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)
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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Memorize
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
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Solution::
N 671488.23 E 585234.45
N 671588.23 E 585724.45
N 671317.08 E 585596.11 500.00-ft
Not to scale 300.00-ft 400.00-ft
B
A
C A’
C’
11.54°
25.33°
53.13°
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2. - Question A surveyor’s total station measured slope distance near station 3245+26.35 is recorded as 437.380-m with a zenith angle of 118º48’07” and a south easterly bearing. The horizontal distance (m) is most nearly:
a. 383.2 b. 383.276 c. 391.324 d. 391.300
Solution: Sketch the statement:
α = 118º48’07” - 90º00’00”
α = 28.80
cos 28.80 = x . 437.38
X = (437.38)(0.8763)
X = 383.276-m (answer is b)
118º48’07”
α
383.276-m 437.380-m
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3. - Question Convert the azimuth reading of 273.34º to a land bearing:
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-yd3
20%
shrinkage
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A soil’s swell factor represents the fact that the volume of soil placed by nature in the ground is not the same as the volume of the same mass of dirt excavated by the
contractor and placed in the dump truck. The same mass of soil occupies more volume in the truck (loose cubic yards) than it does in the ground (bank cubic yards). The swell factor is an adjustment representing this increase in volume. However, the swell factor plays no part in the calculation of an earthwork’s balance. The swell factor is used to determine the subsequent hauling and stockpiling requirements.
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
Clay, dry 2,100 2,650 26 0.79
Clay, wet 2,700 3,575 32 0.76
Clay and gravel, dry 2,400 2,800 17 0.85 Clay and gravel, wet 2,600 3,100 17 0.85
Earth, dry 2,215 2,850 29 0.78
Earth, moist 2,410 3,080 28 0.78
Earth, wet 2,750 3,380 23 0.81
Gravel, wet 2,780 3,140 13 0.88
Gravel, dry 3,090 3,620 17 0.85
Sand, dry 2,600 2,920 12 0.89
Sand, wet 3,100 3,520 13 0.88
Sand and gravel, dry 2,900 3,250 12 0.89 Sand and gravel, wet 3,400 3,750 10 0.91
Shrinkage: A soil decreases in volume when it is compacted:
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 100%
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)
V compacted = [(100% - % shrinkage) ÷ 100%] x V bank
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)
Laboratory maximum 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
DRY MUD
% 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
2V L
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12. - Question For the cross section areas listed in the Table, determine the following. Apply a soil swell of 25% to fills, if required:
1. Is the project’s earthwork balanced?
a. Yes b. No
2. Does it produce waste or require borrow?
a. It produces waste b. It requires borrow
3. In response to question # 2 above, the volume in cubic yards is most nearly:
a. 1350-yds3 b. 1400-yds3 c. 1450-yds3 d. 1500-yds3
[Hint: See CERM page 79-2; Paragraph 5 – CUT and FILL. In highway work, payment is usually for cut, while in dam work it is usually for fill.]
End Area
The precision obtained from the average end area is generally sufficient unless one of the end areas is very small or zero. In that case, the volume should be computed as a pyramid or truncated pyramid using the equation below.
V pyramid = L Abase
3
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Solution:
(a) Since Cut and fill quantities are not same, earthwork is NOT balanced (answer is b)
(b) Since fill quantity is more than cut quantity, it is required to borrow earth from off-site (answer is b)
(c) 4958.0 – 3560.1 = 1398-CY of borrow from off-site is required.
(answer is b)
Use Vpyramid
Use Vpyramid
Vpyramid
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
√(2,000ac x 43,560-ft2/ac) = 9,333-ft (per side)
Calculate volume of “cap” by adding 1’-6” to the base pyramid dimensions [Note the change in dimensions]:
2. Based on in-situ (“in place”) volume, determine the number of acres at the borrow site
2. Based on in-situ (“in place”) volume, determine the number of acres at the borrow site