• •
IVI
Discipline-Specific Review
for the FE/Ell Exam
Third Edition
Robert H. Kim, MSCE,PE and
Thomas A. Spriggs, MSCE
with Michael R. Lindeburg, PE
How to Locate and Report Errata for This Book
At PPJ, we do our best to bring you error-free books. But when errors do occur, we
want to make sure you can view corrections and report any potential errors you find,
so the errors cause as little confusion as possible.
A current list of known errata and other updates for this book is available on the PPJ
website at www.ppi2pass.comlerrata. We update the errata page as often as
necessary, so check in regularly. You will also find instructions for submitting suspected
errata. We are grateful to every reader who takes the time to help us improve the quality
of our books by pointing out an error.
CIVIL DISCIPLINE-SPECIFIC
Third Edition
Current printing of this edition:
REVIEW FOR THE FE/EIT EXAM
1
Printing History
edition
number
printing
number
1
2
4
1
3
1
update
Minor corrections.
New edition. Copyright
New edition. Copyright
update.
update.
Copyright © 2009 by Professional Publications, Inc. (PPI). All rights reserved. No part of this publication
may be.reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic,
mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher.
Printed
in the United States of America
PPI
1250 Fifth Avenue, Belmont,
(650) 593-9119
www.ppi2pass.com
CA 94002
ISBN: 978-1-59126-177-3
Library of Congress Control Number:
2009927833
----------------------Table of Contents
Preface and Acknowledgments
v
Engineering Registration in the Uniled Slates
vii
CommonQuestions About the D5 Exam
.........
. xiii
,
How to Use This Book
xv
Nomenclature ........................
1
Practice Problems
Surveying
. .. . . . . . . .. . ..
. . . . . . . . . . . . . ..
Hydraulics and Hydrologic Systems. . . . . . . ..
.. .. . . . .. .. .. .. .
Soil Mechanics and Foundations
. .
. . . . . . . . . . . . . . . ..
Environmental Engineering .. , . .
. . , . . . . . . . . . . . . . . . . . . . . . ..
Transportation.
. .. .. ..
. . . .. .. .. ..
. . . . . . . . . . . . . . ..
Structural Analysis
. . . . . . . . . . . . . . . . . . . . . . . . ..
Structural Design. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Construction
Management
. . . . .. .. .. . .. ... .
9
12
16
20
23
26
30
35
Materials . . ..
38
....
.. .. .. .. .. .. . .. .. .. .. .. . .. .. .. .. .
Practice Exam 1
Problems
Solutions
.
.
41
50
Practice Exam 2
Problems
Solutions
.
.
65
74
iii
Preface and Acknowledgments
students who are taking a discipline-specific (DS) afternoon session of the Fundamentals of Engineering (FE)
As with all of PPI's books, the problems in this book are
original and have been ethically derived. Although examinee feedback was used to determine its content, this
book contains problems that are only like those that
exam.
are on the exam.
The topics covered in the DS afternoon FE exams are
completely different from the topics covered in the
morning session of the FE exam. Since this book only
in this book.
This book is one in a series intended for engineers and
covers one discipline-specific
exam, it provides exam-
level problems that are like those found on the afternoon half of the FE exam for the civil discipline.
This book is intended to be a quick review of the
material relevant to the afternoon session of the civil
engineering exam. The material presented covers the
subjects most likely to be on the exam. This book is
not a thorough treatment of the exam topics. Its objective is to prepare you with enough knowledge
to pass.
As much as practical, this book uses the notation given
in the NCEES Handbook.
This book consolidates 181 practical review problems,
covering all of the civil discipline-specific exam topics. All problems include full solutions.
In developing this book, the NCEES Handhook and
the breakdown of problem types published by NCEES
were my guide for problem types and scope of coverage. However, as with most standardized tests: there
is no guarantee that any specific problem type will be
encountered.
It is expected that minor variations in
problem content will occur from exam to exam.
There are no actual exam problems
This book was designed to complement my FE Review
Manual, which you will also need to prepare for the FE
exam. The FE Review Manual is PPI's most popular
study guide for both the morning and afternoon general exams. It and the Engineer-In-Training Reference
Manual have been the most popular review books for
this exam since 1980.
You cannot. prepare adequately without your own copy
of the NCEES Handbook. This document contains the
data and formulas that you will need to solve both the
general and the discipline-specific problems. A good way
to become familiar with it is to look up the information,
formulas, and data that you need while trying to work
practice problems.
Exam-prep books are always works in progress. By necessity, a book will change as the exam changes. Even
when the exam format doesn't change for a while, new
problems and improved explanations can always be
added. I encourage you to provide comments via PPI's
errata reporting page, www.ppi2pass.com/errata.
You will find all verified errata there. I appreciate all
feedback.
Best of luck to you in your pursuit of licensure.
This third edition brings the structural code-related
problems in line with the NCEES Handbook, 8th edition, which contains values and calculations
based on
Michael R. Lindeburg, PE
ASCEjSEI7-05, ACI 318-05, and AISC Steel Construction Manual 2005 (13th edition). It also incorporates
corrections to the errata in the second edition and significantly revises numerous problems to more closely
conform to the exam specifications.
The problems in the first and second editions of this
book were developed by Robert H. Kim, PE, and
Thomas A. Spriggs, MSCE, following the format, style,
subject breakdown, and guidelines that I provided.
Rhandi Gallegos,
PE, contributed
significant
updates
for the second edition. Kathleen Sullivan and Jamie
Rana technically reviewed the material.
v
Engineering Registration in the United States
Engineering registration (also known as engineering licensing) in the United States is an examination process
by which a state's board of engineering licensing (i.e.,
Most states have similar registration procedures. However, the details of registration qualifications, experience requirements, minimum education levels, fees, oral
interviews, and exam schedules vary from state to state.
For more information, contact your state's registration
registration board) determines and certifies that you
board (www.ppi2pass.com/stateboards).
ENGINEERING REGISTIlATIOtf
have achieved a minimum level of competence.
This
process protects the public by preventing unqualified
National Council of Examiners for Engineering and Surveying
individuals from offering engineering services.
The National Council of Examiners for Engineering and
Surveying (NCEES) in Clemson, South Carolina, produces, distributes, and scores the national FE and PE
Most engineers do not need to be registered. In particular, most engineers who work for companies that
design and manufacture products are exempt from the
licensing requirement. This is known as the 'industrial
exemption. Nevertheless, there are many good reasons
for registering. For example, you cannot offer consulting
engineering design services in any state unless you are
registered in that state. Even within a product-oriented
corporation) however, you may find that registered engineers have more opportunities for employment and
advancement.
exams.
The individual states purchase the exams from
NCEES and administer them themselves. NCEES does
not distribute applications to take the exams, administer the exams or appeals, or notify you of the results.
These tasks are all performed by the states.
Reciprocity Among States
With minor exceptions, having a license from one state
will be allowed to use the titles Professional Engineer
(PE), Registered Engineer (RE), and Consulting Engineer (CEl.
will not permit you to practice engineering in another
state. You must have a professional engineering license
from each state in which you work. For most engineers,
this is not a problem: but for some, it is. Luckily, it is
not too difficult to get a license fr0111every state you
work in once you have a license from one state.
Although the registration process is similar in all 50
All states use the NCEES exams. If you take and pass
Once you have met the registration
requirements,
you
states, each state has its O\vn registration law. Unless
you offer consulting engineering services in more than
one state, however, you will not need to register in other
states.
The U.S. Registration Procedure
To become a registered engineer in the United States,
you will need to pass two eight-hour written exam-
the FE or PE exam in one state, your certificate
will
be honored by all of the other states. Although there
may be other special requirements imposed by a state,
it will not be necessary to retake the FE and PE exams.
The issuance of an engineering license based on another
state's license is known as reciprocin) or comity.
The simultaneous
administration
of identical exams in
inations. The first is the Fundamentals of Engineering
all states has led to the term uniform examination.
Examination,
However, each state is still free to choose its own minimum passing score and to add special questions and requirements to the examination process. Therefore, the
use of a uniform exam has not, by itself: ensured reciprocity among states.
also known as the Engineer-In- Training
Examination and the Intern Engineer Exam. The initials FE, EIT, and IE are also used. This exam covers
basic subjects from the mathematics, physics, chemistry, and engineering classes you took during your first
four university years. In rare cases, you may be allowed
to skip this first exam.
The second eight-hour exam is the Principles and Practice of Engineering Exam. The initials PE are also used.
This exam is on topics within a specific discipline,
and
only covers subjects that fall within that area of specialty.
TlinETxAM
Applying for the Exam
Each state charges different fees, specifies different requirements,
and uses different fonns to apply for the
vii
r---------------·
viii
Civil Discipline-Spe,i!i, Review for Ihe FE/EIT Exam ------------------------
exam. Therefore, it will be necessary to request an application from the state in which you want to become
registered. Generally, it is sufficient for you to phone for
this application. You'll find contact information (websites, telephone
numbers,
email addresses) etc.)
for
all U.S. state and territorial boards of registration at
www.ppi2pass.com/stateboards.
Keep a copy of your exam application, and send the
original application by certified mail, requesting a delivery receipt. Keep your proof of mailing and delivery
with your copy of the application.
Exam Dates
The national PE and PE exams are administered twice
a year (usually in mid-April and late October), on the
same weekends in all states. For a current exam sched-
ule, check www.ppizpass.oomyfefaqs.
FEExam Format
The NCEES Fundamentals of Engineering examination
has the following format and characteristics.
• There are two four-hour sessions separated by a
one-hour lunch.
• Examination
questions are distributed in a bound
examination booklet. A different exam booklet is
used for each of the two sessions.
• Formulas and tables of data needed to solve ques-
tions in the exams are found in either the NCEES
Handbook or in the body of the question statement itself.
• The morning session (also known as the A.M. session) has 120 multiple-choice questions, each with
four possible answers lettered (A) through (D).
Responses
Morning FEExom Subjec15
percentage of
questions (%)
subject
chemistry
computers
electricity and magnetism
engineering economics
engineering probability
and statistics
engineering mechanics
(statics and dynamics)
ethics and business practices
fluid mechanics
material properties
mathematics
strength of materials
thermodynamics
9
7
9
8
7
10
7
7
7
15
7
7
• There are seven different versions of the afternoon session (also known as the P.M. session), six
of which correspond to specific engineering disciplines: chemical, civil, electrical, environmental,
industrial, and mechanical engineering.
The seventh version of the afternoon exam is a
general examination suitable for anyone, but in
particular, for engineers whose specialties are not
one of the other six disciplines. Though the subjects in the general afternoon exam are similar
to the morning subjects, the questions are more
complex-hence their double weighting. Questions
on the afternoon exam are intended to cover con-
cepts learned in the last two years of a four-year
degree program. Unlike morning questions,
these
questions may deal with more than one basic concept per question.
Each version of the afternoon session consists of
60 questions. All questions are mandatory. Questions in each subject may be grouped into related problem sets containing between two and ten
questions each.
must be recorded with a pencil pro--
The percentages of questions for each subject in
vided by NCEES on special answer sheets. No
the general afternoon session exam are given in
credit is given for answers recorded in ink.
the followingtable.
• Each problem in the morning session is worth one
Afternoon FEExom Subjects
(General Exum)
point. The total score possible in the morning is
120 points. Guessing is valid; no points are subtracted for incorrect answers.
subject
• There are questions on the exam from most of the
undergraduate engineering degree program sub--
jects. Questions from the same subject are all
grouped together, and the subjects are labeled.
The percentages of questions for each subject in
the morning session are given in the following ta--
ble.
PPI. www.ppi2pasl .com
advanced engineering mathematics
application of engineering mechanics
biology
electricity and magnetism
engineeringeconomics
engineeringprobability and statistics
engineeringof materials
fluids
thermodynamics and heat transfer
~~-:'~~_~~-.:..;~-------
percentage of
questions (%)
10
13
5
12
10
9
11
15
15
-------------------Engineering Registration in the United States
Each of the discipline-specific afternoon examinations
covers a substantially different body of knowledge than
the morning exam. The percentages of questions for
each subject in the civil discipline-specific afternoon session exam are as follows.
Afternoon FEExom Subiecls
(Civil DS Exoml
subject
percentage of
questions (%)
surveying
11
hydraulics and hydrologic systems
soil mechanics and foundations
12
environmental
12
12
10
10
engineering
transportation
structural analysis
structural design
construction management
materials
15
10
8
Some afternoon questions stand alone, while others are
grouped together, with a single problem statement that
describes a situation followed by two or more questions
about that situation. All questions are multiple-choice.
You must choose the best answer from among four, lettered (A) through
(D).
• Each question in the afternoon is worth two points,
making the total possible score 120 points.
• The scores from the morning and afternoon sessions are added together to determine your total
score. No points are subtracted for guessing or
incorrect answers. Both sessions are given equal
weight. It is not necessary to achieve any minimum
score OIl either the morning or afternoon sessions.
• All grading is none by computer optical sensing.
ix
correctly. The summation of the estimated fractions for
all test questions becomes the passing score. Because
the law in most states requires engineers to achieve a
score of 70% to become licensed 1 you may be reported
as having achieved a score of 70% if your raw score is
greater than the passing score established by l'\CEES,
regardless of the raw percentage. The actual score may
be slightly more or slightly less than 110 as determined
from the performance of all examinees on the equating
subtest.
About 20% of the FE exam questions arc repeated from
previous exams-> this is the ecuatinq subtest. Since the
scores of previous examinees on the equating subtest
are known 1 comparisons can be made between the two
exams and examinee populations.
These comparisons
are used to adjust the passing score.
The individual states are free to adopt their own passing
score, but all adopt NCEES's suggested passing score
because the states believe this cutoff score can be defended if challenged.
You will receive the results within 12 weeks of taking the
exam. If you pass, you will receive a letter stating that
you have passed. If you fail, you will be notified that
you failed and be provided with a diagnostic report.
Permilled Reference Material
Since October 1993, the FE exam has been what NCEES
calls a "limited-reference" exam. This means that no
books or references other than those supplied by NCEES
may be used. Therefore, the FE exam is really an
"NCEES-publication only" exam. NCEES provides its
own FE Supplied-Reference Handbook for use during
the examination. No books from other publishers may
be used.
Use of Sl Units on Ihe FEExam
Metric questions are used in all subjects, except sorne
civil engineering and surveying subjects that typically use only customary U.S. (i.e., Englisb) units.
Sl
units are consistent with ANSI/IEEE standard 268 (tbe
American Standard for Metric Practice). Non-SI metric
units might still be used when common or where needed
for consistency with tabulated data (e.g., use of bars in
pressure measurement).
Grading and Scoring the FEExam
CALCULATORS
To prevent unauthorized transcription and redistribution of the exam questions, only calculator models approved by NCEES will be permitted in the exam room.
You cannot share calculators with other examinees. For
a list of allowed calculators check www.ppi2pass.eom/
calculators.
It is essential that a calculator used for engineering examinations have the following functions.
The FE exam is not graded on the curve, and there is
no guarantee that a certain percentage of examinees will
pass. Rather, NCEES uses a modification of the Angoff
procedure to determine the suggested passing score (the
• trigonometric
cutoff point or cut score),
• pi
2
• square root and x
• common and natural
With this method, a group of engineering professors and
other experts estimate the fraction of minimally qualified engineers who will be able to answer each question
_________________________________
functions
• inverse trigonometric
functions
• hyperbolic functions
•
logarithms
yX and eX
PPI.www.ppi2pass.com
x
Civil Discipline-Spedfic Review for the FE/Ell Exom
For maximum speed, your calculator should also have
Or be programmed for the following functions.
• extracting
equations
roots of quadratic
and higher-order
• converting between polar (phasor) and rectangular vectors
• finding standard deviations and variances
• calculating
determinants
• take calculations out to a maximum of four significant digits
of 3 x 3 matrices
• linear regression
• economic analysis and other financial functions
STRATEGIES FOR pASSING THE FE ExAM··
• prepare in all exam subjects,
cialty areas
not just your spe-
At the beginning of your review program, you should
locate a spare calculator. It is not necessary to buy a
spare if you can arrange to borrow one from a friend or
the office. However, if possible, your primary and spare
calculators should be identicaL If your spare calculator
is not identical to the primary calculator, spend some
time familiarizing yourself with its functions.
A Few Doys Before the Exam
There are a few things you should do a week or so before the exam date. For example, visit the exam site in
order to find the building, parking areas, examination
room, and rest rooms, You should also make arrange-
The most successful strategy for passing the FE exam
is to prepare in all of the exam subjects. Do not limit
the number of subjects you study in hopes of finding
enough questions in your strongest areas of knowledge
to pass.
ments now for child care and transportation. Since the
exam does not always start or end at the designated
Fast recall and stamina are essential to doing well. You
times, make sure that your child care and transportation arrangements can tolerate a late completion.
must be able to quickly recall solution procedures, formulas, and important data. You will not have time during the exam to derive solutions methods-you must
know them instinctively.
This ability must be maintabled for eight hours. Be sure to gain familiarity with
the NCEES Handbook by using it as your only reference for some of the problems you work during study
sessions.
In order to get exposure to all exam subjects, it is imperative that you develop and adhere to a review schedule. If you are not taking a classroom review COurse
(where the order of your preparation is determined by
Next in importance to your scholastic preparation is the
preparation of your two examination kits. The first kit
consists of a bag or box containing items to bring with
you into the examination room.
[
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[]
the lectures), prepare YOUI' own review schedule.
[I
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There arc also physical demands on your body during
the exam. It is very difficult to remain alert and atten-
[]
[]
tive for eight hours or more. Unfortunately, the more
time you study, the less time you have to maintain your
physical condition. Thus, most examinees arrive at the
exam site in peak mental condition but in deteriorated
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physical condition. While preparing for the FE exam is
not the on Iy good reason for embarking on a physical
[J
conditioning
[I
program, it can serve as a good incentive
to get in shape.
It will be helpful to make a few simple decisions prior
to starting your review. You should be aware of the
different options available to you.
should decide early on to
For example, you
• use 81 units in your preparation
• perform electrical calculations with effective (rms)
or maximum values
PPI. www.ppi2pass.com
[]
letter admitting you to the exam
photographic identification
main calcnlator
spare calculator
extra calculator batteries
unobtrusive snacks
travel pack of tissues
headache remedy
$2.00 in change
light, comfortable sweater
loose shoes or slippers
handkerchief
cushion for your chair
small hand towel
earplugs
wristwatch wit.h alarm
wire coat hanger
extra set of car keys
The second kit consists of the following it.ems and
should be left in a separate bag or box in your car in
case you need them.
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copy of your application
proof of delivery
this book
other references
regular dictionary
scientific dictionary
-~'!'"'------~--~-""":'i~~'H'~
EngineeringRegistration in the United States
[
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[]
course notes in three-ring binders
instruction booklets for all your calculators
light lunch
beverages in thermos and cans
sun glasses
extra pair of prescription glasses
raincoat, boots, gloves, hat, and umbrella
street map of the exam site
note to the parking patrol for your windshield
explaining where you are, what you are doing,
and why your time may have expired
battery-powered
desk lamp
The Day Before the Exam
Take the day before the exam off from work to relax.
Do not cram the last night. A good prior night's sleep
is the best way to start the exam. If you live far from
the exam site: consider getting a hotel room in which
to spend the night.
Make sure your exam kits are packed and ready to go.
The Day of the Exom
You should arrive at least 30 minutes before the exam
starts. This will allow time for finding a convenient
parking place, bringing your materials to the exam
room: and making room and seating changes. Be prepared: though, to find that the examination room is not
open or ready at the designated time.
Once the examination
ing suggestions.
xi
has started, consider the follow-
• Set your wristwatch alarm for five minutes before
the end of each four-hour session) and use that
remaining time to guess at all of the remaining
unsolved problems. Do not work up until the very
end. You will be successful with about 25% of you!'
guesses, and these points will more than make up
for the few points you might earn by working during the last five minutes.
• Do not spend more than two minutes per morning
question. (The average time available per problem
is two minutes.) If you have not finished a question in that time, make a note of it and move on.
• Do not ask your proctors technical questions. Even if they are knowledgeable in engineering) they
will not be permitted to answer your questions.
• Make a quick mental note about any problems for
which you cannot find a correct response or for
which you believe there are two correct answers.
Errors in the exam are rare, but they do occur.
Being able to point out an error later might give
you the margin you need to pass. Since such problems are almost always discovered during the scoring process and discounted from the exam, it is not
necessary to tell your proctor, but be sure to mark
the one best answer before moving on.
• Make sure all of your responses on the answer
sheet are dark and completely
_~
fill the bubbles.
• PPI. www.ppi2pDSs.mm
Common Questions About the DS Exam
Q:
Do I have to take the OS exam?
A: Most people do not have to take the DS exam and
may elect the general exam option. The state boards
do not care which afternoon option you choose, nor do
employers. In some cases: an examinee still in an undergraduate degree program may be required by his or
her university to take a specific DS exam.
Q: Do all mechanical, civil, electrical, chemical, industrial, and environmental
engineers take the DS exam?
A: Originally, the concept was that examinees from the
"big five" disciplines would take the DS exam, and the
general exam would be for everyone else. This remains
just a concept, however. A majority of engineers in all
of the disciplines apparently take the general exam.
Q:
Q: Will my OS FE certificate
states?
be recognized by other
A: Yes. All states recognize passing the FE exam and
do not distinguish between the OS and general afternoon portions of the FE exam.
Q: Is the OS FE certificate
FE certificate?
"better"
than the general
A: There is no difference. No one will know which
option you chose. It's not stated on the certificate you
receive from your state.
Q: What is the format of the OS exam?
A: The DS exam is 4 hours long. There are 60 problems, each worth 2 points. The average time per problem is 4 minutes. Each problem is multiple choice with 4
answer choices. Most problems require the application
of more than one concept (i.e., formula).
When do I elect to take the OS exam?
A: You will make your decision on the afternoon of the
FE exam, when the exam booklet (containing all of the
OS exams) is distributed to you.
Q: Where on the application for the FE exam do I
choose which DS exam I want to take?
A: You don't specify the OS option at the time of your
application.
Q: fs there anything
exam is administered?
special about
the way the OS
A: In all ways, the DS and general afternoon exam are
equivalent. There is no penalty for guessing. No credit
is given for scratch pad work. methods, etc.
Q: Are the answer choices close or tricky?
A: Answer choices are not particularly close together
in value, so the number of significant digits is not going
to be an issue. Wrong answers, referred to as "distractors" by NCEES are credible. However, the exam is
not "tricky"; it does not try to mislead you.
1
Q: After starting to work on either the OS or general
exam, can I change my mind and switch options?
A: Yes. Theoretically, if you haven't spent too much
time on one exam, you can change your mind and start
a different one. (You might need to obtain a new answer
sheet from the proctor.)
Q: Are any problems in the afternoon session related
to each other?
A: Several questions may refer to the same situation
or figure. However, l\CEES has tried to make all of t.he
questions independent. If you make a mistake on (HlP.
question, it shouldn't, carryover to another.
Q: After I take the OS exam, does anyone know that
I took it?
Q: Is there any minimum
exam?
A: After you take the FE exam, only NCEES and your
state board will know whether you took the OS or general exam. Such information mayor may not be retained by your state board.
A: No. It is the total score from your morning and
afternoon sessions that determines your passing, not the
individual session scores. You do not have to "pass"
each session individually.
passing
score for the DS
xiii
----------------xiv
Civil Discipline-Specific Review for the fE/EIT Exom
Q: Is the general portion easier, harder, or the same
as the DS exams?
A: Theoretically, all of the afternoon options are the
same. At least, that is the intent of offering the specific
options-to reduce the variability. Individual passing
rates, however, may still vary 5% to 10% from exam
to exam. (PPI lists the most recent passing statistics
for the various DS options on its website at
www.ppi2pass.com/fepassrates.)
Q: Do the DS exams cover material at the undergraduate or graduate level?
A: Like the general exam, test topics come entirely
from the typical undergraduate degree program. However, the emphasis
is primarily on material from the
third and fourth year of your program. This may put
_
Q: Is everything in the DS portion of the i\CEES Handbook going to be on the exam?
A: Apparently, there is a fair amount of reference material that isn't needed for every exam. There is no way,
however, to know in advance what material is needed.
Q: How long does it take to prepare for the DS exam?
A: Preparing for the DS exam is similar to preparing
for a mini PE exam, Engineers typically take two to
four months to complete a thorough review for the PE
exam. However, examinees who are still in their degree
program at a university probably aren't going to spend
more than two weeks thinking about, worrying about,
or preparing for the DS exam. They rely on their recent
familiarity with the subject matter.
examinees who take the exam in their junior year at a
disadvantage.
Q: If I take the DS exam and fail, do I have to take
the DS exam the next time?
Q: Do you need practical work experience to take the
DS exam?
A: No. The examination process has no memory.
A:
No.
Q: Where can I get even more information about the
DS exam?
Q: Does the DS exam also draw on subjects that are
in the general exam?
A: Yes. The dividing line between general and DS topics is often indistinct.
A: If you have internet access, visit the Exam FAQs
and the Engineering Exam Forum at PPI's website
(www,ppi2pass.com/fefaqs
and www.ppizpass,
com/ fornms, respectively).
Q: Is the DS exam in customary U.S. or SI units?
A: The DS exam is nearly entirely in SI units. A few
exceptions exist for some engineering subjects (survey-
ing, hydrology, code-hased design, etc.) wbere current
common practice uses only customary U.S. units.
Q: Does the NCEES Handhook Cover everything that
is on the DS exam?
A: No. You may be tested on subjects that are not
in the NCEES Handbook. However, NCEES has apparently adopted an unofficial policy of providing any
necessary information,
data, and formulas in the stem
of the question. You will not be required to memorize
any formulas.
PPI. WWW.ppi2pcss.ccm
-----------------~-:-+~~-~~~~
How 10 Use This Book
HOW EXAMINEES CAN USE THIS BOOK
This book is divided into three parts: The first part
consists of 61 representative practice problems covering all of the topics in the afternoon DS exam. You
may time yourself by allowing approximately 4 minutes
per problem when attempting to solve these problems,
out that was not my intent when designing this book.
Since the solution follows directly after each problem
in this section, I intended for you to read through the
problems, attempt to solve them on your own, become
familiar with the support material in the official NCEES
Handbook, and accumulate the reference materials you
think you will need for additional study.
The second and third parts of this book consist of two
complete sample examinations that you can use as
sources of additional practice problems or as tirned diagnostic tools. They also contain 60 problems, which
corresponds to the number of problems in the afternoon
DS exam. The number of problems in each subject
corresponds to the breakdown of subjects published by
NCEES. Since the solutions to these parts of the book
are consolidated at the end) it was my intent that you
would solve these problems in a realistic mock-exam
mode.
You should use the NCEES Handbook as your only reference during the mock exams.
The morning general exam and the afternoon DS exam
essentially cover two different bodies of knowledge. It
takes a lot of discipline to prepare for two standardized
exams simultaneously.
Because of that (and because
of my good understanding of human nature), Lsuspect
that you will be tempted to start preparing for your chosen DS exam only after you have become comfortable
with the general subjects.
If: however; you are limited in time to only two or three
months of study, it will be quite difficult to do a thorough DS review if you wait until after you have finished
your general review. With a limited amount of time;
you really need to prepare for both exams in parallel.
HOW INSTRUctORS CANlJSfTHIS
11001<
The availability of the discipline-specific FE exam has
greatly complicated the lives of review course instructors and coordinators.
The general consensus is that
it is essentially impossible to do justice to all of the
general FE exam topics and then present a credible 1'8view for each of the DS topics. Increases in course cost,
expenses, course length, and instructor pools (among
many other issues) all conspire to create quite a difficult situation.
One-day reviews for each DS subject are subjectoverload from a reviewing examinee's
standpoint.
Efforts to shuflJe FE students over the parallel PE review courses meet with scheduling conflicts. Another
idea, that of lengthening lectures and providing more
in-depth coverage of existing topics (e.g., covering transistors during the electricity lecture), is perceived as a
misuse of time by a majority of the review course attendees. Is it any wonder that virtually every FE review
course in the country has elected to only present reviews
for the general afternoon exam?
But, while more than half of the examinees elect to take
the general afternoon exam, some may actually be required to take a DS exam. This is particularly the case
in some university environments where the FE exam has
become useful as an "outcome assessment tool. l' Thus,
some method of review is still needed.
Since most examinees begin reviewing approximately
two to three months before the exam (which corresponds to vvhen most review courses begin); it is impractical to wait until the end of the general review to
start the DS review'. The DS review must proceed in
parallel with the general review.
In the absence of parallel DS lectures (something that
isn 't yet occurring in too many review courses), you may
want to structure your review course to provide lectures
only on the general subjects. Your DS review could be
assigned a." "independent study," using chapters and
problems from this book. Thus, your DS review would
consist of distribnting this book with a schedule of assignments. Your instructional staff could still provide
assistance on specific DS problems, and completed DS
assignments could still be recorded.
The final chapter on incorporating DS subjects into
review courses has yet to be written.
Like the landscape architect who waits until a well-worn path appears through the plants before placing stepping stones,
we need to see how review courses do it before we can
give any advice.
xv
Nomenclature
SURVEYING
a
c
d
D
E
E
g,
g,
I, t.
L
LC
M
r
R
T
x
Xm
parabola constant
length of sub chord
angle of subchord
ft
ft/ft
ft/ft
deg
ft
ft
ft
-/ft
ft
ft
m
rad
rad
m
m
m/m
m/m
deg
m
m
m
-/m
m
m
ft
ft
m
m
cross section of area of flow
watershed area
runoff coefficient
ft'
ft2 ac
m'
ha
ft
ft
ft
ft
ft
m
m
m
ft/sec'
m/s'
ft
ft
ft
ft/hr
m
m
m
m/h
ft/sec
ft
tti]«
m
Ibf/ft'
ft
ft
ft' [sec.
ft
ft' /sec
Pa
m
m
m3/s
m
m'/s
ft
deg
deg
degree of curve, arc definition
external distance (horizontal)
tangent offset at PVI
grade of back tangent
grade of forward tangent
intersection angle (angle between two tangents)
length of curve (horizontal)
length of long chord from PC to PT
length of middle ordinate
rate of change of grade
radius
tangent distance
horizontal distance from PVC (or point of tangency)
to point on curve
horizontal distance to min/max
elevation on curve
ft
HYDRAuLIcs AND HVDRO[OGICsvsTEMs .
A
A
C
d
d
D
DH
n
depth
diameter of capillary tube
pipe diameter
hydraulic diameter
roughness factor for pipe (specific roughness)
friction factor
acceleration of gravity
specific gravity
capillary rise
head
head loss due to friction
rainfall intensity
Manning's equation constant
hydraulic conductivity
length
cotangent of side slope angle
Manning roughness coefficient
p
pressure
p
precipitation
wetted perimeter
discharge rate
runoff
e
f
g
G
h
h
/>.f
i,I
k
K
L
171
P
Q
Q
Q
volumetric flow rate
1
m
m
2
(iyil Discipline-Spedfil Reyiew for the fEIEIl Exam
R,Ru
Re
S
S
t
hydraulic radius
Reynolds number
maximum basin retention
slope of energy grade line
time
velocity
width
v
w
W
power
z
elevation
f3
-y
angle made by liquid with the wetted tube wall
specific weight
1)
efficiency
(J
surface tension
p
density
It
m
It
It/It
sec
It/sec
It
ft-Ibf/sec
It
deg
Ibf/ft3
m
m/m
s
m/s
m
m
deg
N/rn3
Ibf/It
lbm/ft-'
N/m
kg/m3
W
SOIL MECHANICS AND FOUNDATIONS
A
B
c
area
cohesion
C,
C,
C"
Co
DIO
coefficient of curvature or gradation
compression index
uniformity coefficient
coefficient of consolidation
grain diameter corresponding to 10%
passing by weight or mass
grain diameter corresponding to 30%
passing by weigbt or mass
grain diameter corresponding to 60%
passing by weight or mass
D30
DG(J
Da
relative density
Df
depth of footing below surface
relative density
D,.
e
FS
factor 01 safety
specific gravity
specific gravity of water
layer thickness
height or thickness
hydraulic head
length 01 drainage path
settlement
hydraulic gradient
G
c;
h
H
H
»;
~H
i
J(
eoefficient of permeability (hydraulic conductivity)
I<a
coefficient of active earth pressure
coefficient of earth pressure at rest
coefficient of passive earth pressure
]\0
K1J
L
LL
length 01 slip plane
liquid limit
m
mass
11
number 01 drainage layers
porosity
bearing-capacity factors
number of potential drops in a flow net
n
n;
.!\lfil
N,
number of flow paths in a flow net
PPI. www.ppi2poss,com
jI,
-- - -
m'
III
Pa
ft2/hr
m'/h
It
III
ft
III
It
%
It
m
ft
ft
ft
It
ft
ft/ft
It/sec
m
m
m
m
rn
m/m
m/s
It
m
Ibm
kg
III
%
void ratio
original void ratio
Co
Na
N!
1t2
It
Ibf/lt2
width of a footing
Nomencloture
p
pressure
P
force or point load
r,
active resultant force per unit width
PI
PL
plasticity index
plastic limit
q"
unconfined compressive strength
ultimate fonndation bearing capacity
flow rate (per nnit width in a flow net)
quLt
Q
S
S
S
degree of saturation
shear force
SI
SL
shrinkage index
t
time
time factor for consolidation
pore water pressure
volumes of air, water, solids) and voids respectively
water content
settlement
shrinkage limit
T
'U
Val V
W1
V,;, V
W
Ww,Ws,Wt
a
')
1d
1,
1w
V
weights of water and solids and total weight, respectively
angle of failure plane
dry unit weight of soil
unit dry weight of soil
unit weight of solids
unit weight of water
(J
normal stress
(J
total stress
(J'
effective stress
p
density
T
shear stress or strength
¢
angle of internal friction
Ibf/ft2
lbf
lbf/ft
Pa
N
N/m
lbf/ft'
Ibf/ft2
ft3/sec
Pa
Pa
m3/s
lbf
ft
N
m
hr
h
lbf/ft'
ft3
Pa
m"
%
%
Ibf
deg
lbf/fts
Ibf/ft3
lbf/ft3
lbf/ft3
lbf/ft'
Ibf/ft'
lbf/ft2
lbm/ft"
lbf/ft2
deg
N
deg
N/m"
N/m"
N/m"
N/m3
Pa
Pa
Pa
g/cm3
Pa
deg
ft'
ft'
ft2
mg/L
days-l
days"!
m'
m'
ft
mg/L
ft3/lbmol
Ibm/day
ft3/sec
ft" /sec
ft" /sec
ft3/sec
ft3/sec
ft3/sec
mg/L
days
ft"
ft3
mg/L
mg/L
m
ENViRONNIENTAL·ENGINEERING··
Ax
surface area of unit
surface area of media in fixed-film reactor
cross-sectional area of channel
BOD
biochemical oxygen demand
k
kd
BOD exertion reaction rate constant) base e
microorganism endogenous decay rate
K,p
L
L
solubility product constant
linear length of weir
ultimate BOD
NI
M
molar volume
A
AM
Q
Qe
Qo
Q,
Qw
R
So
t
V
sludge production rate (dry basis)
flow rate
effluent flow rate
influent flow rate
sludge volumetric flow rate
waste sludge flow rate
recycle flow rate
influent substrate concentration (typically BOD)
time
tank volume
VA
aeration basin volume
X
suspended solids concentration
mixed liquor suspended solids concentration (MLSS)
XA
___________________________________
3
m2
mg/L
d-1
d-1
mg/L
L/gmol
kg/d
L/s, m3/s
L/s, m3/s
L/s, m3/s
L/s, m3/s
L/s, m3/s
L/s, m" /s
mg/L
d
m3
m3
mg/L
mg /L
PPI-www.ppi2pasuam
4
Civil Discipline-Speti!it Review for the FE/Ell Exom
X,
Xw
effiuent suspended solids concentration
waste sludge suspended solids concentration
the amount of BOD exerted at time t
solids residence time
wet sludge density
V,
8,
P.
mg/L
rng/L
rng/L
days
Ibm/It3
mg/L
mg/L
mg/L
d
kg/m3
ft/sec'
m/s"
%
%
ft
ft
deg
m
m
deg
TRANSPORTATION
a
A
C
d
D
e
e
E
J
J
9
G
G
GJ
C,
hI
h,
HP
I
K
L
L.
M
N
PC
PI
PVC
PVI
PVT
R
R
s
s
S
T
T
TE
v
v
v
v
W
acceleration
algebraic difference in road grades
vertical clearance for an overhead structure (underpass)
total stopping distance
degree of curve
efficiency of railroad diesel-electric drive system
superelevation
equilibrium elevation of outer railroad track rail
coefficient of friction between tires and roadway
side friction factor
acceleration of gravity
effective railroad track gage (center-to-center of rails)
grade of road
grade from which tbe stationing starts
grade toward which the stationing heads
height of driver's eyes above the roadway surface
height of object above the roadway surface
rated horsepower of a diesel-electric locomotive unit
interior angle
air resistance coefficient
length of curve
length of spiral
middle ordinate of curve
number of train axles
point of curvature (where back tangent ends
and curve begins)
point of intersection of back and forward tangents
point of vertical curvature
point of vertical intersection
point of vertical tangency
level tangent resistance
radius of curve (use minimum radius for
superelevation calculations)
braking/passing
sight distance
stopping sight distance
slope of road
(sem i-) tangent distance from PVI to PC
or from PVI to PT
driver reaction time
tractive effort of a locomotive unit
vehicle speed
velocity
design speed
initial speed
average load per train axle
PPI • www.ppi2poss.com
%
%
ft
m
ft/sec'
It
decimal
mis'
m
decimal
%
%
%
%
ft
ft
m
lip
deg
deg
ft
m
ft
ft
m
m
ft
ft
m
m
m
m
m
ft
ft
ft
lbf/ton
m
ft
ft
m
m
ft
III
Ft
sec
lbf
mi/hr
ft/se«
mi/hr
mi/hr
ton
m
s
N
km/h
m/s
km/h
km/h
------------""!'~+--~HH_+~.;~~~
1
Nomencloture
STRUCTURAL ANALYSIS
A
d
E
FEM
10
r;
I
L
Tn
]vI
n
p
p
r
ft
Ibl/lt'
It-Ibl
Ibl
Ibl
It4
It
Ibl-It
reaction
R
T
V
temperature
shear force
uniformly distributed load
weight
coefficient of thermal expansion
deflection
shear stress
ow
W
a
D.
T,V
m2
m
Pa
N·m
N
N
It'
cross-sectional area
distance
modulus 01 elasticity
fixed-end moment
member force due to a unit load
member force due to external loads
moment of inertia
span length
bending moment due to unit load
bending moment due to external loads
number
pressure
point load or force
radius of gyration
5
4
TTl
m
N·m
lbf-ft
N·m
Ibl/lt'
lbf
It
Ibl
of
Fa
N
m
N
Ibl
N
N/m
N
°e
lbf/ft
Ibl
l/'F
i/'e
ft
lbl/lt
m
Fa
STRUCTI.IALOESiGN
Reinforced (onerele Design
a,
depth of rectangular compressive stress block
Aconcrete
area of concrete
gross cross-sectional area
area of tension reinforcement
area of shear reinforcement within a distance,
along a member
section width in compression
effective flange width in compression
Ag
As
Av
web width
distance from extreme compression
fiber to neutral axis
effective depth
dead load
modulus of elasticity of concrete
compressive strength of concrete
modulus of rupture (tensile stress in bending)
yield stress of reinforcement
T -beam flange thickness
moment of inertia of cracked transformed section
effective moment of inertia
gross moment of inertia
effective length factor
live load
column moment
cracking moment
maximum service moment
nominal moment
factored moment
d
D
Ee
I~
IT
Iy
hf
fer
Ie
Ig
k
L
Me
M"
M
Olax
Mn
Mu
____
5,
-.-
""'!'
in2
in
in
in
in
in
Ibl, lbl/It,
Ibf/in'
lbf/in'
Ibf/in'
Ibl/in'
in
lbl/ft2
jn4
in4
in4
Ibf, lbljft,
ft-Ibf
ft-lbf
ft-Ibf
ft-Ibf
ft-lbf
Ibl/lt'
PPI. www.ppi2polS.com
6
f3,
P
Pb
Pg
r/J
Civil Dislipline-Specifil Review for the FE/Elf Exom
nominal axial load at given eccentricity
nominal P" for axially loaded column
factoral axial load
shear reinforcement spacing
nominal shear strength provided by concrete
nominal shear strength of reinforced section
nominal shear strength provided by reinforcement
factored shear force
concrete density used for calculating modulus of elasticity
uniform loading
distance from neutral axis to extreme fiber of concrete
section in tension
ratio of depth of rectangular stress block, a, to the
depth to the neutral axis, C
compressive strain in concrete
net tensile strain in extreme reinforcing steel
at nominal strength
reinforcement ratio for tension steel
reinforcement ratio for balanced stress condition
gross reinforcement ratio
capacity reduction factor
_
Ibf
Ibf
in
Ibf/in2
Ibf
lbf/in?
lbf
Ibf/ft3
Ibf/ft
III
Slrullurol Sleel Design
.4.
Ag
Agt
A,v
An.
Ant
AnY
b
Cc
db
D
D.
E
Fa
F"
F,
F.
r;
Fy
9
I
k
L
L
effective net area
gross area
gross area in tension
gross area in shear
net area
net area in tension
net area in shear
member width
critical slenderness ratio
bolt diameter
dead load
hole diameter
modulus of elasticity of concretc
allowable axial compressive stress
critical stress
allowable tensile stress
specified minimum ultimate strength
allowable shear stress
specified minimum yield stress
transverse distance between hole centers (
.'
moment of inertia
gage spacing]
P"
effective length factor for compression member
unbraced member length
live load
cri tical Euler axial loading
nominal axial strength
r
required axial strength
radius of gyration
r;
r.
Rn
s
SR
t
U
(sum of factored loads]
nominal block shear rupture strength
longitudinal distance between hoi
t
(p:
e cen ers pItch' spacll1g
. )
slenderness ratio
member thickness
reduction coefficient
PPI. www.ppi2pass.com
in2
in2
in2
in2
in2
in2
in2
in
Ibf, Ibf/ft,
in
lbf/in'
Ibf/in'
lbf/in?
lbf/in'
lbf/in'
Ibf/in'
Ibf/in'
mrn''
mm'
mm"
rnrn"
mm'
mrrr'
mnr'
mm
lbf/ft2
III
N, N/m, Pa
mm
Pa
Pa
Pa
Pa
Pa
Pa
Pa
mm
in"
rnm"
ft
lbf, Ibf/ft,
Ibf
Ibf
Ibf
III
Ibf
in
III
Ibf/ft2
m
N, N/m,
N
N
N
mm
N
mm
Pa
mm
-------~-:t~ti7~~~~~~~~~
___________________________________
PPI. www.ppi2poSl.[om
Practice Problems
For sta 20+10.50 to 20+21.50,
SlJRVE'i1NG
Problems 1 and 2 are based on the following information.
L = 21.50 m - 10.50 m
= 11.00 m
Earthwork quantities for a section of roadway indicate
a transition from fill to cut. The following areas are
scaled from the print cross sections.
(This is the transition from fill to cut, so use the formula
for pyramid volume to calculate cut area.)
AI+A2)
2
station
fill volume = L (
(m)
20+00
20+10.50
20+21.50
20+28.45
20+40
14.32
6473
187.42
173.21
43.56
9.63
2
2)
( 43.56111 ; 9.63 m
= (11.00 m)
= 292.5
area of base)
cut volume = h (
In the region where there is a transition [Tom fill to cut,
the fill area and cut area are both triangular in shape
on the road cross section.
m'
3
14.32 m2)
3
= (11.00)m (
= 52.5 rn3
Problem 1
The total volume of fill required for this section of road
is most nearly
For sta 20+21.50 to 20+28.45,
L = 28.45 m - 21.50 m
(A) 1430 m3
1450 m3
'(c) 1730 m3
(D) 1780 m3
~
REb
~;;.
=695m
(This is the transition from fill to cut, so use the formula
",1 (Q..,O:C}\D~orpyramid volume to calculate fill area.)
CL,
Solution
Y:>!:JOI
Earthwork volumes for fill areas and cut areas can be
calculated using the average end area formula. Since
the cut and fill areas are triangular in shape as given
in the problem statement, earthwork volumes in the
transition region from fill to cut can be calculated from
area of base)
fill volume = h (
3
9.63 m')
3
= (6.95 m) (
= 22.3 m3
the formula that gives the volume of a pyramid.
L = 10.50 111 - 0 m
= 10.50 m
AI
fill volume = L (
+
A2)
2
= (10.50 m ) (
AI + A2)
cut volume = L (
For sta 20+00 to 20+ 10.50,
2
14.32 m2 + 64.73 m2)
2
= (6.95 m) (
= 274.7 m3
173.21 m2 + 43.56 rn2)
2
= 1138.0 m3
9
10
Civa Discipline-Specific Review for the FE/EIT Exam
_
Problems 3-5 are based on the following information
and illustration.
For sta 20+28.45 to 20+40,
L = 40 m - 28.45 m
The proposed vertical profile for a transport
= 11.55 m
airport>
runway is shown.
cut volume = L (Al+A')
2
-0.5%
0.8%
B
64.73 m' + 187.42 m')
= (11.55 m ) (
2
z grade
1---=---1
o
= 1456.2 m3
A table chat summarizes earthwork volumes is now
Problem 3
made.
The minimum required length of vertical curve at the
point of vertical intersection (PVI) A is most nearly
cut
station
(m)
cut
fill
area
fill
volume
volume
(m')
(m3)
(m")
20+00
173.21
20+10.50
43.56
1138.0
52.5
20+21.50
14.32
20+28.45
6473
20+40
187.42
292.5
Solution
22.3
At point A, the grade change is from -0.5% to 0.8%.
Therefore, the absolute value of the total percent grade
change at PVI A is
9.63
274.7
(A) 300 ft
(B) 390 ft
(C) 1000 ft
~1300ft
1456.2
AA = 1-0.5% - 0.8%1
total
1783.4
Therefore, the total volume of fill required for this section of road is 1452.8 m" (1450 m").
The answer
= 1.3%
1452.8
For transport airports, the minimum required length of
vertical curve at the point of vertical intersection is
is B.
L.A--(
Problem 2
1000ft)A
1% change
1000 ft
The total volume of cut required for this section of road
= ( 1% change
is most nearly
= 1300 ft
1430 m3
(A)
(B) 1450 m3
~1730m3
~1780m3
Solution
From Sol. I, the total volwne of cut reqnired for this
section of road is 1783.4 m3 (1780 m3).
The answer
is D.
~A
)
(1.3%)
The answer is D.
Problem 4
The maximum allowed longitudinal grade (maximum
allowed downward slope) to the right of PVI B is most
nearly
(AJ -2.0%
(B) -1.5%
~-1.2%
(~-07%
Solution
Two criteria must be checked for this problem. First,
the maximum allowed longitudinal grade change is 1.5%
for
.
.
, transport air por t s. Second, the maximum
longitudinal grade is 1.5% for transport airports.
Practice Problems
Checking the first criteria, the absolute value of the
total grade change at PYI B can be rearranged to give
the maximum allowed longitudinal grade (maximum allowed downward slope), z , as
~max,B
11
Problem 6
CN~1800m
\E~900m
\
\
= 1.5%
= 0.8%-z
z = 0.8% - 1.5%
\
c
= ~0.7%
390.512 m \ a
\
\
\
The second criteria is that Izi not exceed 1.5%. which it
does not. Therefore, the maximum allowed longitudinal
grade (maximum allowed downward slope), z , is -0.7%.
\
\
\
\
\
A
The answer is D.
---------------------
If the grade to the right of the point of vertical intersection B is -0.4% (i.e., Z = -0.4%), the minimum
required distance between PYIs A and B is most nearly
N~?
E~'
The north and east coordinates of point B for the situation shown are most nearly
(A)
(A) 250 ft
(B) 630 ft
~
1000 ft
\~2500
ft
\
.. B
b
N~1300m
E~850m
Problem 5
\
430.116m
250 Ill, 350 m
(E) 350 Ill, 250 III
~
~
1200 m, 1550 m
1550 rn, 1200 m
Solution
Sol'ution
From Sol. 3, the grade change at the point of vertical
intersection A is ~A = 1.3%. For a -0.4% grade at
To solve this problem, begin by inversing
known coordinates of A and C.
between the
PYI B,
Ll.B = [0.8% - (-04%)1
(Cflorthing
c=
= 1.2%
Therefore, for transport airports, the minimum required
distance between PYIs A and B is
D = (
= (
Anorthing)2
+ (Ceasting - Aeasting)2
(1800 m - 1300 m)2
+ (900 m -850
1000 ft ) (Ll.A + Ll.B)
1% change
1000 ft ) (1.3% + 1.2%)
J% change
-
m)
2
= 502.494 m
.
h
t
aaimut AC = a .an
= 2,500 ft
= atan
Ceasting
-
AeastinK
Cnorthing
-
Anorthing
900m-850m
1800 m - 1300 m
= 5°42'38"
The answer is D.
Using law of cosines, angle A is
cos LA =
b2 + c2 _ a2
2bc
(430.116 m)' + (502.494 m)' - (390.512 m)'
(2)(430.116 m)(502.494 Ill)
= 0.6593
LA = 48°45'06"
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12
avll DisciplineoSpe,Hi' Review for the FE/!IThom ----------------------
The direction from A to B is found using the direction
of line AC and the calculated angle at A.
Solution
The gage pressure at point B is 0 kPa. The mass density
of water is Pw = 1000 kg/m3 at standard conditions.
5'42'38" + 48'45'06" = 54°27'44"
0
azimuthxn = 54 27'44"
Now use the known distance and direction of line AB
to find the latitudes and departures for the line.
latitudeAB = (430.116 m)( cos 54'27' 44")
departursjj,
pipe fluid
= 250 m
= (430.116 m)(sin54°27'44/1)
= 350 m
mercury
Use the coordinates of point A and the latitude and
departure for line AB to find the coordinates of point B.
Bnorthing = Anorthing
Therefore, from equilibrium,
PB=OkPa
+ latitudejj,
= 'Yoilhp + ')'Hghm
= 1300 m + 250 m
= 1550 m
Bea.'lting
= Aeasting
+ departureAB
= 850 m+350 m
= 1200 m
The answer
= GoilPwghp
+ PA
+ GHgPwghm + PA
This can be rearranged to solve for gage pressure at
point A.
+ GHgPwghm)
PA = -(GoilPwghp
is D.
(0.8) (1000 ~~) (9.81 ;)
~yDRAUUcs.A~D.HXD~OLOGIC·SYstEMs·
=-
C~~
~)
C~~
:)
+ (13.6) (1000 ~~) (9.81 ;)
Problem 7
A manometer is shown with heads of hp = 25 em and
hm = 63 em. The pipe fluid is oil with a specific gravity
of 0.8. Mercury has a specific gravity of 13.6. Assume
standard conditions.
= -86014 Pa (-86 kPa)
The answer is C.
Problem8
The rational formula runoff coefficient of a 300 ill long
hy 200 m wide property with a 3% slope is 0.35. The
rainfall intensity is 116 mm/h,
The discharge from this property is most nearly
mercury
(A) 2200 m' /h
(B) 2400 m' /h
(C) 3800 m3/h
(D) 7000 m3/h
Solution
The gage pressure at point A is most nearly
(A) -90 kPa
SJQ -88 kPa
((gJ-86
kPa
(D) -80 kPa
The discharge from this property is
Q=CiA
= (0.35) (116 ~)
= 2436 rn' /h
(
1 m ) (300 m)(200 rn]
h
1000 mm
(2400 rn' /h)
The answer is B.
Practice Problems
A concrete sanitary sewer is 150 ill long and has a pipe
diameter of 1.25 IIl. The inlet elevation is 50.0 Ill, and
the outlet elevation is 49.0 m. The Manning roughness
coefficient, assumed to be constant with depth of flow 1
is 0.012. During heavy rainfalls, the sewer pipe flows
full with no surcharge.
During heavy rainfalls) the capacity of the sewer is most
nearly
(JA)
~
(el
(D)
3.1 m' /5
3.8 m'/s
4.7 m3/s
5.7 m3/s
tion.
Water is pumped from a lake with a pipe inlet with a
head of 200 m to a tank with a head of 205 m. The
pipeline from the lake to the tank is 300 m long and
is cast iron: with a 30 em inside pipe diameter. The
pump efficiency is 80%. Minor losses, entrances losses:
and exit losses are negligible.
The flow rate through
the piping is 1.25 m3/s. Assume steady, incompressible
6
2
flow. The kinematic viscosity of water is 1 x 10- m /s.
The specific roughness for cast iron is e = 0.25 rnm.
Problem 10
Using the Darcy equation,
the head loss in the piping
iS~;~~Ym
Solution
The slope is
(E) 310 m
50.0 m - 49.0 m
s = Zinlet - Zoutlet
150 m
L
=000667
(C) 320 m
(D) 330 m
Solution
Since the pipe flows full during heavy rainfalls, the wetted perimeter is the entire perimeter of the pipe. The
hydraulic
10 and 11 are based on the following informa-
Problems
Problem 9
13
radius is
The specific roughness for cast iron is e
The relative roughness is
relative roughness = ~
R=:4:=
P
rr(1.25 m)'
0.25 mm
4
rr(1.25 m)
(30 em) ( 10-mm)
em
= 0.000833
= 0.3125 m
From the Manning equation) the velocity of flow is
The area of flow is
A = rrD2
= rr (30
4
= (~1_)
0.012
(0.3125 m)'/3(0.00667)1/2
= 3.13 m/s
The flow capacity
0.25 mm.
= 3.84 m' /s
C~o:m)Y
4
= 0.07069 m'
The Reynolds
number is
Q
is
Q = vA = (313
em)
:')
(rr(1.2:
)TI)')
vD
AD
I)
v
Re=-=--
3
(3.8 m /s)
The answer is B.
6
= 5.305 X 10
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14
Civil Discipline-Specific Review for the FE/EITExam
From the Moody diagram for the calculated relative
roughness and Reynolds number, the friction factor is
J '" 00188
Therefore, from the Darcy equation,
the piping is
"'f(~) (:;) =f(~)
the head loss in
(qn
300 m
= (0,0188) (
_
Problem 12
A reservoir with a water surface level at an elevation
of 200 rn drains through aIm
diameter pipe with the
,outlet at an elevation of 180 m. The pipe outlet discharges to atmospheric pressure. The total head losses
in the pipe and fittings are 18 m. Assume steady, ineompressihle flow.
The flow rate out of the pipe outlet is most nearly
er;;} 4,9 m /s
3
(B) 6,3 m' /s
(e) 31 m3/s
(D) 39
[s
J
m'
(30 em) (~)
100 em
Solution
1.25 ~
Using the pipe outlet as the datum, the variables in the
energy equation are as follows.
) 2
S
( 0.07069 m2
x
21
= 200 rn
water has negligible
]
[ velocity at reservoir surface
0 m/s
vi '"
= 299,6 m
reservoir free surface is ]
[ at atmospheric pressure
P, = 0 kPa
(2) (9.81 :~)
(300 m)
pipe outlet discharges ]
[ to atmospheric pressure
P2 = 0 kPa
The answer is A.
Z2
= 180 ill
Problem 11
The power required by the pump to raise water from
the lake to the tank at the flow rate indicated is most
nearly
The energy equation can be rearranged
velocity at the pipe outlet,
(A) 3.0 MW
3'8MW
) 4,7 MW
~ ) 5.4 MW
VI _ P2
PI
to solve for the
v~
~, + ZI + -2 9 - -i + Z2 + -2g' + hI total
29 (PI--:; + + vi2g - P2--:; ZI
Z2 -
hl,total
)
Solution
The total head required to lift the fluid is
h = (Zt"nk -
Zlak,)
2g (Pl-
+ hI
pg
+ZI
= (205 m - 200 m) + 300 m
=305
=
m
(2) (9,81
ill)
8'
The input power required by the pump to provide the
reg uired head is
9
-
hito"al )
,
The flow rate out of the pipe outlet is
7[d)
(1000~)
(9.81~)
Q = V2A = v, ( "4
(305 m)
0.80
106 W
pg
(0 m + 200 ill + 0 m
)
- 0 m - 180 rn - 18 ill
'7
(125~)
= 4675 x
"i
-2 - P2
- -Z2
= 6,26 m/s
w = Qiwh = QPwgh
']
+
2
= (626~)
(7[(I;n)2)
= 4.92 m' /s
(4,9 m3/s)
(4,7 MW)
The answer is C.
The answer
....
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..
is A.
~.":i~~~~~-~.!.!..!!.....:...".!..--_l.
Practice Problems
Problem 13
A pump discharges 1000 kPa water into a 90 m long,
0.1 m diameter steel pipe at 2 ta]«. The maximum elevation the water could reach if friction is neglected is
most nearly
1S
Information about conditions at the pump are not used
because point 1 comes after the pump.
The answer is C.
Problem 14
The name [or the flow in which quantity does not vary
with time at any location along the channel is
2
(A)
steady
flow
(8) critical flow
(e)
(D)
..
Solution
v = 2 rn/s
Steady flow is the term most often used to descrihe
a flow quantity that does not vary with time at any
location along an open channel. If the flow cross section
does not vary with location along the channel) it is said
to be uniform flow. Steady flow can also be nonuniform
flow, as in the case of a river with a varying cross section
or on a steep slope.
(A) 50 m
(8)
uniform flow
nonuniform flow
79 rn
(C) 100 m
(D) 120 m
Solution
This can be solved using the energy equation
2
2
VI
P2
V2
- + Zl + - = - + Z2 + - + h
I
29
I
29
PI
If Z, = 0, the potential energy is 0 J Ikg at the pump's
discharge. The pressure energy and velocity energy arc
(1000 kPa) (1000
P
Ep =p
The answer is A.
&a)
Problem 1S
A developer intends to build single-family
dwellings on
a 3.02 km" site that is currently pastureland,
The site
is to be broken into 50 lots of O.Oll km2 with a 186
m2 home on each lot. The remaining 0.02 krrr' will be
devoted to roadways. Assume a runoff time of ] 0 mill
and a rainfall intensity of 10 ern/h.
1000 k~
m
(2 :)
v2
e; = 2 = ~-2"--'-= 2 J/kg
Et,l = Ep + E;
J
=1000-+2kg
discharge
J
kg
Because the pipe is frictionless, the total energy is converted to potential energy at point 2.
= E',l
Z29 =
1002 J Ikg
1002 kJg
Z2=--=
m
9
9.81 "2
s
= 102.1 m (100 m)
Et,2
___________________________________
04
0.2
0.9
is
= 1002 J/kg
Et,2
C
0.1
006
pasture
forest
single-family dwellings
lawns
asphalt roadways
2
The total energy at the pump's
runoff coefficient,
land use
= 1000 J/kg
The storm sewer system
most nearly what flow?
should
he designed
to carry
(A) 062 m3/s
(8) 2.7 m3/"
(C) 3.1 m3/s
(D) 17 m3/s
Solution
Determine the areas of each classificat ion. There are
three: single-family
dwellings; lawns, and asphalt road-
ways.
Each 0.06 km2 lot contains
59814
m2
a ]86 m2 bouse,
leaving
of lawn per lot.
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16
Civil Discipfine-SpecificReview for the fE/lIT Exam
The total areas of each classification are
_
The pressure head at point B in the soil sample is mos
nearly
Adw,lling,
= (186
::)
(50 lots)
(A)
(B)
(C)
(D)
= 9300 m2
Alawn, = (59814
::)
= 2990700
(50 lots)
m'
Solution
2
Aroadways = 20 000 m
Atotal
6.00 III
6.33 m
6.67 III
6.75 III
= Adwellings + Alawns + Aroadways
= 9300 m' + 2990700 m' + 20000 m'
= 3020000 m'
The total head, ht, is the sum total of the elevation an,
This can lx
pressure heads as given by ht = he + hp.
rearranged to give hp = h, - he.
Tabulate the known heads with respect to the datum.
Find the average runoff coefficient for the entire site.
point
hp (m)
A
B
C
13
?
3
(9300 m') (0.4)
+ (2990700 m2)(0.2)
C
_ I:CiA, _
+ (20000 m2)(0.9)
ave A,otal
- ---';3"0"'2"'0"'0"'0"'0-m'-'o2~= 0.205
Use the rational equation to find the flow.
Since the soil sample is saturated,
homogeneous,
and
isotropic, and since the steady-state continuity equation requires constant flow velocity through the soil.
the head gradient through the soil is linear.
a-s ct«
3m
= (0.205) (em)
10 h (3020000 m')
h,
x
(llll)(lh)
100 em
3
= 17.20 m /s
3600 s
2m
(17 m' Is)
The answer is D.
-A
.....
---'13m
At point B, the pressure head is
Problem 16
hp,B
A constant-head permeameter is shown. The soil is homogeneous, isotropic, and saturated.
=3m+ (
1m
1m+2m
) (13 m - 3 m)
= 6.33 m
The answer is B.
drain""':
soil sample
===
water C
F1
3m
=1m
datu m - ~ _ . f---~"..J
-.l2 m
Problems 17 and 18 are based on the following informaticn and illustration.
At·
.
re ammg wall extends from the top of bedrock to
the ground surface. A resisting force is located on the
opposite side of the wall. A frictionless hinge at point A
Phreventsthe base of the wall from sliding. The soil is
omogeneous " isso t topic,
.
and cohesionless.
,~-----------------Prallice Problems
/"
The total force is the resultant, Pa, as determined
by
finding the total area under the active earth pressure
retaining wall
f
11
ground surface
profile.
2
Pa _
'21 L''( a"Y H' -- '2lK apg H
F
=
q, ~ 35°
(0.27) (1834
= 21860 Nym
p ~ 1834 kg/m'
1.5 m
G)
bedrock
~~)
(981
~~) (3 m)'
(22 kN/m)
The answer is B.
Problem 18
Prablem 11
Using the Rankine theory, the total active resultant lateral earth force per unit length of retaining wall is most
Assuming that wall friction is negligible, the minimum
required force per unit length of retaining wall to resist
the overturning moment is most nearly
(A)
(D)
(C)
15 kN/rn
22 k"l/m
44 kN/m
(D) 82 kN/m
nearly
(A)
(B)
(C)
15 kN/rn
22 k'J/m
44 kN/m
(D) 82 kN/m
Solution
The resultant,
Solution
From the Rankine theory: the coefficient of active lateral
earth pressure for cohesionless soils is given by
n; = tar? (45
Pal acts at a height above the bedrock
H
3m
-=--=lm
3
:1
From Prob. 17, P" = 22 kN per meter of wall length.
Summing moments about point A gives
0
-~)
= tan2 ( 45° -
of
35D)
2
P H
(22 kN) (1 m)
u
m
3
F = -= -"------::---::"--y
1.5111
= 0.27
= 14.67 kN/m
The active latera] earth pressure distribution
(15 kN/m)
is linear.
The answer is A.
/'
retaining
,
wall
ground surface
Problem 19
~0. W
A soil sample has a total mass of 23.3 g, a volume of
12 cm3, an oven-dry mass of 21.2 g, and a specific grav-
active earth
pressure
distribution
F
ity of 2.5 for the solids.
H=3m
The void ratio of this soil sample is most nearly
(A) 0.42
(B) 0.53
(C) 0.62
(D) 0.71
The active lateral earth pressure at any depth,
the ground surface C8Jl be found by
aa
= KalJv
h, below
= !(a"Yh
= Kapgh
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18
avif Discipline-Specific Review for the fE/EIT Exam
_
100
Solution
f'\
Par this problem, soil is modeled as a three-phase system,
80
air
V.
-
Vv
Vw
vt
l
water
v,
solid
m,
0
20
01,
\
r-,
10
0.1
0.01
0.001
grain size (mm)
ln this model,
Vi = total volume
m,
= total mass
Vv = volume of voids mw == mass of water
Va = volume of air
m. = mass of solids
Vw = volume of water
V. = volume of solids
The volume of soil solids is given by the followingequation, in which G is the specific gravity of the solids and
Pw is the density of water.
v _ m. _
21.2 g
, - Gp., - (2.5)
cm3
= 8.48 em'
(1 ...L)
The volume of the voids is found by subtracting the
volume of solids from the total volume.
= 12 em' - 8.48 cm3
= 3,52 em'
The void ratio is
3.52 em3
V,
8.48 em'
= 0.415 (0.42)
V,
The uniformity coefficient is most nearly
(A) 1.6
(B) 2.1
(C) 2.6
(D) 32
Solution
As read from the distribution
and DlO = 0.19 mm.
curve, DBD
= 0.49 mn
The uniformity coefficient is
D60
0.49 mm
Cu = = ::-::-DIO
0,19 mm
= 2.58 (2.6)
The answer is C.
Problem 21
Vv=v,-V,
e =-
Problem 20
The coefficient of gradation is most nearly
(A) 0.17
(B) 0.44
(e) 1.6
(D) 3.0
= -=c-:::--c-
The answer is A.
Solution
As read from the distribution curve, D3D = 0.39 mn
The values of DBD and DlO are from Prob. 20.
The coefficient of gradation is
Problems 20 and 21 are based on the fOllowinginforma_
C; =
A soil's grain-size distribution curve is as shown.
D10
D60DlO
tion and illustration.
=
(0.39 mm)'
(0.49 mm)(0.19 mm)
= 1.63 (1.6)
The answer is C.
Practice Problems
•
19
Problem 22
Problem 23
The specific gravity of a saturated soil sample is 2.70
with a total unit mass of 2400 kg/rn". The dry unit
An undisturbed sample of clay has a wet mass of 30 kg.
a dry mass of 23 kg, and a total volume of 0.014 m3
with a specific gravity of 2.05. The void ratio is
mass is most nearly
(A) 031
(A) 470 kg/m'
(B) 1500 kg/m3
(0) 2200 kg/m3
(D) 2500 kg/m3
(B)
050
(0) 061
(D) 1.00
Solution
Solution
-
3
Assume the total volume to be 1 rn .
ml
2400 kg
I
water
-
v;
mw = 7 kg
water
1 m'
1J
soil
m,
voids
30 kg
1-
m$ = 23 kg
soil
L---
Pw = 1000 kg/m'
Vt = Vw + V, = 1 rn3
lll, = lllw + lll, = 2400 kg
Use the relationship between the mass of water and the
mass density of water to find the volume of water.
rna = GV~pw
'fI1.w
v =~
= VwPw
(1000
Vw + (2.7) (1000
~~)
~~)
V,
(1000
~~)
(1 m
-
V,) + (2.7) (1000 ~~)
7 kg
1000 k~
m
= 0.007 m3
= 2400 kg
3
=
Pw
w
V,
Solve for the volume of solids using the relationship between the specific gravity of tbe sample, the solids mas".
and the mass density of water.
= 2400 kg
V, = 0.824 m3
Vw = 1
m" - 0.824 m'
23 kg
= 0.176 m3
(2.65) (kg)
JOOO m"
1718 = GVsPw
= 0.0087
= (2.7)(0.824
3
m ) (1000
m'
~~)
Find the volume of voids.
= 2225 kg/m'
The answer
(2200 kg/m')
Vv=V,-Vs
is C.
= 0.014
m' - 0.0087 Ill'
= 0.0053
m'
The void ratio can now be found.
Vv
0.0053
=
V,
0.0087
e= -
= 0609
The answer
____
~--~-------------------------
m'
1113
(0.61)
is C.
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20
CivilDislipline-Spelilic Review lor the FE/Ell hom
_
Problem 24
clay II
ground surface
L
5m
water
fill
kg) 9
un = (15 m) (kg
1890 3 - 1000 -3
3
+ (10 m) (2322
clay [
Psat == 1890 kg/m3
eo = 1.2
C, = 0.414
15 m
k~_
1000
m'
k~)
m
9
kg) 9
+ (20 m) (kg
2082 m3 - 1000 m
3
= 482109
10 m
Pa
U, = 48 210g Pa + (5 m)
(2002
~~)
9
= 58220g Pa
clay n
PSill
2082 kg/m3
m ) (1.6) log (5822099 Papa)
0=;
40m
ill
ill
r.,..."t~ab;:;,le~~""C"",...,.,~p;w~ot;=:;;2;:;0;;;0:2
;:;k9:::/::::nm
40
Sn = ( 1 + 0.79
eo"'" 0.79
Cc = 1.6
48210
= 2.93 m
Combining the two settlements
ment of the soil due to the fill.
Given the factors shown in the illustration, the settlement of the fill due to the compression of both clay
layers normally loaded is most nearly
(A)
2 m
gives the total
settle-
S'o'al = 1.12 m + 2.93 m
=4.05m
(4m)
The
is B.
answer
(E) 4 m
(C) 6 m
(D)
8 m
ENYilltlNl'iIENTAf.ENGINEEIlING
Problem25'
Sohrtion
To solve this problem,
each clay layer.
begin by finding the stress in
clay I
The chloride (CI-) concentration
in a lake is found to
2
be 10- M. The HgCI,(aq) concentration
is found to be
7
10- M. The following chemical equations and equilibrium constants apply.
ao = ~Hlayer (Psat - PH 0) 9
2
= (7.5 m) (1890
~~ - 1000 ~~)
= 66759 Pa+
(5 m) (2002
~~)
9
equation
is solved for this layer.
}fa
) Cclog- u,
-( 1 + eo
ou
= (~)
1+1.2
K2 = 3.0 X 106
5.9 x 10-17 M
(B) 3.3 x 10-12 M
(C) 3.0 X 106 M
(D) 5.6 x 106 M
(A)
= 166859 Pa
Sf =
+ CI- .=" HgC12
The concentration of Hg2+ is most nearly
= ero + HfillPwet9
Now the settlement
K, = 5.6 X 106
HgCl+
9
= 66759 Pa
(/1
Hg2+ + Cl- .=" HgGl+
pa)
(0.414) log (166859
6675g Pa
= 1.12 m
Solution
K2
=
[HgC12]
[HgCl+j[CI
6
] = 3.0 x 10
[HgCl+j = [HgC12]
1(2[Cl-]
10-7
Similarly, the stress and settlement
are found.
in the clay II layer
= "(3;-:. 0::-X-=-:1OC:;C6'""')
("-'10--=2)
= 3.33
X
10-12
M
[given]
PracticeProblems
21
Problem 27
The 10 000 gal aeration basin shown maintains
1
a. con-
stant 1500 mg(L mixed liquor suspended solids (MLSS)
concentration
(5.6 x 106)(10 2)
= 5.9 X 10-17
and treats 25,000 gal of liquid waste per
day. The suspended solids arc separated in a clarifier
with recycle of separated sludge. The recycle flow rate
is 5000 gal per day. Each day, 500 gal of recycle are
wasted.
M
The effluent from the clarifier contains a con-
stant 30 mg(L MLSS. Assume steady-state flow conditions.
The answer is A.
influent
Problem 26
Nickel is removed by hydroxide precipitation from water
with a pH of 9. The atomic weight of nickel is 58.70.
The chemical equation and solubility-product constant
for this reaction are
NiH + 20H- ~
Ni(OHMs)
aeration
basin
16
K,p = 5.54 X 10-
effluent
10,000 gal
clarifier
The solubility of NiH in this water is most nearly
(A) 0.0060 mg(L
(B) 0.33 mg(L
(e) 0.55 mg(L
(D) 0.59 mg(L
The solids residence time is most nearly
(A) 2 days
(B) 3 days
(C) 4 days
(D) 5 days
Solution
The given pH of 9 can be used to find the OH- concentration.
pH+pOH = 14
pOH=14-pH
-log[OW]
recycle
...:..::.=.:.::...._-.L..---+_
waste
'--
Sol7dion
= 14-pH
= 14-9
The variables for flow rates, Q, and MLSS concentrations, X can be shown on the illustration to help organize the solution to this problem.
1
=5
[OH-] = 1 x 10-5 mol(L
Qin
influent
From the chemical equation and solubility-product constant,
K,p = [Ni2+1I0H-f
= 5.54 x 10-10
[given]
Oe = (qn + (Or - Q~J) - 0
1
aeration
basin
[NiH] = [;~'PF
The solubility of nickel is
qn - Ow
effluent
VA; XA
5.54 x :10-16
= (1 x 10-5)2
= 5.54 X 10-6 mol(L
=
clarifier
~
Or
Xr
recycle
'---'-"..,--_==.::....
......_.1Qr-
Ow
X,
.. waste
Qw
Xw = Xr
[NiH] x MW
= (554
X 10-6 m~l) (5870 ;01) (1000 n~g)
= 0.325 mg(L
(0.33 mg(L)
The answer is B.
___________________________________
PPI. WWI'I.ppi2pcss.wm
22
Civil Discipline-Sped/it Review for the FE/lIT Exom
_
The theoretical minimum amount of chlorine required
to destroy the NaCN waste is most nearly
VA = 10,000 gal
XA = 1500 mg/L
(A)
(B)
(C)
(D)
X, = 30 mg/L
Qin = 25,000 gal/day
Q w = 500 gal/day
80 kg/d
160 kg/d
170 kg/d
200 kg/d
Qr = 5000 gal/day
Xw = Xr
[unknown]
To determine X" a solids balance must be taken at
the clarifier. Since the total solids entering the clarifier
must be equal to tbe total solids leaving the clarifier for
steady-state
= Q,X,
+ Q,X,
= (Qin - Qw)Xe
X
T
The total mass of NaCN flowing past a given point each
day is
flow conditions,
(Qin + (Q, - Qw))XA
= (Qin + (Q, - Qw))XA
Qr
-
+ QrXr
(Qin - Qw)X,
+ (5000 gal _ 500 gal))
(25,000 dgal
ay
day
day
Solution
The 225 mg/L concentration of cyanide can be expressed as 225 parts per million (ppm) = 225 parts/
106 parts.
m = Vp(concentration)
(1500 mg)
L
g
- (
- 500 -gal ) ( 30- m )
day
day
L
=_---'----_-"'=-_--'5~-----::~-e
2v,000
-gal
5000 gal
day
225 kg )
( i x 106 kg
1000 L
m3
= 21.4 kg/d
Relevant molecular weights are
MWNaCN
= 22.98977
= 8703 mg/L
Since Xw = XT)
I~~)
(95000 ~) (1000
~
mol
mo
.s.
mol
+14.0067
the solids residence time is
+ 12.011 ~l
= 49.007 g/mol
MWC1,
= (2) (35.453
~)
mol
= 70.906 g/mol
(10,000 gal) (1500 ~g)
(500 ::~)
(8703 n~g)
+ (25,000 gal _ 500 gal)
day
day
= 2.95 days
The answer
The number of moles of NaCN flowing past a given
point each day is
(30 mg)
L
214 kg
. -,d'-..- __
n=
(3 days)
(49007 ~)
mol
is B.
~
(~)
1000 g
= 436.7 mol/d NaCN
Problem 28
A water sample from a stream with an average flow of
95000 L/d contains 225 mg/L of cyanide waste in the
form of sodium cyanide (NaCN). Chlorine can be added
to the stream to destroy this NaCN waste according to
From the given chemical equation the destruction of
2 mol of NaCN requires 5 mol of chlorine (CI2). Therefore, th e amount of Cl required to destroy the gIven
.
2
amount of NaCN each day is
the reaction
2NaCN + 5Cl, + 12NaOH ---;
N, + Na,C03
mol Cl2 )
( 2 5mol
NaCN
+ lONaCl + 6H20
X
Atomic weights are found from a periodic table of the
chemical elements to be Na = 22.98977, C = 12.011
N = 14.0067. CI = 35.453, 0 = ] 5.9994, and B = 1.0079'
PPI.www.ppi2poss.com
(
436.7
mol NaCN)
d
(70.906 ~)
( 1 kg )
mol
1000 g
= 77.4 kg Chid
(80 kg/d)
The answer is A.
------~~r-Hib~tiiitM~Hi~*i_:+~~~
Practice Problems
23
Problem 29
Solution
A proposed landfill is to be 400 m x 200 m in plan
area and 25 m deep. The average daily filling pattern is
expected to be 15 m x 10 m x 3 m deep, and the daily
cover to be used is 0.2 m thick. Assume that the landfill
will be operational
every week from Monday through
The solid loading rate is determined using the solid flux
rate and the area of the clarifier. The solid loading rate
is
QMGoBOD
A
solid loading rate =
1fD2
A=-
Friday.
4
The projected
life of t.he landfill is most nearly
The solid flux equation
(A)
16 yr
(B) 17 yr
(C) 20 yr
(D) 23 yr
is
solid flux = (5MGD)
(150 n~g) (3.7854
g~l)
= 2839 kg/d
Solution
Combining the two equations gives
The total volume of the proposed
Vtotal
landfill is
(2839
= (400 m)(200 m)(25 m)
solid loading rate =
= 2 x 106 m"
k:)
(4)
2
1f (30.5
m)
2
= 3.9 kg/d·rn
The rate of trash into the landfill is
vtrash
=
(15 m)(10 m)(3 In)
1d
The answer is C.
=450m3/d
The rate of fill from use of the daily cover is
TRANSPORTATION
Problems
~over
31 and 32 are based on the following informa-
(15 m)(10 m)(0.2 m)
=
1d
tion.
= 30 m3/d
The connection matrix shown represents a road transportation network between six locations.
Since the landfill is to be operational
for 5 days each
week and there are 52 weeks in the year, the projected
node
life of the landfill is
tlandfill
=
vtrash
+ V.cover
node
6
= (
3
2 X 10 m
.)
m3
m"
450 d+30 d
= 16.03 yr
(1 Wk) (~)
5d
52 wk
1
2
3
4
5
6
1
2
3
4
5
6
0
1
0
0
1
1
1
0
1
0
0
1
0
1
0
1
-1
0
0
0
1
0
1
0
1
0
1
1
0
1
1
1
0
0
1
0
(16 yr)
Problem 31
The answer is A.
Problem 30
The solids loading rate for a 30.5 m diameter clarifier
with a flow rate equal to 5 MGD and an influent BOD5
equal to 150 mg/L is most nearly
(A) 1.1 kg/d·rn'
(B) 2.2 kg/d·m2
(C) 3.9 kg/dm2
(D) 4.2 kg/d·m2
_ ...
... ~
The total number
of links in the network
is
(A) 7
(B) 8
(C) 9
(D)
10
Solution
Connecting the nodes in accordance with the connection
matrix produces a graphical representation
of the road
transportation network.
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24
Civil Discipline-Specifi, Review for the fE/EIT Exom
2
_
(A) 15 veh/rni
(E) 45 veh/rni
(e) 75 veh/rni
(D) 230 veh/rni
6
3
Solution
The mean velocity relationship
mi
illl
60 --v
k =
-45-
hr
hr
mi2
0.2--
veh-hr
A link is defined by the nodes that exist at both its ends
and does not specify direction. Therefore, there are nine
links in this network.
answer
60 -
0.2--
~------<4
The
is rearranged
to give
mi
hr
mi 2
veh-hr
= 75 veh/rni
The answer
is C.
Problem 34
is C.
The maximum capacity of overall traffic volume for this
road is most nearl y
Problem 32
The total number of arcs in the network is
0
(E) 1
(e) 2
(D) 3
(A)
(A) 3400 veh/hr
(E) 4300 veh/hr
(C) 4500 veh/hr
(D) 5000 veh/hr
Solution
Solution
An arc is a link with a specific direction
assigned to it.
JIl the connection matrix, a negative number indicates
The mean speed relationship
can be substituted
into
the traffic flow relationship resulting in a quadratic relationship (i.e., a parabolic curve).
that the direction of travel from one node to another
goes against the direction assigned to that arc. From
lhe graphical representation
of the network shown in
Sol. 31, there is One arc.
The
answer
2
mi ))
q = kv = k ( 60 -mi - ( 0.2 --hr
veh-hr
k
2
= ( 60 -mi) k - ( 0.2 --rni )
hr
veh-hr
is B.
k2
5000
Problems
33 and 34 are based on the fOllowing informa-
tion.
capacity
.c~
A traffic flow relationship is given by q::::: kv, where q is
the traffic volume in veb/hr, k is the traffic density in
veh /mi, and v is the mean speed in lui/hl". The mean
speed on a road in mijlu· is given by the relationship
4000
s:
• 3000
~
cr
'~"
E 2000
a>
1000
2
v = 60 -mi - ( 02 --rai )
lu
veh-hr
k
50
100
150
200
250
300
density, k (veh/mi)
Problem 33
Jf the mean speed on a road during the rush hom is
45 mi/hr, the maximum capacity of traffic density for
this road during rush hour is most nearly
To determine the traffic volume capacity it is necessary
to find the maximum point on the parabolic curve (i.e.,
the location where the slope of the curve equals 0).
dq
-=0
dk
-----"""!'-"""!'-~~it'~r'H*H~~~~~'!""'~~
PPI. www.ppi2poss.(Q111
Procti,e Problems
d ( (60 ~) k - (02 ~)
k')
25
Problems 36 and 37 are based on the following illustration of a vertical sag curve.
dk
.
hr
= 60 m' _
(0.4 ~ ., )
veh-hr
k
~IPVC-2.0%
=0
PVI ~ st. 87 +00
== 743 m
60 mi
hr
k =
PVJ elevation
'2
04~
veh-hr
Problem 36
= 150 veh/mi
The low point station
Substituting k = 150 veh/rni into the traffic flow relationship gives
mi)
~
~6%
(
q = ( 60 hr
k -
= (60 ::)
(150
mi')
0.2 veh-hr
k
2
v::)
- ( 0.2 --mi')
veh-hr
(A) sta 78+55
(B) sta 87+00
(C) sta 87+44
(D) sta 91+00
Solution
( 150- veh)2
In
L
8 sta
PVC = PVl - - = sta 87+00 - -2
2
= sta 83+00
= 4500 veh/hr
The answer is C.
for the vertical curve sbown is
located at
Use the low point formula to solve [or the distance from
the PVC to the low point.
Problem 35
A sag vertical curve has an elevation of 443.15 ft at
sta 7+65 and an elevation of 441.16 ft at the point of
vertical intersection (PYI). For an overpass crossing the
vertical curve's roadway at sta 7+65, the required clearance is 15 ft 4 in. What is most nearly the minimum
Xm.
(-2%)(8 sta)
G,L
=
-2% -16%
G, -G,
= 4.44 sta
=
low point = PVC + Xm
= sta 83+00 + 4.44 sta
elevation of the overpass?
= sta 87+44
(A) 4278 ft
(B) 4280 ft
(e) 4565 ft
(D) 459.0 ft
The answer is C_
Problem 31
Using the values in the illustration, the elevation of the
Solution
low point on the vertical curve is most nearly
15 ft + (4 in) (1ft)
-.12 III
= 15.33 ft
The overpass can be no lower than 443.15 ft + 15.33 ft
= 458.48 ft. Option D is the only option that is greater
than or equal to this value.
The fact that the curve is a sag curve does not change
the solution procedure. The elevation at the PYI is not
used.
(A) 743 m
(B) 747 rn
(e) 755 m
(D) 758 m
Solution
Determine the low point gradient elevation along the
originating tangent. Working from the PYI at sta 87+00
for 44 m,
The answer is D.
elevgn"u~" = 743.00 m - (0.02)(44 01) = 742.12 III
,,;,;...
.:.:0
PPI. www.ppi2pO\s.com
26
Civil Discipline-Spedfic Review for the FE/EITExam
_
Soluiior:
from the PVC, the low point is located at
Divide the beam into two shapes
as shown
L
800 m
-+44m=--+44m
2
2
=444
Determine
offset, y, at the low point,
the tangent
y=
m
(0, - GI)x'
2L
_ (0016 - (-002))(444
-
2
m)'
(2)(800 m)
= 4.44 m
eleVlow point = 742,12 01 + 4.44 m
= 746,56 m
(747 m)
Al = (0,08 m)(0,15 m)
The answer is B.
= 0,012 m'
A, = (0,05 m)(0.15 m)
= 0.007501'
STRueriiRAL'ANALYSIS"
The distance from the top of section 2 to the centroid is
Problem 38
The v-coordinate
(measured [rom the top) of the centroid for the T-shaped beam is
,
_ "A,y
i..J
I, e,l
Yc ~Ai
0,05 m 0.05 m 0,05 m
I-r-------I-'
((0.012
x
(
=
ill))
ill')
)
m') (0.0~
+ ((0.0075
(~+
0,08 m))
0.012 m' + 0,0075 m'
= 0,08 m
0,23 m
The answer
is A.
Problem 39
The force in member HE from the truss shown is most
nearly
y
G
(A)
0.08 m
(B) 0,16 m
(e) 0,30 m
(D) 038 m
12,0 m
(A) 1110 N compression
(B) 1110 N tension
(e) 2490 N compression
(D) 2490 N tension
Procli,e Problems
Solution
Problem 40
Sum moments
The magnitude of the vertical reaction
port A is most nearly
about A to find the vertical reaction at E.
LMA=O
dEA
(D)
_ (2224 N)(6 m)
16 kN
12 m
E,» -
Tbe free-body
= 1112 N
Solution
about point E is
Since support D is a roller support, the horizontal reaction force, RAx, is 0 kN. To find the vertical reaction
at support A, RAy, a free-body diagram is drawn of the
entire truss and moments are summed about support D.
diagram
~H
DE
r'
RA'I(15 m) - (20 kN)(5 m) = 0 kN
R
_ (20 kN)(5 m)
15 m
Ay -
= 6.67 kN
Summing forces in the horizontal
horizontal reaction at E as 0 N.
direction
The
(A) 4.2 kN
(E) 6.7 kN
(C)
FEH,h = (2)(1112 N)
Solution
force in member EH is
(FEH,v)2
= J(1112
From Sol. 40, the reaction at support A is 6.67 kN.
Next, a free-body diagram is drawn for support A with
member forces and their force components. Using the
Pythagorean theorem, the relative magnitudes of each
force and each force's horizontal and vertical components can be found. (In this case, 6.403, 5, and 4 are
the relative magnitudes of the member AB force and
the forces's horizontal and vertical components; respec-
+ (FEH,h)'
N)2 + (2224 N)2
= 2487 N compression
(2490 N compression)
The answer is C.
Problems
85 kN
(D) 11 kN
= 2224 N
J
is B.
The magnitude of the compressive force in member AB
is most nearly
The horizontal component of the force in member EH is
FEH =
answer
Problem 41
= 0 lbf
= 1112 N compression
The resultant
(6.7 kN)
gives the
The sum of forces in the vertical direction is
FEH,v - RE,v
force at sup-
(A) 3.3 kN
(E) 67 kN
(e) 10 kN
RE,u
R
27
tively. )
40~42 are based on the following information
and illustration.
A plane truss is loaded as shown.
RAy
_"!"'
"""!
~
6.67 kN
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28
Civil Discipline-Specific Review for the FE/Ell Exom
For equilibrium,
all forces on a free body must sum to
o kN. Summation of vertical forces gives
This can be rearranged to give
FA By = -RAy
= -6.67 kN
_
Tbe positive sign in the calculated member AE foro
means that the assumed direction of the force on thr
free-body diagram, which indicates tension. is in tht
same direction as the calculated force.
A free-body diagram of joint E will show that tbe vel
tical member BE is unable to sustain any horizonta
force. Therefore, the force in member EF is the sam,
as the force in member AE.
Recall that joints in trusses are frictionless, so no bending moments exist.
FEF = FAE
= 8.33 kN
The force and its components are proportional to the
geometric lengths of the triangle sides.
FAB
6.403
4 ill
= (
m)
FABY}
m) (-6.67 kN)
= (
4m
6.403
= -10.68 kN
(-11 kN)
The answer is negative. This means that the calculated
force is in the opposite direction to the assumed force direction on the free-body diagram. Since the member AB
force was assumed to apply tension on the free-body diagram, the negative answer means that member AB is
In compression.
The answer is D.
Problem43
If the truss members are made of steel and the crosssectional area of each member is 1000 nun', tbe magnitude of the vertical deflection at joint E is most nearly
(Al 0.70 mm
(B) 1.6 mm
(C) 2.3 mm
(D) 2.8
mm
Solution
Use the principle of virtual work.
The answer is D.
Problem 42
The magnitude of the force in member EF is most nearly
(A)
(B)
(C)
(D)
(8.3k'l)
4.2 kN
5.3 kN
67 kN
83 kN
The actual forces in each member can be determined
by applying the equations of equilibrium to each truss
joint and are shown as follows. (Some round-off error
exists in these calculated numbers.)
Truss member lengths in meters are as follows.
Solution
From Sol. 41, FASy is -6.67 kN. The horizontal component of the member AB force is
FAB"
= (: :)
FAS"
The actual forces in kilonewtons are shown.
(: :)
(-6667 kN)
= -8.33 kN
For equilibrium at support A, the sum of the horizontal
forces must be equal to 0 kN. The force in member AE is
8.33
6.87'1~
14~
FilE = FAEx
= -FABx
= -(-833
= 833
kN
~!G'.--~-.".~~~=~~D
8.33
E
8.33
F
16.67
kN)
6.67
13.33
Practice Problems
•
Application of a vertical unit load at joint E results
in the virtual member forces in kilonewtons as follows.
(Some round off error exists in these calculated numbers. )
Problems
29
44 and 45 are based on the following illustra-
tion.
p~ 15kN
w « 5 kNlm
rTTlllD
f_%------=-_,I_~~f I
~
0.83
,.
0.67[A
5m
5m
5m
~--="",,"_f--!- __--::-::-::-""":~-::-:-:""-HD
0.83
E~ 1
0.83
F
0.42
Problem44
0.33
0.67
It is recommended
that a table be used to keep all vari-
ables organized.
FQ
Fp
L
member
(kN)
(kN)
(m)
AB
10.67
16.67
21.34
8.33
8.33
16.67
0.00
6.67
10.67
1.07
0.42
0.53
0.83
0.83
0.42
1.00
0.33
0.53
6.4
5.0
6.4
5.0
BC
CD
AE
EF
FD
BE
CF
BF
73.1
35.0
72.4
34.6
34.6
35.0
0.0
8.8
36.2
5.0
5.0
4.0
4.0
6.4
The magnitude of the vertical reaction
port A is most nearly
Solution
Since support C is a roller support, there is no horizontal reaction force at that point. The vertical reaction
force at support A, RAy, can be found by converting
the uniformly distributed load, W, into a resultant point
load.
W = wL = ( 5 -kN) (5 m)
m
= 25 kN
This resultant load is located at the centroid
formly distributed load.
resultant
point E can be found by
~E
(~15kN
= "DFQJL = "FpL
DFQ AE
= (AlE)
LFQFpL
329.7 kN'·m
,(
(1000 mm)
1 m )'
1000 mm
x (2.1 x 10 11 Pa )(lkN)
1000 N
x (1000 m:)
= 1.57 mm
(1.6 mrn)
lAY
I"
1--
The answer is B.
___________________________________
W = 25 kN
~
t ,,, 1.r1
o
B
-----=--_.1.
-.
5m
of the uni-
lTTTTl
t
-----0--
5m
l ~l
5m
The support A vertical reaction can then be found by
summing moments about support C. This results in a
vertical reaction at support A of
RAy(lO m) - (\5 kN)(5 m)
+ (25 kN)(2.5 m) = 0 kN·m
Since the unit load in the virtual force system was downward and the answer is positive in sign, the actual deflection is also downward.
SllP--
(Al 1.3 kN
(B) 5.0 kN
(C) 13 kN
(D) 20 kN
2: = 329.7
The modulus of elasticity of steel is E = 2.1 X 1011Pa.
Since the area and modulus of elasticity are the same
for all truss members, their product, AE, is common to
all members and can be taken outside of the summation
for simplification. Therefore, the vertical deflection at
force at
RAy
,(1_:.:5..,:k=-N~)
:.:(
5...:,=-n
)~-c--'-=(2:.:5..,:k=-Nl.:)
(:.:2::::.
5--.:I::.c.Il)
=10m
= 1.25 kN
(1.3 kN)
PPl www.ppi2pass.com
e
30
Civil DiscipJine-Spe,i1i, Review lor the FE/EIT Exom
_
Since the answer is positive in sign, the direction of the
calculated reaction is the same as that 01 the assumed
reaction; that is, the direction of the reaction is upward.
STRUCTURAL DESIGN
Problems 46 and 47 are based on the following information and illustration.
The answer is A.
The cross section of a reinforced concrete beam with
tension reinforcement is shown. Assume that the beam
is underreinforced.
Problem 45
The magnitude of the maximum vertical shear in the
beam is most nearly
(A)
(B)
(C)
(D)
I~= 3000 lbl jin 2
1.3 kN
14 kN
25 kN
39 kN
,I
12in
w ~ 5 kN/m
23 in
I
3-No. 9 bars
ID
B
1.25
[three no. 9 bars]
5m
cITTTTl
Al
VlkN)
A, = 3 in2
20in
5m
,I
~p~ 15kN
LAV ~
Ibljin2
I•
Solution
One V'lay to determine the answer to this problem is to
construct a shear diagram. The change in shear is the
area under the applied loading. Although a moment
diagram is not required for this problem, it follows that
the change in moment is the area under the shear diagram, so a moment diagram is usually included.
5m
Iv = 40,000
I
I
1.25kN II
I
I
I
I
I
•••
(
Lcv ~
38.75kN
I
I
25
A,
Problem 46
a
In accordance with American Concrete Institute (ACI)
strength design, the allowable moment capacity 01 the
beam is most nearly
(A) 160 It-kips
(B) 180 It-kips
(C) 200 ft-kips
(D) 210 It-kips
M IkN'm) Or-~=====:::::::O""---_i_---=~
-62.5
Solution
A,
3 in2
P = -bd = "(1"'2-'-in-C:)"(2:::-0-'-in-C:)
As can be seen: the maximum value of vertical shear is
25 kN at support C.
= 0.0125
The answer is C.
In an actual design and analysis situation. a check
should always be made to see that the actuai reinforcmg steel ratio falls between the allowable maximum and
allowable 1"
k
.
m mmum steel ratios, even though this cbec
IS not required t
. problem.
o so.Ive t hi18 specific
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~---
~~~
..
~---~----"
Praelice Problems
The minimum allowable
following two equations.
A
=
. S,[Jlin
d wf'
3b w V Jc
steel ratio is the larger of the
f
=
Ibf
= ¢ ( 085 f~ab ( d ~ ~) )
In
40,000 ~
y
III
(0.85)
= 0.986 in'
~=
(
III
= 1.2 in'
(3000
::;)
(3.92 in)
)
= (0.90)
(200)(12 in)(20 in)
lbf
40,000 ~
200bwd
As:ffiin=
sect.ions, rjJ = 0.90. The
For flexure in tension-controlled
allowable moment capacity is
lbf
3000 ~
(3)(12 in)(20 in)
31
. (
3.92 in)
x (12 Ill) 20 in - ~2-
1 kip ) ( I ft )
x ( 1000 lbf
12 in
[controls]
= 162.3 It-kips
(3, is 0.85 because the concrete stress is less than 4000
lbf'/in". With a single layer of no. 9 bars (diameter =
1.125 in), the distance to the extreme steel tension face,
dt, is 20 in + (1.128 in/2) = 20.56 in. The maximum
allowable st.eel in a singly reinforced beam is
(160 ft-kips]
The answer is A.
Problem 47
If the dead load shear force in the beam is 5 kips and
A
_ 0.85f~(3,d
s,max
fy
(3dt)
7
(085)
(3000 ~)
-
(0.85)(12 in))
lbf
40,000 ~
(
(A)
in))
= ,5.73 in'
0.986 in'
Solution
< 3 in' < 5.73 in'
For shear, ¢ = 0.75 as specified by ACI.
Therefore, the actual reinforcing steel ratio falls within
allowable limits and conforms t.o ACI specifications.
The depth
of t.he concrete
0.00l0 in'
(B) 0.0012 in'
(C) 0 135 in'
(D) 0.18 in'
In
x C3)(20~56
t.he live load shear force in t.he beam is 15 kips, t.ben t.he
minimum amount of shear reinforcement needed for a
center-to-center stirrup spacing of 12 in based on ACI
strength design is most nearly
compressive
The ultimate shear force in the beam is
Vu = 1.2Vdead + 1.6Vjive
= (1.2)(5 kips)
stress block is
+ (1.6)(15 kips)
= 30 kips
given by
A,fy
The nominal concrete shear strength is
a = 0.85f;b
Ibf)
(3 in2) ( 40,000 in'
= _---,,...--O--c;-~~(0.85) (3000
::;)
=2
(12 in)
fuf
3000 i112(12 i11)(20 in)
(
lk~
)
1000 lb!
= 26.29 kips
= 3.92 in
¢Vc = (0.85)(26.29
2
kips)
2
= 11.2 kips
Since
_______________________________
"~l>
¢Vc/2, shear reinforcement
is required.
PPl www.ppi2pass.cam
o
....
32
Civil Discipline-Specific Review for Ihe FE/ElYExom
_
The ACI minimum required shear reinforcement for a
stirrup spacing,
'I
of 12 in is
81
(50)(12 in)(12 in)
Ibf
40,000 '"2
a = 18 in
I-I
D 1'8in
m
= 0.18 in'
The nominal shear strength provided by reinforcement
is given by V, = Avfyd/ s, The amount of shear reinforcement, A required by this equation can be found
by using q'>(Vc + V,) ;0: Vu and rearranging.
8ft
V1
_
Av -
v" -q'>Vc
t
r/!fy-
F
column load: D = 200 kips
L ~ 100 kips
d
s
.
30 kips - (0.85 )(26.29 kips)
=----;----'---;~__;-'-:-;-:--C'_7;cc;;-c-__,__
(0.85) (40,000 :~;)
G~::)
CO~~i~bf)
",.
= 0.135 in'
"-
reinforcing steel centroid
The larger value for shear reinforcement controls. Therefore, Au = 0.18 in2
Note: Although not required for this problem, in an
actual design and analysis situation, a check should be
made to ensure that V, does not exceed the ACI-allowed
maximum
8./l:M
shear
reinforcement
given
by
Vs,max
=
The answer is D.
In accordance with ACI strength design, the controlling
(maximum) factored shear stress is most nearly
(A) 25 Ibf/in'
(B) 30 Ibf/in'
(C) 35 Ibf/in'
(D) 43 lbf/in 2
Solution
Problem 48
A square column is supported by a square reinforced
concrete footing with depth to reinforcement of d =
33 in as shown. The column supports a dead load of
200 kips and a live load of 100 kips. The ACI code
requires that the loaded area of footing for beam shear
starts at distance d away from the column face and that
the loaded area of footing for punching shear starts at
distance d/2 away from the column face,
The ultimate applied load is
r; = l.2D + 1.6£
= (1.2)(200 kips)
+ (1.6)(100 kips)
= 400 kips
The net ultimate soil pressure is
r;
400 kips
qu = A = (8 ft)(8 ft)
= 6,25 kips/ft'
Check beam shear.
PPI • WWW.PPi2POSS.lOm
.".!..__
~--.!._~.".!..~..!!....~~~..!_~~.!.!..!.....!.!..~.....!.-~!....-."....!~~
Practice Problems
33
The punching shear stress is larger than the beam shear
stress. Therefore, punching shear controls.
loaded area for beam shear (Shaded)~
The answer is D.
Q
Problems 49-51 are based on the following information
and illustration.
A floor system consists of 20 reinforced concrete beams
and a continuous 3 in deck slab. (A typical section is
shown lor the deck and two 01 the beams.) Assume the
beams are underreinforced.
jl-6in
I~in
9 in--
I--
3in
~~~~~~~_l
The factored shear stress lor beam shear is
20in
lbl
6 in ) (8 It) (1000 k )
ip
( 12 In
It
kiPS)
( 625 It'
(8 It)
I_I
12in
I_------cc~--I
48in
(12 ;) (33 in)
2
I; = 3000 lbf /in'
= 7.89 lbl/in
Iy = 60,000 lbf lin'
Check punching shear.
L = 30 It
[simple span length]
loaded area for punching shear (shaded) ~
Problem49
For each beam in the floor system, the ACI-specified
effective top flange width is most nearly
8ft
a+ d
.1
(A)
(B)
(C)
(D)
36 in
48 in
60 in
90 in
Solution
The effective flange width is
m
The factored shear stress lor punching shear is
(beam span) =
(8
= 42.6 lbllin'
(30 It) (12 ~:)
= 90 in
Vu = 4(a+d)d
kiPS)
( 6.25 1t2
m
It)(8 It)
)
b; = min
bw + (16)(slab depth) = 12 in + (16)(3 in)
= 60 in
_ (51 in)(51 in)
(1000 lbl)
In'
kIp
(
144 It"
(4) (18 in + 33 in) (33 in)
2
(431bl/in )
beam centerline
.
spacmg
Therelore,
48 .
In
the effective flange width is 48 in.
The answer
___________________________________
=
is B.
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34
Civil Disdpline-Spedfll
Review for the FE/Ell Exam
Problem SO
Assume the effective flange width for this beam is 48 in.
[f the area of reinforcing steel per beam is 7.25 in2, the
nominal moment capacity of each beam based on ACI
strength design is most nearly
(A) 680 ft-kips
(B) 770 It-kips
(C) 800 It-kips
(0) 880 It-kips
Solution
This problem asks for the nominal moment capacity,
lvln1 not the allowable moment capacity, ¢Mn.
Therefore) the reduction factor, cPl is not needed.
The depth of the concrete compressive stress hlock must
be checked to see whether or not it exceeds the 3 in deck
thickness. If the depth of this compressive stress block
exceeds the dcck thickness, then ea.ch beam is a Tvbearn
and T - beam formulas apply for determination
of the
nominal moment capacity. If, howevcr, the depth of the
concrete compressive stress block does not exceed the
deck thickness, then each beam is a rectangular beam
ann rectangular beam formulas apply for determination
of the nominal moment capacity.
_
To find the correct depth a, sum horizontaJ forces in tb
T-beam to show that the upper (above the neutral axis
compressive concrete stress block force is equal to th,
lower (below the neutral axis) maximum tensile fore
sustained by tbe reinforcing bars, By dividing tbe en
tire concrete compressive stress block section into thre
parts (a rectangular part and two overhanging flanges)
depth a can be found.
A f is tbe area of overhanging flanges. Ac is tbe total
area of concrete compressive stress block. Ar is the area
of concrete compressive stress block in the rectangular
part of the T-beam between the overhanging flanges.
Af = (b, - bw)hf
= (48 in-12
in)(3 in)
= 108 in'
From equilibrium
of horizontal
A f
A - -----!!.:....1L _
c -
0.85!,
forces,
lbf)
( 7,2"".,Ill) ( 60,000 '""2
In
-
(Jbf)
(0,85) 3000 '""2
c
III
= 170.59
b = be = 48 in
in2
Ar = b'wa = Ac - Ai
j
a = Ac - Af _ 170.59 in' - 108 in'
b.;
12 in
at
= 5.22 in
20in
A,
I
• .I
bw= 12 in
First. assume that each beam is rectangular
with a
width of b = be = 48 in. The depth of the concrete
compressive stress block is
Alternatively,
used.
a =
=
a redefined
A,Iy
_ hf(he - bw)
0.851'b
c w
bw
(7.25 in') (60'000
block depth
could be
Ib;)
III
Ibf)
(0,85) ( 3000 in'
(7.25 in") (60,000 Jbf)
A,Iy
in?
a = -= --i--'--;-;:c<;--'---''0,85f~b
) (
Ibf)
(0,85
3000 in2 (48 in)
stress
(12 in)
(3 in)(48 in -12
in)
12 in
= 5.22 in
= 3.55 in
Since
0.
> slab dcpth of 3 in, the beam is a T-be3.1n.
Since a = 3.55 in was found by assuming that the beam
was a rectangular beam with width b = 48 in, this deptb
only indicates wbether or not the beam is a T-beam and
is not the correct depth for determining the nominaJ
moment capacity, The correct depth is now found b
applying the concepts of static equilibrinm to the bear:.
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---------~-~""!"~""!"~~~O:-~'!-~-
•
e
Prallice Problems
The nominal
moment
u; = 0.85f;hf(b
e
capacity
bw)
-
of the T-beam
is
(d _ h;)
Since a < slab depth of 3 in, the nominal moment
capacity of the beam is the same as for a rectangular
singly reinforced concrete beam. Using b = be = 48 in,
it is
+ o 85f;abw (d -~)
(0.85) ( 3000 lbf)
in2
M; = o 85f;ab (d - ~)
(3 in)
(0.85)
(
x (48 in - 12 in) (23 in _ 3 ~n)
Ibf) (5.22 in)
+ (0.85) ( 3000 in
2
(3000
:~;)
(2.94 in)
)
2.94 in)
x (48 in) ( 23 in - -21ft
) ( 1 kip )
x ( 12 in
1000 lbf
5.22 in)
x (12 in) ( 23 in - -2-
= 646 It-kips
(650 ft-kips)
This can also be calculated by using the following equation in accordance with rectangular singly reinforced
concrete beam theory.
1ft
) ( 1 kip )
x ( 12 in
1000 lbf
= 765 ft- kips
3S
(770 ft- ki ps)
Even though the problem statement assumes that the
beam is underreinforced, the actual reinforcing steel ratio and its limits should always be checked in real design/ analysis problems.
The answer is B.
Even though the problem statement assumes that the
beam is underreinforced, the actual reinforcing steel ratio and its limits should always be checked in real design/analysis
problems.
The
is D.
answer
Problem SI
Assume the effective flange width for this beam is 48 in.
If the area of reinforcing steel per beam is 6.00 in", the
nominal moment capacity of each beam based on ACI
strength design is most nearly
(A)
CONSTRUCrIONMANAGEMENT'
Problem S2
What is the following type of chart called?
150 ft-kips
(B) 160 ft-kips
(C) 590 It-kips
(D) 650 ft- kips
operation
I
I"'"
~~_""TC,!--":~'-----,~""TC,----+-:------.----!--ye:::a;:,r~2'----;--,-------j
Jul
Feb Mar
Apr
excavation
Solution
This problem asks for the nominal moment capacity,
Mn, not the allowable moment capacity, ¢Mw Therefore , the reduction factor, ¢, is not needed.
First. assume that each beam is rectangular
with a
width b = be = 48 in. The depth of tbe concrete compressive stress block is
a=
A,fy
085f~b
. 2) ( 60,000 ;;:;2
lbf)
(6.00 III
(0.85) (3000
:~;)
formwork
legend:
scheduled work ~
actual work ~
(A) critical path chart
(B) rectangular-bar
progress schedule
(C) PERT chart
(D) triangular-bar
progress schedule
(48 in)
= 2.94 in
___________________________________
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36
CivilDiscipline-Specific Review for Ihe FE/Ell Exom
Solution
The chart shown is a triangular-bar
_
progress schedule.
D
9
The answer is D.
F-------i
~}----(
E
3
Problem 53
What is the following type of model called?
c )--------1
7
H
1
B 4
~
The critical path for this project is
I
A
1
C
2
start
I
E
3
~
\
end
F
\
D 5
(A)
(B)
(C)
(D)
a bubble (activity-an-node) network
an arrow (activity-an-arrow) network
a PERT chart
a bar (Gantt) chart
Solution
This is called an arrow (activity-an-arrow)
(A) start-A-F-cnd
(B) start-C-H-end
(C) start-B-D-F-end
(D) start-B-E-H-end
SolHtion
Problem 55 can be solved with critical path method
(CPM) calculations. Since tbe project is to start on January 1, it is easy to define the actual start time as the
end of the previous day, December 31, and designate it
as day 0, with January 1 designated as day l.
network.
The answer is B.
Determine the earliest start time (EST) and earliest
finish time (EFT) for an activity by making a forward
pass through the diagram. The duration, D, of each
Problem 54
activity is known.
What is the following type of model called?
start f---W---<!c
The EST of an activity is calculated as the maximum
of the EFTs of the activities preceding it. For example,
activity A has no activities preceding it, so it has an
EST of day 0. The EFT of this activity is
end
F
(A)
(B)
(C)
(D)
a bubble (activity-an-node) network
an arrow (activity-an-arrow) network
a PERT chart
a bar (Gantt) chart
Solution
This is called a bubble (activity-an-node)
network.
The answer is A.
Problem 55
A construction
project is composed of activities A
through H with the durat.ions, in days, given for each
activity in the diagram shown. This project is on a strict
schedule that must be maintained and is scheduled to
start at the beginning of .Ianuary 1. Work can only be
performed during the day and must be done on every
day of the week (Sunday through Saturday).
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EFT = EST + D = day
= day 8
°+
8d
Similarly, activity B has an EST of day 0 and an EFT
of day 5.
Activity D is preceded only by activity B, so its EST is
the EFT of activity B. That is, the EST of activity D
IS day 5 and the EFT is day 14.
Activity F is preceded by two activities, A and D, so
the EST of activity F is the maximum EFT of the two.
That is, the EST of activity F is day 14 and its EFT is
day 17.
Similar calculations show that the minimum project duration IS 17 days (i.e., the EFT for activity F is day 17).
D
etermination of the latest start time (LST) and latest
finish
tune (LFT) f
".
d
Or an actrvity IS done by a backwar
pass through the dl'ag
.
h
. t
.
ram using t e maximum projec
duratmu as the st ti
.
. h
startm.
. t c ar mg point. That IS , day 17 IS t e
g pam ror these calculations.
----------------~i-+~*~~~+_~~~
Procti,e Problems
The LFT of an activity is the nummum LST of the
activities following it. For example, the LFTs of the
activities precediug the project fiuish are all day 17 siuce
no activities follow them. Accordingly, the LFTs of
acti vities F and G are each day 17. The LST of activity
F is
LSF = LFT - D = day 17 - 3 d
37
Use the following diagram for Probs. 56 and 57.
A
E
1
2
D
F }-_...-{
5
end
3
= day 14
G
6
B
Similarly, the LST of activity G is
4
LST = LFT - D = day 17 - 5 d
Problem 56
= day 12
Activity D is preceded ouly by activity F, so its LFT is
the LST of activity F. That is, the LFT of activity Dis
day 14 and the LST of activity D is day 5.
Activity B is preceded by two activities, D and G, so
its LFT is the minimum LST of the two. That is, the
LFT of activity B is day 5.
The total float time (TF) of an activity is determined
by either subtracting the EFT from the LFT or subtracting the EST from the LST. For example, the TF
of activity
G is
TF = LFT - EFT = day 17 - day 10
Making a summary table helps to organize these results.
duration
EST
EFT
LST
activity (d)
(day no.) (day no.) (day no.)
E
F
G
H
0
0
0
8
5
7
9
3
3
5
1
(A) 0 d
(B) 1 cl
(C) 6 d
(D)
25 d
Solution
Solve this problem using the critical path method
(CPM), The critical path for this project is start-.I\.-CD-G-encl. Because activity G is along the critical path,
the float time for this activity is a d.
The answer is A.
=7d
A
B
C
D
A construction project h3.5 activities A through G. Each
activity's duration is given in days. The float time for
act.ivi ty G is
5
5
14
5
8
8
5
7
14
8
17
10
9
6
0
9
5
13
14
12
1G
LFT
(day no.)
TF
(d)
14
5
16
14
16
17
17
17
6
0
9
0
8
0
7
8
The critical path is the path that passes through the
activities that result in a TF of 0 d. From the summary
table, this passes through nodes B-D- F.
~\-------::J~
start}---{ ~~-------{
E
3
Problem 57
The EFT for activity F is
(A) 3 d
(B) 13 d
(C) 16 cl
(D) 19 d
Solution
A summary table is used to organize the characteristics
of each activity, including the EFT and EST.
activity
duration
(d)
EST
(clay no.]
EFT
(day no.)
A
B
C
1
4
7
a
a
1
D
5
E
F
G
2
3
6
1
8
1
13
13
8
13
3
16
19
4
~\--------{~
critical path -
The answer
is C.
_~ __ "'!"~~
~
~
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r
38
Civil Discipline-Specifi( Review for the FE/Ell Exam
Determine the EST by making a forward pass through
the diagram. The EST is calculated as a maximum of
the EFTs of the activities preceding it. For activity A,
no activities precede it so it has an EST equal to day
O. The EFT is found using the equation EFT = EST
+ duration. For activity A
EFTA = day 0 + 1 d
= day 1
Activity C is preceded by two activities, start and activity A, so the EST of C is the maximum EFT of the
two. The EST of activity C is day 1 and its EFT is
day 8.
The minimum project duration is 19 d.
_
MATERIALS
Problem 59
A steel rail was installed when its temperature was 5°C
(L = 20.0 m, A = 60 X 10-4 m"). The rail was installed without allowance for expansion and the ends
were constrained by adjacent rails. The coefficient for
linear expansion is 11.7 x 10-6 1rC. What is the compressive force in the rail at 25°C?
(A) 281 kN
(B) 285 kN
(C) 291 kN
(D) 295 kN
Solution
The thermal strain is found as
Activity F is preceded only by activity D. The EFT of
activity D is day 13, so the EST of activity F is day 13.
The EFT of activity F is
EI.I,
EFT F = day 13 + 3 d
= day 16
Note: A backward pass is not necessary because the
LST and LFT will not affect the EFT for activity F.
= a(T2 -
T,)
= (11.7
x 10-6
o~)
(25°C - 5°C)
= 2.34 X 10-4 m/m
The modulus of elasticity of steel is 20 x 1010 Pa and
the compressive stress is given by Hooke's Law a.'S
ath
The answer is C.
= EEth
= (20 X 10'0 Pa) (234
x 10-4
:)
= 4.68 x 10' Pa
Problem 58
The maximum time that an activity can be delayed
without causing the project to fall behind schedule is
(A) critical path
(B) float time
(C) earliest start time
(D) latest start time
The compressive force is
F = "thA
= (4.87 x 10' Pa) (60 x 10-4 m")
= 280800 N
(281 kN)
The answer is A.
Solution
The critical path connects all of a project's activities
that have a minimum of zero slack time. It is the longest
direct path through the network.
The slack time, also called float time, is the maximum
amount of time an activity can be delayed without causing a delay in the schedule.
The earliest time at which an activity may begin in the
schedule of a project is its EST. The LST is the latest
time an activity may begin without causing a delay.
The
answer
is B.
Problem 60
The engineering stress in a solid tension member is
324056 kPa at failure. The reduction in area is 80%.
The true stress at failure is
(A) 16 kPa
(B) 160 kPa
(C) 1.6 MPa
(D) 1.6 GPa
Solution
True stress is
a=
s
1-
RA
= 1620278
The answer
PPI. www.ppi2poss.[Qm
324056 kPa
= ----'---'=-=
1 - 0.80
kPa
(1.6 GPa)
is D.
~~---~----------
.. -
Practice Problems
39
Problem 61
The engineering
stress in a solid tension member is
324 MPa at failure. The reduction of area is 80%. The
true strain at failure is most nearly
(A) 40%
(B) 60%
(e) 140%
(D) 160%
Solution
The true strain is
S=lnC_1RA)
= 1.61 x 100%
= 161%
(160%)
The answer is D.
___________________________________
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Practice Exam 1
PROBLEMS
(A) sta 5+32.4
(B) sta 9+92.6
(e) sta 11+27.9
(D) sta 13+92.4
1. The value of ¢ is most nearly
y
110,11)
3, A boundary and traverse line bordering an irregular
area are shown.
boundary\
I
I
I
I
I
Ih2=8.1m
I
I
I
I
I
14,31
I
I
Ih,=4.2m
I
I
I
x
13°09'07/1
23°27'53/1
(e) 36°52'12/1
(D) 53°07'48/1
(A)
(B)
I
traverse line~
5m
5m
Using Simpson's
2. A 6° curve has forward and back tangents that intersect at sta 14+87.33. The station of the point of
beginning curvature (Be) is most nearly
I
I
I
I
Ih3=6.7m
I
I
I
I
.I•
5m
Ih4=7.6m
I
I
I
I
I
I
I
I
I
I
I
I
I
hs = 8.3 mr
I
I
I.
5m
1/3 rule, the total area between the
boundary and traverse line is most nearly
(A) 141 m2
(B) 143 m2
(e) 148 Ill2
(D) 151
m"
PI at sta 14+87.33
6 = 11<'21'35'1
Problems 4 and 5 arc based on the following information.
EC
BC
R
R
R
A back tangent with a 7% grade meets a forward tangent with a -5% grade on a vertical alignment. A 350 m
(10 sta) horizontal length of vertical curve is placed
such that the point of vertical curvature (PVC) is at
sta 10+35 at an elevation of 60.0 m.
4. The vertical curve elevation at sta 11+35 is most
nearly
65 m
67 m
(e) 69 m
(D) 71 m
(A)
(B)
41
42
Civil Discipline-Specific Review for the FE/EITExom
5. The tangent elevation at the point of vertical intersection (PVI) is most nearly
(A)
(B)
(C)
(D)
66 m
68 m
70 m
72 m
Problems 6 and 7 are based on the following information
and illustration.
A horizontal curve is laid out with the point of curve
(PC) station and the length of long chord (LC) as shown.
_
9. A reservoir with a water surface at an elevation of
200 m drains through aIm
inside diameter pipe with
the outlet at an elevation of 180 m. The pipe outlet
empties to atmospheric pressure. The total head losses
in the pipe and fittings are 18 m. Assume a steady,
incompressible flow of 4.92 m3/s.
A turbine is installed at the pipe outlet. The chosen
turbine has an efficiency of 85% and does not add any
head loss to the system. The expected power output of
the turbine is most nearly
(A) 82 kW
(B)
96 kW
(C) 100 kW
(D) 120 kW
Problems
tion.
10 and 11 are based on the following informa-
A circular sewer with a 1.5 m inside diameter is designed
for a flow rate of 15 m" /s when flowing full. Assume
that the Manning roughness coefficient and Darcy friction factor are constant with depth of How.
PC 10+46
10. The How rate when the depth of How is 0.50 m is
most nearly
(A) 1.7 m-' /s
6. The radius of the curve is most nearly
(8)
(C)
(D)
(A) 158 m
(B) 160 m
(C) 316 m
(D) 320 m
7. The point of tangent
(PT) is most nearly at
3.4 m' /s
50 m' /s
7.5 m' [s
11. The velocity of flow when the depth of How is 0.50 m
is most nearly
(A) 2.8 rn/s
(B) 4.2 ui]«
(C) 4,8 m/s
(D) 6.6 m/s
(A) sta 13+27.1
(B) sta 13+28.5
(C) sta 13+34.4
(D) sta 13+39.2
8. Given the cross section of a triangular channel as
shown, the wetted perimeter is most nearly
12. A property consists of 7500 m? of lawn area with
a runoff coefficient of 0.20, 2000 m2 of gravel roadway
with a runoff coefficient of 0.15, and 500 m2 of roof
surfaces with a runoff coefficient of 0.80.
The overall runoff coefficient for the entire property
most nearly
(A) 015
(A)
2,4 m
(B) 3.0 m
(C) 4,3 m
(D)
(B) 0.22
(C) 038
(D) 0.80
6.7 rn
PPI • WWW.PPi2pass.com----
lIIIII!----~---!.....-...!....".!.....!...-~!.....lIIIII!!.....-"".!.---
is
Praltiu Exam1
Problems 13 and 14 are based on the following information.
A housing development has a population of 20000 people. The average sewage flow for the development is
8000 m'/d.
impermeable
upstream
dam
H,
13. The estimated minimum sewage flow for this housing development is most nearly
soil
(A) 2000 m'ld
(B) 2600 m'ld
(C) 4000 m'ld
(D) 6000 m3/d
I'
impermeable
14. The estimated peak sewage flow for this housing
development is most nearly
17. The rate of flow per lineal meter of dam width is
most nearly
(D)
15. A river has a continuous water flow of 10 m'ls between two bridges that are 1000 m apart. At bridge A,
upstream, the river has a cross-sectional area of 150 m",
while at bridge B, downstream, the river has a crosssectional area of 100 m2. The increase in water velocity
between the two bridges is most nearly
(A) 0,033 so]«
(B) 0,067 mls
(C) 0.D75 mls
0,130
I I
bedrock
(Al 3 x 10-4 m'ls
(B) 5 x 10~3 m' /5
(C) 2 X 1O~2 m' /s
(A) 8000 m'ld
(B) 20000 m'ld
(C) 27000m'/d
(D) 32000 m'ld
(D)
43
m/s
16. The shrinkage of a soil that has a density of 5200
kg/m' naturally and 7500 kg/m3 when compacted is
2 x 1O~1 m' /s
18. If the dam is 10 m wide, the total flow rate of water
passing underneath the entire dam is most nearly
(A) 3 x 10-3 tn'' Is
(B) 5 x 10-2 m'ls
(C) 2 X 1O~1 m3 Is
(D) 2 m"/s
Problems 19-21 are based on the following information
and illustration.
A 10 m thick clay layer lies between
two soil layers as shown. The coefficient of consolidation
is 0.004 m'/d. It is predicted that the total consolidation settlement of the clay layer will be 8 ern.
most nearly
(A)
0%
(B) 30%
(C) 70%
(D)
Problems
10 m
clay
90%
17 and 18 are based on the following informa-
leliI'
sand (for Probs. 19 and 20)
III--. impervious
bedroc~(for Prob. 21)
tion and illustration.
A flow net is drawn for seepage through homogeneous,
isotropic soil beneath an impermeable concrete dam as
shown. Beneath the soil lies impermeable bedrock. The
upstream water level, H i, is 3 m above the top of the
soil, and the downstream water level, H21 is 1 ill above
the top of the soil. The soil's coefficient of permeability,
k, is 3 X 10-'
__________
The following table relates the average one-dimensional
consolidation of a uniform clay layer to its corresponding time factor.
csa]«.
""'!'
PPI. www.ppi2poss.mm
44
Civil Discipline-Specific Review for the FE/Elf Exom
Uavg (%)
T
0
10
20
30
40
50
0.000
0.008
0.031
0.071
0.126
0.197
19. Assuming that the clay layer lies between two sand
layers, the amount of time required for 20% of the total
settlement to occur is most nearly
(A)
(B)
(C)
(D)
39 d
78 d
190 d
780 d
_
station
23.
fill (rn")
10+00
20+00
30+00
150
25
75
25
o
100
The table represents the area, of cut and fill at eaeh
roadway station along a rural road project. The amount
of borrow or waste between the stations is most nearly
(A) 925 m3 borrow
(B) 925
waste
(C) 25000 m' borrow
(D) 25000 m' waste
m'
24. A saturated sample of undisturbed clay has a wet
mass of 318 kg. The total volume is 0.193 m", The soil
has a dry mass of 204 kg. What is the specific gravity
of the soil?
20. Assuming that the clay layer lies between two sand
layers, the total amount of settlement after 1 yr is most
nearly
(A)
(B)
1.0 em
2.1 em
(C) 3.2 em
(D) 4.3 ern
21. Assuming that the clay layer is bounded by a sand
layer above and impervious bedrock below, the total
amount of settlement after 1 yr is most nearly
(A) 2.35
(E) 258
(0) 2.65
(D) 290
25. A concrete basement wall extends below the
water table as shown in the following illustration. The
total lateral pressure against the wall at a point 2.4 m
below the surface is most nearly
(A) 1.0 em
(B)
2.1 cm
."iJ ~.:~
1.2 m
(C) 3.2 ern
.,
p = 1922 kg/m3
".<:1 •.
(D) 4.3 cm
......
"p
<>
22. The base of a 2 ill wide continuous footing is 1 m
below the ground surface. The cohesionloss soil under
the footing has the following properties.
oj
•
0";";, .t-----~ ...
c- water table
204m
II
-.
,<).
D· .'.
p ~ 961 kg/m'
1.2 m
Ka = 0.4
." ..
oj
•
.D." •. :<;;.....
v«
p = 1835 kg/m'
N; = 9.6
1> = 10°
Nq = 2.7
(A) 2.6 kPa
(B) 14 kPa
c = 0.0
(C) 25 kPa
N7 = 1.2
(D) 34 kPa
If a factor of safety of three is required, the allowable
bearing capacity
nearly
(A) 17kPa
(B)
23 kPa
(C) 49 kPa
(D)
70 kPa
PPI. www.ppi2poss.wm
of the soil under the footing is most
26. A 0.2 m layer of soil-bentonite is to be placed beneath the secondary geomembrane liner of a proposed
landfill. The layer is to be constructed in two 0.1 m
lifts. The bentonite content is to be 5% (dry weight
basis}. The compacted dry density of the soil/bentonite
mixt ure is 1630 kg/ m' .
..
.. _ ....
_
P,actite Exam 1
The amount of dry bentonite that has to be mixed into
each lift is most nearly
temperature
(0C)
oxygen solubility
(mg/L)
21
22
23
24
90
8.8
8.7
8.5
8.4
(A) 82 kg/m'
(B) 8.6 kg/m'
(C) 150 kg/m'
(0) 160 kg/m'
25
27. The number of covalent bonds formed by a carbon
atom under normal conditions is
The percent saturation of dissolved oxygen in the water
sample is most nearly
(B) 4
(A)
(B)
(C)
(C) 94%
(A)
3
5
(D)
(D) 6
45
63%
66%
96%
28. Most frequently. water hardness is caused by ions
32. A rectangular channel is 1.5 m wide, 3 m deep, and
of calcium and
25 ill long. The design flow rate of the wastewater in
the channel is 0,5 m3/s. The approach velocity in the
channel is most nearly
(A) Iron
(B) magnesium
(C) manganese
(0) sodium
Problems 29 and 30 are based on the following inforrna-
(A) 01 m/s
(B) 0.2 m/s
(C) 0,3 m/s
(D) 0.5 m/s
tion.
A sample of wastewater is incubated for 7 d at a temperature of 20°C, After this incubation period, the BOD
is found to be 211 mg /L. Assume that the reaction rate
constant is 0.14 d-1 (base e).
29. The ultimate BOD of this sample is most nearly
(A)
(B)
(C)
(D)
0.01 krn
0.03 km
(C) 004 km
(D) 0.06 km
(A)
(B)
225 mg/L
310 mg/L
3,10 mg/L
560 mg/L
34. A freeway route bas a horizontal curve with a PI
at sta 11+01.86, an intersection angle, ~, of 12°24'00"
right, and a radius of 537 m. The PC station is lo-
30. The five-day BOD of this sample is most nearly
(A)
(B)
(C)
(D)
33. A driver with a reaction time of 0.5 s is driving a
car down a 5% grade at 89 km/h when a deer Tuns into
the road 0.04 km in front of the car. The road is dry
with a coefficient of friction of f = 0.7. The distance
the car travels from the moment the deer appears until
the car comes to a stop is no less than
110 mg/L
170 mg/L
180 mg/L
200 mg/L
cated at
sta 10+44
(B) sta J.l +60
(C) sta 12+84
(D) sta 14+08
(A)
31. A freshwater sample has a dissolved oxygen concentration of 5.7 mg/L when the temperature is 23.3 C
and the atmospheric pressure is 730 mm Hg. A partial
listing of the solubility of dissolved oxygen in freshwater at equilibrium with dry air containing 20.9% oxygen
and at an atmospheric pressure of 760 mm Hg is as folD
35. A one-lane rural road includes a 10° curve extending for 700 It. The road is 15 ft wide with 9 ft wide
shoulders. The design speed [or this road is 45 rni/hr.
lows.
___
.. _"'!""'!"!' __
'!"!"--~----------------------
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46
CivilDiscipline-Specifi, Review for the FE/Ell Exom
The minimum required length of spiral transition
tween the curve and road is most nearly
(A)
(E)
(C)
(D)
_
be-
28 ft
36 ft
44 ft
50 ft
are within
su-
an accept-
(E) No, the VMA is excessive.
(C) No, the dust-to-asphalt
ratio is too high.
(D) No, Gmm at Nmax is too high.
design speed = 80 kmjh
coefficient of friction = 0.35
driver reaction time = 2.0 s
driver eye height = 1.2 m
(to be avoided) height = 0.2 m
The downhill design braking
is most nearly
distance
for this highway
(A) 45 III
(E) 75 m
(C) 100 m
(D) 120 m
39. A road leading to a stone quarry is traveled by
40 trucks, with each truck making an average of 10 trips
per day. When fully loaded, each truck consists of a
front singie axle transmitting
a force of 10,000 lbf and
two rear tandem axles, each axle transmitting a force
of 20,000 lbf. The load eqnivalency factor for the front
single axle is 0.0877. The load eqnivalency factor for
each rear tandem axle is 0.1206.
The 18,000 Ibf equivalent single axle load (ESAL) for
the truck traffic on this road for 5 yr is most nearly
0.33 ESAL
(E) 130 ESAL
(C) 48,000 ESAL
(D) 240,000 ESAL
(A)
37. A crest on a section of highway consists of a vertical
curve with a 1500 m radius and a positive 1% grade followed by a negative 3% grade. The design requirements
are as follows.
Problems
tion.
40 and 41 are based on the following illustra-
tr
design speed = 80 knr/h
driver eye height = 1.2 III
object (to be avoided) height = 0.2 m
stopping sight distance = 300 m
~
The minimum required length of vertical curve needed
to satisfy the design stopping sight distance is most
nearly
(A)
(E)
(C)
(D)
satisfy their corresponding
(A) Yes, all the parameters
able range.
36. The design requirements for a section of highway
with a 1.5% grade are as follows.
object
Do these characteristics
perpave requirements?
680 m
700 m
760 m
840 In
8
1_. ~---J
5m
15 kN
w~5kN/m
lffilll'D
>---;o-_;:I '
••
1
5m
40. The magnitude of the maximum
in the beam is most nearly
38. A superpave design mixture for a highway with
ESALs < 107 has a nominal maximum aggregate size
of 19 mm. The mixture has been tested and has the
following characteristics:
air voids = 4.0%
VMA = 13.2%
VFA = 70%
dust-to-asphalt
ratio = 0.97
at N = 8 gyrations, Cmm = 87.1 %
at N = 174 gyrations, Gmm = 97.5%
5m
.
Lending moment
(A) 6.3 kN·m
(El 14 kN·m
(C) 25 kN·m
(D) 63 kN·m
41. If the beam is made entirely of steel and the whole
beam has a moment of inertia about the axis of bending
8
of 2.0 x 10 mrn", the magnitude of the vertical deflection at point D is most nearly
(A) 020 rnm
(E) 2.3 mm
(C) 23 mm
(D) 50 mrn
---------------~"'!"-!'"'-~~_i~_+~-!'"'
PPI • www.ppi2poss.com
q
Pradi,e Exam 1
Problems 42 and 43 are based on the following information and illustration.
A truck is facing in its intended direction of travel along
the beam as shown.
L;~
J
71.17 kN
(rear axle) ,
4.27 m
44. In the x-direction as shown, the maximum influence line ordinate for tensile force in member BF is most
nearly
(A)
(B)
(C)
(D)
0.18 kN/kN
0.36 kN/kl\
0.53 kN/kN
071 kN/kN
45. In the x-direction as shown, the maximum influence line ordinate for compressive force in member BF
is most nearly
J 17.79 kN
, (front axle)
I
47
I
(A) -0.71 kN/kN
(B) -0.53 kl\ /kN
(C) -036 kN /kN
(D) -0.18 kN/kN
Problems 46 and 47 are based on the following information and illustration.
42. For a truck traveling in the direction shown, the
maximum vertical live load shear at support C is most
The cross sections of two short; concentrically
reinforced concrete columns are shown.
loaded
nearly
f~= 4000 Ibf/in2
'v ~
(A)
27 kN
(D) 76 kN
(C) 100 kl\
(D) 110 kN
60,000 Ibf/in2
18in
I'
18in
-
43. For a truck traveling in the direction shown, the
maximum live load bending moment at support C is
.
,I
eight bars
most nearly
longitundinal
(A) 80 kN·m
reinforcement
(B)
(C)
90 kN·m
140 kN·m
(D) 360 kN·m
round spiral column
(cross section)
IProb.461
~=::::=::::=-J
square tied column
(cross section)
IProb. 471
Problems 44 and 45 are based on the following information and illustration.
A plane truss span is shown. The roadway behaves, as
simply supported beam spans between the supportmg
lower chord truss joints. By convention positive forces
are tensile forces, and negative forces are compressive
1
46. For the short round spiral column, the applied axial
dead load is 150 kips, and the applied axial live load is
350 kips. Assuming that the longitudinal reinforcing
bars are all the same size, the minimum required size of
each longitudinal reinforcing bar is
(A) no. 7
forces.
B
C
(B)
•
no. 8
(C) no. 9
(D)
no 10
4m
rOadWay)
1
~
~
E
I '
,
5m
______
'!""'
F
1
~
' ,
•I
1
5m
47. For the short square tied column, the applied axial
dead load is 150 bps, and t.he applied axial live load
is 250 kips. Assuming r.hat the longitudinal reinforcing
5m
~----~-----------------
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48
Civil Discipline-Specifj( Review for the FE/EITExam
_
bars are all the same size, the minimum required size of
each longitudinal reinforcing bar is
Problems 50 and 51 are based on the following information and illustration.
(A) no. 3
(B) no. 4
(C) no. 5
(D) no. 6
A bolted steel tension member is shown. The total applied design load consists of a dead load of 15 kips and
an unspecified live load.
~ in diameter
Problems 48 and 49 are based on the following information and illustration.
Fy = 36 kips/ina
for ~ in bolts
Fu = 58 kips/ina
4in
Ii
A steel compression member has a fixed support at one
end and a frictionless ball joint support at the other
as shown. The total applied design load consists of a
dead load of 7 kips (which includes the weight of the
member) and an unspecified live load. Recommended
effective lengths are to be used.
0
0
0
0
holes
-t2-l- in
J3in
_t2-l- in
t in thick steel plate
50. What is the effective net area in tension for this
plate?
fixed wall
frictionless
rollers
frictionless
ball joint
steel compression
member
properties:
2.25 in2
2.5 in2
(C) 3.0 in2
(D) 3.2 in2
(A)
(E)
member
51. In accordance with AISC LRFD specifications,
t, = 533 in4
Iy = 174 in4
A = 19.1 in2
10 ft
50 kips
(A)
(B)
(C)
(D)
Fy = 50 kips/in2
'?l~7?7
fixed base
(built-in) ~
56 kips
65 kips
70 kips
52. The progress of a construction project is monitored
as follows.
48. In accordance with American Institute of Steel
Construction (AISC) load and resistance factor design
(LRFD) specifications, this compression member is a
year 1
operation
Jul
(A)
(B)
(C)
(D)
pier
short column
intermediate column
long col umn
Aug Sep
I
Feb Mar
Apr
l
I
legend:
Jan
l
excavation
49. Tn accordance with AISC LRFD specifications, the
maximum allowable design live load is most nearly
340 kips
490 kips
550 kips
650 kips
year 2
Oct Nov] Dec
formwork
(A)
(B)
(C)
(D)
the
maximum allowable design live load is most nearly
scheduled workl
I
I
I actual work I
I
This is an example of a
(A) critical path chart
(B) rectangu lar- bar progress schedule
(C)
PERT chart.
(D) triangular-bar
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...
progress schedule
---~------,;.~'"'!"--ioiI--
Practice Exam 1
Problems 53-55 are based on the following information
and illustration.
A construction
project is composed of activities A
through H with the durations, in days, given for each
activity in the diagram shown. This project is on a
strict schedule that starts at the beginning of January l.
Work can only be performed during the day and must
be done on every day of the week (Sunday through Saturday).
G
B
5
5
end
E
3
H
1
C
7
53. The earliest date this project can be completed is
(A) January
(B) January
(C) January
(D) January
8
9
11
17
54. The total float for activity A is
(A)
0 d
(B) 3 d
(Cl 6 d
(D)
(A) $596
(B) $597
(C) $598
(D) $599
58. At the elastic limit of a copper specimen, the stress
62 MPa
600 MPa
(C) 6 GPa
(D) 62 GPa
59. A steellintc! has a modulus of elasticity of 2.0 x 1011
Pa, The lintel is subjected to a strain of 8.9 x 10-3 m/m.
The area of the lintel is 0.0206 m 2. What is most nearly
the compressive force applied to the lintel?
(A) 0.016 kN
(B)
450 kN
(C) 37 MN
(D) 140 MN
60. The American Association of State and Highway
and 'Iransport.atlon Officials' (AASHTO) design structural number for a road is 4. The material specifications
are as follows.
material
layer
thickness
(in)
experience
coefficient
sandy gravel subbase
crushed stone base course
10
6
011
0.14
9 d
55. The latest day that activity E can start is
(Al
(B)
January
January
(C) January
(D) January
8
13
14
16
56. A highway department has $375,000 in the present budget set aside for transportation improvement
projects. It has the option of constructing a bridge now
for a cost of $350,000 or waiting 5 years and constructing the same bridge when there is $400,000 in the budget. Which option is best, assuming an average inflation
rate of 5%?
(A)
(B)
(C)
(D)
~
The elastic modulus
(Al
(B)
D
9
start
57. If $500 were deposited into a bank savings account,
bow much would be in the account in 3 yr if the bank
pays 6% interest compounded annually?
is 310 266 kPa with a 5% strain.
for this specimen is most nearly
F
3
A
8
49
1f a high-stability
plant mix asphalt concrete surface
course with an experience coefficient of 0.44 is to be
placed on top of the specified subbase and base course
materials, the required surface course thickness is most
nearly
(A) 3 in
(B)
4 in
(C) 5 in
(D)
6 in
Build tbe bridge now.
Wait 5 years to build the bridge.
Both options are equally affordable.
There is not enough iuformation to solve.
PPlowww.ppi2poS..com
SO
Civil Dis<iplineoSpe<ifi<Review for the FE/EIl Exam
SOLUTIONSTOPRAcrICEEXAMI··
1. </> is found from trigonometry using the basic relationships for slopes.
</> = arctan
(Yl -Y2)
Xl -X2
=arctan
3)
(1110-4
_
4. From the PVC at sta 10+35 to the location of desired
elevation at sta 11+35, .r = 1135 m - 1035 m = 100 m.
The horizontal vertical curve length, from PVC to PVT,
is given in the problem statement as L = 350 m. The
grades are given as gl = 7% and g2 = -5%. The elevation of the PVC is given as YPVC = 60.0 m.
11+35 is
Therefore, the curve elevation at sta
--
YIH35
= uvvc + glX + (g, 2~gl ) x
2
= 53°07'48"
= 60.0 m + (7%) (~)
The answer is D.
(100 m)
100/0
+((-5%-(7%))(wk))
2. First, find the radius of the curve.
2
(2)(350 m)
R = 5729.58 ft_o
(100 In)
D
572958 ft-°
6°
= 954.93 ft
The answer is A.
5. ThePVI islocatedatx = L/2= (350m)/2
from the PVC. The tangent elevation is
Next, solve for the tangent.
T= Rtan-
= 65,29 m (65 m)
t,.
= 175m
Y = uvvc: + g,X
2
= 60.0 m + (7%) (_1_)
= (954.93 ft) (tan 1l02~'35")
(175 m)
100%
= 72.25 m (72 m)
= 94.98 it
The answer is D.
PC is the same as station BC in this problem,
BC = PI-T
= 1487,33 ft - 94.98 ft
6.
A
I
/
= 1392,35 ft
(sta 13+92.4)
'
I
,-,
"
I
I
The answer is D.
I
"
II
-,
' -,
L
-,
/
I
3. By Simpson's '/3 rule,
I
'
I
I.
(2:)Od<l')
+ 4 (.2::>"v<",) + k
h, + 2
5
A=w
3
The intersection
angle, 1 is
1
I = 104° + (36') (~)
4,2 m + (2)(6.7 m)
)
= (5 m)
+ (4)(8.1 m; 7.6 m) + 8.3 m
(
60'
= 104.6'
The radius of the curve is
R=~-
= 1478 m'
(148 m')
The answer is C.
PPI • www.ppi2poll.lom
I -
2 sin 2'
= 157.98 m
The answer is A.
250m
2 sin 1O~,6
o
(158 m)
------------!---""!'~:7r~--'i_~~~.;;."'l_"'!"'--~
...- ..
Practice Exom 1 Solution.
7. From Sol. 6, R = 157.98
from PC to PT is
The length of curve
ill,
The full-flow velocity is
1TI~~
Q
L = RI
15-
Vf=-L=
s
7[(1.5 m?
Af
= (15798
m)(104.6
(7[ rad)
0
)
4
1800
= 8.49 mls
= 288.4 m
PT = PC+L
The ordinate
The
to be located
graph is
= 1046.0 m + 288.4 m
= 1334.4 m
rl
(sta 13+34.4)
D
is C.
answer
8. The wetted
SI
perimeter
for a triangular
p = 2rlv'1
+ m2
channel is
=
on the hydraulic
elemcnt.s
0,50 m
1.5 m
= 0,333
From the relevant curves on the hydraulic elements graph
for circular sewers, the QIQf and vIv : rat.ios are
,--------~--+ (2)2
= (2) (1.5 m) v'1
= 6.7m
The
is D.
answer
9. The head supplied to the turbine can be calculated
from the energy equation. Assume outlet velocity head
is insignificant.
2
§,~ 0.225
2
PI
V2
~
+ 21 + -V1 = P2
- + 22 + -2 + hi + hturbine
'I
'I
29
= -
'I
=
VI
P2
+ Zl + -2 - -
°+
9
200 ill + 0 -
'I
2
V2
-
°-
Z2 --
2
9
180 m -
- hi
°-
The flow rat.e with a depth of flow of 0,50 111 is
Q
18 m
Qf = 0.225
=2m
The power output
v,
9
2
PI
hturbiDe
'!.- ~ 0,78
Q = 0,22.5Qf
of the turbine
is
= (0,225) (15
P
= QrhturbinefJ
= (4.92
= QpghturbinerJ
~3)
= 82051 W
(1000 ~~)
(9.81
= 3,375 m3
:Z) (2 m)(085)
The
answer
~3)
Is
(3.4 111 M
3
is B.
(82 kW)
11. The flow velocity when the dept.h of flow is 0.50 m i,
The answer is A.
v
- = 078
vf
10. The hydraulic elements graph for circular sewers in
the civil engineering section of the NCEES Handbook
can be used to avoid having to calculate the hydraulic
radius and cross-sectional
area of flow for partial flow
v = 0.78vf
= (0.78) (8.49
= 6.62 m/s
in circular pipes.
The
___________________________________
answer
1:1)
(6,6 m/s)
is D.
PPI. www.ppi2polS.(om
52
Civil DiSlipline-Specilic Review for the FE/EIThom
_
12. The overall runoff coefficient is determined by weighting each runoff coefficient by the area it covers.
15. At bridge A,
m3
= A1awli
Atotal
Q
+ Aroildway + Aroof
2
= 0.067 ui]«
= 10000 m2
The overall runoff coefficient
AlawnClawn
(
Coverall
At bridge B,
is
m3
Q
10 ~
vn = :4 = 100 m2
+ AmndWayCmadWay)
+ AroofCroof
= 0.100
= --'-----==--==--=--------'-Atotal
(7500
(
+ (2000 m2)(0.15))
m')(0.20)
10 ~
vA=:4=150m2
= 7500 m + 2000 m + 500 m2
2
sa]«
The change in velocity between the bridges is
.6.v = VB -
+ (500 m2)(0.80)
VA
m
10,000 rn2
m
= 0.100 - - 0.067 s
s
= 0.033 mls
= 0.22
The answer is B.
The
13. From the sewage flow ratio curves in the civil engineering section of the NCEES Handbook, the minimum
flow ratio is conservatively determined from curve E21
unless otherwise specified, to be about 0.32 for a population of 20 000. The estimated minimum sewage flow is
16. Shrinkage for the soil is found from the ratios of
weight to volume for both the natural and compacted
soil.
answer
is A.
shrinkage = (1 -
Qm'n = 0.32
Qavg
V;,;cal ) x 100%
Vcompacted
Qmin = 0.32Qavg
The
answer
1 -
= (0.32) (8000
~3)
= 2560 m3/d
(2600 m3/d)
(
Pna'uml
)
Pcompacted
1 - 5200 ~)
(
is B.
X 100%
x 100%
7500 ~~
= 30%
14. From the sewage flow ratio curves in the civil engineering section of the NCEES Handbook, the maximum flow ratio is conservatively determined from curve
C, unless otherwise specified, to be about 3.4 for a population of 20000. Therefore) the estimated maximum
sewage flow is
The answer is B.
17. For the flow net shown, Nt = 3 and Nd = 6. The
flow per lineal meter of darn width is given by
Qmax = 3.4
o.;
Qrnax = 3.4Qavg
= (3.4) (8000
= 27200 m"/d
The answer is C.
PPI. www.ppi2pass.com
~3)
(27000
= ( 3 x 10 -2 -em)
s
m3/d)
( ---1 m ) (3 m - 1 m) (3)_
100 em
6
= 3 X 10-4 m2 Is
The
answer
is A.
"""!~!'!"'_~~_~~":'!'~...~'!"!" ... ~'!""'!""!'~"!"!'
Prod;,. Exam 1 Solutions
18. From Sol. 17, Q = 3 X 10-4 m' /s per lineal meter
of dam width. The total flow rate under the dam is
Q'Deal = Qw
=
(3
4
10-
X
Linear interpolation of the time factor table for T =
0.0146 gives Uavg = 13.0%. Based on a total settlement of 8 em, the amount of settlement after one year
is approximately
(10 m)
:')
S = (130%) (10~%)
= 3 x 10-3
S3
m3/s
= 1.G4 em
The answer is A.
(8 em)
(1.0 em)
The answer is A.
19. For 20% of the total settlement, T = 0.031. For the
donbly drained clay layer with sand layers bounding it
10 m
H=--=5m
2
22. The soil is cohesion less (c = 0). The ultimate bearing capacity of the soil is
qultimaLe
+ jDfNq + O..5,BN-y
= eN, + pgDfNq + 0.5pgBNo
= c~Nc
The amount of time required for 20% settlement is
T,oH'
= (0)(9.6) + (1835 ~~)
(0.031)(5 m)'
t,o = --
=
~)
(1 m)(2.7)
+ (05) (1835 ~~) (981
~)
(2 m)(1.2)
,
c;
0.004 ~
= 193.75 d
(981
(190 d)
= 70205 Pa
The answer
is C.
qnet
20. For doubly drained clay,
10m
H=--=5m
2
= qultimate
pgD f
-
= 70205 Pa-
(1835 ~~)
= 52203 Pa
(52.2 kPa)
~)
(1 Ill)
The allowable bearing capacity is
The time factor is
qultimate
qa =
Cvt
(981
0.004 m')
(
d
T = H' =
(365 d)
FS
52.2 kPa
3
= 17.4 kPa
(5 m)'
(17 kPa)
= 0.058
Linear interpolation of the time factor table for T =
0.058 gives Uavg = 26.8%. Based on a total settlement
of 8 em, the amount of settlement after 1 yr 15
s = (26.8%) CO~%)
(8 em)
The answer is A.
23. Use the average end arca method to solve.
out:
sta 1 to 2 =
150 Ill' + 0
2
m'
= 75 Ill'
= 2.1 em
The answer is B.
sta 2 to 3 =
o Ill' + 100 m'
2
m'
150 m' + 0 m'
= 50
21. The impervious
bedrock bounding one side of the
.
.
I d . d lay laver
for which
clay makes this a sing Y rame c
0
H = 10 m. The time factor is
0.004 m')
Cvt
T = H2 =
(
fill:
stalt02=
= 75
(365 d)
d
(10 m)2
2
sta2to3=
75
m'
III' +2 25 m'
= 50 m'
______-!-_~-------------------------= 0.0146
PPl • www.ppi2posl.com
S4
CivilDiscipline-Spedfil Review for the FE/Ell Exam
Now determine the earthwork volumes.
V = (75 rn2) (1000 m) + (50 rn2) (1000 rn)
cut:
_
26. Each lift is 0.1 m thick.
meter of lift is
= 125000 rn'
(0.1 m) (1630 ~~)
V = (50 m2) (1000 rn) + (50 rn2) (1000 m)
fill:
m'
= 100000
There is 125000 rn3 - 100000
waste than borrow.
rn'
25000 rn' more
The mass of each square
= 163 kg/rn"
The amount of bentonite in the soil is to be 5% dry
basis. Therefore, the amount of bentonite in each lift is
(0.05) (163 ~~)
= 8.15 kg/m2
The answer is D.
(8.2 kg/rn")
The answer is A.
24. The specific gravity
following equations.
of the soil is found using the
21. The number of covalent hands formed by a carbon
atom under normal conditions is four.
rnw = 318 kg - 204 kg
= 114 kg
The answer is B.
114 kg
Vw =
kg
1000 ""
m
28. Although iron and manganese ions contribute to
water hardness, their presence is less commonplace than
calcium and magnesium ions. Sodium ions do not contribute to water hardness.
= 0.114 rna
Vs = vt - lI,l!
= 0.193 rn3 - 0.114 m3
The answer is B.
= 0,079 rna
The specific gravity is
G = /8 =
29. Substitute the seven-day values into the rearranged
equation for BOD exertion.
'm.~
v'iw
204 kg
fW
y, = L(I - e-k')
(0.079 m-') (1000
r~~)
L =
Yt
I- e
= 2.58
The answer is B.
1- e
25. Using the equations
water pressure gives
CTh
= CTh + U
CTh
= KaO'v
(Iv
= pgh
for effective
stress
= 337,8 rng/L
and pore
The
answer
:,~)
(981
+ (961 ~~)
= 339387
~)
Yt = L(1 - e-k')
(L2 m)
(9.81 ~)
Pal
(25 kPa)
~~)
(981
~)
(L2 m)
31. The dissolved oxygen content (DO) at 760 nun Hg
can be found by linear interpolation from the table given
in the problem statement.
mg
DO 23,3"C - 8.7 -L
is C.
8.5 n~g _ 8.7 ~g
PPI. www.ppi2poss.mm
d-')(5dl)
= 170 mg/L
Pa
= 25347 Pa
~g)(l_e-(014
The answer is B.
= 13575 Pa + (1000
answer
(L2 m)
Pa
= (0.4)(33938,7
= 13575.5
The
(340 rng/L)
is C.
Y" = (337.8
IJ"h
(0.14 d ')(7 d)
30. From Prob. 29, L = 338 mg/L. Substituting
appropriate values into the equation for BOD exertion gives
= (1922
IJ"h
let:
211 mg
L
=
23,3'C - 23'C
24'C - 23'C
----------------------"""'!-- ...-----iooi---
q
Practice Exam 1 Solutions
SS
34. The tangent of the curve is
6-
T=Rtan= 8.64 mg/L
2
= (537 m) tan
Oxygen is only slightly soluble in water and does not
react with water chemically. Therefore, Henry's law is
applicable, and oxygen's solubility is directly proportional to its partial pressure. The percent saturation
1240 )
( -~-
= 58.33 m
PC = PI-T
= 1102 m -
of oxygen in the given freshwater sample at an atmo-
58.33 m
= 1043.67 m
(sta 10+44)
spheric pressure of 730 mm Hg is
The answer is A.
730 mm Hg) _5_.7_m",,~=g)
x 100%
% saturation = (
H
760 mm g ( 864
L
= 63.4%
(63%)
35. The degree of curvature is converted into radians,
and the radius of curvature is
s
700 ft
R = - = --.,---------,.,.-
The answer is A.
¢
32. The cross-sectional area of the channel is
(~;;n
= 4010.7 ft
A=wd
= (1.5 m)(3 m)
= 4.5
(10°)
The minimum required length of spiral transition onto
this road is
m'
Therefore, the approach velocity is
L. ~ '"~
m3
0.5 = _----"s'"
4.5 m'
= 0.11 m/s
The answer
~ "OJ
= 36.35 ft
The answer
(0.1 m/s)
is A.
(
\::,~rJ
(36 ft)
is B.
36. A vehicle will require a longer braking distance if
it is traveling downhill. Therefore, stopping sight distance, S, has to be subtracted from the denominator
to result in the largest design braking distance. The
braking distance is found to be
v'
(1000
(2) (9.81 ~)
:rJ
(3600 ~/
5% )
x ( 0.7 - 100%
= 0.0479 km
h
= 0.06 km
((80~)
(1000~)
(Ill)
(2) 9.81 s'
(~))'
( 0.35 - 100%
1.5%)
+ (2.0 s) (80 km) (1000 ~)
h
km
s=vT+d
= (89 km)
d = 2g(J ± S) + tv
(~)
3600 s
(0.5 s) + 0.0479 km
= 119.6 III
(~)
3600 s
(120 m)
The answer is D.
The answer is D.
-;.,;-~~":'~.;.-~4':~~~-----~
...
-.._-----------
PPI. www.ppi2poss.(om
56
Civil Discipline-Specific Review for the FE/EITExom
37. There are two equations to check.
_
For 5 yr, the total ESAL is
First, where stopping sight distance, S, is less than the
y;:-
ESAL5 yr = (5 yr) (days)
365
curve length, L:
( 131.56 ESAL)
day
AS'
L - -------"
= 240,097 ESAL
- (100%) (.j2Hr + .j2H2)2
The answer is D.
(1% - (-3%)) (300 mf
(100%) (/(2)(1.2
(240,000 ESAL)
m) + 1(2)(0.2
m))'
40. Construct shear and moment diagrams. The change
in shear is the area under the applied loading, and the
= 756.4 m
Second, where stopping sight distance is greater than
the curve length:
change in moment is the area under the shear diagram.
om
(200%) (,JH; + /H2)'
A
L=2S-
I'
I
llTITlo
w~5kN/m
= (2)(300 m) -
(200%) (vT.2Ii1 + .j['2""ffi) 2
1% _ (-3%)
= 481.0 m
Lev ~
Since the previous two equations show that the stopping sight distance is less than the curve length, the
minimum required vertical curve length is
L = 756.4 m
The answer
38.75kN
1
1
25
(760 m)
is C.
V (kNI 1.2~p:====I---
__
=_....jL_!!..':::::::,.j
-13.75"-1
_""';:'':'''-_..J
38. This mixture would be designated as a 19 mm superpave mixture.
1
1
1
The limits for such a mixture are
air voids = 4.0%
minimum VMA = 13%
VFA = 65-75%
dust-to-asphalt ratio = 0.6-1.2
= 8 gyrations, maximum Gmm = 89%
6.25
at Ninit
at Jv~max = 174 gyrations, maximum Gmrn = 98%
All parameters
specifications.
in this mixture
are within superpave
-62.5
The
largest
magnitude
of
bending
moment
The answer is A.
The answer is D.
39. The total ESAL per truck for each trip is
ESAL"uck = (1 single axle)(0.0877)
41. Use the principle of virtual work to find the vertical
deflection at point D.
+ (2 tandem axle8)(0.1206)
= 0.3289 ESAL/truck-trip
The total daily ESAL for 40 trucks,
10 trips a day, is
x (0.3289
~D
each making
ESALJay = (40 trucks) (10 ~:s)
=
L U (;~)dJ)
ni
The moment functions in the directions indicated by
the local x-coordinate for each beam segment under the
actual loading are shown on the following moment diagram.
ESAL )
truck-trip
= 131.56 ESAL/day
PPI • WWW.PPi2poss.com
is
-62.5 kN-m (63 kN-m) at support C.
"".!.
...:..._~-~.-!.~~~~...!..!!..!..~~..!.!..~
Pradice Exam 1 Solutions
r~ cmTTl
5m
I
15 kN
Al
The modulus of elasticity of steel is E = 2.1 X lOll Pa,
5m
5m
I
w > 5 kN/m
I
From the principle of virtual work, the vertical deflection at point D is
jD
A
B
t
I
I
I
I
I
I
I
I
LAY ~ 1.25 kN
x
~x
M~ 1.25x
I
I
I
6.25
=
UD
fo5 (-0.57;)(1.25x)
EI
o
x
r (-25 - O.5x)(6.25 - 13.75x) d
+ Jo
EI
r (-x)(-2.5x')x
5
x
+ Jo
d
5
-O.625X315
3
0 +
+
EI
M = -2.5x2
.X
EI
1
M~ 6.25-13.75X~
d
5
RCY ~ 38.75 kN
I
I
I
I
I
I
I
I
I
I
51
-62.5
1
- 15.625x 0
31.25x215
6.875X315
2
+
3
o
0
4
2.5X \5
+--
4
The moment functions in the directions indicated by the
IDeal x-coordinate for each beam segment under the virtual unit loading at point Dare shown on the following
lllOlnent diagram.
0
_ 26.042 kN'·m3
-
78.125 kN2m3
+ 390.625 kN' ·m"
+ 286.458 kN' ·m3
5m
=
+ 390.625 kN' ·m3
Al
B
I
I
tAY~ -0.5 kN I
I
I
I
I
I
x (1000 :')
= 22.9 mm
~x
x
I
I
m IkN'ml
Ok(=-'~~~Il--_-2.--C5~~~"'::f--~~~7"i
m ~ -0.5x
(
~5
(23 mm)
Since the unit load in the virtual force system was downward and the answer is positive in sign, the actual de.-flection is also downward.
\. m ~ -x
The answer is C.
m = -2.5 - O.5x
A table smnmarizes the moment functions as follows.
region
function
O<x<5
region
B-C
0<xo;5
M
m
1.25x
-O.5x
6.25 - I3.75x
-2.5 - O.5x
-2.5x'
-x
EI
EI
£1
EI
region
A-B
_____________________________________
C-D
Oo;xo;5
PPlowww.ppi2poss.mffi
S8
CivilDiscipline-Specific
Reviewfor IhefE/ElfExom
_
42. Construct and use an influence line for vertical shear
at support C. The influence line is constructed by plotting the change in response on a free-body diagram of
a section of beam at support C as a unit load travels
across the structure.
Sm
Sm
•
Sm
~'
C
AI
ty
ID
ol-----I------"<.,..--------i
ty
B
I
I
I
I
I
I
I
I
I
I
influence line for Me
[kN'm/kN)
I
I
I
I
load position
for maximum
bending moment
1.S
1.0 I
I
o
-5
71.17 kN
(rear axle)
17.79 kN
(front axle)
1-.;-;;:;-_1
4.27 m
influence line for Vc (kN/kN)
load position
for maximum
vertical shear
71.17kN
(rear axle)
17.79 kN
(front axle)
I • 4.27 m .1
The load position shown gives the maximum response
in the beam at support C for the specified direction of
travel. The maximum bending moment can be found
by superposition
as
Ve,max =
(1.5 ~~) (7117 kN)
= 106.8 kN
(110 kN)
This problem asked for the magnitude of maximum vertical shear at support C for a truck traveling in the
direction shown. If this problem had asked for the magnitude of maximum vertical shear at support C for the
given truck axle configuration, the axle load positions
would have to be switched around and both axle loads
placed on the influence line with the heavier axle load
on tbe larger influence line ordinate and the lighter axle
load on the smaller influence line ordinate. This would
result in a larger numerical answer.
The answer is D.
I(
-5 kN.m)
kN
(71.1 7 kN)
= 355.8 kNom
(360 kN·m)
Me,max =
The load position shown results in the maximum response in the beam at support C for the specified direction of travel. The maximum vertical shear can be
found by superposition as
I
This problem asked for the magnitude
of maximum
bending moment at support C for a truck traveling in
the direction shown. If this problem had asked for the
magnitude of maximum bending moment at support C
[or the given truck axle configumtion, the axle load positions would have to be switched around and both axle
loads placed on the influence line with the heavier axle
load on the larger influence line ordinate and the lighter
axle load on the smaller influence line ordinate. This
would result in a. larger numerical answer
The
answer
is D.
44. Construct an influence line for the force in member BF. The influence line is constructed
by plotting
the change 1I1 response in member BF as a unit load
travels across the structure.
43. Construct and use an influence line for bending moment at support C. The influence line is constructed by
plotting the change in response on a free-body diagram
of a section of bea.m at support C as a unit load travels
across the structure.
PPJ • WWW.PPi2poss.com------------.!..."~-._!."".!.~...:.~~~~..i_:~~.;.~.!..!.~..!..!!.!_.!...~
Practice Exam 1 Solutians
I•
5m
I•
5m
5m
•
.Y1SJSl_4m
[}
~
influence
line for FBF
(kN/kN)
I : !~I
~
i
~
59
It is required that ¢Pn, 2: Pu. For axial compression
with spiral reinforcement, 1> = 0.70. Setting 1>Pn = Pu
and solving for the area of longitudinal reinforcing steel
gives
Pu
,
0:85¢ - 0.S5fcAg
~
f1) - 0.S5J~
a~
(740 kips) (1000 lbf)
kip
-0.53::
(0.S5)(0.70)
When moving the unit load across a truss structure, this
load must be distributed to the two joints. It is stated
that the roadway acts as simple beam spans between
truss joints. Therefore, when a load is placed between
truss joints, the two joints adjoining the beam span act
as simple beam support.s and the magnitudes of the
loads applied to the two truss joints are the same as for
t.he calculated reaction forces of this beam.
The maximum ordinate for tensile force in member BF
is 0.53 kN /kN.
_ (0.S5) (4000 ~)
("(lS in)2)
4
lbf - (0.S5) (ibf)
60,000 :-;;
4000:-;;
m
m
= 6.69 in2
As = 6.69 in2, required for the given applied axial com2
pressive loads, is greater than As = 2.54 in based on
the minimum allowed reinforcement ratio, pg = 0.01.
The minimum required area of reinforcement is As
6.69 in"
The answer is C.
45. From the influence line in Sol. 44, the maximum
ordinate for compressive force in member BF is -0.53
kN/kN.
The column has six longitudinal reinforcing bars. The
required area of each longitudinal reinforcing bar is
= 6.69 in2
A = ~
6 bars
nbars
The answer is B.
= 1.11
46. Determine the amount of reinforcing steel required
by the minimum required reinforcement ratio, Pg) of
0.01. The minimum area of reinforcing steel required is
in2/bar
A bar area of 1.11 in? is satisfied hy a no. 10 har,
which has a nominal area of 1.27 in2.
In an actual design/analysis
A, = pyAg
= (0.01) ("(lS in)2)
4
= 2.54 inz
situation, a check should
also be made to see that the actual longitudinal reinforcement ratio does not exceed the maximum allowable
ratio of O.OS.
The answer is D.
Determine the required amount of reinforcing steel based
47. Determine the amount of reinforcing steel required
by the minimum required reinforcement ratio, Pgl of
0.01. The minimum area of reinforcing steel required is
on the factored axial load, Pu·
r; = 1.2Pdead + 1.6Plive
= (1.2)(150 kips) + (1.6)(350 kips)
= 740 kips
The nominal axial compressive load capacity is given by
P; = 0.S5Po
= (0.S5)(0.S5J~Aconc,ete+ JuA,)
= (0.S5)(0.S5J;(Ag
-
As) + f.As)
A, = pyAa
= (O.01)(lS in)2
= 3.24 inz
Determine the required amount of reinforcing steel based
on the factored axial load, Pu·
P« = 1.2Pdead + 1.6Plive
= (1.2)(150 kips) + (1.6)(250 kips)
= ,5S0 kips
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60
CivilDisdpline-Spe,ili, Review for the FE/Elf Exam
_
The nominal axial compressive load capacity is given by
Pn = 0.8Po
= (0.8)(0.85f~A,on"ete
= (0.8)(0.85j~(Ag
The unbraced length of the compression member is the
same about the x-axis as about the y-axis. Therefore,
+ fyA,)
i; = ly = (10 It) (12 ~)
- As) + fyA,)
It is required that 1JPn 2: Pu. For axial compression
with tied reinforcement, 1J = 0.65. Setting 1JPn = Pu
and solviog for the area of longitudinal reinforcing steel
= 120 in
The radius of gyration about the x-axis is
gives
~
A.~=
Tx
-0.85f~Ag
=
fi
=
533 in4
19.1 in2
= 5.28 in
fy - 0.85f~
kiP
The slenderness ratio about the z-axis is
(580 kips) ( 1000 Ibf)
= kxlx = (0.80)(120 in)
SR
(0.8)(0.65)
x
5.28 in
Tx
- (0.85) (4000 ~)
(18 in)'
= 18.2
=
The radius of gyration about the y-axis is
60,000 Ib; _ (0.85) (4000 Ib;)
In
In
= 0.244 in'
A, = 0.244 in' required for the given applied axial
compressive loads is less than As = 3.24 in2 based on
the minimum allowed reinforcement ratio, Pg = 0,01.
Therefore, the minimum required area of reinforcement
is A, = 3.24 in"
Ty
reinforcing bars.
nbars
=
174 in4
19.1 in2
in
The slenderness ratio about the y-axis is
SRy = kyly = (0.80)(120 in)
Ty
3.02 in
The required area of each
= 31.8
longitudinal reinforcing bar is
A =~
fi
= 302
The square tied column has eight uniformly sized longitudinal
=
= 3.24 in'
8 bars
The larger SR controls. Use SR = SRy = 31.8.
The SR is now compared with the critical slenderness
= 0.405 in' /bar
ratio.
8021
802.1
Cc = ---..:..-= -----F;;;:~ = 113.4
'0 kips
A bar area of 0.405 in' corresponds to a no. 6 bar, which
p;
has a nominal area of 0.44 in2.
In an actual design and analysis situation, a cbeck should
also be made to see that the actual longitudinal rein-
o
-in2
forcement ratio docs not exceed the maximum allowable
Since SR = 31.8 is less than C, = 113.4. the column is
intermediate and fails by inelast.ic buckling.
ratio of 0.08.
The answer
The answer
48. From
is C.
is D.
the
civil
NCEES Haudbook,
engineering
section
of
the
the design value for the effective
column length factors about the x-axis and y-axis, re-
spectively, are kx = 0.80 and ky
of elasticity of steel is
= 0.80. The modulus
E = (29 X 106 .Ib;) ( 1 kip )
tll
1000lbf
= 29,000 kips/in'
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49. From Prob. 48, the controlling slenderness ratio is
31.8, and SR S 802.1/
Therefore the column
strength is
'
p;.
.
rpPcr = 0.9Py
(F(SR)'
0.658 )iii,.,,,
)r".'J')
= (0 .9) (50 ~
kiPS) ( 0.658 ('" ';;;'T
286,200
m
= 41.79 kips/in'
""!"~~~~!""-~-"""!"!'"'"'!"'!~~-~ ...........
Prollice Exom I Solutions
This value could have been read directly from ATSC
Table 4-22.
The column capacity is
. 2
(19.1 III )
= 798.2 kips
By setting Pwtal = <p,P" = 798.2 kips, the allowable
design live load can be found.
Ld
= (2.25 in + 3 in + 2,25 in) - (2 I10Ies)(0.75in)
= 6 in
For line ABEF across the member width, the net width
is
bn=b-'\'d+-
LS'
LJ
+ 1.6.Rive
Ptotal = 1.2Pdead
p
For line ABCD across the member width, the net width
is
bn=b-
kiPS)
P = ¢F"A = ( 41.79 in'
61
4g
= (2,25 in + 3 in + 2,25 in) - (2 I10Ics)(0.75in)
_ PtotaL - 1.2Pdea.d
1.6
live -
798.2 kips - (1.2)(7 kips)
1.6
= 493.6 kips (490 kips)
+
(1 spacc)(4 tn)'
(4)(3 in)
= 7.33 in
The smaller net width controls. Use bn = 6 in.
The answer is B.
The net area is
50. It is necessary to check both yielding on the gross
area and fracture on the effective net area.
For yielding, the gross area is
Ag = (0.5 in)(2.25 in + 3 in + 225 in)
= 3.75 in'
An = bu.t = (6 in)(0.5 in)
= 3 in2
This is less than 85% of the gross area, and the shear
lag factor, U, is 1.0. So, the effective net area is
A, = U An = (1.0)(3 in')
The allowable tensile stress for yielding is
= 3.0 in'
F, =0.6Fy
The answer is C.
kiPS)
= (0.6) ( 36 in'
51. It is necessary to cbeck both yielding on the gross
= 21.6 kips/in'
area and fracture on the effective net area.
The allowable tensile force for yielclingis
For yielding, the gross area is
Ptuta! = Fe Ag
Ag = (0.5 in)(2.25 in + 3 in + 2.25 in)
= (216 ~:,s)
(375 in')
= 81.0 kips
= 3,75 in'
The allowable tensile force for yielding is
For fracture, in order to determine the net areal th~con-
trolling net width of the member must be determmed.
¢tPn = 0.9FyAg
kiPS)
= (0.9) ( 36 in'
~ in diameter
holes
. 2
(3 75 III )
= 121.5 kips
for .2. in bolts
B
1
I
B«
0
1/
"--lin
2
I
OE
/
CO
For fracture, in order to determine the net area. the con-
trolling net width of the member must be determined,
From Sol. 50, the effective net area is Ae = 3.0 in'.
thick st~1 plate
___________________________________
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62
Civil Discipline-Specific Review lor the FE/EITExom
_
The allowable tensile force for fracture is
Activity F is preceded by two activities, A and D, so
the EST of activity F is the maximum EFT of the two.
That is, the EST of activity F is day 14 and its EFT is
day 17.
<P'Pn = 0.75FuAe
= (0.75) (58 ~:,s) (3.0 in2)
Similar calculations show that the minimum project duration is 17 d (EFT for activity F is day 17).
= 130.5 kips
<P'Pn = 130.5 kips for fracture is larger than <p,P"
=
121.5 kips for yielding, so use 1>,p', = 121.5 kips.
By setting P'otal = 1>'Pn = 121.5 kips, the allowable
design live load can be found by
PtotaJ = 1.2Pdead
p.
+ 1.6Plive
The LFT of an activity is the minimum LST of the
activities following it. For example, the LFTs of the
activities preceding the project finish are all day 17 since
no activities follow them. Accordingly, the LFTs of
activities F and G are each day 17. The LST of activity
F is
LST=LFT-D
_ Ptotal - 1.2Pdead
1.6
live -
Determine of the latest start time (LST) and latest finish time (LFT) for an activity with a backward pass
through the diagram using the minimum project duration as the starting point. That is, day 17 is the starting
point for these calculations.
121.5 kips - (1.2)(15 kips)
1.6
= 64.7 kips (65 kips)
=
= day 17-3d
The answer is C.
= day 14
52. The chart shown IS a rectangular-bar progress
schedule.
Similarly, the LST of activity G is day 17-5 d = day 12.
The answer is B.
Activity D is preceded only by activity F, so its LFT is
the LST of activity F. That is, the LFT of activity D is
day 14 and the LST of activity D is day 5.
Problems 53-55 can be solved with critical path method
(CPM) calculations. Since the project is to start on
Activity B is preceded by two activities, D and G, so
its LFT is the minimum LST of the two. That is, the
LFT of activity B is day 5.
January 1, it is easy to designate the actual start time
as the end of the previous day, December 31, (day 0)
and January 1 as day I.
Determine the earliest start time (EST) and earliest
finish time (EFT) for an activity from a forward pass
through the diagram. The duration, D, of each activity
The total float time (TF) of an activity is determined
either by subtracting the EFT from the LFT or by subtracting the EST from the LST. For example, the TF
of activity G is
TF= LFT - EFT
is given in the arrow network.
The EST of an activity is calculated as the maximum
of the EFTs of the activities preceding it. For example,
activity A has no activities preceding it, so it has an
EST of day O. The EFT of this activity is
= day 17 - day 10
=7d
Making a. summary table is a good wayto organize these
resuJts.
EFT = EST+D
duration
= day 0 + 8 d
= day 8
(d)
A
8
5
B
Similarly, activity B has an EST of day 0 and an EFT
of day 5.
Activity D is preceded only by activity B, so its EST is
the EFT of activity B. That is, the EST of activity D
is day 5 and the EFT is day 14.
PPI. www.ppi2pcss.com
activity
'""'!'_ ....
C
D
E
F
G
H
7
9
EST
EFT
LST
LFT
TF
(day no.) (day no.) (day no.) (day no.) (d)
0
0
0
5
5
8
5
7
14
3
3
14
5
17
5
1
10
8
9
8
6
14
6
0
9
5
0
16
14
16
17
9
5
13
14
12
16
0
8
0
17
7
17
8
'""'!'---'!-~~_++'!""---'""'!'~----~
...
Practile Exam1 Salulions
63
58. The elastic modulus equation is the stress divided
by the strain of the material.
~)-------<
E='!.
E
310266 kPa
0.05
= 6205320 kPa
E
3
~)------:::::::0~)
(6 GPa)
critical path --
The answer is C.
53. The earliest date this project can be completed is
the maximum EFT of the entire project. The EFT of
activity F is day 17. Therefore, using January 1 as
day 1, day 17 falls on January 17.
The answer is D.
59.
E='!.
E
,,= EE
= (2.0 X lOll Pal (8.9 x 10-3 :)
= 1.78 x 109 MPa
54. The total float for activity A is read from the summary table as 6 d.
Determine the force.
F = "A
= (1.78 x 109 Pa)(0.0206 m')
The answer is C.
55. The LST for activity E is day 13. Therefore, using
January 1 as day 1, the latest day that activity E can
start is January 13.
1 kN )
x ( 1000 N
= 36668 kN
(37 MN)
The answer is C.
The answer is B.
56. The equation F = p(1+i)n is used to calculate the
future worth of the money based on the present value.
Use that to determine the more cost-effective solution.
F = P (1 + i)n
= ($350,000)(1
+ 0.05)'
= $446,698
60. The AASBTO structural number equation can be
rearranged to solve for surface thickness, Di·
SN = aID!
+ a2D2 + a3D3
_ SN - a2D, - a3D3
D J-
a,
= 4 - (0.14)(6 in) - (0.11)(10 in)
In 5 years, the project will ccst $446,698 to build, yet
there will only be $400,000in the budget. The highway
department should build the bridge now.
= 4.68 in
0.44
(5 in)
The answer is C.
The answer is A.
57. Use the followingequation to determine the future
worth of the money.
F=P(1+it
= ($500) (J + 006)3
= $595.51
($596)
The answer is A.
_________________________________
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Practice Exam 2
3. The length of curve is most nearly
PROBLEMS
(A) 370 rn
1. Boundary and traverse lines bounding an irregular
area are shown.
(B) 380 m
(C) 390 m
(D) 410 m
boundary ~
i
--'-"""I
I
I
I
I
I
I
I
I
I
I
I
7.1ml
I
I
6.5 m I
I
I
9m
9m
I
10.3 m I
I
I
I'
J
I
I
11
I
9m
I
9m
J •
I
I •
9m
I
I
I
I
I
I
I
J
J
J
boundary
I
I
I
I
I
6.9 m I
9.6ml
8.8 mr
I
I
I
I
4. The boundary and traverse line of an irregular area
are shown.
traverse/'
line
I 8.1 m
I
I
I
I
14.2 m
The total area between the irregular boundary and the
traverse line is most nearly
(A)
(B)
(C)
(D)
m'
330
350 rn'
370 rn'
390
I,
I,
I
I
I
I
I
I
I
I
8.3m
5m
. I,
I
I
I
I
I
5m
I
I
I
I
I
17.6m
I
I
I
I
I
I
5m
I
I
16.7 m
I
I
I
l~er5e
5m
line
Using the trapezoidal rule, the total area between the
irregular boundary and the traverse line is most nearly
m'
Problems 2 and 3 are based on the following illustration.
PI
PC:~~
I
I
I
I
I
I
~
~LC::.-:::..:::32::.:5:..:m::-
..-;"PT
(A)
5. A reading of :3.50 III is taken on a 4 III leveling rod
that is 0.50 m out of plumb at the top of the rod. The
correct reading, when the rod is truly vertical, is most
nearly
(A)
1=110'45'
141 m'
(B) 143 rn'
(C) 148 Ill'
(D) 151 Ill'
3.06 rn
(B) 347 III
(C) 3.53 III
(D) 3.94 III
2. The tangent distance is most nearly
(A)
(B)
(C)
(D)
195 m
197 III
284 III
286 III
65
66
Civil Discipline-Specific Review for the FEIEITExam
6. What is the length of the curve with an intersection
angle of 11.25° and a radius of 352 Ill?
PI = sta 4+53
A
B
R~ 352 III
R
170 m
(C) 310 m
(0) 350 m
.5:00 p,m,-5;15 p,m.-650
veb
5; 15 p.m.-5;30 p,m,-920
veh
5;30 p.rn.c-to 5:45 p,m.-1l40
veh
5;45 p.m.c-to 6:00 p.IO.-790
veh
Most nearly, what is the peak hour factor?
(A) 030
(B) 077
(C) 1.3
(0) 17
(A)
(B)
(C)
(0)
330 kW
350 kW
480 kW
500 kW
11 and 12 are based on the following infor-
has an 8 ha asphalt parking lot with
a runoff coefficient of 0,85, a 2 ha building roof with
a runoff coefficient of 0.75, and 1.5 ha of lawn with a
runoff coefficient of 0,20. The time of concentration
for
the watershed is 30 min. A 30 min, 20-year storm with
an intensity of 80 mrn/h occurs. A concrete sewer pipe,
flowing full at peak runoff, carries the runoff from the
site at a velocity of 10 m/s .
11. The peak runoff for the store property
(A)
(B)
(C)
(0)
75 rn/s
8.9 m/s
9.8 m/s
10 m/s
9. Water flows at 20°C through 10 m of 8 mm inside
diameter smooth glass pipe at 2,0 m/s. The friction factor for glass is 0.0180, The head loss caused by friction
is most nearly
(A)
(B)
(C)
(0)
is most nearly
0,50 m" /s
0,75 m" /s
1.9 m3/s
22 m3/s
12. The minimum
most nearly
8. Irrigation water flows at a depth of 4 m in a 15 m
wide, concrete, rectangular open channel with a 0.5%
slope, The Manning roughness coefficient for this channel is 0,015, The water velocity is most nearly
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w. A centrifugal pump lifts groundwater 100 m vertically to a surface storage tank at a rate of 0,25 Ill3/S,
The pump has a 75% efficiency. The power required to
drive this pump is most nearly
A store property
7. A highway used for evening commuters experiences
peak hour traffic of 3500 passenger equivalent vehicles
per bour between 5:00 p.rn. and 6:00 p.m. The traffic
during the 15-minute segments is
(A)
(B)
(C)
(0)
(A) 17 m
(B) 2,4 m
(C) 4,6 m
(0) 6,5 m
Problems
mation.
(A) 69 m
(B)
_
pipe diameter
to handle this flow is
0.50 m
0,84 m
1.4 m
1.7 m
13. A capillary tube 3,8 mm in diameter is placed in a
beaker of 40°C distilled water. The surface tension is
0.0696 N/m, and the angle made by the water with the
wetted tube wall is negligible. The specific weight of
water at tbis temperature
is 9.730 kN/m3, The height
to which the water will rise in the tube is most nearly
(A) 1.2 mm
(B) 3,6 mm
(C) 7,5 mm
(0) 9,2 mm
-------"'"'!--------~---~':"-~~h:_.-..-"!'!"-~--
Prallice Exam 2
14. Water flows through a 30.0 em inside diameter pipe
at an initial velocity of 1.9 m/rnm. The pipe diameter
subsequently reduces to 15.0 em before discharging into
an open channel. The discharge velocity is most nearly
(A)
67
18.
3.8 m/rnin
(B) 7.5 m/rnin
0.9 m
(C) 8.6 m/rnin
(D) 9.3 m/rnin
Problems 15-17 are based on the following information
and illustration.
An impervious dam on pervious soil above an impervious rock layer has piezometric: data as shown.
The 1.2 m x 1.2 m footing shown is 0.9 m below the
sand's surface. Assuming the water table is at the base
of the footing, the allowable bearing pressure with a
factor of safety of 3 is most nearly
piezometer
(A) 240 kPa
(Bl 360 kPa
(C) 600 kPa
(D) 790 kPa
25 m
A
Problems
19-21 are based on the following information.
water seepage -..-
An undisturbed
sample of clay has a weight of 29 kg,
3
a dry weight of 26 kg, and a total volume of 0.014 m .
Clay solids have a specific gravity of 2.65.
15. The total pressure head above atmospheric
at point
A is most nearly
is most
nearly
the water
content
of the
(A) 1.2%
(B) 4.3%
(Cl 12%
(D) 81%
(A) 9 m
(B)
19. What
sample?
10 TIl
(C) 20 TIl
(D) 30 m
16. The total pressure head above the tailwater
at point
A is most nearly
20. The degree of saturation
is most nearly
(A) 19%
(B) 24%
(C) 62%
(D) 75%
(A) 9 m
(Bl 10 m
(C) 14 m
(D) 15 m
21. The void ratio is most nearly
17. The uplift pressure
at point A is most nearly
(A)
10 kPa
(B) 30 kPa
(e) 70 kPa
(D) 190 kPa
___________________________________
(A) 0.2
(B)
0.3
(C) 0.4
(D) 07
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68
CivilDisdpline·Spedfic Review for the FE/Ell Exam
Problems
_
22-24 are based on the following information.
A soil specimen has a unit weight of 17.6 kN/m3, the
specific gravity of the solids is 2.70, and the water content is 30%.
22. The degree
most nearly
of saturation
of the soil specimen
is
(A) A-2-4
(B) A-2-5
(C) A-4
(D) A-6
27. Using the Unified Soil Classification System,
sify a soil with the following characteristics.
clas-
(A) 57%
(B)
(C)
F200 = 0.69
64%
71%
liquid limit (LL) = 72
(D) 84%
plasticity
23. The void ratio of the soil specimen
is most nearly
(A) 0.67
(B) 0.74
(C) 086
(D)
OW
ML
(C) MH
(D)
CH
0.96
24. The porosity
of the soil specimen
is most nearly
(A) 023
(B)
(C)
(D)
(A)
(B)
index (PI) = 48
28. Using the AASHTO Soil Classification
System,
identify the classification for the soil with the following characteristics.
036
0.49
0.57
grain size
% passing
no. 10
no. 40
no. 200
72
55
41
plasticity
index (PI) = 15
25. An excavated slope in a uniform soil is shown.
liquid limit. (LL) = 32
(A) A-2
(B) A-4
(C) A-5
(D) A-6
'The soil properties are unit weight, I = 17.3 kN /m·3,
cohesion c = 19.2 kPa, and friction angle, 1> = 15°.
The factor of safety for slope stability
is most nearly
(A) 15
(B)
(C)
(D)
29. An open tank contains 8.0 m of water beneath
1.5 m of kerosene. Kerosene has a specific weight of
8.0 kN /m'- The pressure at the kerosene/water
interface is most nearly
(A) 3.5 kPa
(B) 5.0 kPa
(e) 8.0 kPa
2.3
2.9
3.4
(D)
26. Using the American Association of the State Highway and Transportation
Officials (AASHTO) Soil Classification System, determine the classification of soil
with the following characteristics.
F200 = 0.:14
12 kPa
30. A municipal water system requires an effluent. chlorine residual of 0.15 mg/L. The chlorine demand placed
upon the syst.em is 0.45 mg/L, and t.he treatment. plant.
processes a daily wat.er flow of 15,000,000 gal. The ch 10nne required is most nearly
liquid limit (LL) = 39
plastic
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limit (PL) = 29
"-!__ ..
~~~---~--~~
PracticeExam 2
69
34. The elevation at the PVT is most nearly
(A) 17 kg/d
(B) 26 kg/d
(C) 34 kg/d
(D) 43 kg/d
(A)
740 ft
(B) 750 It
(C)
760 ft
(D) 770 ft
31. One liter of a solution is made by adding 3 g of
acetic acid (HAc) to distilled water. The molecular
weight of acetic acid is 60.0S2 g/mol. The acid-dissocia5
tion constant for acetic acid is K A = 1.7S X 10- . The
chemical equation for the dissociation of acetic acid into
hydrogen ions and acetate ions is HAc ~
H+ + Ac-.
The percentage of acetic acid ionized in solution is most
nearly
Problems 35 and 36 are based on the following information.
A car traveling at 15 m/s accelerates uniformly at a rate
of 1.5 mis' until it reaches a speed of 20 tn]»:
35. What distance does the car travel in reaching a
final speed of 20 m/s?
(A) 040%
(B)
(A) 44 m
(B) 47 m
(C) 50 m
(D) 58 m
1.9%
(C) SO%
(D) 9.3%
32. A continuous flow stirred tank reactor treats 0.25
m3/s of settled wastewater having 2S0 mg/L BODs at
200C. The design mean cell resistance time,
is 10 d.
e~,
36. How long does it take the car to reach its final
cruising speed?
(A) 1 s
y =
li;;s,mg
BOD5,mg
(B)
= 0.5
The effluent BOD5 is 6.2 mg/L. MLVSS = 3S00 mg/L.
1
The endogenous decay coefficient is Kd = 0.06 d-
37. The following illustration
3000 m'' /d
3300 m3/d
3
4700 m /d
SOOO m3/d
60
53
~
$
50
>
40
E
-0
ID
ID
Problems 33 and 34 are based on the following information.
a.
'"
U
:c
30
ID
ID
A vertical curve with Gl = -2.0% and G, = 1.6% has
a PVI at sta 87+00 and an elevation of 743.24 ft. The
relates traffic density to
mean vehicle speed.
The reactor capacity is most nearly
(A)
(B)
(C)
(D)
3 s
(C) 7 s
(D) 10 s
20
>
C
'"E
ID
10
length of the curve is 800 ft.
33. The PVC station for this curve is
10
20
30
40
50
traffic density, k (veh/mi)
60
70
66.25
(A) sta 79+00
(B) sta 83+00
(C) sta 91+00
(D) sta 9S+00
_____
""!'--~--------------------------
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70
Civil Discipline-Specific Review for the FE/EITExom
The traffic flow relationship is given by q = kv, in which
q is the traffic volume in veb /hr. The maximum traffic
volume for this road is most nearly
(A)
760 veh/hr
880 veh/hr
(E)
(C) 900 veh/hr
(0)
_
Problems 41 and 42 are based on the following information and illustration.
The beam shown is loaded with two 1000 N point loads.
The separation is maintained at 2 m, but the loads may
be moved to any location on the bearn.
960 veh/hr
2m
r------I
38. The stopping sight distance is 430 ft for a design
speed of 50 mph on a section of highway. The grades
for this highway section are -1 % followed by 3%. The
required length of vertical curve needed to satisfy the
AASHTO stopping sight distance for this design speed
is most nearly
(A) 270 ft
(E) 380 ft
(C) 4lO ft
(0) 450 ft
1000 N 1000 N
t t
J$-. ----==-'-"'-=---;,;;~~---'-----'--'-'-'-'
I~----.i~---_
10 m
41. The maximum value for shear at support A is most
nearly
39. A one-lane rural road has a 10° curve extending for
230 m along its centerline. The road is 5 m wide with
3 m wide shoulders. The design speed for this road is
75 km/h.
The superelevation needed so that side friction is not
needed is most nearly
(A)
0.00050
(A)
(B)
(C)
(0)
(A) 1.8 kN-m
(E) 8.0 kN·m
(C) lO kN-m
(0)
(A)
2000N
2800 N
3000 N
3800 N
42. The maximum value for moment at support A is
most nearly
(E) 0034
(C) 1.1
(D) 19
40. The worn surface course of a high-volume pavement is being replaced with a design requiring a total
structural number of 6.6. The engineer has decided to
replace 6 in of the surface with recycled-in-place asphalt
concrete having a surface course strength coefficient of
0.42, leaving in place 3 in of sound original pavement
having a strength coefficient of 0.3. Under the original pavement are a 10 in cement-treated base having
a strength coefficient of 0.20, and an 8 in sandy gravel
subbase. What is the minimum strength coefficient for
the subbase?
10 m
18 kN·m
43. A triangular pin-connected
truss carries a load of
4448 N as shown. Each member has the same modulus
of elasticity and cross-sectional
%
area.
M
3.0 m
0.05
(E) OlO
(C) 015
(D) 020
p
N
4448 N
4.6m
PP/. www_ppi2pcss.ccm
_
.,.-------------------------..-.11;
Pra<liceExam 2
The truss member properties are E = 200 X 10' kPa
and A = 2580.6 mrn". The vertical deflection at point
P is most nearly
(A) 0.25 mm
(B) 0.48 mm
(C) 0.51 mrn
(D) 0.75 mm
Problems
71
Problems 47-49 are based on the following information
and illustration.
The span length and cross section of a reinforced concrete beam are shown. The beam is underreinforced.
The concrete and reinforcing steel properties are f~=
3000 lbf /in", j~ = 40,000 lbflin', and A, = 3 in'51t
44-46 are based on the following illustration.
·1
live load, P1ive
t
dead load, wd"d ~ 51bf/lt
;?I;
18i~nI15~i1~
A
3m
I"
3m
10 It
44. If the reaction at support A is 18.75 N, the reaction
at each of the outer supports is most nearly
(A)
(B)
(C)
(D)
5.6 N
7.2 N
11 N
14 N
45. The maximum value of vertical shear at any point
along the beam is most nearly
(A) 4.7 N
(B) 9.4 N
(C) 14 N
(D) 18 N
46. The maximum value of moment at any point along
the beam is most nearly
(A)
(B)
(C)
(D)
12in
I-I
beam
cross
section
47. Neglecting beam self-weight and based only on tbe
allowable moment capacity of the beam as determined
using American Concrete Institute (AC[) strength design specifications, the maximum allowable live load is
most nearly
(A)
23,000 lbf
(B) 29,000 lbf
(C)
35,0001bf
(D) 50,000 lbf
48. The beam supports a concentrated
live load of
50,000 lbf. Neglect beam self-weight. The minimum
amount of shear reinforcement required for a centerto-center stirrup spacing of \2 in under ACI strength
design specifications is most nearly
1.2 N·m
(A)
3.1 N·m
4.6 N·m
5.7 Nun
(B) 036 in'
(C) 067 in'
(D) 0.78 in'
018 in'
49. The balanced reinforcing steel ratio for this beam
in accordance with ACI specifications is most nearly
(A)
0037
(B) 0043
(C)
0.051
(D) 0.058
___________________________________
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72
Civil Discipline-Specific Review fa, the FE/EITExam
_
Problems 50 and 51 are based on the lollowing information and illustration. A solid steel column with a
fixed support and material and geometric properties as
shown is concentrically
52. A steel beam is shown.
A
loaded.
r
dead load, wdead = 2 kips/ft
p
t
9 in
tw = 0.25 in
I----J .J
___
h = 18 in
-E+-I-~in
I
9ft
t
column
cross section
section A-A
The yield strength is 50 kips/in" Neglect beam weight.
In accordance with AlSC LRFD specifications, the maximum allowable live load is most nearly
(A)
(B)
(C)
(D)
Fy = 50 kips/in2
E ~ 29 X 10' Ibflin2
2 kips/It
5 kips/It
6 kips/It
8 kips/It
53. Upon graduating lrom college in 4 years, Irene will
need $800 lor a trip to Europe. If the money is to come
50. In accordance with American Institute 01 Steel
Construction (AISC) load and resistance lactor design
(LRFO) specifications, the available axial compressive
stress for design purposes is most nearly
(A)
(B)
(C)
(0)
(A)
(B)
(C)
(D)
18 klps/in?
26 kips/In?
29 kips/in2
39 kips/in2
51. If the column is braced against buckling in the weak
direction at midheight, the available capacity is
(A)
(B)
(0)
(0)
470 kips
780 kips
940 kips
1400 kips
PPI
• WWW.PPi2poss.com-------------
from a savings account that pays 5% interest annually,
approximately how much does she need to deposit at
the beginning 01 those 4 yr?
$660
$740
$820
$970
54. A man deposits $5000 on January 1 in a savings
account that pays 8% interest compounded annually.
He wishes to withdraw all the money in five equal endof-the-year payments beginning December 31 of the first
year. How much will he withdraw each year?
(A) $1000
(B) $1080
(C) $1252
(D) $1469
..
..;.;
__
o:.;.__
~;.;.;. ... ;.;.;.--.;.;.~.:.!o"i..
Pralli,e Exam 2
73
The critical path consists of the activities
55. What is a surety bond?
(A)
(B)
(C)
(D)
(A) insurance that a project will be completed
(B)
insurance to protect the owner in case the chosen bidder refuses to perform the work
(C) a guarantee of funds equivalent to a cosigned
promissory note
(D) a guarantee of funds to pay subcontractors
A-B-E-G
A-B-D-G
A-B-D-F-G
A-C-F-G
Problems 59 and 60 are based on the following illustra-
tion.
56. Which of the following statements
bond?
is valid for a bid
(Pal
(A) It pays for costs incurred by the bidder if the
bid deadline is missed.
(B) It represents the costs that the owner incurs if
the bidder fails to enter into a contract.
(C) It represents the costs incurred by subcontractors if a project is underbid.
(D) It pays for office overhead costs related to a
40
35
30
25
20
15
10
5
bid.
4
5
59. What is most nearly the ultimate
strength
2
3
E
(%)
57. One of the main reasons to enter into a joint venture on a big construction
(A)
(B)
project is to
Jet bidders know the size of the project for
more accurate cost estimating
better plan future uses for the completed pro-
ject
spread the risks associated with the project
(D) pay for the bidding process
[C]
of the
material?
(A) 17 Pa
(B) 25 Pa
(C) 35 Pa
(D) 40 Pa
60. The yield point of the material is most nearly
58. A construction project network consisting of activities B through F is shown, along with their durations
in days.
B \-
----;~
A '1-------(
c\--
17Pa
26 Pa
(C) 34 Pa
(D) 40 Pa
E
2
4
start
(A)
(B)
D'I-------I
5
~
6
___________________________________
G end
F
3
PPlowww.ppi2poss.(om
74
(ivil Discipline-Specific Review for the FE/Ell Exam
_
SOLUTIONS TO PRACTICE EXAM 2
Calculate the radius of the curve.
LC
1. The trapezoidal rule is
R= -.-1 =
2Slll2
n
A = w (h' ; h + h, + h3 + h4 + ... hn-1)
325 m
(110.750)
2sin
2
= 197.48 m
Convert the intersection angle into radians.
bOundary""
I
I
I
I
I
I
10.3 m I
I
I
,
I •
lead
7.1 m I
6.5 ml
I
9m
I
I
,
I
I
I
9m
I
,
I
I
I
I
I
8.8ml
I
I
9m
I
9.6ml
I
,
6.9 m
I
9m
I
,
= (11075') (1;00)
= 1.9329 rad
I
I
I
I
I
I
I
I
I
9m
I
I
The length of curve from point of curvature (PC) to
point of tangent (PT) can be found using the following
I
equation.
,
' I
L=Rl
traverse /
line
= (197.48 m) (1.9329 rad)
= 381.7 m
A = (9 m) (
10.3 m + 6.9 m + 6.5 m )
----;;:---
(380 m)
The answer is B.
+ 7.1 ;, + 8.8 m + 9.6 m
4. By the trapezoidal rule, the area is
= 365.4 m2
(370 m')
h,
The answer is C.
A =w (
+ hs
2
4.2 ill + 8.3 m
2
+ 8.1 m + 6.7 m + 7.6 m )
2. Convert the intersection angle into decimal format.
= (5 m) (
110° + (45') (;~,)
= 143.2 m'
[=
(143 m')
The answer is B.
= 110.75°
The tangent distance is
T=--l
+ hz + he + h4 )
5. The 4 III long rod can be described as extending from
the center of a 4 ill radius circle. The end of the rod
is at the top of the circle when it is truly vertical and
LC
makes a circular arc when it goes out of plumb.
2C08
2
325 m
y
2 cos (110~750)
0.50 m
lout of plumb)
circular arc
= 285.99 m
(286 m)
---->JJyl
The answer is D.
.s,
- -
3.
)ox
'0 -
line of sight
Convert the intersection angle into decimal format.
1= 110° + (45') (:~, )
"circle center
10 m. -4m)
= 110.75°
PPJ. www.ppi2poss.lOm
+'!-!"'"
..
~
Practice Exam 2 Solution,
With the reference taken at the top of the circle, the
center is at coordinates h = x = 0 m and k = y = -4 1)1.
The equation of a circle with the center at (h, k) and a
radius, Y", is
7. The peak hour factor, phf, is
phf =
(x - h)' + (y - k)' = r'
peak hourly traffic
4(peak 15 minute traffic)
3500 veh
hr
4 periodS) (1140 ~)
(
hr
period
=
(x - 0 ill)' + (y + 4 ill)' = (4 ill)'
Removing units for simplicity (but remembering that
all distances are in meters) and simplifying the equation
gives
y'
+ 8y + x' = 0
7S
= 0768
(0.77)
The answer is B.
The change in vertical distance at the top end of the
rod when it goes 0.50 m out of plumb can be solved by
letting x = 0.50 and solving for y.
8. The Manning equation for open channel How is
y2 + 8y + x2 = 0
y'
+ 8y + (0.50)'
y'
=0
Determine
+ 8y + 0.25 = 0
y = -0.0314
the hydraulic
[the nontrivial solution]
The change in vertical distance
radius.
R= ~ =
(15 m)(4 m)
P
4 m+ 15 m +4 m
of the leveling rod end
= 2.61 m
is YI = -y = 0.0314 m.
A ratio of the smaller circular arc with a radius of 3.50 m
Solve for the velocity.
to the larger circular arc with a radius of 4.00 m can be
used to find y,.
v =
YI
y,
3.50 m = 4.00 m
::R2!31S
n
= (0.~15)
Y2 = (3.50 m) (4.~~ m)
(2.61 m)'!3 VO.005
= 8.94 mls
_ (
) (0.0314 m)
- 3.50 m
4.00 m
The
answer
(8.9 m/s)
is B.
= 0.0275 m
9. Assuming
Since the out-of-plumb reacting was 3.50 ill, the correct
reading with the rod truly vertical is
3.50 m - Y' = 3.50 m - 0.0275
ill
steady. incompressible
head loss is
c:)
h f = f (~)
= 3.47 tn
The answer is B.
6. In highway work, the length of the curve is understood to be the actual curved arc length, and the degree
of the curve is the angle subtended by an arc of 100 ft.
= (0.018)
10 m
(0008 m)
= 4.59 m
(4.6 m)
The answer
flow in a pipe. the
( 2 10)'
)
-;
( (2) (9.81 ~)
is C.
211" )
L = HI ( 360
0
= (352 m)(11.25°)
= 6911
The
answer
ill
(3~~0)
(69 m)
is A.
___________________________________
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,.-
- ~,
76
Civil Dis<ipline-Specific Review for the FE/EITExam
10. The pump power equation is
h
P = Q-y-
= Qpg-
7)
= (0.25
_
13. The capillary rise in hquids is
h= 4aeosj3
h
'Id
1)
~3)
(1000 ~~)
(4) (0.0696 :)
(981 ~)
100 m) ( 1 kW )
x ( 0.75
1000W
= 327 kW
(9730 ~~)
= 7.53 mm
(330 kWj
(1000 k~)
cos 0°
(0.0038 m)
(7.5 mm)
The answer is C.
The answer is A.
14. The continuity equation for one-dimensional
11. The total watershed area is
flow is
A = 80000 m2 + 20000 m' + 15000 m'
Solve for A, and A,.
= 115000 m'
A, = 7fdi = 7f(0.3 m)'
4
4
= 0.071
Determine the weighted runoff coefficient for the total
area.
m'
(80000 rn')(0.85) + (20000 m') (0.75)
A, = 7fdl = 7f(0.15 rn)'
4
4
= 0.018 m'
0._20"")
-'-+--'-(-.:.:15-'-0-'-0-'-0.=m=--2"")(c_
115000 rn2
C =
= 0.748
Since the volumetric flow in the pipe stream is continu-
The rational formula is
ous, solve for the water velocity at the discharge point.
Q= CIA
4 v
(0.071 rn") (1.9 ~)
_"11_
mm
v,--_
A,
0.018 m?
= 7.49 m/rnin
(7.5 m/rnln)
The peak runoff is
Q=CIA
The answer is B.
= (0.748) (80
~m)
(115000 m')
x CO~O:m) C:O~ s)
15. The pressure head (length of water in the standpipe), hp, at point A is
= 1.9 m'/s
hp=h+z=9m+lOm
The answer is C.
= 19 m
The answer
(20 m)
is C.
12. The pipe area is
16. The total head above the tail water is
ht=h=9m
The answer
is A.
= 0.192 m2
A=-
nd2
17. The uplift pressure, p, at point A is
4
d = ~
p = hp'Yw
=
= hpPwg
(4) (0.~92 m')
= (19 m) (1000 kg)
= 0.49 m
The answer is A.
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(0.50 m)
= 186 kPa
rn'
(190 kPa)
(9.81 m) ( 1 kPa )
S2
1000 Pa
The answer is D.
.. __
~ .. ~~--
.. --~
... -
P,ollile Exam2 Solulions
18. Determine the allowable stress by first determining the ultimate
bearing pressure.
Find the saturation.
o
The cohesion, c, is
+ ,,(DfNq + eNe
+ pgDfNq
= 0.5pgBNo
m")"x
Vw
(0.003
s = -v. X 10010 = 0
v
.004 m
assumed to be O.
quit = 0.5"(BNo
77
100%
= 75%
+ eNe
_ ( 2002 m
kg - 1000
. m
kg) ( 9.81 S2
m)
= (0.0)
3
3
x (1.2 m)(30.2) + (2002 ::)
The answer is D.
21. The mass-volume relationships for the soil arc used
(981 ~)
to determine the void ratio.
x (0.9 m)(30.2) + 0
_
2:.;6'----'kg"-----~~
m,
V,----
= 711 917 Pa
ci;
Divide the ultimate bearing pressure by the factor of
= 0.0098 m"
(0.01 m")
Vv = Vi - V, = 0.014 m3 - 0.01 m"
safety, FS, to determine the allowable bearing pressure
of the soil.
quit
711917 Pa
qalLowable = FS =
3
= 237305 Fa
(2.65) (1000 ::)
= 0.004 m3
Vv
0.004 m"
e= - =
V,
0.0l m3
(240 kPa)
= 0.4
The answer is A.
The answer is C.
m.t = mw
19.
+ ms
mw = mt - ms
= 29 kg - 26 kg
22.
Assume a total volume of the sample of LOO m3.
= 3 kg
W,= 0
w = mw x 100%
I
air
m,
-
1
Vv
= 3 kg x 100%
26 kg
= 11.5% (12%)
1\
Ww
W
--
The answer is C.
mt = mw +ms
20.
Jv,
...
W,
ms
=29kg-26kg
water
~I
VW
,•.,u·
1
iifii,~" \~
solid
rnw = mt -
I~
V,
t.<, r ; II ',t >:
= 3 kg
Calculate the water content.
Find the volumes of water} soil, and voids.
mw
3 kg
w
V = 1000 k~ = 1000 k~
m
m
= 0.003 m3
mw
26 kg
V, = Gpw = (2.65) (1000 ::)
= 0.0098 m3
(0.01 m")
"
Vv = Vt - V, = 0.014 m - 0.0l rn
= 0.004 m"
The weight of 1.00 m3 of soil is
"
w = "(V = (176 ~)
(1 m3)
= 176 kN
_________
'!""
__""'!-------'!""--------------
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78
Civil Discipline-Spedfic Review for the FE/EIT Exom
_
23. The void ratio is
Substituting,
v"
0.49 m3
- V, - 0.51 m3
W = W, + Ww = W, + 0.30W,
e--
= 1.30W,
= 0.96
= 17.6 kN
Solve for the weight of the solid and liquid phases.
W
-'
-
The
is D.
answer
24. The porosity
= 17.6 kN = 13.5 kN
is
1.30
V,
Ww = wW, = (0.30)(13.5 kN)
= 4.05 kN
(WOO~)
m'
= 0.49
The volumes of the water, solid,
space, respectively, are determined
gram.
IW = Pwg =
0.49
n--- V - 1.00 m'
(981
air; and total void
from the phase dia-
~)
N
1000 kN
The answer is C.
25. The following diagram shows relationships between
forces on a free-body diagram of the soil wedge.
= 9.80 kN/m3
x
Wsina
Ww
Vw
V
_ Ww _ 4.05 kN
"Iw - 9.80 kN
w -
m'
= 0.41 m3
W,
"18=V,
N = wcos«
V, = W, = W,
"Is
G-yw
L
13.5 kN
kN)
(2.70) ( 980 m
3
The shear force along the assumed
the force that resists sliding.
= 0.51 m3
Va = V - V, - Vw
= 1.00 m' - 0.51 m3 - 0.41 m'
failure plane,
S, is
8 = cL + Ntanq,
= cL + W cos Q tan q,
= 0.08 m3
VU=~,.I+l~u
The force that drives sliding of the soil wedge along the
assumed failure plane is W sin Q.
= 0.08 m' + 0.41 m3
= 0.49 m3
The degree of saturation
The factor of safety against sliding, FS is defined as the
resisting force divided by the driving force.
is
Vw X 100,0
'"
s=V
FS = -;c;-S_
v
= cL+ WCOSQtallq,
= 837%
_
The answer is D.
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Fdriving
0.41 m3
= 0.49 m3 x 100%
(84%)
Wsina
"I
15 m
L ---=-sin o
sin 20°
=43.9 m
~"'!-----..-'"'!'-":'"---~-""'!'--
Practice Exam2 Solulions
15 m
30. Chlorine must be supplied to the system to satisfy
both the demand and the residual.
x = Lcosa ~ ---
tan 40°
= (43.9 m)cos20
o
---
15 m
tan 40°
The chlorine required is
= 23.4 m
C = demand + residual
mg
mg
= 0.45 L + 0.15 L
lV = Asoil wedge")'
=
G)
19
(23.4 m)(15 m) (173
::)
= 0.60 mg/L
= 3036 kN/m
FS=
060mg)(.
.
L
(19.2 kPa) (43.9 m)
+ (3035 ~)
Ig
)(~))
lOOO lmg
1000 g
L
x (15,000,000 ~:y) (3.785 gal)
m = CQ = ((
cL+Wcosatan¢
Wsina
= 34.1 kg/d
cos 20° tan 15°
The answer
kN)
( 3035 -;.;;- sin 20°
(34 kg/d)
is C.
31. The molar concentration
= 1.5
of the acid solution is
m
The answer is A.
[HAc] = MW
V
3g
26. From the AASHTO Soil Classification System,
when F200 = 0.34, the soil is first classified as granular material. The plasticity index is
60.052 --L
rno l
1L
= 0.05 rnol/L
(0.05 M)
PI = LL-PL
By letting x equal the number of moles of acetic acid
that ionize in solution forming hydrogen ions (H+) and
acetate ions (Ac-), the appropriate celation ship can be
= 39 -29
= 10
For an LL of 39 and a PI of 10, the classification
is
A-2-4.
determined to calculate the amount of acetate iOIlH in
solution.
mol
[HAc] = 0.05 L -x
The answer is A.
[H+] = [Ac-I
=x
27. When F200 is 0.69, the soil is first classified as fine
grained. From the plasticity chart, for an LL of 72 and
a PI of 48, the soil is classified as CH.
[W][Ac-]
[HAcl
_ K
A
2
x
___
= 1.75 X 10 -5
0.05 - x
The answer is D.
28. The soil is first classified as silty-clay material because 4] % passes through the no. 200 sieve. With an
LL of 32 and a PI of 15, the soil is classified as A-5.
Solving for x gives the concentration
acetate ions in solution.
of hydrogen
or
=x
The answer is D.
= 927
29. The hydrostatic
[in moll L]
10-4 rnol/L
X
4
(9.27 X 10-
M)
pressure is
The percentage of acetic acid ionized in solution is
P = 'Yh = (
= 12 kPa
The answer is D.
8.0 kN)
m
3
(1.5 m)
9.27
(
4
lIJ0.05 M
X
M)
X
100% = 1.85,1% (1.9%)
The answer is B.
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80
Qvil Discipline-Specifi,Review for the FE/Ell Exam
_
36. Time is a function of the velocity and acceleration.
32. The solids residence time is
e = V/Q
m
The suspended solids concentration
the following equation.
is determined
using
The answer
m
s
1.5.,
(3 s)
is B.
37. From the vehicle speed versus density graph, the
relationship
(0..5) (250 ~
:g) (1+ (0.06
is linear and is found to be
- 62 ~)
mi
(
v=53--
D d))
hr
(10
= 53 - 0.8k
= 0.0544 m3/s
~3)
(86400~)
ml)
53 -
hr
k
66.25 veh
mi
[velocity in mi/hr]
This expression for speed can be substituted into the
relationship for traffic volume and density to give
= 4700 m3/d
The answer
a
ITI
the volume of the reactor can be
(10 d) (0.25 ~)
V = (00544
t=---=
s
= 3.33 s
V = e1QY (So - S)
X (1 + Kdeg)
(3500
20 - -1.5-
v f - Vi
s
x = e1Y(so - S)
e (1 + Kdeg)
Combining equations,
found.
.
q = kv
is C.
= k(53 - 0.8k)
= 53k - O.8k2
33. The PVC station for a vertical curve is found by
half the curve length from the PVI station.
subtracting
From this, the traffic volume is 0 veh/hr at traffic densities of 0 veh/rni and 66.25 veh/rni. This results in a
parabolic curve as shown.
L=800ft
L
.
2 = 400 ft
PVC = PVI -
!:.2
volume,
q
Iveh/hrl
= 8700 ft -
400 ft
= 8300 ft
(sta 83+00)
800
600
The answer is B.
400
34. The elevation at the PVT is
200
L
= PVleLev +022
PVT"cv
= 743.24 ft + (0.016)(400 ft)
= 749.64 ft
(7.50 ft)
The answer is B.
VJ - v;
(20
2a
= 58.3 m
The answer is D.
20
30
40
50
70
~f- ~)2
(15
(2)
(58 m)
(15 ~)
dq _ d(53k - 0.8k2)
dk -
dk
= 53 -1.6k
=0
"3 veh
-
iJ
kmax 1/ = ---illL
,
1.6
= 33.125 veh/rm
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60
traffic density, k{veh/mi)
The maximum traffic volume is where the slope of tbis
curve equals O.
35. Using the basic relationship of velocity. acceleration, and distance to solve for distance gives
d=
10
~------"'!"~-------...
The density is 33.125 veh/rni for maximum traffic volume. Substitution into the traffic flow relationship gives
the maximum traffic volume.
Prodic. Exam2 Solutions
81
For a side-friction factor of f = 0, the superelevation
is
given by
v'
e+1=e= -
gR
q = 53k - 0.8k'
= (53
~:)
=
(33125 :~)
(880 veh/hr)
~)(1000
~~;)(~)y
(981 :~) (1318 111)
2 veh)'
- (08. --mi' ) ( 3315veh-hr
rni
= 878 veh/hr
((75
= 0.0336
(0.034)
The answer is B.
The answer is B.
40. The structural
38. Since there is a negative grade preceding a positive
grade, this is a sag vertical curve. Using the sag vertical curve equations from the civil engineering section of
the NCEES Handbook, the aigebraic difference between
number is the sum of products of
the layer depths (thicknesses) and strength coefficients.
= arecydeDrecycle + aoriginal surfaceDoriginal
SN
surface
+ abaseDbase + asubbaseDsubba.<;e
6.6 = (0.42)(6 in) + (0.3)(3 in) + (0.2)(10 in)
grades is
A=I-l%-3%1
=4%
+ a,ubbas,(8 in)
Where the stopping sight distance, 5, is iess than the
vertical curve length, L,
A5'
L=-~~
400 + 355
(4)(430 ft)'
= 400 + (3.5)(430 ft)
asubbase
= 0.15
The answer is C.
41. Draw an influence line for shear at support A with
loads positioned as shown for maximum shear value.
For example, for shear at support A, using basic beam
statics,
= 388.2 ft
~ movement of unit load
Where the stopping sight distance is greater than the
curve length,
400 + 3.55
L=25A
400 + (3.5)(430 ft)
= (2)(430 It) 4
= 383.8 ft
;17
~
A
20 m
1~=::::::::==========J2.0
influence line for shear at support A
2m
From the two values for length of curve, it can be seen
that the stopping sight distance is greater than the
curve length. Therefore, the required vertical length
of curve is
L = 383.8 ft
(380 ft)
The answer is B.
1000N 1000N
",-===::::::::::===:::~lAd
2.0
loads placed for maximum
230 ill
s
R--- 1> - ( 0) ('IT
10
= 1318 m
shear value at support A
The maximum value for shear at support A is
39. The radius of curvatu re is
..
1--1
fad)
J 800
VA = (1000 N)(1.8) + (1000 N)(2.0)
= 3800 N
The answer is D.
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82
Civil Discipline-SpecificReview for the FE/EIT Exam
_
42. An influence line for bending moment at support A
can be drawn with loads positioned as shown for maximum moment value. As in finding the influence line
By inspection, the a-component of force in member I\.1P
is the same magnitude and opposite direction as the
reaction at point :M.
for shear, a unit load is moved across the heam and the
variation in bending moment at point A is graphed.
t
1
J:
J Ri.,. + R~,"
= )(6820 NJ2 + (4448 NJ2
.
of un,t load
movement
MP = RM =
---
= 8142 N
;;p;;
M
1----.1-, ----I
10 m
-6820
N
10 m
:-::--~:-:----~=======:j-10.0
influence line for moment
m
at support A
8142 N
3.0 m
ON
2m
1--1
1000
N1000
N
------====id-100
loads placed for maximum
p
-
6820 N
'-01;.-----------"J
N
f
.....
6820N
m
-8.0
m
4448N
value at support A
·1
4.6 m
The maximum value for moment at support A (disregard the sign, since the maximum value is wanted)
is
Applying a unit load at point P produces the following
virtual forces.
MA = (1000 N)(8.0 m) + (1000 N)(10.0 m)
= 18000 N'm
M
-1.5
(18 kN·m)
N
The answer is D.
3.0 m
1.8 N
ON
43. The principle of virtual work can be used to find
the deflection at point P.
Choose the positive directions as upward and to the
-
1.5 N _-::-_--_:-:-:-:--
right. Choose positive moments as clockwise.
__
N
-.-;~p
t
1.5 N
LM =ON
1N
N
= RM• (3.0 m) + (4448 N)(4.6 m)
RM• = 6820 N
[to the left]
LMM=ON
The following table summarizes the actual and virtual
= -RNJ3.0
RN• = 6820 N
m) + (4448 N)(4.6 m)
y
-
RM, = 4448 N
prl • www.ppi2pDss.com
forces.
[to the right]
LF =0 N
= RM,
.1
4.6 m
4448 N
[upward]
FQ
Fp
member
(N)
L
FQFpL
(virtual force)
(m)
(Nm)
MP
NP
MN
8142
6820
0
1.8
1.5
0
5.49
80459
4.6
47058
3.0
0
total = 127517
"!'""!""~~!""!~~~~~~~"!""'""!'~~~~':"'!"~~~---"'!"!
P,adice Exam2 Solution.
83
The change in moment is given by the area under the
shear diagram up to that point.
Calculate the vertical deflection at point P.
F'L
.0,.p = LFQoL
= LF'Q-PEA
1
= EALFQFpL
=
6
(5.6 N)(U2
m)
= 3.1 N-m
1
(200 X 10
G)
kPa)
(1.000
:;a) )
M.~3m = M.~1.12 m +
G)
(-9.4 N)(L88 01)
= 3.1 Non - 8.8 N-m
)'
( x (2580.6 mm") lIn
( 1000 mm
= -5.7 N·m
x (127517 Nm) (1000 :~)
= 0.25 mrn
=
t
t
w e 5 N/m
The answer is A.
~
44. From the laws of equilibrium, each reaction at an
outer support is
1"
;;t;
3m
5.6 N
5.6 N
Route< = ~(wL - 18.75 N)
=
G) ((
5 ~)
(6 m) -18.75
N)
x=1,12m
1,88m
l----j
1.88m
9'4
V(N)5LI~
= 5.6 N
t
3m
18.75 N
n
1.12m
-----=:::::J=---5.6
The answer is A.
-9.4
45. The maximum shear can be determined from a
shear diagram. (Although not necessary for this problem, a moment diagram has also been constructed, and
this is used for solving the next problem.)
3.1
3.1
~--i--,£------:>
M (N.m) :L
The change in shear is the area under the load diagram
up to that point. Up to the center of the beam, point A,
v = 5.6 N - (5 :)
(3 m]
-5.7
FrOIn the shear diagram, the maximum
value for shear
occurs at support A.
= -9.4 N
Vmax = 9.4 N
At point A,
v = -9.4 N + 18.75 N
= 9.35 N
The answer
(9.4 N)
The moment is equal to the area under the shear dia-
46. From the moment diagram in SoL 45, the value for
maximum moment occurs at support A.
gram.
Mm"", = 1- 5.7 N·m[
= 5.7 N·m
M=JVdx
The answer
___________________________________
is B.
is D.
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84
Civil Discipline-Specifi( Review for Ihe FE/EITExom
47. The height of the stress block is
. 2
(3 Ill)
Ad.
a = o 85jib
The nominal concrete shear strength is
lbf)
40,000;;;2
(
_
v, = 2 \I!jibd
= 2 3000 lb,f(12 in)(15 in)
Jc
In
= (0.85) (3000 :~;) (12 in)
= 19,718 Ibf
(0.75)(19,718Ibf)
¢iVc
= 3.92 in
2
2
= 7394 lbf
For flexure, the strength reduction factor, 1> is 0.90.
¢iV,
VU>T
Therefore, shear reinforcement is required.
In accordance with ACI specifications, the minimum
required amount of shear reinforcement for a stirrup
III
12 ft
spacing of 12 in is
= 117,360 ft-Ibf
Au = 50bs = (50)(12 in)(12 in)
/y
40000 lbf
,
The maximum bending moment occurs at midspan.
in2
= 0.18 in'
u; = 1.2Mdead + 1.6MI;ve
The amount of shear reinforcement based on factored
loading can be determined as follows.
WDL2
hveL
8 + 16
.
4
-1
-. 2
V, = Au/y-
This can be solved for the maximum allowable live load.
d
s
1>(V, + V,) ::>Vu
4(Mu_1.2W~L2)
PUye =
Au = Vu -
1>\1;,
d
6£
1.
1>/y-;
(4) ((117,360 ft.(lb(fl lbf)
- (1.2)
5 It 8 (10 ft)
_ 40,030 lbf - (075)(19,718 Ibf)
2))
-
(0.75) (40,000 ~bf) (15 in)
Ill'
12 III
= 0.67 in'
(1.6)(10 ft)
The larger value for Au controls. Use Au = 0.67 in"
= 29,:121 lbf
(29,000 lbf)
The answer
is C.
The answer is B.
49_ The ratio of the rectangular stress block depth to
the neutral axis depth is
48. For shear, the strength reduction factor, 1>, is 0.75.
The maximum
factored shear force in the beam is at
either one of the supports and is
.
wDL
v" = 1.2Vdead + 1.6V,ive= 1.2- 2
~ ("I ( (,
= 40,030 lbf
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'*'l
(W "')
,
P
2
+ 1.6- -1ive
{HI ('0":, "")
131 = 085::> (0.85 - 0.05
(f~-
4000))
1000
> 0 65
-'
0.85 - (0.05) (/~ - 4000)
1000
= 0.85 - (0.05)
= 0.90
lbf
3000 --4000in2
(
1000 lbf
lbf )
in2
in2
------""!'"-~-~~~~~~~~~~~~-:+~b~~~4~~~
'I
!
!
Practic. Exam 2 Solutions
Use {3,
=
0.85 since this is its maximum allowed value.
0.85{31f')
c
fy
Pb=
87,000 lb;
(
(
lbf
8S
In the strong direction,
)
In
87,000 ,---, + fy
III
,
(0.85)(0.85)
(
,
(6 in)(9 in)3
= \ _---"-12f-_
(6 in)(9 in)
(3000 ~))
40,000 lb;
m
x (
-
= 2.6 in
87,000 ~
)
Ibf
lbf
87,000 ,---, + 40,000 ,---,
III
= 0.0371
SD
~\..strong
(2.1)(9 ft) (12 in)
_
ft
26 .
-
.
1Il
_
- 87.2
III
(0.037)
SRstrong
The answer is A.
> SRweak
so SRstrong
J
controls.
From AISC Table 4-22, the available strength is 25.9
kips/in2. The capacity is
50. Since the unsupported lengtb of the column is the
same about both the strong and weak axes, the largest
slenderness ratio results from bending about the weak
axis. Therefore, the least radius of gyration applies.
kiPS)
25.9 inz
(
(6 in)(9 in) = 1398.6 kips
(1400 kips)
The answer is D.
i
r=R-=fff
~
(9 in)(6 in)3
12
= \ (9 in)(6 in)
52. The maximum shear, Vmax1 occurs at the supports.
V';nax
(1.2WD + 1.6wLlL
2
=
Rearranging,
2Vmax
-L- -1.2wD
= 1.73 in
'WL =
The design effective length factor, k; is 2.10 for the given
column end support conditions (fixed-free). Therefore,
For the given cross section,
the slenderness ratio is given by
•
kl
h
18 in
-=---
tw
0.25 in
=72
418
418
(2.1)(9 ft) (12 ~)
SR= -r =
I
1.6
1.73 in
,jFyw =
= 131.1
50 kips
in2
= 59.1
The available column strength is read directly from AISC
Table 4-22 as 17.7 kips/in2 (18 kips/In").
The answer
522
,jFyw
is A.
522
50 kips
in2
= 73.8
=
51. The effective length in the weak direction has been
halved, so SRweak also halves.
131.1
c
SRweak= -2= 6~.5
Therefore,
418
h
__...__""!'~-'!""!'!"'...'!' ..,.'''-'!--~--'!' ' -'!' ' ---'' !' ---~'!.' ' ----
c...····
522
--<-<-,jFyw - tw - ,jFyw
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86
CivilDisdpline-Spedfil Review for the FE/ElY Exom
_
For this condition, the normal shear strength is
56. Bid bonds are usually based on the amount of the
bid and can run from 5-20% of the amount of the to-
418
tal bid. This amount represents the damages or costs
incurred by the owner if the bidder fails to enter into a
JF
yw
Vn = 0.6FywAw-- h
tw
= (0.6) (50
~~s)
((0.25 in)(18 in))
c~n
contract and the work has to be readvertised for bids,
or the difference in cost between the low bid submitted by the defaulting bidder and the next responsible
bid where the work must be awarded to the next lowest
bidder.
= 110.8 kips
The answer is B.
¢vVn = (0.9)(110.8 kips)
= 99.7 kips
57. Risk is an important factor in construction.
Set Vm= = Vn to find the allowed WL.
WL
L
2¢vVn
12. WD
2Vmax
---
- -L- -1.2wD
1.6
(2)(99.7 kips) _ (1.2) (2 kiPS)
20 ft
ft
1.6
= 4.7 kipsjft
(5 kips/ft)
1.6
The answer
One way to do this is to enter into a joint venture
with other contractors. A joint venture is a short-term
partnership arrangement in which each of two or more
participating construction companies is committed to a
predetermined percentage of a contract, and each shares
proportionately in the final profit or loss. One of the
participating companies acts as the manager or sponsor
of the project.
The answer
is B.
53. Determine
the present value of the money from its
is C.
58. This can be solved with critical path method (CPM)
calculations.
anticipated future value.
Determination
of the earliest start time
(EST) and earliest finish time (EFT) is done by a forward pass through the network. Determination of the
latest start time (LST) and latest finish time (LFT) is
done by a backward pass through the network.
P=F(l+i)-n
= ($800) (J + 0.05)-4
= $658
There-
fore, it is prudent to spread it as widely as possible.
($660)
The answer is A.
During a forward pass through the network, the EST of
an activity is the maximum of the EFTs of the activities preceding it. The EFT is the sum of the EST and
54. Use the annuity equation to determine the five equal
the duration, D. Therefore, minimum project time to
completion is 12 d.
payments made from the present-value deposit.
A =P (
i(1 + i)n
During a backward pass, the LST is found by starting
with the minimum project time of 12 d. Then, the
LFT of an activity is the minimum LST of the activities
preceding it. The LST is the LFT minus the duration.
)
(1 + i)n - 1
+ 0.08)5)
(1 + 0.08)5 - 1
= ($5000) ((0.08)(1
The critical path is the path that results in a total float
(TF) of 0 d. TF is the LFT minus the EFT.
= $1252
duration
The answer is C.
55. Bonds are not insurance. A surety bond is equivalent to a cosigned promissory note. The principal on
a surety bond is primarily liable for the project.
The
surety is a cosigner who is liable only if the principal
EST
d. no.
activit
start
B
d
0
C
6
0
0
D
.5
2
4
4
3
0
9
12
E
F
G (end)
4
fails to discharge the obligation undertaken.
Therefore, the critical path is A-B-D-F-G.
The answer
is C.
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The answer is C.
~~-~~-~~~~~-.;.--~--..;.~
Practice Exam 2 Solution.
87
59. The ultimate strength is defined as the maximum
stress the material can support without failure. The
diagram shows it is 3.5 Pa.
The answer is C.
60. The yield point is the point at which a material will
experience permanent deformation. It is usually close
to the elastic limit. The diagram shows it is 26 Pa.
The answer is B.
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