the ultimate civil fe practice exam · a. analysis of forces in statically determinant beams,...

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THE ULTIMATE CIVIL

FE PRACTICE EXAM

http://www.civilengineeringacademy.com

http://www.civilengineeringacademy.com/


TABLE OF CONTENTS

WELCOME…………………………………………….

EXAM SPECIFICATION……………………

START TEST ………………………………...

SOLUTIONS………………………………………

SCORE SHEET…............................

Page 6

Page 13

Page 112

Page 225




WELCOME!!

Welcome to The Ultimate Civil FE Practice Exam! Thank you so much

for purchasing this book!

This exam contains 110 questions and solutions following the exact

same format of the NCEES exam. This was built to have the same

look and feel of the real exam. The only difference, of course, is that

this is a paper based exam and the real exam is computer based (it

didn’t use to be!).

Passing the FE exam is typically a requirement for most civil

engineering students to graduate - it is also a must for your

professional career too.

The FE is your first step to eventually obtaining your PE license. After

you’ve finished school and gained the necessary experience, you’ll be

back studying, preparing to take the PE. There are many similarities

and quite a few differences between the exams, but that’s something

to worry about much later down the road.

This exam was built to help you pass by providing material that will

challenge you to think. You are given 6 hours to take the exam with

only one reference at your disposal - the NCEES approved digital

reference book that sits side by side with the exam. It’s searchable

and will be your main resource to pass this exam.




If you don’t have this reference manual you can download it for free

from the NCEE website (www.ncees.org). You need to become

intimately familiar with that and use it when taking this exam.

Good luck on this exam, your career, and eventually become a

professional engineer!

Sincerely,

Isaac Oakeson, P.E.

(You’re going to have that by your name too!)

P.S. Errata for this and any other exam we have made can be found

at www.civilengineeringacademy.com/errata.



http://www.civilengineeringacademy.com/errata


LEGAL INFORMATION

Civil Engineering Academy’s

The Ultimate Civil FE Practice Exam

Isaac Oakeson, P.E.

Rights and Liability:

All rights reserved. No part of this book may be reproduced or

transmitted by photocopy, electronic, recording, or any other method

without first obtaining permission from the author. The information in

this book is in no way endorsed by the NCEES organization and the

author shall not have any liability to any person with respect to any

loss or damage caused by the problems in this book.

In other words, please don’t go copying this thing willy-nilly without

giving credit where it should be given by actually purchasing a copy.

Also, don’t go designing real things based on these problems.

If you find errors in this book (I am human of course), or just want to

comment on things, then please let me know! I can be reached

through the website at www.civilengineeringacademy.com or by email

at [email protected].

ABOUT THE AUTHOR

Isaac Oakeson, P.E. is a registered professional civil engineer in the

great state of Utah. Shortly after passing the PE exam in the Fall of

2012 he started www.civilengineeringacademy.com and

www.civilpereviewcourse.com to help future students pass. He has

authored and helped author various exams with his entire goal of

providing the best resources for engineers to study and pass the PE.




FE CIVIL EXAM SPECIFICATIONS

I. Mathematics (7-11)

A. Analytic geometry

B. Calculus

C. Roots of equations

D. Vector analysis

II. Probability and Statistics (4-6)

A. Measures of central tendencies and dispersions (e.g., mean, mode,

standard deviation)

B. Estimation for a single mean (e.g., point, confidence intervals)

C. Regression and curve fitting

D. Expected value (weighted average) in decision making

III. Computational Tools (4-6)

A. Spreadsheet computations

B. Structured programming (e.g., if-then, loops, macros)

IV. Ethics and Professional Practice (4-6)

A. Codes of ethics (professional and technical societies)

B. Professional liability

C. Licensure

D. Sustainability and sustainable design

E. Professional skills (e.g., public policy, management, and business)

F. Contracts and contract law




V. Engineering Economics (4-6)

A. Discounted cash flow (e.g., equivalence, PW, equivalent annual

worth, FW, rate of return)

B. Cost (e.g., incremental, average, sunk, estimating)

C. Analyses (e.g., breakeven, benefit-cost, life cycle)

D. Uncertainty (e.g., expected value and risk)

VI. Statics (7-11)

A. Resultants of force systems

B. Equivalent force systems

C. Equilibrium of rigid bodies

D. Frames and trusses

E. Centroid of area

F. Area moments of inertia

G. Static friction

VII. Dynamics (4-6)

A. Kinematics (e.g., particles and rigid bodies)

B. Mass moments of inertia

C. Force acceleration (e.g., particles and rigid bodies)

D. Impulse momentum (e.g., particles and rigid bodies)

E. Work, energy, and power (e.g., particles and rigid bodies)

VIII. Mechanics of Materials (7-11)

A. Shear and moment diagrams

B. Stresses and strains (e.g., axial, torsion, bending, shear, thermal)




C. Deformations (e.g., axial, torsion, bending, thermal)

D. Combined stresses

E. Principal stresses

F. Moh’r circle

G. Column analysis (e.g., buckling, boundary condition)

H. Composite sections

I. Elastic and plastic deformations

J. Stress-strain diagrams

IX. Materials (4-6)

A. Mix design (e.g., concrete and asphalt)

B. Test methods and specifications (e.g., steel, concrete, aggregates,

asphalt, wood)

C. Physical and mechanical properties of concrete, ferrous and

nonferrous

metals, masonry, wood, engineered materials (e.g., FRP, laminated

lumber, wood/plastic composites), and asphalt

X. Fluid Mechanics (4–6)

A. Flow measurement

B. Fluid properties

C. Fluid statics

D. Energy, impulse, and momentum equations

XI. Hydraulics and Hydrologic Systems (8–12)

A. Basic hydrology (e.g., infiltration, rainfall, runoff, detention, flood

flows,




watersheds)

B. Basic hydraulics (e.g., Manning equation, Bernoulli theorem, open-

channel

flow, pipe flow)

C. Pumping systems (water and wastewater)

D. Water distribution systems

E. Reservoirs (e.g., dams, routing, spillways)

F. Groundwater (e.g., flow, wells, drawdown)

G. Storm sewer collection systems

XII. Structural Analysis (6–9)

A. Analysis of forces in statically determinant beams, trusses, and

frames

B. Deflection of statically determinant beams, trusses, and frames

C. Structural determinacy and stability analysis of beams, trusses, and

frames

D. Loads and load paths (e.g., dead, live, lateral, influence lines and

moving

loads, tributary areas)

E. Elementary statically indeterminate structures

XIII. Structural Design (6–9)

A. Design of steel components (e.g., codes and design philosophies,

beams, columns, beam-columns, tension members, connections)

B. Design of reinforced concrete components (e.g., codes and design

philosophies, beams, slabs, columns, walls, footings)




XIV. Geotechnical Engineering (9–14)

A. Geology

B. Index properties and soil classifications

C. Phase relations (air-water-solid)

D. Laboratory and field tests

E. Effective stress (buoyancy)

F. Stability of retaining walls (e.g., active pressure/passive pressure)

G. Shear strength

H. Bearing capacity (cohesive and noncohesive)

I. Foundation types (e.g., spread footings, deep foundations, wall

footings, mats)

J. Consolidation and differential settlement

K. Seepage/flow nets

L. Slope stability (e.g., fills, embankments, cuts, dams)

M. Soil stabilization (e.g., chemical additives, geosynthetics)

N. Drainage systems

O. Erosion control

XV. Transportation Engineering (8–12)

A. Geometric design of streets and highways

B. Geometric design of intersections

C. Pavement system design (e.g., thickness, subgrade, drainage,

rehabilitation)

D. Traffic safety




E. Traffic capacity

F. Traffic flow theory

G. Traffic control devices

H. Transportation planning (e.g., travel forecast modeling)

XVI. Environmental Engineering (6–9)

A. Water quality (ground and surface)

B. Basic tests (e.g., water, wastewater, air)

C. Environmental regulations

D. Water supply and treatment

E. Wastewater collection and treatment

XVII. Construction (4–6)

A. Construction documents

B. Procurement methods (e.g., competitive bid, qualifications-based)

C. Project delivery methods (e.g., design-bid-build, design build,

construction

management, multiple prime)

D. Construction operations and methods (e.g., lifting, rigging,

dewatering

and pumping, equipment production, productivity analysis and

improvement, temporary erosion control)

E. Project scheduling (e.g., CPM, allocation of resources)

F. Project management (e.g., owner/contractor/client relations)

G. Construction safety

H. Construction estimating




XVIII. Surveying (4–6)

A. Angles, distances, and trigonometry

B. Area computations

C. Earthwork and volume computations

D. Closure

E. Coordinate systems (e.g., state plane, latitude/longitude)

F. Leveling (e.g., differential, elevations, percent grades)




START TEST




1. What is the length of a straight line connecting two points in

three-dimensional space: A(2, 1, 7) and B(-1, 3, 3)?

a) 20

b) 29

c) 39

d) 21




2. Solve for the roots of the following polynomial equation.

3 7 6 0x x

a) 1, 2, 3 b) -1, 2, -3

c) 1, -2, 3

d) 1, 2, -3




3. Determine the coordinates of a circle’s center which satisfy:

2 2 10 14 49 0x y x y

a) (-10, 14)

b) (10, -14)

c) (-5, 7)

d) (5, -7)




4. A straight line goes through a point (1, 7) and is perpendicular to

the line 5 3 16 0y x . Solve for the equation of the line.

a) 3 5 16 0y x

b) 3 5 16 0y x

c) 3 5 16 0y x

d) 3 5 26 0y x




5. Which of the following equations described a circle having center

at point (2, -5) and passing through the point (8, 3)?

a) 2 2 4 10 29 0x y x y

b) 2 2 4 10 79 0x y x y

c) 2 2 4 10 71 0x y x y

d) 2 2 4 10 100 0x y x y




6. A particle moves in the x-y plane with the following equation path:

9sinx t

2cosy t

Solve for the equation of the path of the particle.

a) 2 281 4 324x y

b) 2 24 81 324x y

c) 2 22 9 18x y

d) 2 29 2 18x y




7. What is the cross product of these two vectors 8 2A i j k and

3 3B i j k ?

a) 30 11i j k

b) 30 11i j k

c) 18 11i j k

d) 18 11i j k




8. Solve for the angle made by these two vectors:

M i j k

N i j k

a) 85°

b) 80°

c) 75°

d) 70°




9. Solve for the inflection point of the following polynomial function:

4 3 212 30f x x x x

a) (-5, 125)

b) (5, -125)

c) (19, 1)

d) (-1, -19)




10. Which expression below is equivalent to 1?

a) 2sin cos

b) 2 2sin cos

c) 1 cos2

2

d) 2 2cos sin




11. A bag contains five red balls, seven blue balls, and three black

balls. What is the probability of picking a red ball and a black ball?

a) 0.07

b) 0.14

c) 0.20

d) 0.34




12. A dice is rolled and it shows number 2. What is the probability of

obtaining the number 2 again if the dice is rolled once more?

a) 1

2

b) 1

3

c) 1

6

d) 1

12




13. A book publisher is doing a quality assurance check of books

produced everyday. The data said that 3% of the total books are

defective. Solve for the probability of having 2 defective books out

of a total of 15 books chosen randomly.

a) 0.004

b) 0.093

c) 0.133

d) 0.064




14. Ten samples of a concrete cube were tested for compressive

strength. The resulting data presented below is in the units of psi.

2995 ; 3005 ; 3001 ; 2991 ; 2984;

3004 ; 3009 ; 3015 ; 2998 ; 3002

Find the mean and the variance of the sample data.

a) 3000.4 ; 79.6

b) 3000.4 ; 71.6

c) 2997.8 ; 87.1

d) 2997.8 ; 78.4




15. A doctor collects data from a group of six pregnant women.

Suppose that the probability of having a baby with brown eyes is 0.3. The probability that at least one woman will have a baby with

brown eyes is nearly:

a) 0.3

b) 0.05

c) 0.7 d) 0.9




16. Which of the following statements is the function to create a

spicific number of loops?

a) If, else, end

b) For, end c) While, do

d) If, elseif, end




17. A student writes several lines of code as follows:

x = 0;

y = 1;

while x <= 4 y = y+1;

x = x+2;

do

What is the value of variable y in the end of the program?

a) 3

b) 4

c) 5 d) 6




18. Which of the following is the benefit of behaving ethically?

a) Better reputation

b) Better feeling about yourself

c) Higher salary d) None of the above




19. You are a city engineer in charge of selecting bids for a big

infrastructure project in the city. There is one contractor who

came in with the high bid and offered you a limited-edition Rolex

watch. You love Rolex watches but you know that the lowest bid

will be selected as the winner of project. What should you do?

a) Accept the Rolex watch and accept the bid

b) Accept the Rolex watch and reject the bid

c) Reject the Rolex watch and accept the bid d) Reject the Rolex watch and reject the bid




20. An engineer-architect works for a big design firm. During the night

(after working hours), he uses some of office’s equipment and materials to make an architectural model/maquette for his side

job. He receives overtime salary because he is considered as

working overtime. Why is this action considered unethical?

a) His contract prohibits misuse and misappropriation of office’s

equipment

b) He may wear out the office’s materials c) He may encourage other employees to do side jobs in the office

d) He can undercut his fee because he has a lower overhead




21. You are aware that a registered engineer intentionally violates a

state’s rule of professional conduct. What should you do?

a) Do nothing

b) Report the violation to the employer c) Report the violation to the affected parties

d) Report the violation to the state’s engineering registration board




22. Which of the followings is the best description of ‘plan stamping’?

a) Legal action of signing off on a project you won on tender b) Legal action of signing off on a project you accepted money for

c) Illegal action of signing off on a project you didn’t design but did

check d) Illegal action of signing off on a project you neither designed nor

checked




23. Mr. Smith buys a new car for $30,000. He pays the down payment

of $8,000, and then borrows the rest from a bank at 6% interest

for four years. Solve for the required monthly payment for the

loan.

a) $460

b) $490

c) $520

d) $550




24. A tower crane was just purchased for $80,000. The life span of the

crane is estimated as 10 years, and the salvage value is $6,500.

Find the depreciation value of the crane.

a) $6,500

b) $7,350

c) $8,000

d) $8,650




25. Mrs. Smith deposits $2,000 in a bank that gives 5% interest

compounded annually. The money she will receive in her account after 15 years is most nearly:

a) $2,100 b) $3,950

c) $4,200

d) $31,500




26. A table manufacturing factory has a total operating cost of

$400,000 per year, including salaries, rent fee, and depreciation costs. Each table needs $60 to produce, and the sale price of the

table is $99. To reach break-even sales, the number of tables

produced per year is most nearly:

a) 4050

b) 5050 c) 6700

d) 10300




27. What is the annual effective interest rate of money that is invested

at 2.5% per year and compounded quarterly?

a) 2.50% b) 2.51%

c) 2.53%

d) 2.55%




28. An engineer plans to receive an annual bonus from money he

invests to a bank. How much should he invest monthly for a year

at a 10% nominal interest rate, compounded monthly, so that he

will receive a $100,000 bonus in a year?

a) $7,500

b) $8,000

c) $8,333

d) $9,170




29. Which of the following loads could be resisted by a fixed support?

I. Moment II. Shear III. Axial

a) I only

b) I and II

c) II and III

d) I, II, and III




30. Three ropes are connected as shown in the figure below. They are

holding a box that weighs 2500 lb. What are the tension forces in

rope 1 and 2 if the system is in equilibrium?

a) 1 2

3125 lb ; 1875 lbT T

b) 1 2

2500 lb ; 2500 lbT T

c) 1 2

2725 lb ; 2225 lbT T

d) 1 2

2925 lb ; 1725 lbT T

8

6

1T

2T

2500 lb




Problems 31 and 32 refer to the following truss.

31. Solve for the support reaction at joint A and B.

a) 600 lb ; 600 lbA B

R R

b) 700 lb ; 500 lbA B

R R

c) 800 lb ; 400 lbA B

R R

d) 900 lb ; 300 lbA B

R R

32. Find the axial forces in members CD and EG.

a) 600 lb tension ; 600 lb compressionCD EG

T T

b) 1200 lb compression ; 900 lb tensionCD EG

T T

c) 0 ; 950 lb compressionCD EG

T T

d) 0 ; 900 lb compressionCD EG

T T

H

G

F

E

D

C

B

8ft

3

8ft

3

AR

8 ft

1200 lb

8 ft 8 ft 8 ft

BR

A




33. Find the horizontal reaction at support C.

a) 0.4 ton

b) 0.6 ton

c) 1.2 ton

d) 1.6 ton

3 ft 6 ft

2 ft

4 ft

B

C

A

2 ton

3 ft




Problems 34 and 35 refer to the following shape.

34. Solve for the coordinate of the centroid.

a) (4 cm, 1.5 cm)

b) (6 cm, 1.342 cm)

c) (4 cm, 1.342 cm)

d) (4 cm, 1.286 cm)

35. Find the moment of inertia about the centroid x-axis of the shape.

a) 3.66 cm4

b) 9.66 cm4

c) 12.5 cm4

d) 16.16 cm4

1

2

3

2 6 8 0 X

Y




36. Solve for the tension force in cable BC.

a) 75 lb

b) 100 lb

c) 150 lb

d) 300 lb

D

C

B

A

150 lb

150 lb




37. A book shelf weighs 200 N in total. It is supported by hinges A and

B on the wall. Solve for the horizontal reaction at hinge B.

a) 120 N

b) 150 N

c) 180 N

d) 240 N

2.5 m

B

A

5 m

3 m

400 N




38. Calculate the centroidal polar moment of inertia with respect to

the z-axis of W14X176 with the following data: 2 4

3

4 3

51.8 in 0.830 in 838 in

15.2 in 1.31 in 320 in

15.7 in 2140 in 163 in

w y

f x

f x y

A t I

d t Z

b I Z

a) 1302 in4

b) 1489 in4

c) 2140 in4

d) 2978 in4




39. A particle moves by the velocity function : 5 3v t t (m/s).

Calculate the distance traveled by the particle from 2 to 6

seconds.

a) 52 m

b) 72 m

c) 82 m

d) 92 m




40. A ball is thrown upward with velocity 8 m/s. If the mass of the ball

is 0.5 kg, solve for the kinetic energy of the ball when it reaches 2

m above the initial point.

a) 4 Joule

b) 6 Joule

c) 8 Joule

d) 10 Joule




41. A 12 kg box starts from rest at the bottom of an inclined plane.

The box is then pulled upward following the inclined plane of 10

to the horizontal plane. Solve for the force required to push the

box until it accelerates 0.5 m/s2 at the inclined plane. Assume that

the coefficient of friction between the inclined plane and box is

0.15.

a) 35 N

b) 40 N

c) 45 N

d) 50 N

10°

F




42. In a friendly baseball match, a 0.15 kg baseball is thrown at the

speed of 10 m/s. The batter hits the ball with 60 N of force. If the

ball touches the bat in 0.2 seconds, what is the speed of ball right

after leaving the bat?

a) 60 m/s

b) 70 m/s

c) 80 m/s

d) 90 m/s




43. A projectile is fired horizontally to a wooden block hang on a

pendulum. The velocity of the projectile is 450 m/s. The projectile

enters the wooden block and they swings together until reaching

the maximum height (see the figure). If the projectile mass is

0.01 kg and the wooden block mass is 1.49 kg, find the maximum

height they can reach. (Use g= 10 m/s2)

a) 35 cm

b) 40 cm

c) 45 cm

d) 50 cm

maxh

'v

450 m/s




44. A 2 kg mass block is released from rest condition on a frictionless

inclined plane, as seen in the figure. The mass block moves down

along the inclined plane, then compresses a spring with a stiffness

of 500 N/m. Calculate the maximum compression of the spring.

a) 50 cm

b) 60 cm

c) 70 cm

d) 80 cm

5 m




45. The following figure shows a typical stress-strain relationship of

steel rebar. Which of the following is the strain hardening part?

a) A-B

b) B-C

c) C-D

d) D-E

D

E C B

A Strain

Str

ess




Problems 46 and 47 refer to the simply-supported beam as follows:

46. Which of the following diagrams is the correct shear force diagram

(unit: kips)?

2.5

2.5 2.5

3 ft 3 ft

5 kips

0

2.5 2.5

2.5 2.5

0

+

-

0

2.5 2.5

0

+

-

-

+ 0

2.5

2.5

0

-

+ 0

2.5

0

a)

b)

c)

d)




47. Which of the following diagrams is the correct bending moment

diagram (unit: kips-ft)?

15

7.5

15

0

7.5

0

0 0

+

+

- 0 0

-

0 0

a)

b)

c)

d)




48. Solve for the maximum shear stress of this two-dimensional

element.

a) 60 MPa

b) 64 MPa

c) 67 MPa

d) 70 MPa

60 MPa

60 MPa

40 MPa

40 MPa

100 MPa 100 MPa




49. A steel rod is stressed by a tension force of 80 lb. It is found that

the rod has a length of 25 ft and a diameter of ¼“. If the modulus

of elasticity of the steel rod is assumed as 10,900 ksi, calculate

the strain of the steel rod due to the applied force.

a) 3.745 in

b) 0.011 in

c) 0.045 in

d) 0.252 in




50. A 2 m long hollow cylindrical rod has the maximum allowable

shear stress 40 MPa and a maximum allowable twist angle of

0.025 radians. The outer and inner diameters of the hollow rod are

75 mm and 50 mm, respectively. Solve for the maximum

allowable torque T to this hollow rod if the shear modulus is

70,000 MPa and the torsion constant is 250 cm4.

a) 2 kN-m

b) 3 kN-m

c) 4 kN-m

d) 5 kN-m




51. An alloy steel pipe has a length of 1 meter when the temperature

is 25°C. If the thermal expansion coefficient is 8x10-6/°C, calculate

the length of steel pipe when the temperature reaches 60°C.

a) 1.3 m

b) 1.03 m

c) 1.003 m

d) 1.0003 m




52. For a simply supported beam, where does the maximum deflection

usually occur due to gravity loading?

a) At the top fibers

b) At the bottom fibers c) At the support

d) At the mid-span




53. Calculate the load P if the concrete beam deflects 8.5 mm.

Assume the modulus of elasticity of concrete is 23,500 MPa and the self-weight of concrete is ignored.

a) 1 ton b) 1.5 ton

c) 2 ton

d) 2.5 ton

30 cm

P = ?

fixed

1.5 m




54. A simply-supported steel beam is loaded by a distributed load as

seen below. The elastic modulus of steel is 10,900 ksi and the moment of inertia about X-axis is 1500 in4. The deflection at point

C (2 ft from the left support) is most nearly:

a) 0.1 in

b) 0.2 in c) 0.3 in

d) 0.4 in

2 ft 8 ft

C

w = 3 lb/ft




55. An H-section steel column is fixed at the base and pinned at the

top. The elastic modulus of steel is 2.1 x 105 MPa. Find the maximum concentric vertical load that the column can support

without buckling. Assume that the steel column has no

imperfection.

a) 2,500 kN

b) 5,000 kN

c) 7,500 kN d) 8,500 kN

8 m Ix = 11500 cm4

Iy = 3880 cm4




56. ASTM (American Standard Testing and Material) provides a lot of

standard testing method for materials. Which of the following testing methods is included for aggregates?

a) Sieve analysis b) Relative density

c) Absorption

d) All of the above




57. A new material is tested in the elongation test. The result shows

that the material has strain of 5.3% (elastic limit) when the stress

of 320 MPa is applied. Find the elastic modulus of the material

based on the result.

a) 17 MPa

b) 170 MPa c) 1700 MPa

d) 6000 MPa




58. Which of the following materials is tested for stability through

Marshall test?

a) Concrete

b) Steel

c) Wood d) Asphalt




59. Which of the following checmical admixtures is used to increase

the workability of concrete?

a) Accelerator

b) Retarder c) Plasticizer / water reducer

d) Air entraining agent




60. Which of the following does affect the kinematic viscocity of a

fluid?

I. Density II. Surface tension

III. Stress IV. Absolute viscocity

a) I and II

b) II and III

c) III and IV d) I and IV




61. A water tank with a height of 5 meters is full of water (density =

1000 kg/m3). Determine the gage pressure at a depth of 50 cm

from the bottom of water tank. Use g = 10 m/s2.

a) 5 kPa

b) 10 kPa

c) 45 kPa

d) 50 kPa




62. A continuous pipe has two different diameter sizes at the inlet and

outlet of pipes. The diameter at the inlet is 1.5 times bigger than

the diameter at the outlet. If the flow velocity at the inlet is 0.5

m/s and the pipe outlet goes freely to the atmosphere, determine

the gage pressure at the inlet.

Note: neglect frictional effects in the pipe

a) 0.2 kPa

b) 0.3 kPa

c) 0.4 kPa

d) 0.5 kPa




63. Water flows in the tube with velocity of 2 m/s. The kinematic

viscocity is found to be 7.3 x 10-7 m2/s. If the inside diameter of

tube is 50 mm, calculate the Reynolds number and determine the

flow type.

a) 1.73 x 103 , laminar

b) 3.71 x 103 , critical

c) 1.37 x 105 , critical

d) 1.37 x 105 , turbulent




Problems 64 and 65 refer to the following case:

3 m3 of water are pumped each minute through a 150 mm diameter

pipe. The pipe length is 50 m, and has a Darcy friction factor of 0.02.

Because of the pump, the water can reach upward until a height of 18

m. The pump efficiency is 80%.

64. Calculate the friction loss for the entire length of pipe.

a) 2.7 m

b) 10 m

c) 27 m

d) 162 m

65. How much power is needed to deliver to the pump?

a) 11 kW

b) 13 kW

c) 14 kW

d) 16 kW




66. Which of the flows has cross section that does not vary with time

at any location along an open channel?

a) Uniform flow

b) Non-uniform flow

c) Steady flow

d) Critical flow




67. Water flows from a reservoir through a 0.5 m diameter pipe. The

elevations of inlet and outlet of the pipe are 221 m and 168 m,

respectively. The flow is found to be steady and incompressible

flow. The pipe outlet discharges to atmospheric pressure. If the

flow rate out of the pipe outlet is 5 m3/s, calculate the total head

loss in the system.

a) 20 m

b) 33 m

c) 46 m

d) 53 m




Problems 68 and 69 refer to the following information:

A pumping system transfers water from a lake to a water tank with a

150 m length and 35 cm diameter pipeline. The flow rate is found to

be 1.34 m3/s and the kinematic viscocity of water is 10-6 m2/s. The

head of pipe inlet is 150 m, while the head of pipe outlet is 175 m. The

pipeline is made of cast iron (specific roughness is 0.25 mm). Assume

that minor losses, entrance losses, and exit losses in the pumping

system can be ignored. Consider the pump efficiency to be 85%. Also

assume the flow is stready and incompressible.

68. Calculate the head losses in the piping by using Darcy equation.

a) 55 m

b) 75 m

c) 95 m

d) 115 m

69. Find the required power to pump the water from the lake to water

tank.

a) 1300 kW

b) 1400 kW

c) 1500 kW

d) 1600 kW




Problems 70 and 71 refer to the following case:

A large complex of apartments has a population of 8000 people. The

average sewage flow for this apartment complex is 4000 m3/day.

70. What is the most nearly minimum sewage flow for this apartment

complex?

a) 1200 m3/day

b) 1500 m3/day

c) 2000 m3/day

d) 2200 m3/day

71. What is the most nearly peak sewage flow for this apartment

complex?

a) 8400 m3/day

b) 10500 m3/day

c) 12200 m3/day

d) 15250 m3/day




72. A sanitary sewer has a length of 60 m and a pipe with diameter of

90 cm. The inlet elevation of the sewer is 1.5 higher than the

outlet elevation. Assume that the Manning’s roughness coefficient

is 0.012 and constant with depth of flow. Determine the

approximate sewer capacity during heavy rainfall if the sewer is

full of water flow with no surcharge.

a) 1 m3/s

b) 2 m3/s

c) 3 m3/s

d) 4 m3/s




73. What is the static determinacy of the following truss?

a) Stable, statically determinate

b) Stable, statically indeterminate c) Unstable

d) None of the above




Problems 74 and 75 refer to the following beam.

74. What is the largest magnitude of the shear force in the beam?

a) 15 kN

b) 20 kN c) 35 kN

d) 40 kN

75. What is the largest magnitude of the bending moment in the

beam?

a) 20 kN-m

b) 25 kN-m

c) 50 kN-m d) 100 kN-m

5 m 5 m 5 m

P = 20 kN

D A B C

w = 4 kN/m




76. Based on ACI 318, calculate the top flange effective width of the

following T-beam.

a) 20 cm b) 40 cm

c) 150 cm

d) 200 cm

35

cm 10 cm

20 cm

6 m




Problems 77 and 78 refer to the following beam case:

A simply supported beam is subjected to an ultimate bending moment

of 110 kips-ft and an ultimate shear force of 40 kips.

The beam section is 10”x20”. The compressive strength of concrete is

3000 psi, the yield strength of longitudinal reinforcement and stirrups

are 60 ksi.

77. Calculate the required number of #6 rebar to resist the bending

moment (design for tension only).

a) 2

b) 3

c) 4 d) 5

78. Solve for the required spacing of #3 stirrups to resist the shear

force.

a) 4”

b) 6” c) 8”

d) 10”




79. Which of the following concepts about ASD and LRFD are correct?

I. ASD is newer than LRFD II. ACI 318 is based on the LRFD concept

III. LRFD uses factored load combinations while ASD doesn’t

IV. ASD uses ultimate strength while LRFD doesn’t V. ASD uses a factor of safety while LRFD doesn’t

a) I, II, IV b) II, III, V

c) II, IV, V

d) II, III, IV




80. A W16x36 of A992 steel (Fy = 50 ksi) is used as a beam to

support a concrete floor slab that provides continuous lateral

support to the compression flange. The applied loading is shown in

the figure.

What is the maximum ultimate distributed loading which can be

resisted by the beam? Assume that the beam is laterally

supported.

Shape b (in) h (in) tf (in) tw (in) Sx (in3) Zx (in

3)

W16x36 6.99 15.9 0.43 0.295 56.5 64.0

a) 2 kips/ft

b) 2.5 kips/ft

c) 3 kips/ft

d) 3.5 kips/ft

Wu

24 ft




81. Find the void ratio of the soil sample, based on the following data:

Mass = 17.4 gram Volume = 10 cm3

Oven-dry mass = 13.4 gram

Specific gravity = 2.15

a) 0.2

b) 0.4 c) 0.5

d) 0.6




82. A sample of saturated soil from the field is taken to a test

laboratory. The specific gravity is found to be 2.6 with a total unit mass of 2250 kg/m3. Find the dry unit mass of the soil sample.

a) 1950 kg b) 2030 kg

c) 2120 kg

d) 2150 kg




83. Six meters of excavation was done through two different soil

layers as shown below. Calculate the total active lateral pressure against the retaining wall at the bottom of retaining wall.

a) 30 kN/m2 b) 35 kN/m2

c) 50 kN/m2

d) 65 kN/m2

3 m

3 m

Layer 1

Layer 2

3

1

'

1

1

19 kN/m

33

0c

3

1

'

1

1

20 kN/m

25

0c




84. A 5’x5’ square footing rests at 4-ft depth below the ground. The

following data about the soil will be used for the calculation:

125 pcf

0

30

c

37.2

19.7

22.5

c

q

N

N

N

1.25

0.85

cS

S

Use Terzaghi equation to calculate the allowable bearing capacity

of the square footing using a safety factor of 3.

Note: The water table is below the footing and neglect the weight

of soil above the footing.

a) 2500 psf

b) 4500 psf

c) 5500 psf d) 16000 psf




85. Which of the following factors doesn’t affect the rate of

consolidation?

a) Permeability

b) Layer thickness c) Compressibility

d) None of the above




Problems 86 and 87 are based on the following vertical curve.

86. Determine the low point station for this vertical curve.

a) Sta 70+00

b) Sta 75+68 c) Sta 76+32

d) Sta 82+00

87. Determine the elevation of the low point.

a) 500 m

b) 502 m c) 504 m

d) 506 m

-2%

PVI

Sta 76+00

Elev = 500 m

EVC PVC

+1.8%

L = 12 sta




88. Based on several studies, what is the perception-reaction time

AASHTO used to design Stopping Sight Distance?

a) 1.5 sec

b) 2 sec c) 2.5 sec

d) 3 sec




89. Traffic flow can be written as q kv where q is the traffic

volume, k is the traffic density, v is the mean speed of vehicles. Based on several studies at a road, the mean speed is given by

the relationship 80 0.5v k .

What is, most nearly, the maximum capacity of total traffic

volume of the road?

a) 3000 veh/hour b) 3200 veh/hour

c) 3400 veh/hour

d) 3600 veh/hour




90. A horizontal curve is designed to have a diameter of 1200 m. The

tangent is 125 meters and the PI is located at sta 10+000 (in

kilometer). Find the stationing of PT.

a) Sta 9+744

b) Sta 9+875

c) Sta 10+121

d) Sta 10+600




Problems 91 and 92 refer to the following case.

A sample of wastewater is placed in the incubator by keeping the

temperature at 23°C. After 5 days, the BOD (Biochemical Oxygen

Demand) is found to be 234 mg/L. Assume that the reaction rate

constant is 0.13 d-1 (base e).

91. Find the ultimate BOD of the sample.

a) 290 mg/L

b) 468 mg/L

c) 490 mg/L

d) 1,170 mg/L

92. Find the BOD of the sample if the incubation period is one week.

a) 290 mg/L

b) 468 mg/L

c) 490 mg/L

d) 1,170 mg/L




93. A tank reactor treats 0.35 m3/s of settled wastewater having 258

mg/L BOD5 at 22°C. The design mean cell resistance time ( d

c ) is

12 days.

,

5,

0.55ss mg

mg

VY

BOD

The effluent BOD5 is 6.4 mg/L. MLVSS = 3600 mg/L. The

endogenous decay coefficient 10.06d

K d . Determine the reactor

capacity per day.

a) 3600

b) 6400

c) 8100

d) 9200




94. A sample of freshwater is taken from a small river. After several

observations, the sample is found to contain a dissolved oxygen

concentration of 6.2 mg/L when the temperature is 24.6°C and

the atmospheric pressure is 740 mmHg. A partial listing of the

solubility of dissolved oxygen in freshwater at equilibrium with dry

air containing 21.7% oxygen and at an atmospheric pressure of

760 mmHg is as follows:

Temperature (°C) Oxygen solubility (mg/L)

22 9.6

23 9.3

24 8.9

25 8.6

26 8.2

Calculate the saturation of dissolved oxygen in the water sample.

a) 50%

b) 60%

c) 70%

d) 80%




95. A municipal has a population of 18,400 people. The monthly

volume of solid waste is 20,000 m3. If the density of solid waste is assumed to be 125 kg/m3, find the average mass of solid waste

generated by a person per day.

a) 1.1 kg

b) 1.8 kg

c) 3.2 kg

d) 4.5 kg




96. Find the mass loading of a 35 MGD wastewater discharge with an

ultimate BOD of 28 mg/L.

a) 8,000 lbm/day

b) 8,200 lbm/day

c) 8,400 lbm/day

d) 8,600 lbm/day




97. A municipal produces wastewater with a BOD5 of 225 mg/L and an

ultimate BOD of 486 mg/L. Find the reaction rate constant.

a) 10.025 d

b) 10.1 d

c) 10.125 d

d) 10.175 d




98. Which of the following is the correct definition of a critical path in

a project?

a) The sequences of tasks in a project

b) The time unaccounted for in a project

c) The longest path of sequential tasks through a project

d) The extra time available in a project




99. Which project scheduling tool is illustrated in the figure below?

a) Bar (Gantt) chart

b) AON diagram

c) AOA diagram

d) PERT chart

1 3

2

2 4

2

5

2

6

2 7

2




Problems 100-102 refer to the following diagram.

100. Find the critical path of this diagram.

a) START – A – C – F – FINISH

b) START – A – D – F – FINISH

c) START – B – D – F – FINISH

d) START – B – E – F – FINISH

101. Find the EST (Earliest Start Time) of activity F.

a) Day 8

b) Day 9

c) Day 10

d) Day 11

102. Find the float time of activity D. a) 0 day

b) 1 day

c) 2 days d) 3 days

Start

A|3

Finish

B|5

C|2

D|3

E|4

F|3




103. Which of the following statements is correct about construction

contracts?

a) In a lump sum contract, a contractor bids a price for each work

item by the cost per unit b) A unit price contract is often used when the exact quantities in a

project are known

c) In a unit price contract, the contractor bids a total price for all work in a project, including the profit

d) A lump sum contract cannot be changed unless there is a

contract modification, while the unit price contract may be

changed depends on the actual quantities




104. Which bond guarantees that the contractor will perform the

specified work in accordance with the contract (typically full value of project)?

a) Bid bond b) Performance bond

c) Payment bond

d) None of all the above




105. According to OSHA, any adequate exits in trench excavation must

be provided when the depth reaches:

a) 0 ft (no such requirement)

b) 2 ft c) 4 ft

d) 5 ft




106. A wall of a single story building has a length of 72 ft and a height

of 16 ft. A contractor shall fill the wall by brick. From previous experience the contractor found that 602 bricks are needed for

every 100 ft2. If the total opening (window and door) area is 250

ft2 and there is 5% waste of material, calculate the required number of bricks for the wall.

a) 3810 b) 3900

c) 4000

d) 5702




Problems 107 and 108 refer to the following information.

The earthwork of a new roadway requires less soil to be thrown out so

that engineers can now calculate the cut-and-fill volume based on the

following information in the table. Assume both cut and fill areas are

triangular for the transition region from fill to cut area.

Station (m) Cut area (m2) Fill area (m2)

100+00 - 153.42

10+4.50 - 32.56

10+12.35 13.67 8.25

10+16.40 52.84 -

10+20 165.14 -

107. Find the total approximate volume of fill work.

a) 500 m3

b) 505 m3 c) 510 m3

d) 515 m3

108. Find the total approximate volume of cut work.

a) 540 m3 b) 560 m3

c) 580 m3

d) 600 m3




109. A borrow pit is figured in the grid below. This area is going to be

leveled to 100 m for all grid corners. Find the total volume of

earth excavated.

Point Elevation

(m)

A1 103.5

A2 105.6

A3 102.1

A4 101.9

B1 103.9

B2 104.0

B3 102.8

B4 103.7

C1 105.2

C2 104.3

C3 100.3

a) 480 m3

b) 1700 m3 c) 2500 m3

d) 6800 m3

4 3 2 1

B

C

A

12 m 12 m 12 m

8 m

8 m




110. What is the correct term for the increasing volume of soil when

earth is excavated?

a) Loose

b) Void c) Swell

d) Shrinkage




PROBLEM SOLUTIONS




Vector Length

Problem 1 Solution:

Length of a straight line AB can be calculated as vector length:

2 2 2

1 2 3 1 3 7 29AB

(Answer B)




Roots of Equation

Problem 2 Solution:

For third-degree polynomial, the number of roots will be three.

3

2

7 6 0

1 6 0

1 2 3 0

x x

x x x

x x x

To satisfy the equation, the roots are 1, 2, -3.

(Answer D)




Circle Center

Problem 3 Solution:

The general equation of a circle is:

2 2 2x a y b r

Where ,a b is the coordinate of its center and r is the radius of

the circle.

Solve the coordinate by modifying the equation into the general

equation:

2 2

2 2

2 2 2

10 14 49 0

10 25 14 49 25

5 7 5

x y x y

x x y y

x y

The coordinate of its center is (5, -7).

(Answer D)




Calculus

Problem 4 Solution:

The slopes of two lines, which are perpendicular, can be written as

follows:

1 21m m

Where 1

m and 2

m are the slopes of the first and second line,

respectively.

First line: Second line:

1

5 3 16 0

5 3 16

3 16 3

5 5 5

y x

y x

y x m

1 2

2

2

1

31

5

5

3

m m

m

m

Solve for the equation of the second line:

5

3m known point , 1,7x y

Use the general equation of a straight line:

57 1

3

16

3

y mx b

b

b

The equation is 5 16

3 3y x or 3 5 16 0y x .

(Answer A)




Circle Equation

Problem 5 Solution:

The general equation of a circle is:

2 2 2x a y b r

Where ,a b is the coordinate of its center and r is the radius of

circle.

Substitute the known point (x,y) into the circle equation:

2 2 2

22 28 2 3 5

100 10

x a y b r

r

r

The equation of the circle is:

2 2 2

22 2

2 2

2 2

2 5 10

4 4 10 25 100

4 10 71 0

x a y b r

x y

x x y y

x y x y

(Answer C)




Calculus

Problem 6 Solution:

Solve the two coordinate equations:

9sin sin9

2cos cos2

xx t t

yy t t

Use the trigonometry identity:

2 2

2 2

2 2

sin cos 1

19 2

4 81 324

t t

x y

x y

(Answer B)




Vector Analysis

Problem 7 Solution:

For the cross product remember to cross out the row and column for

each variable, i, j, k and multiply that respective variable by what’s left over in the matrix. Also, remember that the signs start with a positive

at i, a negative at j, and then a positive at k, and so forth.

8 1 2

3 1 3

1 2 8 2 8 1

1 3 3 3 3 1

30 11

i j k

A B

i j k

i j k

(Answer A)




Vector Analysis

Problem 8 Solution:

Use dot product to get the angle:

2 2 22 2 2

cos

1 1 1 1 1 1 1 1 1 1 1 1 cos

1 3 3 cos

1cos 70.5 70

3

M N M N

(Answer D)




Inflection Point

Problem 9 Solution:

Inflection point occurs when the second derivation of f x is 0.

4 3 2

3 2

2

12 30

' 4 36 60

" 12 72 60

f x x x x

f x x x x

f x x x

Solve for inflection point:

2

2

1 2

" 12 72 60 0

12 6 5 0

12 1 5 0

1 and 5

f x x x

x x

x x

x x

4 3 2

1

4 3 2

2

1 1 1 12 1 30 1 19

5 5 5 12 5 30 5 125

x f

x f

The inflection points are (1, 19) and (5, -125).

(Answer B)




Trigonometry

Problem 10 Solution:

This talks about trigonometry identities.

Option A → 2sin cos sin2

Option B → 2 2sin cos 1

Option C → 21 cos2sin

2

Option D → 2 2cos sin cos2

(Answer B)




Probability


The possible successful outcomes are that either a red ball is

picked then a black ball, or a black ball is picked then a red ball.

Number of red balls = 5

Number of blue balls = 7

Number of black balls = 3

Total number of balls = 5 + 7 + 3 = 15

red then black or 5 3 3 50.143

15 14 15 14black then redP

(Answer B)




Probability


The sample space of rolling dice is {1, 2, 3, 4, 5, 6}.

The event of rolling dice is independent each other, so that the

second rolling dice is not affected by the result from the first

rolling dice.

The event “number 2” of the rolling dice is {2}.

1

6

n EP E

n S

(Answer C)




Probability


Use binomial distribution to solve this problem:

2 15 2

0.03

1 0.97

!

! !

15!2 0.03 0.97 0.064

2! 15 2 !

x n x

p

q p

np x p q

x n x

p

(Answer D)




Mean and Standard Deviation


The mean is determined by dividing the sum of data by the total

number of data.

i ix

2

ix x

1 2995 29.16

2 3005 21.16

3 3001 0.36

4 2991 88.36

5 2984 268.96

6 3004 12.96

7 3009 73.96

8 3015 213.16

9 2998 5.76

10 3002 2.56

Total 30004 716.4

1 300043000.4

10

n

ii

x

xn

Variance is defined as:

2

1 716.479.6

1 10 1

n

ii

x x

n

(Answer A)




Probability


The fastest way to calculate the probability of “at least one” event

is by substract the probability of having baby with no brown eyes from the probability of all events.

zero 0.7 0.7 0.7 0.7 0.7 0.7 0.118

at least one 1 zero 1 0.118 0.882 0.9

P

P P

(Answer D)




Programming - Looping Function


“if ... else ... end” and “if ... elseif ... end” are the functions to

create conditional statements.

“for ... end” is the function to create loopings for a certain specified number.

“while ... do” is the function to create loopings as long as a certain

condition is true.

(Answer B)




Programming – Looping Result


Looping condition can be solved by observing the variable data for

each loop.

Condition x y

Initialize 0 1

First looping x = 0 2 2

Second looping x = 2 4 3

Third looping x = 4 6 4

Fourth looping x = 6 looping stops

The last value of variable y is 4.

(Answer B)




Ethics


By doing ethical behavior, you will not get any benefits because

ethical behavior promises nothing.

(Answer D)




Professional Practice


You should not accept any king of gifts from parties expecting

special consideration, so you should reject the Rolex watch. As long as the contractor satisfies all the bid requirement and follows the

bidding rules, the bid could be accepted.

(Answer C)




Ethical Behavior


Options (A), (B), (C), (D) may all be valid. However, the rationale

for specific ethical prohibitions on using your employer’s equipment for a second job is economic. When you don’t have to

pay the equipment, you don’t have to recover its purchase price in

your fees for services.

(Answer D)



Professional Societies


A violation should be reported to the organization that has

promulgated the rule.

(Answer D)



Professional Liability


It is legal to stamp plans that you personaly designed and/or

checked. It is illegal to stamp plans that you didn’t personally design or check, regardless of whether you got paid. It is legal to

work as a ‘plan checker’ consultant.

(Answer D)



Loan Payment Analysis


Use capital recovery discount factor:

48

48

| , %,

$30,000 $8,000 $22,000

6%0.5%

12 compounding periods per year

4 years 48 months

| ,0.5%,48

0.005 1 0.005$22,000

1 0.005 1

$516.67 $520

A P A P i n

P

i

n

A P A P

(Answer C)



Depreciation Value


Depreciation value is equal to the difference between initial cost C

and salvage value n

S , then divided by the life span n .

$80,000 $6,500$7,350

10n

C SD

n

(Answer B)



Future Value Analysis


15

| ,5%,15 $2,000 1 0.05 $4,157F P F P

(Answer C)



Break Even Point Analysis


Break even point occurs when the total of costs (negative) and

sales (positive) is zero. Let n be the number of tables per year at

break even point.

$400,000 $60 $99 0

$39 $400,000

10257 10300 tables/year

n n

n

n

(Answer D)



Effective Interest Rate Analysis


40.025

1 1 1 1 2.5235%4

mr

im

(Answer C)



Future Value Analysis


Use the uniform series sinking fund discount factor. The interest period is one month, there are 12 compounding periods, and the

effective interest rate per interest perios is 10%/12 = 0.83%.

12

0.0083| ,0.83%,12 0.079597

1 0.0083 1

| ,0.83%,12 $100,000 0.079597 $7,960 $8,000

A F

A F A F

(Answer B)



Basic Concept of Structure’s Support


Fixed support can resist vertical (shear) and horizontal (axial)

forces as well as moment. It can restrain rotation and translation.

(Answer D)



Static Equilibrium


The equilibrium system means all the forces acting to the system

are balanced. Therefore, we will use two equilibrium equations

(vertical and horizontal forces) to solve this problem.

Equilibrium in vertical forces (Y direction):

1

1

0

sin 2500 0

3125 lb

yF

T

T

Equilibrium in horizontal forces (X direction):

1 2

2

0

cos 0

1875 lb

xF

T T

T

(Answer A)



Truss Analysis


To solve the support reaction, equilibrium should be satisfied.

0

1200 lb 8 ft 32 ft 0

300 lb

A

B

B

M

R

R

0

1200 lb 0

300 lb 1200 lb 0

900 lb

y

A B

A

A

F

R R

R

R

(Answer D)



Truss Member Forces


Member CD is zero-force member. Check the equilibrium of

vertical forces at joint D:

0

0

y

CD

F

T

Member EG can be solved by ‘cutting’ the truss as shown below.

Check the equilibrium of system to get the support reaction at B:

0

32 ft 1200 lb 8 ft 0

300 lb

A

B

B

M

R

R

Check the equilibrium of moment at joint F by looking at ‘right

side’ part:

2 2

2 2

0

88 8 316 ft ft 8 ft 0

38 88 8

3 3

948.68 lb 948.68 lb compression

F

B EG EG

EG

M

R T T

T

Answer : C

E

B

8ft

3

8ft

3

AR

8 ft

BR

8 ft 8 ft 8 ft

A

1200 lb

H

G

F D

C



Frame Analysis


This is a statically determinate structure. We have four unknowns

(A

H , A

V , C

H , C

V ), so that we need four equilibrium equations to

solve this problem.

Check equilibrium of horizontal & vertical forces in the system (ton

& ft):

0

(1)

x

A C

F

H H

0

2 (2)

y

A C

F

V V

Check equilibrium of moment at joint A in the system (ton & ft):

0

2 3 2 12 0

6 3 (3)

A

C c

C C

M

H V

H V

AH

3 ft 3 ft 6 ft

B

C

A

2 ton

AV

CV

cH

2 ft

4 ft



Check the moment at pin B for the right part (ton & ft):

0

4 6 0

1.5 (4)

B

C c

C C

M

H V

H V

Substituting equation (4) to equation (3):

6 3 (3)

1.5 6 3

0.4 ton

C C

C C

C

H V

V V

V

Therefore, the horizontal reaction at support C can be calculated

using equation (4):

1.5 (4)

1.5 0.4 0.6 ton

C C

C

H V

H

(Answer B)



Centroids


Shape x (cm) y (cm) A (cm2)

I 4/3 1 3

II 4 2.5 4

III 20/3 1 3

IV 4 0.5 4

Total 14

,

,

4 203 4 4 3 4 4

3 34 cm

14

1 3 2.5 4 1 3 0.5 41.286 cm

14

c n n

c

c n n

c

x Ax

A

y Ay

A

(Answer D)

1

2

3

2 6 8 0 X

II

IV III I

Y



Moment of Inertia


From the previous solution, the centroid of system was found to be (4

cm, 1.286 cm). In order to get the moment of inertia of the system,

we should find the distance of centroid (d ) between each shape and

overall system.

Shape ,c xI (cm4) A (cm2) d (cm) 2Ad (cm4)

I

33 2 3

1.536 36

bh

3 -0.286 0.2454

II

33 4 1

0.3312 12

bh

4 1.214 5.8952

III

33 2 3

1.536 36

bh

3 -0.286 0.2454

IV

33 4 1

0.3312 12

bh

4 -0.786 2.4712

Total 3.66 8.8571

2 4

, , ,3.66 8.8571 12.52 cm

c x c x i i iI I Ad

(Answer C)



Cable Tension Force


Alternative 1:

Use the law of sines to solve the triangle at joint B.

150150 lb

sin120 sin120AB

AB

TT

Alternative 2:

Use equilibrium at joint B.

0

cos30 cos30

x

AB BC

AB BC

F

T T

T T

0

sin30 sin30 150

2 sin30 150

150 lb

y

AB BC

AB

AB

F

T T

T

T

(Answer C)

ABT BC

T

150 lb

120°



Static Analysis


To find the horizontal reaction at support A, check the equilibrium

at support B. Pinned support cannot resist moment, so that

equilibrium of moment at support B is 0.

Assume the horizontal reaction at support A is to the right ().

0

400 1.5 2.5 0

240 N

B

A

A

M

H

H

(Answer D)



Centroidal Polar Moment of Inertia


42140 838 2978 inxx yy

J I I

(Answer D)



Kinematic Function


Velocity is defined in terms of time, so that distance can be

written as follows:

6

2

6

2

2

5 3

5 3

2

92 m

s t v t dt

t dt

t t

(Answer D)



Kinetic Energy


The velocity at 2-m height:

2 2 2

02 8 2 9.8 2 24.8

24.8 4.98 m/s

t

t

V V gh

V

The kinetic energy at 2-m height:

21 10.5 24.8 6.2 Joule

2 2k

E mV

(Answer B)



Friction Force


Define the forces working to the moving box in the inclined plane,

then apply the equation.

coskf N mg

sin cos

12 9.8 sin10 0.15 12 9.8 cos10 12 0.5

43.8 45 N

F ma

F mg mg ma

F

F

(Answer C)

kf

sin10mg

10° F



Impulse & Momentum


2 1

2

2

60 0.2 0.15 10

90 m/s

I P

F t m v v

v

v

(Answer D)



Law of Conservation of Momentum


The law of conservation of momentum states that the total

momentum of the two objects before the collision is equal to the

total momentum of the objects after the collision.

p bp p ' '

' '

p b

p p p p b b

p p

m v m v m v

This case is a perfectly inelastic collision, because both projectile

and wooden block have the same motion after collision. The

velocity after collision (' ' '

p bv v v ) can be obtained as follows:

'

'

'

0.01 450 0.01 1.49

3 m/s

p p p bm v m m v

v

v

Using the law of energy, solve the maximum height as follows:

1

2m 2v m

213 10

2

0.45 m 45 cm

gh

h

h

(Answer C)



Work & Energy


The inclined plane is frictionless, so there is no energy loss in the

system. Use the conservation of energy to solve this.

2

2

1

2

12 9.8 5 500

2

0.62 m

i fE E

mgh k

(Answer B)



Stress-Strain Relationship


The stress-strain relationship has elastic and plastic part. The

elastic part is shown at part A-B, while the rests are plastic

condition. The complete list of this following stress-strain is given

below:

Part A-B : Elastic

Part B-C : Yielding

Part C-D : Strain hardening

Part D-E : Fracture

(Answer C)

C B

A Strain

Str

ess

D

E



Shear Force Diagram


Before drawing ths shear force diagram, first we need to find the

support reactions. For this case, the applied concentrated loading

is in the mid-span, so that the reaction at both supports is half of

the applied loading = 2.5 kips (upward each).

Shear Force Diagram (unit: kips)

(Answer A)

0

2.5 2.5

2.5 2.5

0

+

-



Bending Moment Diagram


For this case, the bending moment increases linearly until the

maximum moment is located when the shear force diagram is

zero.

Bending Moment Diagram (unit: kips-ft)

(Answer A)

0 0

7.5

+



Mohr’s Circle of Element Stresses


100 4070 MPa

2 2

x y

avg

By looking at the Moh’r circle above, the maximum shear stress is

actually the radius of the Moh’r circle.

2 2

max100 70 60 67 MPa

(Answer C)

max

1

2

avg

(-40,-60)

(-100,60)



Strain (Elastic Deformation)


23

in80 lb 25 ft 12

ft0.045 in

110900 10 psi 0.25 in

4

PL

EA

(Answer C)



Torsion


There are two limitations to find the maximum allowable torque:

1. Check the maximum allowable shear stress

0max

6

2540

2500000

4 10 N-mm 4 kN-m

Tr

J

T

T

2. Check the maximum allowable twist

max

20000.025

70000 2500000

2187500 N-mm 2.1875 kN-m

TL

GJ

T

T

The maximum allowable torque is the minimum between those

two values, taken as 2 kN-m.

(Answer A)



Thermal Deformation


0

6

4

0

4

8 10 C 1 m 60 C 25 C

2.8 10 m

1 m 2.8 10 m

1.00028 m

L L T

L L L

(Answer D)



Deflection


For a simply supported beam, the maximum deflection usually

occurs at the mid-span, where the maximum moment occurs.

While at the support, the maximum shear usually occurs.

Compression stress will occur at the top fibers and tension will

occur at the bottom fiber.

(Answer D)



Deflection of Cantilever Beam


The quick formula of deflection in this case is:

3

3

PL

EI

The moment of inertia for this rectangular section is:

3 38 4150 300

3.375 10 mm12 12

bhI

Therefore, the load is:

3

3

8

3

15008.5

3 23500 3.375 10

59925 N 6 ton

PL

EI

P

P

(Answer C)



Deflection of Simply-Supported Beam


The quick formula of deflection in this case is:

3 2 3224

wxL Lx x

EI

Therefore, the load is:

3 2 3

3 2 3

224

3000 2 12 10 12 2 10 12 2 12 2 12

24 10900000 1500

0.3 in

wxL Lx x

EI

(Answer C)



Steel Column Analysis


Fixed-pinned → 0.7k

Effective length for buckling in the weak direction:

' 0.7 8 m 5.6 mL kL

Critical buckling based on Euler’s formula:

2

2

2 5 4

2

'

2.1 10 3880 10

5600

2564 kN

cr

EIP

L

(Answer A)



Standard Testing Methods for Material


ASTM contains:

C136-06 covers the standard test method for sieve analysis

for fine and coarse aggregates.

C127-12 covers the standard test method for density,

relative density (specific gravity), and absorption of coarse

aggregates.

C128-12 covers the standard test method for density,

relative density (specific gravity), and absorption of fine

aggregates.

(Answer D)



Material Properties


Since the material is still in elastic part, we can use the

relationship between stress and strain to get the elastic modulus.

3206038 MPa 6000 MPa

0.053E

(Answer D)



Material Testing Method


Marshall test is a common method used to calculate the load and

flow rate of asphalt specimens.

(Answer D)



Concrete Admixtures


Accelerator is used to shorten the setting time in concrete.

Retarder is used to extend the setting time of cement paste in

concrete.

Plasticizer / water reducer is used to increase the workability of

concrete, allowing the concrete be placed more easily.

Air entraining agent is used to provide space for the water to

expand upon freezing.

(Answer C)



Fluid Properties


Kinematic viscocity ( ) is defined as the ratio of absolute viscocity

( ) to mass density ( ), written as follows:

(Answer D)



Water Pressure


The depth h is calculated from the water surface.

1000 10 5 0.5 45000 Pa 45 kPap gh

(Answer C)



Fluid Continuity


Using continuity equation:

1 1 2 2

1

4

Av A v

2

1 1

1

4D v

2

2 2

2 2

2 2 2

2

1.5 0.5 m/s

1.125 m/s

D v

D D v

v

Using Bernoulli’s equation:

2

1 11

2

p vz

g g

2

2 22

2

p vz

g g

2 2

1 1 2

2 2

2 11

2 2

02 2

2

1.125 0.5 1000

2

507.815 Pa 0.5 kPa

p v v

v vp

(Answer D)



Reynolds Number


Reynolds number determines the fluid type:

Re < 2100 → laminar flow

2100 < Re < 4000 → critical flow

Re > 4000 → turbulent flow

7 2

5

Re

2 m/s 0.05 m

7.3 10 m /s

1.37 10 turbulent

vD vD

(Answer D)



Hydraulic Friction Loss


Flow rate: 3 3m m

3 0.05min s

Q

Flow area: 2 2 21 10.15 0.0177 m

4 4A D

Flow velocity: 3

2

0.05 m /s2.82 m/s

0.0177 m

Qv

A

Friction loss:

22

2

0.02 50 m 2.82 m/s2.7 m

2 2 0.15 m 9.8 m/sf

fLvh

Dg

(Answer A)



Hydraulic Pump Power


Total head by the pump:

18 m 2.7 m 20.7 mz f

h h h

Pump power:

21 min3000 kg/min 9.8 m /s 20.7 m

60 sec

0.8

12678.75 W 13 kW

mghP

(Answer B)



Open Channel Flow


Uniform flow: the flow cross section does not vary with time at

any location along an open channel.

Steady flow: the flow quantity does not vary with time at any

location along an open channel.

(Answer A)



Pipe Head Loss


From the given information, we obtain some known values:

10p (reservoir is at atmospheric pressure)

20p (pipe outlet is at atmospheric pressure)

10v (the water in reservoir doesn’t have velocity)

3

2 2

5 m /s25.5 m/s

0.25 0.5 mv

(flow rate is given)

1220 mz (given in the problem statement)

2180 mz (given in the problem statement)

Using energy equation:

2 2

1 1 2 21 2

2

2 2

25.50 0 221 0 168

2 9.8

20 m

f

f

f

p v p vz z h

g g g g

h

h

(Answer A)



Pipe Head Loss


Specific roughness of cast iron = e =0.23 mm

Relative roughness = 0.23

0.000657350

e

D

Reynolds number is determined by using the following equation.

2

6

6

1.34 0.25 0.35 0.35Re 4.875 10

10

vD

After that, the friction factor can be obtained from the Moody

diagram on the next page: 0.0185f

Pipe head loss (Darcy’s equation):

23

22

2

1.34 m /s0.0185 150 m

0.25 0.35 m78.5 m

2 2 0.35 m 9.8 m/sf

fLvh

Dg

(Answer B)



Pumping Power


From Problem 68 solution, the head loss (f

h ) is 78.5 m.

Total head by the pump:

watertank175 m 150 m 78.5 m 103.5 m

lake fh h h h

Pump power:

21340 kg/sec 9.8 m /s 103.5 m

0.85

1599014 W 1600 kW

mghP

(Answer D)



Minimum Sewage Flow


From the sewage flow ratio curves above, we can find the ratio of

minimum-to-average daily sewage flow for population of 8000

people follows curve A2:

0.2

min

0.2

min

3

min

5

8

4000 5

1213 m /day

avg

Q P

Q

Q

Q

(Answer A)



Peak Sewage Flow


From the sewage flow ratio curves above, we can find that the

ratio of peak-to-average daily sewage flow for a population of

8000 people follows the curve G:

peak

peak

3

peak

18

4

18 8

4000 4 8

12201 m /day

avg

Q P

Q P

Q

Q

(Answer C)



Sewer Capacity


Because the sewer is full of water flow during heavy rainfall, it can

be assumed that the wetted perimeter is equal to the pipe

perimeter. Therefore, the hydraulic radius is:

20.25 0.90.225 m

0.9

AR

P

Sewer slope:

1.50.025

60inlet outlet

z zS

L

Flow velocity using Manning’s equation:

2 1 2 13 2 3 2

1 10.225 0.025 4.87 m/s

0.012v R S

n

Flow capacity:

2 34.87 m/s 0.25 0.9 m 3.1 m /sQ vA

(Answer C)



Static Determinacy (Truss)


Static determinacy categorizes structures into:

m + r = 2j → statically determinate

m + r > 2j → statically indeterminate

m + r < 2j → unstable

where m is number of members, r is number of reactions, j is

number of joints.

For this truss, we have:

m = 18 r = 3 j = 10

Check:

(18 + 3) > (2 x 10)

The structure is stable and statically indeterminate.

(Answer B)

18

17

16 14

15

13

12 11

10

9

8

7

6

5 3

2

4

1



Shear Force


Before we can answer the largest shear force, we must obtain the

support reaction at support A and C.

0

20 kN 5 m 10 m 4 kN/m 5 m 12.5 m 0

35 kN

A

c

c

M

V

V

0

20 kN 4 kN/m 5 m 0

40 kN

35 kN 40 kN

5 kN

y

A c

A c

A

A

F

V V

V V

V

V

Shear Force Diagram (unit: kN)

As seen in the shear force diagram, the largest magnitude of the

shear force is 20 kN.

(Answer B)

5

0 0

20

15 15

5 +

-

+



Bending Moment


Bending Moment Diagram (unit: kN-m)

As seen in the bending moment diagram, the largest magnitude of

the bending moment is 50 kN-m.

(Answer C)

0 0

50

25

+

-



Effective Width of T-Beam


The effective width of T-beam is the minimum of:

center-to-center

1 1600 150 cm

4 4

16 20 16 12 212 cm

600 cmw s

L

b t

L

(Answer C)



Bending Design


Equilibrium between concrete compression block & tension rebars:

'0.85

0.85 3 ksi 10 in 60 ksi

2.35

c

c s y

s

s

C T

f ab A f

a A

a A

Find the required steel rebars:

2

2

0.9 0.5

110 kips-ft 12 in/ft 0.9 60 ksi 20 in 2.5 in 0.5 2.35

1467 kips-in 1050 70.5

1.56 in

u n s y

s s

s s

s

M M A f d a

A A

A A

A

Check minimum and maximum rebar (use ACI equations):

2

min ,min min

2

max ,max max

0.0033 0.5775 in

0.0135 2.3625 in

s

s

A bd

A bd

Required steel rebars area is in between minimum and maximum

rebar, so 1.56 in2 is used for steel rebars area.

Required number of #6 rebar =

2

2

1.56 in3.53 4

60.25 in

8

(Answer C)



Shear Design


Shear strength of RC beam is obtained from concrete and stirrups.

Contribution from concrete:

'2

2 1 3000 psi 10 in 17.5 in

19170 lb 19.17 kips

c c wV f b d

Contribution from stirrups:

40 kips19.17 kips 34.2 kips

0.75u

s c

VV V

Required spacing of stirrups (use #3 rebar):

2

32 0.25 in 60 ksi 17.5 in

86.78 6 in

34.2 kips

v y

s

A f ds

V

Check with maximum spacing:

2

max

17.5 in8.25 in

2 2

32 0.25 in 60000 psi

826.5 in

50 psi 10 in

v y

w

d

s

A f

b

Stirrups use #3 @ 6”

(Answer B)



Basic Concept of Design


(X) I. ASD is newer than LRFD

(V) II. ACI 318 is based on the LRFD concept

(V) III. LRFD uses factored load combinations while ASD doesn’t (X) IV. ASD uses ultimate strength while LRFD doesn’t

(V) V. ASD uses factor of safety while LRFD doesn’t

(Answer B)



Steel Beam Design


Check for compactness:

0.38 8.13 9.15 (OK, compact flange)2

f

f y

b E

t F

3.76 53.9 90.55 (OK, compact web)w y

h E

t F

W16X36 is compact, so that the nominal moment is:

250 ksi 64 in 3200 kips-in 266.7 kips-ftn p y x

M M F Z

Allowable moment using LRFD concept:

0.9 266.7 kips-ft 240 kips-ftb nM

Allowable maximum distributed load:

2

2

1

8

1240 kips-ft 24 ft

8

3.33 kips/ft

b n u u

u

u

M M w L

w

w

(Answer C)



Soil Properties


The sample of soil can be divided into three components: dry soil,

water, and air.

Dry soil properties :

3

3

13.4 gr6.23 cm

2.15 1 gr/cms

s

s w

mV

G

Volume of the void can be defined as the summation between

volume of water and air. Therefore,

310 6.23 3.77 cmv total s

V V V

Void ratio is calculated by using this following equation:

3.770.6

6.23v

s

Ve

V

(Answer D)



Soil Properties


Assume that the total volume of saturated soil is 1 m3.

1s w

V V

Substitute the volume relationship to this equation:

3

2250

2250

2.6 1000 1 1000 2250

2600 1000 1000 2250

0.78 m

s w

s s w w w

s s

s s

s

m m

G V V

V V

V V

V

After obtaining volume of soil, we can calculate dry unit mass

(mass of soil).

2.6 0.78 1000

2031.25 kg

s s s wm G V

(Answer B)



Active Lateral Pressure


Total lateral pressure is the combination between active earth

pressure and pore water / hydrostatic pressure.

At the bottom of retaining wall = depth of 6 m

2 22 25tan 45 tan 45 0.4059

2 2a

K

Active earth pressure at depth = 6 m:

1 1 2 2

2

0.4059 19 3 20 9.8 3

35.56 kN/m

a a wK z z

Pore water / hydrostatic pressure at depth = 6 m:

2

29.8 3 29.4 kN/m

w wz

Total lateral pressure at depth = 6 m:

2

total35.56 29.4 64.96 kN/m

a w

(Answer D)



Bearing Capacity


Using Terzaghi’s equation for a square footing:

0.5

0 125 pcf 4 ft 22.5 0.5 125 pcf 5 ft 19.7 0.85

16482.8125 psf

ult c c f qq cN S D N BN S

Using safety factor of 3:

16482.8125 psf

3

5494 5500 psf

ult

all

qq

FS

(Answer C)



Consolidation Rate


Permeability:

An increase in permeability of the consolidating soil would

lead to an increase in the rate of seepage flow, other factors

remaining constant. With the greater rate of expulsion of

water from the soil the pore pressures will dissipate more

rapidly. This means that a more rapid rate of consolidation

occurs.

Layer thickness:

An increase in the layer thickness leads to a decrease in the

total head gradient during the stage of pore water expulsion.

It also means an increase in the volume of water to be

expelled and both of these effects lead to a lower rate of

consolidation.

Compressibility:

A greater compressibility leads to a greater decrease in the

void space of the soil for a particular stress change. This

means that a greater volume of water must be expelled from

the soil and this will require a longer time. Consequently, a

lower rate of consolidation will result.

All three factors affect the consolidation rate.

(Answer D)



Vertical Curve


12 sta

sta 76 00 sta 70 002 2

LPVC PVI

1

1 2

0.02 12 sta6.32 sta

0.02 0.018m

G Lx

G G

low point sta 70 00 6.32 sta sta 76.32m

PVC x

(Answer C)



Vertical Curve


2

2 1

lowpoint 1

2

2

0.018 0.02 6.32 sta 500 0.02 6 sta 0.02 6.32 sta

2 12 sta

505.68 506 m

m

PVC m

G G xelev elev G x

L

(Answer D)



Driver Perception-Reaction Time


AASHTO’s Green Book Chapter 3 lists 2.5 sec as the value used to

determine the minimum stopping sight distances, and is

appropriate for approximately 90% of the population.

(Answer C)



Traffic Capacity


By combining two equations given in the above information, we

can obtain a new equation about the relationship between the

traffic volume and traffic density.

2

80 0.5

80 0.5

q kv

k k

k k

According to mathematics principle, we can get the maximum

value of q when the first derivation of the equation is equal to 0.

2

max

max

2

max max max

80 0.5

0

80 0

80 veh/km

80 0.5 3200 veh/hr

q k k

dq

dk

k

k

q k k

(Answer B)



Horizontal Curve


Find the central angle:

tan2

125 600tan2

23.53

T R

Find the length:

600 23.53246 m

180 180

RL

Determine the stationing:

sta 10 000 125 sta 9 875PC PI T

sta 9 875 256 sta 10 121PT PC L

(Answer C)



BOD Analysis


Substitute the five-day values into the BOD equation:

0.13 5

1

234 1

489.6 490 mg/L

kt

ty L e

L e

L

(Answer C)



BOD Analysis


From the previous solution, the ultimate BOD was obtained.

Now, substitute the seven-day values into the BOD equation:

0.13 7

1

489.6 1

292.5 290 mg/L

kt

ty L e

e

(Answer A)



Reactor Capacity


Solids residence time:

V

Q

Suspended solids concentration:

0

0

0

3 3

1

1

1

12 0.35 0.55 258 6.4

3600 1 0.06 12

0.094 m /s 8110 m /day

d

c

d

d c

d

c

d

d c

d

c

d

d c

Y S SX

K

QY S SX

V K

QY S SV

X K

(Answer C)



Dissolved Oxygen


The dissolved oxygen (DO) at atmospheric pressure (760 mmHg)

is obtained by using linear interpolation of data from the table.

24.6

24.6

24.6

8.9 24.6 24

8.6 8.9 25 24

8.9 0.18

8.72 mg/L

C

C

C

DO

DO

DO

Oxygen is only slightly soluble in water and does not react with

water. Therefore, Henry’s law is applicable for this case, and the

solubility of oxygen is directly proportional to its partial pressure.

740 6.2% saturation 100% 69.2% 70%

760 8.72

(Answer C)



Solid Waste


Monthly mass of solid waste (for 18,400 people in a month):

3 3125 kg/m 20,000 m 2,500,000 kg

Average mass of solid waste per person per day:

2,500,000 kg4.5 kg/person.day

18,400 people 30 days

(Answer D)



Waste Water


Assume : 1 mg/L = 1 ppm

1 gal = 8.34 lbm

6

6

28 parts gal lbmMass loading 35 10 8.34

day gal10 part

lbm 8173.2

day

(Answer B)



Reaction Rate Constant


5 0

5

5

1

1

225 486 1

2251

486

225ln 1 5

486

225ln 1

4860.124

5

kt

k

k

BOD L e

e

e

k

k d

(Answer C)



Critical Path


Critical Path is the longest path to complete a project.

(Answer C)



Project Scheduling Tool


In the figure, we could see the activity (labeled as number) is on

node. Therefore, the figure above represents AON (Activity-On-

Node) diagram.

(Answer B)

1 3

2

2 4

2

5

2

6

2

7

2



Critical Path


Critical path is defined as the longest path to complete a project.

From the AON diagram, we could determine the total duration of

sequences for each path.

START – A – C – F – FINISH : 3 + 2 + 3 = 8 days

START – A – D – F – FINISH : 3 + 3 + 3 = 9 days

START – B – D – F – FINISH : 5 + 3 + 3 = 11 days

START – B – E – F – FINISH : 5 + 4 = 9 days

Path START – B – D – F – FINISH has the longest duration so it’s

the critical path of this project.

(Answer C)



Scheduling


Activity F is dependent to two activities: C and D. Therefore, the

EST (Earliest Start Time) of activity F can be determined as the

maximum EFT of those two activities.

Activity Duration (days)

EST (day no.)

EFT (day no.)

LST (day no.)

LFT (day no.)

TF (days)

A 3 0 3 3 6 3

B 5 0 5 0 5 0

C 2 3 5 6 8 3

D 3 5 8 5 8 0

E 4 5 9 7 11 2

F 3 8 11 8 11 0

From the above table, we can conclude that the EST of activity F

starts at Day 9, after the EFT of activity D ends at Day 8.

(Answer B)



Scheduling


Activity D is along the critical path (see the Problem 100 Solution),

so that the float time (TF) of activity D is 0 day. It means that

activity D has no spare time to delay.

(Answer A)



Construction Contract


Lump sum contract:

Contractor bids a total price for all work in a project, including

the profit.

Requires contract modification if there are any changes.

Unit price contract:

Contractor bids a price for each work item by cost per unit.

The actual quantities will be used for the final payment.

(Answer D)



Contract Bond


Contractors are required to obtain bonds from a surety company

prior to submitting bid. The types of bond can be seen below:

Bid bond guarantees the contractor will enter into a contract with

the client if selected – typically 5% to 20% of the estimated

project cost.

Performance bond guarantees the contractor will perform the

specified work in accordance with contract – typically full value of

the project.

Payment bond guarantees the contractor will pay for all

materials and labor used on the project – typically full value of the

contract. This is usually used to protect clients from being sued for

payment by subcontractors.

(Answer B)



Construction Safety


OSHA 1926.651(c) about Specific excavation requirement:

(2) Means of egress from trench excavations.

A stairway, ladder, ramp or other safe means of egress shall

be located in trench excavations that are 4 feet (1.22 m) or

more in depth so as to require no more than 25 feet (7.62 m)

of lateral travel for employees.

(Answer C)



Construction Estimating


Gross wall area:

272 ft 16 ft 1152 ftg

A

Net wall area (without opening):

2 2 21152 ft 250 ft 902 ftn g opening

A A A

Bricks required:

2

2

602 brick902 ft 1.05 5702 bricks

100 ftbrick

n

(Answer D)



Earthwork - Fill


As given in problem statement, cut and fill area can be assumed

as triangular in shape for transition area. While they are assumed

as trapezoidal area for other parts.

Station

(m)

Fill area

(m2)

Fill volume

(m3)

10+00 153.42

153.42 20.56

4.50 391.462

10+4.50 20.56

20.56 8.25

7.85 113.082

10+12.35 8.25

8.25

4.05 11.143

10+16.40 -

-

10+20 -

total 515.68

Total volume of fill work is 515.68 m3.

(Answer D)



Earthwork - Cut


Station

(m)

Cut area

(m2)

Cut volume

(m3)

10+00 -

-

10+4.50 -

13.67

7.85 35.773

10+12.35 13.67

13.67 52.84

4.05 134.682

10+16.40 52.84

52.84 165.14

3.6 392.362

10+20 165.14

total 562.81

Total volume of cut work is 562.81 m3.

(Answer B)



Borrow Pit Volumes


It’s easier to solve borrow pit volumes using table.

Point Elevation

(m)

Elev to be

Cut (m)

No. of

Rectangles

Height of

Cut (m)

A1 103.5 3.5 1 3.5

A2 105.6 5.6 2 11.2

A3 102.1 2.1 2 4.2

A4 101.9 1.9 1 1.9

B1 103.9 3.9 2 7.8

B2 104.0 4.0 4 16

B3 102.8 2.8 3 8.4

B4 103.7 3.7 1 3.7

C1 105.2 5.2 1 5.2

C2 104.3 4.3 2 8.6

C3 100.3 0.3 1 0.3

Total = 70.8

3

,

12 m 8 m70.8 m 1699.2 1700 m

4 4i jn

AV h

(Answer B)



Earthwork Terms


Swell: the increase in volume of earth from its natural to

loose state. When earth is excavated, it increases in

volume because of an increase in voids.

Shrinkage: the decrease in volume of earth from its natural to

compacted state.

(Answer C)



SCORE SHEET

Correct Answers: ___________

Percentage: (correct answers)/110 = ____/110 = ______

Congratulations! You’ve finished your practice exam! How did you

do? Are you happy with the results or are there areas that you need

to improve?

If you didn’t do as well as you had hoped don’t give up. Just keep on

practicing more problems and you’ll get there. Practicing a ton of

problems is the key. You’ve either been doing that all throughout your

schooling or you are having to do it to study for this exam. Either

way, you have to practice. You’ll have to do the same thing when the

PE comes around too!

I hope this challenged you and helped to assess where you stand.

Good luck on the FE exam and on your engineering future!

Helpful Tools:

We have built www.civilengineeringacademy.com to help any civil

engineer take and pass the FE and PE exams. We have tons of free

video practice problems there to get you going. We also have plenty

of tips, must have resources, advice on courses, and more. We’ve

even created our own FE review course that can guide you step-by-

step through the entire exam. You can check that out at

www.civilfereviewcourse.com.


http://www.civilfereviewcourse.com/

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