the ultimate - amazon s3...find the ultimate bearing capacity for the square footing given the...

CIVIL PE BREADTH EXAM

Isaac Oakeson, P.E.Volume 1

THE ULTIMATE

HELPING YOU ON YOUR JOURNEY TO BECOMING A PROFESSIONAL ENGINEER.

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 1

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 43

Page 82




WELCOME!!

Welcome to The Ultimate Civil PE Breadth Practice Exam Volume 1!

Thank you so much for purchasing this book!

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

exam makes the effort to be updated to 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. Afterwards, 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 and sub-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!).

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, and any other Civil Engineering Academy exam,

can be found at www.civilengineeringacademy.com/errata.




LEGAL INFORMATION

Civil Engineering Academy’s

The Ultimate Civil PE Breadth Exam Vol. 1

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 1)




1. Choose the correct activity-on-arrow diagram for the following

activities:

Activity Predecessor

A -

B -

C A, B

D B

2. The method of overlapping the design and construction phases of

a project is called:

a) Crashing

b) PERT

c) Fast Tracking

d) Precedence mapping




3. Details of a beam are given below:

If steel reinforcement has a unit weight of 0.283 lbf/in3, how much

does the steel cost for a 3-ft length of beam given that the steel

has a price of $0.75 per lbf?

a) $13.20

b) $18.00

c) $17.45

d) $34.70




4. A 400 ft long fill section is planned for a future embankment on a

construction site. The areas of fill sections are 75 feet apart. The

total volume of earthwork between stations 13+00 to 16+00 is

most nearly:

STATION AREA (ft2)

13+00 347.33

13+75 647.24

14+50 974.50

15+25 702.14

16+00 358.67

a) 7436 yd3

b) 8956 yd3

c) 9041 yd3

d) 3647 yd3




5. A 15 ft wide x 18 ft deep trench in a clay soil is temporarily braced

as shown in the figure below. The longitudinal spacing of the

struts is 6 ft. Calculate the axial load in strut 1. 700 psfucS and

120 pcf .

a) 5.5 kips

b) 18 kips

c) 24.5 kips

d) 33.5 kips




6. A 40-ton crane has four extendable outriggers that weigh 85,000

lbs. The pad of each outrigger has a diameter of 24 in. The soil

has an allowable soil contact pressure of 2500 psf. What area of

cribbing is needed under each outrigger if each outrigger sees the

total load?

a) 60 ft2

b) 65 ft2

c) 70 ft2

d) 75 ft2




7. Which of the two mechanisms listed below are found to be the

major cause of crane failure?

a) Proper use of outriggers and ground conditions

b) Proper ground conditions and leveling

c) Stability and structural capacity

d) All of the above

8. Find the effective overburden pressure at a point 10 ft below the

water level.

a) 2313 psf

b) 2451 psf

c) 2461 psf

d) 3076 psf




9. A 3 ft x 3 ft square footing carries a load of 3000 lbs including its

weight. The base of the footing is at 2 ft below the ground. The

sandy soil has a unit weight of 145 lb/ft3. Find the ultimate

bearing capacity for the square footing given the following bearing

capacity factors:

23 37 20q cN N N

a) 10368 lb/ft2

b) 9351 lb/ft2

c) 12470 lb/ft2

d) 8786 lb/ft2




10. A retaining wall is shown in the figure below. The wall supports a

mass of cohesionless soil with a normal density of 85 lb/ft3, a

saturated density of 100 lb/ft³, a void ratio of 0.65, and an angle

of shearing resistance of 30°. The top of the wall is level with the

horizontal surface of the soil. By ignoring the wall friction,

determine the total earth thrust on the wall.

a) 8800 lbs

b) 5957 lbs

c) 6347 lbs

d) 12720 lbs




11. A set of group piles has been determined to be placed for a

foundation. Which of the following is not true?

a) Piles bearing on rock virtually do not settle.

b) Piles in granular soils experience minimal settlement after

placement.

c) Piles in cohesive soils may experience very little settlement.

d) For piles in clay, the capacity is taken as the sum of the individual

capacities.

12. Slope stability in saturated clay is found using the Taylor Slope

Stability chart. Which of the following statements are not true

when using it to solve for slope stability?

a) There are no surcharges or tension cracks assumed.

b) For slopes less than 53°, toe circle failures occur.

c) For slopes less than 53°, slope circle failures occur.

d) Shear strength is derived from cohesion only and is constant with

depth.




13. An 18 ft layer of clay is found between a sand layer of 3 ft and a

gravel layer of 5 ft. How long will it take to achieve a degree of

consolidation of 80%? Use vC = 4 ft2/year.

a) 3650 days

b) 16,790 days

c) 14,235 days

d) 4198 days




14. A truss system is shown below. Compute the force on member BC.

a) 0 kips

b) 75 kips

c) 70 kips

d) 460 kips




15. A cantilever solid steel shaft 6 inches in diameter and 7 ft long is

subject to torque as shown. The modulus of rigidity is 50109

lb/ft2. What is the maximum shearing stress in the shaft?

a) 2852 kips/ft2

b) 1623 kips/ft2

c) 2037 kips/ft2

d) 3213 kips/ft2




16. The loads on a structure (dead and live) are shown below. Solve

for the closest approximate vertical reaction at B.

a) 138 lbs

b) 107 lbs

c) 55 lbs

d) 90 lbs




17. What is the moment diagram most nearly for the load shown

below?




18. Solve for the maximum bending stress on a beam which has 1 ft

width, 3 ft depth and 20 ft length. Assume the beam is simply

supported and loaded with a 200 kips/ft uniform load.

a) 151 kips/ft2

b) 9995 kips/ft2

c) 5667 kips/ft2

d) 6667 kips/ft2




19. A newly installed building has a fixed span on the second floor that

is subject to a 2 kips/ft load. A W24x84 beam has been sized for

the span (I=2370 in4, E=29,000 ksi). What is the maximum

deflection on the beam if we neglect the self-weight of the beam?

a) 1 in

b) 0.002 in

c) 0.34 in

d) 1.68 in




20. A trapezoidal canal with sides sloping 45° has a base width of 5 ft.

If the flow is has a 7 ft width, what is the hydraulic radius?

a) 0.77 ft

b) 0.81 ft

c) 0.74 ft

d) 0.69 ft




21. What is the theoretical velocity of flow through a sharp edged

orifice located 10 ft below the water surface?

a) 25.38 ft/s

b) 24.87 ft/s

c) 25.72 ft/s

d) 26.74 ft/s

22. Which of the following statement is NOT TRUE about a detention

basin?

a) It is designed to store water temporarily as part of a flood control

management system.

b) If the flow exceeds the storage capacity an uncontrolled structure

provides an outlet for the excess water.

c) They are designed with no conservation pool.

d) It is used to hold water for an extended period of time and have a

conservation pool.




23. Two pipes, 1 and 2, having the following properties are connected

in series:

Pipe 1: length = 4500 ft; diameter = 1 ft; f = 0.025

Pipe 2: length = 6000 ft; diameter = 0.75 ft; f = 0.020

If these two pipes are to be replaced by a single pipe with length

of 7000 ft, what will be the diameter? Use f = 0.018.

a) 4.70 ft

b) 0.73 ft

c) 1.20 ft

d) 0.91 ft




24. Water is pumped through a 1.6 ft diameter circular pipe which is

2500 ft long at a velocity of 7.0 ft/s. Which of the following most

nearly gives the head loss using Manning’s formula if n = 0.013?

a) 52.8 ft

b) 17.4 ft

c) 29.5 ft

d) 31.6 ft




25. A detention basin has been in service for 50 years. What is the

probability that a 2% flood occurred within the 50 year service?

a) 66%

b) 64%

c) 74%

d) 57%

26. The total net precipitation of a watershed was found to be 2.75 in.

The most downstream culvert of the area saw 45,000 ft3 of water

pass through it during the storm event. What is the total acreage

of the watershed area?

a) 3.9 acres

b) 4.1 acres

c) 4.5 acres

d) 0.38 acres




27. A reversed curve of equal radii connects two parallel tangents 30

ft apart. The length of chord from P.C. to P.T. is 420 ft. Determine

the total length of the reversed curve.

a) 419.57 ft

b) 420.35 ft

c) 421.45 ft

d) 426.75 ft




28. A car traveling at a constant speed approached a hazard on the

road and stepped on the brake. If the car travelled 80 ft from

perception of the hazard to the time the driver steps on the brake,

what is the perception-reaction time of the driver given that the

car has a constant speed of 30 mph.

a) 1.8 sec

b) 2.1 sec

c) 3.4 sec

d) 4.5 sec




29. Given the following information, find the stationing at the low

point of the curve.

a) 21+47

b) 22+42

c) 23+67

d) 24+77




30. A field density test on a compacted fill of sandy clay has the

following results:

Volume of test hole = 0.015 ft3

Weight of moist soil from the hole = 2 lbs

Oven-dried weight of the soil = 1.75 lbs

Maximum Dry Density = 118 lb/ft3

Optimum Moisture Content = 10%

Solve for the percent compaction of the soil.

a) 99%

b) 100%

c) 89%

d) 72%




31. Refer to the following test data to solve this problem:

% passing #200 sieve 47

Liquid Limit 57

Plastic Limit 15

The AASHTO classification of this soil is:

a) A-7-5

b) A-7-6

c) A-5

d) A-6




32. Cohesive borrow material has just been transported and spread to

the desired lift thickness at a fill site. What type of compaction

equipment is best suited for this job?

a) Smooth shell roller

b) Mesh roller

c) Sheepsfoot roller

d) Tamping foot roller

33. Stress versus strain curves are shown for various materials. Which

corresponds to a hard and strong material?




34. An engineer is using the following boring log to design a caisson

foundation. What is the standard penetration resistance for the

soil at a depth of 23 ft?

a) 4

b) 8

c) 3.7

d) 44




35. Given the figure below, determine the discharge. Water is flowing

perpendicular to the soil layers.

Soil 1: 0.00065 ft/sKx



Cross sectional area = 2.5 ft2

a) 0.00089 ft3/s

b) 0.00075 ft3/s

c) 0.00066 ft3/s

d) 0.00022 ft3/s




36. The ground makes a uniform slope of 3.2% from Station 1+980 to

Station 2+120. At Station 1+980, the center height of the road is

1 m fill. At Station 2+120, the center height of the road is 3.2 m

cut. Find the grade of the finished road.

a) 0.2%

b) 0.5%

c) 0.01%

d) 0.02%




37. A road with dimensions of 100 ft length, 16 ft width, and 2 ft thick

needs to be constructed. The moisture content of the compacted

road needs to be 15% with a wet density of 110 lb/ft3. If the

borrow pit has a soil with a wet density of 100 lb/ft3 and a

moisture content of 10%, what is the required volume of soil from

the borrow pit that needs to be hauled?

a) 3270 ft3

b) 3520 ft3

c) 3370 ft3

d) 3450 ft3




38. Choose the correct construction sequence:

a) Pre-mobilization, Layout, Excavation, Footing Concreting

b) Layout, Footing Concrete, Excavation, Pre-mobilization

c) Layout, Pre-mobilization, Excavation, Footing Concrete

d) Pre-mobilization, Excavation, Layout, Footing Concrete

39. 200 ft3 of soil from a borrow pit was excavated, transported by

truck to the construction site, and finally compacted. The soil has

a swelling factor of 0.2 and shrinkage factor of 0.15. What was the

volume of the soil while being transported by the truck?

a) 170 yd3

b) 7.6 yd3

c) 8.9 yd3

d) 6.3 yd3




40. If a construction worker is entering a confined space, what is the

amount of oxygen content that must be maintained?

a) 10.5%

b) 15.5%

c) 20%

d) 5.0%




VOLUME 1 SOLUTIONS




PROBLEM 1 SOLUTION:

PROJECT PLANNING

ACTIVITY IDENTIFICATION AND SEQUENCING

This is a complicated problem because of the “dummy activity” that

must be included in order to maintain the correct order and

precedence. Activity D depends on B only (drawn as answer A), but C

depends on A and B, which answer C might confuse you with, but it is

wrong because that means activity C happens twice and it doesn’t. In

order to draw this correctly you have to have a “dummy activity” to

show the activities in order of precedence. Answer D is just wrong.

(Answer B)

PROBLEM 2 SOLUTION:

PROJECT PLANNING

PROJECT SCHEDULES

Besides efficient scheduling, construction time can be compressed with

fast-track scheduling. This particular method overlaps the design and

construction phases of a project.

(Answer C)




PROBLEM 3 SOLUTION:

PROJECT PLANNING

COST ESTIMATING

Obtain the total mass of steel:

For the links:

For every 3-ft, there will be:

Number of links = 3’ length / 8” spacing or (36”/8”)

= 4.5 ≈ 5 links

Total length per link = 8" 2 16" 2 20" 2 4" 2 96"/link

Since there are 5 links on a 3-ft length of beam:

Total length = 96"/link 5 links 480"

To obtain the volume, multiply the total length by the area:

Volume =

2

33480 in in 53.0144 in

4 8

For the main bars:

Since there are 10 bars on a cross section:

Total volume for 3-ft length of steel,

Volume =

2

3510 bars 36 in in 110.4466 in

4 8

Total Volume of steel considering links and main bars:




Total Volume = 3 3 353.0144 in 110.4466 in 163.461 in

Total Mass = Total Volume density

= 3 3163.461 in 0.283 lbf/in 46.2595 lbf

Calculate the price:

Total Price = $0.75/lbf 46.2595 lbf $34.69

(Answer D)




PROBLEM 4 SOLUTION:

PROJECT PLANNING

QUANTITY TAKE-OFF METHODS

You can solve this by using the trapezoidal method or you can solve

this by intuitively thinking about it. I’m using the later here.

First, let’s draw it out in plan view and look at the tributary areas of

each fill zone:

Now just multiply everything out and convert the volume to yd3:

3 3

347.33 37.5 647.24 75 974.50 75 702.14 75 358.67 37.5

200766 ft 7435.8 yd

(Answer A)




PROBLEM 5 SOLUTION:

MEANS AND METHODS

TEMPORARY STRUCTURES AND FACILITIES

For this temporary structure we must first test what type of soil we are

in. If H

c

is greater or equal to 6 then we are dealing with soft clay.

c (cohesion) = 1 1

700 350 psf2 2

ucS

H is total depth of the trench.

120 186.17 6 it is a soft clay

350

H

c

The lateral pressure is linear for the top 4.5 ft4

H of the depth and

then stays at a constant ( maxP ) after that.

max4 120 18 4 350 760 psfP H c

10.5 760 psf 4.5 ft 1710 lb/ftR

2760 psf 4.5 ft 3420 lb/ftR




Make a hinge at strut 2 and take the moment about strut 2. We will

include the 6 ft longitudinal spacing here too:

2

1 2 1

1

1

0

6 2.25 6 0

1710 6 3420 2.25 6 0

2992.5 lb/ft

sM

R R S

S

S

Because the longitudinal spacing is 6 ft, then:

12992.5 lb/ft 6 ft 17955 lb 18 kipsS

(Answer B)




PROBLEM 6 SOLUTION:

MEANS AND METHODS

CONSTRUCTION LOADS

First, find the total load:

Crane = 40 tons x 2000 lbs = 80,000 lbs

Outriggers = 85,000 lbs

Total = 165,000 lbs

We were given a soil pressure of 2500 psf. Therefore, the area of the

cribbing needed is:

2 2165000 lbsArea 66 ft 70 ft

2500 psf

(Answer C)

It should be noted that each outrigger sees the total load. If this

wasn’t the case we would have divided the load by 4 outriggers.




PROBLEM 7 SOLUTION:

MEANS AND METHODS

CONSTRUCTION METHODS

Cranes can fail two different ways which are by stability and structural

capacity; and within those 2 categories you will find every option listed

in the answers.

(Answer D)

PROBLEM 8 SOLUTION:

SOIL MECHANICS

EFFECTIVE AND TOTAL STRESS

At point 10 ft below the water level you are placed 5 ft into the clay

layer. That means you need to calculate the stress in 3 different layers

with 2 of those being in water. The dry layer is total stress and the

saturated layers will be the effective stress.

Total stress = H

Effective stress = 'H where 'w

Most sand layer 3 2130 lb/ft 15 ft 1950 lb/ft

Saturated Sand Layer 3 3 2135 lb/ft 62.4 lb/ft 5 ft 363 lb/ft

5’ into clay layer 3 3 290 lb/ft 62.4 lb/ft 5 ft 138 lb/ft

Total effective overburden pressure 21950 363 138 2451 lb/ft

(Answer B)




PROBLEM 9 SOLUTION:

SOIL MECHANICS

BEARING CAPACITY

Using Terzaghi’s ultimate bearing capacity equation for a square

footing:

1.3 0.425u c f qq cN D N BN

Where:

c = cohesion

fD = depth to base of footing

B = width of footing

1.3 = cN multiplier value for a square footing

0.425 = N multiplier value for a square footing ( 0.85 0.5 )

Remember that sandy soil has no cohesion. Therefore, the first

component of the equation will be zero.

3 3

2

0 145 lb/ft 2 ft 23 0.425 145 lb/ft 3 ft 20

10368 lb/ft

u

u

q

q

(Answer A)

Note:

The value of 3000 lbs mentioned in the problem is meant to confuse

you. You are solving for what loads the soil can take. Because the load

is 23000 lbs

333 lb/ft3 ft 3 ft

, this soil will be fine.




PROBLEM 10 SOLUTION:

GEOTECHNICAL

LATERAL EARTH PRESSURE

1 sin 1 sin30 1

1 sin 1 sin30 3aK

(Don’t forget the active earth pressure!)

aK is the active earth pressure coefficient and is only applied to the

soil, not to the water.

3 3 2

2

1100 lb/ft 62.4 lb/ft 10 ft 125.3 lb/ft

3a

P K H

3 2

362.4 lb/ft 10 ft 624 lb/ft

wP H

Solving for the forces (areas of each):

2

3

1125.3 lb/ft 10 ft 1 ft 626.5 lbs

2F

2

4

1624 lb/ft 10 ft 1 ft 3120 lbs

2F

(Answer B)





SOIL MECHANICS

FOUNDATION SETTLEMENT

Piles in cohesive soils my, in fact, experience significant settling. The

rest of the answers are true.

(Answer C)


SOIL MECHANICS

SLOPE STABILITY

The only statement that is not true (false) in the answer selection is B.

The Taylor chart shows that toe circle failures occur in slopes steeper

than 53°.

(Answer B)





SOIL MECHANICS

SOIL CONSOLIDATION

If a clay layer is between a sand and gravel layer this means that the

soil is double draining. This makes H in the following equation halved.

Using the table for time factors (table 40.1 in CERM) for 80%

compaction you get 0.567vT . Plug everything into the following

equation and solve:

2 20.567 911.5 years or 4198 days

4v

v

T Ht

C

Note:

t = time to consolidate

vT = time factor

H = height of clay layer

vC = coefficient of consolidation

(Answer D)





STRUCTURAL MECHANICS

TRUSSES

Solve for the reactions first and then make a cut at section Z-Z:

0

21 100 14 200 10.5 100 7 420 5 0

100 kips

A

v

v

M

G

G

0

21 100 14 200 10.5 100 7 420 5 0

300 kips

G

v

v

M

A

A

0

420 0

420 kips

H

H

H

F

A

A




0

3 3.5 3 0

420 3 300 3.5 3 0

70 kips

H

H v

M

A A BC

BC

BC

(Answer C)

The sign convention doesn’t matter in this question because we are

just looking for the magnitude of the reaction not whether it is in compression or tension.






SHEAR STRESS

Solve for the torque between each section (start from point D to A):

Torque at point D = -50 kip-ft

Torque at point C = -50 + 40 = -10 kip-ft

Torque at point B = -10 - 60 = -70 kip-ft

Torque at point A = -70 kip-ft

From the torque diagram, the maximum torque = 70 kip-ftABT .

Calculate the maximum shear stress:

First, derive equation for circular shaft design from CERM Eq. 45.50.

maxmax max

4 3

0.5 16

32

D TrT TS

J D D

Therefore, the maximum shear stress is:

2max

3 3

16 16 702852 kips/ft

0.5

TS

D

(Answer A)






LOADS

Solve for the angle and length of EB using Pythagorean Theorem:

2 2 25 10.6

11.72 ft

EB

EB

5

tan 25.2510.6

Solve for the force F:

length of 20 11.72 234.4 lbsF w EB

Now take the moment about point A (CCW as positive):

0

21.2 120 10 sin 7.5 cos 15.9 0

21.2 120 10 234.4sin25.25 7.5 234.4cos25.25 15.9 0

137.8 lbs

A

y

y

y

M

B F F

B

B

(Answer A)

Note: The force F is broken into its X and Y components

for ease of solving for its moment arm about point A.






MOMENT DIAGRAM

Let’s first solve for the reactions (this is the long way so solve this

thing). Take + as counter clockwise:

0

22.5 kips 6 ft 9 ft 0

15 kips

A

B

B

M

F

F

0

15 kips 22.5 kips 0

7.5 kips

Y

A

A

F

F

F

From the shear diagram 0 to 5.2 ft, the

value of shear is decreasingly positive,

so the slope of the moment diagram is

also decreasingly positive (dm/dx=V).

Everything past 5.2 ft on the shear

value is increasingly negative so the

slope of the moment is increasingly

negative. The short way is to look at

the Appendix 44A of the CERM for a

triangular load case It should help with

other scenarios too.

(Answer C)

Note: -22.5 kips is found by taking

the area of a triangle (0.5BH) for the

distributed area and applying it at a

distance 1/3 from the base.

Where the shear is 0, the moment has a

maximum. The area under the curve

leading up to 0 is the maximum value for

the moment. Because the load is linear, the

shear diagram is of the second degree (X2)

and the moment diagram would be the

n+1 degree of the shear (X3).

The distance of 3.46 ft was

found by using the method of

sections and getting the

equation of a line.

Solve for when V=0 by FY

and solve for x:

7.5 – 0.5(5(x/9))x = 0

x = 5.2 ft






BENDING

To solve for the maximum bending stress, first find the max moment,

which is at the mid-span of a simply supported beam loaded uniformly

throughout its length.

2

2

max

200 kips/ft 20 ft10000 kips-ft

8 8

wLM

Then solve for the maximum bending stress using the formula for max

bending stress on a rectangular cross-section:

2max

2 2

6 6 10000 kips-ft6666.67 kips/ft

1 ft 3 ftb

Mf

bd

(Answer D)






DEFLECTION

We are given from the problem statement the following:

w = 2 kips/ft

E = 29,000 ksi

I = 2370 in4

Solve for deflection with this size of beam and a fixed-fixed end

connection:

4

4

max 4

2 kips/ft5 40 ft 12 in/ft

5 12 in/ft1.676 in

384 384 29000 ksi 2370 in

wL

EI

(Answer D)





HYDRAULICS AND HYDROLOGY

OPEN CHANNEL FLOW

Solve for the depth d and y:

7 ft 5 ft1 ft

2x

sin45

1 1 ft

2

1.4142 ft

d

y

y

y

tan45

11 ft

1 ft

d

x

d

d

Solve for Hydraulic Radius (R):

21Area 5 ft 1 ft 1 ft 1 ft 2 6 ft

2

Perimeter 5 ft 2 5 ft 2 1.4142 ft 7.8284 fty

26 ft

0.766 ft7.8284 ft

AR

P

(Answer A)






ENERGY EQUATION

Derived from the Bernoulli’s Equation:

2 (Eq 17.66)tv gH

Sharp edged orifice 0.98vC

→ (Note that Cv only used for actual velocity)

2 2 32.2 10 25.38 ft/stv gH

(Answer A)



DETENTION/RETENTION PONDS

Detention basins are used to store water temporarily as part of a flood

control system. They have no conservation pool. The only false answer

is D. A retention basin is used to hold water for an extended period of

time.

(Answer D)






PRESSURE CONDUIT

For pipes in series:

1 2

1 2

eq

eq

HL HL HL

Q Q Q

Using Darcy’s equation:

2

2

5

2 2

5 5

2

(1)

2

0.025

(2)

0.025 0.025 4500 0.025 0.020 6000

1 0.75

15.45

eq

L VHL f

D g

f L QHL

D

Q QHL

Q

For the equivalent pipe:

2

2

5

0.025 0.018 700015.45

0.73 ft

eqHL HL

QQ

D

D

(Answer B)

(CERM Eq. 17.28)






LOSSES

Using the Manning’s Equation, the slope can be found:

21.64 0.4 ft

1.6

AR

P

2 13 2

2 13 2

1.49

1.497 0.4

0.013

0.01266

V R Sn

S

S

To get the head loss, use the slope and length of the pipe:

head loss

length of pipe

head loss0.01266

2500

head loss 31.64 ft

S

(Answer D)






STORM CHARACTERISTICS

Note (see Eq. 20.20):

1% = 100 year frequency



Probability (50 year flood in 50 years) = 50

1 1 0.02 0.64 64%

(Answer B)



RUNOFF ANALYSIS

Solve by using rational method (Eq. 20.36):

Q CiA

C isn’t given in the problem, so it is assumed to be 1.

I = 2.75 in/time

Q = 45,000 ft3/time

3

2

in 1 ft45,000 ft /time 1 2.75

time 12 in

196363.64 ft 4.5 acres

Q CiA

A

A

(Answer C)





GEOMETRICS

BASIC CURVE ELEMENTS

In triangle (a):

30

sin 4.096420

2 8.192I

In triangle (b):

105sin

105sin4.096

1470 ft

R

R

R

Use the length of curve equation:

1470 ft 8.192

2 2 420.35 ft180 180

c

RIL

(Answer B)





GEOMETRICS

TRAFFIC VOLUME

Even without the knowledge of transportation engineering, this

problem can be solved by equations of motion. Using the equation for

constant velocity (since the driver is yet to step on the brake):

ft 1 hr80 30 mph 5280

mile 3600 sec

1.82 1.8 sec

ox v t

t

t

(Answer A)





GEOMETRICS

BASIC VERTICAL CURVE ELEMENTS

First, find the stationing at the low point of the curve and then solve

for the station at that location:

1 2 1

2 1

1

and

2450 1950 500 5 stations

2.5% 3.5%1.2% per station

5 stations

3.5%Sta 2.9166 291.67 ft

1.2% / station

turn

turn

G G GX R

R L

L

G GR

L

GX

R

Now add the station to BVC to get the location of the low point:

,1950 291.67 2241.67 ft Sta 22 42

turn sta turn staX BVC X

(Answer B)





MATERIALS

COMPACTION

Solve for the dry density:

3

3

1.75 lb116.67 lb/ft

0.015 fts

dry

w

V

Dry Density 116.67Percent Compaction 98.87%

Maximum Dry Density 118

(Answer A)





MATERIALS

SOIL CLASSIFICATION

Using the AASHTO classification system (CERM Table 35.4) you will

find the following:

47% passes the #200 sieve so that limits the choices to A-4, A-5, A-6,

A-7.

The LL is 57 and the PL is 15 so the PI (Plasticity Index) = LL-PL = 42.

That leaves you in the A-7 column.

Now you have to break it down to either A-7-5 or A-7-6.

If the PL is less than 30 it is an A-7-6, and for a PL that is 30 or

greater it would be A-7-5. Remember that the PL = LL-PL.

(Answer B)





MATERIALS

COMPACTION

Sheepsfoot rollers are best suited for cohesive soils. Smooth wheel

rollers are good for all kinds of soils but are mostly used for

compacting asphalt pavements. Mesh rollers are used for rocky soils

and gravel. The tamping foot roller is best used for compacting fine-

grained soils.

(Answer C)


MATERIALS

MATERIAL TESTING

“A” is for a soft and weak material, “B” is for a weak and brittle

material, “C” is for a strong and tough material.

(Answer D)





MATERIALS

BORING LOG INTERPRETATION

Standard penetration resistance, or blow count, is the number of blows

required to drive a sampler (140 lb hammer) a distance of 12 in. On

boring logs, it is the last two values added up (last 12 in of 18 in

bored). Option A is the last number in the blows measured (3-4-4).

Option C is the average of all three measurements and option D is just

a random number.

The purpose of this test is to get the relative density of the soil.

(Answer B)





MATERIALS

SOIL PROPERTIES

You must solve for an equivalent coefficient of permeability (not in the

CERM but see page 21-2).

1 2 3

31 2

1 2 3

equivalent

H H HKy

HH H

K K K

; 1 1 2 2 3 3

1equivalent

Kx H K H K H KH

Since flow is perpendicular to the soil layer us Ky:

4

4 4 4

2 2 27.13 10 fps

2 2 2

6.5 10 6 10 10 10

equivalentKy

30.5

6

HiL

(in this case, H is the difference in water levels)

2.5A (given)

4 37.13 10 0.5 2.5 0.00 ft /s089 Q KiA

(Answer A)





SITE DEVELOPMENT

CUT AND FILL

Consider the slope of the ground = +3.2%

Since the run is 140 m:

rise = run x slope = 140 m x 0.032 = 4.48 m

Cut is 3.2 m in the end station.

Solve for x:

x = 4.48 m – 3.2 m = 1.28 m

Knowing that at the starting station, there is a fill of 1.0 m, and the

end station of the road is 1.28 m, the end station of the road is higher

0.28 m than the starting station.

Therefore, solving for the slope of the road:

Slope of the road rise 0.28 m

0.2%run 140 m

(Answer A)





SITE DEVELOPMENT

EXCAVATION AND EMBANKMENT

Volume of the compacted soil length width height

3

100 16 2

3200 ft

The only constant between the two states of soil even after

compaction is the mass of solids (sM ).

Find the sM on the compacted state:

110 pcf95.65 pcf

1 1 0.15wet

dryw

395.65 pcf

3200 ft

306087 lbs

s sdry

s

M M

V

M

(also equal to the sM of the soil in the borrow pit)

Using the moisture content (w) of the soil in the borrow pit to find

mass of water (wM ):

0.1306087 lbs

30608.7 lbs

w

s

w

w

Mw

M

M

M

Therefore, the total mass of the soil in the borrow pit is s wM M :

306087 30608.7 336695.7 lbstotalM




Note:

It can be noticed that the required volume of soil from the borrow pit

is greater. This is because after hauling, the soil will be compacted,

thus reducing its volume.

Utilizing the wet density of the soil in the borrow pit, the volume of the

soil needed to be hauled can be solved:

3

336695.7 lbs100 pcf

3367 3370 ft

totalwet

M

V

V

V

(Answer C)





SITE DEVELOPMENT

CONSTRUCTION SITE LAYOUT AND CONTROL

The correct answer in this sequence is A. The others are out of order

for proper a proper construction sequence.

(Answer A)


SITE DEVELOPMENT

SEDIMENT CONTROL

The volume of the soil while being transported is larger since it swells.

3

3 3 3 3

Volume in truck volume on the borrow pit 1 swell factor

200 ft 1 0.2

240 ft 8.9 yd (1 yd = 27 ft )

(Answer C)

FYI - The volume of the soil after compaction is smaller since it shrinks

after compaction.

3

3 3 3 3

Volume after compaction volume on the borrow pit 1 shrinkage factor

200 ft 1 0.15

170 ft 6.3 yd (1 yd = 27 ft )

Remember that the given factors are for the original volume and NOT

the factor for the intermediate volumes.





SITE DEVELOPMENT

SAFETY

Based on OSHA 1926.651, if an employee is entering a confined space

there must be 19.5% oxygen content or higher. The correct answer is

C, 20%. See CERM Construction and Job Site Safety as well.

(Answer C)




SCORE SHEET

Correct Answers: ___________

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

Remember that you should shoot for above 90%. Don’t be

discouraged if you aren’t there yet. Just keep practicing and you’ll get

there. I promise. You need approximately 70% to pass the PE exam.

So if you can ace the morning you need only get half right in the

afternoon to get it. Say you got 35 right in the morning then you

would need to get 21 right in the afternoon to get 70% (56/80 =

70%).




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.



the ultimate - amazon s3...find the ultimate bearing capacity for the square footing given the...

Documents