

ee

ee

ee

ee

Figure 22-2 shows that the cost per square foot for a warehouse with galvanized steel siding over a steel frame and an area of 25,000 square feet is $85.05 per square foot. It also shows that the base perimeter is 640 feet, and the adjustment for perimeter length is $2.60 per square foot per 100 feet.

Add for perimeter 

650 ft.  640 ft.  $2.60 s.f.100 ft.

100 ft.

Add for perimeter = $0.26/s.f.

Figure 22-3 shows that the base story height is 24 feet, and the adjustment for story height is $0.45 per square foot per foot of height.

Add for story height   26 ft.  24 ft.  $0.45 per s.f. per ft.  $0.90/s.f.

Base cost  $85.05/s.f.  $0.26/s.f.  $0.90/s.f.  $86.21/s.f.

Base cost  25,000 s.f.  $86.21/s.f.  $2,155,250

Figure 22-2 shows that asphalt costs $7.85 per square yard.

Area of asphalt   50 ft.  125 ft.  50 ft.   50 ft.  200 ft.  50 ft.  125 ft.  200 ft.

Area of asphalt  225 ft.  300 ft.  125 ft.  200 ft.  42,500 s.f.  4,722 s.y.

Asphalt  4,722 s.y.  $7.85/s.y.  $37,068 Total cost = $2,155,250 + $37,068 = $2,192,318 Use $2,190,000 13. The estimated cost for a roof on a new office building is $150,000. It is expected that the roof will need to be replaced in 25 years for $360,000 and again in 50 years for $850,000. The replacement costs include the costs of disposing of the old roof. The 219 Copyright © 2024, 2017, 2010 Pearson Education, Inc. All Rights Reserved.

Chapter 22: Other Estimating Methods

building has a useful life of 75 years and is expected to be decommissioned at the end of its useful life. The cost of disposing of the roof is $25,000. Determine the lifecycle cost of the roof using an interest rate of 6 percent.

Convert the costs to the current period (year zero) and sum them:

C0 

$150,000

1 0.06 

0



$360,000

1  0.06 

25



$850,000

1  0.06 

50



$25,000

1  0.05 

75

C0  $150,000  $83,800  $46,145  $316 C0  $280,341

14. The estimated cost to install solar panels on an office building is $200,000 and occurs in year 0. The estimated utility savings are $10,000 in year 1 and increase by 3 percent annually. In year 2, the savings would be $10,300 and $10,609 in year 3.

The life of the solar panels is 30 years, at which time they will need to be decommissioned at an estimated cost of $20,000. Hint: Use a spreadsheet to perform the calculations.

The costs are shown in the second and fif th columns of the following table. The cost savings are negative numbers and are shown in paratheses. The third and sixth columns show the cost in year zero (the present values). Note that a cost savings of $23,556 and a decommissioning cost of $20,000 occur in year 30. The total cost (the sum of the third and sixth columns) is $11,019. ee

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Chapter 22: Other Estimating Methods

Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Cost 200,000 (10,000) (10,300) (10,609) (10,927) (11,255) (11,593) (11,941) (12,299) (12,668) (13,048) (13,439) (13,842) (14,258) (14,685) (15,126)

Present Value 200,000 (9,434) (9,167) (8,908) (8,655) (8,410) (8,172) (7,941) (7,716) (7,498) (7,286) (7,080) (6,879) (6,685) (6,495) (6,312)

Year 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 30

Cost (15,580) (16,047) (16,528) (17,024) (17,535) (18,061) (18,603) (19,161) (19,736) (20,328) (20,938) (21,566) (22,213) (22,879) (23,566) 20,000

Present Value (6,133) (5,959) (5,791) (5,627) (5,468) (5,313) (5,162) (5,016) (4,874) (4,736) (4,602) (4,472) (4,346) (4,223) (4,103) 3,482

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Appendix F: BidSim

Appendix F—BidSim BidSim is a program designed to simulate receiving supplier and subcontractor bids on bid day using the Real Estate Office plans from Appendix E. The simulation is intended to be run during a 50-minute class. During the class, the students will receive bids via the Internet. The first bid will be available 30 minutes before the closing, and the last will be available five minutes before the bid closing.

Learning Outcomes After this game, the student should be able to: Learning Outcome F.1: Explain the challenges facing estimators on bid day.

Instructional Hints 

Be sure the students come prepared to class on the day of the BidSim. The instructions for the students are found in Appendix F.

Instructional Resources 

BidSim game files: BidSim folder in the IM Resource Files

Bidsim With the Internet The instructions in Appendix F indicate what bids the students will receive from suppliers and subcontractors. The students are responsible for completing all other takeoffs necessary to complete the bid. On the bid day, they must incorporate the bids into their cost estimate to complete the bid.

Before the bid, the instructor should upload the contents of the BidSim folder in the IM Resource Files to the school’s website. The settings on the internet folder should be set to prevent the user from browsing its contents.

The instructor must open the BidSim.htm file in the BidSim folder with Notepad or another text editing program to set the bid time. The file may be opened by right-clicking on the file, selecting Open With from the popup menu, and selecting Notepad. Once open, the file should appear as the following figure:

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Appendix F: BidSim

The bid closing time is determined by the numbers following bidHour = and bidMin =.

The above figure shows that the bid closing time is 13:58 or 1:58 pm. The computer uses a 24-hour clock; therefore, bid hours after noon should be between 13 and 24. To change the bid hour, change the 13 after bidHour = to the hour of the bid closing. To change the bid minutes, change the 58 after bidMin = to the minute of the bid closing.

Be sure to save the file.

On the day of the bid, each bidding team should have a computer on which to receive bids. The bids will be released based on the time set on the computer’s clock; therefore, the instructor should ensure all the clocks are set to the same time. The students are to open the BidSim.htm web page. As the bids become available, they will appear as links on the web page that may be opened to show the bid. At bid closing time, the links will no longer be available.

Bidsim With Stand-Alone Computers The process is the same as using BidSim with the internet, except the folder contents must be loaded on each computer. Because the files are located on the computer, the students can easily view the folders’ contents and bids before they are available.

Bidsim Without Computers The student preparation is the same as above. Before the bid day, the instructor should make copies of all the bids in the BidSim folder. On the day of the bid, the instructor should pass out the bids during the class period. Alternatively, the bids can be emailed to the students.

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Appendix F: BidSim

General Notes 

The instructor’s materials include a bid form (BidSim Bid Form.docx) that may be used with this exercise.

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The bid from The Roofing Company should not be used because the shingles quoted do not meet specifications.

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If the ceramic tile subcontractor bid is used, money must be added for the backing board.

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If the bid from Electrical Inc. is used, money needs to be added for the phone system, which they exclude.

224 Copyright © 2024, 2017, 2010 by Pearson Education, Inc. All rights reserved.