School of
PE
School of PE
Professional Engineer
by George Stankiewicz, P.E., LEED ® A. P.
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 SYSTE M ...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
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
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
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.
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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
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:
C
oncept 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.T
erminologyrefers 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
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
CHAPTER 1 – SURVEYING
Concept
Terminology
Application
Surveying
CHAPT
ER
1
Construction Surveying “State” of Soils Average End Area Earthwork Volume Mass Haul DiagramSwell 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
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.
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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.
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 p
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° p
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 p
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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SOILS - SW ELL 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-yd 3 20% shrinkage p
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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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
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:
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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
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 5-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)
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)
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 122.5-94-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)
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.
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
fast facts
Maximum achievable density for the
compacting effort Target Area
The Laboratory Proctor
MUD
DRY
% Moisture
100%
98%
Dry
Density PCF
A
B
Maximum density is found at point “B” and at the intersection of Optimum Moisture Content point “A”
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
)
(
A
1A
2L
V
p12. - 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
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
EARTHW ORK 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
√(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 S22))
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
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
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.
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.
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
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
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
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.
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
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.
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
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
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
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)
[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’
CHAPTER 2 – CONSTRUCTION MANAGEMENT
Concept
Terminology
Application
Construction
Management
CHAPT
ER
2
Construction Management Quantity Takeoff Productivity Analysis Engineering Economics Takeoff Factor TablesTime 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., 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)
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.
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.
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.
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
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.
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
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.
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
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%
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
Step 3 - Calculate Fully Burdened Cost Rate for Crew Day
Trade QTY U/M Rate/hr Total Foreman ea1 $ 46.35 $ 46.35 Journeyman ea4 $ 38.85 $ 155.40 Laborer ea2 $ 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.
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.
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.
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)
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
