THE ULTIMATE CIVIL FE PRACTICE EXAM · A. Analysis of forces in statically determinant beams,...
Transcript of THE ULTIMATE CIVIL FE PRACTICE EXAM · A. Analysis of forces in statically determinant beams,...
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TABLE OF CONTENTS
WELCOME…………………………………………….
EXAM SPECIFICATION……………………
START TEST ………………………………...
SOLUTIONS………………………………………
SCORE SHEET…............................
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Page 6
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Page 225
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WELCOME!!
Welcome to The Ultimate Civil FE Practice Exam! Thank you so much
for purchasing this book!
This exam contains 110 questions and solutions following the exact
same format of the NCEES exam. This was built to have the same
look and feel of the real exam. The only difference, of course, is that
this is a paper based exam and the real exam is computer based (it
didn’t use to be!).
Passing the FE exam is typically a requirement for most civil
engineering students to graduate - it is also a must for your
professional career too.
The FE is your first step to eventually obtaining your PE license. After
you’ve finished school and gained the necessary experience, you’ll be
back studying, preparing to take the PE. There are many similarities
and quite a few differences between the exams, but that’s something
to worry about much later down the road.
This exam was built to help you pass by providing material that will
challenge you to think. You are given 6 hours to take the exam with
only one reference at your disposal - the NCEES approved digital
reference book that sits side by side with the exam. It’s searchable
and will be your main resource to pass this exam.
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If you don’t have this reference manual you can download it for free
from the NCEE website (www.ncees.org). You need to become
intimately familiar with that and use it when taking this exam.
Good luck on this exam, your career, and eventually become a
professional engineer!
Sincerely,
Isaac Oakeson, P.E.
(You’re going to have that by your name too!)
P.S. Errata for this and any other exam we have made can be found
at www.civilengineeringacademy.com/errata.
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LEGAL INFORMATION
Civil Engineering Academy’s
The Ultimate Civil FE Practice Exam
Isaac Oakeson, P.E.
Rights and Liability:
All rights reserved. No part of this book may be reproduced or
transmitted by photocopy, electronic, recording, or any other method
without first obtaining permission from the author. The information in
this book is in no way endorsed by the NCEES organization and the
author shall not have any liability to any person with respect to any
loss or damage caused by the problems in this book.
In other words, please don’t go copying this thing willy-nilly without
giving credit where it should be given by actually purchasing a copy.
Also, don’t go designing real things based on these problems.
If you find errors in this book (I am human of course), or just want to
comment on things, then please let me know! I can be reached
through the website at www.civilengineeringacademy.com or by email
ABOUT THE AUTHOR
Isaac Oakeson, P.E. is a registered professional civil engineer in the
great state of Utah. Shortly after passing the PE exam in the Fall of
2012 he started www.civilengineeringacademy.com and
www.civilpereviewcourse.com to help future students pass. He has
authored and helped author various exams with his entire goal of
providing the best resources for engineers to study and pass the PE.
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FE CIVIL EXAM SPECIFICATIONS
I. Mathematics (7-11)
A. Analytic geometry
B. Calculus
C. Roots of equations
D. Vector analysis
II. Probability and Statistics (4-6)
A. Measures of central tendencies and dispersions (e.g., mean, mode,
standard deviation)
B. Estimation for a single mean (e.g., point, confidence intervals)
C. Regression and curve fitting
D. Expected value (weighted average) in decision making
III. Computational Tools (4-6)
A. Spreadsheet computations
B. Structured programming (e.g., if-then, loops, macros)
IV. Ethics and Professional Practice (4-6)
A. Codes of ethics (professional and technical societies)
B. Professional liability
C. Licensure
D. Sustainability and sustainable design
E. Professional skills (e.g., public policy, management, and business)
F. Contracts and contract law
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V. Engineering Economics (4-6)
A. Discounted cash flow (e.g., equivalence, PW, equivalent annual
worth, FW, rate of return)
B. Cost (e.g., incremental, average, sunk, estimating)
C. Analyses (e.g., breakeven, benefit-cost, life cycle)
D. Uncertainty (e.g., expected value and risk)
VI. Statics (7-11)
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
VII. Dynamics (4-6)
A. Kinematics (e.g., particles and rigid bodies)
B. Mass moments of inertia
C. Force acceleration (e.g., particles and rigid bodies)
D. Impulse momentum (e.g., particles and rigid bodies)
E. Work, energy, and power (e.g., particles and rigid bodies)
VIII. Mechanics of Materials (7-11)
A. Shear and moment diagrams
B. Stresses and strains (e.g., axial, torsion, bending, shear, thermal)
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C. Deformations (e.g., axial, torsion, bending, thermal)
D. Combined stresses
E. Principal stresses
F. Moh’r circle
G. Column analysis (e.g., buckling, boundary condition)
H. Composite sections
I. Elastic and plastic deformations
J. Stress-strain diagrams
IX. Materials (4-6)
A. Mix design (e.g., concrete and asphalt)
B. Test methods and specifications (e.g., steel, concrete, aggregates,
asphalt, wood)
C. Physical and mechanical properties of concrete, ferrous and
nonferrous
metals, masonry, wood, engineered materials (e.g., FRP, laminated
lumber, wood/plastic composites), and asphalt
X. Fluid Mechanics (4–6)
A. Flow measurement
B. Fluid properties
C. Fluid statics
D. Energy, impulse, and momentum equations
XI. Hydraulics and Hydrologic Systems (8–12)
A. Basic hydrology (e.g., infiltration, rainfall, runoff, detention, flood
flows,
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watersheds)
B. Basic hydraulics (e.g., Manning equation, Bernoulli theorem, open-
channel
flow, pipe flow)
C. Pumping systems (water and wastewater)
D. Water distribution systems
E. Reservoirs (e.g., dams, routing, spillways)
F. Groundwater (e.g., flow, wells, drawdown)
G. Storm sewer collection systems
XII. Structural Analysis (6–9)
A. Analysis of forces in statically determinant beams, trusses, and
frames
B. Deflection of statically determinant beams, trusses, and frames
C. Structural determinacy and stability analysis of beams, trusses, and
frames
D. Loads and load paths (e.g., dead, live, lateral, influence lines and
moving
loads, tributary areas)
E. Elementary statically indeterminate structures
XIII. Structural Design (6–9)
A. Design of steel components (e.g., codes and design philosophies,
beams, columns, beam-columns, tension members, connections)
B. Design of reinforced concrete components (e.g., codes and design
philosophies, beams, slabs, columns, walls, footings)
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XIV. Geotechnical Engineering (9–14)
A. Geology
B. Index properties and soil classifications
C. Phase relations (air-water-solid)
D. Laboratory and field tests
E. Effective stress (buoyancy)
F. Stability of retaining walls (e.g., active pressure/passive pressure)
G. Shear strength
H. Bearing capacity (cohesive and noncohesive)
I. Foundation types (e.g., spread footings, deep foundations, wall
footings, mats)
J. Consolidation and differential settlement
K. Seepage/flow nets
L. Slope stability (e.g., fills, embankments, cuts, dams)
M. Soil stabilization (e.g., chemical additives, geosynthetics)
N. Drainage systems
O. Erosion control
XV. Transportation Engineering (8–12)
A. Geometric design of streets and highways
B. Geometric design of intersections
C. Pavement system design (e.g., thickness, subgrade, drainage,
rehabilitation)
D. Traffic safety
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E. Traffic capacity
F. Traffic flow theory
G. Traffic control devices
H. Transportation planning (e.g., travel forecast modeling)
XVI. Environmental Engineering (6–9)
A. Water quality (ground and surface)
B. Basic tests (e.g., water, wastewater, air)
C. Environmental regulations
D. Water supply and treatment
E. Wastewater collection and treatment
XVII. Construction (4–6)
A. Construction documents
B. Procurement methods (e.g., competitive bid, qualifications-based)
C. Project delivery methods (e.g., design-bid-build, design build,
construction
management, multiple prime)
D. Construction operations and methods (e.g., lifting, rigging,
dewatering
and pumping, equipment production, productivity analysis and
improvement, temporary erosion control)
E. Project scheduling (e.g., CPM, allocation of resources)
F. Project management (e.g., owner/contractor/client relations)
G. Construction safety
H. Construction estimating
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XVIII. Surveying (4–6)
A. Angles, distances, and trigonometry
B. Area computations
C. Earthwork and volume computations
D. Closure
E. Coordinate systems (e.g., state plane, latitude/longitude)
F. Leveling (e.g., differential, elevations, percent grades)
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1. What is the length of a straight line connecting two points in
three-dimensional space: A(2, 1, 7) and B(-1, 3, 3)?
a) 20
b) 29
c) 39
d) 21
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2. Solve for the roots of the following polynomial equation.
3 7 6 0x x
a) 1, 2, 3 b) -1, 2, -3
c) 1, -2, 3
d) 1, 2, -3
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3. Determine the coordinates of a circle’s center which satisfy:
2 2 10 14 49 0x y x y
a) (-10, 14)
b) (10, -14)
c) (-5, 7)
d) (5, -7)
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4. A straight line goes through a point (1, 7) and is perpendicular to
the line 5 3 16 0y x . Solve for the equation of the line.
a) 3 5 16 0y x
b) 3 5 16 0y x
c) 3 5 16 0y x
d) 3 5 26 0y x
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5. Which of the following equations described a circle having center
at point (2, -5) and passing through the point (8, 3)?
a) 2 2 4 10 29 0x y x y
b) 2 2 4 10 79 0x y x y
c) 2 2 4 10 71 0x y x y
d) 2 2 4 10 100 0x y x y
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6. A particle moves in the x-y plane with the following equation path:
9sinx t
2cosy t
Solve for the equation of the path of the particle.
a) 2 281 4 324x y
b) 2 24 81 324x y
c) 2 22 9 18x y
d) 2 29 2 18x y
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7. What is the cross product of these two vectors 8 2A i j k and
3 3B i j k ?
a) 30 11i j k
b) 30 11i j k
c) 18 11i j k
d) 18 11i j k
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8. Solve for the angle made by these two vectors:
M i j k
N i j k
a) 85°
b) 80°
c) 75°
d) 70°
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9. Solve for the inflection point of the following polynomial function:
4 3 212 30f x x x x
a) (-5, 125)
b) (5, -125)
c) (19, 1)
d) (-1, -19)
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10. Which expression below is equivalent to 1?
a) 2sin cos
b) 2 2sin cos
c) 1 cos2
2
d) 2 2cos sin
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11. A bag contains five red balls, seven blue balls, and three black
balls. What is the probability of picking a red ball and a black ball?
a) 0.07
b) 0.14
c) 0.20
d) 0.34
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12. A dice is rolled and it shows number 2. What is the probability of
obtaining the number 2 again if the dice is rolled once more?
a) 1
2
b) 1
3
c) 1
6
d) 1
12
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13. A book publisher is doing a quality assurance check of books
produced everyday. The data said that 3% of the total books are
defective. Solve for the probability of having 2 defective books out
of a total of 15 books chosen randomly.
a) 0.004
b) 0.093
c) 0.133
d) 0.064
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14. Ten samples of a concrete cube were tested for compressive
strength. The resulting data presented below is in the units of psi.
2995 ; 3005 ; 3001 ; 2991 ; 2984;
3004 ; 3009 ; 3015 ; 2998 ; 3002
Find the mean and the variance of the sample data.
a) 3000.4 ; 79.6
b) 3000.4 ; 71.6
c) 2997.8 ; 87.1
d) 2997.8 ; 78.4
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15. A doctor collects data from a group of six pregnant women.
Suppose that the probability of having a baby with brown eyes is 0.3. The probability that at least one woman will have a baby with
brown eyes is nearly:
a) 0.3
b) 0.05
c) 0.7 d) 0.9
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16. Which of the following statements is the function to create a
spicific number of loops?
a) If, else, end
b) For, end c) While, do
d) If, elseif, end
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17. A student writes several lines of code as follows:
x = 0;
y = 1;
while x <= 4 y = y+1;
x = x+2;
do
What is the value of variable y in the end of the program?
a) 3
b) 4
c) 5 d) 6
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18. Which of the following is the benefit of behaving ethically?
a) Better reputation
b) Better feeling about yourself
c) Higher salary d) None of the above
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19. You are a city engineer in charge of selecting bids for a big
infrastructure project in the city. There is one contractor who
came in with the high bid and offered you a limited-edition Rolex
watch. You love Rolex watches but you know that the lowest bid
will be selected as the winner of project. What should you do?
a) Accept the Rolex watch and accept the bid
b) Accept the Rolex watch and reject the bid
c) Reject the Rolex watch and accept the bid d) Reject the Rolex watch and reject the bid
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20. An engineer-architect works for a big design firm. During the night
(after working hours), he uses some of office’s equipment and materials to make an architectural model/maquette for his side
job. He receives overtime salary because he is considered as
working overtime. Why is this action considered unethical?
a) His contract prohibits misuse and misappropriation of office’s
equipment
b) He may wear out the office’s materials c) He may encourage other employees to do side jobs in the office
d) He can undercut his fee because he has a lower overhead
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21. You are aware that a registered engineer intentionally violates a
state’s rule of professional conduct. What should you do?
a) Do nothing
b) Report the violation to the employer c) Report the violation to the affected parties
d) Report the violation to the state’s engineering registration board
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22. Which of the followings is the best description of ‘plan stamping’?
a) Legal action of signing off on a project you won on tender b) Legal action of signing off on a project you accepted money for
c) Illegal action of signing off on a project you didn’t design but did
check d) Illegal action of signing off on a project you neither designed nor
checked
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23. Mr. Smith buys a new car for $30,000. He pays the down payment
of $8,000, and then borrows the rest from a bank at 6% interest
for four years. Solve for the required monthly payment for the
loan.
a) $460
b) $490
c) $520
d) $550
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24. A tower crane was just purchased for $80,000. The life span of the
crane is estimated as 10 years, and the salvage value is $6,500.
Find the depreciation value of the crane.
a) $6,500
b) $7,350
c) $8,000
d) $8,650
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25. Mrs. Smith deposits $2,000 in a bank that gives 5% interest
compounded annually. The money she will receive in her account after 15 years is most nearly:
a) $2,100 b) $3,950
c) $4,200
d) $31,500
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26. A table manufacturing factory has a total operating cost of
$400,000 per year, including salaries, rent fee, and depreciation costs. Each table needs $60 to produce, and the sale price of the
table is $99. To reach break-even sales, the number of tables
produced per year is most nearly:
a) 4050
b) 5050 c) 6700
d) 10300
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27. What is the annual effective interest rate of money that is invested
at 2.5% per year and compounded quarterly?
a) 2.50% b) 2.51%
c) 2.53%
d) 2.55%
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28. An engineer plans to receive an annual bonus from money he
invests to a bank. How much should he invest monthly for a year
at a 10% nominal interest rate, compounded monthly, so that he
will receive a $100,000 bonus in a year?
a) $7,500
b) $8,000
c) $8,333
d) $9,170
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29. Which of the following loads could be resisted by a fixed support?
I. Moment II. Shear III. Axial
a) I only
b) I and II
c) II and III
d) I, II, and III
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30. Three ropes are connected as shown in the figure below. They are
holding a box that weighs 2500 lb. What are the tension forces in
rope 1 and 2 if the system is in equilibrium?
a) 1 2
3125 lb ; 1875 lbT T
b) 1 2
2500 lb ; 2500 lbT T
c) 1 2
2725 lb ; 2225 lbT T
d) 1 2
2925 lb ; 1725 lbT T
8
6
1T
2T
2500 lb
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Problems 31 and 32 refer to the following truss.
31. Solve for the support reaction at joint A and B.
a) 600 lb ; 600 lbA B
R R
b) 700 lb ; 500 lbA B
R R
c) 800 lb ; 400 lbA B
R R
d) 900 lb ; 300 lbA B
R R
32. Find the axial forces in members CD and EG.
a) 600 lb tension ; 600 lb compressionCD EG
T T
b) 1200 lb compression ; 900 lb tensionCD EG
T T
c) 0 ; 950 lb compressionCD EG
T T
d) 0 ; 900 lb compressionCD EG
T T
H
G
F
E
D
C
B
8ft
3
8ft
3
AR
8 ft
1200 lb
8 ft 8 ft 8 ft
BR
A
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33. Find the horizontal reaction at support C.
a) 0.4 ton
b) 0.6 ton
c) 1.2 ton
d) 1.6 ton
3 ft 6 ft
2 ft
4 ft
B
C
A
2 ton
3 ft
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Problems 34 and 35 refer to the following shape.
34. Solve for the coordinate of the centroid.
a) (4 cm, 1.5 cm)
b) (6 cm, 1.342 cm)
c) (4 cm, 1.342 cm)
d) (4 cm, 1.286 cm)
35. Find the moment of inertia about the centroid x-axis of the shape.
a) 3.66 cm4
b) 9.66 cm4
c) 12.5 cm4
d) 16.16 cm4
1
2
3
2 6 8 0 X
Y
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36. Solve for the tension force in cable BC.
a) 75 lb
b) 100 lb
c) 150 lb
d) 300 lb
D
C
B
A
150 lb
150 lb
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37. A book shelf weighs 200 N in total. It is supported by hinges A and
B on the wall. Solve for the horizontal reaction at hinge B.
a) 120 N
b) 150 N
c) 180 N
d) 240 N
2.5 m
B
A
5 m
3 m
400 N
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38. Calculate the centroidal polar moment of inertia with respect to
the z-axis of W14X176 with the following data: 2 4
3
4 3
51.8 in 0.830 in 838 in
15.2 in 1.31 in 320 in
15.7 in 2140 in 163 in
w y
f x
f x y
A t I
d t Z
b I Z
a) 1302 in4
b) 1489 in4
c) 2140 in4
d) 2978 in4
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39. A particle moves by the velocity function : 5 3v t t (m/s).
Calculate the distance traveled by the particle from 2 to 6
seconds.
a) 52 m
b) 72 m
c) 82 m
d) 92 m
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40. A ball is thrown upward with velocity 8 m/s. If the mass of the ball
is 0.5 kg, solve for the kinetic energy of the ball when it reaches 2
m above the initial point.
a) 4 Joule
b) 6 Joule
c) 8 Joule
d) 10 Joule
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41. A 12 kg box starts from rest at the bottom of an inclined plane.
The box is then pulled upward following the inclined plane of 10
to the horizontal plane. Solve for the force required to push the
box until it accelerates 0.5 m/s2 at the inclined plane. Assume that
the coefficient of friction between the inclined plane and box is
0.15.
a) 35 N
b) 40 N
c) 45 N
d) 50 N
10°
F
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42. In a friendly baseball match, a 0.15 kg baseball is thrown at the
speed of 10 m/s. The batter hits the ball with 60 N of force. If the
ball touches the bat in 0.2 seconds, what is the speed of ball right
after leaving the bat?
a) 60 m/s
b) 70 m/s
c) 80 m/s
d) 90 m/s
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43. A projectile is fired horizontally to a wooden block hang on a
pendulum. The velocity of the projectile is 450 m/s. The projectile
enters the wooden block and they swings together until reaching
the maximum height (see the figure). If the projectile mass is
0.01 kg and the wooden block mass is 1.49 kg, find the maximum
height they can reach. (Use g= 10 m/s2)
a) 35 cm
b) 40 cm
c) 45 cm
d) 50 cm
maxh
'v
450 m/s
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44. A 2 kg mass block is released from rest condition on a frictionless
inclined plane, as seen in the figure. The mass block moves down
along the inclined plane, then compresses a spring with a stiffness
of 500 N/m. Calculate the maximum compression of the spring.
a) 50 cm
b) 60 cm
c) 70 cm
d) 80 cm
5 m
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45. The following figure shows a typical stress-strain relationship of
steel rebar. Which of the following is the strain hardening part?
a) A-B
b) B-C
c) C-D
d) D-E
D
E C B
A Strain
Str
ess
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Problems 46 and 47 refer to the simply-supported beam as follows:
46. Which of the following diagrams is the correct shear force diagram
(unit: kips)?
2.5
2.5 2.5
3 ft 3 ft
5 kips
0
2.5 2.5
2.5 2.5
0
+
-
0
2.5 2.5
0
+
-
-
+ 0
2.5
2.5
0
-
+ 0
2.5
0
a)
b)
c)
d)
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47. Which of the following diagrams is the correct bending moment
diagram (unit: kips-ft)?
15
7.5
15
0
7.5
0
0 0
+
+
- 0 0
-
0 0
a)
b)
c)
d)
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48. Solve for the maximum shear stress of this two-dimensional
element.
a) 60 MPa
b) 64 MPa
c) 67 MPa
d) 70 MPa
60 MPa
60 MPa
40 MPa
40 MPa
100 MPa 100 MPa
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49. A steel rod is stressed by a tension force of 80 lb. It is found that
the rod has a length of 25 ft and a diameter of ¼“. If the modulus
of elasticity of the steel rod is assumed as 10,900 ksi, calculate
the strain of the steel rod due to the applied force.
a) 3.745 in
b) 0.011 in
c) 0.045 in
d) 0.252 in
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50. A 2 m long hollow cylindrical rod has the maximum allowable
shear stress 40 MPa and a maximum allowable twist angle of
0.025 radians. The outer and inner diameters of the hollow rod are
75 mm and 50 mm, respectively. Solve for the maximum
allowable torque T to this hollow rod if the shear modulus is
70,000 MPa and the torsion constant is 250 cm4.
a) 2 kN-m
b) 3 kN-m
c) 4 kN-m
d) 5 kN-m
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51. An alloy steel pipe has a length of 1 meter when the temperature
is 25°C. If the thermal expansion coefficient is 8x10-6/°C, calculate
the length of steel pipe when the temperature reaches 60°C.
a) 1.3 m
b) 1.03 m
c) 1.003 m
d) 1.0003 m
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52. For a simply supported beam, where does the maximum deflection
usually occur due to gravity loading?
a) At the top fibers
b) At the bottom fibers c) At the support
d) At the mid-span
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53. Calculate the load P if the concrete beam deflects 8.5 mm.
Assume the modulus of elasticity of concrete is 23,500 MPa and the self-weight of concrete is ignored.
a) 1 ton b) 1.5 ton
c) 2 ton
d) 2.5 ton
30 cm
P = ?
fixed
1.5 m
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54. A simply-supported steel beam is loaded by a distributed load as
seen below. The elastic modulus of steel is 10,900 ksi and the moment of inertia about X-axis is 1500 in4. The deflection at point
C (2 ft from the left support) is most nearly:
a) 0.1 in
b) 0.2 in c) 0.3 in
d) 0.4 in
2 ft 8 ft
C
w = 3 lb/ft
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55. An H-section steel column is fixed at the base and pinned at the
top. The elastic modulus of steel is 2.1 x 105 MPa. Find the maximum concentric vertical load that the column can support
without buckling. Assume that the steel column has no
imperfection.
a) 2,500 kN
b) 5,000 kN
c) 7,500 kN d) 8,500 kN
8 m Ix = 11500 cm4
Iy = 3880 cm4
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56. ASTM (American Standard Testing and Material) provides a lot of
standard testing method for materials. Which of the following testing methods is included for aggregates?
a) Sieve analysis b) Relative density
c) Absorption
d) All of the above
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57. A new material is tested in the elongation test. The result shows
that the material has strain of 5.3% (elastic limit) when the stress
of 320 MPa is applied. Find the elastic modulus of the material
based on the result.
a) 17 MPa
b) 170 MPa c) 1700 MPa
d) 6000 MPa
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58. Which of the following materials is tested for stability through
Marshall test?
a) Concrete
b) Steel
c) Wood d) Asphalt
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59. Which of the following checmical admixtures is used to increase
the workability of concrete?
a) Accelerator
b) Retarder c) Plasticizer / water reducer
d) Air entraining agent
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60. Which of the following does affect the kinematic viscocity of a
fluid?
I. Density II. Surface tension
III. Stress IV. Absolute viscocity
a) I and II
b) II and III
c) III and IV d) I and IV
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61. A water tank with a height of 5 meters is full of water (density =
1000 kg/m3). Determine the gage pressure at a depth of 50 cm
from the bottom of water tank. Use g = 10 m/s2.
a) 5 kPa
b) 10 kPa
c) 45 kPa
d) 50 kPa
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62. A continuous pipe has two different diameter sizes at the inlet and
outlet of pipes. The diameter at the inlet is 1.5 times bigger than
the diameter at the outlet. If the flow velocity at the inlet is 0.5
m/s and the pipe outlet goes freely to the atmosphere, determine
the gage pressure at the inlet.
Note: neglect frictional effects in the pipe
a) 0.2 kPa
b) 0.3 kPa
c) 0.4 kPa
d) 0.5 kPa
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63. Water flows in the tube with velocity of 2 m/s. The kinematic
viscocity is found to be 7.3 x 10-7 m2/s. If the inside diameter of
tube is 50 mm, calculate the Reynolds number and determine the
flow type.
a) 1.73 x 103 , laminar
b) 3.71 x 103 , critical
c) 1.37 x 105 , critical
d) 1.37 x 105 , turbulent
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Problems 64 and 65 refer to the following case:
3 m3 of water are pumped each minute through a 150 mm diameter
pipe. The pipe length is 50 m, and has a Darcy friction factor of 0.02.
Because of the pump, the water can reach upward until a height of 18
m. The pump efficiency is 80%.
64. Calculate the friction loss for the entire length of pipe.
a) 2.7 m
b) 10 m
c) 27 m
d) 162 m
65. How much power is needed to deliver to the pump?
a) 11 kW
b) 13 kW
c) 14 kW
d) 16 kW
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66. Which of the flows has cross section that does not vary with time
at any location along an open channel?
a) Uniform flow
b) Non-uniform flow
c) Steady flow
d) Critical flow
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67. Water flows from a reservoir through a 0.5 m diameter pipe. The
elevations of inlet and outlet of the pipe are 221 m and 168 m,
respectively. The flow is found to be steady and incompressible
flow. The pipe outlet discharges to atmospheric pressure. If the
flow rate out of the pipe outlet is 5 m3/s, calculate the total head
loss in the system.
a) 20 m
b) 33 m
c) 46 m
d) 53 m
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Problems 68 and 69 refer to the following information:
A pumping system transfers water from a lake to a water tank with a
150 m length and 35 cm diameter pipeline. The flow rate is found to
be 1.34 m3/s and the kinematic viscocity of water is 10-6 m2/s. The
head of pipe inlet is 150 m, while the head of pipe outlet is 175 m. The
pipeline is made of cast iron (specific roughness is 0.25 mm). Assume
that minor losses, entrance losses, and exit losses in the pumping
system can be ignored. Consider the pump efficiency to be 85%. Also
assume the flow is stready and incompressible.
68. Calculate the head losses in the piping by using Darcy equation.
a) 55 m
b) 75 m
c) 95 m
d) 115 m
69. Find the required power to pump the water from the lake to water
tank.
a) 1300 kW
b) 1400 kW
c) 1500 kW
d) 1600 kW
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Problems 70 and 71 refer to the following case:
A large complex of apartments has a population of 8000 people. The
average sewage flow for this apartment complex is 4000 m3/day.
70. What is the most nearly minimum sewage flow for this apartment
complex?
a) 1200 m3/day
b) 1500 m3/day
c) 2000 m3/day
d) 2200 m3/day
71. What is the most nearly peak sewage flow for this apartment
complex?
a) 8400 m3/day
b) 10500 m3/day
c) 12200 m3/day
d) 15250 m3/day
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72. A sanitary sewer has a length of 60 m and a pipe with diameter of
90 cm. The inlet elevation of the sewer is 1.5 higher than the
outlet elevation. Assume that the Manning’s roughness coefficient
is 0.012 and constant with depth of flow. Determine the
approximate sewer capacity during heavy rainfall if the sewer is
full of water flow with no surcharge.
a) 1 m3/s
b) 2 m3/s
c) 3 m3/s
d) 4 m3/s
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73. What is the static determinacy of the following truss?
a) Stable, statically determinate
b) Stable, statically indeterminate c) Unstable
d) None of the above
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Problems 74 and 75 refer to the following beam.
74. What is the largest magnitude of the shear force in the beam?
a) 15 kN
b) 20 kN c) 35 kN
d) 40 kN
75. What is the largest magnitude of the bending moment in the
beam?
a) 20 kN-m
b) 25 kN-m
c) 50 kN-m d) 100 kN-m
5 m 5 m 5 m
P = 20 kN
D A B C
w = 4 kN/m
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76. Based on ACI 318, calculate the top flange effective width of the
following T-beam.
a) 20 cm b) 40 cm
c) 150 cm
d) 200 cm
35
cm 10 cm
20 cm
6 m
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Problems 77 and 78 refer to the following beam case:
A simply supported beam is subjected to an ultimate bending moment
of 110 kips-ft and an ultimate shear force of 40 kips.
The beam section is 10”x20”. The compressive strength of concrete is
3000 psi, the yield strength of longitudinal reinforcement and stirrups
are 60 ksi.
77. Calculate the required number of #6 rebar to resist the bending
moment (design for tension only).
a) 2
b) 3
c) 4 d) 5
78. Solve for the required spacing of #3 stirrups to resist the shear
force.
a) 4”
b) 6” c) 8”
d) 10”
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79. Which of the following concepts about ASD and LRFD are correct?
I. ASD is newer than LRFD II. ACI 318 is based on the LRFD concept
III. LRFD uses factored load combinations while ASD doesn’t
IV. ASD uses ultimate strength while LRFD doesn’t V. ASD uses a factor of safety while LRFD doesn’t
a) I, II, IV b) II, III, V
c) II, IV, V
d) II, III, IV
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80. A W16x36 of A992 steel (Fy = 50 ksi) is used as a beam to
support a concrete floor slab that provides continuous lateral
support to the compression flange. The applied loading is shown in
the figure.
What is the maximum ultimate distributed loading which can be
resisted by the beam? Assume that the beam is laterally
supported.
Shape b (in) h (in) tf (in) tw (in) Sx (in3) Zx (in
3)
W16x36 6.99 15.9 0.43 0.295 56.5 64.0
a) 2 kips/ft
b) 2.5 kips/ft
c) 3 kips/ft
d) 3.5 kips/ft
Wu
24 ft
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81. Find the void ratio of the soil sample, based on the following data:
Mass = 17.4 gram Volume = 10 cm3
Oven-dry mass = 13.4 gram
Specific gravity = 2.15
a) 0.2
b) 0.4 c) 0.5
d) 0.6
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82. A sample of saturated soil from the field is taken to a test
laboratory. The specific gravity is found to be 2.6 with a total unit mass of 2250 kg/m3. Find the dry unit mass of the soil sample.
a) 1950 kg b) 2030 kg
c) 2120 kg
d) 2150 kg
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83. Six meters of excavation was done through two different soil
layers as shown below. Calculate the total active lateral pressure against the retaining wall at the bottom of retaining wall.
a) 30 kN/m2 b) 35 kN/m2
c) 50 kN/m2
d) 65 kN/m2
3 m
3 m
Layer 1
Layer 2
3
1
'
1
1
19 kN/m
33
0c
3
1
'
1
1
20 kN/m
25
0c
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84. A 5’x5’ square footing rests at 4-ft depth below the ground. The
following data about the soil will be used for the calculation:
125 pcf
0
30
c
37.2
19.7
22.5
c
q
N
N
N
1.25
0.85
cS
S
Use Terzaghi equation to calculate the allowable bearing capacity
of the square footing using a safety factor of 3.
Note: The water table is below the footing and neglect the weight
of soil above the footing.
a) 2500 psf
b) 4500 psf
c) 5500 psf d) 16000 psf
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85. Which of the following factors doesn’t affect the rate of
consolidation?
a) Permeability
b) Layer thickness c) Compressibility
d) None of the above
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Problems 86 and 87 are based on the following vertical curve.
86. Determine the low point station for this vertical curve.
a) Sta 70+00
b) Sta 75+68 c) Sta 76+32
d) Sta 82+00
87. Determine the elevation of the low point.
a) 500 m
b) 502 m c) 504 m
d) 506 m
-2%
PVI
Sta 76+00
Elev = 500 m
EVC PVC
+1.8%
L = 12 sta
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88. Based on several studies, what is the perception-reaction time
AASHTO used to design Stopping Sight Distance?
a) 1.5 sec
b) 2 sec c) 2.5 sec
d) 3 sec
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89. Traffic flow can be written as q kv where q is the traffic
volume, k is the traffic density, v is the mean speed of vehicles. Based on several studies at a road, the mean speed is given by
the relationship 80 0.5v k .
What is, most nearly, the maximum capacity of total traffic
volume of the road?
a) 3000 veh/hour b) 3200 veh/hour
c) 3400 veh/hour
d) 3600 veh/hour
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90. A horizontal curve is designed to have a diameter of 1200 m. The
tangent is 125 meters and the PI is located at sta 10+000 (in
kilometer). Find the stationing of PT.
a) Sta 9+744
b) Sta 9+875
c) Sta 10+121
d) Sta 10+600
Page 96
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Problems 91 and 92 refer to the following case.
A sample of wastewater is placed in the incubator by keeping the
temperature at 23°C. After 5 days, the BOD (Biochemical Oxygen
Demand) is found to be 234 mg/L. Assume that the reaction rate
constant is 0.13 d-1 (base e).
91. Find the ultimate BOD of the sample.
a) 290 mg/L
b) 468 mg/L
c) 490 mg/L
d) 1,170 mg/L
92. Find the BOD of the sample if the incubation period is one week.
a) 290 mg/L
b) 468 mg/L
c) 490 mg/L
d) 1,170 mg/L
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93. A tank reactor treats 0.35 m3/s of settled wastewater having 258
mg/L BOD5 at 22°C. The design mean cell resistance time ( d
c ) is
12 days.
,
5,
0.55ss mg
mg
VY
BOD
The effluent BOD5 is 6.4 mg/L. MLVSS = 3600 mg/L. The
endogenous decay coefficient 10.06d
K d . Determine the reactor
capacity per day.
a) 3600
b) 6400
c) 8100
d) 9200
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94. A sample of freshwater is taken from a small river. After several
observations, the sample is found to contain a dissolved oxygen
concentration of 6.2 mg/L when the temperature is 24.6°C and
the atmospheric pressure is 740 mmHg. A partial listing of the
solubility of dissolved oxygen in freshwater at equilibrium with dry
air containing 21.7% oxygen and at an atmospheric pressure of
760 mmHg is as follows:
Temperature (°C) Oxygen solubility (mg/L)
22 9.6
23 9.3
24 8.9
25 8.6
26 8.2
Calculate the saturation of dissolved oxygen in the water sample.
a) 50%
b) 60%
c) 70%
d) 80%
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95. A municipal has a population of 18,400 people. The monthly
volume of solid waste is 20,000 m3. If the density of solid waste is assumed to be 125 kg/m3, find the average mass of solid waste
generated by a person per day.
a) 1.1 kg
b) 1.8 kg
c) 3.2 kg
d) 4.5 kg
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96. Find the mass loading of a 35 MGD wastewater discharge with an
ultimate BOD of 28 mg/L.
a) 8,000 lbm/day
b) 8,200 lbm/day
c) 8,400 lbm/day
d) 8,600 lbm/day
Page 101
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97. A municipal produces wastewater with a BOD5 of 225 mg/L and an
ultimate BOD of 486 mg/L. Find the reaction rate constant.
a) 10.025 d
b) 10.1 d
c) 10.125 d
d) 10.175 d
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98. Which of the following is the correct definition of a critical path in
a project?
a) The sequences of tasks in a project
b) The time unaccounted for in a project
c) The longest path of sequential tasks through a project
d) The extra time available in a project
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99. Which project scheduling tool is illustrated in the figure below?
a) Bar (Gantt) chart
b) AON diagram
c) AOA diagram
d) PERT chart
1 3
2
2 4
2
5
2
6
2 7
2
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Problems 100-102 refer to the following diagram.
100. Find the critical path of this diagram.
a) START – A – C – F – FINISH
b) START – A – D – F – FINISH
c) START – B – D – F – FINISH
d) START – B – E – F – FINISH
101. Find the EST (Earliest Start Time) of activity F.
a) Day 8
b) Day 9
c) Day 10
d) Day 11
102. Find the float time of activity D. a) 0 day
b) 1 day
c) 2 days d) 3 days
Start
A|3
Finish
B|5
C|2
D|3
E|4
F|3
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103. Which of the following statements is correct about construction
contracts?
a) In a lump sum contract, a contractor bids a price for each work
item by the cost per unit b) A unit price contract is often used when the exact quantities in a
project are known
c) In a unit price contract, the contractor bids a total price for all work in a project, including the profit
d) A lump sum contract cannot be changed unless there is a
contract modification, while the unit price contract may be
changed depends on the actual quantities
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104. Which bond guarantees that the contractor will perform the
specified work in accordance with the contract (typically full value of project)?
a) Bid bond b) Performance bond
c) Payment bond
d) None of all the above
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105. According to OSHA, any adequate exits in trench excavation must
be provided when the depth reaches:
a) 0 ft (no such requirement)
b) 2 ft c) 4 ft
d) 5 ft
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106. A wall of a single story building has a length of 72 ft and a height
of 16 ft. A contractor shall fill the wall by brick. From previous experience the contractor found that 602 bricks are needed for
every 100 ft2. If the total opening (window and door) area is 250
ft2 and there is 5% waste of material, calculate the required number of bricks for the wall.
a) 3810 b) 3900
c) 4000
d) 5702
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Problems 107 and 108 refer to the following information.
The earthwork of a new roadway requires less soil to be thrown out so
that engineers can now calculate the cut-and-fill volume based on the
following information in the table. Assume both cut and fill areas are
triangular for the transition region from fill to cut area.
Station (m) Cut area (m2) Fill area (m2)
100+00 - 153.42
10+4.50 - 32.56
10+12.35 13.67 8.25
10+16.40 52.84 -
10+20 165.14 -
107. Find the total approximate volume of fill work.
a) 500 m3
b) 505 m3 c) 510 m3
d) 515 m3
108. Find the total approximate volume of cut work.
a) 540 m3 b) 560 m3
c) 580 m3
d) 600 m3
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109. A borrow pit is figured in the grid below. This area is going to be
leveled to 100 m for all grid corners. Find the total volume of
earth excavated.
Point Elevation
(m)
A1 103.5
A2 105.6
A3 102.1
A4 101.9
B1 103.9
B2 104.0
B3 102.8
B4 103.7
C1 105.2
C2 104.3
C3 100.3
a) 480 m3
b) 1700 m3 c) 2500 m3
d) 6800 m3
4 3 2 1
B
C
A
12 m 12 m 12 m
8 m
8 m
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110. What is the correct term for the increasing volume of soil when
earth is excavated?
a) Loose
b) Void c) Swell
d) Shrinkage
Page 113
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Vector Length
Problem 1 Solution:
Length of a straight line AB can be calculated as vector length:
2 2 2
1 2 3 1 3 7 29AB
(Answer B)
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Roots of Equation
Problem 2 Solution:
For third-degree polynomial, the number of roots will be three.
3
2
7 6 0
1 6 0
1 2 3 0
x x
x x x
x x x
To satisfy the equation, the roots are 1, 2, -3.
(Answer D)
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Circle Center
Problem 3 Solution:
The general equation of a circle is:
2 2 2x a y b r
Where ,a b is the coordinate of its center and r is the radius of
the circle.
Solve the coordinate by modifying the equation into the general
equation:
2 2
2 2
2 2 2
10 14 49 0
10 25 14 49 25
5 7 5
x y x y
x x y y
x y
The coordinate of its center is (5, -7).
(Answer D)
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Calculus
Problem 4 Solution:
The slopes of two lines, which are perpendicular, can be written as
follows:
1 21m m
Where 1
m and 2
m are the slopes of the first and second line,
respectively.
First line: Second line:
1
5 3 16 0
5 3 16
3 16 3
5 5 5
y x
y x
y x m
1 2
2
2
1
31
5
5
3
m m
m
m
Solve for the equation of the second line:
5
3m known point , 1,7x y
Use the general equation of a straight line:
57 1
3
16
3
y mx b
b
b
The equation is 5 16
3 3y x or 3 5 16 0y x .
(Answer A)
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Circle Equation
Problem 5 Solution:
The general equation of a circle is:
2 2 2x a y b r
Where ,a b is the coordinate of its center and r is the radius of
circle.
Substitute the known point (x,y) into the circle equation:
2 2 2
22 28 2 3 5
100 10
x a y b r
r
r
The equation of the circle is:
2 2 2
22 2
2 2
2 2
2 5 10
4 4 10 25 100
4 10 71 0
x a y b r
x y
x x y y
x y x y
(Answer C)
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Calculus
Problem 6 Solution:
Solve the two coordinate equations:
9sin sin9
2cos cos2
xx t t
yy t t
Use the trigonometry identity:
2 2
2 2
2 2
sin cos 1
19 2
4 81 324
t t
x y
x y
(Answer B)
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Vector Analysis
Problem 7 Solution:
For the cross product remember to cross out the row and column for
each variable, i, j, k and multiply that respective variable by what’s left over in the matrix. Also, remember that the signs start with a positive
at i, a negative at j, and then a positive at k, and so forth.
8 1 2
3 1 3
1 2 8 2 8 1
1 3 3 3 3 1
30 11
i j k
A B
i j k
i j k
(Answer A)
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Vector Analysis
Problem 8 Solution:
Use dot product to get the angle:
2 2 22 2 2
cos
1 1 1 1 1 1 1 1 1 1 1 1 cos
1 3 3 cos
1cos 70.5 70
3
M N M N
(Answer D)
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Inflection Point
Problem 9 Solution:
Inflection point occurs when the second derivation of f x is 0.
4 3 2
3 2
2
12 30
' 4 36 60
" 12 72 60
f x x x x
f x x x x
f x x x
Solve for inflection point:
2
2
1 2
" 12 72 60 0
12 6 5 0
12 1 5 0
1 and 5
f x x x
x x
x x
x x
4 3 2
1
4 3 2
2
1 1 1 12 1 30 1 19
5 5 5 12 5 30 5 125
x f
x f
The inflection points are (1, 19) and (5, -125).
(Answer B)
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Trigonometry
Problem 10 Solution:
This talks about trigonometry identities.
Option A → 2sin cos sin2
Option B → 2 2sin cos 1
Option C → 21 cos2sin
2
Option D → 2 2cos sin cos2
(Answer B)
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Probability
Problem 11 Solution:
The possible successful outcomes are that either a red ball is
picked then a black ball, or a black ball is picked then a red ball.
Number of red balls = 5
Number of blue balls = 7
Number of black balls = 3
Total number of balls = 5 + 7 + 3 = 15
red then black or 5 3 3 50.143
15 14 15 14black then redP
(Answer B)
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Probability
Problem 12 Solution:
The sample space of rolling dice is {1, 2, 3, 4, 5, 6}.
The event of rolling dice is independent each other, so that the
second rolling dice is not affected by the result from the first
rolling dice.
The event “number 2” of the rolling dice is {2}.
1
6
n EP E
n S
(Answer C)
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Probability
Problem 13 Solution:
Use binomial distribution to solve this problem:
2 15 2
0.03
1 0.97
!
! !
15!2 0.03 0.97 0.064
2! 15 2 !
x n x
p
q p
np x p q
x n x
p
(Answer D)
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Mean and Standard Deviation
Problem 14 Solution:
The mean is determined by dividing the sum of data by the total
number of data.
i ix
2
ix x
1 2995 29.16
2 3005 21.16
3 3001 0.36
4 2991 88.36
5 2984 268.96
6 3004 12.96
7 3009 73.96
8 3015 213.16
9 2998 5.76
10 3002 2.56
Total 30004 716.4
1 300043000.4
10
n
ii
x
xn
Variance is defined as:
2
1 716.479.6
1 10 1
n
ii
x x
n
(Answer A)
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Probability
Problem 15 Solution:
The fastest way to calculate the probability of “at least one” event
is by substract the probability of having baby with no brown eyes from the probability of all events.
zero 0.7 0.7 0.7 0.7 0.7 0.7 0.118
at least one 1 zero 1 0.118 0.882 0.9
P
P P
(Answer D)
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Programming - Looping Function
Problem 16 Solution:
“if ... else ... end” and “if ... elseif ... end” are the functions to
create conditional statements.
“for ... end” is the function to create loopings for a certain specified number.
“while ... do” is the function to create loopings as long as a certain
condition is true.
(Answer B)
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Programming – Looping Result
Problem 17 Solution:
Looping condition can be solved by observing the variable data for
each loop.
Condition x y
Initialize 0 1
First looping x = 0 2 2
Second looping x = 2 4 3
Third looping x = 4 6 4
Fourth looping x = 6 looping stops
The last value of variable y is 4.
(Answer B)
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Ethics
Problem 18 Solution:
By doing ethical behavior, you will not get any benefits because
ethical behavior promises nothing.
(Answer D)
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Professional Practice
Problem 19 Solution:
You should not accept any king of gifts from parties expecting
special consideration, so you should reject the Rolex watch. As long as the contractor satisfies all the bid requirement and follows the
bidding rules, the bid could be accepted.
(Answer C)
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Ethical Behavior
Problem 20 Solution:
Options (A), (B), (C), (D) may all be valid. However, the rationale
for specific ethical prohibitions on using your employer’s equipment for a second job is economic. When you don’t have to
pay the equipment, you don’t have to recover its purchase price in
your fees for services.
(Answer D)
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Professional Societies
Problem 21 Solution:
A violation should be reported to the organization that has
promulgated the rule.
(Answer D)
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Professional Liability
Problem 22 Solution:
It is legal to stamp plans that you personaly designed and/or
checked. It is illegal to stamp plans that you didn’t personally design or check, regardless of whether you got paid. It is legal to
work as a ‘plan checker’ consultant.
(Answer D)
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Loan Payment Analysis
Problem 23 Solution:
Use capital recovery discount factor:
48
48
| , %,
$30,000 $8,000 $22,000
6%0.5%
12 compounding periods per year
4 years 48 months
| ,0.5%,48
0.005 1 0.005$22,000
1 0.005 1
$516.67 $520
A P A P i n
P
i
n
A P A P
(Answer C)
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Depreciation Value
Problem 24 Solution:
Depreciation value is equal to the difference between initial cost C
and salvage value n
S , then divided by the life span n .
$80,000 $6,500$7,350
10n
C SD
n
(Answer B)
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Future Value Analysis
Problem 25 Solution:
15
| ,5%,15 $2,000 1 0.05 $4,157F P F P
(Answer C)
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Break Even Point Analysis
Problem 26 Solution:
Break even point occurs when the total of costs (negative) and
sales (positive) is zero. Let n be the number of tables per year at
break even point.
$400,000 $60 $99 0
$39 $400,000
10257 10300 tables/year
n n
n
n
(Answer D)
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Effective Interest Rate Analysis
Problem 27 Solution:
40.025
1 1 1 1 2.5235%4
mr
im
(Answer C)
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Future Value Analysis
Problem 28 Solution:
Use the uniform series sinking fund discount factor. The interest period is one month, there are 12 compounding periods, and the
effective interest rate per interest perios is 10%/12 = 0.83%.
12
0.0083| ,0.83%,12 0.079597
1 0.0083 1
| ,0.83%,12 $100,000 0.079597 $7,960 $8,000
A F
A F A F
(Answer B)
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Basic Concept of Structure’s Support
Problem 29 Solution:
Fixed support can resist vertical (shear) and horizontal (axial)
forces as well as moment. It can restrain rotation and translation.
(Answer D)
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Static Equilibrium
Problem 30 Solution:
The equilibrium system means all the forces acting to the system
are balanced. Therefore, we will use two equilibrium equations
(vertical and horizontal forces) to solve this problem.
Equilibrium in vertical forces (Y direction):
1
1
0
sin 2500 0
3125 lb
yF
T
T
Equilibrium in horizontal forces (X direction):
1 2
2
0
cos 0
1875 lb
xF
T T
T
(Answer A)
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Truss Analysis
Problem 31 Solution:
To solve the support reaction, equilibrium should be satisfied.
0
1200 lb 8 ft 32 ft 0
300 lb
A
B
B
M
R
R
0
1200 lb 0
300 lb 1200 lb 0
900 lb
y
A B
A
A
F
R R
R
R
(Answer D)
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Truss Member Forces
Problem 32 Solution:
Member CD is zero-force member. Check the equilibrium of
vertical forces at joint D:
0
0
y
CD
F
T
Member EG can be solved by ‘cutting’ the truss as shown below.
Check the equilibrium of system to get the support reaction at B:
0
32 ft 1200 lb 8 ft 0
300 lb
A
B
B
M
R
R
Check the equilibrium of moment at joint F by looking at ‘right
side’ part:
2 2
2 2
0
88 8 316 ft ft 8 ft 0
38 88 8
3 3
948.68 lb 948.68 lb compression
F
B EG EG
EG
M
R T T
T
Answer : C
E
B
8ft
3
8ft
3
AR
8 ft
BR
8 ft 8 ft 8 ft
A
1200 lb
H
G
F D
C
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Frame Analysis
Problem 33 Solution:
This is a statically determinate structure. We have four unknowns
(A
H , A
V , C
H , C
V ), so that we need four equilibrium equations to
solve this problem.
Check equilibrium of horizontal & vertical forces in the system (ton
& ft):
0
(1)
x
A C
F
H H
0
2 (2)
y
A C
F
V V
Check equilibrium of moment at joint A in the system (ton & ft):
0
2 3 2 12 0
6 3 (3)
A
C c
C C
M
H V
H V
AH
3 ft 3 ft 6 ft
B
C
A
2 ton
AV
CV
cH
2 ft
4 ft
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Check the moment at pin B for the right part (ton & ft):
0
4 6 0
1.5 (4)
B
C c
C C
M
H V
H V
Substituting equation (4) to equation (3):
6 3 (3)
1.5 6 3
0.4 ton
C C
C C
C
H V
V V
V
Therefore, the horizontal reaction at support C can be calculated
using equation (4):
1.5 (4)
1.5 0.4 0.6 ton
C C
C
H V
H
(Answer B)
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Centroids
Problem 34 Solution:
Shape x (cm) y (cm) A (cm2)
I 4/3 1 3
II 4 2.5 4
III 20/3 1 3
IV 4 0.5 4
Total 14
,
,
4 203 4 4 3 4 4
3 34 cm
14
1 3 2.5 4 1 3 0.5 41.286 cm
14
c n n
c
c n n
c
x Ax
A
y Ay
A
(Answer D)
1
2
3
2 6 8 0 X
II
IV III I
Y
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Moment of Inertia
Problem 35 Solution:
From the previous solution, the centroid of system was found to be (4
cm, 1.286 cm). In order to get the moment of inertia of the system,
we should find the distance of centroid (d ) between each shape and
overall system.
Shape ,c xI (cm4) A (cm2) d (cm) 2Ad (cm4)
I
33 2 3
1.536 36
bh
3 -0.286 0.2454
II
33 4 1
0.3312 12
bh
4 1.214 5.8952
III
33 2 3
1.536 36
bh
3 -0.286 0.2454
IV
33 4 1
0.3312 12
bh
4 -0.786 2.4712
Total 3.66 8.8571
2 4
, , ,3.66 8.8571 12.52 cm
c x c x i i iI I Ad
(Answer C)
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Cable Tension Force
Problem 36 Solution:
Alternative 1:
Use the law of sines to solve the triangle at joint B.
150150 lb
sin120 sin120AB
AB
TT
Alternative 2:
Use equilibrium at joint B.
0
cos30 cos30
x
AB BC
AB BC
F
T T
T T
0
sin30 sin30 150
2 sin30 150
150 lb
y
AB BC
AB
AB
F
T T
T
T
(Answer C)
ABT BC
T
150 lb
120°
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Static Analysis
Problem 37 Solution:
To find the horizontal reaction at support A, check the equilibrium
at support B. Pinned support cannot resist moment, so that
equilibrium of moment at support B is 0.
Assume the horizontal reaction at support A is to the right ().
0
400 1.5 2.5 0
240 N
B
A
A
M
H
H
(Answer D)
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Centroidal Polar Moment of Inertia
Problem 38 Solution:
42140 838 2978 inxx yy
J I I
(Answer D)
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Kinematic Function
Problem 39 Solution:
Velocity is defined in terms of time, so that distance can be
written as follows:
6
2
6
2
2
5 3
5 3
2
92 m
s t v t dt
t dt
t t
(Answer D)
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Kinetic Energy
Problem 40 Solution:
The velocity at 2-m height:
2 2 2
02 8 2 9.8 2 24.8
24.8 4.98 m/s
t
t
V V gh
V
The kinetic energy at 2-m height:
21 10.5 24.8 6.2 Joule
2 2k
E mV
(Answer B)
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Friction Force
Problem 41 Solution:
Define the forces working to the moving box in the inclined plane,
then apply the equation.
coskf N mg
sin cos
12 9.8 sin10 0.15 12 9.8 cos10 12 0.5
43.8 45 N
F ma
F mg mg ma
F
F
(Answer C)
kf
sin10mg
10° F
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Impulse & Momentum
Problem 42 Solution:
2 1
2
2
60 0.2 0.15 10
90 m/s
I P
F t m v v
v
v
(Answer D)
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Law of Conservation of Momentum
Problem 43 Solution:
The law of conservation of momentum states that the total
momentum of the two objects before the collision is equal to the
total momentum of the objects after the collision.
p bp p ' '
' '
p b
p p p p b b
p p
m v m v m v
This case is a perfectly inelastic collision, because both projectile
and wooden block have the same motion after collision. The
velocity after collision (' ' '
p bv v v ) can be obtained as follows:
'
'
'
0.01 450 0.01 1.49
3 m/s
p p p bm v m m v
v
v
Using the law of energy, solve the maximum height as follows:
1
2m 2v m
213 10
2
0.45 m 45 cm
gh
h
h
(Answer C)
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Work & Energy
Problem 44 Solution:
The inclined plane is frictionless, so there is no energy loss in the
system. Use the conservation of energy to solve this.
2
2
1
2
12 9.8 5 500
2
0.62 m
i fE E
mgh k
(Answer B)
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Stress-Strain Relationship
Problem 45 Solution:
The stress-strain relationship has elastic and plastic part. The
elastic part is shown at part A-B, while the rests are plastic
condition. The complete list of this following stress-strain is given
below:
Part A-B : Elastic
Part B-C : Yielding
Part C-D : Strain hardening
Part D-E : Fracture
(Answer C)
C B
A Strain
Str
ess
D
E
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Shear Force Diagram
Problem 46 Solution:
Before drawing ths shear force diagram, first we need to find the
support reactions. For this case, the applied concentrated loading
is in the mid-span, so that the reaction at both supports is half of
the applied loading = 2.5 kips (upward each).
Shear Force Diagram (unit: kips)
(Answer A)
0
2.5 2.5
2.5 2.5
0
+
-
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Bending Moment Diagram
Problem 47 Solution:
For this case, the bending moment increases linearly until the
maximum moment is located when the shear force diagram is
zero.
Bending Moment Diagram (unit: kips-ft)
(Answer A)
0 0
7.5
+
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Mohr’s Circle of Element Stresses
Problem 48 Solution:
100 4070 MPa
2 2
x y
avg
By looking at the Moh’r circle above, the maximum shear stress is
actually the radius of the Moh’r circle.
2 2
max100 70 60 67 MPa
(Answer C)
max
1
2
avg
(-40,-60)
(-100,60)
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Strain (Elastic Deformation)
Problem 49 Solution:
23
in80 lb 25 ft 12
ft0.045 in
110900 10 psi 0.25 in
4
PL
EA
(Answer C)
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Torsion
Problem 50 Solution:
There are two limitations to find the maximum allowable torque:
1. Check the maximum allowable shear stress
0max
6
2540
2500000
4 10 N-mm 4 kN-m
Tr
J
T
T
2. Check the maximum allowable twist
max
20000.025
70000 2500000
2187500 N-mm 2.1875 kN-m
TL
GJ
T
T
The maximum allowable torque is the minimum between those
two values, taken as 2 kN-m.
(Answer A)
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Thermal Deformation
Problem 51 Solution:
0
6
4
0
4
8 10 C 1 m 60 C 25 C
2.8 10 m
1 m 2.8 10 m
1.00028 m
L L T
L L L
(Answer D)
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Deflection
Problem 52 Solution:
For a simply supported beam, the maximum deflection usually
occurs at the mid-span, where the maximum moment occurs.
While at the support, the maximum shear usually occurs.
Compression stress will occur at the top fibers and tension will
occur at the bottom fiber.
(Answer D)
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Deflection of Cantilever Beam
Problem 53 Solution:
The quick formula of deflection in this case is:
3
3
PL
EI
The moment of inertia for this rectangular section is:
3 38 4150 300
3.375 10 mm12 12
bhI
Therefore, the load is:
3
3
8
3
15008.5
3 23500 3.375 10
59925 N 6 ton
PL
EI
P
P
(Answer C)
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Deflection of Simply-Supported Beam
Problem 54 Solution:
The quick formula of deflection in this case is:
3 2 3224
wxL Lx x
EI
Therefore, the load is:
3 2 3
3 2 3
224
3000 2 12 10 12 2 10 12 2 12 2 12
24 10900000 1500
0.3 in
wxL Lx x
EI
(Answer C)
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Steel Column Analysis
Problem 55 Solution:
Fixed-pinned → 0.7k
Effective length for buckling in the weak direction:
' 0.7 8 m 5.6 mL kL
Critical buckling based on Euler’s formula:
2
2
2 5 4
2
'
2.1 10 3880 10
5600
2564 kN
cr
EIP
L
(Answer A)
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Standard Testing Methods for Material
Problem 56 Solution:
ASTM contains:
C136-06 covers the standard test method for sieve analysis
for fine and coarse aggregates.
C127-12 covers the standard test method for density,
relative density (specific gravity), and absorption of coarse
aggregates.
C128-12 covers the standard test method for density,
relative density (specific gravity), and absorption of fine
aggregates.
(Answer D)
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Material Properties
Problem 57 Solution:
Since the material is still in elastic part, we can use the
relationship between stress and strain to get the elastic modulus.
3206038 MPa 6000 MPa
0.053E
(Answer D)
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Material Testing Method
Problem 58 Solution:
Marshall test is a common method used to calculate the load and
flow rate of asphalt specimens.
(Answer D)
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Concrete Admixtures
Problem 59 Solution:
Accelerator is used to shorten the setting time in concrete.
Retarder is used to extend the setting time of cement paste in
concrete.
Plasticizer / water reducer is used to increase the workability of
concrete, allowing the concrete be placed more easily.
Air entraining agent is used to provide space for the water to
expand upon freezing.
(Answer C)
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Fluid Properties
Problem 60 Solution:
Kinematic viscocity ( ) is defined as the ratio of absolute viscocity
( ) to mass density ( ), written as follows:
(Answer D)
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Water Pressure
Problem 61 Solution:
The depth h is calculated from the water surface.
1000 10 5 0.5 45000 Pa 45 kPap gh
(Answer C)
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Fluid Continuity
Problem 62 Solution:
Using continuity equation:
1 1 2 2
1
4
Av A v
2
1 1
1
4D v
2
2 2
2 2
2 2 2
2
1.5 0.5 m/s
1.125 m/s
D v
D D v
v
Using Bernoulli’s equation:
2
1 11
2
p vz
g g
2
2 22
2
p vz
g g
2 2
1 1 2
2 2
2 11
2 2
02 2
2
1.125 0.5 1000
2
507.815 Pa 0.5 kPa
p v v
v vp
(Answer D)
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Reynolds Number
Problem 63 Solution:
Reynolds number determines the fluid type:
Re < 2100 → laminar flow
2100 < Re < 4000 → critical flow
Re > 4000 → turbulent flow
7 2
5
Re
2 m/s 0.05 m
7.3 10 m /s
1.37 10 turbulent
vD vD
(Answer D)
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Hydraulic Friction Loss
Problem 64 Solution:
Flow rate: 3 3m m
3 0.05min s
Q
Flow area: 2 2 21 10.15 0.0177 m
4 4A D
Flow velocity: 3
2
0.05 m /s2.82 m/s
0.0177 m
Qv
A
Friction loss:
22
2
0.02 50 m 2.82 m/s2.7 m
2 2 0.15 m 9.8 m/sf
fLvh
Dg
(Answer A)
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Hydraulic Pump Power
Problem 65 Solution:
Total head by the pump:
18 m 2.7 m 20.7 mz f
h h h
Pump power:
21 min3000 kg/min 9.8 m /s 20.7 m
60 sec
0.8
12678.75 W 13 kW
mghP
(Answer B)
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Open Channel Flow
Problem 66 Solution:
Uniform flow: the flow cross section does not vary with time at
any location along an open channel.
Steady flow: the flow quantity does not vary with time at any
location along an open channel.
(Answer A)
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Pipe Head Loss
Problem 67 Solution:
From the given information, we obtain some known values:
10p (reservoir is at atmospheric pressure)
20p (pipe outlet is at atmospheric pressure)
10v (the water in reservoir doesn’t have velocity)
3
2 2
5 m /s25.5 m/s
0.25 0.5 mv
(flow rate is given)
1220 mz (given in the problem statement)
2180 mz (given in the problem statement)
Using energy equation:
2 2
1 1 2 21 2
2
2 2
25.50 0 221 0 168
2 9.8
20 m
f
f
f
p v p vz z h
g g g g
h
h
(Answer A)
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Pipe Head Loss
Problem 68 Solution:
Specific roughness of cast iron = e =0.23 mm
Relative roughness = 0.23
0.000657350
e
D
Reynolds number is determined by using the following equation.
2
6
6
1.34 0.25 0.35 0.35Re 4.875 10
10
vD
After that, the friction factor can be obtained from the Moody
diagram on the next page: 0.0185f
Pipe head loss (Darcy’s equation):
23
22
2
1.34 m /s0.0185 150 m
0.25 0.35 m78.5 m
2 2 0.35 m 9.8 m/sf
fLvh
Dg
(Answer B)
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Pumping Power
Problem 69 Solution:
From Problem 68 solution, the head loss (f
h ) is 78.5 m.
Total head by the pump:
watertank175 m 150 m 78.5 m 103.5 m
lake fh h h h
Pump power:
21340 kg/sec 9.8 m /s 103.5 m
0.85
1599014 W 1600 kW
mghP
(Answer D)
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Minimum Sewage Flow
Problem 70 Solution:
From the sewage flow ratio curves above, we can find the ratio of
minimum-to-average daily sewage flow for population of 8000
people follows curve A2:
0.2
min
0.2
min
3
min
5
8
4000 5
1213 m /day
avg
Q P
Q
Q
Q
(Answer A)
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Peak Sewage Flow
Problem 71 Solution:
From the sewage flow ratio curves above, we can find that the
ratio of peak-to-average daily sewage flow for a population of
8000 people follows the curve G:
peak
peak
3
peak
18
4
18 8
4000 4 8
12201 m /day
avg
Q P
Q P
Q
Q
(Answer C)
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Sewer Capacity
Problem 72 Solution:
Because the sewer is full of water flow during heavy rainfall, it can
be assumed that the wetted perimeter is equal to the pipe
perimeter. Therefore, the hydraulic radius is:
20.25 0.90.225 m
0.9
AR
P
Sewer slope:
1.50.025
60inlet outlet
z zS
L
Flow velocity using Manning’s equation:
2 1 2 13 2 3 2
1 10.225 0.025 4.87 m/s
0.012v R S
n
Flow capacity:
2 34.87 m/s 0.25 0.9 m 3.1 m /sQ vA
(Answer C)
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Static Determinacy (Truss)
Problem 73 Solution:
Static determinacy categorizes structures into:
m + r = 2j → statically determinate
m + r > 2j → statically indeterminate
m + r < 2j → unstable
where m is number of members, r is number of reactions, j is
number of joints.
For this truss, we have:
m = 18 r = 3 j = 10
Check:
(18 + 3) > (2 x 10)
The structure is stable and statically indeterminate.
(Answer B)
18
17
16 14
15
13
12 11
10
9
8
7
6
5 3
2
4
1
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Shear Force
Problem 74 Solution:
Before we can answer the largest shear force, we must obtain the
support reaction at support A and C.
0
20 kN 5 m 10 m 4 kN/m 5 m 12.5 m 0
35 kN
A
c
c
M
V
V
0
20 kN 4 kN/m 5 m 0
40 kN
35 kN 40 kN
5 kN
y
A c
A c
A
A
F
V V
V V
V
V
Shear Force Diagram (unit: kN)
As seen in the shear force diagram, the largest magnitude of the
shear force is 20 kN.
(Answer B)
5
0 0
20
15 15
5 +
-
+
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Bending Moment
Problem 75 Solution:
Bending Moment Diagram (unit: kN-m)
As seen in the bending moment diagram, the largest magnitude of
the bending moment is 50 kN-m.
(Answer C)
0 0
50
25
+
-
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Effective Width of T-Beam
Problem 76 Solution:
The effective width of T-beam is the minimum of:
center-to-center
1 1600 150 cm
4 4
16 20 16 12 212 cm
600 cmw s
L
b t
L
(Answer C)
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Bending Design
Problem 77 Solution:
Equilibrium between concrete compression block & tension rebars:
'0.85
0.85 3 ksi 10 in 60 ksi
2.35
c
c s y
s
s
C T
f ab A f
a A
a A
Find the required steel rebars:
2
2
0.9 0.5
110 kips-ft 12 in/ft 0.9 60 ksi 20 in 2.5 in 0.5 2.35
1467 kips-in 1050 70.5
1.56 in
u n s y
s s
s s
s
M M A f d a
A A
A A
A
Check minimum and maximum rebar (use ACI equations):
2
min ,min min
2
max ,max max
0.0033 0.5775 in
0.0135 2.3625 in
s
s
A bd
A bd
Required steel rebars area is in between minimum and maximum
rebar, so 1.56 in2 is used for steel rebars area.
Required number of #6 rebar =
2
2
1.56 in3.53 4
60.25 in
8
(Answer C)
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Shear Design
Problem 78 Solution:
Shear strength of RC beam is obtained from concrete and stirrups.
Contribution from concrete:
'2
2 1 3000 psi 10 in 17.5 in
19170 lb 19.17 kips
c c wV f b d
Contribution from stirrups:
40 kips19.17 kips 34.2 kips
0.75u
s c
VV V
Required spacing of stirrups (use #3 rebar):
2
32 0.25 in 60 ksi 17.5 in
86.78 6 in
34.2 kips
v y
s
A f ds
V
Check with maximum spacing:
2
max
17.5 in8.25 in
2 2
32 0.25 in 60000 psi
826.5 in
50 psi 10 in
v y
w
d
s
A f
b
Stirrups use #3 @ 6”
(Answer B)
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Basic Concept of Design
Problem 79 Solution:
(X) I. ASD is newer than LRFD
(V) II. ACI 318 is based on the LRFD concept
(V) III. LRFD uses factored load combinations while ASD doesn’t (X) IV. ASD uses ultimate strength while LRFD doesn’t
(V) V. ASD uses factor of safety while LRFD doesn’t
(Answer B)
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Steel Beam Design
Problem 80 Solution:
Check for compactness:
0.38 8.13 9.15 (OK, compact flange)2
f
f y
b E
t F
3.76 53.9 90.55 (OK, compact web)w y
h E
t F
W16X36 is compact, so that the nominal moment is:
250 ksi 64 in 3200 kips-in 266.7 kips-ftn p y x
M M F Z
Allowable moment using LRFD concept:
0.9 266.7 kips-ft 240 kips-ftb nM
Allowable maximum distributed load:
2
2
1
8
1240 kips-ft 24 ft
8
3.33 kips/ft
b n u u
u
u
M M w L
w
w
(Answer C)
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Soil Properties
Problem 81 Solution:
The sample of soil can be divided into three components: dry soil,
water, and air.
Dry soil properties :
3
3
13.4 gr6.23 cm
2.15 1 gr/cms
s
s w
mV
G
Volume of the void can be defined as the summation between
volume of water and air. Therefore,
310 6.23 3.77 cmv total s
V V V
Void ratio is calculated by using this following equation:
3.770.6
6.23v
s
Ve
V
(Answer D)
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Soil Properties
Problem 82 Solution:
Assume that the total volume of saturated soil is 1 m3.
1s w
V V
Substitute the volume relationship to this equation:
3
2250
2250
2.6 1000 1 1000 2250
2600 1000 1000 2250
0.78 m
s w
s s w w w
s s
s s
s
m m
G V V
V V
V V
V
After obtaining volume of soil, we can calculate dry unit mass
(mass of soil).
2.6 0.78 1000
2031.25 kg
s s s wm G V
(Answer B)
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Active Lateral Pressure
Problem 83 Solution:
Total lateral pressure is the combination between active earth
pressure and pore water / hydrostatic pressure.
At the bottom of retaining wall = depth of 6 m
2 22 25tan 45 tan 45 0.4059
2 2a
K
Active earth pressure at depth = 6 m:
1 1 2 2
2
0.4059 19 3 20 9.8 3
35.56 kN/m
a a wK z z
Pore water / hydrostatic pressure at depth = 6 m:
2
29.8 3 29.4 kN/m
w wz
Total lateral pressure at depth = 6 m:
2
total35.56 29.4 64.96 kN/m
a w
(Answer D)
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Bearing Capacity
Problem 84 Solution:
Using Terzaghi’s equation for a square footing:
0.5
0 125 pcf 4 ft 22.5 0.5 125 pcf 5 ft 19.7 0.85
16482.8125 psf
ult c c f qq cN S D N BN S
Using safety factor of 3:
16482.8125 psf
3
5494 5500 psf
ult
all
FS
(Answer C)
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Consolidation Rate
Problem 85 Solution:
Permeability:
An increase in permeability of the consolidating soil would
lead to an increase in the rate of seepage flow, other factors
remaining constant. With the greater rate of expulsion of
water from the soil the pore pressures will dissipate more
rapidly. This means that a more rapid rate of consolidation
occurs.
Layer thickness:
An increase in the layer thickness leads to a decrease in the
total head gradient during the stage of pore water expulsion.
It also means an increase in the volume of water to be
expelled and both of these effects lead to a lower rate of
consolidation.
Compressibility:
A greater compressibility leads to a greater decrease in the
void space of the soil for a particular stress change. This
means that a greater volume of water must be expelled from
the soil and this will require a longer time. Consequently, a
lower rate of consolidation will result.
All three factors affect the consolidation rate.
(Answer D)
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Vertical Curve
Problem 86 Solution:
12 sta
sta 76 00 sta 70 002 2
LPVC PVI
1
1 2
0.02 12 sta6.32 sta
0.02 0.018m
G Lx
G G
low point sta 70 00 6.32 sta sta 76.32m
PVC x
(Answer C)
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Vertical Curve
Problem 87 Solution:
2
2 1
lowpoint 1
2
2
0.018 0.02 6.32 sta 500 0.02 6 sta 0.02 6.32 sta
2 12 sta
505.68 506 m
m
PVC m
G G xelev elev G x
L
(Answer D)
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Driver Perception-Reaction Time
Problem 88 Solution:
AASHTO’s Green Book Chapter 3 lists 2.5 sec as the value used to
determine the minimum stopping sight distances, and is
appropriate for approximately 90% of the population.
(Answer C)
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Traffic Capacity
Problem 89 Solution:
By combining two equations given in the above information, we
can obtain a new equation about the relationship between the
traffic volume and traffic density.
2
80 0.5
80 0.5
q kv
k k
k k
According to mathematics principle, we can get the maximum
value of q when the first derivation of the equation is equal to 0.
2
max
max
2
max max max
80 0.5
0
80 0
80 veh/km
80 0.5 3200 veh/hr
q k k
dq
dk
k
k
q k k
(Answer B)
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Horizontal Curve
Problem 90 Solution:
Find the central angle:
tan2
125 600tan2
23.53
T R
Find the length:
600 23.53246 m
180 180
RL
Determine the stationing:
sta 10 000 125 sta 9 875PC PI T
sta 9 875 256 sta 10 121PT PC L
(Answer C)
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BOD Analysis
Problem 91 Solution:
Substitute the five-day values into the BOD equation:
0.13 5
1
234 1
489.6 490 mg/L
kt
ty L e
L e
L
(Answer C)
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BOD Analysis
Problem 92 Solution:
From the previous solution, the ultimate BOD was obtained.
Now, substitute the seven-day values into the BOD equation:
0.13 7
1
489.6 1
292.5 290 mg/L
kt
ty L e
e
(Answer A)
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Reactor Capacity
Problem 93 Solution:
Solids residence time:
V
Q
Suspended solids concentration:
0
0
0
3 3
1
1
1
12 0.35 0.55 258 6.4
3600 1 0.06 12
0.094 m /s 8110 m /day
d
c
d
d c
d
c
d
d c
d
c
d
d c
Y S SX
K
QY S SX
V K
QY S SV
X K
(Answer C)
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Dissolved Oxygen
Problem 94 Solution:
The dissolved oxygen (DO) at atmospheric pressure (760 mmHg)
is obtained by using linear interpolation of data from the table.
24.6
24.6
24.6
8.9 24.6 24
8.6 8.9 25 24
8.9 0.18
8.72 mg/L
C
C
C
DO
DO
DO
Oxygen is only slightly soluble in water and does not react with
water. Therefore, Henry’s law is applicable for this case, and the
solubility of oxygen is directly proportional to its partial pressure.
740 6.2% saturation 100% 69.2% 70%
760 8.72
(Answer C)
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Solid Waste
Problem 95 Solution:
Monthly mass of solid waste (for 18,400 people in a month):
3 3125 kg/m 20,000 m 2,500,000 kg
Average mass of solid waste per person per day:
2,500,000 kg4.5 kg/person.day
18,400 people 30 days
(Answer D)
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Waste Water
Problem 96 Solution:
Assume : 1 mg/L = 1 ppm
1 gal = 8.34 lbm
6
6
28 parts gal lbmMass loading 35 10 8.34
day gal10 part
lbm 8173.2
day
(Answer B)
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Reaction Rate Constant
Problem 97 Solution:
5 0
5
5
1
1
225 486 1
2251
486
225ln 1 5
486
225ln 1
4860.124
5
kt
k
k
BOD L e
e
e
k
k d
(Answer C)
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Critical Path
Problem 98 Solution:
Critical Path is the longest path to complete a project.
(Answer C)
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Project Scheduling Tool
Problem 99 Solution:
In the figure, we could see the activity (labeled as number) is on
node. Therefore, the figure above represents AON (Activity-On-
Node) diagram.
(Answer B)
1 3
2
2 4
2
5
2
6
2
7
2
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Critical Path
Problem 100 Solution:
Critical path is defined as the longest path to complete a project.
From the AON diagram, we could determine the total duration of
sequences for each path.
START – A – C – F – FINISH : 3 + 2 + 3 = 8 days
START – A – D – F – FINISH : 3 + 3 + 3 = 9 days
START – B – D – F – FINISH : 5 + 3 + 3 = 11 days
START – B – E – F – FINISH : 5 + 4 = 9 days
Path START – B – D – F – FINISH has the longest duration so it’s
the critical path of this project.
(Answer C)
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Scheduling
Problem 101 Solution:
Activity F is dependent to two activities: C and D. Therefore, the
EST (Earliest Start Time) of activity F can be determined as the
maximum EFT of those two activities.
Activity Duration (days)
EST (day no.)
EFT (day no.)
LST (day no.)
LFT (day no.)
TF (days)
A 3 0 3 3 6 3
B 5 0 5 0 5 0
C 2 3 5 6 8 3
D 3 5 8 5 8 0
E 4 5 9 7 11 2
F 3 8 11 8 11 0
From the above table, we can conclude that the EST of activity F
starts at Day 9, after the EFT of activity D ends at Day 8.
(Answer B)
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Scheduling
Problem 102 Solution:
Activity D is along the critical path (see the Problem 100 Solution),
so that the float time (TF) of activity D is 0 day. It means that
activity D has no spare time to delay.
(Answer A)
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Construction Contract
Problem 103 Solution:
Lump sum contract:
Contractor bids a total price for all work in a project, including
the profit.
Requires contract modification if there are any changes.
Unit price contract:
Contractor bids a price for each work item by cost per unit.
The actual quantities will be used for the final payment.
(Answer D)
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Contract Bond
Problem 104 Solution:
Contractors are required to obtain bonds from a surety company
prior to submitting bid. The types of bond can be seen below:
Bid bond guarantees the contractor will enter into a contract with
the client if selected – typically 5% to 20% of the estimated
project cost.
Performance bond guarantees the contractor will perform the
specified work in accordance with contract – typically full value of
the project.
Payment bond guarantees the contractor will pay for all
materials and labor used on the project – typically full value of the
contract. This is usually used to protect clients from being sued for
payment by subcontractors.
(Answer B)
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Construction Safety
Problem 105 Solution:
OSHA 1926.651(c) about Specific excavation requirement:
(2) Means of egress from trench excavations.
A stairway, ladder, ramp or other safe means of egress shall
be located in trench excavations that are 4 feet (1.22 m) or
more in depth so as to require no more than 25 feet (7.62 m)
of lateral travel for employees.
(Answer C)
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Construction Estimating
Problem 106 Solution:
Gross wall area:
272 ft 16 ft 1152 ftg
A
Net wall area (without opening):
2 2 21152 ft 250 ft 902 ftn g opening
A A A
Bricks required:
2
2
602 brick902 ft 1.05 5702 bricks
100 ftbrick
n
(Answer D)
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Earthwork - Fill
Problem 107 Solution:
As given in problem statement, cut and fill area can be assumed
as triangular in shape for transition area. While they are assumed
as trapezoidal area for other parts.
Station
(m)
Fill area
(m2)
Fill volume
(m3)
10+00 153.42
153.42 20.56
4.50 391.462
10+4.50 20.56
20.56 8.25
7.85 113.082
10+12.35 8.25
8.25
4.05 11.143
10+16.40 -
-
10+20 -
total 515.68
Total volume of fill work is 515.68 m3.
(Answer D)
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Earthwork - Cut
Problem 108 Solution:
Station
(m)
Cut area
(m2)
Cut volume
(m3)
10+00 -
-
10+4.50 -
13.67
7.85 35.773
10+12.35 13.67
13.67 52.84
4.05 134.682
10+16.40 52.84
52.84 165.14
3.6 392.362
10+20 165.14
total 562.81
Total volume of cut work is 562.81 m3.
(Answer B)
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Borrow Pit Volumes
Problem 109 Solution:
It’s easier to solve borrow pit volumes using table.
Point Elevation
(m)
Elev to be
Cut (m)
No. of
Rectangles
Height of
Cut (m)
A1 103.5 3.5 1 3.5
A2 105.6 5.6 2 11.2
A3 102.1 2.1 2 4.2
A4 101.9 1.9 1 1.9
B1 103.9 3.9 2 7.8
B2 104.0 4.0 4 16
B3 102.8 2.8 3 8.4
B4 103.7 3.7 1 3.7
C1 105.2 5.2 1 5.2
C2 104.3 4.3 2 8.6
C3 100.3 0.3 1 0.3
Total = 70.8
3
,
12 m 8 m70.8 m 1699.2 1700 m
4 4i jn
AV h
(Answer B)
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Earthwork Terms
Problem 110 Solution:
Swell: the increase in volume of earth from its natural to
loose state. When earth is excavated, it increases in
volume because of an increase in voids.
Shrinkage: the decrease in volume of earth from its natural to
compacted state.
(Answer C)
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SCORE SHEET
Correct Answers: ___________
Percentage: (correct answers)/110 = ____/110 = ______
Congratulations! You’ve finished your practice exam! How did you
do? Are you happy with the results or are there areas that you need
to improve?
If you didn’t do as well as you had hoped don’t give up. Just keep on
practicing more problems and you’ll get there. Practicing a ton of
problems is the key. You’ve either been doing that all throughout your
schooling or you are having to do it to study for this exam. Either
way, you have to practice. You’ll have to do the same thing when the
PE comes around too!
I hope this challenged you and helped to assess where you stand.
Good luck on the FE exam and on your engineering future!
Helpful Tools:
We have built www.civilengineeringacademy.com to help any civil
engineer take and pass the FE and PE exams. We have tons of free
video practice problems there to get you going. We also have plenty
of tips, must have resources, advice on courses, and more. We’ve
even created our own FE review course that can guide you step-by-
step through the entire exam. You can check that out at
www.civilfereviewcourse.com.
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