Notes for Mar 22nd 23rd and 24th Classes Part 1

136
School of PE School of PE Professional Engineer by George Stankiewicz, P.E., LEED ® A. P. C IVIL E NGINEER ahmed youssef ([email protected]) This copy is given to the following student as part of School of PE course. Not allowed to distribute to others.

Transcript of Notes for Mar 22nd 23rd and 24th Classes Part 1

Page 1: Notes for Mar 22nd 23rd and 24th Classes Part 1

School of PE

School of PE

Professional Engineer

by George Stankiewicz, P.E., LEED ® A. P.

C I V I L E N G I N E E R ahmed youssef ([email protected])

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2 Contents |

CONTENTS

Contents .....................................................................................................................................2

How to use this Refresher Course Study Guide ...................................................................5

Preface .......................................................................................................................................6

Refresher Course Activity Organization/Administration ........................................................7

Chapter 1 – Surveying ................................................................................................................9

LATITUDE AND DEPARTURES ..................................................................................10

COORDINATE SYSTEM ...............................................................................................11

Soils - Swell and Shrinkage ...............................................................................................17

Average End Area Method .................................................................................................28

Earthwork Volume Calculations .........................................................................................31

Borrow Pit Leveling ............................................................................................................34

Differential Leveling ...........................................................................................................40

Chapter 2 – Construction Management .....................................................................................47

Construction Management - Procurement Methods ...........................................................49

Cost Estimating ..................................................................................................................50

Estimating Takeoff Quantities ............................................................................................51

Cost Estimating – Board Feet ............................................................................................59

Methods of Budgeting ........................................................................................................61

ConstructionHistoric Data ..................................................................................................62

Engineering Economics .....................................................................................................63

Factor Table Quick View Exercise .....................................................................................65

Time Value of Money .........................................................................................................67

Compound Interest – Nominal and Effective ......................................................................70

Solving Engineering Economic Problems ...........................................................................72

Present Worth ....................................................................................................................73

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Contents 3

Future Worth or Value ........................................................................................................74

Annual Cost .......................................................................................................................75

Maintenance Costs ............................................................................................................76

Rate of Return Analysis – Three Alternatives .....................................................................77

Benefit/Cost Analysis .........................................................................................................78

Alternate Project Selection .................................................................................................79

Alternate Selection of Components ....................................................................................80

Contractor Project Financing ..............................................................................................82

Internal Rate of Return .......................................................................................................83

Project Schedule Financial Analysis...................................................................................85

Estimating Activity Durations ..............................................................................................85

Project Scheduling Fundamentals ......................................................................................88

Project Scheduling – Types of Methods .............................................................................89

Precedence Relationships .................................................................................................90

Precedence Diagramming Methods ...................................................................................91

Arrow Diagramming Method...............................................................................................91

Critical Path - Activity on Node ...........................................................................................97

Project Float -- “Free Float” and “Total Float” .....................................................................98

Resource Leveling .............................................................................................................99

Chapter 3 - Materials............................................................................................................... 107

Mechanical Properties of Materials .................................................................................. 108

Actual versus Ultimate strength ........................................................................................ 111

Elastic Stretch .................................................................................................................. 112

Thermal Expansion .......................................................................................................... 113

Lifting Load – Offset ......................................................................................................... 114

Equipment Production ...................................................................................................... 115

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4 Contents |

Daily standard production rate of Equipment .................................................................... 116

Daily standard production rate of a dump truck ................................................................ 117

Productivity Analysis and Improvement ............................................................................ 118

Operating Costs ............................................................................................................... 119

Effects of job size on productivity ..................................................................................... 120

Material Specifications ..................................................................................................... 121

Quality Control Process (QA/QC) ..................................................................................... 122

Concrete Mix Design ........................................................................................................ 123

Concrete Mix Design Ratio 1 : 2: 3 ................................................................................... 125

Water Cement Ratio ........................................................................................................ 127

Concrete Strength Testing ............................................................................................... 128

Asphalt Performance ....................................................................................................... 132

INDEX ..................................................................................................................................... 134

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Contents 5

HOW TO USE THIS REFRESHER COURSE STUDY GUIDE

Throughout the Refresher Course Notes the following symbol

represents references to the NCEES Fundamental of Engineering

Supplied Reference Handbook 8th Edition 2nd Revision and page

locations for further review, self-study, and ease of navigation

through this refresher course:

Sample 1:

Sample 2:

Sample 3:

Sample 4:

This symbol represents topics within the Refresher Course that are part of

the subject matter which will further help your understanding.

The information is intented for self-study and may not be

reviewed during the refresher course.

p

NCEES Reference Handbook, 8thedition

Page number

This example text box shows necessary equations.

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.

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6 Preface |

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.

Application 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

Application

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Preface 7

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 and is outlined in NCEES Fundamental of Engineering

Supplied Reference Handbook 8th Edition 2nd Revision.

The refresher class focus is on interpreting the Civil Engineering afternoon session 60 questions

in nine topic areas. The course provides a graduating series of problem statements to better

the understanding of the content for the Civil Engineering Exam.

CHAPTER ORGANIZATION

I. Surveying 11% = 7/60 A. Angles, distances, and trigonometry B. Area computations C. Closure D. Coordinate systems (e.g., GPS, state plane) E. Curves (vertical and horizontal) F. Earthwork and volume computations G. Leveling (e.g., differential, elevations, percent grades)

II. Construction Management 10% = 6/60

A. Procurement methods (e.g., design-build, design-bid-build, qualifications based) B. Allocation of resources (e.g., labor, equipment, materials, money, time) C. Contracts/contract law D. Project scheduling (e.g., CPM, PERT) E. Engineering economics F. Project management (e.g., owner/contractor/client relations, safety) G. Construction estimating

III. Materials 8% = 5/60

A. Concrete mix design

B. Asphalt mix design

C. Test methods (e.g., steel, concrete, aggregates, and asphalt)

D. Properties of aggregates

E. Engineering properties of metals

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8 Preface |

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Chapter 1 – Surveying 9

CHAPTER 1 – SURVEYING

Concept

Terminology

Application

Surveying

CH

AP

TE

R

1

Construction Surveying

“State” of Soils

Average End Area

Earthwork Volume

Mass Haul Diagram

Swell

Shrinkage

Bank Soil

Stations

Cut

Fill

Borrow Pit

Staking & Layout

Differential Leveling

Benchmark

Back sight

Foresight

Height of Instrument

Terrain

Cumulative Volume

NCEES – FE Civil Engineering Topics I. Surveying 11% = 7/60 A. Angles, distances, and trigonometry B. Area computations C. Closure D. Coordinate systems (e.g., GPS, state plane) E. Curves (vertical and horizontal) F. Earthwork and volume computations G. Leveling (e.g., differential, elevations, percent grades)

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10 Chapter 1 – Surveying |

LATITUDE AND DEPARTURES

fast facts

Latitude of a line is the distance that the line extends in a north or south direction. A line that

runs towards north has positive latitude; a line that runs towards south has negative latitude.

Departure of a line is the distance that the line extends in an east or west direction. A line that

runs towards east has a positive departure; a line that runs towards west has a negative

departure.

p

Memorize

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Chapter 1 – Surveying 11

COORDINATE SYSTEM

fast facts

1. A Benchmark provides only elevation data.

2. Coordinate System provides northing and easting coordinates within a

defined system.

3. Coordinates require a minimum of eight significant digits.

4. The project site coordinates and datum are referenced by the State Plane

Coordinate System

5. State Plane Coordinate System is represented as a grid map of the United

States where coordinates are referenced within the 1st Quadrant.

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12 Chapter 1 – Surveying |

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Chapter 1 – Surveying 13

1. - Question The northing and easting coordinates for point C is most nearly:

a. N 671248.23 E 585554.45

b. N 671188.2 E 585258.45

c. N 671317.08 E 585595.3

d. N 671317.08 E 585596.11

Solution:

Section the triangle and calculate using Pythagorean Theorem and the Law of Sine’s or use the

calculator Pol() Rec() function. See the annotated figure on the next page.

A’~B = (585724.45 - 585234.45) = 490.00-ft A~A’ = (671588.23 - N 671488.23) = 100.00-ft

Pol(490,100) = 500.00 and 11.54°

B~C’ = Rec(300,25.33°) = 271.15-ft C~C’ = Rec(300,64.67°) = 128.34-ft

Apply results to the coordinates (answer=d)

671588.23 – 271.15 = N 671317.08 585724.45 – 128.34 = E 585596.11

N 671488.23 E 585234.45

N 671588.23 E 585724.45

N _________ E

____-ft

Not to scale

300.00-ft

400.00-ft

B

A

C

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14 Chapter 1 – Surveying |

Solution::

N 671488.23 E 585234.45

N 671588.23 E 585724.45

N 671317.08 E 585596.11

500.00-ft

Not to scale

300.00-ft

400.00-ft

B

A

C

A’

C’

11.54°

25.33°

53.13°

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Chapter 1 – Surveying 15

2. - Question A surveyor’s total station measured slope distance near station

3245+26.35 is recorded as 437.380-m with a zenith angle of 118º48’07” and a south

easterly bearing. The horizontal distance (m) is most nearly:

a. 383.2

b. 383.276

c. 391.324

d. 391.300

Solution: Sketch the statement:

α = 118º48’07” - 90º00’00”

α = 28.80

cos 28.80 = x .

437.38

X = (437.38)(0.8763)

X = 383.276-m (answer is b)

118º48’07”

α

383.276-m

437.380-m

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16 Chapter 1 – Surveying |

3. - Question Convert the azimuth reading of 273.34º to a land bearing:

a. N 273.34 W

b. N 3º 20’ 24” W

c. N 86º 39’ 36” W

d. W 3º 20’ 24” N

Solution:

Convert 273.34º to degrees minutes and seconds = 273º 20’ 24”

Next, convert the angle to orient to IV quadrant:

273º 20’ 24” - 360º 0 ’ 0” = 86º 39’ 36”

Assign bearing: N 86º 39’ 36” W

True North Based Azimuths From North

North 0° or 360° South 180°

North-Northeast 22.5° South-Southwest 202.5°

Northeast 45° Southwest 225°

East-Northeast 67.5° West-Southwest 247.5°

East 90° West 270°

East-Southeast 112.5° West-Northwest 292.5°

Southeast 135° Northwest 315°

South-Southeast 157.5° North-Northwest 337.5°

fast facts

In land navigation, azimuth is usually denoted as alpha, α, and defined as a horizontal angle measured clockwise from a north base line or meridian. Azimuth has also been more generally defined as a horizontal angle measured clockwise from any fixed reference plane or easily established base direction line. The reference plane for an azimuth in a general navigational context is typically true north, measured as a 0° azimuth. In any event, the azimuth cannot exceed the highest number of units in a circle – for a 360° circle; this is 359 degrees, 59 minutes, 59 seconds (359° 59' 59").

For example, moving clockwise on a 360° degree circle, a point due east would have an azimuth of 90°, south 180°, and west 270°.

In land surveying, a bearing is the clockwise or counterclockwise angle between north or south and a direction. For example, bearings are recorded as N57°E, S51°E, S21°W, N87°W, or N15°W. In surveying, bearings can be referenced to true north, magnetic north, grid north (the Y axis of a map projection), or a previous map, which is often a historical magnetic north.

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Chapter 1 – Surveying 17

SOILS - SWELL AND SHRINKAGE

4. - 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; in-

situ; 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 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. I & II

b. I, II, & III

c. I, II, III, & IV

d. I, II, III, IV, & V

Solution: 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.)

Bank Loose Compacted

1-yd3

1.25-yd3

25% swell

0.80-yd3 20% shrinkage

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18 Chapter 1 – Surveying |

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.

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: A soil increases in volume when it is excavated.

Swell Density

Swell Volume

Swell (%) = Bank Density x 100 Loose Density

V loose = 100% + % swell x Vbank = Vbank 100% Load Factor

Load Factor = Loose Density Bank Density

Load Factor = (1 + decimal swell) -1

Bank Volume = Loose volume x Load factor

Soil Diagram Soil Phase Diagram

-1

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Chapter 1 – Surveying 19

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

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:

Material

Loose

(lb/cy)

Bank

(lb/cy)

Swell

(%)

Load

Factor

Clay, dry 2,100 2,650 26 0.79

Clay, wet 2,700 3,575 32 0.76

Clay and gravel, dry 2,400 2,800 17 0.85

Clay and gravel, wet 2,600 3,100 17 0.85

Earth, dry 2,215 2,850 29 0.78

Earth, moist 2,410 3,080 28 0.78

Earth, wet 2,750 3,380 23 0.81

Gravel, wet 2,780 3,140 13 0.88

Gravel, dry 3,090 3,620 17 0.85

Sand, dry 2,600 2,920 12 0.89

Sand, wet 3,100 3,520 13 0.88

Sand and gravel, dry 2,900 3,250 12 0.89

Sand and gravel, wet 3,400 3,750 10 0.91

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20 Chapter 1 – Surveying |

Shrinkage: A soil decreases in volume when it is compacted:

Shrinkage Density

Shrinkage Volume

Shrinkage(%) = Bank Density x 100 Compacted Density

V compacted = 100% - % shrinkage V bank 100%

Shrinkage factor = 1 – Shrinkage (% decimal)

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.

The TABLE illustrates

soil in a variety of states.

Initial

Soil Type Soil Condition Bank Loose Compacted

Clay Bank 1.00 1.27 0.90

Loose 0.79 1.00 0.71

Compacted 1.11 1.41 1.00

Common Earth Bank 1.00 1.25 0.90

Loose 0.80 1.00 0.72

Compacted 1.11 1.39 1.00

Rock (blasted) Bank 1.00 1.50 1.30

Loose 0.67 1.00 0.87

Compacted 0.77 1.15 1.00

Sand Bank 1.00 1.12 0.95

Loose 0.89 1.00 0.85

Compacted 1.05 1.18 1.00

Converted to:

-1

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Chapter 1 – Surveying 21

5. - 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)

6. - 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:

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

V compacted = 100% - % shrinkage) V bank

100%

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22 Chapter 1 – Surveying |

7. - Question A proposed building site requires 135,000-ft3 of

imported fill. A borrow site is located 5-miles Northeast of the project site

where the soil has a shrinkage of 16%. The amount of cubic yards of soil

that must be excavated from the borrow site is most nearly:

a. 135,000-ft3

b. 4,310-yds3

c. 5,800-yds3

d. 5,950-yds3

Solution: Apply the equation, calculate the bank volume from the

borrow site using the known components of the equation:

135,000-ft3 = [(100% - 16% shrinkage) ÷ 100%] x V bank

= 135,000-ft3 ÷ .84 = Vbank

V bank = 160,715-ft3 ÷ (3-ft/yd)3 = 5,950-yds3 (answer is d)

8. - Question A contractor was awarded a Contract to excavate

and haul 200,000-yds3 of silty clay (USCS classification ML) with a bulking

factor of 30%. The contractor’s fleet of dump trucks have a capacity of 26-

yds3 and operate on a 23-minute cycle. The job must be completed in 5-

working 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. 48

d. 125

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 ÷ (23-min/cycle ÷ 60-min/hr)) = 67.82-yds3/truck hour 3,250-yds3/hr ÷ 67.82-yds3/truck-hr = 47.92-trucks use 48 trucks (answer is c)

V compacted = [(100% - % shrinkage) ÷ 100%] x V bank

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Chapter 1 – Surveying 23

9. - 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 and compacted to an average dry unit weight of 110-

PCF. The soil shrinkage is most nearly:

a. 3.0% b. 3.5% c. 4.0% d. 5.5%

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

Dry unit weight = Total Unit Weight (1 + water content)

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24 Chapter 1 – Surveying |

10. - 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. 9,500-yd3 b. 10,000-yd3 c. 11,700-yd3 d. 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: 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.

RC = 100 * (field dry density, PCF) Laboratory maximum dry density (PCF)

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Chapter 1 – Surveying 25

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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26 Chapter 1 – Surveying |

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

WATER AIR

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Chapter 1 – Surveying 27

fast facts

Maximum achievable

density for the

compacting effort

Target Area

The Laboratory Proctor

MUD DRY

% Moisture

100%

98%

Dry

Densi

ty P

CF

A

B

Maximum density is found at point “B” and at the

intersection of Optimum Moisture Content point “A”

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28 Chapter 1 – Surveying |

AVERAGE END AREA METHOD

11. - Question Using the information given in Figure 1, the volume

of excavation 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)

Figure 1

A2 = 544-ft2 at station 19+00

A1 = 725-ft2 at station 18+00

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:

2

)( 21 AALV

p

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Chapter 1 – Surveying 29

12. - 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

[Hint: See CERM page 79-2; Paragraph 5 – CUT and FILL. In highway work, payment is usually for cut, while in dam work it is usually for fill.]

End Area Station Cut Fill (ft2) (ft2) 10 + 00 0 11 + 00 168 12 + 00 348 13 + 00 371 14 + 00 146 14 + 60 0 0 15 + 00 142 16 + 00 238 17 + 00 305 18 + 00 247 19 + 00 138 20 + 00 106

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

p

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30 Chapter 1 – Surveying |

Solution:

End Area (ft2) Station Distance

(ft) Cut Fill cut vol fill vol fill vol

+25% (sf) (sf) (cy) (cy) 10+00 0

100 207.4 11+00 168

100 955.6 12+00 348

100 1331.5 13+00 371

100 957.4 14+00 146

60 108.2 14+60 0 0

40 70.1 87.7 15+00 142

100 703.7 879.6 16+00 238

100 1005.6 1256.9 17+00 305

100 1022.2 1277.8 18+00 247

100 713.0 891.2 19+00 138

100 451.9 564.8 20+00 106

TOTAL 3560.1 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)

Use Vpyramid

Use Vpyramid

Vpyramid

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Chapter 1 – Surveying 31

EARTHWORK VOLUME CALCULATIONS

13. - Question A municipality has awarded a contract to cap an existing landfill.

Prior to the landfill’s construction, the base area was surveyed and found to be level and

bound to a neat 2,000 square acres in size. The Contract requirement is to cover the landfill

with 1’-6” of the borrow material. The landfill’s average height is 12 feet with side slopes

groomed 3H:1V. Soil Boring reports at the borrow location limits the excavation to no greater

than 3’–0” depth from existing grade with a requirement for side slopes to be groomed to

3H:1V. Determine the following:

1. The number of cubic yards of in-situ (in-place) borrow needed to cap the landfill is most

nearly:

a. 3,689,330-yd3

b. 4,189,330-yd3

c. 4,789,330-yd3

d. 5,789,330-yd3

2. Based on in-situ (“in place”) volume, the number of acres at the borrow site that will be

disturbed based on the soil boring constraint is most nearly: a. 595-Ac b. 789-Ac c. 856-Ac d. 992-Ac

Solution:

1. The problem can be evaluated using the “truncated pyramid” equation; first, calculate the

base pyramid (the “landfill”); then, second, calculate the “capped” landfill using the given

dimension and subtract the amounts to obtain the net volume.

Calculate volume of base pyramid = V1

Volume = V1 = h/3 (A1 + A2 + √(A1 x A2))

[not to scale]

Plan View

Cap Landfill

Base Landfill

12-ft

13.5-ft

A1

A2

A2

A1 9333’

1

3

36-ft

Cap Volume

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32 Chapter 1 – Surveying |

√(2,000ac x 43,560-ft2/ac) = 9,333-ft (per side) Area of Base Pyramid = A2 = 9333-ft x 9333-ft = 87,120,000-ft2 Area of Top Pyramid = A1 = (9333-ft – 72-ft) (9333-ft – 72-ft) = 85,766,121-ft2 Volume Base Pyramid = V1 = [12-ft/3 (87,120,000-ft2 + 85,766,121-ft2 + √(85,766,121-ft2 x 87,120,000-ft2)] ÷ (27-ft3/yd3) V1 = 38,418,745-yd3

Calculate volume of “cap” by adding 1’-6” to the base pyramid dimensions [Note the change in

dimensions]:

h2 = h + 1.5-ft = 12-ft + 1.5-ft = 13.5-ft Side A2 = 9333 + 1.5 + 1.5 = 9336-ft Side A1 = 9336-ft – 81-ft = 9256-ft [Note: 72’ + 4.5’ + 4.5’ = 81’] V2 = [13.5/3 (9,3362 + 92562 + √(93362 x 92562)] ÷ (27-ft3/yd3) V2 = 43,208,075-yd3

Net in situ volume = V2 – V1 = 43,208,075-yd3 - 38,418,745-yd3 = 4,789,330-yd3 (answer is c)

Solution:

2. Based on in-situ (“in place”) volume, determine the number of acres at the borrow site that will be disturbed based on the soil boring constraint of 3-ft deep.

Use the computed volume from the calculations above.

4,789,330-yd3 x (27-ft3/yd3) = 129,311,910-ft3

129,311,910-ft3 = h/3 (A1 + A2 + √ (A1 x A2))

Substitute known data into the equation; note that A1 = (S2 + 18)2 ; A2 = S22

129,311,910-ft3 = 3/3 ((S2 + 18)2 + S22 + √(( S2 + 18 )2 x S2

2))

Calculate A2 using the SOLVE function on your calculator or use quadratic equation:

S2 side = 6,556-ft; length of one side of the excavation. Compute the number Acres disturbed at the borrow site. Substituting: S1 side = (S2 + 18) = (6556 + 18) = 6,574-ft per side Ac = (6,574-ft x 6,574-ft) ÷ (43,560-ft2/Ac) = 992-Ac (answer is d)

[not to scale]

1

3

3-ft

9-ft

S1

S2

A1

A2

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Chapter 1 – Surveying 33

14. - Question On a 5-acre level terrain building site, an earthwork

contractor has instructed her crew to strip and grub the topsoil of a 60,000-

ft2 proposed building pad to a minus 2-ft sub-grade and limit the stockpile to

75-ft radius. The soil has a swell of 40% and an angle of repose at 30°.

The initial height of the stockpile is most nearly:

a. 20

b. 30

c. 45

d. 50

Solution:

Determine the cubic volume of the cut and the swell of the soil:

60,000-ft2 x 2-ft x 1.40 (40% swell) = 168,000-ft3 or 6,222-yd3

Evaluate the question using the equation for the volume of a cone and the

maximum incline of the sides of the cone are at the natural angle of repose

equal to the angle of internal friction.

Check the maximum height based on the natural angle of repose.

r = h ÷ tan α°

75-ft = h ÷ tan 30° = h = 43-ft

Using the equation to find the Volume of the cone, solve for h, the Height: V = π r2 h 3 168,000-ft3 = (π 752 h) ÷ 3 Solve for: h = 28.53-ft is less than the natural angle of repose therefore the solution checks. (answer is b)

α= 30°

h

r

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34 Chapter 1 – Surveying |

BORROW PIT LEVELING

fast facts Borrow-pit leveling calculates the excavation volume by applying a grid to the excavation area. The grids can be staked to squares of 10, 20, 50, 100, or more feet depending on the project size and the accuracy desired. For each grid square, final elevations are established for each corner of every grid square. These are subtracted from the existing elevations at the same location to determine the depth of cut or height of fill at each corner. For each grid square an average of the depths/heights of the four corners is multiplied by the area of the square to determine the volume of earthwork associated with the grid area. The total earthwork volume for the project is calculated by adding the volumes of each grid square in the excavation area. Follow the following steps to evaluate and calculate the volume of soil at a borrow pit:

Step l Determine by visual study of the site drawing if the net total will be an import (more fill required than cut) an export (less fill required than cut) or a blend (cut and fill about equal) Step 2

Determine the pattern of calculation points or grid size. Step 3

Determine elevations at each calculation location, the corners of each grid.

Step 4

Calculate the cubic yards of cut or fill required in each grid cell. Step 5 Add the individual Grid Cell quantities together to arrive at the total cut, total fill volume and the import or volume export yardage required for the job.

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Chapter 1 – Surveying 35

15. - Question Which of the following statements about unit area or

borrow pit are true?

I. Sometimes called Average Depth

II. Works well for volumes for building sites and surface mines

III. Needs grid survey for best results

IV. Not a good choice for roads

V. Based on the principle of measuring material based on adding or removal from pit, hence, Borrow Pit

a. I & II

b. I, II, & III

c. I, II, III, IV, & V

d. None

Solution:

All the statements are true; (answer is c)

fast facts Note that the refresher course questions force you to focus your

attention on the units. Throughout, the questions make it a point to mix up

the “units”, that is, FT3, CY, PCF, yd3, etc. The purpose for this is for those

that are not as familiar with the terminology to become acutely aware of the

differences.

Strategies for Test Taking

• Rank order for difficulty all the questions and go for the “low hanging fruit” first. • Determine what is given and what is being asked. • Scan all answer choices before answering a question. • When approximation is required, scan answer choices to determine the degree of approximation or precision. • Avoid long computations. Use reasoning instead, when possible. • Scan the set of data to see what it is about. • Try to make visual comparisons and estimate products and quotients rather than perform computations. • Answer questions only on the basis of data given. • Answer “the” question, not “a” question. • Select the “best” answer choice.

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36 Chapter 1 – Surveying |

16. - Question Surveyed elevations are shown in Figure-1 for a 50’ x

50’ grid at a proposed parking lot location. The amount of borrow needed

to bring the area to an elevation of 90-ft is

most nearly:

a. 160-yd3

b. 178-yd3

c. 190-yd3

d. 200-yd3

Solution: Find the average elevation: Average = 87.6-ft + 87.6-ft + 87.6-ft + 88.6-ft 4 Average = 87.85-ft Change = 90-ft – 87.85-ft in elevation = 2.15-ft Fill Required = 2.15-ft x 50-ft x 50-ft = 199-yd3 (answer is d)

Figure 1

Datum

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Chapter 1 – Surveying 37

17. - Question The information given in Figure 1 is translated from

the surveyor’s field book data for a proposed local borrow pit. The volume

of excavation from the marked benchmark in yd3 is most nearly:

a. 2,050

b. 2,250

c. 2,325

d. 2,350

Solution:

Figure 1

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38 Chapter 1 – Surveying |

18. - Question For the project site shown in Figure 1, the amount of

import or export to bring the entire site to elevation 90-ft is most nearly:

a. The site is balanced

b. 526-CY import

c. 656-CY export

d. 833-CY export

Figure 1

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Chapter 1 – Surveying 39

Solution:

By inspection, the project has a combination of

areas where there are imports and exports to

bring the entire site to elevation 90-ft. Further,

the multiple choices given for the answer

should be analyzed. There is a spread of

approximately 20% among the choices which

suggests a broad perspective analysis to

determine the solution.

The figures to the right illustrate a broad based

analysis. Assume one grid:

Existing 90.50-ft

Proposed 90.00-ft

Cut 0.50-ft

Total Export = [150-ft x 300-ft x 0.50-ft] ÷ 27ft3/yd3 = 833-yd3

(answer is d)

fast facts

Many of the problems on the NCEES Civil PE exam will include

“extraneous” information that is not necessary to solve the problem. It is

important to remain focused on the information that is relevant and sift through

the distractions. Use a technique of underlining the relevant information in the

question so as to remain focused and not become distracted by irrelevant

content.

Remember that discrete quantitative questions measure: • basic mathematical knowledge • your ability to read, understand, and solve a problem that involves either an actual or an abstract situation.

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40 Chapter 1 – Surveying |

DIFFERENTIAL LEVELING

fast facts

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.

HB

HA

A’

Earth’s surface

Reference Surface

Horizontal Line

A

A B

B

B

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Chapter 1 – Surveying 41

19. - Question Which of the following statements about differential

leveling are true:

I. Bench mark (BM) – relatively permanent point of known

elevation as indicated on the Contract drawings

II. Back sight (BS) – a sight taken to the level rod held at a point of

known elevation (either BM or TP)

III. Height of Instrument (HI) – the elevation of the line of sight of

the telescope

IV. Foresight (FS) – a sight taken on any point to determine its

elevation

a. I & II

b. I, II, & III

c. I, II, III, & IV

d. None

Solution: 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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42 Chapter 1 – Surveying |

20. - Question Based on the information provided in Figure 1, the

difference in elevation between BM and TP2 is most nearly:

a. +3.60-ft

b. -11.08-ft

c. -7.48-ft

d. +7.48-ft

[not to scale]

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)

Point BS HI FS Elevation

BM 1.27 357.95 356.68

TP1 2.33 355.37 4.91 353.04

TP2 6.17 349.20

Check Sum +3.60 + -11.08 = -7.48

BS 2.33

BS 1.27

FS 6.17

FS 4.91

BM Elev. 356.68

TP1

TP2

BM + BS = HI

HI – FS = TP Elevation

p

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Chapter 1 – Surveying 43

21. - Question The trigonometric leveling from the surveyor’s notes is

shown below. The ground elevation of T (ft) is most nearly:

a. 1667.01

b. 1730.75

c. 1730.97

d. 1731.77

[not to scale]

FIGURE 1

Solution: The elevation of point P can be found from:

Elev P = elev T + HI – (Horizontal Distance) tan α + Rod Reading

Elev T = elev P – HI + HD tan α + RR

= 1703.99-ft – 5.02-ft + (220.85) tan 7°10’10” + 4.22-ft

= 1730.97-ft (answer)

4.22-ft Elev. 1703.99-ft P

α=7°10’10”

T HI=5.02-ft

220.85-ft

p

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44 Chapter 1 – Surveying |

22. - Question 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:

a. 425.3

b. 433.3

c. 439.5

d. 441.2

Not to scale

72-in. O.D.

Concrete Pipe (10-in

Wall Thickness)

STA. 253+65.7

Invert elev=438.33

Proposed MH

STA. 254+80.3

Ground elev=448.33

STA. 256+30.7

Invert elev=429.05

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Chapter 1 – Surveying 45

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)

Samples of various types of Manholes

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46 Chapter 1 – Surveying |

[not to scale] Figure 1

23. - Question A section view of a proposed roadway is shown in

Figure 1 with an existing grade point elevation of 100-ft. The proposed

roadbed finish grade elevation at station 50+00 is 95.00-ft and slopes at

2% grade. The grade rod reading at station 55+00 is most nearly:

a. 13.50-ft

b. 18.00-ft

c. 19.00-ft

d. 23.50-ft

Solution: The Height of Instrument elevation can be calculated: HI = 100.00-ft + 8.5-ft = 108.5-ft The grade rod reading at station 50+00 is calculated by: HI – station 50+00 elevation = 108.50-ft - 95.00-ft = 13.50-ft The distance between station 50+00 and 55+00 is 500-ft The slope is set at 2%, therefore; 2% x 500-ft = 10-ft elevation difference between stations. Add the results of the proposed roadbed slope which is 10-ft to the grade rod reading at station 50+00 to obtain the grade rod reading at station 55+00 which is 23.50-ft. (answer is d)

grade rod grade rod

Roadbed finish grade

Existing grade

ground rod

50+00

55+00

grade point

HI 8.50’

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Chapter 2 – Construction Management 47

CHAPTER 2 – CONSTRUCTION MANAGEMENT

Concept

Terminology

Application

Construction

Management

CH

AP

TE

R

2

Construction Management

Quantity Takeoff

Productivity Analysis

Engineering Economics

Takeoff

Factor Tables

Time Value of Money

Compound Interest

Present Worth

Future Worth

Annual Cost

Rate of Return

Benefit/Cost Ratio

Alternate Project Selection

Internal Rate of Return

NCEES – FE Civil Engineering Topics Construction Management 10% = 6/60 A. Procurement methods (e.g., design-build, design-bid-build, qualifications based) B. Allocation of resources (e.g., labor, equipment, materials, money, time) C. Contracts/contract law D. Project scheduling (e.g., CPM, PERT) E. Engineering economics F. Project management (e.g., owner/contractor/client relations, safety) G. Construction estimating

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48 Chapter 2 – Construction Management |

fast facts

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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Chapter 2 – Construction Management 49

CONSTRUCTION MANAGEMENT - PROCUREMENT METHODS

24. - Question Which of the following statements about the construction contract

process are true:

I. Surety bond assures the project owner a guarantee of funds equivalent to a

promissory note. The surety bond is a promise to pay the owner a certain

amount if the contractor fails in fulfilling the terms of a contract.

II. Performance bond is a surety bond issued by an insurance company or a

bank to guarantee satisfactory completion of a project by a contractor.

Performance bonds are issued upon Contract award and cost approximately

0.50% to 1.25% of the total contract value.

III. Builder’s risk insurance is a special type of property insurance which

indemnifies against damage to buildings while they are under construction.

Builder's risk insurance is coverage that protects a person's or an

organization's insurable interest in materials, fixtures and/or equipment being

used in the construction or renovation of a building or structure should those

items sustain physical loss or damage from a covered cause

IV. A bid bond is issued as part of a bidding process by the surety to the project owner, to guarantee that the winning bidder will undertake the contract under the terms at which they bid. The cash deposit is subject to full or partial forfeiture if the winning contractor fails to either execute the contract or provide the required performance and/or payment bonds. The bid bond assures and guarantees that should the bidder be successful, the bidder will execute the contract and provide the required surety bonds.

V. Bonds are not insurance. Bonds are a guarantee to pay made by a cosigner

who is liable only if the principal fails to discharge the obligations under the

Contract.

a. I & II

b. I, II, & III

c. I, II, III, & IV

d. I, II, III, IV, & V

Solution: All are true.

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50 Chapter 2 – Construction Management |

COST ESTIMATING

25. - 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. $65,771.25

b. $66,416.60

c. $68,402.10

d. $71,065.58

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)

Material Costs:

Aluminum Posts $645.35 -each

Picket Infill Panel $1,985.50- each

Placed Concrete $498.00/CY

Ironworker $78.00/hr

State Sales Tax 7%

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. Also, material cost on a capital

improvement project is non-taxable. Remember that “distracters” are included

in questions to test your engineering judgment.

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Chapter 2 – Construction Management 51

ESTIMATING TAKEOFF QUANTITIES

26. - Question A 450-ft length canal is to be lined with concrete for

erosion control. It is estimated that there will be 10% waste. Unit material

costs of concrete and curing compound based on other recent projects in

the area with similar volume are $98/ yd3 and $40/5-gal, respectively.

Project specifications require an application rate of curing compound at 1-

gal per 300-ft2.

[not to scale]

Determine the following:

a. The total material cost for delivered concrete is most nearly:

a. $85,000.00

b. $90,160.00

c. $99,176.00

d. $109,094.00

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

a. $1,120.00

b. $1,160.00

c. $1,200.00

d. $1,240.00

20-ft

19-ft

2

3

8-in - concrete

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52 Chapter 2 – Construction Management |

Solution: a. Horizontal length of side slope = (20 / 2) x 3 = 30 ft

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

Volume of lining = (60.74 x 450) / 27 = 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 = 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 is enough to

support the manufacturer’s position that the coverage rate was not met. Always

round up in this situation.

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Chapter 2 – Construction Management 53

27. - 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. 105-ft2

b. 134-ft2

c. 266-ft2

d. 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)

Plan View

Section View

1’-6”

40’-6”

1’-6”

3’-2”

Concrete Knee

Wall

Not to scale

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54 Chapter 2 – Construction Management |

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.

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Chapter 2 – Construction Management 55

28. - 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. $8.75 b. $9.78 c. $10.12 d. $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 x9-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 Total labor cost per 90-yd3 = $880.00 Cost per cubic yard $880/90-yd3 = $9.78/yd3 (answer is b)

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56 Chapter 2 – Construction Management |

29. - Question The Owner’s representative requested a cost proposal for the

Architect’s design change. The revised Scope of Work (SOW) is to provide a credit for

the installation of one layer of ½” Gypsum Wallboard (GWB) and provide a new

scope of work consisting of: (1) install two (2) layers of 5/8” Fire Code (FC) GWB

on the inside face of the existing framing system, (2) as part of the building code

requirement mud coat the first layer of the GWB, (3) mud coat and finish coat the

second layer of GWB, (4) and provide sound-batt insulation within the existing metal

stud wall cavity. The room is located within a warehouse with dimensions: 180-ft x 200-

ft; 10-ft high floor to underside of deck dimension; and, eight (8) 6’-0”wide x 10’-0” high

openings. The total cost for the new scope of work is most nearly:

a. $19,500

b. $20,500

c. $21,000

d. $22,500

Use the following excerpt from the Company’s cost standards:

Work Crew 4 Carpenters 2 Laborers Working Foreman

Labor Rates per hour

Carpenter Foreman (working) $46.35 fully burdened Carpenter (journeyman) $38.85 fully burdened Laborer $28.45 fully burdened Work Crew Productivity (based on 8-hr/day) GWB 960 ft2/Work Crew Hr Insulation 1,920 ft2/ Work Crew Hr Tape &Spackle Mud Coat 2,800 ft2/ Work Crew Hr Tape &Spackle Finish Coat 1,800 ft2/ Work Crew Hr

Material Costs price includes all taxes and delivery; add a 10% waste factor to all

materials.

4’-0” x 10’-0” x ½” GWB $0.185/ ft2 4’-0” x 10’-0” x 5/8” GWB (FC) $0.285 / ft2 Sound Batt Insulation (65-ft2/Bag) $0.45/ ft2 Mud Coat (coverage 150- ft2/gal) $0.125/ft2 Finish Coat(coverage 300- ft2 /gal) $0.08/ft2 Contractor Change Order Pricing Contactor’s Overhead 10% Contractor’s Profit 5%

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Chapter 2 – Construction Management 57

Solution: Step 1; Sketch the project,

[not to scale]

Step 2 - Determine GWB surface Area

Walls

200-ft

200-ft

180-ft

180-ft

760-ft Total

760-ft x 10-ft = 7,600-ft2

Outs (delete)

(8) x 6-ft x 10-ft =480-ft2

Surface Area

7,600-ft - 480-ft = 7,120-ft2

200-ft

180-ft

Openings

(typ) (2) layers

GWB Interior

Side Only

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

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58 Chapter 2 – Construction Management |

Step 3 - Calculate Fully Burdened Cost Rate for Crew Day

Trade QTY U/M Rate/hr Total

Foreman 1 ea 46.35$ 46.35$

Journeyman 4 ea 38.85$ 155.40$

Laborer 2 ea 28.45$ 56.90$

TOTAL 258.65$ per crew hour

TOTAL 2,069.20$ per crew day (8-hrs)

Step 5 - Calculate Change Order amount for the new SOW

Description QTY U/M Production U/M Unit Cost U/M Crew Hrs

Total Labor

Cost

Total Material

Cost

5/8" GWB Labor 14,240 ft^2 960 ft^2 258.65 14.83 3,836.64$

Material 15,680 ft^2 includes waste 0.285$ ft^2 4,468.80$

Sound Batt Labor 7,120 ft^2 1,920 ft^2 258.65 3.71 959.16$

Material 7,865 ft^2 includes waste 0.450$ ft^2 3,539.25$

Mud Coat Labor 14,240 ft^2 2,800 ft^2 258.65 5.09 1,315.42$

Material 15,750 ft^2 includes waste 0.125$ ft^2 1,968.75$

Finish Labor 7,120 ft^2 1,800 ft^2 258.65 3.96 1,023.10$

Material 8,100 ft^2 includes waste 0.08$ ft^2 648.00$

Column Totals 27.58 7,134.33$ 10,624.80$

SUMMARY

Labor Cost 7,134.33$

Material Cost 10,624.80$

SUB-TOTAL 17,759.13$

Contractor OH @ 10% 1,775.91$

SUB-TOTAL 19,535.04$

Contractor Profit @ 5% 976.75$

Grand Total 20,511.79$ New Scope of Work

(answer is b)

Step 4 - Determine the Material List for the new Scope of Work

5/8" GWB 7,120 x 2-layers x 1.10 (waste) / 40-SF/BD = 392 boards

Sound Batt 7,120 / 65ft^2/bag x 1.10 (waste) = 121 bags

Mud Coat 14,240 /150-ft^2/gal x 1.10 (waste) = 105 gallons

Finish 7,120 /300-ft^2/gal x 1.10 (waste) = 27 gallons

fast facts 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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Chapter 2 – Construction Management 59

COST ESTIMATING – BOARD FEET

Sample calculations provided in Table below for 2” thick stock.

fast facts

One board foot equals 144 cubic inches. The thickness in inches, times the

width in inches, times the length in inches, divided by 144 cubic inches,

equals total board feet in a piece of stock. For instance, a piece two inch

thick by twelve inches wide by twelve inches long would be 2" x 12" = 24" x

12" = 288 cubic inches ÷ 144 cubic inches = 2 board feet.

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60 Chapter 2 – Construction Management |

30. - Question A project engineer has ordered 100 hardwood boards,

1-1/4-in thick by 8-in wide and 8-ft long. The total board feet ordered is

most nearly:

a. 533

b. 600

c. 667

d. 1,067

Solution:

1.25” x 8” x 8’ x 100-pcs = 666.67

board feet (answer)

12

Or, 1.25” x 8” x (8’x 12”/1’) x 100-pcs

= 666.67 board feet (answer is c)

144

fast facts The questions on the NCEES exam generally do not include units with

the associated answer choices. You are more likely to see answers such as

20, 40, 60, and 80, rather than 20-PSF, 40-PSF, etc.

Read the problem statement carefully to ensure that you know what

units to solve for. Some of the answers are logical distracters and are only

included to test your “engineering judgment”.

Four minutes per question requires focusing your attention to the time

variable of the exam.

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Chapter 2 – Construction Management 61

METHODS OF BUDGETING

31. - Question The total equipment cost for an addition to a

pharmaceutical plant is estimated to be $5,000,000. The percentages

provided in the Table represent the average expenditures in other cost

phases within the project budget. The total cost for the project is most

nearly:

a. $5,000,000

b. $12,500,000

c. $14,000,000

d. $16,700,000

Solution:

Calculate the total percentage of the items provided:

Total percentage = 70%

Apply the equation:

Total Project Cost = $5,000,000 = $16,666,667 (answer is d)

(1 – .70)

Description Percent

Engineering, overhead, and fees 22%

Equipment Storage 5%

Services 2%

Utilities 6%

Piping 20%

Instrumentation 5%

Electrical 6%

Buildings 4%

Estimated Equipment Cost = Total Project Cost

(1 – Percent of Other Costs)

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62 Chapter 2 – Construction Management |

CONSTRUCTIONHISTORIC DATA

32. - Question A concrete specialty company’s uses productivity standards calculated as averages from historical data. On average 100-square 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 = $360.00 Laborer — 5 hr at $22.00/hr = $110.00 Total labor cost for 100 ft2 = $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:

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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Chapter 2 – Construction Management 63

F

ENGINEERING ECONOMICS

P – Present worth of money

F – Future worth of money

A – An end-of-period cash receipt or disbursement in a uniform series

i – Interest rate per interest period

n – Number of interest periods

Single Payment Present Worth

P = F(P/F, i, n) = ni)(1

F

Interest Formulas: Payments

Single Payment Compound Amount Factor

(F/P, i%, n) = (1 + %, i )n

Single Payment Present Worth Factor

(P/F, i%, n) = 1/ (1 + %, i )n= 1/ (F/P, i%, n)

Uniform Series Compound amount Factor

(F/A, i%, n) = (1 + %, i )n - 1 / i

Uniform Series Sinking Fund Factor

(A/F, i%, n) = i / (1 + i)n - 1 = 1 / (F/A, i%, n)

P

fast facts Factor Tables are derived from the equations shown in the introductory pages

of this section. The Factor Tables are a convenience, however, not all % factors are

available and although interpolation can be used, it is strongly recommended to

become as familiar with the equations. FE – Ref Hb. provides a summary of the

varying equations used in engineering economics.

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64 Chapter 2 – Construction Management |

See pages 114 through 120 in the NCEES Supplied Reference

Handbook for Factor Tables

For example, where F/P is the column selector for the Factor Tables, the interpretation is:

Calculate “F” / Given “P”

Where: A = Annual Amount

F = Future Worth

P = Present Worth

G = Uniform Gradient Amount

n = number of compounding periods or life of asset

i = effective rate per period

p

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Chapter 2 – Construction Management 65

FACTOR TABLE QUICK VIEW EXERCISE

The following factor table quick exercise uses the value of $1.00 using 10% for 10-yrs.

Using the appropriate Factor Table found in the NCEES Supplied Reference Handbook, find

the appropriate answer. Use the Calculate “x” / Given “y” study aid.

a. If you want $1.00, 10-yrs from now, deposit $_____ now.

a. 0.3855 b. 6.1446 c. 2.5937 d. 0.1627 e. 0.0627 f. 15.9374 g. 22.8913

b. If you deposit $1.00 at the end of every year for 10-yrs the present value is?

a. 0.3855 b. 6.1446 c. 2.5937 d. 0.1627 e. 0.0627 f. 15.9374 g. 22.8913

c. $1.00 today is worth $_______10-yrs from now in an account yielding 10%

a. 0.3855 b. 6.1446 c. 2.5937 d. 0.1627 e. 0.0627 f. 15.9374 g. 22.8913

d. The annual amount of interest deposited in the account with a starting balance of $1.00 at 10% for 10-yrs is?

a. 0.3855 b. 6.1446 c. 2.5937 d. 0.1627 e. 0.0627 f. 15.9374 g. 22.8913

p

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66 Chapter 2 – Construction Management |

e. If you want to have $1.00 in the bank, deposit $_____every year for 10–years at 10% an annual yielding account.

a. 0.3855 b. 6.1446 c. 2.5937 d. 0.1627 e. 0.0627 f. 15.9374 g. 22.8913

f. If you deposit $1.00 every year in an account yielding 10%, you will have this amount in 10-yrs $__________.

a. 0.3855 b. 6.1446 c. 2.5937 d. 0.1627 e. 0.0627 f. 15.9374 g. 22.8913

g. If you deposit $1 in yr-1; $2-at the beginning of yr-2; $3 at the beginning of yr-3, and so on to the 10th year, the present worth of the deposits is $__________.

a. 0.3855 b. 6.1446 c. 2.5937 d. 0.1627 e. 0.0627 f. 15.9374 g. 22.8913

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Chapter 2 – Construction Management 67

TIME VALUE OF MONEY

33. - Question If you want to have $60,000 in 10 years, the amount that should be

put into a 6.0% (effective annual rate) savings account now is most nearly:

a. $33,503.69

b. $43,000.39

c. $48,475.09

d. $53,500.60

Solution:

This problem could also be stated: What is the equivalent present worth of $60,000 ten

years from now if monthly money is worth 6% per year?

P = F(P/F, I, n) = $60,000(P/F, 6%, 10) = $60,000 * 0.5584 = $33,503.69 (answer is a)

34. - Question The cost of utilities, taxes and maintenance on a home is $3,000

per year. The amount of money that would have to be invested now at 8% to cover

these expenses for the next 5 years is most likely: (Assume no inflation or tax

increase).

a. $10,980

b. $11,980

c. $12,980

d. $13,980

Solution:

Referencing Appendix 86-A (pg. A-136) in

CERM-11 and using the Factor Tables

$3,000 (P/A, 8%, 5) = $3,000 x 3.9927 =

$11,978.10 (answer is b)

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68 Chapter 2 – Construction Management |

35. - Question Your home mortgage is $300,000 for 30 years with a

nominal annual rate of 7%. The monthly payment is most nearly:

a. $1,899.00

b. $1,900.00

c. $1,995.10

d. $2,015.00

Solution:

n = 360 months interest = 7%-annual ÷ 12-months/year = 0.583% per month $300,000(A/P, 0.00583, 360) = Apply the equation: = 0.006650339 A = 300,000 x 0.006650339 = $1,995.10 per month (answer is c)

i(1+i)n (1+i)n - 1

fast facts This question illustrates the importance of interpreting the information provided before running through the computations. Although the nominal annual rate is given as 7%, the monthly rate needs to be computed. A common approach would be to use the CERM Appendix Factor Tables to find the monthly payments which would yield: (A/P, 7%, 30) = $300,000(A/P, 0.0806, 30) = $24,180/12 = $2,015. / month or a 1% error.

Conclusion: Be familiar and comfortable with both the Factor Tables and the Equations which comprise the results in the Tables.

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Chapter 2 – Construction Management 69

36. - Question You wish to buy a house and you can afford to make

a down payment of $50,000. Your monthly mortgage payment cannot

exceed $2,000. If 30-year loans are available at 7.5% annual interest rate

which is compounded monthly, the highest price that you may consider is

most nearly:

a. $332,000

b. $334,000

c. $336,000

d. $338,000

Solution:

n = 360 months interest = 7.5%-annual ÷ 12-months/year = 0.625% per month Apply the equation: This yields (A/P, 0.00625, 360) = 0.00699

Apply the following equation to find the highest price to consider:

P x 0.00699 = (Highest $ - $50,000) x 0.00699 ≤ $2,000

Highest $ ≤ ($2,000/0.00699) + $50,000 = $336,123 (answer is c)

i(1+i)n (1+i)n - 1

fast facts

Alternate method - Use one year as the time period.

Then, n = 30 years, and i = (1 + 0.00625)12 – 1 = 7.763%

Then, (A*/P, 0.07763, 30) = 0.0867 [A* = 0.0867 P per year]

Your effective payment per year is:

A* = $2,000 x (F/A, 0.00625, 12) = $2,000 x12.4212 = $24,842

P ≤ (24,842 / 0.0867) + $50,000 = $336,533 same as before;

answer checks.

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70 Chapter 2 – Construction Management |

37. - Question An engineer deposits a “X” amount every 6 months for

3 years so that she’ll have $10,000 at the end of this period. The interest

rate is 5% per year which is rebalanced every 6-month. The amount

deposited is most nearly:

a. $1,565.50

b. $1,585.50

c. $1,595.50

d. $1,600.50

Solution: Determine the components and apply the equation:

n = 6 deposits i = 2.5% per 6-month period

F = $10,000(A/F, 2.5%, 6) = 0.15655 A = $1,565.50 (answer is a)

COMPOUND INTEREST – NOMINAL AND EFFECTIVE

38. - Question A credit card advertises a nominal rate of 18%

compounded monthly. If no monthly payments are made, the effective

annual interest rate is most nearly:

a. 18.00%

b. 18.25%

c. 18.75%

d. 19.56%

The actual rate is, (18% / 12) = 1.5% per month. The effective annual rate

is:

i = (1 + .015)12 – 1 = 0.1956 or 19.56% if you do not pay anything each

month. (answer is d)

fast facts

This question illustrates that 18% interest per month is

equivalent to 19.56% effective annual rate. The terms are

synonymous as they are dependent upon a “point in time”.

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Chapter 2 – Construction Management 71

fast facts

Compound interest is the concept of adding accumulated

interest back to the principal, so that interest is earned on interest

from that moment on. The act of declaring interest to be principal

is called compounding (i.e., interest is compounded). Interest rates

must be comparable in order to be useful, and in order to be

comparable, the interest rate and the compounding frequency must

be disclosed.

Since most people think of rates as a yearly percentage,

many governments require financial institutions to disclose a

(nominally) comparable yearly interest rate on deposits or

advances.

Compound interest rates may be referred to as annual

percentage rate, annual percentage yield, effective interest rate,

effective annual rate, and by other term. Compound interest may

be contrasted with simple interest, where interest is not added to

the principal (there is no compounding).

Note that the effective interest rate i depends on the

frequency of compounding. The following illustrates the effects of

compounding.

Example: nominal interest rate r = 10%

–Compounded annually: i = r = 10%

–Compounded quarterly: i = (1 + 0.1 / 4)4 - 1 = 10.38%

–Compounded monthly: i = (1 + 0.1 / 12)12 - 1 = 10.471%

–Compounded weekly: i = (1 + 0.1 / 52)52 - 1 = 10.506%

–Compounded daily: i = (1 + 0.1 / 365)365 – 1 = 10.516%

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72 Chapter 2 – Construction Management |

SOLVING ENGINEERING ECONOMIC PROBLEMS

Uniform Series Present Worth Factor

P = A (P/A, i, n) = A

n

n

i)i(1

1i)(1

Uniform Series Future Worth Factor

F = A (F/A, i, n) = A

i

1i)(1 n

Net Present Worth NPW = PW of benefits – PW of costs Benefit-Cost Ratio

costs ofPW

benefits ofPW

C

B

F

A A A A A

P

A A A A A

fast facts Engineering Economics applies the principles reviewed in the previous

section and allows the calculation for the time value of money to be evaluated at a

“common” point in time.

The majority of engineering economic analysis questions are alternative

comparisons. In these questions, two or more mutually exclusive investments

compete for limited funds.

To help with the analysis, cash flow diagrams can be drawn to help visualize

and simplify problems having diverse receipts and disbursements. The most

obvious advantage for the diagram is to clearly see the reference point in time.

Remember to pay particular attention to counting the “zero” position.

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Chapter 2 – Construction Management 73

PRESENT WORTH

39. - Question A heavy equipment rental company uses a high

interest rate of 30% for the rental of a specialty heavy haul dump truck.

The net annual profit for this investment is most nearly:

a. - $3,900

b. - $4,500

c. +$5,000

d. +$6,000

Solution:

Calculate the annual capital recovery with return:

$80,000 (A/P, 30%, 6yrs) = $30,272

Calculate the net annual profit:

55,000 – (30,272 + 28,600) = - $3,872 (net loss) (answer is a)

Total annual income: $55,000

Capital cost: $80,000

Annual operating cost: $28,600

Lifespan: 6 years

fast facts This question illustrates a counterintuitive result for the interpretation of the data presented. At first glance, the business scenario appears that it should be profitable. However, the high rate of 30% is intended to cover uncertainty during the rental period. After computation, the results show that it does not cover uncertainty. However, if the annual income were $60,000, then the net annual profit would be 60,000 - (30,272 + 28,600) = $1,128 and the truck rental would return a net profit.

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74 Chapter 2 – Construction Management |

FUTURE WORTH OR VALUE

40. - Question Two alternatives have the cash flows as shown in

the Table. At a 4% interest rate, the yield for the alternative that should

be selected to maximize its value is most nearly:

a. -$247.60

b. +$247.60

c. -$284.20

d. +$284.20

Solution:

This example will be solved using Future Worth analysis at the end of 3-

years.

NFWA = 800(F/A, 4%, 3) – 2000 (F/P,4%,3) =800(3.122) – 2000(1.125)= +$247.60 NFWB = 1100(F/A, 4%, 3) – 2800 (F/P,4%,3) =1100(3.122) – 2800(1.125)= +$284.20

To maximize Net Future Worth, choose alternative B: + $284.20

(answer is d)

Alternative Year A B 0 -$2000 -$2800 1 +800 +1100 2 +800 +1100 3 +800 +1100

Net Future Worth (NFW) = Future Worth of Benefits – Future Worth of Costs

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Chapter 2 – Construction Management 75

ANNUAL COST

41. - Question A contractor purchased an excavator for $100,000.

The life expectancy is 10-years. With the annual interest rate at 7%, the

annual cost for the excavator is most nearly:

a. $10,000

b. $11,000

c. $13,000

d. $14,000

Solution:

Equivalent Uniform Annual Cost (EUAC) = P (A/P ,i, n) = 100,000(A/P, 7%,10) = $14,200 (answer is d)

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76 Chapter 2 – Construction Management |

MAINTENANCE COSTS

42. - Question A contractor purchased a

Caterpillar D-13 dozer for $100,000 and owned it for

five years. The annual cost for maintenance and repair

is provided in the Table. With annual interest rate at

7% and end-of-year disbursements, the uniform annual

cost for maintenance and repair for the dozer is most

nearly:

a. $10,000

b. $11,000

c. $12,000

d. $13,000

Solution: The Equivalent Uniform Annual Cost (EUAC) can be computed

for this irregular series of payments in two steps:

Step 1: Compute the Present Worth for five years using the single

payment present worth factors.

PW of cost = 4500(P/F, 7%, 1) + 9000(P/F, 7%,2) + 11800(P/F, 7%, 3) +

13500(P/F, 7%, 4) + 22500(P/F, 7%,5)

= 4500(.9346) + 9000(.8734) + 11800(.8163) + 13500(.7629) + 22500(.7130)

= $48,040.29

Step 2: With the PW of cost known, compute the EUAC using the

capital recovery factor.

EUAC = 48,040(A/P, 7%, 5)

= 48,040(.2439)

= $11,717.03 (answer is c)

Year Maintenance and Repair

Costs 1 $4,500

2 $9,000

3 $11,800

4 $13,500

5 $22,500

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Chapter 2 – Construction Management 77

RATE OF RETURN ANALYSIS – THREE ALTERNATIVES

43. - Question Three alternatives are being considered for

improving a street intersection. The annual dollar savings on account

of the improvement is shown below. Assume that the intersection will

last for 25 years and the interest rate is 5%. Each of the three

improvements is mutually exclusive but provides similar benefits. The

alternative that is the most economical is:

a. A

b. B

c. C

d. Cannot be determined without additional information

Alternative Total cost Annual Benefit

A $10,000 $ 800

B $12,000 1,000

C $19,000 1,400

Solution:

Estimate the Net Present Worth (NPW) for each alternative and identify the most economical alternative for construction. Net Present Worth (NPW) = PW of benefits – PW of costs Use the present worth factor for uniform series:

(P/A, i =5%, 25 yrs) = 14.094 (from the Factor Tables)

NPW(A) = ($800 x 14.094) - $10,000 = $1,275.20

NPW(B) = ($1000 x 14.094) - $12,000 = $2,094.00

NPW(C) = ($1400 x 14.094) - $19,000 = $731.60

Therefore, select Alternate B because it has the highest net present worth. (answer is b)

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78 Chapter 2 – Construction Management |

BENEFIT/COST ANALYSIS

44. - Question An Earthwork Contractor is planning the purchase of new equipment. Three options are under review:

Option 1 Option 2 Option 3

Initial Cost 500,000 600,000 700,000 Annual O&M cost 10,000 15,000 20,000 Annual Revenue 100,000 120,000 150,000

The interest rate is 10% and the life of each option is 20 years. The

Benefit/Cost Ratio for the recommended option is most nearly:

a. 1.40 b. 1.46 c. 1.47 d. 1.48

Solution: Find the Benefit /Cost ratio for each option using the conventional formulas found in the CERM-11. Set the equation to: Option X = (P/A, i. 10%) for each option. Benefits (Option1) = Present Worth of $100,000 annual revenue for 20 years @10% From the interest tables or by calculation: PW (Benefits) = 8.514 x $100,000 = $851,400 Costs (Option 1) = Initial Cost + Present Worth of $10,000/year O&M From the interest tables or by calculation: PW (O&M) = 8.514 x $10,000 = $85,140 Therefore: B/C (Option 1) = $851,400 / ($500,000+$85,140) = 1.455 Similarly, B/C (Option 2) = $1,021,680 / ($600,000+$127,710)=1.403 And, B/C (Option 3) = $1,277,100 / ($700,000+$170,280)=1.467 Option 3 brings the most benefit to the contractor. (answer is c)

fast facts All comparisons must be made at a single time reference point. Use PW (Present

Value), FV (Future Value) or AW (Annual Worth) as a basis. (PW and AW are the

most commonly used). The analysis will be performed using Present Worth. If the

Benefit to Cost Ratio (B/C) > 1.0, then the project is a go.

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Chapter 2 – Construction Management 79

ALTERNATE PROJECT SELECTION

45. - Question A County is planning to build a new roadway

connector in the rapidly developing area of the county. The road can be

built at a reduced capacity now for $30 million and can be widened 15

years later for an additional $20 million. An alternative is to construct the

full capacity connector now for $40 million. Both alternatives would provide

the needed capacity for the 25-year analysis period. Maintenance cost

differences are small and may be ignored. At 6% interest, which alternative

should be selected:

a. Two stage construction with a $1,700,000 savings

b. Single stage construction with a savings of $1,700,000

c. Two stage construction with a $300,000 savings

d. Single stage construction with a savings of $300,000

Solution: Analysis of the two-stage construction: PW of cost = $30 million + $20 million (P/F, 6%, 15) = $30 million + $8.3 million = $38.3 million Analysis of the single-stage construction: From the information provided: PW of cost = $40 million Two-stage construction alternative is preferred with a $1,700,000 savings. (answer is a)

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80 Chapter 2 – Construction Management |

ALTERNATE SELECTION OF COMPONENTS

46. - Question A city is trying to decide which of two traffic devices to

install. Device A costs $1000 and has useful life of five years. Device A

can be expected to result in $300 savings annually. Device B costs $2000

and has useful life of eight years. Device B can be expected to result in

$400 savings annually. Both devices have no salvage value at the end of

their lives. With interest at 7%, the benefit cost ratio of the most cost

effective device is most nearly:

a. 1.19 b. 1.23 c. 1.32 d. 1.38

Solution: Device A: PW of cost = $1000 PW of benefits = $300 (P/A, 7%, 5) = $1230 B = 1230 = 1.23

C 1000 Device B: PW of cost = $2000 PW of benefits = $400 (P/A, 7%, 8) = $2388 B = 2388 = 1.19

C 2000 In order to maximize the benefit-cost ratio, select Device A; 1.23 (answer is b )

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Chapter 2 – Construction Management 81

47. - Question The construction of a new bakery requires the property development firm to pay the municipality the cost to expand its sewage treatment plant. In addition to the expansion costs, the developer must pay $60,000 annually toward the plant operating costs. The developer plans to finance the annual costs by placing money into a fund that earns 5% per year to pay its share of the plant operating costs forever. The amount to be paid to the fund is most nearly:

a. $120,000 b. $600,000 c. $1,200,000 d. $2,200,000

Solution: Using CERM-11 equation 86.10 for a present worth of an infinite (perpetual) series of annual amounts is known as capitalized cost. P = A / i = 60,000 ÷ .05 = $1,200,000 (answer is c)

48. - Question An earthwork contractor is considering the purchase of a new excavator. Based on 10% interest, the equivalent uniform annual cost for the excavator is most nearly:

a. $12,000 b. $27,000 c. $28,000 d. $32,000

Solution: EUAC = 80,000 (A/P, 10%, 20) – 20,000 (A/F, 10%, 20) + annual operating cost = 80,000 (0.1175) – 20,000 (0.175) + 18,000 = 9400 – 350 + 18,000 = $27,050 (answer is b)

Initial Cost $80,000

End of useful life salvage value $20,000

Annual Operating Cost $18,000

Useful Life 20-years

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82 Chapter 2 – Construction Management |

CONTRACTOR PROJECT FINANCING

49. - Question A contractor is bidding a construction project that will

take 12 months to complete. The terms of the bid provide that the owner

will pay the Contractor one lump sum at the end of 14-months. The

Contractor secured a bank loan with an interest rate of 1% per month on

the outstanding balance. The Contractor will borrow $100,000 per month to

reconcile costs at the end of months 1 through 12. The cost to finance the

project that will be included as part of the bid is most nearly:

a. $68,000 b. $93,000 c. $94,000 d. $118,000

Solution: At the end of month 12, the contractor will have borrowed 12 payments of $100,000 each. The owner is scheduled to pay the contractor at the end of month 14 at which time the contractor will repay the bank. Use the following equation to calculate the future value then extract the cost of financing the project: F = $100,000 (F/A, 1%, 12) (F/P, 1%, 2) = $1,293,700 Amount Borrowed = $1,200,000 Amount of interest = $ 93,700 (answer is c)

fast facts

Your calculator is an important tool during the exam. Select and use the

calculator approved by the NCEES. Use only your selected calculator for the

next few months to become comfortable with all of its relevant functions, and

give yourself enough time to learn how to use it effectively. There is a learning

curve with any new calculator. You will save time during the exam if you are

familiar with its functions. The SOLVE, POL(), REC() functions are all time

savers. Also, obtain an identical model to bring with you as a backup with

personal distinguishing identification marks.

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Chapter 2 – Construction Management 83

INTERNAL RATE OF RETURN

50. - Question A contractor established a policy to purchase equipment over leasing only if the return is a minimum of 15%. Given the following conditions, the contractor’s decision based on the policy reveals that:

a. The equipment purchase yields a minimum 15% return b. The equipment purchase yields less than a minimum 15% return c. The purchase yields a neutral or $0 return d. Cannot be determined without additional information

Solution: The contractor’s goal is to accept a purchase with Internal Rate of Return (IRR) larger than the discount rate in which he can borrow money. The first step is to identify the rate of return on the investment. An example would be to use a geometrically rising series of values. A typical means of computing IRR is to identify the discount rate that sets the Net Present Value (NPV) to $ 0 (zero dollars). Applying the concept yields an Internal Rate of Return (IRR) that must satisfy the Contractor’s goal.

Find the NPV using a 15% IRR by establishing a zero sum equation

$0 = - $20,000 + $5,600 (P/A, IRR%, 5) + $4,000 (P/F, IRR%, 5) $0 = +$760

Therefore, $0 IRR = $760 which is > 15% IRR; the purchase is justified. (answer is a)

Machine “A” Initial Cost $20,000 Life 5-years Salvage Value $4,000 Annual Receipts $10,000 Annual Disbursements $ 4,400

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84 Chapter 2 – Construction Management |

fast facts

Percentage of a number

A percentage is a way of expressing a ratio or a fraction as a whole number, by using 100

as the denominator. One percent is one per one hundred or one hundredth of a whole

number; notation: 1%.

Below are the statements of main percentage problems and their solutions.

1. Find the number b that makes up p% of a number a. b = a p 100

2. Find the number a whose p% is equal to a number b. a = 100 b p

3. What percentage does a number b make up of a number a? p = 100 b % a

Percent Change

To compute the percent change between two numbers:

The equation is: (old value – new value) divided by old value;

Old – New = Percent Change

Old

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Chapter 2 – Construction Management 85

PROJECT SCHEDULE FINANCIAL ANALYSIS

51. - Question The 6th monthly payment application for the construction of a $70,000,000

University dormitory is analyzed and the summary accounting show: the budgeted cost of work

performed is $14,500,000, while the actual cost of work performed is $14,750,000. The cost

status of the project is most nearly:

a. 1.5% under budget

b. 1.5% over budget

c. 1.7% under budget

d. 1.7% over budget

Solution: Financial project status is analyzed using the following equations:

Cost Variance = BCWP - ACWP

Cost Variance = $14,500,000 – $14,750,000 = - $250,000 (over budget)

(1 - ($14,500,000 ÷ $14,750,000)) x 100 = 1.69% over budget (answer is d) (See page 90 of the Notes for more information on effect of percent)

ESTIMATING ACTIVITY DURATIONS

52. - Question The total labor cost for emergency bridge rehabilitation on an interstate

highway is $5,716,440. The average crew-hour cost is $15,879 for this nonstop project. The

number of work days the roadway was shut down is most nearly:

a. 1

b. 15

c. 30

d. 45

Solution: Assigning duration of actives is the estimated or actual time that it will be required to

complete it. Accordingly:

Total labor cost ÷ crew-hour cost = total crew-hours

$5,716,440 ÷ $15,879 = 360 crew-hours

Nonstop project constitutes work at 24-hr 7-days/week Total crew hours ÷ 24-hrs/day = project days 360 crew-hour ÷ 24-hrs/day = 15 days (answer is b)

BCWS = Budgeted Cost of Work Scheduled = planned costs ACWP = Actual Cost of Work Performed = actual spent BCWP = Budgeted Cost of Work Performed = Earned Value

Schedule Variance = BCWP – BCWS

Cost Variance = BCWP - ACWP

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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:

���������������� = ����

���������������������

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.

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Chapter 2 – Construction Management 87

53. - Question A concrete contractor historical records show that the

construction duration for a cast-in-place foundation formwork is 1,000-

SF/crew-day. The number of days a 200,000-SF formwork for a cast-in-

place concrete foundation using 3-crews is most nearly:

a. 55

b. 60

c. 70

d. 75

Solution: Apply the equation:

���������������� = ����

���������������������

Duration = 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.

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88 Chapter 2 – Construction Management |

PROJECT SCHEDULING FUNDAMENTALS

54. - Question The total duration for the four activity project is most

nearly:

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

a. 1

b. 2

c. 3

d. 4

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.

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Chapter 2 – Construction Management 89

fast facts

PROJECT SCHEDULING – TYPES OF METHODS

Project Scheduling – Distinction between the Major Types of Schedule Diagrams

• Precedence Diagrams (PDM)

– Activity on Node (AON)

– Can have any kind of precedence

• Arrow Diagrams (AOA)

– Activity On Arrow (AOA)

– Activity-on-Branch

– May have dummy tasks

– Finish to start precedence only

Units of Time Convention

Units of time are often stated in terms of days. There are two types of convention:

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. NCEES use both conventions in their scheduling examples.

4. Unless otherwise stated, all relationships are assumed to be finish-to-start relationships.

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90 Chapter 2 – Construction Management |

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

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

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Chapter 2 – Construction Management 91

PRECEDENCE DIAGRAMMING METHODS

• Activity on Node (AON) with arrows for dependencies

• Uses all forms of precedence relationships

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.

• Review Figure 1 Activity-on-Arrow diagram below. Note the dummy activity

showing the relationship between the material procurement and installation of the

components.

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92 Chapter 2 – Construction Management |

3

9

6

8 1 Excavate

5 Foundation

2 Assemble

7

4

Material

Delivery Curtain

Wall

Roof

Window

Doors

Electrical

Painting Finishing

Drainage

Grading Install Fence

Plumbing

Operation

Duration i j EF

LF ES

LS

LEGEND

Figure 1 - Activity on Arrow Diagram

fast facts The network illustrates the importance of procurement and material delivery in the

schedule. The “Material Delivery” 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 components must be installed together.

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Chapter 2 – Construction Management 93

A critical Path is:

• The series of activities which determines the earliest possible date of project

completion

• Usually defined as those activities with minimal or zero float

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

• Primary focus when analyzing the project is on calculating float

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:

ID Task Name

1 Task 1

2 Task 2

3 Task 3

4 Task 4

S M T W T F S S M T W T F S

Dec 3, '00 Dec 10, '00

Activity

(duration)

ES EF

LS LF

ACTIVITY

ES DURATION LS

EF FF TF LF

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94 Chapter 2 – Construction Management |

The forward pass works from left to right, calculating when tasks can end using their early start

date and the expected duration.

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

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

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

THE FORWARD PASS

EF = ES + Duration

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Chapter 2 – Construction Management 95

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.

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

THE BACKWARD PASS

LS = LF – Duration

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96 Chapter 2 – Construction Management |

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.

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

CALCULATING FLOAT

TF = LF - ES - DUR

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Chapter 2 – Construction Management 97

CRITICAL PATH - ACTIVITY ON NODE

55. - Question The project precedence table shown below provides the relationships and durations of all activities. Based on end-of-day calculations for starts and finishes, use a critical path analysis to determine the following: a) The project duration is most nearly:

a. 36 b. 48 c. 41 d. 32

b) The Critical Path is most nearly:

a. A,B,C,D,F,G b. A,B,C,D,E,F,G c. A,D,E,G, d. A,B,C,E,G

Solution: Where there are multiple paths between subsets of activities, analyze the subsets to find the longest time path, then string together the longest subset paths to complete the critical path from start to finish. To clarify the situation, sketch the project network, showing activities, durations, and the critical path.

a) Total days to Finish = 41. (Answer is C). b) The critical path is as follows: START, A, B, C, D, F, G, FINISH (answer is a)

Activity Predecessor Duration (days)

Start - A Start 9 B A 8 C B 2 D A, C 6 E D 7 F C, D 9 G E, F 7

Finish G

Legend

Activity

(duration)

ES EF

LS LF

A(9)

0 9

0 9

B(8)

9 17

9 17

C(2)

17 19

17 19

D(6)

19 25

19 25

E(7)

25 32

27 34

F(9)

25 34

25 34

G(7)

34 41

34 41

START(0)

0 0

0 0

FINISH(0)

41 41

41 41

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98 Chapter 2 – Construction Management |

fast facts

Consider the construction of a building, there are various component activities involved in the project as a whole. Some of these activities can run concurrently, for example: purchase the doors, windows, mechanical components. While others are consecutive, for example: the paint cannot be bought until it has been chosen, the window cannot be painted until the window is installed. Delaying the purchase of the windows is likely to delay the entire project - this activity will be on the critical path and have no float and hence it is a “critical activity”. A relatively short delay in the purchase of the paint may not automatically hold up the entire project as there is still some waiting time for the trim to be installed. There will be some “free float” attached to the activity of purchasing the paint and hence it is not a critical activity. However, a delay in choosing the paint in turn inevitably delays buying the paint which, although it may not subsequently mean any delay to the entire project, it does mean that choosing the paint has no “free float” attached to it. Despite having no free float of its own, choosing the paint is involved with a path through the network which does have “total float”. Therefore, float or slack is the amount of time that a task in a project network can be delayed without causing a delay to: subsequent tasks (free float) or project completion date (total float). Activities on the critical path have zero free and total float. A critical activity typically has free float equal to zero, but an activity that has zero free float may not be on the critical path.

PROJECT FLOAT -- “FREE FLOAT” AND “TOTAL FLOAT”

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Chapter 2 – Construction Management 99

RESOURCE LEVELING

56. - 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. Apply the project network to the following:

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

a. 2

b. 4

c. 6

d. 8

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

a. 2

b. 4

c. 6

d. 8

1

3

4

2 A = 2 days

C = 3 days

B = 5 days

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100 Chapter 2 – Construction Management |

Solution:

1. (answer is b)

2. (answer is c)

A’

B’

C’

A’

B’

C’

B’

C’

B’

B’

1

3

4

2 A = 2 days

C = 3 days

B = 5 days

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.

Operation Activity

Duration

(days)

Total

Float

Resource

(workers) ES EF LS LF

1-2 A 2 3 A' (2) 0 2 3 5

1-4 B 5 0 B' (4) 0 5 0 5

1-3 C 3 2 C' (2) 0 3 2 5

2-4 Dummy 0 3 n/a 2 2 5 5

3-4 Dummy 0 2 n/a 3 3 5 5

Resource usage if all Activities start on day one.

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

Wo

rke

rs

Days

2

4

6

8

1 2 3 4 5

A’

B’

A’

B’

B’

B’

B’

Wo

rke

rs

Days

2

4

6

8

1 2 3 4 5

C’ C’ C’

0 0

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Chapter 2 – Construction Management 101

57. - Question

1. A network diagram method that uses only finish-to-start relationships and may

use dummy activities to define logical relationships is called:

a. Precedence diagramming method

b. Arrow diagramming method

c. Conditional diagramming method

d. A Gantt chart

Solution: (answer is b). the Arrow Diagramming Method is the only scheduling

technique that employs dummy activities.

2. Precedence diagramming is best described as:

a. Arrows represent activities and connect with nodes to show dependencies

b. Nodes represent activities and connect with arrows to show dependencies

c. A method that allows loops and conditional branches

d. Using only finish-to-start activities and dummy activities

Solution: (answer is b). Precedence diagramming is also called Activity On Node

(AON) diagramming.

3. Which of the following statements are true about the Critical Path Method:

I. A network analysis technique used to predict project duration by analyzing

which sequence of activities has the least amount of schedule flexibility

(float)

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

III. Late dates are calculated by a backwards pass using a specified

completion date (usually the early finish date)

a. I

b. I & II

c. I, II, & III

d. none

Solution: (answer is c)

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102 Chapter 2 – Construction Management |

58. - Question

1. When the activity A precedes B but succeeds C, the network logic is

represented in which of the following diagrams:

a. .

b. .

c. .

d. .

Solution: (answer is c)

2. The network flow diagram shown below depicts which of the

following conditions:

a. Activity C can be started only after the completion of activities A

and B

b. Activity D can be started only activity B is completed

c. Activity E can be started only when activities A, B, C and D have

been completed

d. All of the above

Solution: (answer is d)

6 7 6 8 9

A B C

6 7 6 8 9

B C A

6 7 6 8 9

C A B

6 7 6 8 9

C B A

0

1 6 3 4 A

C E

2 B

D

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Chapter 2 – Construction Management 103

59. - Question

1. 1. The free float is equal to:

a. Latest allowable event occurrence time + Early finish time

b. Earliest event occurrence time + early finish time

c. Latest allowable event occurrence time - Early finish time

d. Earliest event occurrence time – Early finish time

Solution: (answer is d)

2. When float or slack of an activity is positive, which one of the following

applies:

a. It represents a situation where extra resources are available and the

completion of the project is not delayed

b. It represents that a project falls behind schedule and additional

resources are required to complete the project on time

c. The activity is critical and any delay in its performance will delay the

completion of the whole project

d. Any one of the above.

Solution: (answer is a)

3. When float or slack of an activity is negative, which one of the following

applies:

a. It represents a situation where extra resources are available and the

completion of the project is not delayed

b. It represents that a project falls behind schedule and additional

resources are required to complete the project on time

c. The activity is critical and any delay in its performance will delay

the completion of the whole project

d. Any one of the above.

Solution: (answer is b)

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104 Chapter 2 – Construction Management |

60. - Question

1) In a project network, activities B and C follows A, activity D follows B, activity E

follows C and activities D and E precede F. The correct network for the project

is:

a. .

b. .

c. .

d. .

Solution: (answer is a)

2

3

4

5 6

B

1 A

C

D

E

F

2

3

4

5 6

B

1 A

D

C

E

F

3

4

5

6 1

C

2 A

D

E

F

B

2

3

4

5 6

D

1 F

C

E

B

A

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Chapter 2 – Construction Management 105

2) The flow net of activities of a project is shown below. The duration of the

activities are written along their arrows. The critical path of activities is:

a. 1-2-4-5-7-8

b. 1-2-3-6-7-8

c. 1-2-3-5-7-8

d. 1-2-4-5-3-6-7-8

Solution: (answer is c)

3) The activity-on-arrow network of a project is shown below. The project duration

is most nearly:

a. 40

b. 41

c. 45

d. 47

Solution: (answer is d) Based on the critical path: 1-2-3-5-7-8

2

3 6 6

8

M

O

4 3

7

5

1

L

4

7

N

2

3

P Q

R

6

3 S

7

2

3 6 6

8

M

O

4 10

4

7

5

1 L

8

14

N

10

8

P

Q

R

12

2 S

7

A

6

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106 Chapter 2 – Construction Management |

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Chapter 3 - Materials 107

CHAPTER 3 - MATERIALS

Concept

Terminology

Application

Materials

CH

AP

TE

R

3

Materials

Properties of Materials

Equipment Production

Actual vs. Ultimate Strength

Breaking Strength

Wire Rope

Elastic Stretch

Design Factors

Lifting Load

Standard Productivity

Operating Costs

Job Size Productivity

NCEES – FE Civil Engineering Topics Materials 8% = 5/60

A. Concrete mix design

B. Asphalt mix design

C. Test methods (e.g., steel, concrete, aggregates, and

asphalt)

D. Properties of aggregates

E. Engineering properties of metals

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108 Chapter 3 - Materials |

MECHANICAL PROPERTIES OF MATERIALS

61. - 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:

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.

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Chapter 3 - Materials 109

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 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.

p

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110 Chapter 3 - Materials |

62. - Question Which of the following stress-strain curves represents a soft and weak material:

Solution: As depicted, stress is on the y-axis while the x-axis represents strain. When a is compared to b, the stress is less. Comparatively, both c and d are characteristically brittle and fail relatively quickly when stress is applied (for example, concrete). (answer is a --- “soft and weak”)

ε

σ

ε

σ

ε

σ

ε

σ

c.

d. b.

a.

p

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Chapter 3 - Materials 111

fast facts

ACTUAL VERSUS ULTIMATE STRENGTH

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, 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, .

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.

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 eff is higher than the ASD which results in stronger structures.

Figure -1: ASD vs. LRFD Strength Comparison Rn/ = ASD Capacity Rn = LRFD Capacity

Rn = Nominal Capacity

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112 Chapter 3 - Materials |

ELASTIC STRETCH

63. - Question A 70-ft long, ¾” diameter, 6 x 7 FC wire rope is

resisting a tension force of 12-kips. The elastic stretch of the steel wire

rope given the following properties is most nearly: E = 10,000-ksi; A =

0.288-in2

a. 0.75-in

b. 3.5-lb

c. 6.3-in

d. 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: (1) 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. Constructional stretch

is provided by the manufacturer and is always given. (2) 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:

Where: P= change in load; L=length; A=area of wire rope; E=modulus of elasticity. Elastic Stretch = 12-kips x 70-ft x 12-in/ft = 3.5-in 0.288-in2 x 10,000-ksi

= 3.5-in (answer is b)

Two popular types of wire rope are: (1) FC or Fiber Core

where there are 7 bundles of 7-strands of steel with a fiber

rope core (see illustration nearby); and, (2) IWRC or

Independent Wire Rope Core where there is an

independent wire rope core inside a wire rope outer wrap.

Elastic Stretch = PL AE

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Chapter 3 - Materials 113

THERMAL EXPANSION

64. - Question A continuous mullion within an aluminum curtain wall is supported on the edge of a spandrel beam at 40-ft vertical intervals on a 20-story building façade. The linear expansion (in) of the aluminum mullion due to the 100°F thermal extremes and using a factor of safety of two is most nearly:

a. 0.625

b. 1.10

c. 1.25

d. 1.55

Solution: Apply the following thermal expansion equation (page 33 – Thermal deformations) using the coefficient of linear expansion (α) from the Materials Table on page 38 for aluminum alloy:

ϵth = (13.1 x 10-6 in / in °F) (100°F – 0°F) (2) ϵth = 2.62 x 10-3 in / in ΔL = ϵth = 2.62 x 10-3 in / in x (40-ft x 12-in/ft) ΔL = 1.2576-in (answer is c)

ϵth = α (T2 – T1) (Factor of Safety)

fast facts Curtain wall is a term used to describe a building façade which does not carry any dead

load from the building other than its own dead load, and to transfer horizontal loads (wind loads)

applied on the curtain wall. These loads are transferred to the main building structure through

connections at floors or columns of the building. A curtain wall is designed to resist air and water

infiltration, wind forces acting on the building, seismic forces (usually only those imposed by the

inertia of the curtain wall), and its own dead load forces.

Curtain walls are typically designed with extruded aluminum members, although the first

curtain walls were made of steel. The aluminum frame is typically in filled with glass, which provides

an architecturally pleasing building, as well as benefits such as day lighting. However, parameters

related to solar gain control such as thermal comfort and visual comfort are more difficult to control

when using highly-glazed curtain walls. Other common infill include: stone veneer, metal panels,

louvers, and operable windows or vents.

Curtain walls differ from storefront systems in that they are designed to span multiple floors,

and take into consideration design requirements such as: thermal expansion and contraction;

building sway and movement; water diversion; and thermal efficiency for cost-effective heating,

cooling, and lighting in the building.

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114 Chapter 3 - Materials |

LIFTING LOAD – OFFSET

65. - Question The rigging configuration shown below will be used

to lift a bridge girder using a 200-ton capacity luffing jib crane onto the

foundation abutment. The tension force (lb.) in Sling A is most nearly:

a. 12,000

b. 20,000

c. 28,000

d. 40,000

Solution:

By inspection, the center of gravity of the load is offset causing an eccentric

load condition. The reaction load is heavier on the left side (RL). The

distribution of the 40,000-lbs load at RL is: (40-ft ÷ 60-ft) ( 40,000-lbs) =

26,667-lbs. The reaction force RL must be adjusted for the slope of the

slings. The vector force length of Sling A is √ 602 + 202 = 63.25-ft. The

force in Sling A is (63.25-ft ÷ 60-ft) (26,667-lbs) = 28,111-lbs (answer=c)

60-ft

Section View

Not to Scale

G

Spreader Beam = 60-ft

Sling A

Sling B

To Crane Hook

C

40,000-lbs

20-ft 20-ft 40-ft 40-ft

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Chapter 3 - Materials 115

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 = Cycle Capacity

Time Rate

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116 Chapter 3 - Materials |

DAILY STANDARD PRODUCTION RATE OF EQUIPMENT

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

a. 2,140-yd3 b. 2,150-yd3 c. 2,160-yd3 d. 2,180-yd3

Solution: The daily standard production rate is as follows:

P��� = (����)(���)(�,���

���

��)

�����= 2,160 − yd�

(answer is c)

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

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Chapter 3 - Materials 117

DAILY STANDARD PRODUCTION RATE OF A DUMP TRUCK

67. - 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 8-

hours per day. The number of trucks needed daily using a swell of 10% for

the soil is most nearly:

a. 5

b. 6

c. 7

d. 8

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

P��� = (����)(���)(�,���

���

��)

�����= 2,160yd�x1.1swell = 2,376yd�/day

������� =(������)(���)(�,���

���

��)

�����= 1,728���

����� = (40���) �������

����� = 347���

������ = 1,728 + 347 + 46 = 2,121���

The daily dump truck productivity is:

������ =(�����)(���)(�,���

���

��)

(�,������)= 353���

Calculate the number of trucks required:

� = (�,������/���)

������= 6.73������

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

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118 Chapter 3 - Materials |

PRODUCTIVITY ANALYSIS AND IMPROVEMENT

68. - Question An excavator with a bucket capacity of 3-yd3

has a standard production rate of 2,160-yd3 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. 1,034-yd3 / day and cycle time of 57-sec b. 1,034-yd3 / day and cycle time of 68-sec c. 1,134-yd3 / day and cycle time of 72-sec d. 1,134-yd3 / day and cycle time of 76-sec

Solution: Note that all the factors are less than 1, as such; the job site

productivity of the excavator per day is given by:

�������,��������(.���)(.���)(.�)(.�)��,������� The actual cycle time can be determined as follows:

��������

�����(.���)(.���)(.�)(.�)

������

(answer is d)

Work Conditions at the Site Factor

Bulk composition 0.954

Soil properties and water content 0.983

Equipment idle time for worker breaks 0.8

Management efficiency 0.7

Soil Compaction 0.83

Cycle time (sec) 40

Dump Truck Volume (CY) 26

Fuel Consumption (gal/hr.) 6.4

Daily Excavator Maintenance (after work hr.) .50

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Chapter 3 - Materials 119

OPERATING COSTS

69. - Question A 160-HP Diesel engine (peak fuel consumption = 0.04-gal/HP-hr) hydraulic excavator operates on a cycle time of 20 seconds during a 50-min/hr. During the filling of the bucket cycle, the excavator’s engine is at full power for 5-seconds. The remainder of the time, the engine operates at half-power. The fuel consumed per hour is most nearly:

a. 3.33-gal/hr b. 3.63-gal/hr c. 4.00-gal/hr d. 4.33-gal/hr

Solution: Step 1: Calculate the Time Factor (TF): Time Factor = 50 x 100 = 83.3% 60 Step 2: Calculate the Engine Factor (EF): Filling the bucket = (5 / 20) x 1 power = 0.25 Rest of Cycle = (15 / 20) x .50 power = 0.375 TOTAL 0.625 Operating Factor = Time Factor x Engine Factor = 0.625 x 0.833 = 0.520 Step 3: Calculate the Fuel Consumed Fuel consumed = 0.52 x 160-HP x 0.04-gal/HP-hr = 3.33-gal/hr Hr (answer is a)

Note: Typical Fuel Consumption Standards:

1. Gas engine = 0.06 gal/HP-hr 2. Diesel engine = 0.04 gal/HP-hr

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120 Chapter 3 - Materials |

EFFECTS OF JOB SIZE ON PRODUCTIVITY

70. - 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. .50

b. .65

c. .78

d. .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:

���������(�.����.�)�

���,�������,������,�������,���

��.��

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

Pro

du

ctiv

ity

5 Labor-hours (00,000)

Figure 1: Linear Interpolation of Productivity

Index and Job Size

4 6

1.1

1.0

.85

7 3

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Chapter 3 - Materials 121

MATERIAL SPECIFICATIONS

71. - 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.

II. 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).

III. 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.

IV. Construction specifications normally consist of a series of instructions or

prohibitions for specific operations.

V. 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.

a. I & II

b. I, II, & III

c. I, II, III, & IV

d. 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

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122 Chapter 3 - Materials |

QUALITY CONTROL PROCESS (QA/QC)

72. - Question Which of the following statements about error analysis are true:

I. The expected value (or: most likely; probable value) of a measurement is the value

that has the highest value of being correct.

II. The most probable values are the observed values corrected by an equal part of

the total error.

III. Measurements of a given quantity are assumed to be normally distributed.

IV. The interval between the extremes is known as the 50% confidence interval.

V. The probable error of a quantity that has a mean [ μ ] and a standard deviation [ s ]

represents that the probability is 50% (or; confidence interval) that a measurement of

that quantity will fall within the range of μ ± 0.6745 s or the probable ratio of

precision is μ / 0.6745 s.

a. I & II

b. I, II, & III

c. I, II, III, IV, & V

d. None of the above

Solution: All are true. (answer is c)

Normal Distribution Curve

73. - Question A survey crew’s field book shows the interior angles of a polygon traverse

were measured as: 69°, 168°, 99°, 99°and 107°. The most probable interior angles are most

nearly:

a. 68°, 168°, 99°, 98°and 107°

b. 68.4°, 168.4°, 99.4°, 98.4°and 107.4°

c. 68.6°, 167.6°, 98.6°, 98.6°and 106.6°

d. 67°, 168°, 99°, 99°and 107°

Solution:

The sum of the interior angles of a polygon with n sides is (n - 2) (180°) = (5-2) (180°) = 540°

69°+ 168°+ 99°+ 99°+107° = 542°

The correction to 540° is -2°. As such, subtract 2/5 from all angles to arrive at the most probable interior angle.

69° – (+2/5) = 68.6° 168° – (+2/5) = 167.6° 99° – (+2/5) = 98.6° 99° – (+2/5) = 98.6° 107° – (+2/5) = 106.6° (answer is c)

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Chapter 3 - Materials 123

CONCRETE MIX DESIGN

74. - Question A concrete mix design is 1 : 1.9 : 2.8 by weight. The

water cement ratio is 7 gallons of water per sack. The aggregates are SSD

and have specific weight of 165 lb/ft3 for both the fine and coarse

aggregate. The concrete yield in cubic feet per sack of cement is most

nearly:

a) 3.18

b) 3.53

c) 4.10

d) 4.26

Solution: Create a table and compute the ratios:

Material Ratio Weight per Sack

Specific Weight (lbf/ft3)

Absolute Volume

(ft3/sack) Cement 1.0 1 x 94 = 94 195 94/195 = .48 Sand 1.9 1.9 x 94 = 179 165 179/165 = 1.08 Aggregate 2.8 2.8 x 94 = 263 165 263/165 = 1.60 Water 7 / 7.48 = .94 Total = 4.10

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124 Chapter 3 - Materials |

75. - Question A highway bridge project requires a concrete mix. The mix design has the proportions 1 : 2.7 : 3.65, on a weight basis. 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. The specific gravity of the cement is 3.15. The water/cement ratio (gal/sack) of the concrete mix is most nearly:

a. 5.8 b. 5.5 c. 5.3 d. 6.2

Solution: From the problem statement the proportions are:

Cement : Sand : Gravel

Work on the basis of 1 yd3 of concrete. The concrete consists of cement,

fine aggregate, coarse aggregate, and water.

Weight of cement = (5.6 sacks/yd3)(94 lb/sack) = 526.4 lb/yd3

Volume of cement = (526.4 lb/yd3)(1/3.15)(1 ft3 / 62.4 lb) = 2.68 ft3/yd3

Volume of fine aggregate = (2.7)(526.4 lb/yd3)(1/2.65)( 1 ft3 / 62.4 lb) = 8.59

ft3/yd3

Volume of course aggregate = (3.65)(526.4 lb/yd3)(1/2.65)( 1 ft3 / 62.4 lb) =

11.62 ft3/yd3

Volume of water = 27.0 – 2.68 – 8.59 – 11.62 = (4.11 ft3/yd3)(7.48 gal/ft3) =

30.74 gal/yd3

Water/cement ratio = 30.74 gal/yd3 / 5.6 sacks/yd3 = 5.49 gal/sack

(answer is b)

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Chapter 3 - Materials 125

CONCRETE MIX DESIGN RATIO 1 : 2: 3

.

Water Cement Sand Gravel Standard Weight

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

94 #/sack [195#/ft3]

165 #/ft3

[2 x 94 = 188#/sack]

165 #/ft3 [3 x 94 = 282#/sack]

Proportions w/c = .50 1 2 3

Material

Mix Weight (lbs)

50# (=47#)

100# (=94#)

200# (=188#)

300# (=282#)

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

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-11 Table

49.2.

5gal

Cem

en

t

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126 Chapter 3 - Materials |

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 water-cement 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.

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Chapter 3 - Materials 127

WATER CEMENT RATIO

76. - 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. 7 Load 2 Ticket 1487-TCC Volume 10-CY Mix Description SOG 3000-psi Truck 41 w/c Date/Time 3/14/11 13:48 Material Target Actual Status Moisture Material Target Actual Status Sand 14,271-lb 14,080-lb Done 3.8 Cl 0-oz 0-oz #57 Agg. 17,800-lb 17,700-lb Done 0.0 WR 0-oz 0-oz Retarder 96-oz 96-oz Done Air Entrain. 48-oz 48-oz Done MR 0-oz 0-oz HRWR 480-oz 480-oz Done Calcium 0-oz 0-oz NC Accel 0-oz 0-oz Type I 4,080-lb 4,045-lb Done Water 1979-lb 1964-lb Done Flyash 720-lb 755-lb Over

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: Type1 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)

p

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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128 Chapter 3 - Materials |

CONCRETE STRENGTH TESTING

77. - 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:

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

(answer = b)

P

12”

P

6” p

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Chapter 3 - Materials 129

78. - Question A 6 in. by 12 in. concrete cylinder resisted a

transverse force of 55,000-lbf, at 28 days, in a split tensile cylinder test.

The concrete tensile strength is most nearly:

a. 560 psi

b. 610 psi

c. 375 psi

d. 490 psi

SOLUTION:

fct = 2P/πDL = (2)(55,000-lbf) / π(6-in)(12-in) = 487-psi (answer = d)

P

12” D=6”

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.

p

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130 Chapter 3 - Materials |

79. - Question The common proportion of ingredients in reinforced cement concrete is:

a. Portland cement (1-part), clean sand (2 to 4 parts) and coarse aggregate

(1 to 2 parts)

b. Portland cement (1-part), clean sand (1 to 2 parts) and coarse aggregate

(2 to 4 parts)

c. Portland cement (1-part), clean sand and coarse aggregate (2 to 4 parts)

d. Any of the above

Solution: b = answer; reinforced concrete proportions are most nearly: [1 to 2 to 3]; [cement, sand, aggregate] 80. - Question Segregation in concrete results in:

a. Honey combing

b. Porous layers

c. Surface scaling

d. All of the above

Solution: d = answer 81. - Question An aggregate which may contain some moisture in the pores but has a dry surface is known as:

a. Very dry aggregate

b. Dry aggregate

c. Saturated surface dry aggregate

d. Moist aggregate

Solution: b = answer

82. - Question The aggregate having all the pores filled with water but has a dry surface is known as:

a. Very dry aggregate

b. Dry aggregate

c. Saturated surface dry aggregate

d. Moist aggregate

Solution: c = answer

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Chapter 3 - Materials 131

83. - Question A moist aggregate is one:

a. Where all the pores are filled with water and also has a wet surface

b. Where all the pores are filled with water but has its surface dry

c. That does not contain any moisture either in the pores or the surface

d. That may contain some moisture in the pores but has a dry surface

Solution: answer is a 84. - Question The surface moisture of aggregates increases the water-cement ratio in the mix and results in which of the following:

a. Increases the strength

b. Decreases the strength

c. Has no effect on the strength

d. Reduces the volume of the mix

Solution: b = answer 85. - Question Which of the following statements is correct:

a. The larger the size of the coarse aggregate, the less is

required of the quantity of fine aggregate and cement.

b. If very dry aggregates are used, the workability of the mix is

likely to be reduced.

c. Bulking is caused due to the formation of a thin film of surface moisture

around the sand particles.

d. All of the above.

Solution: answer is d; note that increase in the volume of sand due to the presence of moisture is referred to as the bulking of sand. The ratio of the volume of moist sand to the volume of dry sand is known as the bulking factor. As such, fine sands bulk more than coarse sands.

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132 Chapter 3 - Materials |

ASPHALT PERFORMANCE

86. - 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. The

material specifications are as follows:

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. 2

b. 3

c. 4

d. 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

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

Material Layer Thickness

(in)

Experience Coefficient

Dense Graded Aggregate

12 0.11

Crushed Stone Base Course

6 0.14

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Chapter 3 - Materials 133

87. - 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 3.2-

mile 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. 0.075

b. 0.102

c. 0.150

d. 0.205

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

Graphic

Representation

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

Material Layer Thickness

D (in)

Layer Coefficient

(a)

Equation Subscript

Topping (Asphalt Concrete) Mill

6 0.42 a1D1

Topping (Asphalt Concrete) Remaining

3 0.30 a2D2

Dense Graded Aggregate subbase

8 Find a3D3

Portland cement Base Course

10 0.20 a4D4

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134 INDEX |

INDEX

Arrow Diagramming Method, 91, 101

average end area, 28

bank-measure, 17

Benefit-Cost Ratio, 72

breaking strength, 108

compacted, 17, 19, 20, 21, 23

compressive strength, 128

concrete mix, 124

critical path analysis, 97

differential leveling, 41

dummy activities, 91, 101

effective annual interest rate, 70

Equivalent Uniform Annual Cost, 75, 76

forward pass, 93, 94, 95, 101

linear interpolation, 120

load factor, 19, 20

loose–measure, 17

Net Present Worth, 72, 77

predecessor, 86, 90

Rate of Return, 77, 83

Relative Compaction, 24

successor, 90

swell factor, 18, 19, 20, 21

Uniform Series Present Worth Factor, 72

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INDEX 135

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136 INDEX |

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