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.
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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)
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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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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 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:
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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
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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)
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
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)
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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)
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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)
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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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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)
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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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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
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
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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:
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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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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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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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 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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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)
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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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:
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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
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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
(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
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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
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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.
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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:
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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.
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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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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”)
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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.
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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.
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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,
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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.
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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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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:
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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)
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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.
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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
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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
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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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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:
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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.