Fe Latest Civil Practice 2023 - Rev

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FE
civil
practice exam
Copyright ©2020 by NCEES. All rights reserved.
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ISBN 978-1-932613-97-1
Printed in the United States of America

1st printing January 2020
CONTENTS
Introduction to NCEES Exams ......................................................... 1

About NCEES
Exam Format
Examinee Guide
Scoring and reporting
Updates on exam content and procedures
Exam Specifications.......................................................................... 3
Practice Exam .................................................................................... 7
Solutions .......................................................................................... 59
iii
About NCEES
NCEES is a nonprofit organization made up of the U.S. engineering and surveying licensing boards in
all 50 states, U.S. territories, and the District of Columbia. We develop and score the exams used for
engineering and surveying licensure in the United States. NCEES also promotes professional mobility
through its services for licensees and its member boards.
Engineering licensure in the United States is regulated by licensing boards in each state and territory.
These boards set and maintain the standards that protect the public they serve. As a result, licensing
requirements and procedures vary by jurisdiction, so stay in touch with your board (ncees.org/licensing-
boards).
Exam Format
The FE exam contains 110 questions and is administered year-round via computer at approved Pearson
VUE test centers. A 6-hour appointment time includes a tutorial, the exam, and a break. You’ll have
5 hours and 20 minutes to complete the actual exam.
In addition to traditional multiple-choice questions with one correct answer, the FE exam uses common
alternative item types such as
• Multiple correct options—allows multiple choices to be correct

• Point and click—requires examinees to click on part of a graphic to answer
• Drag and drop—requires examinees to click on and drag items to match, sort, rank, or label
• Fill in the blank—provides a space for examinees to enter a response to the question
To familiarize yourself with the format, style, and navigation of a computer-based exam, view the demo
on ncees.org/ExamPrep.
Examinee Guide
The NCEES Examinee Guide is the official guide to policies and procedures for all NCEES exams.
During exam registration and again on exam day, examinees must agree to abide by the conditions in
the Examinee Guide, which includes the CBT Examinee Rules and Agreement. You can download the
Examinee Guide at ncees.org/exams. It is your responsibility to make sure you have the current version.
Scoring and reporting

Exam results for computer-based exams are typically available 7–10 days after you take the exam. You
will receive an email notification from NCEES with instructions to view your results in your
MyNCEES account. All results are reported as pass or fail.
Updates on exam content and procedures

Visit us at ncees.org/exams for updates on everything exam-related, including specifications, exam-day
policies, scoring, and corrections to published exam preparation materials. This is also where you will
register for the exam and find additional steps you should follow in your state to be approved for the
exam.
1
EXAM SPECIFICATIONS
•
3
Fundamentals of Engineering (FE)
CIVIL CBT Exam Specifications
Effective Beginning with the July 2020 Examinations
• The FE exam is a computer-based test (CBT). It is closed book with an electronic reference.
• Examinees have 6 hours to complete the exam, which contains 110 questions. The 6-hour time also includes a
tutorial and an optional scheduled break.
• The FE exam uses both the International System of Units (SI) and the U.S. Customary System (USCS).
Knowledge Number of Questions
1. Mathematics and Statistics 8–12

A. Analytic geometry
B. Single-variable calculus
C. Vector operations
D. Statistics (e.g., distributions, mean, mode, standard deviation, confidence
interval, regression and curve fitting)
2. Ethics and Professional Practice 4–6
A. Codes of ethics (professional and technical societies)
B. Professional liability
C. Licensure
D. Contracts and contract law
3. Engineering Economics 5–8
A. Time value of money (e.g., equivalence, present worth, equivalent annual
worth, future worth, rate of return)
B. Cost (e.g., fixed, variable, direct and indirect labor, incremental, average, sunk)
C. Analyses (e.g., break-even, benefit-cost, life cycle, sustainability, renewable energy)
D. Uncertainty (e.g., expected value and risk)
4. Statics 8–12
A. Resultants of force systems
B. Equivalent force systems
C. Equilibrium of rigid bodies
D. Frames and trusses
E. Centroid of area
F. Area moments of inertia
G. Static friction
4
5. Dynamics 4–6
A. Kinematics (e.g., particles, rigid bodies)
B. Mass moments of inertia
C. Force acceleration (e.g., particles, rigid bodies)
D. Work, energy, and power (e.g., particles, rigid bodies)
6. Mechanics of Materials 7–11
A. Shear and moment diagrams
B. Stresses and strains (e.g., diagrams, axial, torsion, bending, shear, thermal)
C. Deformations (e.g., axial, torsion, bending, thermal)
D. Combined stresses, principal stresses, and Mohr's circle
E. Elastic and plastic deformations
7. Materials 5–8
A. Mix design of concrete and asphalt
B. Test methods and specifications of metals, concrete, aggregates, asphalt, and wood
C. Physical and mechanical properties of metals, concrete, aggregates, asphalt, and wood
8. Fluid Mechanics 6–9
A. Flow measurement
B. Fluid properties
C. Fluid statics
D. Energy, impulse, and momentum of fluids
9. Surveying 6–9
A. Angles, distances, and trigonometry
B. Area computations
C. Earthwork and volume computations
D. Coordinate systems (e.g., state plane, latitude/longitude)
E. Leveling (e.g., differential, elevations, percent grades)
10. Water Resources and Environmental Engineering 10–15
A. Basic hydrology (e.g., infiltration, rainfall, runoff, watersheds)
B. Basic hydraulics (e.g., Manning equation, Bernoulli theorem, open-channel flow)
C. Pumps
D. Water distribution systems
E. Flood control (e.g., dams, routing, spillways)
F. Stormwater (e.g., detention, routing, quality)
G. Collection systems (e.g., wastewater, stormwater)
H. Groundwater (e.g., flow, wells, drawdown)
I. Water quality (e.g., ground and surface, basic water chemistry)
J. Testing and standards (e.g., water, wastewater, air, noise)
K. Water and wastewater treatment (e.g., biological processes, softening,
drinking water treatment)
5
11. Structural Engineering 10–15
A. Analysis of statically determinant beams, columns, trusses, and frames
B. Deflection of statically determinant beams, trusses, and frames
C. Column analysis (e.g., buckling, boundary conditions)
D. Structural determinacy and stability analysis of beams, trusses, and frames
E. Elementary statically indeterminate structures
F. Loads, load combinations, and load paths (e.g., dead, live, lateral, influence
lines and moving loads, tributary areas)
G. Design of steel components (e.g., codes and design philosophies, beams, columns,
tension members, connections)
H. Design of reinforced concrete components (e.g., codes and design
philosophies, beams, columns)
12. Geotechnical Engineering 10–15
A. Index properties and soil classifications
B. Phase relations
C. Laboratory and field tests
D. Effective stress
E. Stability of retaining structures (e.g., active/passive/at-rest pressure)
F. Shear strength
G. Bearing capacity
H. Foundation types (e.g., spread footings, deep foundations, wall footings, mats)
I. Consolidation and differential settlement
J. Slope stability (e.g., fills, embankments, cuts, dams)
K. Soil stabilization (e.g., chemical additives, geosynthetics)
13. Transportation Engineering 9–14
A. Geometric design (e.g., streets, highways, intersections)
B. Pavement system design (e.g., thickness, subgrade, drainage, rehabilitation)
C. Traffic capacity and flow theory
D. Traffic control devices
E. Transportation planning (e.g., travel forecast modeling, safety, trip
generation)
14. Construction Engineering 8–12
A. Project administration (e.g., documents, management, procurement,
project delivery methods)
B. Construction operations and methods (e.g., safety, equipment, productivity
analysis, temporary erosion control)
C. Project controls (e.g., earned value, scheduling, allocation of resources,
activity relationships)
D. Construction estimating
E. Interpretation of engineering drawings
6
PRACTICE EXAM
•
•
7
FE CIVIL PRACTICE EXAM
1. The area of the shaded portion of the figure shown below is most nearly:
y = 3x2
x=2 x=5 x
o A. 18
o B. 39
o C. 117
o D. 133
2. The indefinite integral of x3 – x + 1 is:
o A. 3x2 – 1 + C
x4 x2
o B. − +1+ C
3 2
x4 x2
o C. − +1
3 2
x4 x 2
o D. − + x+C
4 2
Copyright © 2020 by NCEES 8 NEXT→

4 1
3. The integral 2 x2 dx equals
o A. 1
4
o B. 3
4
o C. − 3
16
o D. 5
16
 ∂2 y 
4. If y = 4 x3 +3x2 z +5xz 2 +6 z3 + 20, then  2  =
 ∂x 
o A. 12 x + 6 z
o B. 24 x + 6 z
o C. 12 x 2 + 6 xz + 5 z 2
o D. 24 x 2 + 6 xz + 5 z 2

5. The following data have been collected:
Average
Test
Score
1 85
2 87
3 95
4 90
5 85
6 88
7 90
8 90
9 91
Which of the following statements is true?
o A. The median and the mode are equal.

o B. The mean and the median are equal.
o C. The mean and the mode are equal.
o D. The mean is larger than both the mode and the median.
6. Which of the following is a unit vector perpendicular to the plane determined by the
vectors A = 2i + 4j and B = i + j – k?
o A. –2i + j – k
1
o B. (i + 2 j)
5
1
o C. (−2i + j − k )
6
1
o D. (−2i − j − k )
6

7. You throw two 6-sided fair dice. The probability that the sum will be less than 12 is most nearly:
o A. 0.028
o B. 0.083
o C. 0.333
o D. 0.972
8. The water content of a soil volume is measured four times gravimetrically by oven drying. The
mean value of the water content is 23.2%. If the standard deviation of the four measurements is
1.0%, the 99% confidence interval for the soil water content is most nearly:
o A. (18.6%, 27.8%)
o B. (19.5%, 26.9%)
o C. (20.6%, 25.8%)
o D. (20.9%, 25.5%)
9. According to the Model Rules, Section 240.15, Rules of Professional Conduct, licensed
professional engineers are obligated to:
o A. ensure that design documents and surveys are reviewed by a panel of licensed engineers
prior to affixing a seal of approval
o B. express public opinions under the direction of an employer or client regardless of
knowledge of subject matter
o C. practice by performing services only in the areas of their competence and in accordance
with the current standards of technical competence
o D. offer, give, or solicit services directly or indirectly in order to secure work or other
valuable or political considerations

10. An engineer testifying as an expert witness in a product liability case should:
o A. answer as briefly as possible only those questions posed by the attorneys

o B. provide an evaluation of the character of the defendant
o C. provide a complete and objective analysis within his or her area of competence
o D. provide information on the professional background of the defendant
11. As a professional engineer originally licensed 30 years ago, you are asked to evaluate a newly
developed computerized control system for a public transportation system. The owner requires a
currently licensed engineer to evaluate the system. You may accept this project if:
Select all that apply.
□ A. you are competent in the area of computerized control systems

□ B. your professional engineering license has lapsed, but you have two FE interns working
for you
□ C. you took a transportation course in college
□ D. you have regularly attended meetings of a professional engineering society
□ E. you have another licensed engineer work for you who is competent in this area, and he or
she will conduct all related work and stamp the related design
12. A lien is a:
o A. claim on property for payment of a debt

o B. requirement that a contractor secure a performance bond for a project
o C. requirement that a contractor secure a payment bond for a project
o D. claim for damages for lack of specific performance

13. A company borrows $100,000 today at 12% nominal annual interest. The monthly payment of a
5-yr loan is most nearly:
o A. $1,667
o B. $2,200
o C. $3,100
o D. $12,000
14. You must choose between four pieces of comparable equipment based on the costs and salvage
values given below. All four pieces have a life of 8 years.
Equipment
Parameter
A B C D
First cost $25,000 $35,000 $20,000 $40,000
Annual costs $8,000 $6,000 $9,000 $5,000
Salvage value $2,500 $3,500 $2,000 $4,000
The discount rate is 12%. Ignore taxes. The two most preferable equipment pieces and the
approximate difference between their present worth values based on least cost are:
o A. A and C, $170
o B. B and D, $170
o C. A and C, $234
o D. B and D, $234
15. A company can manufacture a product using hand tools. Tools will cost $1,000, and the
manufacturing cost per unit will be $1.50. As an alternative, an automated system will cost $15,000
with a manufacturing cost per unit of $0.50. With an anticipated annual volume of 5,000 units and
neglecting interest, the payback period (years) to invest in the automated system is most nearly:
o A. 2.0
o B. 2.8
o C. 3.6
o D. 15.0

16. A construction company bought a new rubber-tire loader and is performing a risk analysis about
whether to purchase insurance. The construction company paid $100,000 for the loader. The
annual cost for the insurance premium is $2,000, and the deductible is $1,000. The risk options
to purchase or not to purchase insurance are as follows:
• 0.88 probability of no accident

• 0.11 probability of a small accident at a cost of $800
• 0.01 probability of a total loss for the loader
The best option and projected cost savings are:
o A. purchase insurance and save $990

o B. purchase insurance and save $1,088
o C. do not purchase insurance and save $1,010
o D. do not purchase insurance and save $2,098
17. A tractor cost $7,500. After 10 years it has a salvage value of $5,000. Maintenance costs are $500
per year. If the interest rate is 10%, the equivalent uniform annual cost is most nearly:
o A. $500
o B. $750
o C. $1,400
o D. $2,000

18. The magnitude (N) of the resultant of the three coplanar forces, A, B, and C, is most nearly:
5N
y
A
2.12 N
B
105°
2.12 N
x
C
o A. 7.0
o B. 7.8
o C. 9.2
o D. 10.3

19. A heavy roller is held in equilibrium on a frictionless Plane AB by the force F, as shown. Which
diagram correctly shows a vector polygon of the forces acting on the roller?
Option A Option B
Option C
Option D
o A. Option A
o B. Option B
o C. Option C
o D. Option D

20. Select the location on the x-axis where an additional load must be placed in order to achieve
equilibrium of the L-shaped bar.
21. The figure below shows a simple truss. Which members in the truss have no (zero) force in them?
D
6m
4,000 N
C E
4m
4m
A F
5m G 5m
H
o A. BG, CG, CF, CE

o B. BG, CE
o C. CG, CF
o D. CF

22. Consider the following graph:

y
(3,3)
y=x
x2
y= 3
Which of the following expressions gives the distance from the y-axis to the centroid of the shaded
area?
31 3
0 3 x dx
o A.
3 1 2
0  x + 3 x  dx

3 2 1 
0  x − x3  dx
3 
o B.
3 1 2
0  x − 3 x  dx
3 1 2
0  x − 3 x  dx

o C.
3 1 2
0  x − 3 x  dx
3 1 1 
0  2 x
2
+ x3  dx
3 
o D.
3 1 2
0  x − 3 x  dx

23. The moment of inertia (in4) of the area about the x′ axis (I x′ ) is most nearly:
o A. 78.0
o B. 78.6
o C. 118.5
o D. 138.8
24. In the figure below, the coefficient of static friction between the block and the inclined plane is
0.25. The block is in equilibrium. As the inclined plane is raised, the block will begin to slide
when:
o A. sin ϕ = 1.0
o B. cos ϕ = 1.0
o C. cos ϕ = 0.25
o D. tan ϕ = 0.25

25. Three forces act as shown below. The magnitude of the resultant of the three forces (N) is most
nearly:
o A. 140 y
260 N
o B. 191
o C. 370
300 N
o D. 396 12
3
5 4
x
50 N
26. A boat accelerates at a constant rate of 12 ft/sec2. The boat travels 140 ft while its speed changes
to 60 ft/sec. The initial velocity (ft/sec) was most nearly:
o A. 3.7
o B. 5.0
o C. 15.5
o D. 31.0
27. A small rotating robotic arm weighs 8 N and has a mass radius of gyration of 4.0 cm. The mass
moment of inertia (kg·cm2) is most nearly:
o A. 13.0
o B. 32.0
o C. 128
o D. 256

28. A 5-kg block is sliding along a frictionless surface and is an acted on by a constant force P of
20 N. The time (seconds) when the block is moving at 18 m/s is most nearly:
o A. 0.22
o B. 1.83
o C. 4.50
o D. 72.0
29. During impact of two objects, which of the following is true?
o A. Energy is never conserved.

o B. Energy is always conserved.
o C. Momentum is never conserved.
o D. Momentum is always conserved.

30. The shear diagram for a particular beam is shown below. All lines in the diagram are straight. The
bending moment at each end of the beam is zero, and there are no concentrated couples along the
beam. The maximum magnitude of the bending moment (kN·m) in the beam is most nearly:
o A. 8
o B. 16
o C. 18
o D. 26
31. The pressure gauge in an air cylinder reads 1,680 kPa. The cylinder is constructed of a 12-mm
rolled-steel plate with an internal diameter of 700 mm. The tangential stress (MPa) inside the tank
is most nearly:
o A. 25
o B. 50
o C. 77
o D. 100
32. A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation
of 0.009 in. The modulus of elasticity (ksi) of the material is most nearly:
o A. 740
o B. 884
o C. 8,840
o D. 10,000

33. The maximum inplane shear stress (ksi) in the element shown below is most nearly:
20 ksi
10 ksi
40 ksi
o A. 10
o B. 14.1
o C. 44.1
o D. 316
34. Based on the given stress-strain curves, which material has the largest plastic deformation?
o A. Material A
o B. Material B
o C. Material C
o D. Material D

35. The piston of a steam engine is 50 cm in diameter, and the maximum steam gauge pressure is
1.4 MPa. If the design stress for the piston rod is 68 MPa, its cross-sectional area (m2) should be
most nearly:
o A. 40.4 × 10–4
o B. 98.8 × 10–4
o C. 228.0 × 10–4
o D. 323.0 × 10–4
36. Which of the following is true when a circular shaft is subjected to torsion only?
o A. Maximum shear stress occurs at the outermost fibers.

o B. Maximum shear stress occurs at the center of the shaft.
o C. Constant shear stress occurs throughout the shaft.
o D. No shear stress is present throughout the shaft.
37. Ready-mixed concrete being delivered to a jobsite is found to have a slump less than specified.
Without compromising strength, which of the following is the most appropriate corrective action?
o A. Decrease the amount of water in the mix before the truck leaves the ready-mix plant.
o B. Increase the water to the mix in the truck at the jobsite before the concrete is poured.
o C. Add a water-reducing admixture to the mix in the truck at the jobsite before the concrete
is poured.
o D. Increase the rotation speed of the mixing drum while the truck is in transit to the jobsite.

38. The test that measures the energy required to fracture a specimen at a given temperature is the:
o A. Brinell Test
o B. Rockwell Test
o C. Endurance Test
o D. Charpy Test
39. In general, a metal with high hardness will also have:
o A. good formability
o B. high impact strength
o C. high electrical conductivity
o D. high yield strength
40. The following preliminary concrete mix has been designed assuming that the aggregates are in
oven-dry condition. However, the aggregates used are in SSD condition.
Water = 305 lb/yd3

Cement = 693 lb/yd3
Coarse aggregate (SSD) = 1,674 lb/yd3
Fine aggregate (SSD) = 1,100 lb/yd3
The properties of the aggregates are:
Coarse
Property Fine Aggregate
Aggregate
Moisture content at SSD 0.5% 0.7%
Moisture content as used in mix 2.0% 6.0%
The amount of water (lb/yd3) that would be used in the final mix is most nearly:
o A. 206
o B. 222
o C. 305
o D. 388

41. The solid line represents results from a uniaxial tension test. The slope of the dashed line is
associated with which physical property of the material?
o A. Yield strength
o B. Yield strain
o C. Secant modulus
o D. Modulus of elasticity
42. The pitot tube shown below is placed at a point where the velocity is 2.0 m/s. The specific gravity
of the fluid is 2.0, and the upper portion of the manometer contains air. The reading h (m) on the
manometer is most nearly:
AIR
h
2.0 m/s
o A. 20.0
o B. 10.0
o C. 0.40
o D. 0.20

43. If the standard density of water is 1,000 kg/m3, a fluid having a specific gravity of 1.263 and an
absolute dynamic viscosity of 1.5 kg/(m·s) has a kinematic viscosity (m2/s) of most nearly:
o A. 1.19 × 10–3
o B. 1.50 × 10–3
o C. 1.89 × 10–3
o D. 528
44. Archimedes' principle states that:
o A. the sum of the pressure, velocity, and elevation heads is constant

o B. flow passing two points in a stream is equal at each point
o C. the buoyant force on a body is equal to the volume displaced by the body
o D. a floating body displaces a weight of fluid equal to its own weight

45. A 1-in-diameter jet of 50°F water is deflected 90° by an angled chute as shown. The water enters
with a velocity of 25 ft/sec and freely exits into the atmosphere with the same velocity. The
forces (lb) in the x and y directions of the chute are most nearly:
o A. Fx = 0
Fy = 0
o B. Fx = 9.33 lb
Fy = 0
o C. Fx = 0
Fy = 9.33 lb
o D. Fx = 9.33 lb
Fy = 9.33 lb

46. The rectangular homogeneous gate shown below is 3.00 m high × 1.00 m wide and has a
frictionless hinge at the bottom. If the fluid on the left side of the gate has a density of 1,600 kg/m3,
the magnitude of the force F (kN) required to keep the gate closed is most nearly:
F
FLUID 3.00 m
FRICTIONLESS
HINGE
o A. 0
o B. 22
o C. 24
o D. 220

47. Four water tanks are shown with varying water heights H and varying nozzle cross-sectional
areas A0. Assume no minor losses in the discharge and a common coefficient of discharge
C = 0.6 for all the nozzles. Match the discharge velocity (ft/sec) to the correct tank.
Discharge Velocity (ft/sec)

13.6
11.8
14.3
15.2
16.7

48. The value of angle A in the figure is most nearly:

t 39
.00f 0.0
80 0f
4 t
A
720.00 ft
NOT TO SCALE
o A. 30° 18' 47"

o B. 32° 47' 50"
o C. 39° 05' 38"
o D. 42° 35' 09"
49. The area inside the quadrilateral, PC, PI, PT, and O, equals 83,164 ft2. The shaded area (ft2)
between the circular curve and the tangents is most nearly:
PI
36° 48'
PC PT
ft
500
R=
O
o A. 2,879
o B. 3,577
o C. 5,407
o D. 8,286

50. The cross-sectional areas to be excavated (cut) at certain sections of a road project are as follows:
Using the prismoidal method, the volume of earth to be excavated (yd3) between Stations 4+35
and 5+65 is most nearly:
o A. 1,460
o B. 1,840
o C. 1,860
o D. 1,900
51. The term projected as it relates to the state plane coordinate system means that the:
o A. survey is inaccurate and needs to be corrected

o B. survey points from found monuments need to undergo a unit conversion
o C. handwritten coordinates are entered into computer software
o D. earth’s curvature needs to be taken into account for survey calculations

52. A total station is set up 5.00 ft above a benchmark that has an elevation of 820.50 ft. A slope
angle and slope distance of –3°15′ and 645.90 ft, respectively, are measured to a reflector that is
set up 4.25 ft above a hub at Point B. Neglecting curvature and refraction, the elevation (ft) of
the hub at Point B is most nearly:
o A. 785.76
o B. 784.63
o C. 783.88
o D. 779.63
53. A backsight of 7.76 ft is taken on a turning point with an elevation of 2,325.58 ft. If the foresight
taken on the top of a construction pin is 4.25 ft, the elevation (ft) of the top of the pin is most
nearly:
o A. 2,313.57
o B. 2,322.07
o C. 2,329.09
o D. 2,337.59
54. A 20-acre parcel of land has a rainfall intensity of 1.5 in./hr and a rational method runoff
coefficient C of 0.10. The flow rate (cfs) for this site is most nearly:
o A. 3.0
o B. 5.0
o C. 7.5
o D. 9.0

55. A flow of 15.5 cfs enters the pipe system at A as shown, and exits at B and C. Pipe data are given
in the following table.
Length Diameter Velocity

Pipe f (ft) (in.) (fps)
AB 0.03 200 24 2.0
AC 0.03 200 36 1.3
CB 0.03 200 48
OW
FL
15.5 cfs A
FL
OW
Assume the pipes are all at the same elevation. The head loss (ft) in Pipe CB is most nearly:
o A. 0.05
o B. 0.13
o C. 0.18
o D. 0.23
56. A pump station delivers wastewater from a sump at an elevation of 78 ft to a maintenance hole
with a water surface elevation of 112 ft. The static head (ft) for this pump is most nearly:
o A. 17
o B. 34
o C. 44
o D. 78

57. A 24-in. water distribution pipeline carrying 15 cfs flow branches at Point A into two pipelines,
1 and 2, both with Hazen-Williams C of 130. Pipeline 1 has an 18-in. diameter and is 2,400 ft
long. Pipeline 2 has a 12-in. diameter and is 1,200 ft long. The branched pipes join at Point B to
a 24-in. pipe to carry the flow downstream. The flow (cfs) in Pipeline 1 is most nearly:
o A. 5.0
o B. 10.0
o C. 12.9
o D. 15.0
58. A suppressed rectangular weir rises from the bottom of the pond to an elevation of 37 ft. The
water is 50 ft deep, and the weir is 200 ft long. The flow (ft3/sec) of water over the spillway is
most nearly:
o A. 1,500
o B. 12,500
o C. 23,800
o D. 31,200

59. For a new development of 75 acres (0.117 mi2), the peak runoff for a 25-yr storm of 360 ft3/s is
to be limited to 180 ft3/s through the use of a detention basin. Runoff Q is 3.4 in. Assume Type II
rainfall. Using the TR-55 Method and figure below, the preliminary estimate of the storage
volume (ac-ft) is most nearly:
Approximate Detention Basin Routing for Rainfall Types I, IA, II, and III
Urban Hydrology for Small Watersheds TR-55, U.S. Department of Agriculture, National Resource Conservation Service, June 1986.
o A. 0.07
o B. 0.11
o C. 3.80
o D. 5.94

60. A 36-in-diameter sewer is installed at a slope of 0.006 ft/ft. Assuming the roughness coefficient
n is constant with depth and equal to 0.015, the velocity (ft/sec) of water in a half-full pipe is
most nearly:
o A. 4.3
o B. 5.8
o C. 6.3
o D. 10.1
61. Four monitoring wells—A, B, C, and D—lie equidistant (200 ft) from a fifth well, E. The depth
to the water table is measured at each well and is shown below. The datum for the top of the
casing is equal for the five wells. Groundwater flow moves in which direction from Point E?
o A. South
o B. East
o C. North
o D. West

62. Water quality characteristics can be categorized in four different ways. Match the provided
characteristic with an appropriate category.
Water Quality Category Characteristic

Physical Hardness
Chemical Odor
Microbiological Radon
Radiological Giardia
Lead
Turbidity
63. A jar test is used to evaluate the efficiency of a coagulation process and allows the plant operator
to optimize which of the following parameters?
o A. Alum dosage
o B. Temperature
o C. Lime dosage
o D. Dissolved oxygen

64. A concentrated load of 20 kips moves through the truss shown. Neglecting the weight of the truss,
the maximum force (kips) in Member CD due to the moving load is most nearly:
o A. 12.5
o B. 25
o C. 37.5
o D. 50

65. The 40-kip vertical load at Joint C in the steel truss shown below produces the forces given in the
accompanying table. The cross-sectional area of each member is 4.0 in2, and the length of each
member is given in the table. The elastic modulus of steel is 29,000 ksi. The downward vertical
displacement (in.) of Joint C is most nearly:
Force, F Length, L FL FL
Member f f∙
(kips) (in.) AE AE
AB 50.0 240 0.1034 1.25 0.1292
BC 49.2 473 0.2008 1.231 0.2472
CD –75.0 480 –0.3103 –1.875 0.5818
AD –30.0 288 –0.0745 –0.75 0.0559
BD –25.0 240 –0.0517 –0.625 0.0323
40 kips
B
32.0 ft
16.0 ft
A D
24.0 ft 24.0 ft
o A. 1.046
o B. 0.294
o C. 0.132
o D. 0.102

66. The proportional limit and modulus of elasticity for a material are 40 ksi and 30,000 ksi,
respectively. A square column made from this material has a moment of inertia equal to 6.8 in.4
and an area equal to 9 in2. Assume a pin-pin connected column so that the effective length factor
K is equal to 1.0. The maximum column length (in.) based on the Euler formula is most nearly:
o A. 42.2
o B. 74.8
o C. 195.0
o D. 224.3
67. The frame in the figure below is:
o A. unstable
o B. stable and determinate
o C. indeterminate one degree
o D. indeterminate two degrees

68. Which combination of moment diagram and deflection shape most accurately corresponds to the
continuous beam with loading shown?

69. Which of the vertical-load influence lines shown below is correct for Member U2U3 of the truss
shown below?
o A. Option A
o B. Option B
o C. Option C
o D. Option D

70. The W21 × 57 steel beam shown in the figure has its compression flange laterally braced at the
one-third points over its full length. Assume Fy = 50 ksi and Cb = 1.0 for the critical segment. The
maximum factored load wu (kips/ft) that the beam can carry for this length is most nearly:
wu
W21 × 57
LATERAL SUPPORT POINT

L = 30.0 ft
o A. 2.658
o B. 3.360
o C. 4.302
o D. 4.778
71. According to American Concrete Institute (ACI) 318-11, the value of ϕ that should be used in
computing the design moment strength ϕMn for the beam section shown below is most nearly:
o A. 0.80
o B. 0.81
o C. 0.84
o D. 0.90

72. In the truss shown, there is a pin connecting the members at each joint. The force (kips) in Member
CD is most nearly:
o A. –81 (compression)
o B. –113 (compression)
o C. –122 (compression)
o D. –169 (compression)

73. A reinforced concrete beam is subjected to a factored moment Mu = 648 ft-kips. For concrete,
fc′ = 4,000 psi. For steel, fy = 60,000 psi. The beam is reinforced with eight #8 bars in two rows,
positioned as shown in the figure. Assume that ϕ = 0.90. The minimum adequate overall width b
for this beam is most nearly:
o A. 10
o B. 12
o C. 13
o D. 15

74. Mark the area of the Atterberg chart provided that is associated with an elastic silt.
60
For classification of fine-grained soils
and fine-grained fraction of coarse-grained
NE
soils.
50 I
Equation of “A”-line
”L NE
PLASTICITY INDEX (PI)
I
Horizontal at PI = 4 to LL = 25.5,
“U ”L
then PI = 0.73 (LL – 20) OH “A
40 Equation of “U”-line or
Vertical at LL = 16 to PI =7, CH
then PI = 0.9 (LL – 8)
30
OL
20 or
CL MH or OH
10
7 ML or OL
4 CL ML
0
0 10 16 20 30 40 50 60 70 80 90 100 110
LIQUID LIMIT (LL)
75. An undisturbed sample of soil has a specific gravity of solids of 2.70, a moisture content of 10.5%,
and a void ratio of 0.63. The degree of saturation is most nearly:
o A. 25%
o B. 45%
o C. 65%
o D. 85%

76. Direct shear test data of a sand are shown below:
Area of sample = 16 in2

Normal load at failure = 512 lb
Shear stress at failure = 16 psi
The angle of internal friction is most nearly:
o A. 0°
o B. 27°
o C. 30°
o D. 63°
77. Subsurface exploration indicates that a level site has a 10-ft upper layer of sand. The groundwater
table is at the ground surface. The unit weight of the sand is 135.0 pcf. The effective overburden
stress (psf) at a depth of 10 ft is most nearly:
o A. 625
o B. 725
o C. 1,350
o D. 1,975
78. A 12-ft-high retaining wall has backfill of granular soil with an angle of internal friction of 30°
and a unit weight of 125 pcf. The resultant Rankine active force (lb/ft) on the wall is most nearly:
o A. 2,250
o B. 3,000
o C. 9,000
o D. 27,000

79. A consolidated, undrained triaxial shear test was performed on an overconsolidated clay mix
specimen with a diameter of 1.4 in. The test yielded a cohesion of 530 psf and an angle of
internal friction of 18°. If the normal load at failure was 125 lb, the shear strength (psi) of the soil
is most nearly:
o A. 10
o B. 26
o C. 30
o D. 556
80. A strip footing having a width B = 2 ft is to be constructed at ground surface (Df = 0). Underlying
the footing is sand having the following bearing capacity factors: Nc = 0, Nγ = 25, and Nq = 20. The
unit weight of sand γ = 120 pcf. The ultimate bearing capacity qult (psf) of the footing is most
nearly:
o A. 1,200
o B. 2,400
o C. 3,000
o D. 4,800
81. A three-story concrete building will be constructed on a vacant parcel in a city. The soil boring
log shows a 20-ft-thick layer of loose soil over a 5-ft-thick limestone layer. Which of the
following foundations will provide the least settlement for this building?
o A. Spread footings
o B. Mats foundation
o C. Deep foundation
o D. Wall foundation

82. A normally consolidated 10-ft clay layer is surcharged, which causes a decrease in thickness. The
coefficient of consolidation is 0.16 ft2 per day and the time factor is 1.2 for U = 50%. The clay
layer is confined between two layers of dense sand. The time (days) required for 50% consolidation
is most nearly:
o A. 5
o B. 38
o C. 188
o D. 750
83. A slope of clay-mix material experiences failure along a 100-ft-long slip surface at an angle of
27°. The soil above the slip surface weighs 100 tons, has an angle of internal friction of 20°, and
has a cohesion of 1.2 psi. The factor of safety at slope failure is mostly nearly:
o A. 0.7
o B. 0.9
o C. 1.7
o D. 381.3

84. A highway profile is shown in the figure. If the design stopping sight distance is 600 ft, the driver's
eye height above the roadway surface is 3.50 ft, and the height of an object in the roadway to be
avoided by stopping is 1.00 ft, the minimum design length (ft) of the vertical curve is most nearly:
PVI
−5%
+4%
PVC PVT
LENGTH OF VERTICAL CURVE

DIRECTION OF VEHICLE
o A. 3,600
o B. 1,966
o C. 1,136
o D. 1,017
85. A flexible pavement system is to be designed using the AASHTO structural number design
method with the following criteria:
Minimum
Coefficient of
Material Thickness
Structural Layer
(in.)
AC surfacing 2 0.44
Aggregate base 4 0.25
Aggregate subbase 4 0.10
Structural number = 2.50
If the minimum thicknesses of the surfacing and aggregate base are used, the required thickness
(in.) of the aggregate subbase is most nearly:
o A. 4
o B. 5
o C. 6.5
o D. 8.5

86. A freeway lane has a volume of 1,400 vehicles/hr and an average vehicle speed of 45 mph. The
time spacing (sec) between vehicles (center to center) is most nearly:
o A. 2.6
o B. 5.2
o C. 15
o D. 31
87. At an urban pedestrian crosswalk, the initial WALK signal is displayed for 6.0 sec, after which a
flashing DON'T WALK signal is displayed. The pedestrian walking speed is 3.5 ft/sec, and the
street to be crossed is 31.5 ft wide. At the end of the green time, the flashing DON'T WALK signal
is terminated. The minimum length of time (sec) the green must be displayed is most nearly:
o A. 6.0
o B. 9.0
o C. 12.5
o D. 15.0
88. An urban intersection is being reconstructed to address safety problems, and it is estimated that
the two mutually exclusive countermeasures have a crash reduction factor of 0.25 and 0.15,
respectively. If the expected number of crashes per year is 10 and no significant growth in traffic
is anticipated, the expected number of average crashes per year after reconstruction is most nearly:
o A. 3.6
o B. 4.0
o C. 6.0
o D. 6.4

89. At two-way stop-controlled intersections, the sight distance required for minor street movements
is determined by:
o A. approach sight triangles

o B. departure sight triangles
o C. stopping sight distance
o D. decision sight distance
90. For a given loading, soil conditions, and design life, which of the following pavement types will
most likely be the thinnest highway pavement design?
o A. Hot-mix asphalt
o B. Warm-mix asphalt
o C. Portland cement concrete
o D. Composite hot-mix asphalt over Portland cement concrete
91. Which of the following statements about a basic freeway segment are true according to the
Highway Capacity Manual?
Select all that apply.
□ A. Freeway segments with 75-mph free-flow speed experience a reduction in operating

speeds at a lower-volume threshold than a freeway segment with a 55-mph free-flow
speed.
□ B. The free-flow speed of a freeway segment is impacted by the amount of lateral clearance
on the right side of the roadway.
□ C. The capacity of a freeway segment with a 75-mph free-flow speed is higher than a similar
facility with a 55-mph free-flow speed.
□ D. A freeway segment with 13-ft lanes has higher free-flow speeds than a similar facility
with 12-ft lanes.
□ E. Increases in traffic volume lower the estimated operating speed of a freeway segment for
both low- and high-volume ranges.

92. To encourage more carpooling during the peak commute times, a metropolitan area is considering
adding HOV lanes to a busy interstate segment. An existing transit route has a utility of –0.65,
which is unaffected by the proposed change. If the existing non-HOV lanes have a utility of +1.2
and the proposed change introducing a carpool mode has a utility of –0.40, the percentage of trips
expected to carpool is most nearly:
o A. 0%
o B. 15%
o C. 17%
o D. 74%
93. Operational manuals, warranties, guarantees, and as-built drawings are generally provided to the
owner during which phase of the project?
o A. Construction
o B. Procurement
o C. Close-out
o D. Feasibility
94. A loader has a full-bucket capacity of 3 yd3, and the average time required to place one bucketload
of soil into a truck is 1 min. The loader is supported by four trucks with a volume of 15 yd3 each
and a cycle time of 12 min plus the time to load the truck. The ideal productivity (yd3/hr) of this
system is most nearly:
o A. 180
o B. 212
o C. 277
o D. 300

95. A project has a CPI > 1.0 and an SPI < 1.0. This would indicate the project is:
o A. behind schedule with a cost savings

o B. ahead of schedule with a cost savings
o C. behind schedule with a cost overrun
o D. ahead of schedule with a cost savings
96. An embankment having a volume of 320,000 yd3 is to be constructed from local borrow. The dry
unit weight and moisture content of the borrow material were determined to be 106 pcf and 18.2%,
respectively. The embankment material has a total unit weight of 122 pcf and a moisture content
of 16.7%. The volume of borrow (yd3) needed to construct the embankment is most nearly:
o A. 274,100
o B. 315,500
o C. 324,500
o D. 373,600

97. Which set of orthographic views correctly represents the isometric view of the structure shown
below?
o A. Option A
o B. Option B
o C. Option C
o D. Option D

98. The average production of the excavator is the controlling factor in a highway ditch-cleaning
contract. Excavators with four different bucket sizes are available as rental units. The hourly rental
rate is directly proportional to the bucket capacity of the excavator. Assume that production
(loose yd3/hr) is equal to (excavator cycles per hour) × (average bucket payload in LCY per cycle).
The excavator characteristics are as follows:
Minimum Cycle Average Bucket Payload

Excavator Time (min) (LCY)
1 0.25 0.50
2 0.33 1.00
3 0.50 1.75
4 0.58 2.00
The optimally efficient excavator is:
o A. Excavator 1
o B. Excavator 2
o C. Excavator 3
o D. Excavator 4
99. A work task has a current cost of $3,400, with a budget cost at this same point of $3,200. The
total budget for this task is $4,000. The estimated cost at completion for this task is most nearly:
o A. $3,750
o B. $4,000
o C. $4,250
o D. $4,850

100. A drawing of a roadway cross section is shown below. The arrow with a number indicates the:
o A. thickness of the surface coating is 0.025 in.

o B. slope of the roadway is 3 in. per 10 ft across the slope
o C. edge of the roadway will receive a chamfer of 1/4 in.
o D. grooves ground into the surface of the roadway are 1/4 in. deep

SOLUTIONS
•
59
FE CIVIL SOLUTIONS
Detailed solutions for each question begin on the next page.
1 C 26 C 51 D 76 B
2 D 27 A 52 B 77 B
3 A 28 C 53 C 78 D
4 B 29 D 54 A 79 C
5 A 30 D 55 B 80 C
6 C 31 B 56 B 81 C
7 D 32 C 57 B 82 C
8 A 33 B 58 D 83 B
9 C 34 B 59 D 84 B
10 C 35 A 60 C 85 C
11 A, E 36 A 61 A 86 A
12 A 37 C 62 see solution 87 D
13 B 38 D 63 A 88 D
14 B 39 D 64 C 89 B
15 B 40 B 65 A 90 C
16 C 41 D 66 B 91 B, C
17 C 42 D 67 B 92 B
18 A 43 A 68 B 93 C
19 C 44 D 69 B 94 A
20 see solution 45 C 70 B 95 A
21 A 46 C 71 C 96 B
22 B 47 see solution 72 A 97 B
23 D 48 A 73 D 98 C
24 D 49 A 74 see solution 99 C
25 D 50 C 75 B 100 B
60
FE CIVIL SOLUTIONS
1. Refer to the Integral Calculus section in the Mathematics chapter of the FE Reference Handbook.
5 5
A =  3x 2dx = x3 = 53 − 23
2
2
= 117
THE CORRECT ANSWER IS: C
x4 x2
 x − x +1 =
3
− + x+C
4 2
THE CORRECT ANSWER IS: D
4 1 dx = −1 1
2 x2 x
4
2
( )( )
= −1 1 − −1 1
4 2
=1−1
2 4
=1
4
THE CORRECT ANSWER IS: A
Copyright © 2020 by NCEES 61

FE CIVIL SOLUTIONS
4. Refer to the Differential Calculus section in the Mathematics chapter of the FE Reference
Handbook.
 ∂2 y 
y = 4 x3 + 3 x 2 z + 5 xz 2 + 6 z 3 + 20, then  2 =
 ∂x 
∂y
= 12 x 2 + 6 xz + 5 z 2
∂x
∂2 y
= 24 x + 6 z
∂x 2
THE CORRECT ANSWER IS: B
5. Refer to the Dispersion, Mean, Median, and Mode Values section in the Engineering Probability
and Statistics chapter of the FE Reference Handbook.
1 N 801
Mean = 
N i =1
xi =
9
= 89
95
91
90
90 Mode
Median 90
88
87
85
85
The mean of the sample is 89. The median of the sample is 90. The mode of the sample is 90.
Therefore, the median and the mode are equal.

FE CIVIL SOLUTIONS
6. Refer to the Vectors section in the Mathematics chapter of the FE Reference Handbook.
The cross product of vectors A and B is a vector perpendicular to A and B.
i j k
2 4 0 = i ( − 4) − j(−2 − 0) + k (2 − 4) = − 4i + 2 j − 2k
1 1 −1
To obtain a unit vector, divide by the magnitude.
Magnitude = (− 4)2 + 22 + (−2)2 = 24 = 2 6
−4i + 2 j − 2k −2i + j − k
=
2 6 6
7. Refer to the Engineering Probability and Statistics chapter of the FE Reference Handbook.
There is only one throw, 6 and 6, that sums to 12. There are 36 possible rolls of the dice. Therefore,
35/36 will have a sum less than 12.
35/36 = 0.972
8. Refer to the t-Distribution section in the Engineering Probability and Statistics chapter of the FE
Reference Handbook.
For a 99% confidence interval, α = 0.01→ α/2 = 0.005.
Use Student’s t-distribution with ν = 4. Refer to the table, where t0.005,4 = 4.604.
Confidence interval = 23.2% ± 4.604% = (18.6%, 27.8%)

FE CIVIL SOLUTIONS
9. Refer to the Ethics chapter of the FE Reference Handbook.
Section B.1 in the Rules of Professional Conduct states: Licensees shall undertake assignments
only when qualified by education or experience in the specific technical fields of engineering or
surveying involved.
10. Refer to the NCEES Rules of Professional Conduct in the Ethics chapter of the FE Reference
Handbook.
Licensees may express a professional opinion publicly only when it is founded on adequate
knowledge of the facts and a competent evaluation of the subject matter.
11. Refer to the NCEES Rules of Professional Conduct, Section B, in the Ethics chapter of the FE
Reference Handbook.
THE CORRECT ANSWERS ARE: A, E
12. Black’s Law Dictionary defines lien as a claim on property for payment of debt. Examinees are
expected to be familiar with liens.
13. Refer to the capital recovery equation in the Engineering Economics chapter of the FE Reference
Handbook.
A = P (A/P, i%, n)
= 100,000 (A/P, 1%, 60)
= 100,000 (0.0222)
= $2,220/month

FE CIVIL SOLUTIONS
14. Refer to the cash flow formulas in the Engineering Economics chapter of the FE Reference
Handbook.
Look at the present worth of each option.
The present worth values are all given by:

P = First Cost + Annual Cost × (P/A, 12%, 8) – Salvage Value × (P/F, 12%, 8)
= First Cost + Annual Cost × 4.9676 – Salvage Value × 0.4039
Then P(A) = $63,731

P(B) = $63,392
P(C) = $63,901
P(D) = $63,222
The cash flows are all costs, so the two most preferable pieces of equipment, those with the lowest
present worth costs, are B and D, and the difference between them is $170.
15. Refer to the Breakeven Analysis section in the Engineering Economics chapter of the FE Reference
Handbook.
$1.50 (5,000) = $7,500

$0.50 (5,000) = $2,500
Annual savings = $7,500 − $2,500 = $5,000
Additional investment = $15,000 − $1,000 = $14,000
Payback period = $14,000/$5,000 = 2.8 years

FE CIVIL SOLUTIONS
16. Refer to the Probability Functions, Distributions, and Expected Values section in the Engineering
Probability and Statistics chapter of the FE Reference Handbook.
EVpurchase insurance = (C1P1) + (C2P2) + (C3P3)
EVpurchase insurance = (0.88 × 0) + (0.11 × $800) + (0.01 × $1,000) = 0 + $88 + $10 = $98
Cost with annual insurance premium is $2,000 + $98 = $2,098
EVnot to purchase insurance = (C1P1) + (C2P2) + (C3P3)
EVnot to purchase insurance = (0.88 × 0) + (0.11 × $800) + (0.01 × $100,000) = 0 + $88 + $1,000 = $1,088
Compare EVnot to purchase insurance and EV to purchase insurance
EVnot to purchase insurance < EVto purchase insurance = $2,098 – $1,088 = savings of $1,010
17. Refer to the Engineering Economics chapter of the FE Reference Handbook.
$5,000
i = 10%
0 1 2 3
10
Am = $500/yr
$7,500
A  A 
A = Am + P  , i, n  − SV  , i, n 
P  F 
A  A 
= $500 + $7,500  , 10%, 10  − $5, 000  , 10%, 10 
P  F 
= $500 + $7,500 ( 0.16275 ) − $5, 000 ( 0.06275 )
= $1, 407 per year

FE CIVIL SOLUTIONS
18. Refer to the Resolution of a Force section in the Statics chapter of the FE Reference Handbook.
Rx =  Fxi , R y =  Fyi , i = 1, 2,3

Rx = 2.12 + 5 cos 105° = 2.12 − 1.29 = 0.83 N
R y = 2.12 + 5 sin105° = 2.12 + 4.83 = 6.95 N
R = Rx 2 + R y 2 = 0.832 + 6.952 = 6.999 N
19. Refer to the Systems of Forces section in the Statics chapter of the FE Reference Handbook.
Draw a free-body diagram.

W
F
B
N
∴ N Since force Σ vectors = 0

F

FE CIVIL SOLUTIONS
20. Refer to the Systems of Forces section in the Statics chapter of the FE Reference Handbook.
ΣFx = 0 = −6 + 3 10 Therefore, no additional x-force required

5
ΣFy = 0 ≠ 6 − 10 + 4 10 Therefore, additional y-force required

5
ΣFy = 0 = 6 − 10 + 4 10 + Py
5
Py = −4 ↓
ΣM o = 0 = 6 ( 5 ) − 10 ( 3) + 4 10 ( 6 ) − 4 ( x )
5
x = 12
THE CORRECT ANSWER IS SHADED ABOVE.

FE CIVIL SOLUTIONS
21. Refer to the Statically Determinant Truss section in the Statics chapter of the FE Reference
Handbook.
Zero-force members usually occur at joints where members are aligned as follows:
A
B
C
That is, joints where two members are along the same line (OA and OC) and the third member is
at some arbitrary angle create a zero-force member. That member (OB) is a zero-force member
because the forces in OA and OC must be equal and opposite.
For this specific problem, we immediately examine Joints B and E:

C D
B: E: C
B E
A
G F
BG is a zero-force member CE is a zero-force member
Now, examine Joint G. Since BG is zero-force member, the joint effectively looks like:
C
A G F
and, therefore, CG is another zero-force member.
Finally, examine Joint C. Since both CG and CE are zero-force members, the joint effectively
looks like:
D
B
F
and, therefore, CF is another zero-force member. Thus, BG, CE, CG, and CF are the zero-force
members.

FE CIVIL SOLUTIONS
22. Refer to the Centroids of Masses, Areas, Lengths, and Volumes section in the Statics chapter of
the FE Reference Handbook.
The location of the centroid from the y-axis in the direction parallel to the x-axis is given by:
1
A A
x= xdA where dA = ( y2 − y1 ) dx
3  x2  3 2 1 3
 0 x  x − 3  dx  0  x − x  dx
x=   or x=
3 
3 x 
2 3 1 2
 0  x − 3  dx  0  x − 3 x  dx
 
23. Refer to the Moment of Inertia section in the Statics chapter of the FE Reference Handbook.
Ixc = h3(a2 + 4ab + b2)/36(a + b)
= 63[(32 + (4 × 3 × 6) + 62)]/[36 × (3+6)]

= 78
A = h(a + b)/2
A = 6(3 + 6)/2 = 27
I x′ = I xc + d y2 A
= 78 + (1.52)(27)
= 138.85

FE CIVIL SOLUTIONS
24. Refer to the Friction section in the Statics chapter of the FE Reference Handbook.
Normal to the plane:
ΣFn = 0: N − mg cos φ = 0 → N = mg cos φ
Tangent to the plane:
ΣFt = 0: − mg sin φ + µN = 0
∴ −mg sin φ + µmg cos φ = 0
sin φ
= tan φ = µ
cos φ
tan φ = 0.25
12 3
25. Ry = ΣFy = ( 260 ) + ( 300 ) − 50 = 370
13 5
260 N y
5 4
Rx = ΣFx = − ( 260 ) + ( 300 ) = 140 300 N
13 5
12
3
R= R x2 + R y2 2
= 370 + 140 2 5 4 x
50 N
R = 396 N
12 13 5 3
5 4

FE CIVIL SOLUTIONS
26. Refer to the Constant Acceleration section in the Dynamics chapter of the FE Reference
Handbook.
v 2 = vo2 + 2ac ( S − S o )
vo2 = v 2 − 2ac ( S − So )
= (60 ft/sec)2 – 2 (12 ft/sec2)(140 ft)
= 3,600 ft2/sec2 – 3,360 ft2/sec2
= 240 ft2/sec2
vo = 15.5 ft/sec
27. Refer to the Kinetics of a Rigid Body section in the Dynamics chapter of the FE Reference
Handbook.
I
Radius of gyration rm =
m
I = mr2 = [8 N/9.807 m/s2 × (4 cm)2] = 13.0 kg·cm2
28. Refer to the Particle Kinetics section in the Dynamics chapter of the FE Reference Handbook.
F = ma
(20 N) = (5 kg)(a)
a = 4 m/s2
v = at
18 m/s = 4 m/s2 (t)
t = 4.50 s

FE CIVIL SOLUTIONS
29. Refer to the Impact section in the Dynamics chapter of the FE Reference Handbook.
Energy may or may not be conserved. It is conserved only if the coefficient of restitution is 1.00.
Momentum is always conserved.
30. Refer to the Beams section in the Mechanics of Materials chapter of the FE Reference Handbook.
10 m x
=
10 kN 6 kN
x=6m
Area 1 = 13(2) = 26 kN·m
6 (6)
Area 2 = = 18 kN·m
2
Area 3 = 4(4) = 16 kN·m
Maximum magnitude of the bending moment is 26 kN·m.

FE CIVIL SOLUTIONS
31. Refer to the Cylindrical Pressure Vessel section in the Mechanics of Materials chapter of the FE
Reference Handbook.
The cylinder can be considered thin-walled if t ≤ 0.10. In this case, t = 12 mm and di = 700 mm.
di
2
Since t = 12 = 0.034 which is ≤ 0.10, the pipe is thin-walled.
di 350
2
Pr
Thus σt = i
t
r + r 350 + 362
where r = i o = = 356 mm
2 2
(1.680 MPa)(356 mm)
σt = = 49.8 MPa
12 mm
32. Refer to the Uniaxial Loading and Deformation section in the Mechanics of Materials chapter of the
FE Reference Handbook.
π
P = 1,300 lb A = (0.5) 2 = 0.196 in 2 L = 12 in.
4
δ = 0.009 in.
PL
δ= rearranged gives
AE
PL (1,300)(12)
E= = = 8.84 ×106 psi = 8,840 ksi
δA (0.009)(0.196)

FE CIVIL SOLUTIONS
33. Refer to the Mohr’s Circle section in the Mechanics of Material chapter of the FE Reference
Handbook.
From a constructed Mohr’s Circle, the maximum inplane shear stress is τmax = R.
 σx − σ y 
2
R=   + τ xy
2
 2 
2
 40 − 20 
R=   + 10
2
 2 
R = 200
R = 14.1 ksi
34. Material B displays the most plastic deformation beyond a clear yield point.
35. Refer to the Uniaxial Loading and Deformation section in the Mechanics of Materials chapter of the
 π(0.5)2 
(
ΣF = PA = 1.4 × 106  )4
 = Frod
 
Frod = 275 kN = σ A = 68 × 10 6 A
A = 40.4 × 10 −4 m 2

FE CIVIL SOLUTIONS
36. Refer to the Torsion section in the Mechanics of Materials chapter of the FE Reference Handbook.
Tρ
τ=
J ρ
∴ max τ when ρ = r
r
37. Refer to the Concrete section in the Materials Science chapter of the FE Reference Handbook.
Because the concrete mix design specified a water-to-cement ratio, the most appropriate way to
increase the slump of the concrete is to add an admixture that will increase the slump without
affecting the water-to-cement ratio.
38. Refer to the definition of Charpy Impact Test in the Materials Science/Structure of Matter chapter
of the FE Reference Handbook.
39. By definition, a metal with high hardness has a high tensile strength and a high yield strength.
40. Refer to the Concrete section in the Materials Science chapter of the FE Reference Handbook.
The moisture content of each aggregate includes: (1) water that would be needed to bring
aggregates to saturated surface dry (SSD) condition (the absorbed water) and (2) the excess water
that is free to add to the mix water. Since the as-used moisture content is greater than the absorption
for each aggregate, each aggregate contributes the excess water to the mix, thus reducing the water
that must be added to mix. The water added to the mix is the water computed for oven-dry
aggregates (305 lb/yd3) plus the excess water in each aggregate.
Final water = 305 − [(2.0% − 0.5%)/100] × 1,674 − [(6.0% − 0.7%)/100] × 1,100 = 221.6 lb/yd3

FE CIVIL SOLUTIONS
41. Refer to the Uniaxial Stress-Strain section in the Mechanics of Materials chapter of the FE
Reference Handbook.
In accordance with the 0.2% offset method for identifying yield strength, a line parallel to the
elastic portion of the stress-strain curve is passed through the strain axis at a value of 0.002. Where
this line intersects the stress-strain curve is identified as the yield strength. Therefore, the slope of
the dashed line is associated with the modulus of elasticity of the material.
42. Refer to the Pitot Tube section in the Fluid Mechanics chapter of the FE Reference Handbook.
v2
≈
( 2 ) ≈ 0.204 m
2
2 g ( 2 )( 9.8)
43. Refer to the Stress, Pressure, and Viscosity section in the Fluid Mechanics chapter of the FE
Reference Handbook.
Units of absolute dynamic viscosity (μ) are kg/(m·s).
Units of kinematic viscosity (v) are m2/s.
∴ The relationship between the two is:
v = μ/ρ where ρ is the density in kg/m3.
v = 1.5/1.263(1,000) = 0.001188 = 1.19 × 10–3.
44. Refer to the Archimedes Principle and Buoyancy section in the Fluid Mechanics chapter of the FE
Reference Handbook.

FE CIVIL SOLUTIONS
45. Refer to the Continuity Equation section in the Fluid Mechanics chapter of the FE Reference
Handbook.
slug
ρ = 1.94 3
= 1.94 lb-sec2 /ft 4
ft
( ) × 25 secft = 0.136 sec

2 3
Q = Av = π 1 in. × ft ft
4 12 in.
F = ρQ ( v 2 − v1 )
Due to continuity: v2 = v1 = 25 ft/sec

Fx = ρQ ( v 2 x − v1x )
v 2 x = v 2 cos 45°
v1x = v1 cos 45°
Fx = ρQ ( v 2 x − v1x ) = ρQ ( v 2 cos 45° − v1 cos 45° ) = 0
(
Fy = ρQ v 2 y − v1 y )
v 2 y = v 2 sin 45°
v1 y = − v1 sin 45°
3
sec  sec ( sec )
Fy = 1.94 lb-sec 2 /ft 4 × 0.136 ft ×  25 ft × 0.707 − −25 ft × 0.707  = 9.33 lb


FE CIVIL SOLUTIONS
46. Refer to the Characteristics of a Static Liquid section in the Fluid Mechanics chapter of the
The mean pressure of the fluid acting on the gate is evaluated at the mean height, and the center of
pressure is 2/3 of the height from the top; thus, the total force of the fluid is:
H 3
Ff = ρg ( H ) = 1,600(9.807) (3) = 70,610 N
2 2
and its point of application is 1.00 m above the hinge. A moment balance about the hinge gives:
F (3) − Ff (1) = 0
Ff 70,610
F= = = 23,537 N
3 3
47. Refer to the Orifice Discharging Freely into Atmosphere section in the Fluid Mechanics chapter
Q = CA0(2gh)1/2
Q = VA
V = C(2gh)1/2
V A = 0.6(2 × 32.2 ft/sec2 × 8 ft)1/2 = 13.6 ft/sec
V B = 0.6(2 × 32.2 ft/sec2 × 6 ft)1/2 = 11.8 ft/sec
V C = 0.6(2 × 32.2 ft/sec2 × 10 ft)1/2 =15.2 ft/sec
V D = 0.6(2 × 32.2 ft/sec2 × 12 ft)1/2 = 16.7 ft/sec
THE CORRECT ANSWER IS SHOWN.

FE CIVIL SOLUTIONS
48. Refer to the Trigonometry section in the Mathematics chapter of the FE Reference Handbook.
Use Law of Cosines.
a2 = b2 + c2 − 2bc cos A
4802 + 7202 − 3902

cos A =
2(480)(720)
A = 30° 18' 47"
49. Refer to the Horizontal Curve Formulas section in the Civil Engineering chapter of the
Find the area of the fractional part of the circle.

A = πr 2
A = π ( 500 )
2
36°48 ' 36.8°

Fraction of circle = = = 0.1022
360° 360°
Area of fraction of circle = 0.1022 × π × ( 500 ) = 80, 285 ft 2
2
Shaded area = total area − fraction of circle

= 83,164 − 80, 285
= 2,879 ft 2
50. Refer to the Earthwork Formulas section in the Civil Engineering chapter of the FE Reference
Handbook.
Volume to be excavated = 130[322 + (4)(395) + 418]/[(6)(27)] = 1,862 yd3

FE CIVIL SOLUTIONS
51. THE CORRECT ANSWER IS: D
52.
ELEV. at B = 820.50 + 5.00 – 645.90(sin 3.25) – 4.25 = 784.63
53. TP elevation + backsight – foresight = pin elevation
2,325.58 + 7.76 – 4.25 = 2,329.09
54. Refer to the Rational Formula section in the Civil Engineering chapter of the FE Reference
Handbook.
Q = CIA
Q = (0.10)(1.5 in./hr)(20 acres) = 3.0 cfs

FE CIVIL SOLUTIONS
55. Refer to the Head Loss Due to Flow section in the Fluid Mechanics chapter of the FE Reference
Handbook.
 L  v 
2
hL = f    
 D   2g 
 200   2 
2
hL,AB = 0.03   = 0.03(100)(0.062) = 0.186 ft
 2   2 × 32.2 
 200   1.3 
2
hL,AC = 0.03   = 0.03(67)(0.026) = 0.052 ft
 3   2 × 32.2 
hL ,CB = hL ,AB − hL ,AC = 0.186 − 0.052 = 0.134 ft
Since the head loss from A to B is larger than the head loss from A to C, the head at C is greater
than the head at B. This means there is flow from C to B.
56. Refer to the Pressure Field in a Static Liquid section in the Fluid Mechanics chapter of the
Elevation difference:
112 − 78 = 34 ft

FE CIVIL SOLUTIONS
57. Refer to the Circular Pipe Head Loss Equation (ft) section in the Civil Engineering chapter of the
4.73L1
hf 1 = 1.852
Q1.852
C D14.87
Parallel pipe flow: h f 1 = h f 2
(Q1/Q2) 1.852 = (L2/L1) × (D1/D2)4.87 = (1,200/2,400) × (18/12)4.87 = 3.6
(Q 1/Q2) = 3.60.54 = 2
Q 1 + Q2 = 3Q2 = 15 cfs
Q 2 = 5 cfs; Q1 = 10 cfs
58. Refer to the Weir Formulas section in the Civil Engineering chapter of the FE Reference
Handbook.
Spillway head is 50 ft – 37 ft = 13 ft
Q = CLH3/2 where C = 3.33 for rectangular weir (USCS unit)
Q = 3.33(200 ft)(133/2 ft) = 31,216, or 31,200 ft3/sec

FE CIVIL SOLUTIONS
59. Refer to the Hydrology/Water Resources section in the Civil Engineering chapter of the
Solve for storage volume Vs (ac-ft)

V 
Vs = Vr  s 
 Vr 
Drainage area: Am = 0.117 mi2
Frequency: 25 years
Runoff: Q = 3.4 in.
Runoff volume: Vr = QAm53.33
where 53.33 = conversion factor from in-mi2 to ac-ft
Vr = 3.4 in. × 0.117 mi2 × 53.33 = 21.2 ac-ft
Peak inflow: qi = 360 cfs
Peak outflow: qo = 180 cfs
Compute: qo/qi = 0.50
Determine Vs/Vr from graph: 0.28
V 
Vs = Vr  s 
 Vr 
Vs = 21.2 ( 0.28 ) = 5.94 ac-ft

FE CIVIL SOLUTIONS
60. Refer to the Manning's Equation section in the Civil Engineering chapter of the FE Reference
Handbook.
Using Manning’s equation: V =

n H( )
K R 2 3 S 12
(
0.015 )
V = 1.486 0.75 3 0.006 2
2 1
V = 6.3 ft/sec
Note: K = 1.486 for USCS units
RH can be found from the Hydraulic-Elements Graph in the Civil Engineering chapter of the FE
Reference Handbook
d/D = 1.5/3 = 0.5; therefore R/Rfull = 1.0 (from graph)
Rfull = R = A/P = r/2 = 1.5/2 = 0.75
61. Velocity of flow is normal to lines of constant piezometric head. This is the line that connects
points having the same head (A and C). The velocity vector is in the direction of decreasing
head. In the figure, head decreases from B to D (the water table is higher at B, or the depth to the
water table is less). Therefore, groundwater flows to the south.

FE CIVIL SOLUTIONS
62. Water Quality Category

Physical Odor
Turbidity
Chemical Hardness
Lead
Microbiological Giardia
Radiological Radon
Physical characteristics are often associated with the senses (e.g., taste, color, odor). Chemical
characteristics often involve dissolved substances (minerals, metals, nutrients). Microbiological
characteristics are associated with microorganisms and pathogens in particular (bacteria, viruses,
protozoa). Radiological characteristics are associated with radioactive materials (artificial or
natural).
THE CORRECT ANSWER IS SHOWN ABOVE.
63. Refer to the Design Criteria for Sedimentation Basins table in the Environmental Engineering
chapter of the FE Reference Handbook.
Alum dosage is the one parameter the operator has control over during the coagulation process.
Temperature, although it may have minor effect on coagulation efficiency, is typically outside the
control of the operator. Dissolved oxygen plays no role in the efficacy of the coagulation process.
Lime is used in hardness removal and thus not pertinent to coagulation.

FE CIVIL SOLUTIONS
64. Refer to the Structural Analysis section in the Civil Engineering chapter of the FE Reference
Handbook.
The maximum force in CD occurs when the wheel is placed at the location that corresponds to the
peak of the influence line:
Max FCD = 1.875 × 20 kips = 37.5 kips
C D
1.875
Influence line for CD

FE CIVIL SOLUTIONS
65. Refer to the Structural Analysis section in the Civil Engineering chapter of the FE Reference
Handbook.
Apply a downward 1-kip load (unit load) at Joint C and compute forces f in bars. (This can easily
be done by dividing each F force by 40).
FL
Multiply each member’s change in length ΔL = by its force f in the table below (be sure to
AE
FL
use the signed values of ΔL = and f ).
AE
FL
Then, sum f ∙ to get the displacement at Joint C.
AE
FL
∑ f∙ = + 1.0464 in. down
AE
FL FL
Member F (kips) L (in.) f f∙
AE AE
AB 50.0 240 0.1034 1.25 0.1292
BC 49.2 473 0.2008 1.231 0.2472
CD –75.0 480 –0.3103 –1.875 0.5818
AD –30.0 288 –0.0745 –0.75 0.0559
BD –25.0 240 –0.0517 –0.625 0.0323
∑ = 1.0464

FE CIVIL SOLUTIONS
66. Refer to the Columns section in the Mechanics of Materials chapter of the FE Reference
Handbook.
A fundamental assumption of the Euler formulation for buckling is that the critical stress is less
than the proportional limit (elastic buckling). The maximum length can be found by setting the
critical stress equal to the proportional limit stress.
2
σcr = π E 2 ≤ 40 (1)
( )
KL
r
6.8 = 0.869
r=
9
( 2)
Solving (1) for L
L = 74.76 = 74.8

FE CIVIL SOLUTIONS
67. Refer to the Stability, Determinacy, and Classification of Structures section in the Civil
Engineering chapter of the FE Reference Handbook.
Unknown reactions and internal forces at internal pins: 6 × 2 = 12
Rigid body components (shown as FBDs) @ 3 equations per component: 4
Number of equations: 4 × 3 = 12
There are 12 equations and 12 unknowns → determinate (stable by member arrangement).
2 5 3
6
4 8
3 7
4 8
10
9
1 4
10
1 11
2 12
1 RIGID BODY
1 UNKNOWN FORCE

FE CIVIL SOLUTIONS
68. Refer to the Beams section in the Mechanics of Materials chapter of the FE Reference Handbook.
The moment diagram and curvature of the deflected shape in Option B are consistent with the
given convention.
69. (A)
–3/8
(B)
–1.0
5/24
(C)
–5/6
5/12
(D)
–5/8

FE CIVIL SOLUTIONS
70. Method 1 (computation) For bracing @ L/3: Lb = 10.0 ft and Cb = 1.0 (given)
From the Zx table in the Civil Engineering chapter of the FE Reference Handbook, for
W21 × 57: φMp = 484 ft-kips Lp = 4.77 ft Lr = 14.3 ft BF = 20.1
Since Lp < Lb (4.77 ft < 10 ft):
φMn = φMp − BF × (Lb − Lp) = 484 − 20.1 × (10.0 − 4.77) = 378.9 ft-kips
8φM n 8 ×378.9
wu = = = 3.368 kips/ft
L2
(30)2
Method 2 (graphical) For bracing @ L/3: Lb = 10.0 ft and Cb = 1.0 (given)
From the Available Moment vs. Unbraced Length table in the Civil Engineering chapter of the
FE Reference Handbook:
Enter horizontal axis at Lb = 10.0 ft, and read upward to W21 × 57 curve: φMn = 378
8φM n 8 × 378
wu = = = 3.360 kips/ft
L2
(30)2

FE CIVIL SOLUTIONS
71. Refer to the Design of Reinforced Concrete Components (ACI 318-11) section in the Civil
0.85 fc'
0.003
a
2 c
As = 5.08 in (given)
21.5 in.
5.08(60)
a = = 7.47in.
0.85(4)(12) (4) #10
As fy
a 7.47 et
c = = = 8.79 in.
β1 0.85
12 in.
dt − c = 21.5 − 8.79 = 12.71 in.
dt − c 12.71
εt = (0.003) = (0.003) = 0.004338 in./in.
c 8.79
Since 0.004 < εt < 0.005, compute φ :
φ = 0.48 + 83(0.004338) = 0.84004 = 0.84

FE CIVIL SOLUTIONS
72. Cut the truss through members CD, DK, and JK and draw a free-body diagram of the left side.
F
FH
13 2
3 4 ft
A
50 10 30
6 ft
Let F = force in Member CD

FH = the horizontal component of F
Then FH = 3 F
13
Solving for F: F = 13 FH
3
↑  M A = 50 kips ( 6 ft ) − 3 kips (10 ft ) + FH ( 4 ft ) = 0
FH = −67.5 kips ( comp )
F = 13 ( 67.5 kips ) = 81.12 kips ( comp )

3

FE CIVIL SOLUTIONS
73. Refer to the Design of Reinforced Concrete Components (ACI 318-11) section in the Civil
Mu  a  1 As f y 
Mn = = As f y  d −  = As f y  d − × 
φ  2  2 0.85 f c′ b 
where As = 8 × 0.79 in2 = 6.32 in2
d = 30 − 3.5 = 26.5 in. (do not use dt = 27.5 in.)
648 × 12  1 6.32 × 60 
= 6.32 × 60  26.5 − × 
0.9  2 0.85 × 4 × b 
b = 14.98 in. ( use 15 in.)
d = 26.5 in. dt = 27.5 in.

8 - #8 BARS
30 in.
cgs 2 in.
3.5 in. 2.5 in.

FE CIVIL SOLUTIONS
74. Refer to the Retaining Walls section in the Civil Engineering chapter of the FE Reference
Handbook.
MH stands for an elastic silt. Anywhere below the "A" line and to the right of the LL = 50 line will
satisfy this condition.
THE CORRECT ANSWER IS SHADED ABOVE.
75. Refer to the Geotechnical section in the Civil Engineering chapter of the FE Reference Handbook.
Degree of saturation = S = ω GS/e
GS ω = Se
(2.7)(0.105) = S(0.63)
S = 45%

FE CIVIL SOLUTIONS
P 512 lb
σn = = = 32 psi
A 16 in 2
τ = 16 psi
( n= 32, τ = 16)
16
σ
32
φ = tan
_
1
( 1632 ) ≅ 27°
77. Refer to Effective Stress in the Geotechnical section of the Civil Engineering chapter in the
σ′ = σ – u = (135 × 10) – 62.4 (10) = 726 psf

FE CIVIL SOLUTIONS
H = 12 ft
φ = 30°
γ = 125 pcf
KA = tan2(45° – φ/2)
= tan2 (45° – 30/2)
= 0.3333
PA = 1/2 γ H2 KA
= 1/2 (125 pcf) (12 ft)2 (0.333)
= 2,998 plf
79. Refer to the Geotechnical section in the Civil Engineering chapter of the FE Reference
Handbook.
d = 1.4 in., so A = πr2 = 1.54 in2

c = 530 psf = 3.68 psi
φ = 18°
P = 125 lb
σN = P
A
τ F = c + σ N tan φ
σN = 125 lb/1.54 in2 = 81.2 psi
τF = 3.68 psi + (81.2 psi) tan 18 = 30 psi

FE CIVIL SOLUTIONS
80. Refer to the Ultimate Bearing Capacity section in the Civil Engineering chapter of the FE
Reference Handbook.
qult = cNc + γDfNq + 0.5 γBNγ
Since Nc = 0 and Df = 0,
qult = (0.5)(120)(2)(25)
= 3,000 psf
81. The upper soils are so weak that spread footings would be too large. A good rule of thumb for
buildings is that spread footings cease to be economical when the total plan area of the footings
exceeds about one-third of the building footprint area.
t = T Hdr2/cv = 1.2 (5 × 5)/0.16 = 187.5 days

FE CIVIL SOLUTIONS
83. Refer to the Retaining Walls section in the Civil Engineering chapter of the FE Reference
Handbook.
Given:
Ls = 100 ft
αs = 27°
WM = 100 tons = 200,000 lb
φ = 20°
c = 1.2 psi = 173 psf
TFF cL + WM cos α s tan φ

FS = = s
TMOB WM sin α s
FS = (173 psf)(100 ft) + (200,000 lb)(cos 27°)(tan 20°)/(200,000 lb)(sin 27°) = 0.9
84. Refer to the Transportation section in the Civil Engineering chapter of the FE Reference
Handbook.
Try equation for S ≤ L
AS 2
L =
2
100  2h1 + 2h2 
9 × 6002
= 2
100  2 × 3.50 + 2 × 1.00 
= 1,966 ft
S ≤ L as assumed. Therefore, the correct equation was used.

FE CIVIL SOLUTIONS
85. Refer to the Highway Pavement Design section in the Civil Engineering chapter of the
2.50 = 0.44(2) + 0.25(4) + 0.10(D3)
2.50 − 0.88 − 1.00 0.62

D3 = = = 6.2 in.
0.10 0.10
86. Refer to the Transportation section in the Civil Engineering chapter of the FE Reference
Handbook.
3,600 sec/hr
Spacing = = 2.57 sec/vehicle
1, 400 vehicles/hr
87. WALK + width/pedestrian speed = green
6 + 31.5/3.5 = 15.0

FE CIVIL SOLUTIONS
88. Refer to the Traffic Safety Equations section in the Civil Engineering chapter of the FE Reference
Handbook.
Crash factors are not additive, so combined CR
CR = CR1 + (1 – CR1) CR2 (order CRs from highest to lowest)

= 0.25 + (1 – 0.25)0.15
= 0.36
1 since no change in ADT
Crashes prevented = N × CR 
ADT after 

 ADT before 
3.6 = 10 × 0.36
10 – 3.6 = 6.4
89. Departure sight triangles are based on a vehicle stopped at the stop bar and looking at major street
vehicles for acceptable gaps.
90. Asphalt or any combination with asphalt will be thicker under identical conditions.

FE CIVIL SOLUTIONS
91. Refer to the Basic Freeway Segment Highway Capacity section in the Civil Engineering chapter
Option A: Not true. A 75-mph free-flow speed freeway has a breakpoint of 1,000 pc/h/ln, while
a 55-mph freeway has a breakpoint of 1,800 pc/h/ln. The breakpoint is defined by the Highway
Capacity Manual as the volume for which the operating speeds becomes lower than the free-flow
speed.
Option B: True. The free-flow speed is impacted by the lane width, right side lateral clearance,
and the total ramp density (TRD).
Option C: True. The capacity of a 75-mph free-flow speed freeway is 2,400 pc/h/ln, while a
similar facility with a 55-mph free-flow speed has a capacity of 2,250 pc/h/ln.
Option D: Not true. Lane widths equal to 12 ft and greater have the same free-flow speed.
Increasing lane widths only increases speeds when the lane widths are 12 ft or less.
Option E: Not true. The estimated operating speed is impacted only once the volumes exceed the
breakpoint volume; therefore there is no impact to operating speed changes in volumes when
traffic volumes are low.
THE CORRECT ANSWERS ARE: B, C
92. Refer to the Logit Models section in the Civil Engineering chapter of the FE Reference
Handbook.
P ( x) = eUx
 x =1 eUxi
n
Three modes available: drive alone, carpool, and transit
P ( x) = eUcarpool
eUcarpool + eUdrive alone + eUtransit
P ( x) = e −0.40 = 0.67 = 0.15 = 15%

e − 0.40 1.2
+e +e − 0.65 0.67 + 3.32 + 0.52

FE CIVIL SOLUTIONS
93. THE CORRECT ANSWER IS: C
94. Refer to the Materials Handling section in the Industrial and Systems chapter of the FE
Reference Handbook.
Time to load one truck = 15 yd3/3 yd3/min

= 5 min
Four trucks are available.

Cycle time for one truck = 5 min loading plus 12 min to travel, dump, and return = 17 min.
Loading time for four trucks = 4 × 5 = 20 min.
Therefore, an empty truck is always available to load.
 60 min/ hr   yd3 
Ideal production capacity =   15 
 5min/ truck   truck 
yd 3
= 180
hr
95. Refer to the Indices section in the Civil Engineering chapter of the FE Reference Handbook.
A cost performance index (CPI) greater than 1.0 indicates a cost savings, and a schedule
performance index (SPI) of less than 1.0 indicates the project is behind schedule.
 100 
96. Dry unit weight of embankment material = 122  
 (100 + 16.7) 
= 104.5 pcf
104.5
Volume required = ( 320,000)
106
= 315,470 yd3

FE CIVIL SOLUTIONS
97. Orthographic views represent the top, front, and side views, and are arranged as shown below:
Option A: The top view is reversed horizontally, and the side and top views do not match.
Option C: The top view is rotated and reversed, and neither the front or side views match
the top view.
Option D: The top view is reversed vertically, and the side and top views do not match.
98. Refer to the Materials Handling section in the Industrial and Systems chapter of the FE Reference
Handbook.
60 min/hr
Production of Excavator 3 = × 1.75 LCY capacity = 210 LCY/hr
0.50 min cycle
Similar calculations for the other three excavators show they have lower production rates.

FE CIVIL SOLUTIONS
99. Refer to the Earned-Value Analysis section in the Civil Engineering chapter of the FE Reference
Handbook.
BCWP / ACWP = CPI

$3,200 / $3,400 = 0.9412
(BAC – BCWP) / CPI = ETC

($4,000 – $3,200) / 0.9412 = $850
Estimate at completion:
ACWP + ETC = EAC
$3,400 + $850 = $4,250
100. The arrow and number indicate the cross slope of the roadway.
3 in 0.25
= = 0.025 ft/ft 3-in./12 in./ft = 0.25/10 ft = 0.025 ft/ft.
12 in./ft 10 ft
Slopes of road or bridge cross-sections are expressed as ft/ft and normally represented by a
number, in this case 0.025. Slopes can also be expressed as a percent; for this case it could be
expressed as 2.5%.

FE EXAM PREPARATION MATERIAL
PUBLISHED BY NCEES
FE Reference Handbook
FE Practice Exams for all modules:

Chemical
Civil
Electrical and Computer
Environmental
Industrial and Systems
Mechanical
Other Disciplines
For more information about these and other NCEES publications and services,
visit us at www.ncees.org or contact
Client Services at (800) 250-3196.
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Introduction to NCEES Exams ......................................................... 1

Practice Exam .................................................................................... 7

• Multiple correct options—allows multiple choices to be correct

Scoring and reporting

Updates on exam content and procedures

Knowledge Number of Questions

1. Mathematics and Statistics 8–12

2. The indefinite integral of x3 – x + 1 is:

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5. The following data have been collected:

Which of the following statements is true?

o A. The median and the mode are equal.

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10. An engineer testifying as an expert witness in a product liability case should:

o A. answer as briefly as possible only those questions posed by the attorneys

Select all that apply.

□ A. you are competent in the area of computerized control systems

o A. claim on property for payment of a debt

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• 0.88 probability of no accident

The best option and projected cost savings are:

o A. purchase insurance and save $990

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o A. BG, CG, CF, CE

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22. Consider the following graph:

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29. During impact of two objects, which of the following is true?

o A. Energy is never conserved.

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o A. Maximum shear stress occurs at the outermost fibers.

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39. In general, a metal with high hardness will also have:

Water = 305 lb/yd3

The properties of the aggregates are:

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44. Archimedes' principle states that:

o A. the sum of the pressure, velocity, and elevation heads is constant

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Discharge Velocity (ft/sec)

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48. The value of angle A in the figure is most nearly:

o A. 30° 18' 47"

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o A. survey is inaccurate and needs to be corrected