THE ULTIMATE
CIVIL FE PRACTICE EXAM
A COMPLETE 110 QUESTION PRACTICE EXAM
FOR THE FE - BUILT TO HELP YOU PASS!
Volume 1
Isaac Oakeson, P.E.
THE ULTIMATE CIVIL
FE PRACTICE EXAM
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
TABLE OF CONTENTS
Page 3
WELCOME…………………………………………….
EXAM SPECIFICATION……………………
Page 6
START TEST ………………………………...
Page 13
SOLUTIONS……………………………………… Page 112
SCORE SHEET…............................ Page 225
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 3
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.
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 4
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.
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 5
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 (my team and I are human of course), or
you just want to comment on things, then please let me know! I can
be reached through my site at civilengineeringacademy.com or by
email at isaac@civilengineeringacademy.com.
ABOUT ME
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 civilengineeringacademy.com and
civilpereviewcourse.com to help future students pass. His entire
goal has been to provide the best resources for engineers to study and
pass the FE and PE exam. He recently created a FE review course as
well. Find it at civilfereviewcourse.com!
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 6
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
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 7
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 8
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,
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 9
watersheds)
B. Basic hydraulics (e.g., Manning equation, Bernoulli theorem, openchannel
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 10
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
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 11
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
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 12
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 13
START TEST
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 14
1.
a)
b)
c)
d)
What is the length of a straight line connecting two points in
three-dimensional space: A(2, 1, 7) and B(-1, 3, 3)?
20
29
39
21
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 15
2.
Solve for the roots of the following polynomial equation.
x3 − 7x + 6 = 0
a)
b)
c)
d)
1, 2, 3
-1, 2, -3
1, -2, 3
1, 2, -3
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 16
3.
Determine the coordinates of a circle’s center which satisfy:
x 2 + y 2 − 10x + 14y + 49 = 0
a)
b)
c)
d)
(-10, 14)
(10, -14)
(-5, 7)
(5, -7)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 17
4.
A straight line goes through a point (1, 7) and is perpendicular to
the line 5y + 3x − 16 = 0 . Solve for the equation of the line.
a)
3y − 5x − 16 = 0
3y + 5x − 16 = 0
−3y + 5x − 16 = 0
3y − 5x − 26 = 0
b)
c)
d)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 18
5.
Which of the following equations described a circle having center
at point (2, -5) and passing through the point (8, 3)?
a)
x 2 + y 2 − 4x + 10y + 29 = 0
b)
x 2 + y 2 − 4x + 10y − 79 = 0
c)
x 2 + y 2 − 4x + 10y − 71 = 0
d)
x 2 + y 2 − 4x + 10y − 100 = 0
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 19
6.
A particle moves in the x-y plane with the following equation path:
x = 9 sin t
y = 2cos t
Solve for the equation of the path of the particle.
a)
81x 2 + 4y 2 = 324
b)
4 x 2 + 81y 2 = 324
c)
2x 2 + 9y 2 = 18
d)
9x 2 + 2y 2 = 18
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 20
7.
What is the cross product of these two vectors A = 8i + j − 2k and
B = 3i − j + 3k ?
a)
i − 30 j − 11k
b)
i + 30 j − 11k
c)
i + 18 j + 11k
d)
−i − 18 j + 11k
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 21
8.
Solve for the angle made by these two vectors:
M =i− j+k
N = −i − j + k
a)
b)
c)
d)
85°
80°
75°
70°
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 22
9.
Solve for the inflection point of the following polynomial function:
a)
b)
c)
d)
(-5, 125)
(5, -125)
(19, 1)
(-1, -19)
f ( x ) = x 4 − 12x 3 + 30x 2
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 23
10. Which expression below is equivalent to 1?
a)
b)
c)
d)
2 sin cos
sin2 + cos2
1 − cos 2
2
cos2 − sin2
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 24
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)
b)
c)
d)
0.07
0.14
0.20
0.34
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 25
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)
b)
c)
d)
1
2
1
3
1
6
1
12
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 26
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)
b)
c)
d)
0.004
0.093
0.133
0.064
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 27
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)
b)
c)
d)
3000.4
3000.4
2997.8
2997.8
;
;
;
;
79.6
71.6
87.1
78.4
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 28
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)
b)
c)
d)
0.3
0.05
0.7
0.9
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 29
16. Which of the following statements is the function to create a
spicific number of loops?
a)
b)
c)
d)
If, else, end
For, end
While, do
If, elseif, end
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 30
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)
b)
c)
d)
3
4
5
6
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 31
18. Which of the following is the benefit of behaving ethically?
a)
b)
c)
d)
Better reputation
Better feeling about yourself
Higher salary
None of the above
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 32
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)
b)
c)
d)
Accept the Rolex watch and accept the bid
Accept the Rolex watch and reject the bid
Reject the Rolex watch and accept the bid
Reject the Rolex watch and reject the bid
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 33
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)
b)
c)
d)
His contract prohibits misuse and misappropriation of office’s
equipment
He may wear out the office’s materials
He may encourage other employees to do side jobs in the office
He can undercut his fee because he has a lower overhead
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 34
21. You are aware that a registered engineer intentionally violates a
state’s rule of professional conduct. What should you do?
a)
b)
c)
d)
Do nothing
Report the violation to the employer
Report the violation to the affected parties
Report the violation to the state’s engineering registration board
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 35
22. Which of the followings is the best description of ‘plan stamping’?
a)
b)
c)
d)
Legal action of signing off on a project you won on tender
Legal action of signing off on a project you accepted money for
Illegal action of signing off on a project you didn’t design but did
check
Illegal action of signing off on a project you neither designed nor
checked
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 36
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)
b)
c)
d)
$460
$490
$520
$550
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 37
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)
b)
c)
d)
$6,500
$7,350
$8,000
$8,650
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 38
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)
b)
c)
d)
$2,100
$3,950
$4,200
$31,500
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 39
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)
b)
c)
d)
4050
5050
6700
10300
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 40
27. What is the annual effective interest rate of money that is invested
at 2.5% per year and compounded quarterly?
a)
b)
c)
d)
2.50%
2.51%
2.53%
2.55%
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 41
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)
b)
c)
d)
$7,500
$8,000
$8,333
$9,170
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 42
29. Which of the following loads could be resisted by a fixed support?
I. Moment
II. Shear
III. Axial
a)
b)
c)
d)
I only
I and II
II and III
I, II, and III
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 43
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?
8
T1
6
T2
2500 lb
a)
T1 = 3125 lb ; T2 = 1875 lb
b)
T1 = 2500 lb ; T2 = 2500 lb
c)
T1 = 2725 lb ; T2 = 2225 lb
d)
T1 = 2925 lb ; T2 = 1725 lb
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 44
Problems 31 and 32 refer to the following truss.
1200 lb
E
G
C
A
H
F
D
8 ft
8
ft
3
8
ft
3
8 ft
8 ft
B
8 ft
RA
RB
31. Solve for the support reaction at joint A and B.
a)
RA = 600 lb ( ) ; RB = 600 lb ()
b)
RA = 700 lb ( ) ; RB = 500 lb ( )
c)
RA = 800 lb ( ) ; RB = 400 lb ()
d)
RA = 900 lb ( ) ; RB = 300 lb ()
32. Find the axial forces in members CD and EG.
a)
TCD = 600 lb ( tension) ; TEG = 600 lb ( compression)
b)
TCD = 1200 lb ( compression) ; TEG = 900 lb ( tension)
c)
TCD = 0 ; TEG = 950 lb ( compression)
d)
TCD = 0 ; TEG = 900 lb ( compression)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 45
33. Find the horizontal reaction at support C.
2 ton
B
4 ft
C
A
3 ft
a)
b)
c)
d)
0.4
0.6
1.2
1.6
3 ft
6 ft
ton
ton
ton
ton
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
2 ft
Page 46
Problems 34 and 35 refer to the following shape.
Y
3
2
1
0
2
6
8
X
34.
a)
b)
c)
d)
Solve for the coordinate of the centroid.
(4 cm, 1.5 cm)
(6 cm, 1.342 cm)
(4 cm, 1.342 cm)
(4 cm, 1.286 cm)
35.
a)
b)
c)
d)
Find the moment of inertia about the centroid x-axis of the shape.
3.66 cm4
9.66 cm4
12.5 cm4
16.16 cm4
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 47
36. Solve for the tension force in cable BC.
D
A
C
B
150 lb
150 lb
a)
b)
c)
d)
75 lb
100 lb
150 lb
300 lb
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 48
37. A book shelf weighs 400 N in total. It is supported by hinges A and
B on the wall. Solve for the horizontal reaction at hinge B.
A
5m
400 N
2.5 m
B
3m
a)
b)
c)
d)
120
150
180
240
N
N
N
N
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 49
38. Calculate the centroidal polar moment of inertia with respect to
the z-axis of W14X176 with the following data:
a)
b)
c)
d)
1302
1489
2140
2978
A = 51.8 in2 tw = 0.830 in
Iy = 838 in4
d = 15.2 in
tf = 1.31 in
Zx = 320 in3
bf = 15.7 in
Ix = 2140 in4
Zy = 163 in3
in4
in4
in4
in4
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 50
39. A particle moves by the velocity function : v (t ) = 5t + 3 (m/s).
Calculate the distance traveled by the particle from 2 to 6
seconds.
a)
b)
c)
d)
52
72
82
92
m
m
m
m
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 51
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)
b)
c)
d)
4 Joule
6 Joule
8 Joule
10 Joule
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 52
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.
F
10°
a)
b)
c)
d)
35
40
45
50
N
N
N
N
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 53
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)
b)
c)
d)
60
70
80
90
m/s
m/s
m/s
m/s
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 54
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)
450 m/s
hmax
v'
a)
b)
c)
d)
35
40
45
50
cm
cm
cm
cm
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 55
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.
5m
a)
b)
c)
d)
50
60
70
80
cm
cm
cm
cm
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 56
45. The following figure shows a typical stress-strain relationship of
steel rebar. Which of the following is the strain hardening part?
Stress
D
A
a)
b)
c)
d)
B
E
C
Strain
A-B
B-C
C-D
D-E
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 57
Problems 46 and 47 refer to the simply-supported beam as follows:
5 kips
3 ft
3 ft
46. Which of the following diagrams is the correct shear force diagram
(unit: kips)?
2.5
2.5
+
a)
0
0
2.5
2.5
2.5
2.5
+
b)
0
0
-
2.5
2.5
2.5
c)
0
+
-
0
2.5
2.5
d)
0
+
-
2.5
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
0
Page 58
47. Which of the following diagrams is the correct bending moment
diagram (unit: kips-ft)?
7.5
a)
+
0
0
15
+
b)
0
c)
0
0
-
0
7.5
d)
0
0
15
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 59
48. Solve for the maximum shear stress of this two-dimensional
element.
40 MPa
60 MPa
100 MPa
100 MPa
60 MPa
40 MPa
a)
b)
c)
d)
60
64
67
70
MPa
MPa
MPa
MPa
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 60
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)
b)
c)
d)
0.0018 in
0.1495 in
0.00015 in
0.045 in
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 61
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)
b)
c)
d)
2
3
4
5
kN-m
kN-m
kN-m
kN-m
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 62
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)
b)
c)
d)
1.3 m
1.03 m
1.003 m
1.0003 m
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 63
52. For a simply supported beam, where does the maximum deflection
usually occur due to gravity loading?
a)
b)
c)
d)
At
At
At
At
the
the
the
the
top fibers
bottom fibers
support
mid-span
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 64
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.
P=?
30 cm
fixed
1.5 m
a)
b)
c)
d)
1 ton
1.5 ton
2 ton
2.5 ton
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 65
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:
w = 3 lb/ft
C
2 ft
a)
b)
c)
d)
0.1
0.2
0.3
0.4
8 ft
in
in
in
in
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 66
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.
8m
Ix = 11500 cm4
Iy = 3880 cm4
a)
b)
c)
d)
2,500
5,000
7,500
8,500
kN
kN
kN
kN
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 67
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)
b)
c)
d)
Sieve analysis
Relative density
Absorption
All of the above
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 68
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)
b)
c)
d)
17 MPa
170 MPa
1700 MPa
6000 MPa
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 69
58. Which of the following materials is tested for stability through
Marshall test?
a)
b)
c)
d)
Concrete
Steel
Wood
Asphalt
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 70
59. Which of the following checmical admixtures is used to increase
the workability of concrete?
a)
b)
c)
d)
Accelerator
Retarder
Plasticizer / water reducer
Air entraining agent
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 71
60. Which of the following does affect the kinematic viscocity of a
fluid?
a)
b)
c)
d)
I. Density
II. Surface tension
III. Stress
IV. Absolute viscocity
I and II
II and III
III and IV
I and IV
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 72
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)
b)
c)
d)
5 kPa
10 kPa
45 kPa
50 kPa
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 73
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)
b)
c)
d)
0.2
0.3
0.4
0.5
kPa
kPa
kPa
kPa
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 74
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)
b)
c)
d)
1.73
3.71
1.37
1.37
x
x
x
x
103
103
105
105
,
,
,
,
laminar
critical
critical
turbulent
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 75
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)
b)
c)
d)
2.7 m
10 m
27 m
162 m
65. How much power is needed to deliver to the pump?
a)
b)
c)
d)
11
13
14
16
kW
kW
kW
kW
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 76
66. Which of the flows has cross section that does not vary with time
at any location along an open channel?
a)
b)
c)
d)
Uniform flow
Non-uniform flow
Steady flow
Critical flow
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 77
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)
b)
c)
d)
20
33
46
53
m
m
m
m
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 78
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 0.23 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)
b)
c)
d)
55 m
75 m
95 m
115 m
69. Find the required power to pump the water from the lake to water
tank.
a)
b)
c)
d)
1300
1400
1500
1600
kW
kW
kW
kW
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 79
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)
b)
c)
d)
1200
1500
2000
2200
m3/day
m3/day
m3/day
m3/day
71. What is the most nearly peak sewage flow for this apartment
complex?
a)
b)
c)
d)
8400 m3/day
10500 m3/day
12200 m3/day
15250 m3/day
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 80
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)
b)
c)
d)
1
2
3
4
m3/s
m3/s
m3/s
m3/s
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 81
73. What is the static determinacy of the following truss?
a)
b)
c)
d)
Stable, statically determinate
Stable, statically indeterminate
Unstable
None of the above
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 82
Problems 74 and 75 refer to the following beam.
P = 20 kN
B
A
5m
w = 4 kN/m
D
C
5m
5m
74. What is the largest magnitude of the shear force in the beam?
a)
b)
c)
d)
15
20
35
40
kN
kN
kN
kN
75. What is the largest magnitude of the bending moment in the
beam?
a)
b)
c)
d)
20 kN-m
25 kN-m
50 kN-m
100 kN-m
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 83
76. Based on ACI 318, calculate the top flange effective width of the
following T-beam.
10 cm
35
cm
20 cm
6m
a)
b)
c)
d)
20 cm
40 cm
150 cm
200 cm
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 84
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)
b)
c)
d)
2
3
4
5
78. Solve for the required spacing of #3 stirrups to resist the shear
force.
a)
b)
c)
d)
4”
6”
8”
10”
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 85
79. Which
I.
II.
III.
IV.
V.
a)
b)
c)
d)
of the following concepts about ASD and LRFD are correct?
ASD is newer than LRFD
ACI 318 is based on the LRFD concept
LRFD uses factored load combinations while ASD doesn’t
ASD uses ultimate strength while LRFD doesn’t
ASD uses a factor of safety while LRFD doesn’t
I, II, IV
II, III, V
II, IV, V
II, III, IV
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 86
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.
Wu
24 ft
What is the maximum ultimate distributed loading which can be
resisted by the beam? Assume that the beam is laterally
supported.
Shape
W16x36
a)
b)
c)
d)
b (in)
6.99
h (in)
15.9
tf (in)
0.43
tw (in)
0.295
2 kips/ft
2.5 kips/ft
3 kips/ft
3.5 kips/ft
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Sx (in3)
56.5
Zx (in3)
64.0
Page 87
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)
b)
c)
d)
0.2
0.4
0.5
0.6
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 88
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)
b)
c)
d)
1950
2030
2120
2150
kg
kg
kg
kg
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 89
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.
1 = 19 kN/m3
Layer 1
1' = 33
3m
c1 = 0
Layer 2
1 = 20 kN/m3
1' = 25
c1 = 0
a)
b)
c)
d)
30
35
50
65
kN/m2
kN/m2
kN/m2
kN/m2
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
3m
Page 90
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
Nc = 37.2
c =0
N = 19.7
= 30
Nq = 22.5
Sc = 1.25
S = 0.85
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)
b)
c)
d)
2500 psf
4500 psf
5500 psf
16000 psf
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 91
85. Which of the following factors doesn’t affect the rate of
consolidation?
a)
b)
c)
d)
Permeability
Layer thickness
Compressibility
None of the above
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 92
Problems 86 and 87 are based on the following vertical curve.
L = 12 sta
PVC
-2%
+1.8%
EVC
PVI
Sta 76+00
Elev = 500 m
86. Determine the low point station for this vertical curve.
a)
b)
c)
d)
Sta
Sta
Sta
Sta
70+00
75+68
76+32
82+00
87. Determine the elevation of the low point.
a)
b)
c)
d)
500
502
504
506
m
m
m
m
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 93
88. Based on several studies, what is the perception-reaction time
AASHTO used to design Stopping Sight Distance?
a)
b)
c)
d)
1.5 sec
2 sec
2.5 sec
3 sec
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 94
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 v = 80 − 0.5k .
What is, most nearly, the maximum capacity of total traffic
volume of the road?
a)
b)
c)
d)
3000
3200
3400
3600
veh/hour
veh/hour
veh/hour
veh/hour
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 95
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)
b)
c)
d)
Sta
Sta
Sta
Sta
9+744
9+875
10+121
10+600
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 96
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)
b)
c)
d)
290 mg/L
468 mg/L
490 mg/L
1,170 mg/L
92. Find the BOD of the sample if the incubation period is one week.
a)
b)
c)
d)
290 mg/L
468 mg/L
490 mg/L
1,170 mg/L
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 97
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 ( cd ) is
12 days.
Y =
Vss,mg
BOD5,mg
= 0.55
The effluent BOD5 is 6.4 mg/L. MLVSS = 3600 mg/L. The
endogenous decay coefficient K d = 0.06d −1 . Determine the reactor
capacity per day.
a)
b)
c)
d)
3600
6400
8100
9200
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 98
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)
b)
c)
d)
50%
60%
70%
80%
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 99
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)
b)
c)
d)
1.1
1.8
3.2
4.5
kg
kg
kg
kg
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 100
96. Find the mass loading of a 35 MGD wastewater discharge with an
ultimate BOD of 28 mg/L.
a)
b)
c)
d)
8,000
8,200
8,400
8,600
lbm/day
lbm/day
lbm/day
lbm/day
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 101
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)
b)
c)
d)
0.025 d −1
0.1 d −1
0.125 d −1
0.175 d −1
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 102
98. Which of the following is the correct definition of a critical path in
a project?
a)
b)
c)
d)
The
The
The
The
sequences of tasks in a project
time unaccounted for in a project
longest path of sequential tasks through a project
extra time available in a project
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 103
99. Which project scheduling tool is illustrated in the figure below?
1
3
2
4
2
2
6
2
a)
b)
c)
d)
5
2
7
2
Bar (Gantt) chart
AON diagram
AOA diagram
PERT chart
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 104
Problems 100-102 refer to the following diagram.
C|2
F|3
A|3
D|3
Start
B|5
E|4
100.
a)
b)
c)
d)
Find the critical path of this diagram.
START – A – C – F – FINISH
START – A – D – F – FINISH
START – B – D – F – FINISH
START – B – E – F – FINISH
101.
a)
b)
c)
d)
Find the EST (Earliest Start Time) of activity F.
Day 8
Day 9
Day 10
Day 11
102.
a)
b)
c)
d)
Find the float time of activity D.
0 day
1 day
2 days
3 days
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Finish
Page 105
103. Which of the following statements is correct about construction
contracts?
a)
b)
c)
d)
In a lump sum contract, a contractor bids a price for each work
item by the cost per unit
A unit price contract is often used when the exact quantities in a
project are known
In a unit price contract, the contractor bids a total price for all
work in a project, including the profit
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
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 106
104. Which bond guarantees that the contractor will perform the
specified work in accordance with the contract (typically full value
of project)?
a)
b)
c)
d)
Bid bond
Performance bond
Payment bond
None of all the above
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 107
105. According to OSHA, any adequate exits in trench excavation must
be provided when the depth reaches:
a)
b)
c)
d)
0
2
4
5
ft (no such requirement)
ft
ft
ft
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 108
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)
b)
c)
d)
3810
3900
4000
5702
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 109
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)
100+00
10+4.50
10+12.35
10+16.40
10+20
Cut area (m2)
13.67
52.84
165.14
Fill area (m2)
153.42
32.56
8.25
-
107. Find the total approximate volume of fill work.
a)
b)
c)
d)
500
505
510
515
m3
m3
m3
m3
108. Find the total approximate volume of cut work.
a)
b)
c)
d)
540
560
580
600
m3
m3
m3
m3
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 110
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.
12 m
12 m
12 m
A
2
1
3
8m
B
8m
C
Point
A1
A2
A3
A4
B1
B2
B3
B4
C1
C2
C3
a)
b)
c)
d)
Elevation
(m)
103.5
105.6
102.1
101.9
103.9
104.0
102.8
103.7
105.2
104.3
100.3
480 m3
1700 m3
2500 m3
6800 m3
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
4
Page 111
110. What is the correct term for the increasing volume of soil when
earth is excavated?
a)
b)
c)
d)
Loose
Void
Swell
Shrinkage
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 112
PROBLEM SOLUTIONS
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 113
Vector Length
Problem 1 Solution:
Length of a straight line AB can be calculated as vector length:
AB =
( −1 − 2)
2
+ (3 − 1) + (3 − 7) = 29
2
2
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 114
Roots of Equation
Problem 2 Solution:
For third-degree polynomial, the number of roots will be three.
x 3 − 7x + 6 = 0
( x − 1) ( x 2 + x − 6 ) = 0
( x − 1)( x − 2)( x + 3) = 0
To satisfy the equation, the roots are 1, 2, -3.
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 115
Circle Center
Problem 3 Solution:
The general equation of a circle is:
( x − a)
2
+ ( y − b) = r 2
2
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:
x 2 + y 2 − 10x + 14y + 49 = 0
x 2 − 10x + 25 + y 2 + 14y + 49 = 25
( x − 5)
2
+ ( y + 7 ) = 52
2
The coordinate of its center is (5, -7).
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 116
Calculus
Problem 4 Solution:
The slopes of two lines, which are perpendicular, can be written as
follows:
m1 m2 = −1
Where m1 and m2 are the slopes of the first and second line,
respectively.
First line:
Second line:
5y + 3x − 16 = 0
3
m2 = −1
5
5
m2 =
3
−
5y = −3x + 16
y =−
m1 m2 = −1
3
16
3
x+
→ m1 = −
5
5
5
Solve for the equation of the second line:
m=
5
3
known point ( x, y ) = (1,7 )
Use the general equation of a straight line:
y = mx + b
5
7 = 1 + b
3
16
b=
3
The equation is y =
5
16
x+
or 3y − 5x − 16 = 0 .
3
3
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 117
Circle Equation
Problem 5 Solution:
The general equation of a circle is:
( x − a)
2
+ ( y − b) = r 2
2
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:
( x − a) + ( y − b ) = r 2
2
2
(8 − 2) + (3 − ( −5) ) = r 2
2
2
r = 100 = 10
The equation of the circle is:
( x − a) + ( y − b ) = r 2
2
2
( x − 2) + ( y − ( −5) ) = 102
2
2
x 2 − 4 x + 4 + y 2 + 10y + 25 = 100
x 2 + y 2 − 4 x + 10y − 71 = 0
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 118
Calculus
Problem 6 Solution:
Solve the two coordinate equations:
x = 9 sin t
y = 2 cos t
x
9
y
→ cos t =
2
→ sin t =
Use the trigonometry identity:
sin2 t + cos2 t = 1
2
2
x
y
9 +2 =1
2
4 x + 81y 2 = 324
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 119
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.
i
j
k
AB = 8
1
−2
3 −1
=i
1
−2
−1
3
3
− j
8 −2
3
3
+k
8
1
3 −1
= i − 30 j − 11k
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 120
Vector Analysis
Problem 8 Solution:
Use dot product to get the angle:
M N = M N cos
(1)( −1) + ( −1)( −1) + (1)(1) =
1=
( 3 )( 3 ) cos
cos =
2
12 + ( −1) + 12
( −1)
1
→ = 70.5 70
3
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
2
2
+ ( −1) + 12 cos
Page 121
Inflection Point
Problem 9 Solution:
Inflection point occurs when the second derivation of f ( x ) is 0.
f ( x ) = x 4 − 12x 3 + 30x 2
f ' ( x ) = 4 x 3 − 36 x 2 + 60 x
f " ( x ) = 12 x 2 − 72 x + 60
Solve for inflection point:
f " ( x ) = 12 x 2 − 72 x + 60 = 0
(
)
12 x 2 − 6 x + 5 = 0
12 ( x − 1)( x − 5 ) = 0
x1 = 1 and x2 = 5
x1 = 1 → f (1) = (1) − 12 (1) + 30 (1) = 19
4
3
2
x2 = 5 → f (5) = (5) − 12 (5) + 30 (5) = −125
4
3
2
The inflection points are (1, 19) and (5, -125).
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 122
Trigonometry
Problem 10 Solution:
This talks about trigonometry identities.
Option A → 2 sin cos = sin2
Option B → sin + cos = 1
2
Option C →
2
1 − cos 2
= sin2
2
Option D → cos − sin = cos2
2
2
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 123
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
5
P
+
= 0.143
=
black then red 15 14 15 14
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 124
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}.
P (E ) =
n (E )
n (S )
=
1
6
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 125
Probability
Problem 13 Solution:
Use binomial distribution to solve this problem:
p = 0.03
q = 1 − p = 0.97
p x =
p 2 =
n!
pxqn − x
x !( n − x ) !
2
15 − 2
15!
0.03) ( 0.97 )
= 0.064
(
2!(15 − 2 ) !
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 126
Mean and Standard Deviation
Problem 14 Solution:
The mean is determined by dividing the sum of data by the total
number of data.
(x
i
xi
1
2
3
4
5
6
7
8
9
10
Total
2995
3005
3001
2991
2984
3004
3009
3015
2998
3002
30004
i
−x
)
2
29.16
21.16
0.36
88.36
268.96
12.96
73.96
213.16
5.76
2.56
716.4
n
x =
x
i =1
n
i
=
30004
= 3000.4
10
Variance is defined as:
(
n
=
i =1
xi − x
n −1
)
2
=
716.4
= 79.6
10 − 1
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 127
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.
P ( zero ) = 0.7 0.7 0.7 0.7 0.7 0.7 = 0.118
P ( at least one ) = 1 − P ( zero ) = 1 − 0.118 = 0.882 0.9
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 128
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 129
Programming – Looping Result
Problem 17 Solution:
Looping condition can be solved by observing the variable data for
each loop.
Condition
Initialize
First looping x = 0
Second looping x = 2
Third looping x = 4
Fourth looping x = 6
x
y
0
1
2
2
4
3
6
4
looping stops
The last value of variable y is 4.
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 130
Ethics
Problem 18 Solution:
By doing ethical behavior, you will not get any benefits because
ethical behavior promises nothing.
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 131
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 132
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 133
Professional Societies
Problem 21 Solution:
A violation should be reported to the organization that has
promulgated the rule.
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 134
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 135
Loan Payment Analysis
Problem 23 Solution:
Use capital recovery discount factor:
A = P ( A | P , i%, n )
P = $30,000 − $8,000 = $22,000
i =
6%
= 0.5%
12 compounding periods per year
n = 4 years = 48 months
A = P ( A | P ,0.5%, 48 )
0.005 (1 + 0.005 )48
= ($22,000 )
(1 + 0.005)48 − 1
= $516.67 $520
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 136
Depreciation Value
Problem 24 Solution:
Depreciation value is equal to the difference between initial cost C
and salvage value Sn , then divided by the life span n .
D=
C − Sn $80,000 − $6,500
=
= $7,350
n
10
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 137
Future Value Analysis
Problem 25 Solution:
F = P ( F | P,5%,15) = ($2,000)(1 + 0.05)
15
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
= $4,157
Page 138
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 − n$60 + n$99 = 0
n$39 = $400,000
n = 10257 10300 tables/year
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 139
Effective Interest Rate Analysis
Problem 27 Solution:
m
4
r
0.025
i = 1 + − 1 = 1 +
− 1 = 2.5235%
m
4
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 140
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%.
( A | F ,0.83%,12 ) =
0.0083
(1 + 0.0083)
12
−1
= 0.079597
A = F ( A | F ,0.83%,12 ) = $100,000 0.079597 = $7,960 $8,000
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 141
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 142
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):
F
=0
y
T1 sin − 2500 = 0
T1 = 3125 lb
Equilibrium in horizontal forces (X direction):
F
x
=0
T1 cos − T2 = 0
T2 = 1875 lb
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 143
Truss Analysis
Problem 31 Solution:
To solve the support reaction, equilibrium should be satisfied.
M
A
=0
1200 lb 8 ft − RB 32 ft = 0
RB = 300 lb ( )
F
y
=0
RA + RB − 1200 lb = 0
RA + 300 lb − 1200 lb = 0
RA = 900 lb ( )
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 144
Truss Member Forces
Problem 32 Solution:
Member CD is zero-force member. Check the equilibrium of
vertical forces at joint D:
F
y
=0
TCD = 0
Member EG can be solved by ‘cutting’ the truss as shown below.
E
1200 lb
G
C
A
8 ft
8 ft
8 ft
H
F
D
B
8
ft
3
8
ft
3
8 ft
RB
RA
Check the equilibrium of system to get the support reaction at B:
M
A
=0
−RB 32 ft + 1200 lb 8 ft = 0
RB = 300 lb ( )
Check the equilibrium of moment at joint F by looking at ‘right
side’ part:
M
F
=0
−RB 16 ft − TEG
8
2
8
82 +
3
8
ft − TEG
3
8
3
2
8
82 +
3
8 ft = 0
TEG = −948.68 lb = 948.68 lb ( compression) Answer : C
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 145
Frame Analysis
Problem 33 Solution:
This is a statically determinate structure. We have four unknowns
( H A , VA , HC , VC ), so that we need four equilibrium equations to
solve this problem.
2 ton
B
4 ft
Hc
A
C
HA
2 ft
VC
6 ft
3 ft
3 ft
VA
Check equilibrium of horizontal & vertical forces in the system (ton
& ft):
F
x
F
=0
H A = HC
y
(1)
=0
VA + VC = 2
(2)
Check equilibrium of moment at joint A in the system (ton & ft):
M
A
=0
2 3 − HC 2 − Vc 12 = 0
HC + 6VC = 3
(3)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 146
Check the moment at pin B for the right part (ton & ft):
MB = 0
HC 4 − Vc 6 = 0
HC = 1.5VC
(4)
Substituting equation (4) to equation (3):
HC + 6VC = 3
(3)
1.5VC + 6VC = 3
VC = 0.4 ton ( )
Therefore, the horizontal reaction at support C can be calculated
using equation (4):
HC = 1.5VC
(4)
HC = 1.5 0.4 = 0.6 ton ( )
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 147
Centroids
Problem 34 Solution:
Y
3
II
2
1
I
2
0
Shape
I
II
III
IV
Total
III
IV
6
y (cm)
1
2.5
1
0.5
x (cm)
4/3
4
20/3
4
8
X
A (cm2)
3
4
3
4
14
4
20
3) + ( 4 )( 4 ) +
(
(3) + ( 4)( 4 )
xc ,n An 3
3
xc =
=
= 4 cm
14
A
yc =
y A
A
c ,n
n
=
(1)(3) + (2.5)( 4 ) + (1)(3) + (0.5)( 4 ) = 1.286 cm
14
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 148
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
Ic , x (cm4)
A (cm2)
d (cm)
Ad2 (cm4)
I
bh3 (2 )(3)
=
= 1.5
36
36
3
-0.286
0.2454
II
bh3 ( 4)(1)
=
= 0.33
12
12
4
1.214
5.8952
III
bh3 (2 )(3)
=
= 1.5
36
36
3
-0.286
0.2454
4
-0.786
2.4712
3
3
3
bh3 ( 4)(1)
=
= 0.33
12
12
3.66
3
IV
Total
Ic , x =
(I
c , x ,i
8.8571
)
+ Ai di 2 = 3.66 + 8.8571 = 12.52 cm4
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 149
Cable Tension Force
Problem 36 Solution:
TAB
TBC
120°
150 lb
Alternative 1:
Use the law of sines to solve the triangle at joint B.
TAB
150
=
→ TAB = 150 lb
sin120 sin120
Alternative 2:
Use equilibrium at joint B.
F
x
=0
TAB cos30 = TBC cos30
TAB = TBC
F
y
=0
TAB sin30 + TBC sin30 = 150
2TAB sin30 = 150
TAB = 150 lb
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 150
Static Analysis
Problem 37 Solution:
To find the horizontal reaction at support B, check the equilibrium
at support A. Pinned support cannot resist moment, so that
equilibrium of moment at support A is 0.
Assume the horizontal reaction at support A is to the right ( → ).
M
A
=0
−400 1.5 + HB 2.5 = 0
HB = 240 N ( →)
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 151
Centroidal Polar Moment of Inertia
Problem 38 Solution:
J = Ixx + Iyy = 2140 + 838 = 2978 in4
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 152
Kinematic Function
Problem 39 Solution:
Velocity is defined in terms of time, so that distance can be
written as follows:
s ( t ) = v ( t ) dt
6
=
(5t + 3) dt
2
6
5
= t 2 + 3t
2
2
= 92 m
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 153
Kinetic Energy
Problem 40 Solution:
The velocity at 2-m height:
Vt 2 = V02 − 2gh = 82 − 2 9.8 2 = 24.8
Vt = 24.8 = 4.98 m/s
The kinetic energy at 2-m height:
Ek =
1
1
mV 2 = 0.5 24.8 = 6.2 Joule
2
2
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 154
Friction Force
Problem 41 Solution:
Define the forces working to the moving box in the inclined plane,
then apply the equation.
F
mg sin10
10°
fk
fk = N = mg cos
F
= ma
F − mg sin − mg cos = ma
F − 12 9.8 sin10 − 0.15 12 9.8 cos10 = 12 0.5
F = 43.8 45 N
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 155
Impulse & Momentum
Problem 42 Solution:
I = P
F t = m (v2 − v1 )
60 0.2 = 0.15 (v2 − 10 )
v2 = 90 m/s
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 156
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.
pp + pb = pp' + pb'
mpv p = mpv p' + mbvb'
This case is a perfectly inelastic collision, because both projectile
and wooden block have the same motion after collision. The
'
'
'
velocity after collision ( v p = vb = v ) can be obtained as follows:
(
)
mpv p = mp + mb v '
0.01 450 = ( 0.01 + 1.49) v '
v ' = 3 m/s
Using the law of energy, solve the maximum height as follows:
1
mv 2 = m gh
2
1
32 = 10 h
2
h = 0.45 m = 45 cm
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 157
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.
Ei = Ef
1
k 2
2
1
2 9.8 5 = 500 2
2
= 0.62 m
mgh =
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 158
Stress-Strain Relationship
Problem 45 Solution:
Stress
D
B
A
E
C
Strain
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 159
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).
2.5
2.5
+
0
0
2.5
2.5
Shear Force Diagram (unit: kips)
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 160
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.
7.5
+
0
0
Bending Moment Diagram (unit: kips-ft)
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 161
Mohr’s Circle of Element Stresses
Problem 48 Solution:
max
(-100,60)
1
avg
2
(-40,-60)
avg =
x + y
2
=
−100 − 40
= −70 MPa
2
By looking at the Moh’r circle above, the maximum shear stress is
actually the radius of the Moh’r circle.
max =
( −100 + 70)
2
+ (60) = 67 MPa
2
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 162
Strain (Elastic Deformation)
Problem 49 Solution:
in
80 lb 25 ft 12
PL
ft
=
=
= 0.045 in
2
EA
1
3
10900 10 psi (0.25 in)
4
(
=
=
L
)
0.045 in
= 1.5 10−4
in
25 ft 12
ft
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 163
Torsion
Problem 50 Solution:
There are two limitations to find the maximum allowable torque:
1. Check the maximum allowable shear stress
Tr0
J
T 25
40 =
2500000
T = 4 106 N-mm = 4 kN-m
max =
2. Check the maximum allowable twist
max =
TL
GJ
T 2000
70000 2500000
T = 2187500 N-mm = 2.1875 kN-m
0.025 =
The maximum allowable torque is the minimum between those
two values, taken as 2 kN-m.
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 164
Thermal Deformation
Problem 51 Solution:
L = L0 T
(
)
= 8 10−6 C (1 m)(60C − 25C )
= 2.8 10−4 m
L = L0 + L
(
= (1 m) + 2.8 10−4 m
)
= 1.00028 m
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 165
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 166
Deflection of Cantilever Beam
Problem 53 Solution:
The quick formula of deflection in this case is:
=
PL3
3EI
The moment of inertia for this rectangular section is:
bh3 150 3003
I =
=
= 3.375 108 mm4
12
12
Therefore, the load is:
PL3
=
3EI
P 15003
8.5 =
3 23500 3.375 108
P = 59925 N 6 ton
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 167
Deflection of Simply-Supported Beam
Problem 54 Solution:
The quick formula of deflection in this case is:
=
wx
L3 − 2Lx 2 + x 3
24EI
(
)
Therefore, the load is:
=
=
wx
L3 − 2Lx 2 + x 3
24EI
3000 (2 12 )
(
)
((10 12)
24 10900000 1500
3
− 2 (10 12 )(2 12 ) + (2 12 )
= 0.3 in
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
2
3
)
Page 168
Steel Column Analysis
Problem 55 Solution:
Fixed-pinned → k = 0.7
Effective length for buckling in the weak direction:
L ' = kL = 0.7 8 m = 5.6 m
Critical buckling based on Euler’s formula:
Pcr =
=
2EI
( L ')
2
2 (2.1 105 ) (3880 104 )
56002
= 2564 kN
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 169
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 170
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.
E =
320
=
= 6038 MPa 6000 MPa
0.053
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 171
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 172
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 173
Fluid Properties
Problem 60 Solution:
Kinematic viscocity ( ) is defined as the ratio of absolute viscocity
( ) to mass density ( ), written as follows:
=
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 174
Water Pressure
Problem 61 Solution:
The depth h is calculated from the water surface.
p = gh = 1000 10 (5 − 0.5) = 45000 Pa = 45 kPa
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 175
Fluid Continuity
Problem 62 Solution:
Using continuity equation:
A1v1 = A2v2
1
1
2
2
D1 v1 = D2 v2
4
4
(1.5D2 )
2
0.5 m/s = D22 v2
v2 = 1.125 m/s
Using Bernoulli’s equation:
p1 v12
p
v2
+
+ z1 = 2 + 2 + z2
g 2g
g 2g
v12
v 22
+
=0+
2
2
p1
v 2 − v12
p1 = 2
2
1.1252 − 0.52
= 1000
2
= 507.815 Pa = 0.5 kPa
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 176
Reynolds Number
Problem 63 Solution:
Reynolds number determines the fluid type:
Re < 2100
→ laminar flow
2100 < Re < 4000
→ critical flow
Re > 4000
→ turbulent flow
Re =
=
vD vD
=
(2 m/s )(0.05 m)
7.3 10−7 m2 /s
= 1.37 105 → turbulent
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 177
Hydraulic Friction Loss
Problem 64 Solution:
m3
m3
= 0.05
min
s
Flow rate:
Q=3
Flow area:
A=
1
1
D2 = 0.152 = 0.0177 m2
4
4
Flow velocity:
v =
Q 0.05 m3 /s
=
= 2.82 m/s
A 0.0177 m2
Friction loss:
fLv 2 0.02 (50 m) (2.82 m/s )
hf =
=
= 2.7 m
2Dg
2 ( 0.15 m) 9.8 m/s2
2
(
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
)
Page 178
Hydraulic Pump Power
Problem 65 Solution:
Total head by the pump:
h = hz + hf = 18 m + 2.7 m = 20.7 m
Pump power:
P =
mgh
1 min
2
3000 kg/min 60 sec 9.8 m /s 20.7 m
=
0.8
= 12678.75 W 13 kW
(
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
)
Page 179
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 180
Pipe Head Loss
Problem 67 Solution:
From the given information, we obtain some known values:
p1 = 0 (reservoir is at atmospheric pressure)
p2 = 0 (pipe outlet is at atmospheric pressure)
v1 = 0 (the water in reservoir doesn’t have velocity)
v2 =
5 m3 /s
0.25 (0.5 m)
2
= 25.5 m/s (flow rate is given)
z1 = 220 m (given in the problem statement)
z2 = 180 m (given in the problem statement)
Using energy equation:
p1 v12
p
v2
+
+ z1 = 2 + 2 + z2 + hf
g 2g
g 2g
0 + 0 + 221 = 0 +
25.52
+ 168 + hf
2 9.8
hf = 20 m
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 181
Pipe Head Loss
Problem 68 Solution:
Specific roughness of cast iron = e =0.23 mm
Relative roughness =
e 0.23
=
= 0.000657
D
350
Reynolds number is determined by using the following equation.
Re =
vD
(1.34 (0.25 0.35 )) 0.35 = 4.875 10
=
2
6
10
−6
After that, the friction factor can be obtained from the Moody
diagram on the next page: f = 0.0185
Pipe head loss (Darcy’s equation):
2
1.34 m3 /s
0.0185 (150 m)
2
2
fLv
0.25 (0.35 m) = 78.5 m
hf =
=
2Dg
2 (0.35 m) 9.8 m/s2
(
)
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 182
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 183
Pumping Power
Problem 69 Solution:
From Problem 68 solution, the head loss ( hf ) is 78.5 m.
Total head by the pump:
h = ( hwatertank − hlake ) + hf = (175 m − 150 m) + 78.5 m = 103.5 m
Pump power:
P =
=
mgh
(1340 kg/sec ) (9.8 m2 /s ) 103.5 m
0.85
= 1599014 W 1600 kW
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 184
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:
Qmin P 0.2
=
Qavg
5
Qmin
80.2
=
4000
5
Qmin = 1213 m3 /day
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 185
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:
Qpeak
=
Qavg
Qpeak
4000
Qpeak
18 + P
=
4+ P
18 + 8
4+ 8
= 12201 m3 /day
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 186
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:
A 0.25 0.92
R=
=
= 0.225 m
P
0.9
Sewer slope:
S=
zinlet − zoutlet 1.5
=
= 0.025
L
60
Flow velocity using Manning’s equation:
v =
2
1
1 23 12
1
R S =
( 0.225) 3 ( 0.025)2 = 4.87 m/s
n
0.012
Flow capacity:
(
Q = vA = 4.87 m/s 0.25 ( 0.9 m)
2
) = 3.1 m /s
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
3
Page 187
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:
6
2
1
3
5
4
m = 18
7
8
10
15
14
9 11 12
16
13
18
17
r=3
j = 10
Check:
(18 + 3) > (2 x 10)
The structure is stable and statically indeterminate.
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 188
Shear Force
Problem 74 Solution:
Before we can answer the largest shear force, we must obtain the
support reaction at support A and C.
M = 0
(20 kN)(5 m) − (V )(10 m) + ( 4 kN/m 5 m) (12.5 m) = 0
A
Vc = 35 kN ( )
F
y
c
=0
VA − 20 kN + Vc − ( 4 kN/m)(5 m) = 0
VA + Vc = 40 kN
VA + 35 kN = 40 kN
VA = 5 kN ( )
5
0
20
+
5
+
0
15
15
Shear Force Diagram (unit: kN)
As seen in the shear force diagram, the largest magnitude of the
shear force is 20 kN.
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 189
Bending Moment
Problem 75 Solution:
25
+
0
0
-
50
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 190
Effective Width of T-Beam
Problem 76 Solution:
The effective width of T-beam is the minimum of:
1
1
4 L = 4 600 = 150 cm
bw + 16t s = 20 + 16 12 = 212 cm
L
= 600 cm
center-to-center
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 191
Bending Design
Problem 77 Solution:
Equilibrium between concrete compression block & tension rebars:
Cc = T
0.85fc'ab = As fy
0.85 (3 ksi)( a)(10 in) = ( As )(60 ksi)
a = 2.35 As
Find the required steel rebars:
Mu = Mn = 0.9 As fy ( d − 0.5a )
110 kips-ft 12 in/ft = 0.9 As (60 ksi ) ( (20 in − 2.5 in ) − 0.5 (2.35 As ) )
1467 kips-in = 1050 As − 70.5 As2
As = 1.56 in2
Check minimum and maximum rebar (use ACI equations):
min = 0.0033 → As,min = minbd = 0.5775 in2
max = 0.0135 → As,max = max bd = 2.3625 in2
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 =
1.56 in2
2
6
0.25 in
8
= 3.53 4
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 192
Shear Design
Problem 78 Solution:
Shear strength of beam RC is obtained from concrete and stirrups.
Contribution from concrete:
Vc = 2 fc' bw d
= 2 (1)
(
)
3000 psi (10 in)(17.5 in)
= 19170 lb = 19.17 kips
Contribution from stirrups:
Vs =
Vu
− Vc =
40 kips
− 19.17 kips = 34.2 kips
0.75
Required spacing of stirrups (use #3 rebar):
2
3
2 0.25 in (60 ksi)(17.5 in)
Av fy d
8
s=
=
= 6.78 6 in
Vs
34.2 kips
Check with maximum spacing:
smax
d 17.5 in
= 8.75 in
2 =
2
2
=
3
2 0.25 in (60000 psi)
A f
8
v y =
= 26.5 in
bw
(50 psi)(10 in)
Stirrups use #3 @ 6”
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 193
Basic Concept of Design
Problem 79 Solution:
(X)
(V)
(V)
(X)
(V)
I.
II.
III.
IV.
V.
ASD is newer than LRFD
ACI 318 is based on the LRFD concept
LRFD uses factored load combinations while ASD doesn’t
ASD uses ultimate strength while LRFD doesn’t
ASD uses factor of safety while LRFD doesn’t
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 194
Steel Beam Design
Problem 80 Solution:
Check for compactness:
bf
E
0.38
→ 8.13 9.15 (OK, compact flange)
2tf
Fy
h
E
3.76
→ 53.9 90.55 (OK, compact web)
tw
Fy
W16X36 is compact, so that the nominal moment is:
Mn = Mp = Fy Zx = 50 ksi 64 in2 = 3200 kips-in = 266.7 kips-ft
Allowable moment using LRFD concept:
bMn = 0.9 266.7 kips-ft = 240 kips-ft
Allowable maximum distributed load:
1
wu L2
8
2
1
240 kips-ft = wu (24 ft )
8
wu = 3.33 kips/ft
bMn = Mu =
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 195
Soil Properties
Problem 81 Solution:
The sample of soil can be divided into three components: dry soil,
water, and air.
Dry soil properties :
Vs =
ms
13.4 gr
=
= 6.23 cm3
3
Gs w 2.15 1 gr/cm
Volume of the void can be defined as the summation between
volume of water and air. Therefore,
Vv = Vtotal − Vs = 10 − 6.23 = 3.77 cm3
Void ratio is calculated by using this following equation:
e=
Vv 3.77
=
= 0.6
Vs 6.23
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 196
Soil Properties
Problem 82 Solution:
Assume that the total volume of saturated soil is 1 m3.
Vs + Vw = 1
Substitute the volume relationship to this equation:
ms + mw = 2250
GsVs w + Vw w = 2250
(2.6 )(Vs )(1000) + (1 − Vs )(1000) = 2250
2600Vs + 1000 − 1000Vs = 2250
Vs = 0.78 m3
After obtaining volume of soil, we can calculate dry unit mass
(mass of soil).
ms = GsVs w
= 2.6 0.78 1000
= 2031.25 kg
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 197
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
25
Ka = tan2 45 − 2 = tan2 45 −
= 0.4059
2
2
Active earth pressure at depth = 6 m:
a = K a ( 1 z1 + ( 2 − w ) z2 )
= 0.4059 (19 3 + (20 − 9.8 ) 3)
= 35.56 kN/m2
Pore water / hydrostatic pressure at depth = 6 m:
w = w z2 = 9.8 3 = 29.4 kN/m2
Total lateral pressure at depth = 6 m:
total = a + w = 35.56 + 29.4 = 64.96 kN/m2
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 198
Bearing Capacity
Problem 84 Solution:
Using Terzaghi’s equation for a square footing:
qult = cNc Sc + Df Nq + 0.5 BN S
= 0 + (125 pcf )( 4 ft )(22.5 ) + 0.5 (125 pcf )(5 ft )(19.7 )(0.85)
= 16482.8125 psf
Using safety factor of 3:
qult
FS
16482.8125 psf
=
3
= 5494 5500 psf
qall =
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 199
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 200
Vertical Curve
Problem 86 Solution:
PVC = PVI −
xm =
L
12 sta
= ( sta 76 + 00 ) −
= sta 70 + 00
2
2
( −0.02)(12 sta) = 6.32 sta
G1L
=
G1 − G2
−0.02 − 0.018
low point = PVC + xm = ( sta 70 + 00) + 6.32 sta = sta 76.32
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 201
Vertical Curve
Problem 87 Solution:
elevlowpoint = elev PVC + G1 xm
G2 − G1 ) xm2
(
+
2L
(0.018 + 0.02 )(6.32 sta)
= (500 + 0.02 6 sta) + ( −0.02 )(6.32 sta) +
2 (12 sta)
2
= 505.68 506 m
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 202
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 203
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.
q = kv
= k ( 80 − 0.5k )
= 80k − 0.5k 2
According to mathematics principle, we can get the maximum
value of q when the first derivation of the equation is equal to 0.
q = 80k − 0.5k 2
dq
=0
dk
80 − kmax = 0
kmax = 80 veh/km
qmax = 80kmax − 0.5kmax2 = 3200 veh/hr
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 204
Horizontal Curve
Problem 90 Solution:
Find the central angle:
T = R tan
2
125 = 600 tan
2
= 23.53
Find the length:
L=
R
600 23.53
=
= 246 m
180
180
Determine the stationing:
PC = PI − T = ( sta 10 + 000) − 125 = sta 9 + 875
PT = PC + L = ( sta 9 + 875) + 256 = sta 10 + 121
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 205
BOD Analysis
Problem 91 Solution:
Substitute the five-day values into the BOD equation:
(
yt = L 1 − e − kt
(
)
− 0.13 5
234 = L 1 − e ( )( )
)
L = 489.6 490 mg/L
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 206
BOD Analysis
Problem 92 Solution:
From the previous solution, the ultimate BOD was obtained.
Now, substitute the seven-day values into the BOD equation:
(
yt = L 1 − e − kt
(
)
− 0.13 7
= 489.6 1 − e ( )( )
)
= 292.5 290 mg/L
(Answer A)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 207
Reactor Capacity
Problem 93 Solution:
Solids residence time:
=
V
Q
Suspended solids concentration:
X =
X =
V =
=
cdY ( S0 − S )
(1 + Kdcd )
cd QY ( S0 − S )
(
V 1 + Kdcd
)
cd QY ( S0 − S )
(
X 1 + Kdcd
)
12 0.35 0.55 (258 − 6.4 )
3600 (1 + 0.06 12 )
= 0.094 m3 /s = 8110 m3 /day
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 208
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.
DO24.6C − 8.9 24.6 − 24
=
8.6 − 8.9
25 − 24
DO24.6C − 8.9 = −0.18
DO24.6C = 8.72 mg/L
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 209
Solid Waste
Problem 95 Solution:
Monthly mass of solid waste (for 18,400 people in a month):
125 kg/m3 20,000 m3 = 2,500,000 kg
Average mass of solid waste per person per day:
2,500,000 kg
= 4.5 kg/person.day
18, 400 people 30 days
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 210
Waste Water
Problem 96 Solution:
Assume :
1 mg/L = 1 ppm
1 gal = 8.34 lbm
28 parts
lbm
6 gal
Mass loading = 6
8.34
35 10
day
gal
10 part
lbm
= 8173.2
day
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 211
Reaction Rate Constant
Problem 97 Solution:
(
BOD5 = L0 1 − e − kt
(
)
−k 5
225 = 486 1 − e ( )
)
225
−k 5
=e ()
486
225
ln 1 −
= − k (5 )
486
1−
225
ln 1 −
486
k =
= 0.124 d −1
−5
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 212
Critical Path
Problem 98 Solution:
Critical Path is the longest path to complete a project.
(Answer C)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 213
Project Scheduling Tool
Problem 99 Solution:
1
3
2
2
4
2
6
2
5
2
7
2
In the figure, we could see the activity (labeled as number) is on
node. Therefore, the figure above represents AON (Activity-OnNode) diagram.
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 214
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 215
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
A
B
C
D
E
F
Duration
(days)
3
5
2
3
4
3
EST
(day no.)
0
0
3
5
5
8
EFT
(day no.)
3
5
5
8
9
11
LST
(day no.)
3
0
6
5
7
8
LFT
(day no.)
6
5
8
8
11
11
TF
(days)
3
0
3
0
2
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 216
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 217
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 218
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 219
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 220
Construction Estimating
Problem 106 Solution:
Gross wall area:
Ag = (72 ft )(16 ft ) = 1152 ft2
Net wall area (without opening):
An = Ag − Aopening = 1152 ft2 − 250 ft2 = 902 ft2
Bricks required:
nbrick = 902 ft2
602 brick
1.05 = 5702 bricks
100 ft2
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 221
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
Fill area
Fill volume
(m)
(m2)
(m3)
10+00
153.42
153.42 + 20.56
4.50
= 391.46
2
10+4.50
20.56
20.56 + 8.25
7.85
= 113.08
2
10+12.35
8.25
8.25
4.05
= 11.14
3
10+16.40
-
10+20
total
515.68
Total volume of fill work is 515.68 m3.
(Answer D)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 222
Earthwork - Cut
Problem 108 Solution:
Station
Cut area
Cut volume
(m)
(m2)
(m3)
10+00
-
10+4.50
-
13.67
7.85
= 35.77
3
10+12.35
13.67
13.67 + 52.84
4.05
= 134.68
2
10+16.40
52.84
52.84 + 165.14
3.6
= 392.36
2
10+20
165.14
total
562.81
Total volume of cut work is 562.81 m3.
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 223
Borrow Pit Volumes
Problem 109 Solution:
It’s easier to solve borrow pit volumes using table.
Point
A1
A2
A3
A4
B1
B2
B3
B4
C1
C2
C3
V =
Elevation
(m)
103.5
105.6
102.1
101.9
103.9
104.0
102.8
103.7
105.2
104.3
100.3
Elev to be
Cut (m)
3.5
5.6
2.1
1.9
3.9
4.0
2.8
3.7
5.2
4.3
0.3
(
A
( h ) 4 = 70.8 m
No. of
Rectangles
1
2
2
1
2
4
3
1
1
2
1
12 m 8 m)
i , jn
4
Height of
Cut (m)
3.5
11.2
4.2
1.9
7.8
16
8.4
3.7
5.2
8.6
0.3
Total = 70.8
= 1699.2 1700 m3
(Answer B)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 224
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)
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 225
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-bystep through the entire exam. You can check that out at
www.civilfereviewcourse.com.
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
Page 226
www.civilengineeringacademy.com | isaac@civilengineeringacademy.com |
