School Of Pe Construction Notes

RI

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CHAPTER 0 TABLE OF CONTENTS Preamble .............................................................................................................................. vi How to use this Refresher Course Study Guide ...................................................................... vii Preface ....................................................................................................................................viii Refresher Course Activity Organization/Administration ........................................................... ix References ................................................................................................................................ x Chapter 1 Construction Earthwork .............................................................................................1-1 Soils - Swell and Shrinkage ................................................................................................1-2 Relative Compaction .........................................................................................................1-12 The Laboratory Proctor .....................................................................................................1-14 Soil Compaction................................................................................................................1-16 Excavation & Embankment – Means and Methods ..........................................................1-17 Excavation and Embankment – Visual DICTIONARY ......................................................1-20 Earthwork Volume Computations .....................................................................................1-22 Angle of Repose – Internal Angle of Friction ....................................................................1-23 CARTESIAN COORDINATE SYSTEM ............................................................................1-24 Calculating Percent Grade ................................................................................................1-25 OSHA Classification of Soils .............................................................................................1-26 OSHA Classification of Soils: Subpart P – Excavations Appendix B ................................1-27 LAYERED SOILS .............................................................................................................1-28 Average End Area Method ...............................................................................................1-31 Topographical Contour Data - Cut and Fill Volumes ........................................................1-36 Differential Leveling ..........................................................................................................1-39 Surveying with Construction Applications .........................................................................1-42 Trigonometric Leveling .....................................................................................................1-44 Chapter 2 Estimating Quantities and Costs ...............................................................................2-1

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Cost Estimating...................................................................................................................2-3 Quantity Take-Offs ..............................................................................................................2-4 Estimating Takeoff Quantities .............................................................................................2-7 Estimating Formwork ..........................................................................................................2-9 Building Materials – Roof Surface Materials .....................................................................2-11 Estimating Brick Masonry .................................................................................................2-14 Brick Veneer Quantities ....................................................................................................2-16 Computing Geometric Properties .....................................................................................2-17 Painting Quantities............................................................................................................2-20 Estimating Labor Costs .....................................................................................................2-21 Equipment Production ......................................................................................................2-22 Labor Productivity .............................................................................................................2-23 Chapter 3 Construction Operations and Methods .....................................................................3-1 A Guide to Crane Safety .....................................................................................................3-2 Mechanical Properties of Materials .....................................................................................3-4 Properties of Metals ............................................................................................................3-7 Simple Beam Analysis ......................................................................................................3-11 Design Factor Comparison ...............................................................................................3-12 Wire Rope Stretch ............................................................................................................3-15 Conditions for a Rigid-Body in Equilibrium .......................................................................3-18 Crane Selection, erection, and stability ............................................................................3-19 Center of Gravity – Class Review .....................................................................................3-21 Crane Working Range Diagram ........................................................................................3-22 Tipping Fulcrum ................................................................................................................3-26 Crane Outrigger Stability ..................................................................................................3-31 Equipment Production ......................................................................................................3-32

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Daily standard production rate of Equipment ....................................................................3-33 Daily standard production rate of a dump truck ................................................................3-34 Productivity Analysis and Improvement ............................................................................3-35 Effects of job size on productivity .....................................................................................3-37 Actual versus Ultimate strength ........................................................................................3-38 National Pollutant Discharge Elimination System (NPDES) .............................................3-39 Types of soil erosion .........................................................................................................3-41 Common SWPPP Objectives ...........................................................................................3-42 SWPPP Development—Selecting Erosion and Sediment Control BMPs .........................3-43 Chapter 4

Project Scheduling .................................................................................................4-1

Calculating Project Duration ...............................................................................................4-2 Project Sequencing and Scheduling ...................................................................................4-4 Project Scheduling – Types of Methods .............................................................................4-6 Units of Time Convention ...................................................................................................4-7 Precedence Relationships ..................................................................................................4-8 Lead Lag Relationships ......................................................................................................4-8 Project Scheduling - Defining the Terms ............................................................................4-9 Arrow Diagramming Method .............................................................................................4-10 Precedence Definitions for Activity on Arrow (AOA) Network Diagrams ..........................4-11 Activity Relationships – Precedence Tables .....................................................................4-15 CRITICAL PATH NETWORK ANALYSIS..................................................................4-17 Activity Sequencing ..........................................................................................................4-21 LEVELING ......................................................................................................................4-23 Glossary of Scheduling Terms ..........................................................................................4-26 Chapter 5 Material Quality Control and Production ...................................................................5-1 Material Specifications ........................................................................................................5-2

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Slump Test..........................................................................................................................5-4 AGGREGATE TERMS AND DEFINITIONS ...............................................................5-6 Concrete w/c ratio ...............................................................................................................5-7 Concrete Mix Design - Proportions .....................................................................................5-8 Air Entrained Concrete .......................................................................................................5-9 CONCRETE STRENGTH TESTING – COMPRESSIVE STRENGTH .................5-11 CONCRETE STRENGTH TESTING – TENSILE STRENGTH ..............................5-12 Hot Mix Asphalt - Short Course ........................................................................................5-14 Typical asphalt concrete pavement structure ...................................................................5-16 ASPHALT CEMENT: GRADING SYSTEMS AND PROPERTIES....................................5-17 How Asphalt Concrete Pavements Fail ............................................................................5-18 Fog Seal ...........................................................................................................................5-19 FUNCTION OF A FOG SEAL ...........................................................................................5-19 Asphalt Performance ........................................................................................................5-20 CONCRETE MIX DESIGN – CONCRETE ESSENTIALS ......................................5-26 CONCRETE PROTECTION FOR REINFORCEMENT ...........................................5-28 REINFORCEMENT TYPE ............................................................................................5-31 ACI – Joints in Concrete Construction ..............................................................................5-33 Concrete Mix Design - Ratios ...........................................................................................5-35 Concrete Admixtures ........................................................................................................5-36 Chapter 6

Temporary Structures ............................................................................................6-1

Falsework ...........................................................................................................................6-2 Formwork ............................................................................................................................6-3 Concrete Formwork ............................................................................................................6-4 Load Paths - Defined ..........................................................................................................6-6 Load Paths..........................................................................................................................6-8

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Tributary Area ...................................................................................................................6-11 Temporary Loads – Factor of Safety ................................................................................6-14 Torque ..............................................................................................................................6-16 Balancing Loads ...............................................................................................................6-17 Concentrated Loads Unsymmetrically Placed ..................................................................6-19 Objects at Rest .................................................................................................................6-21 Factor of Safety – Concrete WallForms ............................................................................6-23 Temporary Bracing ...........................................................................................................6-24 Chapter 7 Worker Health, Safety, and Environment .................................................................7-1 OSHA regulations ...............................................................................................................7-2 Fall Protection .....................................................................................................................7-3 OSHA and the NCEES Exam .............................................................................................7-4 How to Read the OSHA Standards 29 CFR 1926 – Construction ......................................7-5 OSHA Subpart P -- Excavations .........................................................................................7-6 OSHA Subpart L --

Scaffolds ...........................................................................................7-7

OSHA Subpart N -- Cranes, Derricks, Hoists, Elevators, and Conveyors ..........................7-8 OSHA Subpart U -- BLASTING and the Use of Explosives ................................................7-9 Scaffolding ........................................................................................................................7-11 Protective Equipment (PPE) .............................................................................................7-12 Safety management ..........................................................................................................7-13 Experience modification Rate ...........................................................................................7-14

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PREAMBLE

The material provided in the refresher course is intended for instructional use only. The design code reference and solution techniques are a guide for instruction. The reference material included herein should not be used as a sole source for the PE Exam and/or engineering practice. The NCEES provides updated design Code standards that should be the source for your use. Visit the NCEES website for the most current information regarding the PE Exam and confirm the design standards used for the test construction All solution steps have been vetted and are provided for ease of instruction. There are many methods that can be used to arrive at a solution which fit your specific educational background and experience. Alternate methods and computational techniques based on your familiarity should be used.

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HOW TO USE THIS REFRESHER COURSE STUDY GUIDE

Throughout the Refresher Course Notes the following symbol represents references to the Civil Engineering Reference Manual for the PE Exam (CERM16) and page locations for further review and study: Sample 1 This symbol provides the reference to similar subject matter in the CERM-16. The Chapter  reference guides you to the general area to help aid Page number  your self study. Often the subject matter and course material are not an exact match but the   location points to a general area to find additional information about the subject matter. Sample 2:

fast facts This example text box contains subject material that is supplemental to the subject matter and/or enhances its knowledge. The information is intended for self-study.

Sample 3:

This is an example text box shows necessary equations.

Sample 4:

CERM-16 reference This symbol represents CERM-16, page reference materials which should be inserted for review. Sample 5:

This symbol represents topics within the Refresher Course that are part of the subject matter which will further help your understanding. The information is intended for self-study and WILL NOT be reviewed during the refresher course.

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PREFACE

fast facts Each of us has different study habits and a preferred way of learning. The material in the Refresher Course uses a technique which helps quicken the pace of understanding of the subject matter. The arrangement of the material follows a hierarchical pattern of learning engaging three basic components:

Concept is a cognitive unit of meaning— an abstract idea or a mental symbol sometimes defined as a "unit of knowledge" which is built from other units. A concept is typically associated with a corresponding representation, for example, the concept of Trigonometry with Triangles. Often, a concept is not a single thought, but a composite of simpler concepts.

Terminology refers to the typical words used in connection with a concept. For example, the elements of the Law of Sine’s: sin a, sin b, sin n.

 refers to the typical manner in which the theory is used in connection with a concept. For example, find the hypotenuse of a right triangle when one side is 4-units with an angle of 53° (4 ÷ sin 53° = 5).

Concept Terminology



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REFRESHER COURSE ACTIVITY ORGANIZATION/ADMINISTRATION

The refresher course is organized in seven chapters as outlined below. Each chapter covers materials which parallels the outline provided by the NCEES Exam Specifications for the Construction Exam. The refresher class focus is on interpreting the breadth and depth review materials. The course provides a graduating series of problem statements to better the understanding of the content for the Construction Exam. CHAPTER ORGANIZATION

1.

Earthwork Construction and Layout

2.

Estimating Quantities and Costs

3.

Construction Operations and Methods

4.

Scheduling

5.

Material Quality Control and Production

6.

Temporary Structures

7.

Worker Health, Safety, and Environment

Workshop Outline •

• •

The instructor will announce the workshop questions and the time allotted for working a solution for the question before the question’s review. Lunch break will be determined by consensus. The workshop solutions will be posted on the School of PE website 24hours after the class has ended. Important Note

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REFERENCES The following list provides the references used in preparation of the Construction Review Notes. The bold face acronyms are the NCEES design standards used for the exam’s construction. NCEES effective beginning with the April 2018 exam. ASCE 37

Design Loads on Structures During Construction, 2014, American Society of Civil Engineers, Reston, VA, www.asce.org.

CMWB

Standard Practice for Bracing Masonry Walls During Construction, 2012, Council for Masonry Wall Bracing, Mason Contractors Association of America, Lombard, IL, www.masoncontractors.org.

AISC

Steel Construction Manual, 14th ed., 2011, Parts 1–3, 8, 16.1 (Chapters M, N) and 16.2, American Institute of Steel Construction, Inc., Chicago, IL, www.aisc.org.

ACI MNL-15 Field Reference Manual, 2016, American Concrete Institute, Farmington Hills, MI, www.concrete.org ACI 347R

Guide to Formwork for Concrete, 2014, American Concrete Institute, Farmington Hills, MI, www.concrete.org (in ACI SP-4, 8th edition appendix).

ACI SP-4

Formwork for Concrete, 8th ed., 2014, American Concrete Institute, Farmington Hills, MI, www.concrete.org.

OSHA

Construction Industry Regulations: 29 CFR Parts 1903, 1904, and 1926 (US Federal version, January 2017), US Department of Labor, Washington, DC.

MUTCD-Pt 6 Manual on Uniform Traffic Control Devices – Part 6 Temporary Traffic Control, 2009, US Federal Highway Administration, www.fhwa.dot.gov. Removed from previous exams, ACI 318; and NDS

Civil Engineering Reference Manual for the PE Exam (CERM16), 16th Edition, Michael R. Lindeburg, PE , 2018.

References used to build the refresher course.

Ratay, Robert T.(1996). Handbook of Temporary Structures in Construction, (2nd ed.), McGraw Hill, New York, NY. Rossnagel, W.E., Higgins, L.R. & MacDonald, J.A. (1988). Handbook of rigging for construction and industrial operations (4th ed.). New York: McGraw-Hill. Parker, H., & Ambrose, J. (1993). Simplified engineering for architects and builders (8th ed.). New York: Wiley Inter-Science. Hicks, Tyler G. Editor (1994). Standard Handbook of Engineering Calculations. McGraw-Hill, Inc.

Remember to check for any changes made by the NCEES at their website address:

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The refresher course prepares you to develop a study and exam taking strategy.

Review the NCEES exam specifications found at their website for your module and follow the subject matter instruction. Become familiar with the NCEES design reference standards that are used to construct the exam. Adult learners bring experiences and self-awareness to learning that younger learners do not. Adults are ready to learn when the need arises. Adults are task-oriented. Adult learners develop “a need to know”. Adults are motivated to put time and energy into learning if they know the benefits of learning and the costs of not learning.

Surprise! Prepare to be surprised—surprised that something unanticipated, and not prepared for, is part of or the central concept in a problem. Do not allow this to cause a negative emotional judgment about the entire exam. Negative reaction is counterproductive and requires too much time to move away from (This is out of the ordinary! Why is this on the exam?) … Don’t let them become distractors! The NCEES exam is a “skills level” test; increase your skills through practice to achieve success. 10/18

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SOH…. CAH….TOA

The Basics Measures Board Foot = 144-in3 27-ft3 = 1-yd3 5,280-ft/mile 1760-yd/mile 43,560-ft2/acre 10,000-sqm/hectare 1-ton = 2,000-lbs 1440-min/day 1 U.S. survey ft =(12÷39.37)m 1 international ft = 0.3048 m 1 in. = 25.4 mm 1 mile = 1.60935 km 1 ha = 10,000 m2 = 2.47104 acres

1 rad = 180÷ π

Water 1-psi pressure = 2.31-ft head 1-ft (water) = 0.433-psi 1.122-ft water/in of mercury 1 U.S. gallon of water weighs = 8.34-lbs 1-ft3 = 7.48-gallon 1 ft3 of water weighs 62.4-lb Detention time = Volume ÷ Flow Rate Velocity = Flow rate ÷ Cross sectional area 1 kg = 2.2046 lb 1 L = 0.2624 gal

http://www.khanacademy.org/

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Construction Earthwork

Concept Terminology



CHAPTER

CHAPTER 1 CONSTRUCTION EARTHWORK

1

Construction Earthwork

“State” of Soils Average End Area Earthwork Volume

Swell Shrinkage Bank Soil Stations Cut Fill Staking & Layout Differential Leveling Benchmark Back sight Foresight Height of Instrument Terrain

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SOILS - SWELL AND SHRINKAGE

1. - Question Which of the following statements about construction earthwork are true: I. The volume of earth known in its natural state is known as bank-measure; insitu; in-place; virgin soil. II. The volume during transport is known as loose–measure; fluffed; swell; bulk. III. The volume after compaction is known as compacted-measure. IV. The change in volume of earth from its natural to its loose state is known as swell. Swell is expressed as a percentage of the natural volume. V. The decrease in volume from its natural state to its compacted state is known as shrinkage. Shrinkage is expressed as percent increase from the natural state. a. b. c. d.

I & II I, II, & III I, II, III, & IV I, II, III, IV, & V

  

Solution: Refer to CERM-80 (page 80-1 Section 3) Item V – “Shrinkage is expressed as percent decrease from the natural state”. (answer is c)

fast facts An example of the relationships of a cubic yard of soil in three states: bank, loose, and compacted. Swell and shrinkage are always measured in relation to the bank condition. The numerical values are examples and are different for each type of soil. (Note the inverse relationship between loose and compacted states of soil.)

1.25-yd3 1-yd3

Bank

25% swell

Loose

0.80-yd3 20% shrinkage Compacted

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

A soil’s swell factor represents the fact that the volume of soil placed by nature in the ground is not the same as the volume of the same mass of dirt excavated by the contractor and placed in the dump truck. The same mass of soil occupies more volume in the truck (loose cubic yards) than it does in the ground (bank cubic yards). The swell factor is an adjustment representing this increase in volume. However, the swell factor plays no part in the calculation of an earthwork’s balance. The swell factor is used to determine the subsequent hauling and stockpiling requirements.   

SoilSoil Diagram Diagram

Soil Phase Diagram Soil Phase Diagram

Swell is the percentage increase in volume caused by the excavation of soil. Physically, the act of excavation breaks up the soil into particles and clods (lump of earth) of various sizes. This creates more air pockets and results in an effective increase in the soil’s void volume. An increase in volume also results in a decrease in density. This decrease in density and increase in volume varies between soil types and is not proportional due to the initial, natural void volume of the bank soil. The swell factor equations are found in the Table below:

Swell:

The most probable weight of a cubic yard of soil is most nearly: a. b. c. d.

500 1,000 5,000 10,000

A soil increases in volume when it is excavated. Swell Density

Swell (%) =

Bank Density -1 Loose Density

Load Factor = Loose Density Bank Density



Swell Volume x 100

V loose = 100% + % swell x Vbank = Vbank 100% Load Factor Load Factor = (1 + decimal swell) -1 or, Load Factor = 1 ÷ (1 + decimal swell)

.

Bank Volume = Loose volume x Load factor

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Applying the equation, soil with a swell of 25% would have a load factor of 80% (the inverse of 1.25). The load factor can be used to show the relationship between Loose and Bank density by dividing the loose density by the load factor (i.e., 2100 / .79 = 2650). Using dry clay (from the Table below) as an example, the calculations are derived as follows: 2650-lb/CY x .79 = 2100-lb/CY; or, 2100-lb/CY x 1.26 = 2650-lb/CY Material Clay, dry Clay, wet Clay and gravel, dry Clay and gravel, wet Earth, dry Earth, moist Earth, wet Gravel, wet Gravel, dry Sand, dry Sand, wet Sand and gravel, dry Sand and gravel, wet

Loose Bank (lb/cy) (lb/cy) 2,100 2,650 2,700 3,575 2,400 2,800 2,600 3,100 2,215 2,850 2,410 3,080 2,750 3,380 2,780 3,140 3,090 3,620 2,600 2,920 3,100 3,520 2,900 3,250 3,400 3,750

Swell (%) 26 32 17 17 29 28 23 13 17 12 13 12 10

Load Factor 0.79 0.76 0.85 0.85 0.78 0.78 0.81 0.88 0.85 0.89 0.88 0.89 0.91

Exact values will vary with grain size, moisture content, compaction, etc. Test to determine exact values for specific materials.

In addition to the swell factor and its associated load factor, soil also has a shrink factor. While the first two relate the volume of an equal mass of bank soil in the ground with the loose mass deposited in stockpiles or dump trucks by excavation, the shrink factor relates the initial bank soil with the volume resulting from subsequent placement and compaction of the loose soil into earthen structures. Often this ratio is not a result of natural characteristics but is based on the construction specifications. For example, clay soils used to construct a high density/low permeability containment layer for landfills are typically constructed in controlled lifts of a certain spread thickness which are then compacted to a final desired thickness. Typically, the soil is spread out over the work area in loose lifts about 8 inches thick. Multiple passes with a compacting roller (sheep foot roller or vibratory smooth drum roller) are then performed to compact and knead the loose clay into a tight layer of about 6 inches thickness. This results in a post-compaction volume that is approximately 25% smaller than that of the initial loose placement volume. The resultant shrink factor equations are found in the following Table:

BCY

LCY

CCY

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A soil decreases in volume when it is compacted:

Shrinkage:

Shrinkage Density Shrinkage(%) =

1-

Shrinkage Volume

Bank Density Compacted Density

x 100

Shrinkage factor = 1 – Shrinkage (% decimal)

V compacted = 100% - % shrinkage 100%

V bank

Compacted Volume = Bank Volume x Shrinkage Factor

The preceding can be applied to an example of an earthwork operator excavating wet clay. Assume its initial bank density to be 3,500 pounds per cubic yard and its excavated loose density to be 2,800 pounds per cubic yard. One ton of this soil (2,000 pounds) would occupy 0.57 (2000-lb / 3500-lb = .57) bank cubic yard in the ground while its hauled or stockpiled volume would be 0.71 (2000 / 2800 = .71) loose cubic yard. This analysis results in a swell factor of 25% (2800 / 3500 = 0.80; 0.80 -1 = 25%). Its related load factor would be 0.80 (remember that 0.80 x 3500 = 2800). Suppose further that this clay is used to construct a landfill cover using compaction as described above thereby reducing its volume to 0.53 cubic yard (given). The shrink factor, then, would be 0.93 (0.53 / 0.57 = 0.93). For planning purposes, the earthwork contractor will have to assume that for every 100 cubic yards he excavates he will need to haul 125 cubic yards so that he will be able to place 93 cubic yards. All of these numbers affect his bottom line. The first determines the amount of the excavation effort, the second determines his hauling requirements and the third determines the overall cost of the finished project. Initial Soil Condition Bank Loose Compacted Common Earth Bank Loose Compacted Rock (blasted) Bank Loose Compacted Sand Bank Loose Compacted Soil Type Clay

Bank 1.00 0.79 1.11 1.00 0.80 1.11 1.00 0.67 0.77 1.00 0.89 1.05

Converted to: Loose Compacted 1.27 0.90 1.00 0.71 1.41 1.00 1.25 0.90 1.00 0.72 1.39 1.00 1.50 1.30 1.00 0.87 1.15 1.00 1.12 0.95 1.00 0.85 1.18 1.00

The TABLE illustrates soil in a variety of states.

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CERM-16 reference Chapter 80, pg-3 Table 80.1 Summary of Excavation Soil Factors Notes:  Review the descriptions of the quantity.  Become familiar with the acronyms listed in the Table (e.g., LCY; BCY; CCY; etc.)  Familiarize yourself with the organization of Table formulas and distinguish between the VOLUME and DENSITY functions; use the formulas appropriately in the question being asked; remember that the formulas are not interchangeable between, for example, CF and Ton.

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Great Advice!!!

Download and print out a copy of the INDEX

for the CERM from the PPI2PASS website (same web location where you purchased your CERM). Bind it separately and use the bound copy to look-up the referenced material while studying and taking the exam. This will help avoid constantly flipping around in the 1500 + pages in the CERM manual between the index and the chapters. This saves lots of time and aggravation working from multiple places in the CERM.

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2. - Question An earthwork contractor encountered a location within the borrow area where the geological conditions changed. Instead of encountering 100 cubic yards of wet clay, the contractor excavates 100 cubic yards of loose sand and clay having a bank density of 3,400 pounds per cubic yard and a loose density of 2,700 pounds per cubic yard. A ton of this material will occupy nearly how many cubic yards in the ground? and, in the truck? a. 0.49-yd3 in the ground; and, 0.75-yd3 in the truck b. 0.59-yd3 in the ground; and, 0.76-yd3 in the truck c. 0.59-yd3 in the ground; and, 0.74-yd3 in the truck d. 0.69-yd3 in the ground; and, 0.93-yd3 in the truck Solution: Two-thousand pounds of this material would occupy 0.59 (2000 / 3400 = 0.59) cubic yard in the ground and 0.74 (2000 / 2700 = .74) cubic yard in the truck. This results in a swell factor of 26%. The contractor will have to haul 126 cubic yards of this material for every 100 cubic yards in the ground. [Be attentive to the units.] (answer is c) 3. - Question 30,000-yd3 of banked soil from a borrow pit is stockpiled before being trucked to the jobsite. The soil has 28% swell and shrinkage of 18%. The final volume of the compacted soil is most nearly: a. 24,600-yd3 b. 25,400-yd3 c. 35,400-yd3 d. 38,400-yd3 Solution: Shrinkage is measured with respect to the bank condition. Apply the equation: V compacted = 100% - % shrinkage) 100%

V bank

Vcompacted = 100% -18% (30,000-yds3) = 24,600-yd3 (answer is a) 100%

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4. - Question A contractor was awarded a Contract to excavate 3 and haul 200,000-yds of silty clay (USCS classification ML) with a bulking factor of 30%. The contractor’s fleet of dump trucks have a capacity of 26yds3 and operate on a 25-minute cycle. The job must be completed in 5working days with the fleet working at two 8-hour shifts per day. The number of trucks required is most nearly: a. 24 b. 37 c. 52 d. 55 Solution:  Apply a bulking factor (swell) of 30% to the total volume.  200,000-yds3 x 1.30 = 260,000-yds3 (Volume to be trucked off-site)  5-wd x 2-shifts x 8-hrs = 80-hrs (Total trucking hours)  260,000-yds3 ÷ 80-hrs = 3,250-yds3/hr (Haulage rate per hour)  (26-yds3/truck ÷ (25-min/cycle ÷ 60-min/hr)) = 62.40-yds3/truck hour  3,250-yds3/hr ÷ 62.40-yds3/truck-hr = 52.08-trucks  Answer__________

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BULLETIN: What does “Most Nearly” really mean?

See answer … Must Read: CERM 16 page xxiii

Note that in the previous problem rounding down to 52-trucks would remove 259,584-CY or 416-CY remaining loads or approx. 16-dump trucks loads necessary to remove all the soil.

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5. - Question Soil at a borrow area has a total unit weight of 120-PCF and a water content of 15 percent. The soil from the borrow area will be used as structural fill to an average dry unit weight of 110-PCF. The soil shrinkage is most nearly: a. b. c. d.

3.0% 3.5% 4.0% 5.5%

Solution: At the borrow area, the dry unit weight is determined from the equation: Dry unit weight =

Total Unit Weight (1 + water content)

Dry Unit Weight = 120 / (1 + 0.15) = 104-PCF The shrinkage factor is the ratio of the volume of compacted material to the volume of borrow material (based on dry unit weight), or: Shrinkage factor = 104-PCF / 110-PCF = 0.945 Convert the shrinkage factor to a percentage: Percent shrinkage = (110 - 104) / 110 = 0.055 = 5.5% (answer is d)

.

fast facts Step 1-- Be certain to make comparisons based on the “state” (bank, loose, compacted) of soil first, then - Step 2 -- analyze the soil using the equations for swell and shrinkage using bank or compacted densities or volumes. Don’t mix up the “units”. Bank soil is not the same as dry unit weight as it may have water content and comparisons cannot be made until the soil’s common denominator is found.

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RELATIVE COMPACTION

6. - Question Project specifications require a relative compaction of 95% (modified Proctor). Construction of a highway embankment requires 10,000-yd3 of fill. The borrow soil has an in-situ dry density of 94-PCF and a laboratory maximum dry density of 122.5-PCF. The total volume of soil that must be excavated from the borrow area is most nearly: a. b. c. d.

9,500-yd3 10,000-yd3 11,700-yd3 12,380-yd3

Solution: The most common method of assessing the quality of field compaction is to calculate the Relative Compaction (RC) of the fill, defined as: RC = 100 * (field dry density, PCF) Laboratory maximum dry density (PCF) Apply the equation using the given data: RC = 100 x 94-PCF = 76.73% 122.5-PCF Calculate the required volume of soil that must be excavated from the borrow area: (Required Fill) x (Compaction %) x (Relative Compaction)-1 = Excavated Volume (borrow) 10,000-yd3 of fill x (95%) x (76.73%)-1 = 12,380-yd3 (answer is d)

fast facts The most common type of nondestructive field test is the nuclear density test method. In this method, the wet density of soil is determined by the attenuation of gamma radiation. The water content is determined by the thermalization or slowing of fast neutrons and direct probe readings over the in-place test area. The nuclear density test uses the laboratory dry density and optimum moisture content to determine the in-place soil density.

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fast facts Although earthwork optimization is related with both swelling and compaction behavior of fill material, it is possible to combine these characteristics by a unique swelling/shrinkage ratio that accounts for field densities measured before excavation and after compaction. Compaction is a soil densification process achieved by the application of mechanical energy and improves several engineering properties of soils. Commonly, it is essential to control certain compaction parameters, namely, dry density and water content, with field tests conducted throughout the earthwork construction. It is desirable that fill material has a field unit weight as close as possible to the maximum dry unit weight obtained by the laboratory Proctor test. The measure of the closeness is defined as the relative compaction (RC), which is required to be higher than a threshold value determined by the project specifications. In order to determine the swelling/shrinkage behavior of a material, field and laboratory tests should be performed to measure field dry unit weight and maximum dry unit weight. Swelling/shrinkage parameters can then be calculated using these test results based on the project compaction criterion and the construction equipment being used. However, soil behavior is inherently ambiguous and the actual compaction control process is usually carried out while earthwork construction is continuing. Therefore, for most of the highway designs, swelling/shrinkage factors are selected from predetermined tables according to specific soil types being considered. The swelling/shrinkage behavior of soils can also be characterized based on their particle size classifications (either fine or coarse grained based on the amount passing No. 200 sieve). In this context, gradation (well or poor) determined by the coefficient of curvature and coefficient of uniformity parameters, can be taken into consideration for coarse grained soils, whereas the plasticity index is the primary distinguishing variable for expressing the swelling/shrinkage behavior of fine grained soils (silts and clays). Natural water content is also a significant factor influencing the shrinkage/swelling potential of both fine and coarse-grained soils. For fine-grained soils, an increase in the plasticity index reduces the swelling/shrinkage potential. At a certain applied energy level, the dry unit weight of a soil reaches to the maximum level for optimum water content. Therefore, the natural water content (either at wet or dry of optimum) should also be considered to characterize swelling/shrinkage behavior.

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THE LABORATORY PROCTOR

fast facts Compaction is achieved by inputting energy to expel the air and water in the soil’s voids. The reduction of the voids creates the following changes in the material: • Increase in unit weight • Decrease in Compressibility • Decrease in Permeability

ENERGY

AIR

WATER

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fast facts

The Laboratory Proctor Target Area

100%

B Maximum achievable density for the compacting effort

Dry Density PCF

98%

MUD

DRY A % Moisture

Maximum density is found at point “B” and at the intersection of Optimum Moisture Content point “A”

Moisture content of the soil is vital to proper compaction. Moisture acts as a lubricant within soil, sliding the particles together. Too little moisture means inadequate compaction—the particles cannot move past each other to achieve density. Too much moisture leaves waterfilled voids and subsequently weakens the load-bearing ability. The highest density for most soils is at a certain water content for a given compaction effort. The drier the soil, the more resistant it is to compaction. In a water-saturated state the voids between particles are partially filled with water, creating an apparent cohesion that binds them together. This cohesion increases as the particle size decreases (as in clay-type soils).

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SOIL COMPACTION

Which of the following in a list of construction equipment can be used most efficiently to compact fine grained clay soils for a large roadway project: a. b. c. d.

Self-propelled Sheepsfoot Roller Smooth Drum Vibratory Roller Pneumatic-tired roller Mid-size tamping-foot (padfoot) roller

The purpose of compaction is to obtain the optimum density of the soils properties, as close as the zero-air-voids as possible. Common types of earthwork compactors include: Sheepsfoot rollers, which run static and are typically towed; the sheepsfoot roller is most effective for compaction of plastic soils like clay or silt, the sheepsfoot compacts from the bottom of each lift towards the top. High contact pressures cause the feet to penetrate through the loose material and actually compact the material directly with the foot tip. Pneumatic-tired rollers, which use rubber tires to provide the familiar kneading action of soil or subgrade; Pneumatic-tired rollers generally compact from the top of the lift downward. The relationship between the tire contact area and the ground contact pressure causes a kneading action, which helps seek out soft spots that may exist. Light- to medium-weight self-propelled units are used primarily for compaction of granular base as well as hot mix asphalt. Vibratory rollers (smooth drum), typically used for granular and mixed soil materials; and Vibratory rollers work on the principle of particle rearrangement resulting from dynamic forces generated by the vibrating drum hitting the ground. As particles in the soil rearrange themselves, voids between particles become smaller, causing an increase in material density. The best vibratory application is the compaction of granular and mixed soils. It is often recommended that a vibratory smooth-drum roller be used on materials having up to 10% cohesive content. Tamping foot, which combines the advantages of a vibratory roller with a sheepsfoot. A tamping foot roller has feet, or pads, that penetrate the soil, compacting from the bottom to the top for uniform density. The forces of gravity and vibratory impact simultaneously compact from the top down. Due to the foot shape, and in combination with vibration, these rollers achieve a kneading and impact effect while the imprints left contribute to a reduction of water content. A tamping foot, or padfoot roller can compact soils having as much as 50% cohesive content. (Answer – a)

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EXCAVATION & EMBANKMENT – MEANS AND METHODS

Frequency and Amplitude Large vibratory rollers usually offer a choice of two amplitudes and two frequencies. That allows the contractor to adjust to job-site conditions. Use high frequency/low amplitude on granular material or thinner lifts—and low frequency/high amplitude on cohesive material or thicker lifts. There is little advantage to running a vibratory roller in static mode on soils or base material. A sheepsfoot roller, on the other hand, always runs in static mode and uses manipulation and impact to achieve compaction. Frequency is a measure of the number of complete cycles or revolutions of the eccentric weights around the axis of rotation in a given length of time. Frequency is usually expressed in units of vibrations per minute (vpm) or hertz (Hz). Amplitude is a measure of the vertical movement of the drum during a vibration. The relationship between frequency and working speed is sometimes simplified to a simple rule of thumb which states that frequency and working speed should be adjusted to yield approximately one impact per inch or 25 mm. Too high a working speed can cause “washboarding” with impacts spaced too far apart; and too low a working speed negatively impacts machine productivity. There is an optimum working speed and frequency for each compaction application, but they may not yield exactly one impact per inch. That is due to many variables of soil mechanics and composition. An operator can sense when over-compaction has occurred. When a soil becomes dense and t nears its maximum density, it deflects some of the impact energy from the drum and transmits it

back into the drum. The drum will rebound high enough for an entire cycle of the eccentric weight to occur without the drum making an impact. In essence, the drum makes a “double-jump.” This

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phenomenon is called “decoupling,” or “double-jumping,” is recognized by the operator because it causes violent vibrations throughout the machine. To stop it, the operator will adjust amplitude down.

Cohesive soils Cohesive soils have the smallest particles. Clay has a particle size range of .00004" to .002". Silt ranges from .0002" to .003". Clay is used in embankment fills and retaining pond beds.

Characteristics Cohesive soils are dense and tightly bound together by molecular attraction. They are plastic when wet and can be molded, but become very hard when dry. Proper water content, evenly distributed, is critical for proper compaction. Cohesive soils usually require a force such as impact or pressure. Silt has a noticeably lower cohesion than clay. However, silt is still heavily reliant on water content.

Granular soils Granular soils range in particle size from .003" to .08" (sand) and .08" to 1.0" (fine to medium gravel). Granular soils are known for their waterdraining properties.

Characteristics Sand and gravel obtain maximum density in either a fully dry or saturated state. Testing curves are relatively flat so density can be obtained regardless of water content. Two factors are important in determining the type of force a compaction machine produces: frequency and amplitude.

Frequency is the speed at which an eccentric shaft rotates or the machine jumps. Each compaction machine is designed to operate at an optimum frequency to supply the maximum force. Frequency is usually given in terms of vibrations per minute (vpm).

Amplitude (or nominal amplitude) is the maximum movement of a vibrating body from its axis in one direction. Double amplitude is the maximum distance a vibrating body moves in both directions from its axis. The apparent amplitude varies for each machine under different job site conditions. The apparent amplitude increases as the material becomes more dense and compacted.

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7. - Question Which of the following list of equipment should be used within 3-ft of a 250-ft long basement masonry wall when placing ABC fill material to achieve 95% compaction? a. b. c. d.

Vibratory plate compactor Sheepsfoot Roller Vibrating compacting roller Pneumatic-tired roller

Solution: Sheepsfoot roller is most effective for compaction of plastic soils like clay or silt; Vibrating compacting roller typically used for granular and mixed soil materials, are very heavy and too large to be used next to a building; Pneumatic-tired roller – has rubber tires to provide a kneading action of soil or subgrade, Pneumatic-tired rollers generally compact from the top of the lift downward, too large for close to building operation; Therefore, little compaction is required for Aggregate Base Course (ABC) as the stone is compacted when placed and spread. The ABC component particles will vary in size from 3/4 inch down to dust. Typically, a small hand operated tamper, vibratory plate compactor may be used. Answer a.

fast facts In place of performing a Proctor test on soils, soil bearing capacities can be confirmed through a variety of simple field tests to help determine the bearing capacity and moisture content, for example: a. For moisture content use the hand test. Squeeze a ball of soil in your hand. If it's powdery and won't hold a shape, it's too dry; if it molds into a ball then breaks into a couple of pieces when dropped, it's about right; if it leaves moisture on your hand and doesn't break when dropped, it's too wet. b. Clay that you can push your thumb a few inches into with moderate effort has a bearing strength in the range of 1000 to 2500-PSF c.

Loose sand that you can just barely push a #4 rebar into by hand has a bearing capacity of 1000 to 3000-PSF

d. Sand that you can drive a #4 rebar into about 1 foot with a 5-pound hammer has a bearing capacity over 2000-PSF

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EXCAVATION AND EMBANKMENT – VISUAL DICTIONARY

C = Cut = Excavation = any adjective describing the removal of earth F = Fill = Embankment = any adjective describing the placement of earth

100+00

200+00

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Angle of Repose - The maximum slope or angle at which a granular material, such as loose rock or soil, will stand and remain stable.

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EARTHWORK VOLUME COMPUTATIONS

8. - Question On a 5-acre level terrain building site, an earthwork contractor has instructed her crew to strip and grub the topsoil of a 60,000ft2 proposed building pad to a minus 2-ft sub-grade and limit the stockpile to 60-ft radius. The soil has a swell of 40% and an angle of repose at 30°. The stabilized height of the stockpile is most nearly: a. 30 b. 35 c. 40 d. 45 Solution: Step 1: Calculate the cubic volume of the cut and add the swell to the soil volume: 60,000-ft2 x 2-ft x 1.40 (40% swell) = 168,000-ft3 or 6,222-yd3 Step 2: Evaluate the question using the equation for the volume of a cone and compare it to the maximum incline of the sides of the cone. The natural angle of repose is equal to the angle of internal friction for the soil. Calculate the maximum height of the cone based on the natural angle of repose (see sketch). r = h ÷ tan α°

h

∴ , 60-ft = h ÷ tan 30° h = 34.6-ft Step 3: Using the equation to find the volume of a cone, solve for h, the height:

α= 30°

r

V = π r2 h 3 168,000-ft3 = (π 602 h) ÷ 3 h = 44.6-ft Select the answer which follows the physical properties of soil.

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ANGLE OF REPOSE – INTERNAL ANGLE OF FRICTION

A

B

C

Soil Properties If you pour dry sand out of a bucket into a heap on the floor it will form a conical “hill”. The angle of the slope of the hill (i.e. the angle that the side of the hill makes with the horizontal) is called the “angle of repose” or the “angle of internal friction” and is a constant for that particular sand under those conditions (e.g. moisture content). The sand forms a hill in this manner because of the existence of friction between the sand grains. From this concept comes the concept of the angle of internal friction and the lower limit (used for long term stability) of the angle of internal friction is equal to the angle of repose. This approach is simple and generally acceptable for granular (or frictional) soils like sands and gravels. This theory assumes that a slope with an angle below that of the angle of repose will be stable to any height; which is true for clean, dry (or clean, submerged) sands. For cohesive soils (e.g. clays) the limiting slope is a function of the height of the hill and time.

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CARTESIAN COORDINATE SYSTEM

A point on the line is labeled by a single coordinate x, a point in the plane is fixed by 2 coordinates (x, y) and a point in space is determined by three coordinates (x, y, z). Depending on which coordinates are positive, one can divide the line, the plane or the space into half lines, quadrants or octants. 1D space = line = 2 half lines 2D space = plane = 4 quadrants 3D space = space = 8 octants

IV

I

(-,+)

(+,+)

III

II

(-,-)

(+,-)

Three axial planes (x=0, y=0, z=0) divide space into eight equal octant domains, each with a coordinate signs from (-,-,-) to (+,+,+).

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CALCULATING PERCENT GRADE

9. - Question

A 235-HP diesel engine bulldozer is running at full throttle and pushing earth to contour and groom an earthen berm with a 1-½ : 1 uphill grade. The power (%) required to push the same work volume at level ground is most nearly: a. b. c. d.

25 50 67 86

μN

Solution: Slope = rise/run = 1 / 1 ½ = 67% = arctan (0.67) = 33º; Therefore, the force of a dozer on a 33º (sin 33º = .54) slope would exert nearly twice the force required to push the same volume of earth on flat ground. Answer b

N

h

W cos σ

Calculating Percent Grades

σ

W sin σ

The larger the absolute value of a slope, the steeper the line. A horizontal line has slope 0, a 45° rising line has a slope of +1, and a 45° falling line has a slope of −1. The equations for converting a slope as a percentage into Newton's second law of motion states that an angle in degrees and vice versa are: the acceleration of an object as produced by Angle = arctan (%slope/100) And, Slope = 100 tan (angle) where angle is in degrees and the trigonometric functions operate in degrees. Therefore a 100% slope is 45º The expression for the slope of a line is: m = Δy / Δx = rise / run A 100% change (degrees) of: a. b. c.

in grade represents a slope angle

30 45 60 d. 90 Answer: Arctan (100%/100) = 45º

a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. Since the problems states that the dozer is at full power going uphill, the friction forces would be less when they are parallel and perpendicular on level ground. The % change would then be proportional to the work done by the applied force. The work done against gravity is the same, but the work done against friction is greater because the incline is longer and the normal force is greater; however, the applied force is less on level ground. Similar to when walking up a hill takes more power than walking on a level surface.

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OSHA CLASSIFICATION OF SOILS

APPENDIX B TO SUBPART P OF PART 1926—SLOPING AND BENCHING (a) Scope and application. This appendix contains specifications for sloping and benching when used as methods of protecting employees working in excavations from cave-ins. The requirements of this appendix apply when the design of sloping and benching protective systems is to be performed in accordance with the requirements set forth in § 1926.652(b)(2). (b) Definitions. Actual slope means the slope to which an excavation face is excavated.

 

Distress means that the soil is in a condition where a cave-in is imminent or is likely to occur. Distress is evidenced by such phenomena as the development of fissures in the face of or adjacent to an open excavation; the subsidence of the edge of an excavation; the slumping of material from the face or the bulging or heaving of material from the bottom of an excavation; the spalling of material from the face of an excavation; and ravelling, i.e., small amounts of material such as pebbles or little clumps of material suddenly separating from the face of an excavation and trickling or rolling down into the excavation. Maximum allowable slope means the steepest incline of an excavation face that is acceptable for the most favorable site conditions as protection against cave-ins, and is expressed as the ratio of horizontal distance to vertical rise (H:V). Short term exposure means a period of time less than or equal to 24 hours that an excavation is open. (c) Requirements—(1) Soil classification. Soil and rock deposits shall be classified in accordance with appendix A to subpart P of part 1926. (2) Maximum allowable slope. The maximum allowable slope for a soil or rock deposit shall be determined from Table B-1 of this appendix. (3) Actual slope. (i) The actual slope shall not be steeper than the maximum allowable slope. (ii) The actual slope shall be less steep than the maximum allowable slope, when there are signs of distress. If that situation occurs, the slope shall be cut back to an actual slope which is at least 1/2 horizontal to one vertical (1/2H:1V) less steep than the maximum allowable slope. (iii) When surcharge loads from stored material or equipment, operating equipment, or traffic are present, a competent person shall determine the degree to which the actual slope must be reduced below the maximum allowable slope, and shall assure that such reduction is achieved. Surcharge loads from adjacent structures shall be evaluated in accordance with § 1926.651(i). (4) Configurations. Configurations of sloping and benching systems shall be in accordance with Figure B-1 (SEE OSHA - Occupational Safety and Health Standards for the

Construction Industry, 29 CFR Part 1926)

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OSHA CLASSIFICATION OF SOILS: SUBPART P – EXCAVATIONS APPENDIX B Type

A Soil

"Type A" means cohesive soils with an unconfined, compressive strength of 1.5 ton per square foot (tsf) (144 kPa) or greater. Examples of cohesive soils are: clay, silty clay, sandy clay, clay loam and, in some cases, silty clay loam and sandy clay loam. Cemented soils such as caliche (crust or layer of hard subsoil encrusted with calcium-carbonate occurring in arid or semiarid regions) and hardpan (A distinct layer of soil that is largely impervious to water) are also considered Type A. However, no soil is Type A if: (i) The soil is fissured; or (ii) The soil is subject to vibration from heavy traffic, pile driving, or similar effects; or (iii) The soil has been previously disturbed; or (iv) The soil is part of a sloped, layered system where the layers dip into the excavation on a slope of four horizontal to one vertical (4H:1V) or greater; or (v) The material is subject to other factors that would require it to be classified as a less stable material. Type

B Soil

"Type B" means: (i) Cohesive soil with an unconfined compressive strength greater than 0.5 tsf (48 kPa) but less than 1.5 tsf (144 kPa); or (ii) Granular cohesionless soils including: angular gravel (similar to crushed rock), silt, silt loam, sandy loam and, in some cases, silty clay loam and sandy clay loam. (iii) Previously disturbed soils except those which would otherwise be classed as a Type C soil. (iv) Soil that meets the unconfined compressive strength or cementation requirements for Type A, but is fissured or subject to vibration; or (v) Dry rock that is not stable; or (vi) Material that is part of a sloped, layered system where the layers dip into the excavation on a slope less steep than four horizontal to one vertical (4H:1V), but only if the material would otherwise be classified as Type B. Type

C Soil

"Type C" means: (i) Cohesive soil with an unconfined compressive strength of 0.5 tsf (48 kPa) or less; or (ii) Granular soils including gravel, sand, and loamy sand; or (iii) Submerged soil or soil from which water is freely seeping; or (iv) Submerged rock that is not stable; or (v) Material in a sloped, layered system where the layers dip into the excavation or a slope of four horizontal to one vertical (4H:1V) or steeper.

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LAYERED SOILS When the excavation contains layers of different types of soils, the general sloping requirements do not apply. The excavation must be sloped according to the Table. The figures provide examples of excavations made in layered soils.

Sloping Requirements for Layered Soils Slope Required For Each Soil Layer Layered Soil Type Type A Layer Type B Layer Type C Layer B over A

3/4:1

C over A

3/4:1

C over B

1-1/2:1 1:1

A over B

1:1

A over C

1-1/2:1

B over C

1:1 1-1/2:1

1:1 1-1/2:1 1-1/2:1

1-1/2:1

Excavation in layered soil (Type C over Type A). The layer of Type C soil is sloped at 1-1/21, while the layer of Type A soil is sloped at 3/4:1.

Excavation in layered soil where Type A soil tops Type C soil. Both the Type A and Type C soils in the excavation must be sloped at 1-1/2: 1.

OSHA design specifications apply only to trenches that do not exceed 20 feet. The soil type in which the excavation is made must be determined in order to use the OSHA data. The specifications do not apply in every situation experienced in the field; the data were developed to apply to most common trenching situations. For example, this figure illustrates timber shoring in a trench approximately 13 feet deep and 5 feet wide in Type B soil. Using OSHA specifications TIMBER TRENCH SHORING - MINIMUM TIMBER REQUIREMENTS will describe the construction as, the 6 x 6 cross braces have been placed at 6 feet horizontally and 5 feet vertically; the 8 x 8 wales are positioned at five feet vertically; and the 2 x 6 uprights are placed every two feet.

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10. - Question

An excavation is planned for the installation of a 96-in O.D. RCP (10-in wall thickness) sanitary pipeline. The soil adjacent to the building is classified as Type C according to OSHA Subpart P - Excavations. The property’s utility easement grants that an undisturbed earth perimeter be maintained around the building equal to half its height. Work crews require a minimum 3-ft access path on either side of the pipe during installation. The distance (ft) from the face of the building to the centerline of the pipe is most nearly: a. b. c. d.

7 10 18 35

Building

CL 20-ft

12-ft

Not to Scale

Solution: Step 1: Determine the OSHA soil classification and the Maximum Allowable Slope (H:V) from OSHA Subpart P - Excavations, Appendix B. Confirm that Type C soils require a 1-½ : 1. Step 2: Calculate the horizontal slope distance Horizontal slope distance Centerline to toe of slope Distance Building Perimeter Distance Pipe centerline to face of building

= 12-ft x 1.5 = 8-ft/2 + 3-ft = 20-ft/2 = 18 + 7 + 10

= 18-ft = 7-ft = 10-ft = 35-ft answer

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11. - Question An earthwork contract was awarded to excavate and backfill the foundation of a proposed 50-ft by 50-ft office building. The existing grade elevation is 437.5-ft while the sub-base for the below grade basement is 432.5-ft. The concrete contractor requires a 3-ft perimeter walkway in order to place the concrete formwork. Soil conditions are classified as OSHA Type B soils. The bank volume (yd3) to be stockpiled and used for backfill is most nearly: a. b. c. d.

110 227 463 690

Solution: Step 1: Determine the OSHA soil classification and the Maximum Allowable Slope (H:V) from OSHA Subpart P - Excavations, Appendix B. Confirm that Type B soils require a 1:1 maximum slope. Step 2: The foundation excavation can be described as an inverted truncated pyramid. Compute the earthwork volume using the buildings dimensions and add a 3-ft perimeter walkway around the building for the workers erecting the concrete formwork. Equation for the volume of a truncated pyramid: Volume = V1 = h/3 (A1 + A2 + √(A1 x A2)) Compute depth of foundation = h = 437.5-ft – 432.5-ft = 5.0-ft A1 = Area of the base of truncated pyramid = (50 + 3 + 3) (50 + 3 + 3) = 3,136-ft2 A2 = Area of the top of truncated pyramid = (50 + 3 + 3 + 5 + 5) (50 + 3 + 3 + 5 + 5) = 4,356-ft2

V1 = 5/3 (3,136 + 4,356 + (√(3,136 x 4,356) V1 = 18,647-ft3 = 690-yd3 Step 3: Compute the volume of the basement and subtract this from the total excavation to determine the volume of backfill material. Building Volume = 50’ x 50’ x 5’ = 12,500-ft3 = 463yd3 Backfill stockpile required = 690-yd3 - 463-yd3 = 227-yd3

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AVERAGE END AREA METHOD

fast facts Average end area method is the most widely used method to calculate the volume of soil between stations in a roadway. The format of the equation is shown below:

V =

L( A1 + A2 ) 2

A2 = 544-ft2 at station 19+00

A1 = 725-ft2 at station 18+00

Figure 1

12. - Question Using the information given in Figure 1, the volume of embankment in yd3 is most nearly: a. 1,350-yd3  b. 2,050-yd3  c. 2,250-yd3 d. 2,350-yd3 Solution: Use the average end area method: Volume (yd3) = [(A1 + A2) ÷ (2)] x [(L ÷ 27)] Volume (yd3) = [725 + 544) ÷ (2)] x [(100 ÷ 27)] Volume (yd3) = 2,350-yd3 (answer is d)

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13. - Question The roadbed cross sections at the given stations are shown. The total excavated material (CY) from Station 11+00 to Station 13+00 is most nearly: a. b. c. d.

0 140 1065 1200

CL CL C=350-SF F=0-SF STA 12+00

C=150-SF F=0-SF

CL

STA 11+00

CL

C=0-SF F=75-SF

STA 13+00

STA 14+00

C=0-SF F=275-SF

Solution: Use the Average End Area Method; set up a table to express the calculation steps; solve for the requested information.

Stationing 11+00

Distance (ft) Start

Cut (SF) 350

Fill (SF) 0

12+00

100

150

0

350 + 150 x 100 = 926 2 27

13+00

100

0

75

150 + 0 x 100 = 278 2 27

Total

Cut Volume (CY)

1204

Fill Volume (CY)

Zero

0 + 75 x 100 = 139 2 27

139

Total Cut Volume = 1204-CY; Total Fill Volume = 139-CY; Amount of Export Material = 1204-CY – 139-CY = 1065-CY Amount of borrow volume needed = 0-CY

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14. - Question The roadbed cross sections at the given stations are as shown. The amount of borrow material (CY) from Station 7+00 to Station 9+00 is most nearly: a. b. c. d.

450 1150 1600 2750

CL C=80-SF F=175-SF STA 7+00

CL

STA 8+25

C=165-SF F=210-SF

CL C=265-SF F=310-SF STA 9+00

Solution: Use the Average End Area Method; set up a table to express the calculation steps; solve for the requested information.

Stationing

Cut (SF) 80

Fill (SF) 175

Cut Volume (CY)

7+00

Distance (ft) Start

8+25

125

165

210

80 + 165 x 125 = 567 2 27

175 + 210 x 125 = 891 2 27

9+00

75

265

310

165 +265 x 75 = 597 2 27

210 + 310 x 75 = 722 2 27

Total

1164

Fill Volume (CY)

1613

Total Excavated Material = 1164-CY Total Fill Material = 1613-CY Total Borrow Material = 1164-CY – 1613-CY = - 449-CY (Total borrow material required) Cumulative Amount of Earthwork Moved = 1164-CY + 1613-CY = 2777-CY

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15. - Question For the cross-section areas listed in the Table, determine the following. Apply a soil swell of 25% to fills, if required: 1. Is the project’s earthwork balanced? a. Yes b. No 2. Does it produce waste or require borrow? a. It produces waste b. It requires borrow 3. In response to question # 2 above, the volume in cubic yards is most nearly: a. 1350-yds3 b. 1400-yds3 c. 1450-yds3 d. 1500-yds3

Station 10 + 00 11 + 00 12 + 00 13 + 00 14 + 00 14 + 60 15 + 00 16 + 00 17 + 00 18 + 00 19 + 00 20 + 00

End Area Cut Fill (ft2) (ft2) 0 168 348 371 146 0 0 142 238 305 247 138 106  

[Hint:

See CERM page 80-2; Paragraph 5 – CUT and FILL. In highway work, payment is usually for cut, while in dam work it is usually for fill.]

fast facts The precision obtained from the average end area is generally sufficient unless one of the end areas is very small or zero. In that case, the volume should be computed as a pyramid or truncated pyramid using the equation below. V pyramid = L Abase 3

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Solution:

Station

10+00 11+00 12+00 13+00 14+00 14+60 15+00 16+00 17+00 18+00 19+00 20+00

End Area (ft2) Distance Cut Fill cut vol fill vol (ft) (sf) (sf) (cy) (cy) 0 100 207.4 168 100 955.6 348 100 1331.5 371 100 957.4 146 60 108.2 0 0 40 70.1 142 100 703.7 238 100 1005.6 305 100 1022.2 247 100 713.0 138 100 451.9 106 TOTAL 3560.1

fill vol +25% Use Vpyramid

Use Vpyramid

87.7

Vpyramid

879.6 1256.9 1277.8 891.2 564.8 4958.0

(a) Since Cut and fill quantities are not same, earthwork is NOT balanced (answer is b) (b) Since fill quantity is more than cut quantity, it is required to borrow earth from off-site (answer is b) (c) 4958.0 – 3560.1 = 1398-CY of borrow from off-site is required. (answer is b)

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TOPOGRAPHICAL CONTOUR DATA - CUT AND FILL VOLUMES

 From the topographic data survey of the site, different alternative site plans can be evaluated. The primary goal is to find a balance with cut and fill volumes.  The basic volume calculation is the difference between the desired elevation and the original elevations  Traditionally, elevations for a topographic survey are collected using some type of regular grid system  Using the elevation data and the measured grid system, a threedimensional model of the cut-and-fill volumes can be constructed. Examine the following. Cross Section view of a topographical contour. Examine the cut and fill areas. Existing Elevation

Proposed Elevation

CUT AREAS

FILL AREAS

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Compute the total volume (CY) found in the following contour grid with a reference point of 0.00-ft. 105’ 102’ 105’

102’

100’ 98’ 98’

100’ 50-ft x 50-ft Grid

3-D View of Grid Cell

Plan View

Step 1: Compute the overall area. 2500-ft2 = 50-ft x 50-ft Step 2: Compute the average height 100’ +102’ + 105’ + 98’ 4

= 101.25-ft

Step 3: Compute the total volume of material from the reference point of 0.00-ft. 101.25-ft x 2,500-ft2 = 253,125-ft3 or 9,375-yd3 say 9,500-yd3 27-ft3/yd3

105’

101’

101’

102’

101’

100’

101’

98’

Existing Elevations

Proposed Elevations

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Compute the total volume (CY) of material required in a 50-ft x 50-ft contour grid from the existing contour to the proposed contour point of 101.00-ft.

Step 1: Compute the overall area. 2500-ft2 = 50-ft x 50-ft 105’ Step 2: Compute the average height

Proposed 101’

100’ +102’ + 105’ + 98’ = 101.25-ft 4 Step 3: Compute the difference in height to the required 101-ft and volume.

102’

Cut Areas 100’ 98’

101.25-ft – 101-ft = 0.25-ft 0.25-ft x 2500-ft2 = 625-ft3 or 23.15-yd3 27-ft3/yd3 Excess Material

3-D View of Grid Cell

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Fill Areas

DIFFERENTIAL LEVELING

fast facts A

Horizontal Line

B B

A

Earth’s surface

HB

HA Reference Surface

B

A

Consider two points A and B, and consider that the elevation of A {HA} is known and the elevation of B {HB} is required. What can be done to find HB when HA is given? Utilize differential leveling, the process used to determine the elevation difference between two points. Using a level, the optical line of sight forms a horizontal plane which is the same elevation as the telescope crosshairs. By reading a graduated rod held at a point of known elevation (benchmark) a difference in elevation can be measured and a height of instrument (HI) calculated by adding the rod readings to the elevation benchmark. Once the height of instrument is established, rod readings can be taken on subsequent points and their elevations calculated by subtracting the readings from the height of instrument.

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16. - Question leveling are true: I.

Which of the following statements about differential

Benchmark (BM) – relatively permanent point of known elevation as indicated on the Contract drawings Back sight (BS) – a sight taken to the level rod held at a point of known elevation (either BM or TP (Turning Point)) Height of Instrument (HI) – the elevation of the line of sight of the telescope Foresight (FS) – a sight taken on any point to determine its elevation

II. III. IV.

a. b. c. d. Solution:

 

I & II I, II, & III I, II, III, & IV None All are true, (answer is c)

A total station is an electronic/optical instrument used in surveying.

A theodolite is an optical instrument for measuring both horizontal and vertical angles, as used in triangulation networks.

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17. - Question Based on the information provided in Figure 1, the difference in elevation between BM and TP2 is most nearly: a. b. c. d.

+3.60-ft -11.08-ft -7.48-ft +7.48-ft

 

BS 1.27

FS 4.91 BS 2.33

BM Elev. 356.68

FS 6.17

TP1

[not to scale] TP2

FIGURE 1

Solution: Set up a Table as shown below and insert the known information. Calculate the BM at each point to arrive at the answer of 349.20. Note that the column totals BS and FS provide the total difference in elevation; use this as a check. -7.48 (answer is c) BM + BS = HI HI – FS = TP Elevation Point

BS

HI

BM

1.27

357.95

TP1

2.33

355.37

TP2 Check Sum

+3.60

+

FS

Elevation 356.68

4.91

353.04

6.17

349.20

-11.08

=

-7.48

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SURVEYING WITH CONSTRUCTION APPLICATIONS

The development of a planned community requires the addition of a sanitary manhole to be located at STA. 254+80.3. The proposed doghouse manhole top of pipe elevation (ft.) is most nearly: 18. - Question

a. b. c. d.

425.3 433.3 439.5 441.2

STA. 253+65.7 Invert elev=438.33

Not to scale

Proposed MH STA. 254+80.3 Ground elev=448.33

STA. 256+30.7 Invert elev=429.05

72-in. O.D. Concrete Pipe (10-in Wall Thickness)

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Solution: Compute the horizontal and vertical distance between the existing and proposed. Existing Conditions: Δ Horizontal = (253+65.7) – (256+30.7) = (25,365.7) – (25,630.7) = 265.0-ft Δ Vertical = 438.33 – 429.05 = 9.28-ft Proposed Construction: Δ Horizontal = (254+80.3) - (253+65.7) = (25,480.3) - (25,365.7) = 114.6-ft Δ Vertical = 114.6 x 9.28 = 4.0-ft 265.0 Invert elevation = 438.33 – 4.0 = 434.33-ft Adjust for top of pipe. The top of pipe will be above the calculated invert elevation. Adjustment must include the thickness of the pipe. (72-in – 10-in) ÷ 12-in/ft = 5.17-ft 434.33 + 5.17 = 439.5-ft (answer=c)

Sketch the solution:

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TRIGONOMETRIC LEVELING

19. - Question The surveyor’s notes for the traverse are shown below. The ground elevation (ft) of R is most nearly: a. b. c. d.

432.01 432.58 433.05 433.52 380.56-ft

α = 6°56’40”

R HI=4.85-ft BS = 5.32-ft

Elev. 386.23-ft BM

[not to scale] FIGURE 1

Solution: The elevation of point R can be found from the general equation: Elev BM = elev R + HI – [(Horizontal Distance) tan α] - Backsight Rearrange the equation to find the elevation of point R Elev R = elev BM – HI + HD tan α + BS = 386.23-ft – 4.85-ft + (380.56) tan 6°56’40” + 5.32-ft = 433.05-ft (answer) Note the elevation difference between the BS and HI of 0.47-ft which requires the adjustment to be made to find the ground elevation at point R.

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Conversion, DMS to Decimal: Example coordinate is 37° 36' 30" (DMS) Divide each value by the number of minutes or seconds in a degree: 36 minutes = 0.60 degrees (36/60) 30 seconds = 0.00833 degrees (30/3600) Add up the degrees to get the answer: 37° + 0.60° + 0.00833° = 37.60833 DD

ROUNDING OF NUMBERS - An important feature stressed in the EXAM is the understanding of the impact of the use of significant figures and the rounding-off of numbers. The number of significant digits recorded and used in computations is important in conveying the collected or computed data. A distance measured with a 100-ft. steel tape graduated to tenths of a foot can be given as 236.47. It has 5 significant figures, the “7" being an estimated or interpolated number. A distance recorded as 376.2 does not mean 376.20 unless the value has been measured to hundredths, in which case the zero should have been recorded. SIGNIFICANT DIGITS - The number of digits in a computed result depends upon the certainty of the digits in the measurements. Thus, the sum of 428.61, 25.13, and 4.2 can be carried only to 457.9. In a calculation involving multiplication, division, trigonometric functions, etc., the number of significant digits in an answer should equal the least number of significant digits in any one of the numbers being multiplied, divided, etc. The product of 16.71 and 5.8 provides only 2 significant figures in the answer if the last digit in each factor is an estimated one. The same is true if 5.8 is divided by 16.71. Two units means 2.00000 units, carried to any number of decimal places. Conversely, a distance measured as 2-ft does not mean 2.00000-ft unless some sort of scale which could provide that number of places was used - and again, in that case, the number should be recorded with the proper number of decimal places, not just at “2.” Leading zeros are not significant. For example, 0.00052 has two significant figures: 5 and 2. Trailing zeros in a number containing a decimal point are significant. For example, 12.2300 has six significant figures: 1, 2, 2, 3, 0 and 0.

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CHAPTER 2 ESTIMATING QUANTITIES AND COSTS

Concept

2

CHAPTER

Estimating Quantities and Costs

Estimating

Quantity Takeoff Productivity Analysis

Terminology



(10/2018)

Takeoff Productivity Geometry Cost Analysis

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fast facts

(10/2018)

I.

Estimating is a complex process involving collection of available and pertinent information relating to the scope of a project, expected resource consumption, and future changes in resource costs.

II.

The estimating process involves a combination of evaluating information through a mental process of visualization of the constructing process for the project. This visualization is mentally translated into an approximation of the final cost.

III.

At the outset of a project, the estimate cannot be expected to carry a high degree of accuracy, because little information is known. As the design progresses, more information is known, and accuracy should improve.

IV.

Estimating at any stage of the project cycle involves considerable effort to gather information. The estimator must collect and review all of the detailed plans, specifications, available site data, available resource data (labor, materials, and equipment), contract documents, resource cost information, pertinent government regulations, and applicable owner requirements. Information gathering is a continual process by estimators due to the uniqueness of each project and constant changes in the industry environment.

V.

Unlike the production from a manufacturing facility, each product of a construction firm represents a prototype. Considerable effort in planning is required before a cost estimate can be established. Most of the effort in establishing the estimate revolves around determining the approximation of the cost to produce the one-time product.

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COST ESTIMATING

20. - Question A capital improvement project requires the installation of a property line fence along the 250-ft Northern boundary line. The decorative aluminum fence is constructed of posts spaced at 10-ft centers and an ornate picket infill panel. The material costs for this scope of work is most nearly: a. b. c. d.

$65,771.25 $66,416.60 $68,402.10 $71,065.58

Material Costs: Aluminum Posts

$645.35 -each

Picket Infill Panel

$1,985.50- each

Placed Concrete

$498.00/CY

Ironworker

$78.00/hr

Solution:

Develop a Bill of Materials and multiply quantities by the costs:

Aluminum Posts:

26 x $645.35 = $16,779.10

Picket Infill Panel

25 x $1,985.50 = $49,637.50

 

Grand Total $66,416.60 (answer is b)

fast facts The most common blunder during quantity take–off estimating is to omit the “zero” position during the count. To help with the analysis, sketch the work so as to better visualize the quantity take-off. Remember that “distracters” are included in questions to test your engineering judgment.

(10/2018)

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QUANTITY TAKE-OFFS

21. - Question

The figure represents Revision No. 6 to the foundation wall which was released after the continuous footing measuring 1’-4” high was placed. The B.O.F. elevation is 415’-8”. Apply ACI 318-14, where all footing reinforcement steel exposed to earth requires a minimum 3-inch cover. The total weight of reinforcing steel (lb) to be placed to the uppermost construction joint for this 16’-0” long foundation wall is most nearly: a. b. c. d.

879 890 920 947

Not to Scale

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Solution: Determine height of wall: BOF Elevation + Footing Height = Bottom Elevation of Wall 415’-8” + 1’-4” = 417’-0” Top Of Wall Elevation - Bottom Elevation of Wall = Height of Wall 429’-5” - 417’-0”= 12’-5” Compute total weight of rebar using the Table of Weights provided in the problem. Remember to adjust length of rebar to allow for the design requirement of a minimum of 3-inch of concrete cover.

Quantity Take-Off Item #5 @12” o.c. E.F.E.W #5 @12” o.c. E.F.E.W #5 @12” o.c. E.F.E.W #5 @12” o.c. E.F.E.W #4 @12: o.c.

(10/2018)

Location

Quantity

Length (ft) 15’-6”

Unit Weight (lb/ft) 1.043

Total Weight (lb) 210.17

Horizontal Face 1 Horizontal Face 2 Vertical Face 1 Vertical Face 2 Longitudinal Top of Wall

13 13

15’-6”

1.043

210.17

17

12’-2”

1.043

215.79

17

12’-2”

1.043

215.79

17

6’-0”

0.668

68.14

TOTAL WEIGHT

920.06

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fast facts Cast-In-Place Structural Concrete Wall In the previous example, we calculated the total weight of rebar for the wall. Let’s extend this exercise and determine the weight of rebar per cubic yard of concrete. Typically, the theoretical measure is weight of rebar in pounds per cubic yard of concrete. Compute the volume of concrete: Wall Dimensions 12’-5” high x 16’-0” long x 1’-0” thick = 198.67 cubic feet of concrete. Next, 198.67 cubic feet of concrete / 27CF/CY = 7.35-CY Total weight of concrete at 150-lb/CF: 198.67-CF x 150-lb/CF = 29,800-lb of concrete Total weight of steel (#5 bar at 12-in center to center; each face - each way) = 920lbs. The concrete converts to 7.35-CY, therefore, 920-lb ÷ 7.35-CY = 125-lbs/CY which is a typical structural concrete “rule of thumb” which is in the range of 125-lbs to 150-lbs of reinforcement steel per CY concrete or about 3+%. The numbers vary, but this is a backcheck to the overall computations for structural concrete. The NCEES number for unit weight of a CY of concrete is 150-lbs/CF, or 4050-lbs/CY. Therefore, Rebar @ 920-lbs ÷ Concrete @ 7.35-CY = 125-lb/CY Finally, percent weight of steel in CIP Structural Concrete: 920-lbs / 29800-lbs = 0.03087; Say 3%. The most probable weight (lbs) of rebar in a 25,000-CY cast in place structural concrete foundation project is most nearly: a. 500,000 b. 1,000,000 c. 2,000,000 d. 3,000,000 Solution: 25,000-CY x 120-lbs/CY = 3,000,000-lbs

(10/2018)

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ESTIMATING TAKEOFF QUANTITIES

22. - Question A 450-ft length canal is to be lined with concrete for erosion control. With10% waste and overbreak, the material unit costs of concrete and curing compound based on other recent projects in the area with similar volumes are: $98/yd3; and, $40.00 per 5-gal, respectively. Project specifications require an application rate of curing compound at 1gal per 300-ft2.

8-in - Concrete

20-ft

2 3

[not to scale]

19-ft

Determine the following: a. The total material cost for delivered concrete is most nearly: a. b. c. d.

$85,000.00 $90,160.00 $99,176.00 $109,094.00

b. The total material cost for the concrete curing compound is most nearly: a. b. c. d.

$1,120.00 $1,160.00 $1,200.00 $1,240.00

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Solution: a.

Horizontal length of side slope = 20-ft x (3 ÷ 2) = 30-ft Slope length = √ [(20-ft)2 + (30-ft)2] = 36.06-ft Cross-sectional area of lining = [(2 x 36.06-ft)+19-ft] 8-in/12-in/ft = 60.74-ft2

Volume of lining = (60.74-ft2 x 450-ft) / 27ft3/yd3 = 1,012-yd3 Delivered volume (add waste) = 1,012-yd3 x 1.10 = 1,113-yd3 Material Cost = $98.00/yd3 x 1,113-yd3 = $109,094. (answer is d) b.

Surface Area of Canal = (19-ft + 36.06-ft x 2) x 450-ft = 41,004 ft2 Quantity of Curing Compound = 41,004 ft2 / (300 ft2/gal.) = 136.68gal. Calculate waste: 136.68-gal. x 1.10 = 150.35-gal Convert to purchase within 5-gal containers:150.35-gal/ 5-gal = 30.07 containers Material Cost = 31-containers x $40.00/container = $1,240.00 (answer is d)

fast facts The manufacturer’s specification cannot be deviated from. This question illustrates the importance of rounding up to meet the product specifications. The seven-hundredths of a 5-gallon container in the example are enough to support the manufacturer’s position that the coverage rate was not met. Always round up in this situation.

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ESTIMATING FORMWORK

23. - Question A A concrete crew will use the available steel form panels measuring 2’-6” wide x 4’-0” high to construct a 40’-6” long by 3’-2” high by 1’-6” wide concrete knee wall. The square foot of contact area for the formwork is most nearly: a. b. c. d.

105-ft2 134-ft2 266-ft2 368-ft2

Solution: The square foot of contact area consists of the surface of the formwork “touched” by the concrete. Therefore, apply the equation to calculate the contact area: (40.5-ft + 1.5-ft + 40.5-ft + 1.5-ft) x 3.167-ft = 266.03-ft2 (answer is c) 40’-6”

1’-6”

Plan View 1’-6”

Concrete Knee Wall

3’-2”

Not to scale Section View (10/2018)

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fast facts Cost per Square Foot of Contact Area The cost of concrete formwork is influenced by three factors: 1. Initial cost or fabrication cost, which includes the cost of transportation, materials, assembly, and erection. 2. Potential reuse which decreases the final total cost per square foot of contact area. The data in Table indicates that the maximum economy can be achieved by maximizing the number of reuses. 3. Stripping costs, this also includes the cost of cleaning and repair. This item tends to remain constant for each reuse up to a certain point at which the total cost of repairing and cleaning start rising rapidly. In deciding to use a specific formwork system, the initial cost should be evaluated versus the available budget for formwork cost. Some formwork systems tend to have a high initial cost, but through repetitive reuse, they become economical. For example, slip forms have a high initial cost, but the average potential reuse (usually over 100 times) reduces the final cost per square foot of contact area for the type of formwork. In the case of rented formwork systems, the period of time the formwork is in use has a great effect on the cost of the formwork.

(10/2018)

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BUILDING MATERIALS – ROOF SURFACE MATERIALS

24. - Question The plan view for a roof on a wood frame house is as shown. Using a waste factor of 15%, the quantity of plywood sheets (4-ft x 8-ft) required for sheathing the hip roof having a 6/12 pitch is most nearly: a. b. c. d.

48 54 62 64 48-ft

32-ft

Plan View Not to Scale

Solution: Based on the 16-ft center to the roof ridge, the ridge height of the roof is (16-ft) (6/12) = 8-ft 12 6

32-ft

The length of the roof slope = √ (16-ft)2 + (8-ft)2 = 17.89-ft Since the area of a hip roof is the same as the area of a gable roof then: (2-sides) (48-ft) (17.89-ft) = 1,717.44-SF Plywood sheathing is available in 4-ft x 8-ft = 32-SF/Sht (1,717.44-SF + 15% waste) ÷ 32-SF/Sht = 61.72 = 62-sheets (answer)

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Roof pitch is determined by finding the amount of rise per foot run. It is represented by a triangular shaped drawing and expressed in inches, 4/12, 5/12, etc. the higher the number the steeper the pitch or angle of incline.

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Orientation of 4’ x 8’ Sheets

The illustrations on this page show a 3-dimensional perspective of the plan view of a roof drawing.

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ESTIMATING BRICK MASONRY

Brick Dimensions There are three different sets of dimensions used with brick:

1. Nominal 2. Specified 3. Actual Each must be used with care and accuracy to avoid confusion during design and construction.

Brick positions in a wall Nominal dimensions. Nominal dimensions apply to modular brick and are the result of the specified dimension of the brick plus the intended thickness of its intended mortar joint. Generally, these dimensions will fall into “round” numbers to produce modules of 4 in. or 8 in.

Specified dimensions. Specified dimensions are the anticipated manufactured dimensions of the brick, without consideration for mortar, which are to be used in project specifications and purchase orders. They are also used by the structural engineer in rational design of brick masonry. In non-modular construction, only the specified dimensions are used.

Actual dimensions.

Actual dimensions are the measurements of the brick as manufactured. Generally the actual dimensions will be within a tolerance of the specified dimensions. The allowable tolerances are dependent upon the type and size of the brick and are given within the applicable ASTM standard specifications, such as those in ASTM C216, Standard Specification for Facing Brick and C652, Standard Specification for Hollow Brick.

Modular and Non-Modular Brick Modular brick are sized such that the specified dimension plus the intended mortar joint thickness equal a modular dimension. Generally, modular dimensions are whole numbers without fractions that result in modules of 4 in. or 8 in. A modular brick has a set of nominal, specified and actual dimensions as referenced above. A non-modular brick has a set of specified and actual dimensions but does not have nominal dimensions. Brick are available in many sizes and are referred to by many different names, depending on region. In addition, the name of a brick and its size, whether modular or non-modular, can vary depending on the manufacturer.

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ESTIMATING BRICK MASONRY There are various methods to estimate material quantities on a project. Hand calculations are used because of its simplicity and accuracy, the most widely used estimating procedure is the “wall-area” method. It consists simply of multiplying the net wall area (gross areas less areas of openings) by known quantities of material required per square foot (square meter). Determining the area of brick and mortar within each unit area of wall depends on both brick size and joint width. For non-modular masonry, both dimensions must be known to make accurate estimates. In modular masonry, mortar joint sizes are dictated by the size of the brick, simplifying the estimating process.

Bond Pattern For most brick sizes, one-half running bond is the basic pattern when laying a wall or pavement; i.e., approximately half of the brick’s length overlaps the brick below. This pattern is the most frequently used pattern in homes, schools and offices. However, some sizes lend themselves best to other bond patterns. As an example, a utility-sized brick has a nominal length three times its nominal thickness.

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BRICK VENEER QUANTITIES

25. - Question A brick veneer is planned for the front elevation of a structure 126-ft wide x 12-ft high and has 26% of the exterior surface allotted for openings. The specified size of the selected brick is: 2-5/8”h x 3-5/8”w x 7-5/8” l; and, the owner selected a 3/8” concave tooled mortar joint. The quantity of bricks needed (count) for a running bond pattern is most nearly: a. 6,700 b. 7,500 c. 9,000 d. 9,500 Solution: STEP 1: Compute the length of the brick and mortar joint to determine the area of each brick. Mortar

7-5/8” + 3/8” = 8-in long 2-5/8” + 3/8” = 3-in high Brick

STEP 2: Compute the area of the brick veneer. Total Area = 126-ft x 12-ft = 1,512-ft2 Total Area of Brick = 8-in x 3-in = 24-in2 ÷ 144-in2/ft2 = 0.167-ft2 Number of Bricks = 1,512-ft2 ÷ 0.167-ft2 = 9,054-bricks Reduce amount for 26% deducts: 9,054-bricks x 0.74 = 6,700-bricks (answer)

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COMPUTING GEOMETRIC PROPERTIES

Y inside Y outside thickness

X inside

Plan View

X outside

Equations to find geometric properties of a dimensional shape: Xinside = Xoutside – 2 thickness Yinside = Youtside – 2 thickness Outside Perimeter = 2(Xoutside + Youtside) Inside Perimeter = 2(Xinside + Yinside) To calculate the Cross Section Area use the following: Cross Sectional Area = 2(thickness)(Xoutside + Yinside) = 2(thickness)(Xinside + Youtside) To calculate the outside perimeter of a shape with recesses along a face, use the following: Outside Perimeter = (2) (Length + Width + Recess) Inside Perimeter = Outside Perimeter + [(4) (2 x ( - thickness))] Use the mean perimeter to determine the volume of the shape: Mean Perimeter = Outside Perimeter – [(4) (2 x (thickness/2))]

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QUESTION A – C: The plan view for the foundation of a proposed building construction is as shown. The volume of concrete required for the construction of the cast in place foundation wall designed to be 10-inch thick and 10-ft high is most nearly: 20-ft 15-ft 45-ft

25-ft 65-ft 25-ft Plan View Not to Scale

25-ft

20-ft

50-ft

20-ft

8-ft

Question A: The outside perimeter of the wall is most nearly: a. b. c. d.

277 301 326 347

Outside Perimeter = 2 x (length + width + recess along the face) = 2 x (90-ft + 65-ft + 8-ft) = 326-ft

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Question B: The theoretical volume of concrete (CY) required to construct the wall is most nearly: a. b. c. d.

85 90 91 100

Outside Perimeter = 2 x (90-ft + 65-ft + 8-ft) = 326-ft Mean Perimeter = 326-ft – (4) [2 x (10/12 ÷ 2)) = 322.67-ft Volume of Concrete = 322.67-ft x (10-ft x 10/12) = 2,685.5 / 27 = 99.58-CY Question C: The required cast in place formwork (SFCA) needed for the foundation wall construction is most nearly: a. b. c. d.

3,260 4,520 6,450 8,450

Outside Perimeter = 2 x (90-ft + 65-ft + 8-ft) = 326-ft Inside Perimeter = 326-ft + (4) (2 x (- 10/12)) = 319.33-ft Outside Forms = 326-ft x 10-ft = 3,260-SFCA Inside Forms = 319.33-ft x 10-ft = 3,193.33-SFCA Total Formwork = 3,260-SFCA + 3,193.33-SFCA = 6,453-SFCA Inside Perimeter Mean Perimeter

Outside Perimeter

Formwork

Concrete (10/2018)

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PAINTING QUANTITIES

26. - Question A painting contractor uses a productivity standard table to calculate a cost proposal. The total cost ($) to paint a rectangular shaped floorplan conference room that is 120-ft by 200-ft with a 2% reduction for wall openings and a $2,000 allowance for specialty trim work is most nearly: a. b. c. d.

9,500 10,500 11,500 13,500

Productivity Standard for Surfaces -- 12-ft high Activity

Ceiling Surface Wall Surfaces Bookcases Tables Wall Openings

Material Coverage (SF/GAL) 230 265 125 100 125

Material Cost ($ per Gallon) 38.50 45.60 66.80 75.00 65.00

Labor Productivity (SF/HR) 225-SF/HR 245-SF/HR 125-SF/HR 100-SF/HR 50-SF/HR

Burdened Labor Cost ($/HR) 38.50 42.50 72.85 86.44 64.50

Solution: Compute areas; both wall and ceiling. Wall Surface = 120-ft + 200-ft + 120-ft + 200-ft = 640-ft x 12-ft = 7,680-SF Less 2% factor for openings or minus 153.6-SF = 7,526.4-SF Ceiling Surface Area = 120-ft x 200-ft = 24,000-SF Use the computed surface to determine the required quantities. Answer d

Productivity Standard for Surfaces -- 12-ft high Activity

Material Coverage (SF/GAL)

Material Cost ($ per Gallon)

Material ($/SF)

Labor Productivity (SF/HR)

Burden Labor Cost ($/HR)

Labor ($/SF)

Total $/SF (Labor + Material

Ceiling

230

38.50

0.1674

38.50

0.1711

0.3385

24,000

8,124.00

Wall Surfaces Wall Openings

265

45.60

0.1721

42.50

0.1735

0.3456

7,526.4

2,654.21

125

65.00

0.52

225SF/HR 245SF/HR 50-SF/HR

64.50

1.29

1.81

Total SF

153.6 Grand Total

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Total Cost

278.02 $11,056.23 plus $2,000 allowance

ESTIMATING LABOR COSTS

27. - Question A contractor will place 90-cubic yards of concrete for a housekeeping pad on the rooftop of a 36-ft tall building. Site conditions dictate that the safest and best method of placement is to use a crane and a 2-cubic-yard bucket. To perform the task efficiently, five Union laborers are needed — one at the concrete truck, three at the point of placement, and one on the portable internal vibrator. The wage rate for laborers is $22.00/hr (Union overtime rate is 1.5 times the wage rate after 8-hour day). Time needed for the operation is: Setup: 15-min; Cycle time consists of Load = 3-min., plus Swing, dump and return = 6-min. which allows a total cycle time = 9-min; Demobilize operation, 10-min. Supervision is done by the superintendent. Allow a 10% factor for inefficiencies during the cycle time. Crane rental cost is $1,800 per 8-hr day. The total labor cost per cubic yard for the concrete crane placement of the 90 cubic yards is most nearly: a. b. c. d.

$8.75 $9.78 $10.12 $10.65

Solution: Identify (by underlining) relevant cost items and calculate summary quantities. No. of cycles 90-CY/2-CY/Bucket = 45 cycles Total cycle time 45-cycles x 9-min/cycle = 405 min Inefficiency(labor,delays,etc.)10%of cycle time = 41-min Setup and demobilize: Sub-total = 25 min Total operation time: 405 + 41 + 25 = 471-min or 7.85-hrs Amount of time needed (adjusted to workday) = 8 hr Laborers — five for 8 hours at $22.00/hr = $880.00 3 = $880.00 Total labor cost per 90-yd 3 = $9.78/yd3 Cost per cubic yard $880/90-yd (answer is b)

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EQUIPMENT PRODUCTION

28. - Question An earthwork excavation operation is under contract to transport a quantity of 3,000-CY of borrow material. The excavation crew consists of an excavator with a production rate of 200 CY/day, a loader production of 250 CY/day, and 3-dump trucks moving a total amount of 150-CY/day. The crew formation required to minimize the duration of the operation will need to add which of the following: a. b. c. d.

1-excavator 1-loader 2-loaders 1-dump truck

Solution: Examine all of the work activities and establish the baseline provisions issued in the problem statement, namely: Examine using one excavator: Duration = 3,000-CY / 200-CY/Day = 15-days Examine using one loader: Duration = 3,000-CY / 250-CY/Day = 12-days Examine using 3-trucks: Duration = 3,000 / 150-CY/Day = 20-days Conclusion - the activity duration is governed by the lowest production rate of a total of 20-days. By inspection, this is an unbalanced workflow crew where the loader is not working with full capacity; the production rate of this crew could be adjusted by increasing the number of trucks from 3 to 4 trucks. The outcome would allow a balanced mix of resources: whereby the adjusted crew becomes: use 1-loader, 1-excavator, and 4-trucks (answer). Accordingly, the activity duration for the 4-trucks = 3,000-CY / 200-CY/Day = 15-days.

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LABOR PRODUCTIVITY

29. - Question A concrete specialty company’s uses productivity standards calculated as averages from historical data. On average 100square feet of formwork requires 6 hours of carpenter time and 5 hours of common laborer time. The wage rate for a carpenter is $38.97/hr plus a 54% burden rate. The wage rate for common laborers is $14.87/hr plus a 48% burden rate. The total square foot cost is most nearly: a. $3.08/ft2 b. $3.25/ft2 c. $4.25/ft2 d. $4.70/ft2 Solution: Burden Rates are amounts charged over and above the actual costs for labor, materials and/or taxes. Add the allotted burden rate to the trade labor rate to determine the total SF cost. The unit cost may be calculated as follows: Carpenter — 6 hr at $60.00/hr Laborer — 5 hr at $22.00/hr Total labor cost for 100 ft2

= $360.00 = $110.00 = $470.00

Labor cost per ft2 = $470.00 ÷ 100-ft2 = $4.70-ft2

(answer is d)

fast facts Labor burden is the cost to a company to carry their labor force aside from salary actually paid to them. Simply stated, burden is the benefits and taxes that a company must pay on their payroll. It is important to stress that burden typically should not include any profit, markup or expenses unrelated to employee compensation, but should be the actual cost to carry the labor. These can include, but are not limited to, all of the following: • • • • • • • • •

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Payroll Taxes – both Federal and State (Statutory) When applicable, Union Fringe Benefits Package Vacation Pay allocation Retirement/Pension Costs Health Care Life/Accidental Death & Dismemberment Insurance (AD&D) Worker’s Compensation Costs Long-Term Disability Insurance Short-Term Disability Insurance

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fast facts Common Estimating Blunders: • Count the “0” position • Take the “Outs” Out • OH&P are cumulative not additive • Round Up material quantities • Follow mfg’s application rate • Include the given waste amount • Use product coverage Qtys’ • Follow bid document info • Calculate work crew rates • Use burdened labor rates • Sketch the work • Construction Estimating calculates the total fully burden cost for labor, material and equipment. Once the “raw” costs are determined, the contractor’s Over Head and Profit (plus Bond) are added to the bottom line costs.

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Construction Operations and Methods

Concept Terminology



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CHAPTER

CHAPTER 3 CONSTRUCTION OPERATIONS AND METHODS

3

Construction Operations

Properties of Materials Crane Safety Equipment Production

Actual vs. Ultimate Strength Breaking Strength Wire Rope IWRC; FC Elastic Stretch Design Factors Lifting Load Crane Stability Standard Productivity Operating Costs Job Size Productivity NPDES SWPPP

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A GUIDE TO CRANE SAFETY The character and magnitude of construction loads produced during heavy lifting, rigging, and handling operations may be significantly different from the service loads for which a particular installation has been designed. These differences mandate the necessity for an evaluation of their effects to ensure that construction will proceed in a safe and expeditious manner, consistent with the project plan, and will result in a highquality installation upon its completion. To guard against failure of a rigging component due to shock load, overload, wear, etc., the load being lifted should not exceed the Safe Working Load (SWL). The SWL is calculated as a fraction of the weakest component’s actual breaking strength. Breaking strength is the measured load required to “break” the component. SWL is calculated by dividing that breaking strength, as identified by the manufacturer or a professional engineer, by a “factor of safety”.

Breaking Strength = Safe Working Load Factor of Safety

Source: http://www.nclabor.com/osha/etta/indguide/ig20.pdf

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Note that the Table is applicable to Symmetrical Loads ONLY

The figures illustrate the various types of lifting configurations during the hoisting of loads. Critical to all lifts is the balance of the load on the rigging equipment and the load itself. Source: http://www.nclabor.com/osha/etta/indguide/ig20.pdf

(10/2018)

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MECHANICAL PROPERTIES OF MATERIALS

fast facts Knowledge of the mechanical properties is obtained by testing materials. Results from the tests depend on the size and shape of material to be tested (specimen), how it is held, and the way of performing the test. The most common procedures, or standards, that are used in Construction are published by the ASTM. Strength, hardness, toughness, elasticity, plasticity, brittleness, and ductility and malleability are mechanical properties used as measurements of how metals behave under a load. These properties are described in terms of the types of force or stress that the metal must withstand and how these are resisted. Common types of stress are compression, tension, shear, torsion, impact, or a combination of these stresses, such as fatigue. Compression stresses develop within a material when forces compress or crush the material. A column that supports an overhead beam is in compression, and the internal stresses that develop within the column are compression. Tension (or tensile) stresses develop when a material is subject to a pulling load; for example, when using a wire rope to lift a load or when using it as a guy to anchor an antenna. "Tensile strength" is defined as resistance to longitudinal stress or pull and can be measured in pounds per square inch of cross section. Shearing stresses occur within a material when external forces are applied along parallel lines in opposite directions. Shearing forces can separate material by sliding part of it in one direction and the rest in the opposite direction.

30. - Question The breaking strength of a material is also known as its: a. Ultimate Strength b. Yield Point  c. Proportional Limit  d. Elastic Limit Solution: This question aids to further Review Stress Strain Curves of Mechanical Properties of Materials as shown in Figures 1, 2 and 3 below:

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Stress is force per unit area and is usually expressed in pounds per square inch. If the stress tends to stretch or lengthen the material, it is called tensile stress; if to compress or shorten the material, a compressive stress; and if to shear the material, a shearing stress. Tensile and compressive stresses always act at right-angles to (normal to) the area being considered; shearing stresses are always in the plane of the area (at right-angles to compressive or tensile stresses). Unit strain is the amount by which a dimension of a body changes when the body is subjected to a load, divided by the original value of the dimension. The simpler term strain is often used instead of unit strain. Proportional limit is the point on a stress-strain curve at which it begins to deviate from the straight-line relationship between stress and strain. Elastic limit is the maximum stress to which a test specimen may be subjected and still return to its original length upon release of the load. A material is said to be stressed within the elastic region when the working stress does not exceed the elastic limit, and to be stressed in the plastic region when the working stress does exceed the elastic limit. The elastic limit for steel is for all practical purposes the same as its proportional limit. Yield point is a point on the stress-strain curve at which there is a sudden increase in strain without a corresponding increase in stress. Not all materials have a yield point. Yield strength, Sy, is the maximum stress that can be applied without permanent deformation of the test specimen. Ultimate strength, Su, (also called tensile strength) is the maximum stress value obtained on a stress-strain curve. (answer is a) Modulus of elasticity, E, (also called Young's modulus) is the ratio of unit stress to unit strain within the proportional limit of a material in tension or compression. Modulus of elasticity in shear, G, is the ratio of unit stress to unit strain within the proportional limit of a material in shear.

Poisson's ratio, is the ratio of lateral strain to longitudinal (axial) strain for a given material subjected to uniform longitudinal stresses within the proportional limit. The term is found in certain equations associated with strength of materials. The ratio of lateral strain to the axial strain for most metals is approximately 0.3. Poisson’s ratio applies only to elastic strain. When the stress is removed, the lateral strain disappears along with the axial strain.

(10/2018)

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FATIGUE: The normative view of the stress/strain is found in the example of the paper clip wire experiment. Take a paper clip wire and try to break it by pulling on it by hand. You are unable to break it because you need to exceed the ultimate stress of the material. Bend a paper clip between your fingers and the following is observed: when the wire rebounds to its original shape the proportional/elastic limit is observed; bend it farther to a point where the wire is permanently deformed, the yield point is observed; and, when the wire is bent one way then the other a few times, the wire breaks easily. The phenomenon is known as fatigue. All materials have microcracks. The cracks are not critical and are averaged as ultimate strength for the bulk material in a tension test. However, when the material is subjected to cyclic loading, the microcracks can grow until some of the cracks reach critical length, at which time the material breaks. The stress value at cyclic loading is significantly lower than ultimate stress of the material. Failure due to cyclic loading at stress levels significantly lower than the static ultimate stress is called fatigue. All engineered systems account for fatigue failure. At low levels of stress the failure may occur at millions or billions of cycles. Endurance tests determine the cyclic loadings and fatigue strength.

Ductility is the ability of a material to deform before it fractures. A material that experiences very little or no plastic deformation upon fracture is considered brittle. A reasonable indication of a fastener’s ductility is the ratio of its specified minimum yield strength to the minimum tensile strength. The lower this ratio the more ductile the fastener will be.

Toughness is a materials ability to absorb impact or shock loading. Impact strength toughness is rarely a specification requirement. A material is Brittle if, when subjected to stress, it breaks without significant deformation (strain). Brittle materials absorb relatively little energy prior to fracture, even those of high strength. Breaking is often accompanied by a snapping sound. Brittle materials include most ceramics and glasses (which do not deform plastically). Many steels become brittle at low temperatures.

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PROPERTIES OF METALS

AA. Ductile cast iron and gray cast iron both contain 4% carbon. However, ductile cast iron has a higher tensile strength and is considerably more ductile. Which of the following statements is the major difference that accounts for superior properties of ductile iron: a. The gray cast iron contains iron carbide, while the ductile iron contains graphite b. The gray cast iron contains flakes of graphite, while ductile iron contains spheroids of graphite c. The ductile iron is tempered to give better properties d. The ferrite grains in the gray cast iron are excessively large Answer B, Gray cast iron contains flakes of graphite while ductile cast iron contains spheroids. The major difference in the shape of the graphite gives cast iron approximately twice the tensile strength and 20-times the ductility of the gray cast iron. BB.

Which of the following statements regarding steel composition is FALSE?

a. High strength, low alloy steels are not as strong as non-alloy, low carbon steels b. Small amounts of copper increase the tensile strength of steels c. Small amounts of silicon in steels have little influence on toughness or fabricability d. Addition of small amounts of silicon to steel can cause a marked decrease in yield strength of the steel.

Answer: D is false as the addition of small amounts of silicon to steel increases both the yield strength and tensile strength.

CC.

Which of the following statements regarding stainless steel is FALSE?

a. Stainless steels contain large amounts of chromium b. There are three basic types of stainless steels: martensitic, austenitic, and ferritic. c. The non-magnetic stainless steels contain large amounts of nickel d. Stabilization of the face-centered cubic crystal structure of stainless steels imparts a non-magnetic characteristic of the alloy.

Answer: B, as there are only two types of stainless steels, namely: (1) magnetic (martensitic or ferritic) and, (2) Non-Magnetic (austenitic)

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DD. Low-carbon steels are generally used in the “as rolled” or “as fabricated” state when producing shapes. The primary reason for this method of production is: a. b. c. d.

There are many types of shapes and thicknesses Their strength generally cannot be increased by heat treatment They degrade severely during heat treatment Their chromium content is very low

Answer: B, since low-carbon steels strength cannot be increased by heat treatment, low-carbon steels are used as fabricated without further treatments. In metallurgy, there are many terms that have very specific meanings within the field, such as "hardness," "impact resistance," "toughness," and "strength". Some of the terms encountered, and their specific definitions are: Strength: also called rigidity, this is resistance to permanent deformation and tearing. Strength, in metallurgy, is usually divided into yield strength (strength beyond which deformation becomes permanent), tensile strength (the ultimate tearing strength), shear strength (resistance to transverse, or cutting forces), and compressive strength (resistance to elastic shortening under a load). Toughness: Resistance to fracture, as measured by the Charpy test. Toughness often increases as strength decreases, because a material that bends is less likely to break. Hardness: Hardness is often used to describe strength or rigidity but, in metallurgy, the term is usually used to describe a surface's resistance to scratching, abrasion, or indentation. In conventional metal alloys, there is a linear relation between indentation hardness and tensile strength, which eases the measurement of the latter. Brittleness: Brittleness describes a material's tendency to break before bending or deforming either elastically or plastically. Brittleness increases with decreased toughness, but is greatly affected by internal stresses as well. Plasticity: The ability to mold, bend or deform in a manner that does not spontaneously return to its original shape. This is proportional to the ductility or malleability of the substance. Elasticity: Also called flexibility, this is the ability to deform, bend, compress, or stretch and return to the original shape once the external stress is removed. Elasticity is inversely related to the Young's modulus of the material. Impact resistance: Usually synonymous with high-strength toughness, it is the ability resist shock-loading with minimal deformation. Wear resistance: Usually synonymous with hardness, this is resistance to erosion, ablation, spalling, or galling. Structural integrity: The ability to withstand a maximum-rated load while resisting fracture, resisting fatigue, and producing a minimal amount of flexing or deflection, to provide a maximum service life.

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Processing of Steels Tempering is most often performed on steel that has been heated above its upper critical temperature and then quickly cooled, in a process called quenching, using methods such as immersing the red-hot steel in water, oil, or forced-air. The quenched-steel, being placed in or very near its hardest possible state, is then tempered to incrementally decrease the hardness to a point more suitable for the desired application. The hardness of the quenched-steel depends on both cooling speed and on the composition of the alloy. Steel with a high carbon-content will reach a much harder state than steel with a low carbon-content. Likewise, tempering high-carbon steel to a certain temperature will produce steel that is considerably harder than low-carbon steel that is tempered at the same temperature. The amount of time held at the tempering temperature also has an effect. Tempering at a slightly elevated temperature for a shorter time may produce the same effect as tempering at a lower temperature for a longer time. Tempering times vary, depending on the carbon content, size, and desired application of the steel, but typically range from a few minutes to a few hours. Tempering quenched-steel at very low temperatures, between 66 and 148 °C (151 and 298 °F), will usually not have much effect other than a slight relief of some of the internal stresses. Tempering at higher temperatures, from 148 to 205 °C (298 to 401 °F), will produce a slight reduction in hardness, but will primarily relieve much of the internal stresses. Tempering in the range of 260 and 340 °C (500 and 644 °F) causes a decrease in ductility and an increase in brittleness, and is referred to as the "tempered martensitic embrittlement" (TME) range. Except in the case of blacksmithing, this range is usually avoided. Steel requiring more strength than toughness, such as tools, are usually not tempered above 205 °C (401 °F). Instead, a variation in hardness is usually produced by varying only the tempering time. When increased toughness is desired at the expense of strength, higher tempering temperatures, from 370 to 540 °C (698 to 1,004 °F), are used. Tempering at even higher temperatures, between 540 and 600 °C (1,004 and 1,112 °F), will produce excellent toughness, but at a serious reduction in the strength and hardness. At 600 °C (1,112 °F), the steel may experience another stage of embrittlement, called "temper embrittlement" (TE), which occurs if the steel is held within the TE temperature range for too long. When heating above this temperature, the steel will usually not be held for any amount of time, and quickly cooled to avoid temper embrittlement. Embrittlement occurs during tempering when, through a specific temperature range, the steel experiences an increase in hardness and a reduction in ductility, as opposed to the normal decrease in hardness that occurs to either side of this range.

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One-step embrittlement usually occurs in carbon steel at temperatures between 230 °C (446 °F) and 290 °C (554 °F), and was historically referred to as "500 degree [Fahrenheit] embrittlement." This embrittlement occurs due to the precipitation of needles or plates, made of cementite, in the interlath boundaries of the martensite. Impurities such as phosphorus, or alloying agents like manganese, may increase the embrittlement, or alter the temperature at which it occurs. This type of embrittlement is permanent, and can only be relieved by heating above the upper critical temperature and then quenching again. However, these microstructures usually require an hour or more to form, so are usually not a problem in the blacksmithmethod of tempering. Two-step embrittlement typically occurs by aging the metal within a critical temperature range, or by slowly cooling it through that range, for carbon steel, this is typically between 370 °C (698 °F) and 560 °C (1,040 °F), although impurities like phosphorus and sulfur increase the effect dramatically. This generally occurs because the impurities are able to migrate to the grain boundaries, creating weak spots in the structure. The embrittlement can often be avoided by quickly cooling the metal after tempering. Twostep embrittlement, however, is reversible. The embrittlement can be eliminated by heating the steel above 600 °C (1,112 °F) and then quickly cooling

Annealing consists of heating a metal to a specific temperature and then cooling at a rate that will produce a refined microstructure, either fully or partially separating the constituents. The rate of cooling is generally slow. Annealing is most often used to soften a metal for cold working, to improve machinability, or to enhance properties like electrical conductivity. In ferrous alloys, annealing is usually accomplished by heating the metal beyond the upper critical temperature and then cooling very slowly, resulting in the formation of pearlite. In both pure metals and many alloys that cannot be heat treated, annealing is used to remove the hardness caused by cold working. The metal is heated to a temperature where recrystallization can occur, thereby repairing the defects caused by plastic deformation. In these metals, the rate of cooling will usually have little effect. Most non-ferrous alloys that are heat-treatable are also annealed to relieve the hardness of cold working. These may be slowly cooled to allow full precipitation of the constituents and produce a refined microstructure. Ferrous alloys are usually either "full annealed" or "process annealed." Full annealing requires very slow cooling rates, in order to form coarse pearlite. In process annealing, the cooling rate may be faster; up to, and including normalizing. The main goal of process annealing is to produce a uniform microstructure. Non-ferrous alloys are often subjected to a variety of annealing techniques, including "recrystallization annealing," "partial annealing," "full annealing," and "final annealing." Not all annealing techniques involve recrystallization, such as stress relieving.

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SIMPLE BEAM ANALYSIS

The net effect on the amount of moment and shear between a uniformly distributed load on a simple beam and an equivalent concentrated load on a simple beam is most nearly:

31. - Question

a. b. c. d.

0.25 1 2 2.5

Solution: Review and apply the moment equations between a simple beam uniformly distributed load, that is: calculate wl2/8; and, a simple beam concentrated load at the center, that is: calculate Pl/4; or at any point, that is: Pab/l; and, the answer is that the moment and shear are doubledanswer. w = 800-lb/ft

Simple Beam – Uniform Load Case 1

4-ft R

R P= 3200-lb 2-ft

Simple Beam – Concentrated Load Case 2

2-ft

R

R P= 3200-lb

Simple Beam – Concentrated Load On Any Point Case 3

R

R

3.5-ft 6-in

Loading Condition Reaction or Shear at RL, lb Reaction or Shear at RR, lb Maximum Bending Moment (in-lb)

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Case 1

Case 2

Case 3

1600 1600 19,200

1600 1600 38,400

2800 400 16,800

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DESIGN FACTOR COMPARISON

32. - Question A rigging contractor is surveying the crane lift of mechanical equipment to the building’s rooftop 100-ft above grade level. The allowable construction lifting load capacity (lb) on a 6 x 19 classification ½” wire rope (IWRC; plow steel) in a vertical lifting configuration is most nearly: a. b. c. d.

4,400 6,200 20,000 21,600

 

Breaking strength of wire rope is the allowable load placed on the wire rope differs in the factor of safety imposed by code and enforced by regulatory agencies. Note the difference of the capacity of the wire rope shown in Table 11.2 and compare it to the allowable loads in Table 11.6 with the design factor = 5:1. As in the example, the breaking strength of a ½” wire rope found in Table 11.2 shows 21,600-lb (IWRC) (plow steel), however the same ½” wire rope in a vertical lifting configuration is allowed 4,400-lb which includes the 5:1design factor. (answer is a) Solution:

fast facts Rope is classified by the construction type. For example, a 6x7 FC consists of 6 strands of 7 wires each, wrapped around a fiber core. A 6 x 19 IWRC has (6) 19 wire strands wrapped around an independent wire rope core. Strength of wire rope is dependent on the components materials. Grades include: traction steel (TX), mild plow steel (MPS), plow steel (PS), improved plow steel (IPS), and extra improved plow steel (XIPS). Wire rope is protected against corrosion by lubrication carried in the saturated fiber core. The wire may be constructed of iron, stainless steel, monel, or bronze, but for construction uses it is nearly always high carbon steel. The core, which is the foundation for the wire rope, is made of either fiber or steel. The most commonly used cores are: fiber core (FC); independent wire rope core (IWRC); and wire strand core (WSC).

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Source: http://www.nclabor.com/osha/etta/indguide/ig20.pdf

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Source: http://www.nclabor.com/osha/etta/indguide/ig20.pdf

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WIRE ROPE STRETCH

33. - Question A 70-ft long, ¾” diameter, 6 x 7 FC wire rope is resisting a tension force of 12-kips. The total stretch (in) of the wire rope given the following properties is most nearly: E = 10,000-ksi; constructional stretch = 0.75%; As = 0.288-in2 a. b. c. d.

0.75-in 3.5-lb 6.3-in 9.8-in

Solution: The elongation or “stretch” of wire ropes must be considered in designing temporary bracing and lifting configurations. Elongation comes from two sources: constructional stretch is dependent on the classification and results primarily from a reduction in diameter as load is applied and the strands compact against each other. Elastic stretch is caused by deformation of the metal itself when load is applied. Use the following equation to establish a value for elastic stretch: Elastic Stretch = PL AE Where: P = change in load; L= length; As = area of the steel in the wire rope; E = modulus of elasticity. Step 1: Determine Constructional Stretch: Constructional stretch % x length of wire rope = constructional stretch length 0.0075 x 70-ft x 12-in/ft = 6.3-in Step 2: Elastic Stretch = 12-kips x 70-ft x 12-in/ft = 3.5-in 0.288-in2 x 10,000-ksi Step 3: Total = 6.3-in + 3.5-in = 9.8-in (answer is d)

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34. - Question The selected rigging technique is to use a field assembled multiple leg wire-rope slings that are attached to a spreader beam to lift a 40,000-lb HVAC unit to the rooftop dunnage. The configuration of the hoisting equipment for the crane pick is shown in the figures below. Using the Safe Working Load limits, the total rated tension (lbs) capacity in Sling B is most nearly: a) b) c) d)

14,142 20,000 28,280 40,000

CL

To Crane Hook

Sling A Sling B

30-ft Spreader Beam = 60-ft

Sling F (H)

Sling E (G) HVAC 40,000-lbs

Section View Spreader Beam = 60-ft Sling D

Sling C

Sling B Sling A

Plan View (Top)

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Solution: The load in 3- and 4-leg slings may only be supported by 2 legs while the others are only balancing the load. Unequal length sling legs may be one reason, off-center or buckling loads another. Same capacity sling legs will stretch unequally if loaded unequally. Hence, the calculations are as follows: Use the Pythagorean Theorem to find the Length of Sling B and apply the equation: Load x (Height or Altitude of the Triangle ÷ Length of Sling B) = Tension (lbs) 40,000-lbs x (30-ft / 42.43-ft) = 28,280-lbs in tension (answer is c)

Alternate Method: 40,000-lb/2 = 20,000-lb /sin 45° = 28,280-lb

Sketch a free body diagram of the force vectors represented in the problem statement and solve for the tension in sling A.

Free Body Diagram 40k lbs.

Sling B 30-ft 28k lbs.

28k lbs. 30-ft

30-ft

Notice the coincidence between the centerline of the symmetrical loading and the center of gravity. The centerline of gravity will always prevail.

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CONDITIONS FOR A RIGID-BODY IN EQUILIBRIUM

z

fi In contrast to the forces on a particle, the forces on a rigid-body are not usually concurrent and may cause rotation of the body (due to the moments created by the forces).

Fi i ri

Indeterminate Solution. y

O x

F2 z

F1 For a rigid body to be in equilibrium, the net force as well as the net moment about any arbitrary point O must be equal to zero.

i

F4

∑F = 0 and ∑MO = 0

F3 y

O x

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CRANE SELECTION, ERECTION, AND STABILITY

fast facts Crane Classification: The following is provided as a basic understanding for a 30-ton crane (for illustration only):  Will lift 60,000 pounds at 10 feet from the center pin of the crane  Based on level surface, no wind, and outriggers fully extended  At 25 feet from the center pin with an 80-foot boom, the capacity is only 14,950 pounds  At 74 feet from the center pin, the capacity is only 4,800 pounds The radius over which a crane has to lift will have a significant effect on the loads that can be lifted and to what height they can be lifted.

Figure 1: Outline of a Crane’s Component parts.

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fast facts The main boom rotates over each of the fully extended outriggers and each outrigger must be capable of taking the full load capacity of the crane and hoist load.

Boom Outrigger

Crane center of gravity

Crane rotates over each outrigger

Figure 1: Schematic diagram showing a plan view of the main components of a crane.

Resisting Soils Surface Pressures -- A 200 lb. man with a size 10 shoe walking down a road is about 11 PSI. A pickup truck with an empty bed passing him is about 25 PSI. And a fully loaded tractor trailer, "road legal" by weight, going down the road is about 75 PSI.

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CENTER OF GRAVITY – CLASS REVIEW

40-lbs

60-lbs

30-lbs

1-ft 8-ft 15-ft datum The distances from the center of the main object and the two additional weights form the datum: •

Center of beam = 8 feet away from datum.

•

Left Weight = 1 foot away from datum

•

Right Weight = 15 feet away from datum

Multiply each object’s distance from the datum by its weight to find its moment. •

Beam Weight: 30 lb. x 8 ft. = 240 ft. x lb.

•

Left Weight = 40 lb. x 1 ft. = 40 ft. x lb.

•

Right Weight = 60 lb. x 15 ft. = 900 ft. x lb.

Add up the three moments. 240 ft. x lb. + 40 ft. x lb. + 900 ft. x lb = 1180 ft. x lb. The total moment is 1180 ft. x lb. Add the weights of all the objects. 30 lbs. + 40 lbs. + 60 lbs. = 130 lbs. Divide the total moment by the total weight. 1180 ft. x lb. by 130 lbs. •

1180 ft. x lb. ÷ 130 lbs = 9.08 ft.

•

The center of gravity is 9.08 feet from the datum, or measured 9.08 feet from the end of the left side which is where the datum was placed.

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CRANE WORKING RANGE DIAGRAM

CG

CG

Working Range Diagram EXAMPLE: To plan a lift for a load of 5-tons (10,000 pounds) at a distance of 45 feet the following two step process. Placement of the crane is determined by the distance measured from the center pin of the crane to the center of the load. Determine the distance and position on the diagram. The crane configuration shows that the length of the boom is required at 98-ft with a 60° offset. The next step is to evaluate the permitted in the load chart. The Chart permissible load is 23,200-lb.

Working Range Diagram Showing Superimposed Building & Crane Boom Reach

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Crane Safety Plan --

Crane operations are complex and can present hazards. The wellprepared Crane Safety Plan reduces the complexity to manageable and understandable elements. Procedures for assuring compliance with the plan are essential. The procedures will vary from site to site, but what will not vary are the operational criteria for the crane and rigging equipment.

OSHA requires that cranes be operated in accordance with the manufacturer’s instructions. Load charts and the operator’s manual supplied by the manufacturer have become the primary resource for safe operation of the crane. In addition, the ANSI standards provide guidance in developing a crane safety plan.

Load charts are complex, reflecting different boom types, boom lengths, rigging configurations, jib configurations, as well as other information in a document which consists of many pages of illustrations. Manufacturers have typically provided their load charts data in different formats and in many cases, the same manufacturer utilizes different formats for different models. The Manufacturer’s crane manual must be modeled for the exact site conditions and cannot be deviated from.

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35. - Question A 450-ton hydraulic crane with a 134-ft boom is used to place an 8-ton HVAC unit on the roof of the building as shown. Due to jobsite conditions, the minimum standoff point (Point C) to the centerline of the boom is indicated. The maximum centerline distance (ft) from the edge of the building that the load can be placed on the roof is most nearly: a. 20 b. 28 c. 36 d. 53

C

8-ft

Building 8-ft

CL

Crane

60-ft Boom Base

12-ft

45-ft 53-ft Not to Scale

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Solution: Using similar triangles, orient the location of Point C from the Boom base to determine the horizontal angle and expand the reach of the boom to 134ft by using the calculated horizontal angle. 60 + 8 – 12 = 56-ft – Vertical height of Point C above Boom Base 45 – 8 = 37-ft – Horizontal distance of Point C from Boom Base tan (x) = O = 56 x = 57° A 37

134-ft

cos (57°) x 134 = 73-ft 73 – 45 = 28-ft answer

67-ft

56-ft

57° 37-ft 73-ft

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TIPPING FULCRUM

36. - Question A truck crane with 143-feet of boom length at a 125-feet radius is lifting a load with fully extending the outriggers for the jobsite conditions. The tipping load (kips) based on the crane data provided is most nearly: a. b. c. d.

10 15 18 22

Given Parameters: a. Weight of crane 220,000-lb b. Weight of boom 24,000-lb c. Boom center of gravity radius of 52-ft from the tipping fulcrum d. 17-ft from crane center of gravity to tipping fulcrum (centerline of outrigger) e. 114.5-ft, distance from lifting load to tipping fulcrum

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Boom Weight

Load

Boom Radius Length from Load to Tipping Fulcrum

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Crane CG Crane Tipping Fulcrum

Length from Crane CG to Tipping Fulcrum

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Solution: Step 1: Develop a free body diagram

y

Load

Wb

α

x

CG

W2

W1 W

Free Body Diagram

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Step 2: Identify the components shown on the figure:

Wc = Weight of crane 220,000-lb Wb = Weight of boom 24,000-lb Bcg = Boom center of gravity of 52-ft CGtf = 17-ft from crane center of gravity to tipping fulcrum (centerline of outrigger) Ldis = 114.5-ft, distance from load to tipping fulcrum Determine the Stability Relationship for the given diagram. Stability = (Lifting Load x Ldis) + (Wboom x Bcg) = (Wcrane x CGtf) Rewrite the equation to find the Lifting Load (Convert the weights to kips) Lifting Load = ( 220-kips x 17-ft) – (24-kips x 52-ft) = 21.7-kips 114.5-ft Discuss the selection of the question’s letter answer.

Tipping Line

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Boom Weight

Crane CG Length from Load to Tipping

Crane Tipping Fulcrum Length from Crane CG to Tipping Fulcrum

Free Body Diagram - Overlay Arrows show resolution of the Forces to achieve equilibrium

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CRANE OUTRIGGER STABILITY

37. - Question The jobsite placement of the hydraulic crane is based on the data provided in the manufacturer’s specifications. The crane outriggers are equipped with steel pontoons with contact area of 307-in2. The main jack reaction is 63,500-lbs which exerts 213-psi ground bearing pressure. The crane is set up on soil with bearing pressures given as: gravel soils compacted, well graded, pressure, tons/ft2 = 10-tons. The contact area (ft2) of cribbing needed under each outrigger while lifting a 60,000-lb load is most nearly: a. 2.13 b. 2.67 c. 3.07 d. 3.26 [Hint: Calculate the minimum bearing pad size for the given allowable soil bearing pressure. The allowable Bearing Pressures are either given or can be obtained from a variety of published sources to determine, without tests, the jobsite conditions (for example, IBC, OSHA, etc.). Once the soil bearing pressure is determined, outrigger stability can be calculated. ALSO, the crane manufacturer’s content in the crane operator’s manual cannot be deviated from, and must be strictly followed. The items in this problem are distractors and are not relevant to the solution steps, that is -63,500-lbs; 307-in2; 60,000-lbs.]

Solution: Determine Bearing force: 213psi x 144-in2/ft2 = 30,672-lb/ft2 Pressure exerted below outrigger pad contact area: 307- in2 /144-in2/ft2 = 2.13-ft2 Allowable soil bearing pressure given as 10-tons / ft2 or 20,000-lb/ft2 Determine the required contact area ratio: 30,672-lb/ft2 / 20,000-lb/ft2 = 1.53 Increase the size of the bearing contact area using timber pads with a minimum size of: 2.13-ft2 x 1.53 = 3.26-ft2 (answer is d)

Crane outriggers extended with bearing pad below.

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EQUIPMENT PRODUCTION

fast facts In order to affect job-site productivity, it is necessary to select equipment with proper operating characteristics and a size based on site conditions. The following is a listing of factors which can affect the selection and operation of equipment. a. Size of the job: Determines size of equipment and quantity. b. Activity time constraints: Dependent on the project schedule. c. Availability of equipment: Affected by specialty equipment. d. Cost of transportation of equipment: Mobilize and demobilize e. Type of job needed to be performed. Based on equipment capacities f. Workflow: Coordinated to the project sequence g. Work crowding: Effect of too much activity in one location h. Space constraints: The performance of equipment is influenced by the spatial limitations for the movement of excavators. i.

Location of dumping areas: Effect on cycle time

j.

Weather and temperature: Rain, snow and severe temperature conditions affect the job-site productivity of labor and equipment.

Dump trucks are usually used as haulers for excavated materials as they can move freely with relatively high speeds on city streets as well as on highways. The cycle capacity C of a piece of equipment is defined as the number of output units per cycle of operation under standard work conditions. The capacity is a function of the output units used in the measurement as well as the size of the equipment and the material to be processed. The cycle time T refers to units of time per cycle of operation. The standard production rate R of a piece of construction equipment is defined as the number of output units per unit time. Rate = Cycle Capacity Time

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Time = Cycle Capacity Rate

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DAILY STANDARD PRODUCTION RATE OF EQUIPMENT

38. - Question An excavator with a bucket capacity of 3-yd3 has a standard operating cycle time of 40 seconds. The 8-hr daily standard production rate of the excavator is most nearly: a. b. c. d.

2,140-yd3 2,150-yd3 2,160-yd3 2,180-yd3

Solution: The daily standard production rate is as follows: sec ) hr

(3yd3 )(8hr)(3,600

Pday =

40sec

= 2,160 − yd3

(answer is c)

Excavator works in tandem with a dump truck to remove spoils off-site.

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DAILY STANDARD PRODUCTION RATE OF A DUMP TRUCK

39. - Question A dump truck with a capacity of 26 cubic yards is used to dispose of excavated materials at a dump site 6 miles away. The load time is 40-seconds using a 3-yd3 bucket. The average speed of the dump truck is 25 mph and the dumping time is 46 seconds. A fleet of dump trucks of this capacity is used to dispose of the excavated materials in 8hours per day. The number of trucks (count) needed daily using a swell of 10% for the soil is most nearly: a. b. c. d.

5 6 7 8

Solution: Calculate the daily standard production rate of a dump truck: Pday =

sec ) hr

(3yd3 )(8hr)(3,600

𝑇𝑇𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =

40sec

(2𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)(6𝑚𝑚𝑚𝑚)(3,600 25𝑚𝑚𝑚𝑚ℎ

𝑇𝑇𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = (40𝑠𝑠𝑠𝑠𝑠𝑠) �

26 𝑦𝑦𝑦𝑦𝑦𝑦3 3𝑦𝑦𝑦𝑦 3

= 2,160yd3 x 1.1 swell = 2,376yd3 /day

𝑠𝑠𝑠𝑠𝑠𝑠 ) ℎ𝑟𝑟

= 1,728 𝑠𝑠𝑠𝑠𝑠𝑠

� = 347 𝑠𝑠𝑠𝑠𝑠𝑠

𝑇𝑇𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 1,728 + 347 + 46 = 2,121 𝑠𝑠𝑠𝑠𝑠𝑠

The daily dump truck productivity is: 𝑃𝑃𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 =

𝑠𝑠𝑠𝑠𝑠𝑠 ) ℎ𝑟𝑟

(26𝑦𝑦𝑦𝑦 3 )(8ℎ𝑟𝑟)(3,600 (2,121 𝑠𝑠𝑠𝑠𝑠𝑠)

= 353 𝑦𝑦𝑦𝑦3

Calculate the number of trucks required: 𝑃𝑃 =

(2,376𝑦𝑦𝑦𝑦 3/day ) 353𝑦𝑦𝑦𝑦 3

= 6.73 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

Therefore, 7 trucks should be used. (answer is c)

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PRODUCTIVITY ANALYSIS AND IMPROVEMENT

40. - Question An excavator with a bucket capacity of 3-yd3 has a standard production rate of 2,160yd3 for an 8-hour day. The job site productivity and the actual cycle time of this excavator under the work conditions at the job site that may affect its productivity as shown in the table, is most nearly: a. b. c. d.

Work Conditions at the Site Bulk composition (soil) Soil properties and water content Equipment idle time for worker breaks Management efficiency Soil Compaction Cycle time (sec) Dump Truck Volume (CY) Fuel Consumption (gal/hr.) Daily Excavator Maintenance (after work hr.)

Factor 0.954 0.983 0.8 0.7 0.83 40 26 6.4 .50

1,034-yd3 / day and cycle time of 57-sec 1,034-yd3 / day and cycle time of 68-sec 1,134-yd3 / day and cycle time of 72-sec 1,134-yd3 / day and cycle time of 76-sec

Solution: Note that all the factors are less than one indicating a reduction in productivity, as such, the job-site productivity of the excavator per day is given by (answer is d):

𝑃𝑃𝑑𝑑𝑑𝑑𝑑𝑑=�2,160𝑦𝑦𝑦𝑦𝑦𝑦3�(.954)(.983)(.8)(.7)=1,134𝑦𝑦𝑦𝑦𝑦𝑦3

The actual cycle time can be determined as follows:

𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎=

40𝑠𝑠𝑠𝑠𝑠𝑠 = 76𝑠𝑠𝑠𝑠𝑠𝑠 (.954)(.983)(.8)(.7)

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The supplemental material found in this section has been found on previous exams and are important for you to review as you prepare for the exam.

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EFFECTS OF JOB SIZE ON PRODUCTIVITY

41. - Question A general building contractor has established that under a set of "standard" work conditions for building construction, a job requiring 500,000 labor hours is considered standard in determining the base labor productivity. All other factors being the same, the labor productivity index will increase to 1.1 or 110% for a job requiring only 400,000 labor-hours. Assume that a linear relation exists for the range between jobs requiring 300,000 to 700,000 labor hours, the labor productivity index for a new job requiring 650,000 labor hours under otherwise the same set of work conditions is most nearly: a. b. c. d.

0.50 0.65 0.78 0.85

Solution: Illustrate the Relationship between Productivity Index and Job Size The labor productivity index “I” for the new job can be obtained by linear interpolation of the available data as follows:

𝐼𝐼

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙=1+(1.10−1.0)�

500,000−650,000 � = .85 500,000−400,000

Productivity

The result implies that labor is 15% less productive on the large job than on the standard project.

1.1 1.0 .85

7 5 6 Labor-hours (00,000) Figure 1: Linear Interpolation of Productivity Index and Job Size 3

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4

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ACTUAL VERSUS ULTIMATE STRENGTH

fast facts The major distinction between ASD and LRFD is that the Allowable Stress Design (ASD) compares actual and allowable stresses. Load and Resistance Factor Design (LRFD) compares required strength to actual strengths. The difference between designing for strengths vs. stresses does not present much of a problem since the difference is normally just multiplying or dividing both sides of the limit state inequalities by a section property. Figure-1 illustrates the member strength levels computed by the two methods on a typical mild steel load vs. deformation diagram. The combined force levels, Load, Moment, and Shear (Pa, Ma, Va) for ASD are typically kept below the yield load for the member by computing member load capacity as the nominal strength, Rn, divided by a factor of safety, Ω, that reduces the capacity to a point below yielding. For LRFD, the combined force levels (ultimate) Load (Pu), Moment (Mu), and Shear (Vu) are kept below a computed member load capacity that is the product of the nominal strength, Rn, times a resistance factor, φ. When considering member strengths, the governance is to always keep the final design's actual loads below yielding so as to prevent permanent deformations in the structure.

Figure -1:

ASD vs. LRFD Strength Comparison Rn/ Ω= ASD Capacity φRn = LRFD Capacity Rn = Nominal Capacity

Consequently, if the LRFD approach is used, then load factors greater than 1.0 must be applied to the applied loads to express them in terms that are safely comparable to the ultimate strength levels. This is accomplished in the load combination equations that consider the probabilities associated with simultaneous occurrence of different types of loads. For structures subjected to highly unpredictable loads (live, wind, and seismic loads for example) the LRFD Ωeff is higher than the ASD Ω which results in stronger structures.

(10/2018)

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NATIONAL POLLUTANT DISCHARGE ELIMINATION SYSTEM (NPDES)

fast facts Federal Requirements for Stormwater Runoff from Construction Sites The Clean Water Act and associated federal regulations (Title 40 of the Code of Federal Regulations [CFR 123.25(a)(9), 122.26(a), 122.26(b)(14)(x) and 122.26(b)(15)) require nearly all construction site operators engaged in clearing, grading, and excavating activities that disturb one acre or more, including smaller sites in a larger common plan of development or sale, to obtain coverage under a National Pollutant Discharge Elimination System (NPDES) permit for their stormwater discharges. Under the NPDES program, the U.S. Environmental Protection Agency (EPA) can authorize states to implement the federal requirements and issue stormwater permits. Today, most states are authorized to implement the NPDES program and issue their own permits for stormwater discharges associated with construction activities.

(10/2018)

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AA . Which of the following terms are similar to a SWPPP which is required by the EPA NPDES permit? I. II. III. IV. V.

Construction Best Practices Plan Sediment and Stormwater Plan Erosion, Sediment, and Pollution Prevention Plan; Erosion and Sediment Control Plan Construction Site Best Management Practices Plan Erosion Control Plan and Best Management Practices a. b. c. d.

I & II I, II, & III I, II, III, & IV I, II, III, IV, & V

If the link is broken, Google the title

  



Answer; all are true. The Stormwater Pollution Prevention Plan (SWPPP) is part to the EPA NPDES. See page ii of the (web reference below) EPA, Developing Your Stormwater Pollution Prevention Plan May 2007, Updated February 2009. https://www3.epa.gov/npdes/pubs/industrial_swppp_guide.pdf

BB. Which of the following NPDES records should be kept at the project site available for EPA inspectors to review are: I. Dates of grading, construction activity, and stabilization II. A copy of the construction general permit III. The signed and certified NOI form or permit application form IV. A copy of the letter from EPA or/the state notifying you of their receipt of your complete NOI/application V. Inspection reports; Records relating to endangered species and historic preservation a. b. c. d.

I & II I, II, & III I, II, III, & IV I, II, III, IV, & V

Important Download Files on this page

Answer; all are true. See page of 26 of EPA SWPPP Template, Version 1.1, September 17, 2007 (reference below) (Notice of Intent) www.epa.gov/npdes/pubs/sw_swppp_template_authstates.doc

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TYPES OF SOIL EROSION

Raindrop erosion: Dislodging of soil particles by raindrops Sheet erosion: The uniform removal of soil without the development of visible water channels

Rill erosion:

Soil removal through the formation of concentrated runoff that creates many small channels

Gully erosion: The result of highly concentrated runoff that cuts down into the soil along the line of flow

Stream bank erosion: Flowing water that erodes unstable stream banks

 

Erosion versus Sedim entation Erosion is the process by which the land surface is worn away by the natural action of water or wind.

Sedimentation is the movement and settling out of suspension of soil particles.

It is usually easier and less expensive to prevent erosion than it is to control sediment from leaving a construction site

(10/2018)

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COMMON SWPPP OBJECTIVES A SWPPP can have different names throughout the country as guided by the general construction permit process. A SWPPP may also be called a “construction best practices plan,” “sediment and stormwater plan,” “erosion, sedimentation, and pollution prevention plan,” or similar term. The SWPPP is generally required to comply with EPA’s or the state’s stormwater construction general permit. The SWPPP outlines the steps to take to comply with the terms and conditions of the construction general permit. The following objectives are required to be incorporated to develop a SWPPP: • Stabilize the site as soon as possible. Get your site to final grade and either permanently or temporarily stabilize all bare soil areas as soon as possible.

• Protect slopes and channels. Convey concentrated stormwater runoff around the top of slopes and stabilize slopes as soon as possible.

• Reduce impervious surfaces and promote infiltration. Reducing impervious surfaces will ultimately reduce the amount of runoff leaving the site.

• Control the perimeter of the site. Divert stormwater coming on to the site by conveying it safely around, though, or under the site.

• Protect receiving waters adjacent to your site. Erosion and sediment controls are used around the entire site, but operators should consider additional controls on areas that are adjacent to receiving waters or other environmentally sensitive areas. The primary purpose of erosion and sediment controls is to protect surface waters. Follow pollution prevention measures.

• Provide proper containers for waste and garbage at the site. • Minimize the area and duration of exposed soils. Clearing only land that will be under construction in the near future, a practice known as construction phasing, can reduce off-site sediment loads significantly. Additionally, minimizing the duration of soil exposure by stabilizing soils quickly can reduce erosion dramatically.

Question - Which of the following construction site operators are responsible for obtaining

NPDES permit coverage for their stormwater discharges? a. General Contractor b. Owner c. Earthwork Contractor d. Chief Operating Officer Solution: See page 6 NPDES EPA, Developing Your Stormwater Pollution Prevention Plan-A Guide for Construction Sites (b)

Question - Revulsion is the opposite of: a. accretion b. avulsion c. erosion d. alluvium Solution: Definition of revulsion: a strong pulling or drawing back or away; withdrawal. A. Accretion (correct answer). The act of growing to a thing; usually applied to the gradual and imperceptible accumulation of land by natural causes. B. Avulsion. The act performed by a stream when it suddenly breaks through its banks in an unexpected manner and forms another channel or cuts off a large quantity of land from one owner and adds it to another. C. Erosion. Wearing away of the lands or structures by running water D. Alluvium. Sediments deposited by streams as a result of markedly decreased current velocity

(10/2018)

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SWPPP DEVELOPMENT—SELECTING EROSION AND SEDIMENT CONTROL BMPS

.

Silt fencing to slow runoff and trap sediment primarily for sheet flow, not channelized flow

Straw bale fencing to slow runoff and trap sediment for sheet flow or channelized flow.

Most Helpful advice, download reference material and bring to exam!!! A sediment trap to slow runoff and trap sediment for channelized flow.

(10/2018)

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PROJECT SCHEDULING

Project Scheduling

Concept Terminology



(10/2018)

CHAPTER

CHAPTER 4

4

Project Scheduling

Precedence Diagrams Arrow Diagrams Critical Path Method

Activity Durations Critical Path Units of Time Activity Sequence Resource Leveling Network Analysis

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CALCULATING PROJECT DURATION

fast facts Scheduling procedures rely upon estimates of the durations of the various project activities, as well as, the definitions of the predecessor relationships among tasks. A straightforward approach to the estimation of activity durations is to keep historical records of particular activities and rely on the average durations from this experience in making new duration estimates. Since the scope of activities is unlikely to be identical between different projects, unit productivity rates are typically employed for this purpose. For example, the duration of an activity such as concrete formwork assembly might be estimated as: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎

=

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑥𝑥 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆

where the Area of the formwork divided by the productivity times the crew size. This formula can be used for nearly all construction activities. The calculation of a duration as is only an approximation to the actual activity duration for a number of reasons. Further, productivity rates may vary. An example of productivity variation is the effect of learning on productivity. As a crew becomes familiar with an activity and the work habits of the crew, their productivity will typically improve. The result is that productivity is a function of the duration of an activity or project.

(10/2018)

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42. - Question A concrete contractor historical records show that the construction duration for a cast-in-place foundation formwork is 1,000SF/crew-day. The number of days a 200,000-SF formwork for a cast-inplace concrete foundation using 3-crews is most nearly: a. 55 b. 60 c. 70 d. 75 Solution: Apply the equation:

𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = Duration =

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴

𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑥𝑥 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆

200,000-SF = 66.67-days 1000-SF/crew-day x 3 crews

(answer is c)

Formwork set with rebar placed and ready for inspection. (10/2018)

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PROJECT SEQUENCING AND SCHEDULING

43. - Question is most nearly: a. 1 b. 2 c. 3 d. 4

The total duration (day(s)) for the four-activity project

Durations and Predecessors for a Four Activity Project Illustration Activity

Predecessor

Duration (Days)

Excavate trench Place formwork Place reinforcing Pour concrete

--Excavate trench Place formwork Place reinforcing

1.0 0.5 0.5 1.0

Solution: Scheduling work activity has associated time duration. The durations shown in the Table were estimated for the project diagrammed below. The entire set of activities would then require 3-days, since the activities follow one another directly and require a total of 1.0 + 0.5 + 0.5 + 1.0 = 3-days (answer is c). If another activity proceeded in parallel with this sequence, the 3-day minimum duration of these four activities is unaffected. More than 3 days would be required for the sequence if there was a delay or a lag between the completion of one activity and the start of another.

(10/2018)

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Work Breakdown Structure (WBS)

A Work Breakdown Structure (WBS) is a representation of a workflow organization used to coordinate all the work necessary to complete a project. A WBS is arranged in a top down hierarchy and constructed to allow for clear and logical groupings, either by activities or project deliverables. The WBS typically represent the work identified in the approved Project Scope of Work (SOW) and serves as the foundation for effective schedule development and cost estimating. The goals of developing a WBS and WBS Dictionary are: 1) for the project team to proactively and logically plan out the project to completion; 2) to collect the information about work that needs to be done for a project; and 3) to organize activities into manageable components that will achieve project objectives. The WBS and WBS Dictionary are not the schedule, but rather the building blocks to it. A sample of a WBS is found in the following: WBS - Construction Level 1

1.0

Construct 22-Story Mixed Use Structure

Level 2 1.1

1.2

Design Development

Level 3

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1.3

1.4

1.5

1.6

Schematic Design

Construction Documents

Bid-Build

Construction

1.1.1 Create Plan

1.2.1 Prepare Drawings

1.3.1 Solicit Bidders

1.4.1 Solicit Biddders

1.5.1 Break Ground

1.6.1 Secure Occupants

1.1.2 Make Budget

1.2.2 Validate Layout

1.3.2 Publicize Bid

1.4.2 Receive Bids

1.5.2 Superstructure

1.6.2 Tenant Fit Out

1.1.3 Prepare Market Share

1.2.3 Finance Options

1.3.3 Bid Proposal Received

1.4.3 Award Contract

1.5.3 M/E/P/F Trades

1.6.3 Launch Model Plan

1.1.4 Coordinate Activities

1.2.4 Secure Loan

1.3.4 Value Engineering

1.4.4 Execute Finance Plan

1.5.4 Interior Finishes

1.6.4 Secure Lease

1.2.5 Start Procurement Process

1.3.5 Prepare Building Depart Documents

1.5.5 Building Turnover

1.6.5 Review Model

Operation

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PROJECT SCHEDULING – TYPES OF METHODS

Project Scheduling – There are two types of scheduling methods for project schedule analysis with the major distinctions detailed as follows. •

Method 1 - Precedence Diagrams (PDM) –

Activity on Node (AON)

–

Can have any kind of precedence

 

Computer Oriented

Finish to Start Finish to Finish

Node

Start to Start Start to Finish

•

Method 2 - Arrow Diagrams –

Activity On Arrow (AOA)

–

Activity-on-Branch

–

May have dummy tasks

–

Finish to start precedence only

 

People Oriented

AOA

(10/2018)

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UNITS OF TIME CONVENTION Units of time are often stated in terms of days. There are two types of convention used in scheduling: 1. “Beginning-of-day” - means that the finish date of the activities is defined in terms of the beginning of the following day. For example, if the first activity in a network has a duration of two days and if its start date is set at the beginning of Day 1, the following calculation would be made: The activity will end on the beginning of Day 3 or, (Early start time of Day 1) + (2-day duration) = (Early start time of Day 3) Clearly, the activity will actually be completed at the end of Day 2, but hand calculations are easier to make if beginning-of-day convention is used. The project’s “early-start” date is for the first activity is 1. 2. “End-of-day” – the early start is assigned 0, meaning that the end-of-day convention is being used. The project’s “early start” date is for the first activity 0. 3. Both methods of project scheduling, Precedence Diagrams and Arrow Diagrams: ALL relationships are considered to be finish-to-start (FS).

fast facts Network analysis system equations:

ES = Earliest date that an activity can start EF = Early finish date for an activity LS = Late Start for an activity without affecting the schedule duration LF = Late Finish for an activity without affecting the project duration TF = Total Float or flexibility in an activity start date EF = ES + Duration LS = LF – Duration

MON

TUE

WED

Time Convention

0

1

2

End of Day

1

2

3

Beginning of Day

(10/2018)

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PRECEDENCE RELATIONSHIPS

•

Finish-to-Start – the most common, the predecessor activity must finish before the successor activity can start

•

Finish to Finish – The predecessor activity must finish before the successor activity can finish

•

Start-to-Start – The predecessor activity must start before the successor activity can start

•

Start-to-Finish – the predecessor activity must start before the dependent activity can finish; or, the completion of the successor depends upon the initiation of the work of the predecessor. A B

s

f

s

A

f

FS – Excavation must Finish before Concrete footings can be Started.

FF – Wallboard must be Finished before painting can be Finished

B SS – Roadway milling must Start before the Roadway paving can Start

A B

B

A

SF- is rarely used and should generally be avoided. E.g., Electrical power connection to the Elevator Must Start before the Elevator construction can be Finished

LEAD LAG RELATIONSHIPS

•

A lead is when the successor task begins before the predecessor task is complete

•

A lag is when a successor task does not start immediately upon the completion of the predecessor task

(10/2018)

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PROJECT SCHEDULING - DEFINING THE TERMS

Project:

Set of activities or tasks with a clear beginning and ending points. The amount of available resources (time, personnel and budget) to carry out the activities is usually limited.

Project Scheduling: Sequential process to allocate resources to execute all activities in the project.

Considerations for evaluating project networks: a. Define activities or tasks according to the project objectives in a Work Breakdown Structure (WBS). b. A task is an individual unit of work with a clear beginning and a clear end. c. Identify precedence relationships or dependencies (FS; FF; SS; SF) d. Estimate time required to complete each task. e. Draw an activity-on-arrow diagram inserting dummy activities if required. f.

Apply CPM to calculate earliest and latest starting times, earliest and latest completion times, float (slack) times, critical path etc.

g. Develop a GANTT chart. h. Reallocate resources and resolve if necessary. i.

Continuously monitor/revise the time estimates along the project duration.

Total Float: Total float is measured as the difference between the early and late start dates (LS – ES) or the early and late finish dates (LF – EF). Total float is shared between the activities in a sequence. A sequence is defined as the activities between a point of path divergence and path convergence. Total float represents the amount of time an activity can be delayed without delaying the overall project duration and is also called “float” or “slack”.

Free Float: Free float is measured by subtracting the early finish (EF) of the activity from the early start (ES) of the successor activity. Free float represents the amount of time that a schedule activity can be delayed without delaying the early start date of any immediate successor activity within the network path. Free float is only calculated on the last activity in an activity sequence.

Example: If activity Install Roof has a duration of 6 days and is occurring concurrently with activity Install Curtain Wall which has a duration of 9 days, activity Install Roof has 3 days of total float. Meaning, it can be delayed up to three days without any effect on the project. However, if activity Install Roof is delayed by 5 days, there is now a negative float situation: that is, -2 days. This reflects the fact that the project will now take two days longer than anticipated.

(10/2018)

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ARROW DIAGRAMMING METHOD

•

Activity-on-Arrow (Branch) with nodes as dependencies

•

Uses finish-to-start precedence only. The solution method for an activity on arrow problem is essentially the same as for the activity-on-node problems requiring forward and backward passes to determine the earliest and latest dates.

•

May require use of dummy activities to maintain proper logic of various construction activities. If two activities have the same starting and ending events, a dummy node is required to give one activity a uniquely identifiable completion event.

•

A dummy is treated as an activity (represented by a dotted line on the arrow network diagram), that indicates that any activity following the dummy cannot be started until the activity or activities preceding the dummy are completed. The dummy activity does not consume time or any resource.

•

See CERM Figures 86.6 and 86.7 for a discussion of the Dummy Node.

•

Review Figure 1 Activity-on-Arrow diagram below. Note the dummy activity showing the relationship between the material procurement and installation of the components.

(10/2018)

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PRECEDENCE DEFINITIONS FOR ACTIVITY ON ARROW (AOA) NETWORK DIAGRAMS

Predecessor

Successor

B A

A must finish before either B or C can start C

A A and B must finish before C can start C B

A

C A and B must finish before either C and D can start

B

D

A

B Dummy

C

(10/2018)

A must finish before B can start; both A and C must finish before D can start due to the dummy activity as identified as a dashed-line

D

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Fabricate & Deliver Structural Steel

Procure Rebar

Excavate Footings

1

2

Form & Pour Footings

Pour Slab on Grade

3

Erect Building

4

5

Install Finishes

6

Turnover

Install Roof Install Curtain Wall Gazing

Procure Roof

7

Procure Curtain Wall

8

Install Curtain Wall Frame

9

LEGEND

i

ES LS

Operation Duration

EF LF

j

Activity on Arrow Network Diagram

The Activity on Arrow network provides the simplest of scheduling formats in that all activities are a Finish-to-Start relationship ONLY. The dummy activity, represented by the dashed line with arrowhead, identifies that there is a relationship between the activities although there are no resources or time associated with the activity. The network illustrates the importance of procurement and material delivery in the schedule. The procurement component in the network tracks the delivery of the Roof and Curtain Wall while the “dummy” activity places a logic link indicating that the two activities must be completed before the installation of the roof can begin.

(10/2018)

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Class Exercise

Legend Activity Duration (Days)

O M 7

L 2

1 4

3

2

6 R 3

S 7

N 3

Q

P 4

3

5

8 7

6

Activity on Arrow Network Diagram Determine the number of paths:

Determine the earliest possible completion day: _____ Days

Determine the critical path: _ _ _ _ _ _

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44. - Question For the project network shown, the total number of paths through the network; number of critical paths; and, project duration (days) is most nearly: a. b. c. d.

7; 2; and, 26 6; 1; and, 26 5; 1; and, 25 5; 3; and, 24

Legend Activity Duration (Days)

7 R M

3

O 2

1

4

2

3

11

S 3 Q

P 4

3

V 9

6

8 N

9

2

7 L

T 2

5

U

10

7

6

Solution: List the number of paths and calculate the duration for each. LMORTV LMOSU LMOSV LMQV LMQU LNPQU LNPQV

(10/2018)

= 4 + 7 + 2 + 2 + 2 + 9 = 26-days -- Critical = 4 + 7 + 2 + 3 + 7 = 23-days = 4 + 7 + 2 + 3 + 9 = 25-days = 4 + 7 + 6 + 9 = 26-days – Critical = 4 + 7 + 6 + 7 = 24-days = 4 + 3 +3 + 6 + 7 = 23-days = 4 + 3 + 3 + 6 + 9 = 25-days

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ACTIVITY RELATIONSHIPS – PRECEDENCE TABLES

45. - Question The following activity relationship tables provide the detail for the project network as shown in the diagram. The activity relationship table that provides the best fit is most nearly: a. b. c. d.

1 2 3 4

11

6

1

2

3

4

5

8

7

12

9

10

Project Network Diagram

(10/2018)

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13

1:

2: Number Activity Predecessor 1 Start 2 B 1 3 C 2 4 D 3 5 E 4 6 F 4 7 G 5,6 8 H 4 9 I 4 10 J 3,8 11 K 6 12 L 7,9,10 13 Finish 11,12

Number Activity Successor 1 Start 2 2 B 3,4 3 C 4,10 4 D 5,6,8,9 5 E 7,8 6 F 7,11 7 G 12 8 H 10 9 I 11,12 10 J 12 11 K 13 12 L 13 13 Finish

3:

4: Number Activity Successor 1 Start 2 2 B 3 3 C 4,10 4 D 5,6,8,910 5 E 7 6 F 7,11 7 G 12 8 H 10 9 I 12 10 J 12,13 11 K 13 12 L 13 13 Finish

Number Activity Predecessor 1 Start 2 B 1 3 C 2 4 D 3 5 E 3,4 6 F 4 7 G 5,6 8 H 4 9 I 4,5,6 10 J 3,8 11 K 6 12 L 7,9,10 13 Finish 11,12

Solution: A project network is a flow chart depicting the sequence in which a project's events are to be completed by showing the events and their dependencies. This question arranges the network diagram and requests selection for the best fit to the given tables. Inspection of the tables identifies either a successor or predecessor relationship. A predecessor is an activity that must be completed (or be partially completed) before a specified activity can begin. Dependencies are created in project scheduling as the Predecessor and Successor fields. The Predecessor is the task completed “prior to” the current task, and the Successor is the task completed “after” the current task. Examination reveals that Table 2 is arranged to match the network diagram. (answer = b)

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CRITICAL PATH NETWORK ANALYSIS

A critical Path is: •

The series of activities which determines the earliest possible date of project completion or the longest path through the project network.

•

Usually defined as those activities with zero float or slack.

•

Gantt Chart - A Gantt chart is a type of bar chart, developed by Henry Gantt, that illustrates a project schedule. Gantt charts illustrate the start and finish dates of the elements and summary elements of a project.

Critical Path Method -- Highlights •

A network analysis technique used to predict project duration by analyzing which sequence of activities has the least amount of schedule flexibility (float)

•

Early dates are calculated by a forward pass using a specified start date

•

Late dates are calculated by a backwards pass using a specified completion date (usually the early finish date)

•

Uses deterministic dates (for example, the most likely date the activity will occur)

•

Single duration estimate for each activity

•

Start date and calculated Early Finish and Late Finish dates for each activity

Note: There are many sign conventions used to display the type of information in schedule analysis. Always interpret the project schedule using the legend provided. Samples are shown below: Activity ACTIVITY Early Early (duration) Duration Start Finish ES EF ES DURATION LS EF

Task Name Late Start

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Float

FF

TF

LF

LS

LF

Late Finish

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THE FORWARD PASS

EF = ES + Duration

The forward pass works from left to right, calculating when tasks can end using their early start date and the expected duration.

Task

Early Start

Early Finish

Duration

A

1

11

10

B

11

16

5

C

16

24

8

D

11

17

6

E

17

26

9

F

26

29

3

Note that Task F cannot begin until both C and E are finished. 1

2

3

4

5

6

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7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

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27

28

29

THE BACKWARD PASS

LS = LF – Duration

The early finish for the forward pass is also the late finish for the project. In the backward pass we move from right to left using the late finish and the duration to determine the late start.

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Task

Late Finish

Late Start

Duration

F

29

26

3

C

26

18

8

B

18

13

5

E

26

17

9

D

17

11

6

A

11

1

10

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CALCULATING FLOAT

TF = LF - ES - DUR

Float for a task is the difference between the early start and the late start. In this case, only tasks B and C have any float, 2 days in both cases. However, if task B is two days late in starting, task C loses its float. Float is a property of a network fragment.

Note that the NCEES uses the above equation. The most common format is LS-ES or LF-EF

Task

Duration

Early Start

Early Finish

Late Start

Late Finish

Float

A

10

1

11

1

11

0

B

5

11

16

13

18

2

C

8

16

24

18

26

2

D

6

11

17

11

17

0

E

9

17

26

17

26

0

F

3

26

29

26

29

0

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ACTIVITY SEQUENCING

46. - Question An activity and relationship table is provided for the network diagram. The workday(s) when the maximum number of workers are present is: a. b. c. d.

3 6 7, 8 12

B

A

C

F

H

E

G

D

Activity A B C D E F G H

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Successor C F F,E,D G G H Finish Finish

Duration Days 2 1 3 2 1 3 1 2

Workers per Day 2 4 6 4 4 4 2 4

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Solution: Determine the number of paths and the critical path duration. BFH

=1+3+2=6

ACFH

= 2 + 3 + 3 + 2 = 10 CRITICAL PATH

ACEG

=2+3+1+1=7

ACDG

=2+3+2+1=8

Extend the activity table to determine the number of workers needed per day. By inspection, the project schedule uses the end-of-day time convention. Worker usage can be assembled and presented in the table format below. Identify the critical path.

Activity A B C D E F G H Total

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1 2 4

2 2

3

Workers Needed per Work Day Work Day 4 5 6 7 8 9

Critical 6

6

6

Critical 4 4 4

6

2

10

6

6

6

12

4 4

8

4 2 6

Critical 4

4

4

4

Critical

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LEVELING

fast facts Resource Leveling is a project management tool used to examine a project for an unbalanced use of resources (usually people) over time, and for resolving over-allocations of resources or resource conflicts. When performing project planning activities, the project manager will attempt to schedule certain tasks simultaneously. When more resources, such as equipment or people are needed than there are available, the tasks will have to be rescheduled concurrently or even sequentially to manage the constraint. Project planning resource leveling is the process of resolving these conflicts. It can also be used to balance the workload of primary resources over the course of the project, usually at the expense of one of the traditional constraints which are: time, cost, or scope of work. Project management resource leveling requires delaying tasks until resources are available. Leveling could result in a later project finish date if the tasks affected are in the critical path. The objective of resource leveling is to analyze the free float or total float of the activities in order to avoid delaying the project. Resource-leveling can take the "work demand" and balance it against the resource pool availability for the given duration.

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47. - Question The project network with activities A, B, and C and durations are as shown. Activity A has 3 days of total float and activity C has 2 days of total float. Activity A requires 2 workers, B requires 4 workers, and C requires 2 workers. 2 Apply the project network to the A = 2 days following:

1 C = 3 days

B = 5 days

4

3

1. If all the activities start on day one, the number of workers needed on day 4 is most nearly: a. b. c. d.

2 4 6 8

 

2. If activity C is delayed 2 days, its total float, the number of workers needed on day 4 is most nearly: a. b. c. d.

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2 4 6 8

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The project network with activities A, B, and C and durations are as shown. Activity A has 3 days of total float and activity C has 2 days of total float. Activity A requires 2 workers, B requires 4 workers, and C requires 2 workers.

2

A = 2 days

B = 5 days

1 C = 3 days

4

3

Path Duration Critical Total Float A 2 5–2=3 B 5 Critical 5 – 5 = 0 C 3 5–3=2 Use the histogram to evaluate the individual questions:

8

8 A’

Workers

Workers

A’ 6 4

B’

B’

B’

2

B’ C’

0

C’

1

6

2

B’

A’

A’

C’

C’

C’

B’

B’

B’

B’

B’

4

C’

2

3

4

Days Resource usage if all Activities start on day one.

5

0

1

2

3

4

5

Days Resource usage if Activity C is delayed 2 days, its total float.

Solution: 1. (answer is b) 2. (answer is c)

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GLOSSARY OF SCHEDULING TERMS Activity — A distinct and identifiable operation within a project that will consume one or more resources during its performance is an activity. The concept of the distinct activity is fundamental in network analysis. The level of detail at which distinct activities are identified in planning depends largely on the objectives of the analysis. An activity may also be referred to as an operation, or work item. A dummy activity, which does not consume a resource, can be used to identify a constraint that is not otherwise apparent. Activity duration — The estimated time required to perform the activity and the allocation of the time resource to each activity define activity duration. It is customary to express the activity duration in worktime units; that is, workday, shift, week, and so on. An estimate of activity duration implies some definite allocation of other resources (labor, materials, equipment, capital) necessary to the performance of the activity in question. Constraints — Limitations placed on the allocation of one or several resources are constraints. Critical activities — Activities that have zero float time are critical. This includes all activities on the critical path. Critical path — The connected chain, or chains, of critical activities (zero float), extending from the beginning of the project to the end of the project make up the critical path. Its summed activity duration gives the minimum project duration or the longest path through the network. Several may exist in parallel. Dummy Activity — A fictitious activity used in an activity on arrow project network to show a constraint between activities on the logic diagram when needed for clarity is called a dummy. It is represented as a dashed arrow with an arrowhead and has duration of zero. In an activity on arrow network, a dummy activity is used to show that an activity A cannot start until activity B is finished. Float (also known as - slack) — In the calculation procedures for any of the critical path methods, an allowable time span is determined for each activity to be performed within. The boundaries of this time span for an activity are established by its early start time and late finish time. When this bounded time span exceeds the duration of the activity, the excess time is referred to as float time. Float can be classified according to the delayed finish time available to an activity before it affects the starting time of its following activities. It should be noted that once an activity is delayed to finish beyond its early finish time, then the network calculations must be redone for all following activities before evaluating their float times. Total float — The total time available between an activity’s early finish time and late finish time as determined by the time calculations for the network diagram. If the activity’s finish is delayed more than its number of days of total float, then its late finish time will be exceeded, and the total project duration will be delayed. Total float also includes any free float available for the activity. Free float — The number of days that an activity can be delayed beyond its early finish time without causing any activity that follows it to be delayed beyond its early start time is called free float. The free float for many activities will be zero, because it only exists when an activity does not control the early start time of any of the activities that follow it. Resources — These are things that must be supplied as input to the project. They are broadly categorized as manpower, material, equipment, money, time, and so on. It is frequently necessary to identify them in greater detail (draftsmen, carpenters, cranes, etc.).

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Material Quality Control and Production

CHAPTER

CHAPTER 5 MATERIAL QUALITY CONTROL AND PRODUCTION

5

Concept Material Specifications

Terminology



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Material Testing Quality Control Process

Construction Testing QA/QC Concrete Mix Design w/c ratio “1 : 2 : 3” Batch Plant Building Code Concrete Strength HMA Compressive Strength Tensile Strength

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MATERIAL SPECIFICATIONS

48. - Question Which of the following statements regarding construction material testing are correct: I.

Construction specifications of required quality and components represent part of the necessary documentation to describe a project. General specifications of work quality are available in numerous fields and are issued in publications of organizations such as the American Society for Testing and Materials (ASTM), the American National Standards Institute (ANSI), or the Construction Specifications Institute (CSI). Distinct specifications are formalized for particular types of construction activities, such as welding standards issued by the American Welding Society (AWS), or for particular facility types, such as the Standard Specifications for Highway Bridges issued by the American Association of State Highway and Transportation Officials (AASHTO). These general specifications must be modified to reflect local conditions, policies, available materials, local regulations and other special circumstances. Construction specifications normally consist of a series of instructions or prohibitions for specific operations. Performance specifications have been developed for many construction operations. They specify the required construction process. These specifications refer to the requirements of the finished facility. The exact method by which this performance is obtained is left to the project owner.

II.

III.

IV. V.

a. b. c. d.

I & II I, II, & III I, II, III, & IV I, II, III, IV, & V

 

Solution: Statement V should read “Rather than specifying the required construction process, these specifications refer to the required performance or quality of the finished facility. The exact method by which this performance is obtained is left to the construction contractor.” For example, traditional specifications for asphalt pavement specified the composition of the asphalt material, the asphalt temperature during paving, and compacting procedures. In contrast, a performance specification for asphalt would detail the desired performance of the pavement with respect to impermeability, strength, etc. How the desired performance level was attained would be up to the paving contractor. In some cases, the payment for asphalt paving might increase with better quality of asphalt beyond some minimum level of performance. (answer is c )

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49. - Question IBC stipulates in Section 1704 that where application is made for construction of Class 1 buildings, the owner or the person in responsible charge acting as the owner’s agent shall employ one or more special inspectors to provide inspections during construction on the types of work listed under Section 1704. Which of the following statements satisfy the requirements of: TABLE 1704.4 REQUIRED VERIFICATION AND INSPECTION OF CONCRETE CONSTRUCTION: I.

II. III. IV. V. VI.

At the time fresh concrete is sampled to fabricate specimens for strength tests, perform slump and air content tests, and determine the temperature of the concrete. Inspection of reinforcing steel, including pre-stressing tendons, and placement. Verifying use of required design mix. Inspect bolts to be installed in concrete prior to and during placement of concrete where allowable loads have been increased. Inspect formwork for shape, location and dimensions of the concrete member being formed. Inspection for maintenance of specified curing temperature and techniques.

Solution: All are correct

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SLUMP TEST

Slump 12-in

Concrete

Apparatus

The concrete slump test is an empirical test that measures the workability of fresh concrete. In the slump test the plastic concrete specimen is formed into a conical metal mold as described in ASTM Standard C-143. It measures the consistency of the concrete in that specific batch. This test is performed to check the consistency of freshly made concrete. Consistency is a term very closely related to workability. It is a term which describes the state of fresh concrete. It refers to the ease with which the concrete flows. It is used to indicate the degree of wetness. Workability of concrete is mainly affected by consistency i.e. wetter mixes will be more workable than drier mixes, but concrete of the same consistency may vary in workability. It is also used to determine consistency between individual batches. The test is popular due to the simplicity of apparatus used and simple procedure. 1. As water content increases, concrete strength decreases 2. The higher the slump the greater the water content the weaker the mix 3. Admixtures have the ability to increase the slump of the concrete mix without affecting the strength 4. Superplasticizers are a class of admixtures which increase the slump of the concrete while allowing the concrete to be pumped

1 sack Cement = 94-lb

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Portland cement Chemical Composition Typical constituents of Portland cement Cement

Mass %

Calcium oxide, CaO

61-67%

Silicon oxide, SiO2

19-23%

Aluminum oxide, Al2O3

2.5-6%

Ferric oxide, Fe2O3 Sulfate

See CERM Table 44.2 (p 44-4):

0-6%

Average Coefficient of Linear Thermal Expansion (multiply all values by 10-6) Substance 1/ºF Concrete 6.7 Steel 6.5

1.5-4.5%

Field Testing Concrete Slump Procedure per ASTM C-143

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AGGREGATE TERMS AND DEFINITIONS

Bulk Volume (of an Aggregate)—the volume of water displaced by aggregate in a SSD condition, and includes both the volume of the impermeable portion of the aggregate particles and the volume of the permeable voids in the particles. Saturated Surface-Dry SSD —the condition of the aggregate when all permeable pores of each particle are completely saturated with water and its surface has no free moisture; neither absorbing water from nor contributing water to the concrete mix. Saturated Surface-Dry Specific Gravity—the ratio of the mass of SSD aggregate to the mass of an equal volume of water. Absorption Moisture Content—the moisture content at saturated surface-dry condition in contrast to its oven-dry condition. Bulking – Bulking occurs when earth materials, such as, fine aggregates, silts, clay soils, etc. are handled. Bulking is the undesirable increase in volume caused by surface moisture holding the particles apart. For example, sand is normally delivered in batch quantities in a damp condition; however due to bulking, actual sand and cement can vary widely in batch volume which is often not in proportion to the moisture content of the sand. As such, when batch mixing is performed it uses the weight of the material. Moisture condition of aggregates State of aggregate Oven Dry

None

Air Dry

Saturated Surface Dry

Less than potential absorption

Equal to potential absorption

Damp or Wet

Greater than potential absorption

Total Moisture

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CONCRETE W/C RATIO

50. - Question The concrete truck jobsite delivery ticket is found below. The actual w/c ratio is most nearly: a) .41 b) .48 c) .49 d) .52 Batch Plt. Volume Truck

7 10-CY 41

Material Sand #57 Agg.

Target 14,271-lb 17,800-lb

Type I-II Flyash

4,080-lb 720-lb

Load 2 Mix Description w/c Actual 14,080-lb 17,700-lb

4,045-lb 755-lb

Status Done Done

Moisture 3.8 % 0.0 %

Done Over

Ticket

1487-TCC

Date/Time

3/14/11 13:48

Material Cl WR Retarder Air Entrain. MR HRWR Calcium NC Accel

Target 0-oz 0-oz 96-oz 48-oz 0-oz 480-oz 0-oz 0-oz

Actual 0-oz 0-oz 96-oz 48-oz 0-oz 480-oz 0-oz 0-oz

Status

Water

1979-lb

1964-lb

Done

SOG 3000-psi

Done Done Done

Solution: Calculate the total weight of the water: Water = 1964-lb + (Sand 14,080-lb x 3.8%) = 2499-lb Calculate the total Weight of Cement: Type-1 4045-lb + FlyAsh 755-lb = 4,800-lb Calculate the w/c ratio, since the units are the same, the ratio can be directly calculated: w/c = 2,499-lb ÷ 4,800-lb = .5206 (answer is d)

fast facts ASTM C 150 defines Portland cement as "hydraulic cement (cement that not only hardens by reacting with water but also forms a water-resistant product) produced by pulverizing clinkers consisting essentially of hydraulic calcium silicates, usually containing one or more of the forms of calcium sulfate as an inter ground addition." Clinkers are nodules (diameters, 0.2-1.0 inch [5-25 mm]) of a sintered material that is produced when a raw mixture of predetermined composition is heated to  high temperature. The low cost and widespread availability of the limestone, shale’s, and other  naturally occurring materials make Portland cement one of the lowest-cost materials widely used over the last century throughout the world. Concrete becomes one of the most versatile construction   materials available in the world. Fly ash is one of the residues generated in the combustion of coal.

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CONCRETE MIX DESIGN - PROPORTIONS

51. - Question A highway bridge project specifications require a specialty concrete mix design. The mix design has the proportions 1 : 2.7 : 3.65, on a weight basis. Type I-II cement content was specified at 5.6 sacks/yd3. The aggregates are SSD and have specific gravities of 2.65 for both the fine and coarse aggregate and the specific gravity of the cement is 3.15. The water/cement ratio (gal/sack) of the concrete mix is most nearly: a. b. c. d.

5.8 5.5 5.3 6.2

 

Solution: Reference CERM Chapter 77-2 for the strength relationships between w/c ratios.  

From the problem statement the design mix proportions are: Cement : Sand : Gravel

Analyze the components on the basis of -yd3 of concrete. The concrete consists of cement, fine aggregate, coarse aggregate, and water. Material

Cement

Ratio

1.00

Sand Aggregate

2.70 3.65

Sacks per

Weight per Sack

yd3

lb/yd3

5.6

5.6 x 94-lb/sack = 526 2.70 x 526 3.65 x 526

= 1421 = 1920

Specific Gravity

Convert to

Volume

ft3/lb

ft3/yd3

x 3.15 -1

x 62.4-1

= 2.68

-1

62.4-1

= 8.59 = 11.62

x 2.65 x 2.65 -1

x x 62.4-1

27-ft3/yd3 – 2.68 – 8.59 – 11.62 = 4.11-ft3/yd3 x 7.48gal/ft3

Water: Summary:

Water/cement ratio = 30.74 gal/yd3 / 5.6 sacks/yd3 =

= 30.74gal/yd3 5.49 gal/sack

(answer is b)

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AIR ENTRAINED CONCRETE

52. - Question

The mix design for a specialty concrete is found in the table. The amount (lb) of fine aggregate (S.G. 2.65) required is most nearly: a. a. b. c.

1120 1185 2000 2100

Water Type I Cement Coarse Aggregate (S.G. 2.70) Air Entrainment

280-lb/CY 700-lbs/CY 1701-lbs/CY 6%

Solution: Use the absolute volume method and determine the cubic volume. Material Water Type I Cement Coarse Aggregate Air Entrainment

Weight 280-lb/CY 700-lbs 1701-lbs (S.G. 2.70) 6%

Absolute Volume Method 280-lbs/62.4-lb/CF 700-lbs/(3.15 x 62.4-lb/CF 1701-lb (2.70 x 62.4-lb/CF

Volume 4.49-CF 3.56-CF 10.10-CF

6% x 27-CF/CY Total

1.62-CF 19.77-CF

Therefore, the fine aggregate volume must complete the CY and the weight is found using the following computations: 27-CF – 19.77-CF = 7.23-CF The weight of the fine aggregate is: 7.23-CF x 2.65 x 62.4-lb/CF = 1196-lb

Air entrainment

is the

intentional creation of tiny air bubbles in concrete. The bubbles are introduced into the concrete by the addition to the mix of an air entraining agent, a surfactant (surface-active substance, a type of chemical that includes detergents). The air bubbles are created during mixing of the plastic (flowable, not hardened) concrete, and most of them survive to be part of the hardened concrete. The primary purpose of air entrainment is to increase the durability of the hardened concrete, especially in climates subject to freeze-thaw; the secondary purpose is to increase workability of the concrete while in a plastic state. While hardened concrete appears solid, it is porous, having small capillaries resulting from the evaporation of water beyond that required for the hydration reaction. Water: cement ratio (w/c) is required for all the cement particles to hydrate. Water beyond the required w/c amount is surplus and is used to make the plastic concrete more workable or flowable. Most concrete has a w/c of 0.45 to 0.60, which means there is substantial excess water that will not react with cement. Eventually the excess water evaporates, leaving little pores in its place. Environmental water can later fill these voids. During freeze-thaw cycles, the water occupying

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those pores expands and creates stresses which lead to tiny cracks. These cracks allow more water into the concrete and the cracks enlarge. Eventually the concrete spalls; or chunks break off. The failure of reinforced concrete is most often due to this cycle, which is accelerated by moisture reaching the reinforcing steel. Steel expands when it rusts, and these forces create even more cracks, letting in more water. The air bubbles are typically 10 to 500 micrometers in diameter (0.0004 to 0.02 in) and are closely spaced. The air bubble can be compressed a little and act as a cushion as a result, the bubbles act to reduce or absorb stresses from freezing. Most concrete if subjected to freezing temperatures, is air-entrained. The bubbles contribute to workability by acting as a lubricant for all the aggregates and large particles in a concrete mix. Air entrained concrete is different from concrete which may contain entrapped air. Entrapped air is larger bubbles and is typically less evenly distributed than entrained air. Entrapped air is considered to not make a positive contribution to durability and is undesirable though not entirely avoidable. QC methods to assure against entrapped air include the code requirement for the use of internal vibrators immersed no more than four-feet during placement of concrete. When concrete is exposed to moisture, water moves through the concrete in these pores. When the temperature drops below freezing, the water turns to ice. Ice occupies 9 percent more volume than water. The expanding ice forces the water through the capillaries as it freezes. The repeated freezing and thawing can damage concrete and deterioration is increased with the use of deicing salts. Air voids act as a pressure relief for the expanding water and ice. The resistance to freeze-thaw damage depends upon the size and distribution of the voids, along with the degree of saturation. Many smaller voids separated a short distance provides the greatest degree of protection. Air entraining agents are added to aid in the

stabilization of the air voids by reducing surface tension in the water. Also, the one end of the air agent molecule is attracted to air and the other is attracted to water and thus to cement grains, thereby allowing a coating of calcium to form around each air bubble, making them more stable in plain water.

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CONCRETE STRENGTH TESTING – COMPRESSIVE STRENGTH

53. - Question A 6 in. by 12 in. cylinder failed at an axial compressive force of 120,000-lbf, at 28 days. The ultimate compressive strength is most nearly:  

a. 25,480 psi b. 4,250 psi c. 3,800 psi d. 3,250 psi SOLUTION: CERM page 48-5, Equation 48.1

 P

f’c = P/A = 120,000 lbf ÷ (π/4)(6 in)2 = 4,247-psi (answer = b)

12” 6” Sketches of Types of Fracture

P

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

CONCRETE STRENGTH TESTING – TENSILE STRENGTH

54. - Question A 6 in. by 12 in. concrete cylinder resisted a transverse force of 55,000-lbf, at 28 days, using a ASTM C496 split tensile cylinder test. The concrete tensile strength is most nearly: a. 560 psi b. 610 psi c. 375 psi d. 490 psi SOLUTION: Use the equation provided in CERM page 48-6, Equation 48.4 fct = 2P/πDL = (2)(55,000-lbf) / π(6-in)(12-in) = 487-psi (answer = d) P   

D=6”



12”

P

fast facts The extent and size of cracking in concrete structures are affected by the tensile strength of the concrete. The maximum load, P, that causes the cylinder to split in half is used to calculate the split tensile strength. The tensile strength of concrete is relatively low, about 10% to 15% of the compressive strength.

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55. - Question

A rebar is imbedded 1-ft-6-in deep in the center of a block of normal weight 4000-psi concrete which has a shear strength of 600-psi that will be used as a deadman anchor. The rebar is bonded to the concrete with a 30º cone of failure. The concrete block base size is 4-ft square and 3-ft high. The factor of safety of the pullout strength of the rebar is most nearly: a. b. c. d.

10 25 50 100

Rebar Anchor 30º

Cone of Failure

Solution: Sketch the problem statement Step 1: Compute the weight of the block of concrete Volume = 4-ft x 4-ft x 3-ft = 48-ft3 Weight = 150-lb/ft3 x 48-ft3 = 7200-lb Step 2 Calculate the pull out capacity of the anchor rod using the surface area of the cone of failure. Radius of the cone = r = 1.5-ft tan 30º = 0.866 Compute the surface area of the cone: Surface Area = 𝞹𝞹 x r √ (r2 + h2) = x 0.866 x √ (0.8662 + 1.52) = 4.712-ft2

Step 3: The capacity of the anchor is a function of the surface area of the cone of failure times the shear strength of the concrete which is usually 10% to 15% of the compressive strength. (4.712-ft2 x 144-in2/ft2) x 600-psi = 407,117-lbs Factor of Safety = 407,117-lb ÷ 7200-lb = 56.5 Answer c

(10/2018)

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HOT MIX ASPHALT - SHORT COURSE

Typical Asphalt Batch Plant Schematic

  



Hot Mix Asphalt (HMA) – Asphalt Concrete (AC) is a mixture of asphalt binder and graded mineral aggregate mixed at an elevated temperature and compacted to form a relatively dense pavement layer. Composition: ≈ 5% binder and ≈ 95% aggregate Hot mix asphalt concrete (commonly abbreviated as HMAC or HMA) is produced by heating the asphalt binder to decrease its viscosity, and drying the aggregate to remove moisture from it prior to mixing. Mixing is generally performed with the aggregate at about 300 °F (150 °C) for virgin asphalt and 330 °F (166 °C) for polymer modified asphalt, and the asphalt cement at 200 °F (95 °C). Paving and compaction must be performed while the asphalt is sufficiently hot. There are restrictions during the winter months as the compacted base will cool the asphalt too much before it is able to be packed to the required density. HMAC is the form of asphalt concrete most commonly used on high traffic pavements such as those on major highways, racetracks and airfields. Superpave, short for "superior performing asphalt pavement," is a pavement system designed to provide longer lasting roadways. Key components of the system are careful selection of binders and aggregates, volumetric proportioning of ingredients, and evaluation of the finished product. (10/2018)

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Polymer-Modified Binders - The term “polymer” refers to a large molecule formed by chemically reacting many (“poly”) smaller molecules (monomers) to one another in long chains or clusters. Physical properties of a specific polymer are determined by the sequence and chemical structure of the monomers from which it is made.

Classification System (label position example)

HMA 12.5 H 64 Surface Course HMA

12.5

Hot Mix Asphalt

H

Binder Grade (based on climate avg. max. 7-day temperature (°C))

64

Surface Course Location within Placement

Nominal Maximum Aggregate Size (mm) Compaction Level (Low Medium High)

HMA Components

Aggregate Base Course (ABC)

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TYPICAL ASPHALT CONCRETE PAVEMENT STRUCTURE

1.5-3-in 2-4-in

4-12-in

4-12-in

Typical asphalt concrete pavement structure. In many cases, the intermediate course is omitted; full-depth asphalt pavements do not include a granular subbase.

The term “hot-mix asphalt” is used generically to include many different types of mixtures of aggregate and asphalt cement that are produced at an elevated temperature in an asphalt plant. Most commonly HMA is divided into three different types of mix—dense-graded, open graded and gap-graded—primarily according to the gradation of the aggregate used in the mix.

Dense-Graded Hot-Mix Asphalt - Dense-graded HMA is composed of an asphalt cement binder and a well or continuously graded aggregate. Conventional HMA consists of mixes with a nominal maximum aggregate size in the range of 12.5 mm (0.5 in.) to 19 mm (0.75 in.). This material makes up the bulk of HMA used in the United States. Large-stone mixes contain coarse aggregate with a nominal maximum size larger than 25 mm (1 in.). Sand asphalt (sometimes called sheet asphalt) is composed of aggregate that passes the 9.5-mm (0.375-in.) sieve. The binder content of the mix is higher than that of conventional HMA because of the increased voids in the mineral aggregate in the mixture

Open-Graded Mixes Open-graded mixes consist of an aggregate with relatively uniform grading and an asphalt cement or modified binder. The primary purpose of these mixes is to serve as a drainage layer, either at the pavement surface or within the structural pavement section. Gap-Graded Mixes Gap-graded mixes are similar in function to dense-graded mixes in that they provide dense impervious layers when properly compacted. Conventional gap-graded mixes have been in use for many years. Their aggregates range in size from coarse to fine, with some intermediate sizes missing or present in small amounts.

(10/2018)

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ASPHALT CEMENT: GRADING SYSTEMS AND PROPERTIES

Penetration and Viscosity Grading Systems The penetration of an asphalt cement (indentation measured by a standard needle in units of 0.1 mm) is determined at 25°C (77°F). The stiffer the asphalt (i.e., the lower its penetration), the stiffer will be the mix containing the material at a given temperature. For example, at a given temperature, a mix containing 60–70 penetration grade asphalt cement typically will be stiffer and may require somewhat more compaction effort by the rollers to achieve the desired density than will a mix made using a 120–150 penetration grade asphalt cement. Grading of asphalt cements by viscosity is defined by a viscosity measurement at 60°C (140°F) on the material in its original (as received from the refinery) condition (termed AC) or on a binder considered to be comparable to the binder after it has passed through the hot-mix process (termed AR). In the AC grading system, a mix containing an AC-20 will be stiffer than a mix containing an AC-10 at the same temperature. Similarly, in the AR grading system, a mix containing an AR-4000 will be stiffer than one containing an AR­2000 at the same temperature.

Superpave Mixture Design Performance Grading System Superior Performing Asphalt Pavements. While grading systems based on penetration and viscosity have worked satisfactorily for many years, requirements have been based on tests performed at prescribed loading times and at standard temperatures not necessarily representative of in-service conditions. Limits for the tests have been based on agency experience. To provide an improved set of asphalt specifications, SHRP developed the PG system. Included in this new set of specifications are tests used to measure physical properties that can be related directly to field performance by engineering principles. Moreover, the tests are performed at loading times, temperatures, and aging conditions that represent more realistically those encountered by in-service pavements. The PG specifications help in selecting a binder grade that will limit the contribution of the binder to low-temperature cracking, permanent deformation (rutting), and fatigue cracking of the asphalt pavement within the range of climate and traffic loading found at the project site. An important difference between the PG specifications and those based on penetration or viscosity is the overall format of the requirements. For the PG binders, the physical properties remain constant; however, the temperatures at which those properties must be achieved vary depending on the climate in which the binder is expected to serve. An example of the binder designation in this system is PG 64-22. This binder is designed to resist environmental conditions in which the average 7-day maximum pavement design temperature is 64°C (147°F) or lower, and the minimum pavement design temperature is −22°C (−8°F) or higher.

4.75

19

(10/2018)

9.5

25

12.5

37.5

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HOW ASPHALT CONCRETE PAVEMENTS FAIL

Rutting Rutting (often referred to as permanent deformation) is a common form of distress in flexible pavements. When truck tires move across an asphalt concrete pavement, the pavement deflects a very small amount. These deflections range from much less than an 1/8—in in cold weather—when the pavement and subgrade are very stiff—to a 1-in or more in warm weather—when the pavement surface is hot and very soft. After the truck tire passes over a given spot in the pavement, the pavement tends to spring back to its original position. Often, however, the pavement surface will not completely recover. There will be a very small amount of permanent deformation in the wheel path. After many wheel loads have passed over the pavement this rutting can become significant. Rutting is a serious problem because the ruts contribute to a rough riding surface and can fill with water during rain or snow events, which can then cause vehicles traveling on the road to hydroplane and lose control. Rut depths of about 3/8-in (10 mm) or more are usually considered excessive and a significant safety hazard. Other related forms of permanent deformation include shoving and wash boarding. Shoving occurs at intersections when vehicles stop, exerting a lateral force on the surface of the hot mix causing it to deform excessively across the pavement, rather than within the wheel ruts. Wash boarding is a similar phenomenon but, in this case, the deformation takes the form of a series of large ripples across the pavement surface.

Fatigue Cracking Like rutting, fatigue cracking results from the large number of loads applied over time to a pavement subject by traffic. However, fatigue cracking tends to occur when the pavement is at moderate temperatures, rather than at the high temperatures that cause rutting. Because the HMA at moderate temperatures is stiffer and more brittle than at high temperatures, it tends to crack under repeated loading rather than deform. When cracks first form in an HMA pavement, they are so small that they cannot be seen without a microscope. The cracks at this point will also not be continuous. Under the action of traffic loading, these microscopic cracks will slowly grow in size and number, until they grow together into much larger cracks that can be clearly seen with the naked eye. Severe fatigue cracking is often referred to as “alligator cracking,” because the pavement surface texture resembles an alligator’s back. These large cracks will significantly affect pavement performance, by weakening the pavement, contributing to a rough riding surface, and allowing air and water into the pavement, which will cause additional damage to the pavement structure. Eventually fatigue cracking can lead to extensive areas of cracking, large potholes, and total pavement failure.

Low-Temperature Cracking Temperature has an extreme effect on asphalt binders. At temperatures of about 150°C (300°F) asphalt binders are fluids that can be easily pumped through pipes and mixed with hot aggregate. At temperatures of about 25°C (77°F), asphalt binders have the consistency of a stiff putty or soft rubber. At temperatures of about −20°C and lower, asphalt binders can become very brittle.

Moisture Damage Water does not flow easily through properly constructed HMA pavements, but it will flow very slowly even through well-compacted material. Water can work its way between the aggregate surfaces and asphalt binder in a mixture, weakening or even totally destroying the bond between these two materials. This moisture damage is sometimes called stripping. Moisture damage can occur quickly when water is present underneath a pavement, as when pavements are built over poorly drained areas and are not properly designed or constructed to remove water from the pavement structure. Even occasional exposure to water can cause moisture damage in HMA mixtures prone to it because of faulty design or construction or poor materials selection.

Raveling Raveling occurs when tires dislodge aggregate particles from the surface of an HMA pavement. Many of the same factors that contribute to poor fatigue resistance will also contribute to raveling, including low asphalt binder contents and poor field compaction. Because the pavement surface is exposed to water from rain and snow, poor moisture resistance can also accelerate raveling in HMA pavements.

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FOG SEAL Fog seals are a method of adding asphalt to an existing pavement surface to improve sealing or waterproofing, prevent further stone loss by holding aggregate in place, or simply improve the surface appearance. However, inappropriate use can result in slick pavements and tracking of excess material. The Asphalt Emulsion Manufacturers Association (AEMA) defines a fog seal as “a light spray application of dilute asphalt emulsion used primarily to seal an existing asphalt surface to reduce raveling and enrich dry and weathered surfaces” and adds that fog seals can seal small cracks. Fog seals are often referred to as enrichment treatments since they add fresh asphalt to an aged surface and lengthen the pavement surface life Fog seals are also useful in chip seal applications to hold chips in place in fresh seal coats. This can help prevent vehicle damage arising from flying chips.

FUNCTION OF A FOG SEAL A fog seal is designed to coat, protect, and/or rejuvenate the existing asphalt binder. The addition of asphalt will also improve the waterproofing of the surface and reduce its aging susceptibility by lowering permeability to water and air. The fog seal material (emulsion) must fill the voids in the surface of the pavement. Therefore, during its application it must have sufficiently low viscosity so as to not break before it penetrates the surface voids of the pavement. This is accomplished by using a slow setting emulsion that is diluted with water. Emulsions that are not adequately diluted with water may not properly penetrate the surface voids resulting in excess asphalt on the surface of the pavement after the emulsion breaks, which can result in a slippery surface.

(10/2018)

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ASPHALT PERFORMANCE

56. - Question

A new local highway exit ramp is under construction. According to the project specification, the American Association of State Highway and Transportation Officials (AASHTO) design structural number for the road is 4.0. The material specifications are as follows: Material

Dense Graded Aggregate Crushed Stone Base Course

Layer Experience Thickness Coefficient (in) 12 0.11 6

0.14

The Superpave plant mix asphalt concrete surface course with an experience coefficient of 0.44 is to be placed on the top of the specified subbase and base course materials. The required surface course thickness (in.) to meet the AASHTO project specification requirements is most nearly: a. b. c. d.

2 3 4 5

 

Solution: The AASHTO structural number can be used to find the structural Number (SN) to solve for the surface thickness. The SN is the sum of products of the layer thicknesses and strength coefficients (ai = layer coefficient; Di = thickness of layer (inches)): SN = a1D1 + a2D2 + a3D3 Rearrange the equation to determine D1 D1 = SN - a2D2 - a3D3 A1 D1 = 4 – (0.14)(6-in) - (0.11)(12-in) 0.44

Surface Course 5-inch DGA 12-inch

Stone Base 6-inch

D1 = 4.18-in (answer is 5-in)

(10/2018)

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57. - Question

The Highway Authority has scheduled for the next 3-months to replace the wearing surface of a major Interstate roadway. The scope of work for the overlay project is to mill and pave while maintaining the AASHTO structural number of 6.6. The original pavement consists of 10-in Portland cement treated base having a strength coefficient of 0.20, and an 8-in dense graded aggregate subbase. For a 3.2mile portion of the project, the paving contract specification is to mill and replace 6-in of the surface with recycled asphalt concrete having a surface course strength coefficient of 0.42 with the remaining 3-in of the original pavement having a strength of 0.30. The minimum strength coefficient for the subbase is most nearly: a. b. c. d.

0.075 0.102 0.150 0.205

Solution: Restate the information in the question and outline the numerical terms: Graphic Representation

Material

Topping (Asphalt Concrete) Mill Topping (Asphalt Concrete) Remaining Dense Graded Aggregate subbase Portland cement Base Course

Layer Layer Thickness Coefficient D (in) (a) 6 0.42

Equation Subscript a1D1

3

0.30

a2D2

8

Find

a3D3

10

0.20

a4D4

The AASHTO pavement structural number can be used to find the structural Number (SN) to solve for the surface thickness (ai = layer coefficient; Di = thickness of layer (inches):  

SN = a1D1 + a2D2 + a3D3 + a4D4 Substitute terms and use the equation to determine D1 6.6 = (0.42)(6-in) + (0.30)(3-in) + (a3)(8-in.) + (0.20)(10-in) a3 = 0.1475

(10/2018)

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The supplemental material found in this section has been found on previous exams and are important for you to review as you prepare for the exam.

(10/2018)

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A Concrete Primer Portland Production

(essentials for self-study)

Production of Portland cement starts with two basic raw ingredients: a calcareous material and an argillaceous material. The calcareous material is a calcium oxide, such as limestone, chalk, or oyster shells. The argillaceous material is a combination of silica and alumina that can be obtained from clay, shale, and blast furnace slag. These materials are crushed and then stored in silos. The raw materials, in the desired proportions, are passed through a grinding mill, using either a wet or dry process. The ground material is stored until it can be sent to the kiln. Modern dry process cement plants use a heat recovery cycle to preheat the ground material, or feed stock, with the exhaust gas from the kiln. In addition, some plants use a flash furnace to further heat the feed stock. Both the preheater and flash furnace improves the energy efficiency of cement production. In the kiln, the raw materials are melted at temperatures of 2500°F to 3000°F, changing the raw materials into cement clinker. The clinker is cooled and stored. The final process involves grinding the clinker into a fine powder. During grinding, a small amount of gypsum is added to regulate the setting time of the cement in the concrete. The finished product may be stored and transported in either bulk or sacks. A standard sack of cement is 94-lb, which is approximately equal to loose cement when freshly packed. The cement can be stored for long periods of time, provided it is kept dry.

Chemical Composition of Portland Cement

The raw materials used to manufacture Portland cement are lime, silica, alumina, and iron oxide. These raw materials interact in the kiln, forming complex chemical compounds. Calcination in the kiln restructures the molecular composition of the base elements.

Fineness of cement particles is an important property that must be carefully controlled.

Since hydration starts at the surface of cement particles, the finer the cement particles, the larger the surface area and the faster the hydration. Therefore, finer material results in faster strength development and a greater initial heat of hydration. Increasing fineness beyond the requirements for a type of cements increases production costs and can be detrimental to the quality of the concrete.

Specific Gravity The specific gravity of cement is needed for mixture proportioning calculations. The specific gravity of Portland cement (without voids between particles) is about 3.15 and can be determined according to ASTM C188. The density of the bulk cement (including voids between particles) varies considerably, depending on how it is handled and stored. For example, vibration during transportation of bulk cement consolidates the cement and increases its bulk density. Thus, cement quantities are specified and measured by weight rather than volume.

Hydration

is the chemical reaction between the cement particles and water. The features of this reaction are the change in matter, the change in energy level, and the rate of reaction. Since Portland cement is composed of several compounds, many reactions are occurring concurrently. The hydration process occurs through two mechanisms: through-solution and topochemical. The through-solution process involves the following steps: 1. dissolution of anhydrous compounds into constituents 2. formation of hydrates in solution 3. precipitation of hydrates from the supersaturated solution The through-solution mechanism dominates the early stages of hydration. (10/2018)

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Topochemical hydration is a solid-state chemical reaction occurring at the surface of the cement particles.

Structure Development in Cement Paste The sequential

development of the structure in a cement paste is summarized in the following: The process begins immediately after water is added to the cement. In less than 10 minutes, the water becomes highly alkaline (neutralizes acids). As the cement particles hydrate, the volume of the cement particle reduces, increasing the space between the particles. During the early stages of hydration, weak bonds can form, particularly from the hydrated elements. Further, hydration stiffens the mix and begins locking the structure of the material in place. Final set occurs when the chemical phase has developed a rigid structure, all components of the paste lock into place and the spacing between grains increases as the grains are consumed by hydration. The cement paste continues hardening and gains strength as hydration continues. Hardening develops rapidly at early ages and continues, as long as un-hydrated cement particles and free water exist. However, the rate of hardening decreases with time.

Setting

Setting refers to the stiffening of the cement paste or the change from a plastic state to a solid state. Although with setting comes some strength, it should be distinguished from hardening, which refers to the strength gain in a set cement paste. Setting is usually described by two levels: initial set and final set.

Mixing Water

Any potable water is suitable for making concrete.

Cementitious Materials Fly Ash

Fly ash is the most commonly used pozzolan in civil engineering structures. Fly ash is a by-product of the coal industry. Combusting pulverized coal in an electric power plant burns off the carbon and most volatile materials. However, depending on the source and type of coal, a significant amount of impurities passes through the combustion chamber. The carbon contents of common coals ranges from 70 to 100 percent. The noncarbon percentages are impurities (e.g., clay, feldspar, quartz, and shale), which fuse as they pass through the combustion chamber. Exhaust gas carries the fused material, fly ash, out of the combustion chamber. The fly ash cools into spheres, which may be solid, hollow (cenospheres), or hollow and filled with other spheres (plerospheres). Particle diameters range from to more than 0.1 mm, with an average of 0.015 mm to 0.020 mm, and are 70% to 90% smaller than 0.045 mm. Fly ash is primarily a silica glass composed of silica alumina iron oxide and lime (CaO). Fly ash is classified (ASTM C618) as follows:

Class N—Raw or calcined natural pozzolans, including diatomaceous earths, opaline cherts and shales, ruffs and volcanic ashes or pumicites, and some calcined clays and shales Class F—Fly ash with pozzolan properties Class C—Fly ash with pozzolan and cementitious properties Class F fly ash usually has less than 5% CaO but may contain up to 10%. Class C fly ash has 15% to 30% CaO.

The spherical shape of fly ash increases the workability of the fresh concrete. In addition, fly ash extends the hydration process, allowing a greater strength development and reduced porosity. Studies have shown that concrete containing more than 20% fly ash by weight of cement has a (10/2018)

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much smaller pore size distribution than Portland cement concrete without fly ash. The lower heat of hydration reduces the early strength of the concrete. The extended reaction permits a continuous gaining of strength beyond what can be accomplished with plain Portland cement.

Ground Granulated Blast Furnace Slag (GGBFS)

Ground granulated blast furnace slag (GGBF slag) is made from iron blast furnace slag. It is a nonmetallic hydraulic cement consisting basically of silicates and aluminosilicates of calcium, which is developed in a molten condition simultaneously with iron in a blast furnace. The molten slag is rapidly chilled by quenching in water to form a glassy sand like granulated material. The material is then ground to less than 45 microns. The specific gravity of GGBF slag is in the range of 2.85 to 2.95. The rough and angular-shaped ground slag in the presence of water and an activator, NaOH or CaOH, both supplied by Portland cement, hydrates and sets in a manner similar to Portland cement. Ground slag has been used as a cementitious material in concrete since the beginning of the 1900s. Ground granulated blast furnace slag commonly constitutes between 30% and 45% of the cementing material in the mix. Some slag concretes have a slag component of 70% or more of the cementitious material.

Silica Fume Silica fume is a byproduct of the production of silicon metal or ferrosilicon

alloys. One of the most beneficial uses for silica fume is as a mineral admixture in concrete. Because of its chemical and physical properties, it is a very reactive pozzolan. Concrete containing silica fume can have very high strength and can be very durable. Silica fume is available from suppliers of concrete admixtures and, when specified, is simply added during concrete production either in wet or dry forms. Placing, finishing, and curing silica fume concrete require special attention on the part of the concrete contractor. Silicon metal and alloys are produced in electric furnaces. The raw materials are quartz, coal, and woodchips. The smoke that results from furnace operation is collected and sold as silica fume. Silica fume consists primarily of amorphous (noncrystalline) silicon dioxide. The individual particles are extremely small, approximately 1/100th the size of an average cement particle.

Pozzolans A pozzolan is a siliceous and aluminous material which, in itself, possesses

little or no cementitious value but will, in finely divided form and in the presence of moisture, react chemically with calcium hydroxide at ordinary temperatures to form compounds possessing cementitious properties (ASTM C595). Naturally occurring pozzolans, such as fine volcanic ash, combined with burned lime, were used and the high content, silica fume is a very reactive pozzolan when used in concrete. The quality of silica fume is specified by ASTM C 1240 and AASHTO M 307. In addition to producing high-strength concrete, silica fume can reduce concrete corrosion induced by deicing or marine salts. Silica fume concrete with a low water content is highly resistant to penetration by chloride ions.

Natural cementitious materials were discovered about 2000 years ago and are used for building construction and pozzolan continues to be used today.

Note: The content found in the pages of the Concrete Primer is an important part of your self-study program. Questions on previous exams have included many of the topics in this section. This is “need to” study material for all PE exam candidates.

(10/2018)

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CONCRETE MIX DESIGN – CONCRETE ESSENTIALS 1. Concrete mix is designated as the ratio of cement to fine aggregate to coarse aggregate 2. Ratios can be in terms of weight or volume 3. Amount of water is usually quantified in terms of gallons of water per 94 lbs. sack of cement (or water-cement ratio) 4. Amount of concrete (in-place) = Sum of solid volumes of cement, sand, aggregate, and water (Absolute Volume Method). 5. Densities of material are given in terms of specific gravity. 6. The easiest way to solve problem statements are with ratio in terms of weight, but problem statements may give ratio in terms of volume (best to convert to weight ratio) 7. Sand and aggregate which are not at the Saturated Surface Dry (SSD) condition (changes weights of sand and aggregate which changes water requirements) 8. Air entrainment (increases volume)by allowing microscopic “bubbles” in the mix to allow for expansion during the freeze-thaw cycle 9. Avoid adding water to concrete mix during placement at the construction site 10. Often water is erroneously added to improve workability (liquefaction) which leads to weakened strength. Adding water will increase w/c ratio thereby decreasing strength. 11. Use of admixtures will improve concrete workability without changing the w/c ratio specified in the mix design. Admixtures are liquid chemicals which do not affect the w/c ratio (strength). 12. If workability (or project LEED) requirements are anticipated, add supplementary cementitious materials (10/2018)

(e.g., fly ash, blast furnace slag, pozzolans and the like) during mix design 13. Silica fume Silica fume is used to increase strength and durability of concrete. A by-product of the production of silicon and ferrosilicon alloys. Silica fume is similar to fly ash, 14. Slag by-product of steel production is used to partially replace Portland cement (by up to 80% by mass). It has latent hydraulic properties is a byproduct of electric-arc furnaces 15. Fly ash a by-product of coal-fired electric generating plants; it is used to partially replace Portland cement (by up to 60% by mass). 16. Do not use Calcium Chloride (CaCl3) as an admixture - deleterious effect as it causes corrosion. 17. What happens in hot weather? a. Add Ice at the batch plant b. Increase size of aggregate which decreases the amount of cement c. Paint formwork white(reflective) d. Admixtures using retarders 18. What happens in cold weather? a. Admixtures which accelerate the heat of hydration reaction are added to the mix b. Add hot water to the batch mix c. Heat the placement area d. Paint formwork dark color e. Use insulated blankets f. Cover with membrane for curing 19. Monitor concrete maturity to determine when concrete is able to be loaded to design strengths 20. Concrete is an age hardening product – Becomes stronger as it gets older 21. Curing compounds strengthen concrete by slowing the moisture escape to aid the cement hydration by using chemical agents or water on the surface

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Types of Portland Cement (PC) 

Type I Normal PC – General Purpose



Type II Modified PC – Moderate sulfate resistance, used in hot weather for larger structures



Type III High Early Strength



Type IV Low Heat – Large structures



Type V Sulfate Resistant

CURING DEFINED

Curing can be defined as a procedure for insuring the hydration of the Portland cement in newly-placed concrete. It generally implies control of moisture loss and sometimes of temperature. The hydration of Portland cement is the chemical reaction between grains of Portland cement and water to form the hydration product, cement gel: and cement gel can be laid down only in water-filled space. Hydration can proceed until all the cement reaches its maximum degree of hydration or until all the space available for the hydration product is filled by cement gel, whichever limit is reached first.

(10/2018)

Honeycombing Honeycombs are voids in the concrete caused by a lack of mortar using coarse aggregate agglomerates. This defect normally occurs when the concrete segregates because of an obstruction blocking flow when the concrete is laid, an excessive drop height, high reinforcement density or because of paste escaping from the formwork. It also tends to happen because of the following circumstances: difficult access to lay the concrete, making it necessary for the concrete to cover large distances; difficulty in vibration because of high mortar density or the distance between the access point and the concrete; high ratio between the maximum aggregate size and reinforcement spacing; and the concrete not properly studied for workability requirements.

Flyash Class C flyash offers higher early-age strength than Class F flyash. However, the strength contribution of the Class F flyash is higher than the Class C; leading to higher long term compressive strength of the concrete.

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CONCRETE PROTECTION FOR REINFORCEMENT

AA. Which of the following statements inhibit concrete corrosion: I. II. III. IV. V.

Use air-entrained concrete with a w/c of 0.40 or less. Use a minimum concrete cover of 1.5 inches and at least 0.75 inch larger than the nominal maximum size of the coarse aggregate. Increase the minimum cover to 2 inches for deicing salt exposure, and to 2.5 inches for marine exposure. Ensure that the concrete is adequately cured. Use of Silica fume, fly ash, and blast-furnace slag reduce the permeability of the concrete to the penetration of chloride ions. a. I & II b. I, II, & III c. I, II, III, & IV d. I, II, III, IV, & V

Solution: All are true (answer = d) ASTM terminology defines corrosion as “the chemical or electrochemical reaction between a material, usually a metal, and its environment that produces a deterioration of the material and its properties.” For steel embedded in concrete, corrosion results in the formation of rust which has two to four times the volume of the original steel and none of the good mechanical properties. Corrosion also produces pits or holes in the surface of reinforcing steel, reducing strength capacity as a result of the reduced cross-sectional area. Steel in concrete is usually in a non-corroding, passive condition. However, steel reinforced concrete is often used in severe environments where sea water or deicing salts are present. When chloride moves into the concrete, it disrupts the passive layer protecting the steel, causing it to rust and pit. Carbonation of concrete is another cause of steel corrosion. When concrete carbonates to the level of the steel rebar the normally alkaline environment, which protects steel from corrosion, is replaced by a more neutral environment. Under these conditions the steel is not passive and rapid corrosion begins. The rate of corrosion due to carbonated concrete cover is slower than chloride-induced corrosion. Corrosion Protection Systems – Increased corrosion resistance can also come about by the use of concrete additives. Silica fume, fly ash, and blast-furnace slag reduce the permeability of the concrete to the penetration of chloride ions. Corrosion inhibitors, such as calcium nitrite, act to prevent corrosion in the presence of chloride ions. In all cases, they are added to quality concrete at w/c less than or equal to 0.45.

(10/2018)

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BB. Ready-mixed 4000-psi concrete being delivered to a jobsite is found by the third-party inspector to have a slump less than the 6-in. specified. Which of the following is the most appropriate corrective action? a. Decrease the amount of water in the mix before the truck leaves the ready-mix plant. b. Increase the water to the mix in the truck at the jobsite before the concrete is poured. c. Increase the rotation speed of the mixing drum while the truck is in transit to the jobsite. d. Add an admixture to the mix in the truck at the jobsite before the concrete is poured.

Answer = d, Analysis: The w/cm ratio cannot be altered, while the number of drum revolutions can affect slump and entrained air content. Therefore, keeping a consistent number of revolutions (typically limited to 300-revolutions) on a specific mixture and project helps the ready-mix producer consistently compensate for any such changes. More importantly, operating a drum at high speeds when driving is a serious safety risk. Many other ASTM requirements–such as for temperature, water addition, and jobsite testing–are in place to ensure concrete quality at the time of discharge and make a limit on revolutions redundant and in many cases unnecessary. Concrete producers have a wide variety of options to ensure concrete quality well beyond 300 revolutions. For example, during hot weather, producers can add chemical admixtures (retarders or hydration control admixtures) and reduce concrete temperature (chilled water, liquid nitrogen, or ice). For long hauls, they can maintain slump by adding superplasticizers with long slump retention or add superplasticizer on the truck during transit or at the jobsite.

During your self-study, take note of the various non-quantitative exam questions and build your reference library on the topics focused on in the School of PE Refresher Courses.

(10/2018)

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CC.

Compressive strength test results are primarily used to determine that the

concrete mixture as delivered meets the requirements of the specified strength, f’c, in the job specification. Strength test results from cast cylinders may be used for: 1. quality control 2. acceptance for concrete placement 3. estimating the concrete strength in a structure 4. the purpose of scheduling construction operations such as form removal 5. evaluating the adequacy of curing 6. frost protection afforded to the structure a. b. c. d.

1,3,4,5 3,4,5,6 1,2 4,5,6 1,2,3,4,5,6

Answer: All are true. Analysis: Cylinders tested for acceptance and quality control are made and cured in accordance with procedures described for standard-cured specimens in ASTM C 31 Standard Practice for Making and Curing Concrete Test Specimens in the Field. For estimating the in place concrete strength, ASTM C 31 provides procedures for field-cured specimens. Cylindrical specimens are tested in accordance with ASTM C 39, Standard Test Method for Compressive Strength of Cylindrical Concrete Specimens. A test result is the average of at least two standard-cured strength specimens made from the same concrete sample and tested at the same age. In most cases strength requirements for concrete are at an age of 28 days.

DD. The project’s specification for cast-in-place concrete requires a f’c of 3000-psi. Which of the following individual test breaks (psi) indicate(s) a failure? a. b. c. d.

2900 3000 3400 None of the above

Answer is d. Analysis: It is important to understand that an individual test falling below ƒ´c does not necessarily constitute a failure to meet specification requirements. When the average of strength tests on a job are at the required average strength, f’cr, the probability that individual strength tests will be less than the specified strength is about 10% and this is accounted for in the acceptance criteria. When strength test results indicate that the concrete delivered fails to meet the requirements of the specification, it is important to recognize that the failure may be in the testing, not the concrete. This is especially true if the fabrication, handling, curing and testing of the cylinders are not conducted in accordance with standard procedures.

(10/2018)

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REINFORCEMENT TYPE

CC. Which of the following statements regarding the use of steel as a reinforcement material in reinforced concrete is/are true: I.

Steel costs less than aluminum.

II.

Aluminum reinforcement is allowed for use in highly corrosive environments such as sewerage treatment plants.

III.

The thermal properties of steel and concrete are closely matched.

IV.

Lead (Pb) coated steel reinforcement rebar is used for the construction of nuclear power plants. a. b. c. d.

I, II, III II, III, IV I ONLY III ONLY

Solution:

(10/2018)

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Aluminum (Al) Embedded aluminum roof flashing, aluminum water stops, aluminum electrical conduit, introduced aluminum powder (sometimes used to foam concrete), or embedded structural aluminum shapes may all corrode in concrete or mortar. In all cases, a reaction that forms aluminum hydroxide and hydrogen gas occurs, and may cause expansion and cracking of the concrete or mortar. The common use of calcium chloride (or other alkali compounds), and dampness of the concrete increases the reaction rate. Usually, coating the aluminum with bituminous paint, impregnated paper or felt, plastic, or an alkali-resistant coating will prevent or sharply reduce the corrosion. Aluminum reacts with the alkalis (OH) found in Portland cement concrete. When these two chemicals are combined, the reaction produces hydrogen gas. When the reaction occurs in wet concrete, tiny bubbles coming to the surface of a slab will be noticed. Significant corrosion of aluminum embedded in concrete can occur. The corrosion can cause expansion of the concrete and subsequent cracking of hardened concrete. Also, if the aluminum is coupled with any ferrous metals, galvanic corrosion will occur also. In both cases the presence of calcium chloride greatly accelerates the corrosion process.

Copper (Cu) Copper embedded in concrete and/or mortar is usually roof flashing. Embedded copper is practically immune to reaction with corrosive alkalis, even if exposed to constant moisture. Copper will not react with dry, hardened concrete and/or mortar. Rainwater leaching, however, may bring chlorides in contact with the metal. Corrosion may occur and result in a green discoloration or runoff. Consequently, chloride admixtures should not be used in concrete if contact with copper is expected.

Lead (Pb) Lead will always corrode when in contact with fresh concrete and/or mortar. The high pH from calcium hydroxide is the cause of the corrosion. Cured, seasoned concrete or mortar will not react with lead. Corrosion of embedded lead flashing in mortar joints will usually result in the production of a lead oxide, a white discoloration. A special case of lead corrosion, called differential aeration, occurs when a lead strip is partially embedded in concrete so that part of the strip is exposed to air. The embedded section has a different electrical potential than the section exposed to air. The result is that the strip will become polar in the presence of moisture. Gradual corrosion and disintegration of the embedded lead will then follow. In such a case, and in all other cases, the embedded portion should be coated with epoxy, varnish, asphalt, or pitch.

Zinc (Zn) Zinc is highly reactive with alkalis and will deteriorate to some degree upon contact with fresh concrete and/or mortar. The reaction is limited due to a corrosive film that forms on the outer layer of the zinc. It protects the underlying metal from further reaction. Zinc will not react with dry, seasoned concrete and/or mortar. Embedded zinc will react with moisture and calcium hydroxide to produce calcium zincate. Zinc corrosion may also occur when galvanized iron, in the form of flat or corrugated sheets and rebar, comes in contact with fresh concrete and/or mortar. Galvanized iron is coated with zinc, and will react with moisture and chlorides in the concrete and/or mortar to produce zinc chloride. The result is expansion and cracking of the concrete and/or mortar. The metal should be protected with epoxy, varnish, asphalt, or pitch. Answer - d (10/2018)

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ACI – JOINTS IN CONCRETE CONSTRUCTION

DD. The concrete pavement slab on grade cross section in the figure represents most nearly: a. b. c. d.

Expansion and/or Contraction Joint Control Joint Construction Joint Isolation Joint

Solution: Reference ACI 224.3 Joints in Concrete Construction:  Control Joints typically are created with a saw cutting the surface of the concrete and are placed at approximately max distance of 18-ft o.c. and is used as a method to control cracking. Steel reinforcement may be present at a control joint, however rebar dowels will not.  Isolation joints separate concrete slabs from walls, columns, and/or footings. The joints isolate and separate the elements that may have varying loading and settlement (usually placed on a soil subgrade). No connection between the various structural elements should exist.  Expansion and contraction joints can be doweled; however, a bond breaker is used to allow the slab to allow for it to slide along the dowel bar as the slab expands or contracts.  Construction joints are doweled to ensure complete load transfer when subsequent portions of the concrete slab are poured. Construction Joints are required between all concrete pours separated by more than 90-minutes in time.( Answer c)

(10/2018)

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EE. Cement Hydration is a very exothermic process, leading to a rise in temperature at the core of very large pours. Massive foundations, bridge piers, thick slabs, structural columns and the like require special attention during construction. If the surface temperature is allowed to deviate greatly from that of the core, thermal cracking will develop. Most codes require a temperature (ºF) differential from the surface to the core of the section is most nearly: a. b. c. d.

10 35 80 120

Solution: From the ACI Manual of Concrete Practice - Cement hydration produces a

rise in internal temperature. The outer surface cools faster than the core of the section. By thermal expansion/contraction, the temperature differential induces thermal (tensile) stresses at the surface. Stresses increase Tensile Strength which results in Thermal Cracking Temperature rise varies by many parameters: Cement composition, fineness, and content; Aggregate content and the Coefficient of Thermal Expansion; Section geometry; Placement & ambient temperature. Most of the concrete mass temperature rise occurs in first 1-3 days after placement. The coarser (larger) the aggregate the cooler the concrete becomes due to a reduced amount of concrete. Most codes require a temperature differential of less than 36ºF from the surface to the core of the section.

1-3 days after placement

Cement Fineness; Cement with a lower fineness with slow hydration, and reduce temperature rise. Cement Content; Mass Concrete mixtures should contain as low of a cement content as possible to achieve the desired strength. This lowers the heat of hydration and subsequent temperature rise. Aggregate Content: Coarse Aggregate should have a greater surface area if possible. A higher coarse aggregate content (7085%) in the design mix can be used to lower the cement content thereby reducing the mix temperature rise. Normal Concrete = Larger Aggregate will allow for better packing which will allow for a use of less cement with the outcome of less heat generation. (Answer b)

(10/2018)

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CONCRETE MIX DESIGN - RATIOS

fast facts The following illustrates an easy way to remember the components and basics of concrete mix design. The Table shows an “approximation” technique for a utility concrete mix design based on using a single bag cement field mixer. The illustration is for a utility . grade jobsite mix based on the proportions of 1 : 2 : 3 with a w/c = .50. See CERM-16 Table 49.2.

Water Cement Sand Gravel Standard Weight

8.34 #/gal

94 #/sack

165 #/ft3

165 #/ft3

[94#/sack ÷ 8.34#/gal = [195#/ft3] 11.26gal/sack]

Proportions w/c = .50

1

[2 x 94 = [3 x 94 = 188#/sack] 282#/sack]

2

3

≅ 200# (=188#)

≅ 300# (=282#)

5gal

Mix Weight (lbs)

≅ 50# (=47#)

Cement

Material

≅ 100# (=94#)

Volume: Mix = 50# + 100# + 200# +300# = 650# Normal weight of concrete ≅ 150#/ft3 650# ÷ 150#/ft3 ≅ 4-ft3 x 7-sacks ≅ 28-ft3 ≅ 1-yd3

(10/2018)

 

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



CONCRETE ADMIXTURES

fast facts Admixtures in concrete are known as ingredients added to concrete immediately before or during mixing (other than Portland cement, aggregates, and water). Six types of concrete admixtures are described below: 1. Accelerators (ASTM C494, Type C): Accelerate setting and enhance early strength (helpful in cold weather concreting). Example: calcium chloride (ASTM D98). However, because of its corrosion potential, calcium chloride—especially in prestressed concrete—has been strictly limited in use. ACI Committee 222 (1988) has determined that total chloride ions should not exceed 0.08% by mass of cement in prestressed concrete. Many specifying agencies strongly recommend that calcium chloride should never be added to concrete containing embedded metals. Although calcium chloride is an effective and economical accelerator, its corrosion-related problem limited its use and forced engineers to look for other options, mainly non-chloride accelerating admixtures. A number of non-chloride compounds—including sulfates, formates, nitrates, and triethanolamine—are being used to conform to the project specifications. 2. Air entraining (ASTM C260): Improves durability and workability. Example: salts of wood resins (vinsol resins). Usually specified for exterior applications in cold weather climates (typical air range of 5% to 6%). Air pockets are formed in the concrete which provide areas where the concrete can expand into during the freeze-thaw cycle without damaging the concrete. 3. Retarders (ASTM C494, Type B): Retard the setting time to avoid difficulties with placing and finishing (typically used in hot weather). Example: lignins. 4. Superplasticizers (ASTM C1017, Type 1): Make high-slump concrete (required for flowing or pumping concrete) from concrete with normal to low watercement ratios, allow for easy placing, and reduce and sometimes eliminate the need for vibration. Example: lignosulfonates. 5. Water reducers (ASTM C494, Type A): Reduce water requirement to produce concrete of a certain slump. Example: lignosulfonates. 6. Pozzolans (ASTM C618): Improve the properties of concrete by changing the properties of the various types of cement; substituted for certain amounts of cement; reduce temperature rise, alkaliaggregate expansion, and harmful effects of tricalcium aluminate. Examples are: fly ash, blast furnace slag, ground pumice.

(10/2018)

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TEMPORARY STRUCTURES

Temporary Structures

Concept Terminology



(10/2018)

CHAPTER

CHAPTER 6

6

Construction Loads

Load Paths

Gravity Loads Dead Loads Live Loads Hydrostatic Pressure Concrete Form Ties Tributary Area Torque Design Loads

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FALSEWORK

Falsework consists of temporary structures used in construction to support spanning or arched structures in order to hold the component in place until its construction is sufficiently advanced to support itself.

Falsework also includes temporary support structures for formwork used to mold concrete to form a desired shape, scaffolding to give workers access to the structure being constructed, and shoring which is temporary structural reinforcement used during repairs. The standard definition for falsework is "Any temporary structure used to support a permanent structure while it is not self-supporting."

Falsework is any temporary structure used to support a permanent structure while it is not self-supporting, either in new construction or refurbishment. Any failure of falsework may lead to the collapse of the permanent structure. This could cause injury or death to those working on or near to it, as well as loss of time and money. Temporary structures are utilized in various construction operations, such as: a. concrete formwork construction; b. scaffolding; c. falsework/shoring; d. cofferdams; e. underpinning; f. diaphragm/slurry walls; g. earth-retaining structures; and h. Construction dewatering. Temporary structures are critical elements of the overall construction plan. A temporary structure in construction affects the safety of the workers on the job and the general public and there is also the relationship of the temporary structure to the finished structure.

(10/2018)

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FORMWORK

Common types of form ties

(10/2018)

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CONCRETE FORMWORK

Design Loads on Forms: The American Concrete Institute (ACI) publishes a document called “ACI 347-14 –Guide to Formwork for Concrete.” 1. Form Dead Loads – the actual weight of the forms, plus the weight of fresh (i.e., wet) concrete. 2. Form Live Loads – the weight of workers, equipment and material storage. The minimum live load is 50 psf, while a live load of 75 psf should be used if motorized buggies are used. 3. Lateral Loads on Formwork – wet concrete is like water – it exerts a lateral pressure which increases with the depth of the form casting a triangular load distribution.

(10/2018)

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(10/2018)

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LOAD PATHS - DEFINED

Every load applied to the building will travel through the structural system until it is transferred to the supporting soil exerting equal and opposite reaction forces, thereby, attaining static equilibrium. The analysis of the travel is known as a load path.

HVAC

WIND

Loads

Snow Rain Wind Dead Live Seismic

a. The building dead load is the only known load. b. All other forces will vary in magnitude, duration, and location. c. The building is designed for design load possibilities that may never occur. The contributory (or tributary) loads imposed on structural analysis of a building concentrate on three kinds of loads that are generally used: 1. Concentrated loads that are single forces acting over a relatively small area, for example vehicle wheel loads, column loads, or the force exerted by a beam on another perpendicular beam. 2. Line loads that act along a line, for example the weight of a partition resting on a floor, calculated in units of force per unit length. 3. Distributed (or surface) loads that act over a surface area. Most loads are distributed or are treated as such, for example wind or soil pressure, and the weight of floors and roofing materials. (10/2018)

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Static Equilibrium

Newton’s third law of motion, the law of action and reaction, states that for every force acting on a body, the body exerts a force having equal magnitude in the opposite direction along the same line of action as the original force.

(10/2018)

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LOAD PATHS

58. - Question Which one of the following figures below most nearly depicts the load path for lateral resistance upon the static load case (Arrowheads depict the direction of force only): a) b) c) d)

A B C D

a. Load Path Diagram

b. Load Path Diagram

(10/2018)

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c. Load Path Diagram

d. Load Path Diagram

Class exercise; determine the answer by examining the forces that are transferred and in equilibrium. (10/2018)

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59. - Question ASTM A-36 structural steel shapes are being fabricated to accept steel wire rope connectors to construct a temporary structure. Which of the following figures properly depict the location of the neutral axis where the through holes will be drilled: a) b) c) d)

A B C D Flexural Load

Flexural Load B

A Neutral Axis

Neutral Axis

Flexural Load

Flexural Load C

D

Neutral Axis

Neutral Axis

Steel Construction Manual, 13th ed., American Institute of Steel Construction. CHAPTER F DESIGN OF MEMBERS FOR FLEXURE, pg 281.

(10/2018)

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TRIBUTARY AREA

60. - Question

An extract from the structural drawings for the first floor rigid frame structure is given. The floor system must support its own weight of 60-psf and a live load of 125 psf. The unfactored column load for column B3 is most nearly: a. b. c. d.

11,000 33,300 45,000 66,600

Solution: Determine the equivalent surface area (tributary area) surrounding column B3. The structural bay spacing is 18’-0” x 20’-0”. The combined live and dead load is 185-lbs and is considered the unfactored load. (18’-0”) x 20’-0” x 185-lb/SF = 66,600-lb

(10/2018)

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61. - Question

Which of the following load combinations produces maximum uplift on the base plate at Footing A: a. b. c. d.

LIVE

Dead + Live Dead + Wind Wind Dead + Live + Wind

DEAD

WIND

Solution:

Footing A

By inspection the structure is a rigid body and the cantilever will contribute to the uplift force, answer is d.

A view of a castellated steel beam representing that the further the web flanges are from the neutral axis increases the load capacity of the beam by creating a greater moment of inertia.

(10/2018)

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AA. A hinge can support: a. b. c. d.

A force but no moment A force and a moment Neither a force or a moment Two forces and a moment

Answer is a

BB.

A pulley is used to move a load for all the following EXCEPT:

a. b. c. d.

Reduce force Reduce energy Change direction of an applied tensile force Gain a mechanical advantage

Answer is b

CC. Which of the following statements is NOT a characteristic of statically indeterminate rigid body force system a. It can be solved for all unknows which are usually reactions supporting the body b. One or more of the members can be removed or reduced in restraint without affecting the equilibrium position c. The number of redundant members is the degree of indeterminacy d. A statically indeterminate body requires additional equations to supplement the equilibrium equations Answer is a

DD. Which of the following statement about axial members is NOT true: a. A horizontal member caries only horizontal loads. It cannot carry vertical loads b. A vertical member caries only vertical loads. It cannot carry vertical horizontal loads c. The vertical components of an axial member’s force is equal to the vertical components of the applied load applied to the member d. The end moments must sum to zero Answer is d

(10/2018)

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TEMPORARY LOADS – FACTOR OF SAFETY

62. - Question A normal weight concrete deadman restrained from sliding with dimensions 3-ft by 3-ft and is 3-ft long is used to support a formwork diagonal brace which has a compressive load of 1500-lb. The brace attaches to a center-pin clevis which is 8-in above the block and is located 12-ft away from the face of the formwork and is attached to a wale 16-ft above the center pin. The factor of safety from overturning is most nearly: a) 1.2 b) 1.5 c) 2.2 d) 2.9 Solution: Sketch the problem statement.

Fv

1500-lb

1200-lb

16’

Not to scale

Fh 900-lb

12’ 8” 3’ 3’ A

3’

CG 4050-lb

(10/2018)

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Step 1: Determine weight of deadman weight using normal weight of 150lb/ft3 Weight Weight

= 150-lb/ft3 x 3-ft x 3-ft x 3-ft = 4050-lb

Step 2: Resolve the axial load force of into the horizontal and vertical components. Note the brace geometry forms a 3-4-5 triangle. Fh Fv

= 1500 cos(53º) = 900 = 1500 sin (53º) = 1200

Step 3: Determine the factor of safety Summing the moments about point “A” on the Deadman, the factor of safety against overturning is calculated as follows: Factor of Safety =

Resisting Moments = 4050 (1.5’) + 1200 (1.5’) Overturning Moments 900 (3.67’)

Factor of Safety = 2.384

(10/2018)

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TORQUE

A workers hand grip force of 35-lb is applied perpendicular to the top of the handle of the specialty spud wrench in the configuration as shown. The torque (in-lb) applied to the ¾” - 10 UNC steel bolt along the vertical bolt axis is most nearly:

63. - Question

a) b) c) d)

250 360 390 420

35-lb Bolt Axis 60º 1-ft

Not to scale

6”

¾”

Section View

Plan View

Solution: Torque is equivalent to moment, find the moment along the vertical axis of the bolt. Notice that the 35-lb force is applied at the top of the wench in a forward motion to move the bolt. Moment Moment

= (Force) (perpendicular distance) = 35-lb (0.75-in + 12-in sin 60º) = 390-in-lb 12 6

60°

¾”

10

(10/2018)

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BALANCING LOADS

64. - Question A horizontal W21 x 132 beam is temporarily supported using a wire rope in tension as shown. The downward force (lb) at point A is most nearly: a. b. c. d.

5280 10560 15840 31680

20-ft

50-ft

A

10-ft

B

Wire Rope

Not to Scale

(10/2018)

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10-ft

40-ft

20-ft

B

A

NB

CG NA Solution:

Draw a free body diagram to outline the forces.

Weight of the beam is 80-ft x 132-#/ft = 10560-lb Center of gravity is located 40-ft to the center of measure. Find the Moment about point B:

y

+ ΣMB = 0: (10560-lb)(30-ft) - NA(20-ft) = 0 NA = 15840-lb

(10/2018)

x

+

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CONCENTRATED LOADS UNSYMMETRICALLY PLACED

65. - Question A 1-1/4” stainless steel wire rope (IWRC) is carrying the traffic signal loads as shown in the figure. The deflection at point B is 18-in. and the signal weighs 225-lb. The signal at point C weighs 125-lb. Neglecting the elongation of the wire rope, the tension force (lb) in the cable between C and D is most nearly: a. 122 12-ft 40-ft 16-ft b. 228 c. 3255 C D B A d. 3263

Solution: Draw a free body diagram to outline the forces and compute the horizontal forces at the posts and the reaction at point D. 40-ft

12-ft

16-ft

HA

HD A

18-in

B

C

D

RA

RD 125-lb 225-lb

RA = (225-lb x 28-ft) + (125-lb x 16-ft) 68-ft RD = (225-lb x 40-ft) + (125-lb x 52-ft) 68-ft HA = 40-ft x RA = 3254.93-lb 1.5-ft FCD =

=

= 122.06-lb = 227.94-lb

HD

√227.942 + 3254.932

FCD = 3262.91-lb

Hint: Use your Beam Diagrams and Formulas found in the reference materials to establish your solution steps. (10/2018)

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Sample of a beam diagrams and formulas.

http://www.awc.org/pdf/codes-standards/publications/design-aids/AWC-DA6-BeamFormulas-0710.pdf

(10/2018)

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OBJECTS AT REST

66. - Question The diagrams below show forces applied to a crane pulley that weighs 20-kips (represented by “W”). The diagram(s) which represent the pulley in equilibrium is: a. b. c. d.

A A and C C D and C 20k

20k

10k

10k

10k

10k

B.

A.

W

W

10k

10k

10k

C. W

D. 10k

Geo

Solution: Objects are at rest, static, and/or in equilibrium when the ∑F = 0; and ∑torque = 0. Effectively, there is no translation of forces and no rotation portrayed in C.

(10/2018)

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AA. Calculate the weight (k-lbs) that can be lifted in the pulley configuration as shown: a. b. c. d.

1 2 3 4

 

Solution: Sketch a free body diagram of the pulley. Starting from the known weight, determine the reactions at the areas cut by the diagram.

Σy = 0; Weight = 4P = 4 x 1000 = 4000-lb = 4-k lbs

(10/2018)

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FACTOR OF SAFETY – CONCRETE WALLFORMS

67. - Question The design for a concrete wall formwork consists of vertical strong backs (studs) spaced 24-inches which are supported by horizontal wales spaced every 18-inches. Eight strands of number 9 wire formwork ties are used to resist the hydrostatic pressure of the 4000-psi concrete exerting 480-PSF on the 1- ¾” plyform sheathing (wire tie tensile bre aking strength = 700-lb per strand). The formwork ties are anchored every fourth stud. The factor of safety at each location of the form wall tie is most nearly: a. b. c. d.

1.0 1.3 2.3 2.5

Solution: Determine the loading area per tie wire. The wall dimensions are not provided, however, by description, the spacing indicates a rectangular area which is 18inches tall by 72-inches wide (the wire is anchored at every fourth stud holding three spans). Area = (72-in) (18-in) ÷ 144-in2/ft2 = 9-ft2 Force = (480-PSF) (9-SF) = 4,320-lb Each computed area of the formwork is known as its contributory which is held by eight wire ties per each corner or contributory area. The factor of safety is computed by: FS = (8-strands) (700-lb/strand) ÷ 4320-lb FS

= 1.29 answer

Tie Wire 18”

72”

Section View

Loading Area (typical)

Schematic Partial View for Illustration (10/2018)

Wale

Studs

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Plyform

TEMPORARY BRACING

68. - Question

Local building code specifies a minimum lateral force of 20-psf for the wind load on concrete wall forms 18-ft high. Steel tubular braces are spaced at 8-ft on center along the 120-ft wall as shown in the cross-section figure. Mechanical analysis of the connection at the base of the form can be considered a hinge. The axial force (lb) resisted by the brace is most nearly: a. b. c. d.

2,592 3,666 4,175 9,500

20-psf 18-ft

10-ft 45º

Not to scale

(10/2018)

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Extend the sketch to simulate the 120-ft long wall and analyze the contributory area for the loading on the 8-ft wide panel.

CL

Brace Tributary Area Loading

8-ft

Per Linear Foot 20-psf 1-ft

18-ft

120-ft 10-ft 45º

A Solution: Determine load on brace at each 8-ft o.c. location Brace = (20-lb/ft2) (8-ft) = 160-lb/vertical ft. per brace location Compute the reaction force at the brace connection to the formwork ∑ Ma = 0 ∑ Ma = (160-lb/ft)(18-ft)(18-ft ÷ 2) – 10-ft (Rx) = 0 Rx = 2,592-lb Compute the axial load in the steel tubular brace Axial load on brace = (2,592-lb) √ 2 = 3,666-lb (answer is b) 1 Alternate solution: (2,592-lb) ÷ sin 45º (10/2018)

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Static Equilibrium

Beam in static equilibrium. The free-body diagram graphically resolves the sum of the forces and the sum of the moments equal to zero.

(10/2018)

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CHAPTER 7 WORKER HEALTH, SAFETY, AND ENVIRONMENT

Concept Terminology



(10/2018)

7

CHAPTER

Worker Health, Safety, and Environment

OSHA

Health Safety Environment

Competent Person Qualified Person Fall Protection Scaffolding Excavations Job Hazard Analysis Personal Protective Equip. Safety Management Safety Statistics Work Related Injuries Blasting Crane Safety EMR

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OSHA REGULATIONS 69. - Question Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926 identifies the one who is capable of identifying existing and predictable hazards in the surroundings or working conditions which are unsanitary, hazardous, or dangerous to employees, and who has authorization to take prompt corrective measures to eliminate them as:  a. Competent person  b. Authorized person c. Qualified Person d. Designated Person Solution: Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926.32(f) "Competent person" means one who is capable of identifying existing and predictable hazards in the surroundings or working conditions which are unsanitary, hazardous, or dangerous to employees, and who has authorization to take prompt corrective measures to eliminate them. (answer is a) 70. - Question Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926 defines “Qualified” person as: a. one who, by possession of a recognized degree, certificate, or professional standing, or who by extensive knowledge, training, and experience, has successfully demonstrated his ability to solve or resolve problems relating to the subject matter, the work, or the project. b. one who, by possession of a recognized by OSHA as having a degree, certificate, or professional standing, or who by extensive knowledge, training, and experience, has successfully demonstrated his ability to solve or resolve problems relating to the subject matter, the work, or the project. one who, by possession of a recognized degree or associate certificate has successfully demonstrated his ability to solve or resolve problems relating to the subject matter, the work, or the project. c. one who, by possession of a recognized ABET degree, 10-hr OSHA training certificate, and has successfully demonstrated his ability to solve or resolve problems relating to the subject matter, the work, or the project. Solution: Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926.32(m) "Qualified" means one who, by possession of a recognized degree, certificate, or professional standing, or who by extensive knowledge, training, and experience, has successfully demonstrated his ability to solve or resolve problems relating to the subject matter, the work, or the project. (answer is a)

(10/2018)

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71. - Question Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926 defines "Safety factor" as the ratio of the ultimate breaking strength of a member or piece of material or equipment to the: a. b. c. d.

actual working stress or safe load when in use safe working load when tested fatigue stress when in use yield point when in use

Solution: Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926.32(n) (answer is a) "Safety factor" means the ratio of the ultimate breaking strength of a member or piece of material or equipment to the actual working stress or safe load when in use.

FALL PROTECTION 72. - Question Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926.501 protects construction workers working over 6 feet. It provides guidance for fall protection requirements, how to select the most appropriate fall system and installation of fall protection systems. Which of the following statements are true regarding fall protection: I.

II.

III.

IV.

V.

Leading edge construction 6-ft or more above lower levels. Personal Fall Arrest system shall be rigged in such a manner that an employee cannot fall more than 6 feet nor contact any lower level. … (b)(4)(i) CFR 1926.501 Personal Fall Protection Top height of top rails shall be 42inches plus or minus 3 inches above the walking/working level. The top edge shall not deflect lower than 39 inches when a 200 pound test load is applied. CFR 1926.502 (b) (3) The minimum tensile load of 5,000 pounds for self-retracting lifelines and lanyards that do not limit free fall to 2 feet. CFR 1926.502 (13) Anchorages used for attachment of personal fall arrest equipment shall be capable of supporting at least 5000 pounds. CFR 1926.502 (15) The employer shall verify compliance with OSHA training by preparing a written certification record. CFR 1926.503.(b) (1) a. b. c. d.

I & II I, II, & III I, II , III, & IV I, II , III, IV, & V

 

Solution: Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926; all are true (answer is d)

(10/2018)

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OSHA AND THE NCEES EXAM

The best approach for preparation with OSHA exam questions is to be familiar with the organization of the book. During the exam, use the index, find the subject matter in the exam question, and make certain to review all the material related to the topic. The following sections are included in Construction Industry - 29 CFR 1926 29 CFR 1903 Inspections, Citations, and Proposed Penalties 29 CFR 1904 Recording and Reporting Occupational Injuries and Illnesses Selected 1910 General Industry Standards 29 CFR 1926 Safety and health regulations for construction

OHSAS 18000 is the name given to the family of international standards relating to occupational health and safety management. It consists of two separate parts: OHSAS 18001 - The Occupational Health and Safety Management Systems Specification OHSAS 18002 - Provides guidelines for the implementation of OHSAS 18001

The OHSAS 18000 standards provide organizations with the elements of an effective safety management system which can be integrated with other management systems and help organizations achieve better occupational health and safety performance and economic objectives.

Needed by ALL Candidates

https://www.amazon.com/1926-OSHA-Construction-IndustryRegulations/dp/1599597748/ref=sr_1_1?ie=UTF8&qid=1500506797&sr=81&keywords=osha+construction+standards+%26+regulations+%2829+cfr+1926

(10/2018)

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HOW TO READ THE OSHA STANDARDS 29 CFR 1926 – CONSTRUCTION UNDER TITLE 29, CHAPTER XVII, THE OSHA REGULATIONS ARE BROKEN DOWN INTO PARTS. PART 1926, FOR EXAMPLE, IS COMMONLY KNOWN AS THE OSHA CONSTRUCTION STANDARDS. PART 1910 COVERS OSHA GENERAL INDUSTRY STANDARDS AND PARTS 1915, 1917 AND 1918 INCLUDE THE OSHA STANDARDS FOR THE MARITIME INDUSTRY. SUBPARTS

SECTIONS

UNDER EACH PART, SUCH AS PART 1926, MAJOR BLOCKS OF INFORMATION ARE FURTHER BROKEN INTO SUBPARTS. THE MAJOR SUBPARTS IN 1926 STANDARDS INCLUDE: Subpart C

General Safety and Health Provisions

Subpart D Subpart E Subpart F Subpart G Subpart H Subpart I Subpart J Subpart K Subpart L Subpart M Subpart N Subpart O Subpart P Subpart Q Subpart R Subpart S Subpart T Subpart U Subpart V Subpart W Subpart X Subpart Y Subpart Z

EACH SUBPART IS FURTHER BROKEN DOWN INTO SECTIONS. FOR EXAMPLE, SUBPART C – GENERAL SAFETY AND HEALTH PROVISIONS, HAS SECTIONS 1926.20 THROUGH 1926.35.

Occupational Health and Environmental Controls Personal Protective and Life Saving Equipment Fire Protection and Prevention Signs, Signals and Barricades Materials Handling, Storage, Use, and Disposal Tools – Hand and Power Welding and Cutting Electrical Scaffolds Fall Protection Cranes, Derricks, Hoists, Elevators, and Conveyors Motor Vehicles, Mechanized Equipment, Marine Excavations Concrete and Masonry Construction Steel Erection Underground Construction, Caissons, Cofferdams, Demolition Blasting and the Use of Explosives Power Transmission and Distribution Rollover Protective Structures; Overhead Protection Ladders Commercial Diving Toxic and Hazardous Substances



1926.20 – General safety and health provisions.

            

1926.21 – Safety training and education.

 

1926.34 – Means of egress.

1926.22 – Recording and reporting of injuries. 1926.23 – First aid and medical attention. 1926.24 – Fire protection and prevention. 1926.25 – Housekeeping. 1926.26 – Illumination. 1926.27 – Sanitation. 1926.28 – Personal protective equipment. 1926.29 – Acceptable certifications. 1926.30 – Shipbuilding and ship repairing 1926.31 – Incorporation by reference. 1926.32 – Definitions. 1926.33 – Access to employee exposure and medical records. 1926.35 – Employee emergency action plans.

EXAMPLE: READING OSHA STANDARD NUMBERS STANDARD: 29 CFR 1926.552(c)(17)(iv)(e)

In standing ropes, more than two broken wires in one lay in sections beyond end connections or more than one broken wire at an end connection. BREAKING DOWN THE NUMBER: TITLE

29

CODE OF FED. REG.

CFR

(10/2018)

PART

1926

LOWER SECTION CASE ALPHA

.552

(c)

ARABIC NUMBER

(17)

LOWER CASE ROMAN

(iv)

CAPITAL/UPPER CASE ALPHA (OPT).

(e)

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OSHA SUBPART P -- EXCAVATIONS

73. - Question The load (lbs.) that a 2-inch inside diameter hydraulic cylinder used as part of an aluminum hydraulic shoring system for trenches can safely support is most nearly: a. 10,000 b. 18,000 c. 23,000 d. 30,000 Solution: According to Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926 • Subpart Title: Excavations • Standard Number: 1926 Subpart P App D • Title: Aluminum Hydraulic Shoring for Trenches

(2) Hydraulic cylinders specifications. (i) 2-inch cylinders shall be a minimum 2-inch inside diameter with a minimum safe working capacity of no less than 18,000 pounds axial compressive load at maximum extension. Maximum extension is to include full range of cylinder extensions as recommended by product manufacturer. While 3-inch cylinders can support 30,000 pounds. 1. Find source material in INDEX

2. Goto Reference Section

3. Hunt for the answer

(10/2018)

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OSHA SUBPART L --

SCAFFOLDS

74. - Question For a scaffolding system to safely hold two 200-pound workers and 50 pounds of equipment, how much weight must the scaffolding planks and suspension ropes be capable of holding? a. The planks must hold 1,800 pounds, and the suspension ropes must hold 2,700 pounds b. Both the planks and suspension ropes must hold 1,800 pounds c. Both the planks and suspension ropes must hold 2,700 pounds d. The planks must hold 2,700 pounds, and the suspension ropes must hold 1,800 pounds Solution: According to Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926 • • • •

Subpart: Subpart Title: Standard Number: Title:

L Scaffolds 1926.451 General requirements.

1926.451(a)(1)

Except as provided in paragraphs (a)(2), (a)(3), (a)(4), (a)(5) and (g) of this section, each scaffold and scaffold component shall be capable of supporting, without failure, its own weight and at least 4 times the maximum intended load applied or transmitted to it. 1926.451(a)(3)

Each suspension rope, including connecting hardware, used on non-adjustable suspension scaffolds shall be capable of supporting, without failure, at least 6 times the maximum intended load applied or transmitted to that rope.

(10/2018)

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OSHA SUBPART N -- CRANES, DERRICKS, HOISTS, ELEVATORS, AND CONVEYORS

75. - Question During a crane operation, if a crane is used to hoist workers then the angle between the base of the crane and the ground must not exceed (%): a. 5 b. 2 c. 1 d. 0.5 • Subpart: • Subpart Title: • Standard Number: • Title:

CC Cranes & Derricks in Construction 1926.1431 Hoisting personnel.

1926.1431(c)(1)

The equipment must be uniformly level, within one percent of level grade, and located on footing that a qualified person has determined to be sufficiently firm and stable. If a crane is used to hoist workers, the angle between the base of the crane and the ground must not exceed one percent. In addition, the total weight of the personnel and personnel platform must be less than 50 percent of the crane's capacity, and all brakes and locking devices on the crane must be engaged when the personnel platform stops. A crane should be used to hoist workers only when more conventional means, such as ladders and scaffolding, are not feasible.

76. - Question The number of wires in one strand of a hoisting wire rope lay that are allowed to be broken before the wire rope is removed from service is(are) most nearly: a. 0 b. 2 c. 3 d. 6 Answer: C OSHA 1926.552(a)(3) Wire rope shall be removed from service when any of the following conditions exists: 29 CFR 1926.552(a)(3)(i) In hoisting ropes, six randomly distributed broken wires in one rope lay or three broken wires in one strand in one rope lay;

§1926.1401 — Definitions Wire rope means a flexible rope constructed by laying steel wires into various patterns of multi-wired strands around a core system to produce a helically wound rope.

(10/2018)

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OSHA SUBPART U -- BLASTING AND THE USE OF EXPLOSIVES

77. - Question During a rock blasting operation, blast holes shall be checked prior to loading to determine depth and conditions. Where a hole has been loaded with explosives but the explosives have failed to detonate, the distance (feet) that drilling is prohibited within the loaded hole is most nearly: a. b. c. d.

10 25 50 75

Solution: According to Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926 According to OSHA 29 CFR 1926.905(k) Where a hole has been loaded with explosives but the explosives have failed to detonate, there shall be no drilling within 50 feet of the hole. Answer is C

78. - Question An "Explosives" is any chemical compound, mixture, or device, the primary or common purpose of which is to function by explosion; that is, with substantially instantaneous release of gas and heat, unless such compound, mixture or device is otherwise specifically classified by the U.S. Department of Transportation. The Class of explosive possessing flammable hazard, such as propellant explosives is labeled as: a. b. c. d.

Class A Class B Class C Class D

According to OSHA 1926.914(n)(3) Class B Explosives. Possessing flammable hazard, such as propellant explosives, including some smokeless propellants. Answer is B "Blasting agent" - A blasting agent is any material or mixture consisting of a fuel and oxidizer used for blasting, but not classified an explosive and in which none of the ingredients is classified as an explosive provided the furnished (mixed) product cannot be detonated with a No. 8 test blasting cap when confined. A common blasting agent presently in use is a mixture of ammonium nitrate (NH(4)NO(3)) and carbonaceous combustibles, such as fuel oil or coal, and may either be procured, premixed and packaged from explosives companies or mixed in the field. "Blasting cap" - A metallic tube closed at one end, containing a charge of one or more detonating compounds, and designed for and capable of detonation from the sparks or flame from a safety fuse inserted and crimped into the open end. "Explosives" - Any chemical compound, mixture, or device, the primary or common purpose of which is to function by explosion; that is, with substantially instantaneous release of gas and heat, unless such compound, mixture or device is otherwise specifically classified by the U.S. Department of Transportation.

(10/2018)

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1926.914(n)(2)

All material which is classified as Class A, Class B, and Class C Explosives by the U.S. Department of

Transportation.

Classification of explosives by the U.S. Department of Transportation is as follows: Class A Explosives. Possessing detonating hazard, such as dynamite, nitroglycerin, picric acid, lead azide, fulminate of mercury, black powder, blasting caps, and detonating primers. Class B Explosives. Possessing flammable hazard, such as propellant explosives, including some smokeless propellants. Class C Explosives. Include certain types of manufactured articles which contain Class A or Class B explosives, or both, as components, but in restricted quantities.

79. - Question A construction employee receives the following exposure to noise on a construction site during an 8-hour period: 3 hours @ 92 dBA 2 hours @ 95 dBA 3 hours @ 90 dBA The percent (%) of the PEL for this exposure level is most nearly: A. 95.5% B. 100.0% C. 108.2% D. 137.5% Solution: The OSHA Permissible Exposures Level (PEL) at these sound levels are: Sound Level Permissible Duration 90 dBA 8 hours 92 dBA 6 hours 95 dBA 4 hours (3/6) + (2/4) + (3/8) = 137.5% Answer • Part Number: 1926 • Part Title: Safety and Health Regulations for Construction • Subpart: D • Subpart Title: Occupational Health and Environmental Controls • Standard Number: 1926.52 • Title: Occupational noise exposure. Protection against the effects of noise exposure shall be provided when the sound levels exceed those shown in Table D-2 of this section when measured on the A-scale of a standard sound level meter at slow response. When the daily noise exposure is composed of two or more periods of noise exposure of different levels, their combined effect should be considered, rather than the individual effect of each. Exposure to different levels for various periods of time shall be computed according to the formula set forth in paragraph (d)(2)(ii) of this section.

(10/2018)

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SCAFFOLDING

80. - Question According to Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926.451 requires employers to provide scaffolding at heights of 10 feet or more. Which of the following statements are true regarding scaffolding: I. II. III.

IV. V.

Scaffolds must be at least 10 feet from electric power lines at all times. Scaffold must not be erected, moved, dismantled or altered except under the supervision of a competent person. Scaffold must be equipped with guardrails, mid-rails and toe boards. Scaffold accessories such as braces, brackets, trusses, screw legs or ladders that are damaged or weakened from any cause must be immediately repaired or replaced. Scaffold platforms must be tightly planked with scaffold plank grade material or equivalent. A “competent person” must inspect the scaffolding and, at designated intervals, re-inspect the scaffold. a. b. c. d.

I & II I, II, & III I, II , III, & IV I, II , III, IV, & V

 

Solution: According to Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926.451, all are true (answer is d).

(10/2018)

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PROTECTIVE EQUIPMENT (PPE)

81. - Question According to Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926 which of the following statements are true regarding PPE: a. I b. I & II c. I, II, & III d. None I.

Eye and Face Protection a. Safety glasses or face shields are worn anytime work operations can cause foreign objects getting into the eye such as during welding, cutting, grinding, nailing (or when working with concrete and/or harmful chemicals or when exposed to flying particles). b. Eye and face protectors are selected based on anticipated hazards. c. Safety glasses or face shields are worn when exposed to any electrical hazards including work on energized electrical systems. Foot Protection a. Construction workers should wear work shoes or boots with slipresistant and puncture-resistant soles. b. Safety-toed footwear is worn to prevent crushed toes when working around heavy equipment or falling objects. Hand Protection a. Gloves should fit snugly.

II.

III.

Solution: Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926; all statements are true (answer is c)

According to the OSHA standard excavation safety rules, the depth (ft) of excavation which requires shoring is most nearly:

AA

a. 4 b. 5 c. 6 d. 7 Answer = 5-ft: “Four-five-six” is a good mnemonic device for OSHA safety rule memory; four is for ladder access, five is for shoring, and six is for fall protection.

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SAFETY MANAGEMENT

82. - Question According to Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926 which of the following statements are true regarding contractor safety management responsibility: I.

“Where joint responsibility exists, both the prime and their subcontractor or subcontractors, regardless of tier, shall be considered subject to the provisions of the Act”.

II.

“With respect to subcontracted work, the prime contractor and any subcontractor or subcontractors shall be deemed to have joint responsibility”

III.

“Where joint responsibility exists, only the prime shall be considered subject to the provisions of the Act”.

IV.

“With respect to subcontracted work, the subcontractor or subcontractors shall be deemed to have joint responsibility” a. b. c. d.

I & II I, II, & III I, II , III, & IV None

Solution: According to Occupational Safety and Health Standards for the Construction Industry, 29 CFR Part 1926: (answer is a) Contractor Management Responsibility a. 29 CFR 1926.16(d) “Where joint responsibility exists, both the prime and their subcontractor or subcontractors, regardless of tier, shall be considered subject to the provisions of the Act”. b. 29 CFR 1926.16 (c) “With respect to subcontracted work, the prime contractor and any subcontractor or subcontractors shall be deemed to have joint responsibility”

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This copy is given to the following student as part of School of PE course. Not allowed to distribute to others. ivan anderson ([email protected])

EXPERIENCE MODIFICATION RATE

83. - Question Experience Modification Rate (EMR) is an adjustment to the “Manual Premium”, calculated by an advisory organization such as NCCI (National Council on Compensation Insurance), and based on historic loss and payroll data of a particular insured. Which of the following statement are true regarding the EMR: I.

The Experience Modification Rate (EMR) is established by the subcontractor’s worker’s compensation insurance carrier, and is based on the subcontractor’s loss history.

II.

For a newly formed Company, the Experience Modification Rate (EMR) is assigned a rating of 1.0.

III.

In workers compensation experience rating, the actual payroll and loss data of the individual employer are analyzed over a period of time. Usually, the latest available three years of data are compared to similarly grouped risks to calculate the experience modification.

IV.

The experience modification rate is derived by dividing adjusted actual losses by the adjusted expected losses.

V.

The essentials of experience rating are: a. It is mandatory for all insured’s that meet the premium eligibility requirements b. The formula measures how the performance of an employer differs predictably from similarly classified employers. a. b. c. d.

I & II I, II, & III I, II , III, & IV I, II , III, IV, & V

Solution: all are true (answer is d).

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This copy is given to the following student as part of School of PE course. Not allowed to distribute to others. ivan anderson ([email protected])

A good safety record can strengthen profitability and reduce insurance premiums. EMR or Experience Modification rate was originally designed to calculate insurance premiums for each type of industry and employee classification. Since the 1980’s, a company’s EMR has been used as a guide by private industry and Government to evaluate the overall expertise and quality of company and its commitment to safety. Within the insurance industry, there are calculations for auto, general liability, and Worker’s Compensation. However, a Company’s Workers’ Compensation EMR will be evaluated by clients to decide whether to award a Contract. Most Federal and Government projects will disqualify a Contractor based on the EMR rating if it is too high. EMR’s are calculated by comparing actual losses to Industry average losses in a Company’s classification. Generally, premiums will be higher when average losses are lower than a company’s safety performance. Premiums will be lower when Industry average losses are higher than a company’s performance. There are two organizations that control classification rates. The first is the State Workers Compensation Rating Bureau, and the second is the National Council on Compensation Insurance, which calculates EMR. It takes three years of experience and generally around $5,000.00 in annual premiums to obtain an EMR. The EMR revolves around a three-year plan. The first year of business is excluded. Companies who perform average in their industry will generally see an EMR rate of 1.0. Companies with many claims will have a higher than average rate, for example, 1.6. Typically, most private clients want a company EMR to be below the average threshold, for example, 0.75. The rating dictates what a company will pay in premiums. As an example, 1.0 pay 100%, 1.6 pay 160%, and .75 pay 75% of average. A company can lower their EMR with a comprehensive safety plan which should include NDA’s safety manual, employee training for all OSHA requirements, drug testing, preemployment physicals, and incentives for employees to work safe.

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This copy is given to the following student as part of School of PE course. Not allowed to distribute to others. ivan anderson ([email protected])

Success in life is not reserved for the talented, and it’s not in the high IQ, nor in the gift at birth. It’s not always in ability, or in the best equipment. Success is almost totally dependent upon drive and persistence – the extra energy required to make another effort, to try another approach, to take a new tack. This is the secret of winning." Secret of Winning . . . . . Denis Waitley

The Best of Success on the exam, and in your future career as a Professional Engineer!

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This copy is given to the following student as part of School of PE course. Not allowed to distribute to others. ivan anderson ([email protected])