THE ULTIMATE - Amazon S3...THE ULTIMATE HELPING YOU ON YOUR JOURNEY TO BECOMING A PROFESSIONAL...

CIVIL PE BREADTH EXAM

Isaac Oakeson, P.E.Volume 2

THE ULTIMATE

HELPING YOU ON YOUR JOURNEY TO BECOMING A PROFESSIONAL ENGINEER.

VO

LUM

E 2

Isaac Oakeson, P.E. is a professional engineer working in the great state of Utah. Having passed the PE Exam in the fall of 2012, he is driven to help others

do the same.

His passion has led him to create CivilEngineeringAcademy.com, CivilPEReviewCourse.com and CivilFEReviewCourse.com.

Each website provides the tools necessary to ace the FE and PE exams. Many have used the resources he has created and now he wants to help you too!

www.civilengineeringacademy.com | [email protected]|

THE ULTIMATE CIVIL PE BREADTH EXAM

VOLUME 2

http://www.civilengineeringacademy.com

http://www.civilengineeringacademy.com/


TABLE OF CONTENTS

WELCOME…………………………………………….

EXAM SPECIFICATION……………………

START TEST ………………………………...

SOLUTIONS………………………………………

LAST SECOND TIPS AND ADVICE…..

Page 6

Page 9

Page 47

Page 88




WELCOME!!

Welcome to The Ultimate Civil PE Breadth Practice Exam Volume 2!

Thank you so much for purchasing this book!

This book includes a full 40-question/solution breadth exam. This

exam has been updated to meet the latest NCEES specifications (2015

as of this writing). Each problem is designed to have a similar look

and feel to the real exam.

The key to passing the PE exam is to absolutely crush the breadth

portion of it. If you can get 90 percent or more in the morning section

then the afternoon section will be that much easier on you. All in all,

this is to help you gain more practice, more experience, and ultimately

more confidence in passing.

This test is not endorsed by the NCEES organization. These are

problems that my team and I have written to help you succeed in

passing the PE exam. I would encourage you to take this timed to see

how long it takes you. Afterward, you can take note of the areas that

you might need to work on. I have spent some time getting all of the

information here for you so that it is easy to use. Each problem is

labeled, so you know what problem and area you are dealing with.

There are multiple ways of taking practice exams – you can work

problems like homework, you can take it like the real exam - it’s really

up to you! I would make sure to do at least one practice exam as if it

were the real deal though so you gain that experience.




As always, I value your feedback and any constructive criticism you

might have on this exam or anything else we produce to help others

pass. You can also get many more resources on the sites we run at

www.civilengineeringacademy and www.civilpereviewcourse.com,

including step by step video practice problems.

I know that with a lot of practice you will become much more

proficient at working problems and doing them with less and less

assistance. Keep at it and you will be prepared to pass the PE.

I don’t need to tell you about the benefits of obtaining your PE license

because I’m sure you already know them. You must get it to have a

great career in civil engineering (and a lot of other fields!).

As always, I wish you the best of luck!

Sincerely,

Isaac Oakeson, P.E.

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

P.S. Errata for this or any other exam we created can be found at

www.civilengineeringacademy.com.




LEGAL INFORMATION

Civil Engineering Academy’s

The Ultimate Civil PE Breadth Exam Vol. 2

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.



|www.civilengineeringacademy.com|[email protected]|

CIVIL – BREADTH EXAM SPECIFICATION

I. Project Planning (4)

A. Quantity take-off methods

B. Cost estimating

C. Project schedules

D. Activity identification and sequencing

II. Means and Methods (3)

A. Construction loads

B. Construction methods

C. Temporary structures and facilities

III. Soil Mechanics (6)

A. Lateral earth pressure

B. Soil consolidation

C. Effective and total stresses

D. Bearing capacity

E. Foundation settlement

F. Slope stability

IV. Structural Mechanics (6)

A. Dead and live loads

B. Trusses

C. Bending (e.g., moments and stresses)

D. Shear (e.g., forces and stresses)




E. Axial (e.g., forces and stresses)

F. Combined stresses

G. Deflection

H. Beams

I. Columns

J. Slabs

K. Footings

L. Retaining walls

V. Hydraulics and Hydrology (7)

A. Open-channel flow

B. Stormwater collection and drainage (e.g., culvert, stormwater

inlets, gutter flow, street flow, storm sewer pipes)

C. Storm characteristics (e.g., storm frequency, rainfall measurement

and distribution)

D. Runoff analysis (e.g., Rational and SCS/NRCS methods,

hydrographic application, runoff time of concentration)

E. Detention/retention ponds

F. Pressure conduit (e.g., single pipe, force mains, Hazen-Williams,

Darcy-Weisbach, major and minor losses)

G. Energy and/or continuity equation (e.g., Bernoulli)

VI. Geometrics (3)

A. Basic circular curve elements (e.g., middle ordinate, length, chord,

radius)

B. Basic vertical curve elements

C. Traffic volume (e.g., vehicle mix, flow, and speed)




VII. Materials (6)

A. Soil classification and boring log interpretation

B. Soil properties (e.g., strength, permeability, compressibility, phase

relationships)

C. Concrete (e.g., nonreinforced, reinforced)

D. Structural steel

E. Material test methods and specification conformance

F. Compaction

VIII. Site Development (5)

A. Excavation and embankment (e.g., cut and fill)

B. Construction site layout and control

C. Temporary and permanent soil erosion and sediment control (e.g.,

construction erosion control and permits, sediment transport,

channel/outlet protection)

D. Impact of construction on adjacent facilities

E. Safety (e.g., construction, roadside, work zone)




(START TEST - VOLUME 2)




1. Suppose that 2000 ft3 of concrete is required to pour 40% of the

slabs of the third floor of the Black Crown Hotel. To construct a

3500 psi concrete, the batching plant produced a 1:2:4 by volume

mixture. Compute the bulk volume of the cement required for the

40% of the third floor slabs.

a) 10.6 yd3

b) 74.1 yd3

c) 29.6 yd3

d) 37.0 yd3




2. Specifications on a job required a fill using borrow soil to be

compacted at 95% of its standard proctor maximum dry density.

Tests indicate that the maximum dry density is 124 pcf with 12%

moisture content. The borrow material has a void ratio of 0.6 and

a solid specific gravity of 2.65. Compute the minimum volume of

borrow soil required per 1 ft3 of fill.

a) 2.25 ft3

b) 1.00 ft3

c) 3.05 ft3

d) 1.14 ft3




3. The following table tabulates the activities for a new project’s

network diagram. Based on the data provided below, determine

the duration of the project?

a) 14 weeks

b) 20 weeks

c) 10 weeks

d) 12 weeks

Path Duration

(weeks)

0-1 (A) 3

0-2 (B) 4

0-3 (D) 2

1-2 (C) 5

2-3 (E) 3

3-4 (F) 3




4. A road is to be constructed and the cut is shown in the following

figure. Determine the cost of the excavation process if it is worth

$2.5/ft3. Use the end area method.

a) $187,500

b) $185,800

c) $178,500

d) $185,700

A2 = 700 ft2

A1 = 800 ft2

100 ft




5. A 28 ft deep braced cut in a sandy soil is shown. The struts are

placed 5 ft center-to-center. Compute the reaction at strut C using

Peck’s empirical pressure diagram.

a) 37,160 lbs

b) 38,990 lbs

c) 35,660 lbs

d) 31,590 lbs




6. A sling system can carry 2000 lbs using 90° basket hitch. What is

the maximum load the sling system can carry using 60° basket

hitch as shown below?

a) 1,155 lb

b) 2,310 lb

c) 2,000 lb

d) 1,732 lb

60° 60°




7. A crane is caring a crate weighing 1000 lb. It is held by four cables

attached to the corners. The attachment point is located directly

above the center of the load. What, most nearly, is the tension in

each cable?

a) 250 lb

b) 300 lb

c) 340 lb

d) 410 lb

15 ft 5 ft

1000 lb

6 ft




8. A vertical retaining wall has a height of 20 ft that supports a

horizontal backfill. Its dry unit weight is 100 lb/ft3 and a saturated

unit weight of 140 lb/ft3. Ground water was found at 8 ft below

the ground surface. Compute the total Rankine active force per

unit length of wall.

a) 9,005 lb/ft

b) 8,544 lb/ft

c) 7,920 lb/ft

d) 6,785 lb/ft




9. From the soil layer shown, determine the closest effective stress

at point C. Layer A is dry, layer B and C are saturated.

a) 1,087 lb/ft2

b) 1,095 lb/ft2

c) 1,054 lb/ft2

d) 1,025 lb/ft2




10. A normally consolidated clay layer is 15 ft (one way drainage).

From the application of a given pressure, the total anticipated

primary consolidation settlement will be 3 inches. Given that the

coefficient of consolidation is 0.00323x10-3 ft2/s, how many days

will it take for 50% settlement to occur if the time factor from a

table is 0.197?

a) 159 days

b) 124 days

c) 170 days

d) 205 days

11. Settling on a structure has three distinct periods. Which of the

following statements is not true about settling?

a) Elastic settling occurs immediately after a structure is constructed.

b) Secondary consolidation occurs in clay soil and happens at a much

slower rate after primary consolidation has finished.

c) Primary consolidation occurs gradually in clay because water is

being extruded from the voids.

d) An influence chart can be used to determine the primary

consolidation rate.




12. What is the factor of safety for the following slope stability

analysis? The unsaturated clay soil has the following properties:

120 pcf , 600 psfc .

a) 1.1

b) 1.8

c) 2.0

d) 2.5




13. A 5 ft x 5 ft foundation is embedded 3 ft into sand. The sand has a

density of 120 pcf and bearing capacity factors of 51c

N , 38q

N

, 44N . Determine the maximum allowable column load, P (in

lbs), to maintain a factor of safety of 3. Ignore shape factors.

a) 8,860 lb

b) 8,960 lb

c) 221,500 lb

d) 224,000 lb




14. A square footing with a depth of 20 inches carries a dead load of

100 kips and a live load of 80 kips. From the figure shown,

determine the approximate dimension of the footing given the

allowable soil bearing capacity of 4000 psf. Use a concrete unit

weight of 150 pcf.

a) 8 ft x 8 ft

b) 7 ft x 7 ft

c) 7.5 ft x 7.5 ft

d) 8.5 ft x 8.5 ft

20”

12”x12” Column

5’ gsoil = 110 pcf




15. In the figure shown, the beam is simply supported at point A and

5 ft from point C. Determine the maximum positive moment.

a) 550 lb-ft

b) 650 lb-ft

c) 750 lb-ft

d) 1500 lb-ft

800 lbs 500 lbs

5 ft 5 ft 5 ft

A C B




16. After analyzing a beam, its moment diagram is shown below.

Given the diagram, determine the maximum positive bending

stress. The section modulus of the beam, S = 8.14 in3.

a) 16.58 ksi

b) 15.36 ksi

c) 12.89 ksi

d) 29.84 ksi

135 kips-in

-125 kips-in

105 kips-in

4 ft 4 ft 3 ft Moment

diagram




17. A column carries a concrete flat plate with a 5 inch thickness. The

slab carries a dead load of 80 psf. Determine the compressive

stress of the column. Use a concrete unite weight of 150 pcf

(tributary area 13 ft x 15 ft).

a) 21.16 psi

b) 48.24 psi

c) 35.21 psi

d) 27.40 psi

13 ft

15 ft

5 in

Column

4 ft

1 ft




18. A 20 ft beam, loaded as shown, is fixed at both ends. Given the

properties of the beam, determine the deflection at the midpoint.

a) 0.87 in

b) 1.5 in

c) 1.3 in

d) 0.12 in

A B

Beam Properties E = 29000 ksi I = 118 in⁴

W=250lbs/ft

3000 lbs




19. Determine the moment diagram of the loaded beam given.

a)

b)

c)

d)

4 ft 8 ft

A B




20. In the figure shown below, design the diameter of pipe CD given a

headloss of 14.5 ft using Manning’s Equation (n=0.013).

a) 20 in

b) 45 in

c) 30 in

d) 60 in




21. Water flows from point B to A. In the system shown, the pressure

from A to B is 1200 psf when the gate valve is closed, and when

the valve is opened, the pressure from B to A is measured 1000

psf. Determine the headloss from point B to A.

a) 19.2 ft

b) 35.2 ft

c) 20.2 ft

d) 10.2 ft




22. Determine the hydraulic radius from the channel section shown.

a) 2.87 ft

b) 3.00 ft

c) 2.50 ft

d) 2.75 ft




23. Given the pipe network with flow rates shown, determine the flow

rate at pipe BE.

a) 2.771 ft3/s

b) 2.935 ft3/s

c) 2.823 ft3/s

d) 2.698 ft3/s

B

3.528 ft3/s QBE = ?

CC

0.176 ft3/s

0.529 ft3/s

E

E

A B

D

C

E




24. An oil with density = 1.7 slugs/ ft3 flows through a cast iron

pipe at a velocity of 3 ft/s. The pipe is 150 ft long and has a

diameter of 6 in. Given an absolute viscosity = 0.0017 lbs - s /

ft2. Determine the head loss due to the friction factor. Use

Reynold’s number.

a) 1.897 ft

b) 1.542 ft

c) 1.675 ft

d) 1.786 ft




25. In a non-uniform flow, water has a discharge of 705 ft3/s through

a rectangular channel measuring 13 ft wide. The water runs from

a steep slope to a mid-slope creating a hydraulic jump. The

upstream depth of flow is 3.94 ft. Determine the downstream

depth of flow.

a) 4.98 ft

b) 5.23 ft

c) 4.81 ft

d) 5.12 ft




26. A simplified water supply system is shown in the figure. The

pressure supplied to the townhomes is 10,000 psf and the

discharge of the system is 4 ft3/s. Determine the head that the

pump will need for the given situation. Use f=0.012. The water

tower is open to the atmosphere.

a) 40 ft

b) 45 ft

c) 50 ft

d) 55 ft

El. 150 ft

El. 20 ft

Length = 3000 ft Diameter = 12”

P




27. A back tangent of a simple curve intersects the forward tangent at

station 67+62 which is 345 ft from the PC. Compute the external

distance of the curve when the radius of the curve is 300 ft.

a) 146 ft

b) 187 ft

c) 157 ft

d) 45 ft




28. A train passing point A at a speed of 65 fps accelerates at 2.46

fps2 for one minute along a straight path, and then decelerates at

3.3 fps2. How far from point A will it be 2 minutes after passing

point A?

a) 15,144 ft

b) 13,879 ft

c) 13,855 ft

d) 13,760 ft

29. A vertical parabolic curve has a back tangent of -5% and a

forward tangent of 3%. If the stationing of PC is at 80+50, locate

the stationing of the lowest point of the curve. The length of the

curve is 820 ft.

a) 86+62

b) 81+62

c) 85+62

d) 88+62




30. Solve the porosity, n, knowing that the sample has a SG=2.55,

weighs 32 lbs total, a total volume of 0.4 ft3, and a water content

of 17%.

a) 0.43

b) 0.47

c) 0.50

d) 0.57

Soil

Water

Air




31. Given a concrete beam that has a dimension of 12”x25” (with 3”

cover) and a tension steel area of 3.33 in2. Determine the depth of

the equivalent stress block. Assume that the steel yields.

a) 4.35 in

b) 6.73 in

c) 5.41 in

d) 11.0 in

a

22 in

3.33 in²

12 in

fc’ = 3 ksi fy = 40 ksi




32. The angle of friction of a cohesive soil which was tested using a

tri-axial shear test is equal to 29°. Failure occurred when the

shear stress reached 45 psi and the normal stress reached 65 psi.

Determine the cohesion of soil.

a) 10 psi

b) 16 psi

c) 9 psi

d) 20 psi




33. The laboratory compaction test of a certain type of soil gives a

maximum dry density of 120 pcf with an optimum moisture

content of 14%. Sand Cone Method was used to determine the

field unit weight and the results are shown below. Determine the

relative compaction.

a) 90.2

b) 92.5

c) 94.6

d) 95.4

V

Soil Properties

Volume of soil excavated from the hole = 0.017 ft³

Mass of moist soil removed

from the hole = 2.2lbs Field moisture content = 13%

Sand Cone Method




34. A given soil has the following particle size distribution: 8% sand,

20% gravel, 48% silt, and 24% clay. Classify the type of soil using

the USDA method below.

a) Silt Loam

b) Loamy Sand

c) Sandy Clay Loam

d) Silty Clay Loam




35. Given the location of the neutral axis of the structural steel tee

section shown, determine the maximum bending stress if the

moment is 150 kips-ft.

a) 11 ksi

b) 40 ksi

c) 26 ksi

d) 51 ksi

Neutral

axis

WT 13.5 x 108.5 properties: d = 14.215 in

y = 3.11 in

Ix = 502 in4




36. A square lot is divided into 60 ft2 and the corners are numbered I-

IV vertically and A-D horizontally. The heights removed after

excavation are tabulated below for a borrow pit. Compute the

volume of earthwork excavated.

AI = 20.3 BI = 19.0 CI = 20.5 DI = 15.1

AII = 18.4 BII = 14.1 CII = 17.1 DII = 12.5

AIII = 12.5 BIII = 8.4 CIII = 14.1 DIII = 8.8

AIV = 6.9 BIV = 6.23 CIV = 12.8 DIV = 5.9

a) 395,460 ft3

b) 459,366 ft3

c) 436,014 ft3

d) 345,690 ft3

B C D A

I

II

III

IV

60 ft

60 ft

60 ft

60 ft 60 ft 60 ft




37. A surveyor measures the elevation on a construction site. Based

on the data below, find the difference in elevation between BM1

and BM2.

a) 4.23 ft

b) 6.30 ft

c) 10.63 ft

d) 4.10 ft

Station BS FS Elev

BM1 8.56 540.51

TP1 9.45 4.23

BM2 3.15 ?

BM1

BM2

TP 1




38. Two levelling rods placed at a distance of 160 ft and 320 ft from

the instrument gives an intercept of 1.6 ft and 3.2 ft respectively.

Determine the stadia interval factor, K.

a) 150

b) 120

c) 100

d) 160




39. Hearing safety is of high importance on a construction site.

According to OSHA, what is the maximum “all-day” 8 hour noise

level limit?

a) 90 dBa

b) 92 dBa

c) 100 dBa

d) 115 dBa

40. A surveyor is setting up new control points for a building layout.

The tangents have a bearing of N40°E and S65°E respectively.

Determine the angle of intersection.

a) 105°

b) 115°

c) 65°

d) 75°




VOLUME 2 SOLUTIONS




PROBLEM 1 SOLUTION:

QUANTITY TAKE OFF METHODS

Use the proportions to compute the bulk volume. The ratios are given

in this sequence cement:fine aggregate:coarse aggregate. Write the

ratios down and multiply by the volume required to see what you need

of each type.

The cement required is 285.7 ft3 or 10.6 yd3.

(Answer A)

Material Computation Bulk Volume

Gravel

(coarse aggregate)

42000

7 1142.9 ft3

Sand

(fine aggregate)

22000

7 571.4 ft3

Cement 1

20007

285.7 ft3




PROBLEM 2 SOLUTION:

QUANTITY TAKE-OFF

First, for the borrow soil, solve for the dry unit weight of the soil:

2.65 62.4103.35 pcf

1 1 0.6s w

dry

G

e

Then, solve for the compacted soil at 95% compaction:

0.95 124 117.8 pcfdry

The weight of the compacted soil = the weight of soil needed as borrow:

3

117.8 1 103.35

1.14 ft

borrow

borrow

V

V

(Answer D)




PROBLEM 3 SOLUTION:

ACTIVITY IDENTIFICATION AND SEQUENCING

First, draw the paths given and then find the critical path (longest

duration, blue arrows).

Path DF = 2 + 3 = 5 weeks

Path ACEF = 3 + 5 + 3 + 3 = 14 weeks (Answer A)

Path BEF = 4 + 3 + 3 = 10 weeks

In order to finish a project we should complete all activities from the

starting node to the finishing node (0 to 4). But all activities should be

done to complete a project.

Example: In order for node 2 to be completed, activities at path AC

should be accomplished since it has the longest duration of activity

inside node 2.

Path Duration (weeks)

0-1 (A) 3

0-2 (B) 4

0-3 (D) 2

1-2 (C) 5

2-3 (E) 3

3-4 (F) 3

0

1

2

3

4

D (2 weeks)

B (4 weeks)

A (3 weeks) C (5 weeks)

F (3 weeks)




PROBLEM 4 SOLUTION:

COST ESTIMATING

First, compute the volume of soil excavated (cut) using the end area

method:

1 2 3800 700 100

Volume 75000 ft2 2

A A L

Now solve for the cost of the excavation:

Cost 75000 2.5 $187,500

(Answer A)




PROBLEM 5 SOLUTION:

TEMPORARY STRUCTURES

First, solve for the active earth pressure coefficient:

1 sin30 1

1 sin30 3a

K

Find the maximum lateral pressure for braced cuts in sand (Peck’s):

1

0.65 0.65 100 28 5 3033.3 lb/ft3

a aP K H

Then isolate BC to solve for the reaction at point C:

0

3033.3 14 7 8 0

37158 37160 lb

B

C

C

M

R

R

(Answer A)




PROBLEM 6 SOLUTION:

CONSTRUCTION LOADS

This problem can be resolved into a simple statics problem. A sling

system can carry 2000 lbs using a 90° basket hitch, so the capacity of

one leg is half of 2000 lbs, which is 1000 lbs.

So, we need to calculate the vertical component of each leg, then sum

the total capacity of both legs using this 90° basket hitch at 60°.

Draw a force triangle and solve for X:

Vertical component of one leg 1000 lbs sin60 866 lbs

Vertical component of two legs 2 866 lbs 1732 lbs

1732 lbs is the max load this sling system can carry using a 60° angle.

(Answer D)

1000 lbs 1000 lbs

60° 60°

1732 lbs

90° 90°

2000 lbs

1000 lbs 1000 lbs

90° basket hitch goes all

the way around the load.




PROBLEM 7 SOLUTION:

CONSTRUCTION LOADS

Because the problem is 3 dimensional we are dealing with a X, Y, and

Z coordinate. In order to get the length of each cable we must do the

following:

Remember that the pickup point is half of 15 ft and that the pickup

point is half of 5 ft in width too. Solve for the magnitude of L:

2 2 27.5 2.5 6 9.92 ftL

Next you need to solve for the vertical

component in each cable. It can be written

as the following:

6 / 9.92 0.605y

F F F

Next, take the sum of forces in the y direction and solve for the force

in the cable:

4 1000 0

4 0.605 1000

413.2 410 lbs

yF

F

F

(Answer D)

You can see that each cable sees a much larger load that you would

think based on the angle you are pulling up at.

6 ft

L

7.5 ft

2.5 ft




PROBLEM 8 SOLUTION:

SOIL MECHANICS

LATERAL EARTH PRESSURE

First, solve for the active earth pressure coefficient:

1 sin32

0.3071 sin32

aK

Then draw the pressure diagram using the given unit weight and

pressure coefficient:

1st soil layer 12 100 0.307 368.4 psf

2nd soil layer 8 140 62.4 0.307 190.6 psf

Pore water pressure 8 62.4 1 499.2 psf

Note: P1, P2, and P3 are all from the effective pressure of the soil,

while P4 is from pore water pressure.




Then solve for the forces by parts from the diagram and add for the

total force:

1

1368.4 12 2210.4 lb

2P

2368.4 8 2947.2 lbP

3

1559 368.4 8 762.4 lb

2P

4

1500 8 2000 lb

2P

1 2 3 47920 lb per ft

totalP P P P P

(Answer C)




PROBLEM 9 SOLUTION:

EFFECTIVE AND TOTAL STRESS

First, solve for the dry unit weight and saturated unit weight of the

soil:

2.7 62.4

104.65 pcf1 1 0.61

s wdry

G

e

2.7 0.61 62.4

128.3 pcf1 1 0.61

s w

sat

G e

e

Then solve for the effective stress at point C:

' 104.65 6 128.3 62.4 6 1023.3 psf

(Answer D)




PROBLEM 10 SOLUTION:

SOIL CONSOLIDATION

Solve for the time to consolidate:

2 2

3

0.197 1513722910 sec 159 days

0.00323 10v

v

T Ht

C

(Answer A)


FOUNDATION SETTLEMENT

All of the answers to this are true about settlement, except option D.

The influence chart (Newmark) is used to find the vertical pressure

underneath a foundation. It is not used to determine the primary

consolidation rate. See Settling in the CERM.

(Answer D)





SLOPE STABILITY

To solve this, use the Taylor slope stability chart (CERM Fig. 40.6):

Soil Mechanics, NAVFAC Design Manual DM-7.1

To find d, you solve 85 ft

3.27 326 ft

Dd

H and the slope was given

as 40°. See where these two intersect to find the slope stability

number, oN . Say 5.6

oN .

Solve for the safety factor:

6005.6 1.07 1.10

120 26o

cF N

H

This is an unacceptable safety factor. You want between 1.3-1.5 for

design.

(Answer A)





BEARING CAPACITY

Solve for the ultimate bearing capacity of the soil:

0.5ult c q

Q cN DN BN

0c for sands

D depth into sand layer

B width of footing

Groundwater is neglected because it is greater than footing

D B

0 120 3 38 0.5 120 5 44 26880 psfult

Q

Solve for allowable load given the safety factor of 3:

268808960 psf

3ult

allow

QQ

FS

8960 psf 5 ft 5 ft 224000 lballow allow

P Q A

(Answer D)





FOOTINGS

Solve for the total Unfactored Load:

100 80 180 kipsunfactored D L

P P P

Effective stress:

20 20' 4000 110 5 150 3383.3 psf

12 12

2

'

1800003383.3

53.2 ft

unfactoredP

A

A

A

Since the footing is square, the dimension of length and width should be equal.

Length width 53.2 7.29 7.5 ftA

Dimension = 7.5 ft x 7.5 ft

(Answer C)





BEAMS

First, use the equilibrium equations to solve for the reactions:

0

800 5 10 500 15 0

1150 lb

A

B

B

M

R

R

0

1300 1150 0

150 lb

y

A

A

F

R

R

Draw the shear and moment diagrams; the maximum positive

moment is 750 lb-ft.

(Answer C)

Note:

To determine the maximum positive

moment, draw the shear force

diagram and the moment diagram.

1150 lbs

Shear Diagram:

Moving along the beam you start at +150 until you hit 800 lbs,

-650 is next (150-800 = -650) followed by a positive 500 lbs

(-650+1150 = 500).

Moment Diagram:

Because the moment is the integral of the shear, every time the

shear crosses 0 you should have a max or a min point. To find

the max and min’s find the area under the shear curve: 150

lb*5 ft = 750, and linear because shear slope is 0, then 650

lb*5 ft = 3250, so take 750-3250 = -2500 and then up to 0.

800 lbs 500 lbs

150 lbs

-650 lbs

500 lbs

750 lbs-ft

-2500 lbs-ft

150 lbs

5 ft 5 ft 5 ft





BENDING

The maximum positive bending moment is 135 kips-in. The maximum

negative bending moment is -125 kips-in. Therefore, we use the larger

of the two values: 135 kips-in.

Solve for the bending stress:

135

or 16.58 ksi8.14

b

Mc Mf

I S

(Answer A)

135 kips-in

-125 kips-in

105 kips-in

4 ft 4 ft 3 ft Moment diagram





COLUMNS

First, convert all pressure loads to concentrated loads:

Dead load 80 13 15 15600 lb

Slab weight 5

150 13 1512

12187.5 lb

TOTAL 27787.5 lb

Then use the formula for compressive stress:

27787.56946.9 psf 48.24 psi

4 1a

Pf

A

(Answer B)

80 psf

Dead load + Slab weight

4 ft 1 ft

15 ft

13 ft

5 in

Column





DEFLECTION

+

=

3

192p

PL

EI

4

384w

wL

EI

total p w

First, solve for the deflection of the

concentrated load and uniform

load.

3

3

192

300020 12

1000 0.063 in192 29000 118

p

PL

EI

4

4

384

250 120 12

1000 12 0.053 in

384 29000 118

w

wL

EI

Then use superposition to solve for

the total deformation.

0.063 0.053 0.12 in

total p w

(Answer D)





BENDING MOMENT DIAGRAM

From the information given,

we can now draw the shear and

moment diagram without any

given values.

The reaction at A starts the shear

up and then carries on until the

reaction at B. Because the shear

has a zero slope the moment is

linear until you hit the internal

moment at 4 ft. This drops the

moment to some value and then it

increases linearly until you hit zero

because the shear still has a zero

slope. Remember that the shear is

the derivative of the moment.

(Answer D)

The reactions from the supports at A

and B should create a resisting

moment in order to maintain the

equilibrium. Since the given load is

counterclockwise, the resisting

moment from the reactions should be

clockwise.

V

M





PRESSURE CONDUIT

Parallel

First, solve for the flow rate along the pipe:

330 20 50 ft /s

A B CDQ Q Q

Then solve for diameter of pipe using the head loss

(use Manning’s equation, CERM Eq. 39.30b):

2 2

43

2

2

2

43

2 2

163

2.208

0.25

2.208 0.25

4.66

f

Ln vh

R

QLn

D

D

n LQ

D

2 2

163

2 2

163

4.66

4.66 0.013 950 5014.5

2.487 ft 29.85 in 30 in

f

n LQh

D

D

D

(Answer C)

Note: When pipes are parallel,

flow rates should be added.





PRESSURE CONDUIT

Solve using Bernoulli’s equation: 2 2

1 1 2 21 2

2 2

P V P Vz HL z

g g

Solve for elevation Z at point B. Solve by using the Bernoulli equation

given when the gate valve was closed. Notice that velocity head is

cancelled out on both sides when it is closed because the flow is not

moving. There is also no head loss so we don’t need to account for

that. You are left with the following:

1200

62.5

19.2 ft

B AP P

z

z

z

Now solve for head loss. Use the Bernoulli equation given when the

gate valve is now open. Now that we have the elevation at B it should

be easy:

100019.23

62.4

35.26 ft

B A

B A

P PHL z

P PHL z

HL

HL

(Answer B)





OPEN CHANNEL FLOW

First, solve for the area and the wetted perimeter of the channel:

Area:

2

1 2

13 4 15 4 66 ft

2A A A

Wetted Perimeter:

1 2 3

5 15 4 24 ftP L L L

Then solve for the Hydraulic Radius:

66

2.75 ft24

AR

P

(Answer D)





PRESSURE CONDUIT

B

3.528 ft3/s QBE = ?

C

0.176 ft3/s

0.529 ft3/s

E

E

A B

D

C

E

QAB

First, solve for the flow rates depending on the given discharges in each

pipe before it:

33.528 0.812 0.777 1.939 ft /s

AB A AC ADQ Q Q Q

30.812 0.529 0.283 ft /s

CB AC CQ Q Q

30.777 0.176 0.601 ft /s

DB AD DQ Q Q

Then solve for QBE by adding all flow rates towards point B:

32.823 ft /s

BE CB AB DBQ Q Q Q

(Answer C)





PRESSURE CONDUIT

First, solve for the Reynolds number and check to see if the flow is

laminar or turbulent:

61.7 3

12 15000.0017

r

DVN

2000r

N the flow is laminar

Then solve for the friction factor:

64 64

0.04261500

r

fN

Lastly, solve for the head loss using the Darcy Weisbach formula:

2 20.0426 150 3

1.786 ft62

2 32.212

fLVHL

gD

(Answer D)

Nr > 2000 → Laminar Flow

Nr < 2000 → Turbulent Flow

Note: The formula used if the flow

is laminar →





OPEN CHANNEL FLOW

First, solve for the flow rate per foot of the jump:

3705 ft /s

54.231 13 ft

Qq

b

Then solve for the downstream depth of flow:

2

1 2 1 2

2

2 2

2

1

2

54.231 13.94 3.94

32.2 2

5.12 ft

qd d d d

g

d d

d

(Answer D)





CONTINUITY EQUATION

El. 150 ft

El. 20 ft

Length = 3000 ft Diameter = 12”

P

Note: Since the tank is open to the

atmosphere, initial pressure head is

zero and because of the large area of

the reservoir its velocity head is

neglected (zero).

First, solve for the velocity and head loss of the system using the given.

2

45.09 ft/s

14

QV

A

2 20.012 3000 5.09

14.48 ft122

2 32.212

fLVHL

gD

Then use Bernoulli’s theorem from the Elev. 150 ft to Elev. 20 ft point:

2 2

1 1 2 21 2

2

2 21 2

2

2 2

2

10000 5.09150 14.48 20

62.4 2 32.2

45.1 ft

P V P Vz HL HP z

g g

P Vz HL HP z

g

HP

HP

(Answer B)





BASIC CIRCULAR CURVE ELEMENTS

First, solve for the angle of the simple curve:

tan / 2

345 300 tan / 2

98

T R I

I

I

Then, solve for the external distance, E:

cos / 2

300cos 98 / 2

300

157.28 157 ft

RI

R E

E

E

(Answer C)





TRAFFIC VOLUME

First, instead of constructing a v-t graph and s-t graph, you can

employ this approach (see graph). o

v t is zero if the vehicle starts from

rest. In this problem 65 fpso

v .

D area under the a-t graph

area distance from the a-t graph

65 120 2.46 60 90 3.3 60 30

15144 ft

oD v t Q Q

(Answer A)

a (ft/s²)

t (s²)

2.46

-3.3

120 secs

Acceleration – Time graph (a-t graph)

90

30

60 secs





HORIZONTAL CURVE

First, solve for the location of the lowest point in the curve:

1

1 2

0.05820 512.5 ft

0.05 0.03

gx L

g g

Solve for the station of the lowest point:

Sta 80+50 512.5 or 8050 512.5

Sta 85+62.5 or 8562.5

(Answer C)

-5% +3%

820 ft

S

STA 80 + 50





SOIL PROPERTIES

From CERM Table 35.7, find the mass of solids first, then solve for the

porosity:

3227.35 lbs

1 1 0.17t

s

mm

w

27.351 1 0.57

2.55 0.4 62.4s

t w

mn

SG V

(Answer D)





CONCRETE

In concrete, the resisting force from the steel and effective concrete

block should be in equilibrium. Therefore, C T .

C internal force from the effective compression block of concrete

T internal force from the tension of the steel rebar

'0.85

0.85 3 12 3.33 40

4.35 in

c s y

C T

f ab A f

a

a

(Answer A)

a

22 in

3.33 in²

12 in

C

T

Resisting moment

a

0.85fc’

Asfy

Given:

fc’ = 3 ksi

fy = 40 ksi





MATERIAL TEST METHODS

Mohr’s Circle

65 psi

45 psi

X

First, draw the Mohr’s circle of the problem then solve for x:

tan

45tan29

65

16.18 psi

x

x

x

Then solve for the cohesion:

tan

tan2916.18

8.97 9 psi

c

x

c

c

(Answer C)

Y

Normal stress (σ)

Shear stress (τ)

C 29°





COMPACTION

First, we solve for the field dry unit weight:

2.2

129.41 pcf0.017

moist

W

V

129.41

114.52 pcf1 1 0.13

moistdry

w

Then we solve for the relative compaction:

,max

114.520.9543 95.43%

120d

c

d

R

(Answer D)





SOIL CLASSIFICATION

In classifying the soil using USDA, we need to exclude the percentage

of gravel.

New sand 8%

10%100% 20%

New silt 48%

60%100% 20%

New clay 24%

30%100% 20%

Then draw the lines along the USDA chart. The soil falls in the SILTY

CLAY LOAM area.

(Answer D)





STRUCTURAL STEEL

WT 13.5 x 108.5:

d = 14.215 in

y = 3.11 in Ix = 502 in4

Bottom fiber

Top fiber

Solve using the bending stress formula:

b

Mcf

I

Bending stress on top fiber:

150 12 3.11

11.15 ksi502

b

Myf

I

Bending stress on bottom fiber:

150 12 14.215 3.11

39.82 40 ksi502

b

M d yf

I

The maximum bending stress is taken as the larger of the two

values. Therefore, the bending stress on bottom fiber is the

maximum one.

(Answer B)

Neutral axis





EXCAVATION

Use the formula for excavations:

1 2 3 42 3 4

4

AV h h h h

Where:

1

2

3

4

20.3 15.1 6.9 5.9 48.2 ft

18.4 12.5 19 20.5 12.5 8.8 6.23 12.8 110.73 ft

0 ft

14.1 17.1 8.4 14.1 53.7 ft

h

h

h

h

Solve for the volume:

1 2 3 4

3

2 3 44

60 60 48.2 2 110.73 3 0 4 53.7

4

436,014 ft

AV h h h h

(Answer C)

B C D A

I

II

III

IV

60 ft

60 ft

60 ft

60 ft 60 ft 60 ft

4 2 4

4 2

2

2 4

1 2 2 1

1 2 1 2





CONSTRUCTION SITE LAYOUT

First, plug all the data given into the appropriate field. You can do this

using the figure or by making a table. I’ve solved this using the figure:

Station BS FS Elev

BM1 8.56 540.51

TP1 9.45 4.23

BM2 3.15 ?

8.56 ft 4.23 ft

Elevation= 540.51 ft

BM1

BM2

TP 1

3.15 ft 9.45 ft

HI1=8.56+549.07

HI2=9.45+544.84=554.29

Elevation = 549.07 - 4.23 = 544.84 ft

Elevation = 554.29 - 3.15 = 551.14 ft

After figuring out the elevations at each location, take the elevation

at BM2 and subtract it from BM1’s elevation to find the difference:

551.14 540.51 10.63 ft

(Answer C)





CONSTRUCTION SITE CONTROL

160 ft 160 ft

R2-R1=1.6 ft R2-R1=3.2 ft

Use the equation for stadia reading to find the interval factor, K:

x K R c

160 1.6 Eq. 1

320 3.2 Eq. 2

K c

K c

Where:

X = distance from telescope

K = interval factor

R = reading on rod (between top and bottom reading)

C = instrument factor

Substitute the first equation into the second and solve for K:

100K

(Answer C)





SAFETY

Typical permissible noise exposure levels range, but according to

OSHA, sec 1910.95, for 8 hours it is 90 dBa. Chapter 82 of the CERM

covers some construction and jobsite safety issues with OSHA

standards included.

(Answer A)


CONSTRUCTION SITE LAYOUT AND CONTROL

From the figure below, the angle of intersection is:

180 40 65 75

(Answer D)

N

E

S

W

40°

65°

65°

θ

40°




SCORE SHEET

Correct Answers: ___________

Percentage: (correct answers)/40 = ____/40 = ______

Again, shoot for above 90% and you’ve got it. You’ve now finished 2

test and you’re more prepared to pass the real thing! Keep practice as

much as possible to get more exposure to all kinds of problems.

That’s the secret sauce. Good luck on the real deal and in your career

as a civil engineer. If you need more tips, tricks, and even a review

course then check out www.civilpereviewcourse.com and

www.civilengineeringacademy. We would also love to hear about your

successes, failures, or any way we can improve things to help future

students. Good luck on your exam and in your career as a

professional civil engineer!




LAST SECOND ADVICE AND TIPS

I wanted to wrap up this exam with some tips that I found helpful

when I took the exam. Hopefully, they will help you:

1) Make sure you fully understand what your state board and the

NCEES requires of you to take the PE exam and to receive your

PE. Comply with all rules.

2) You typically need about 3-4 months to really study for the PE.

Map out a schedule and study your depth section first. This will

allow you to make any adjustments as you get closer to test

time. I personally broke down the specification into the 5 major

categories of: water resources, transportation, geotechnical,

structural, and construction, and spent a couple weeks on each

topic with more time devoted to my depth section.

3) Practice everything with the calculator you are going to use on

the real exam. You need to become intimately familiar with it.

4) Know where your exam is, where to park, and where you will get

food (if you don’t plan on bringing your own). Don’t be stuck

trying to figure this out on test day. You’ll regret it.

5) Review courses help. If you can’t get motivated, or need the

extra help and accountability they offer, then consider taking

one. It’s worth it to get your PE and get that boost to your

career. If you’re wondering which one, refer to our helpful tools

below.

Helpful Tools:

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

engineer become a PE. 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 practice exams that cover

your depth section. In addition to this, we have created a civil PE

review course that can guide you step-by-step through the entire

exam. You can check that out at www.civilpereviewcourse.com.



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