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STRUCTURAL DEPTH EXAM

Isaac Oakeson, P.E.

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!

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THE ULTIMATE

STRUCTURAL DEPTH EXAM

http://www.civilengineeringacademy.com


TABLE OF CONTENTS

WELCOME…………………………………………….

EXAM SPECIFICATION……………………

START TEST ………………………………...

SOLUTIONS………………………………………

LAST SECOND TIPS AND ADVICE…..

Page 6

Page 9

Page 50

Page 96



WELCOME!!

Welcome to The Ultimate Structural Depth Exam! Thank you so much

for purchasing this book!

This exam covers 40 questions and solutions that are found on the

structural depth exam. This exam is built to meet the NCEES

specifications. It is designed to have a similar look and feel to the real

exam.

The depth exam is typically a much harder exam than the breadth.

You should be spending quite a bit of your time in your depth section

so that you can score well. Keep in mind though, that if you can crush

the morning portion then the afternoon will be much easier on you as

they add both scores for your final score. Having said that though,

you should be spending 60 to 70% of your time in your depth section.

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 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 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!)

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LEGAL INFORMATION

Civil Engineering Academy’s

The Ultimate Structural Depth Exam

Isaac Oakeson, P.E.

Rights and Liability:

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

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

giving credit where it should be given by 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

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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.



Civil – Structural Depth Exam NCEES Specification

I. Analysis of Structures 14 Questions

A. Loads and load applications 4 Questions

1. Dead loads

2. Live loads

3. Construction loads

4. Wind loads

5. Seismic loads

6. Moving loads (e.g., vehicular, cranes)

7. Snow, rain, ice

8. Impact loads

9. Earth pressure and surcharge loads

10. Load paths (e.g., lateral and vertical)

11. Load combinations

12. Tributary areas

B. Forces and load effects 10 Questions

1. Diagrams (e.g., shear and moment)

2. Axial (e.g., tension and compression)

3. Shear

4. Flexure

5. Deflection



6. Special topics (e.g., torsion, buckling, fatigue,

progressive collapse, thermal deformation, bearing)

II. Design and Details of Structures 20 Questions

A. Materials and material properties 5 Questions

1. Concrete (e.g., plain, reinforced, cast-in-place, precast,

pre-tensioned, post-tensioned)

2. Steel (e.g., structural, reinforcing, cold-formed)

3. Timber

4. Masonry (e.g., brick veneer, CMU)

B. Component design and detailing 15 Questions

1. Horizontal members (e.g., beams, slabs, diaphragms)

2. Vertical members (e.g., columns, bearing walls, shear

walls)

3. Systems (e.g., trusses, braces, frames, composite

construction)

4. Connections (e.g., bearing, bolted, welded, embedded,

anchored)

5. Foundations (e.g., retaining walls, footings, combined

footings, slabs, mats, piers, piles, caissons, drilled shafts)

III. Codes and Construction 6 Questions

A. Codes, standards, and guidance documents 4 Questions

1. International Building Code (IBC)

2. American Concrete Institute (ACI 318, 530)



3. Precast/Prestressed Concrete Institute (PCI Design

Handbook)

4. Steel Construction Manual (AISC)

5. National Design Specification for Wood Construction

(NDS)

6. LRFD Bridge Design Specifications (AASHTO)

7. Minimum Design Loads for Buildings and Other

Structures (ASCE 7)

8. American Welding Society (AWS D1.1, D1.2, and D1.4)

9. OSHA 1910 General Industry and OSHA 1926

Construction Safety Standards

B. Temporary structures and other topics 2 Questions

1. Special inspections

2. Submittals

3. Formwork

4. Falsework and scaffolding

5. Shoring and reshoring

6. Concrete maturity and early strength evaluation

7. Bracing

8. Anchorage

9. OSHA regulations

10. Safety management



START TEST



1. A truss system is simply supported over a 45 ft span as shown in

the figure below.

The modulus of elasticity for all members is 30,000 ksi and the

cross-sectional area of all members is 10 in2. Determine the axial

force occurred in member BD and solve for the elongation of

member BD due to this axial force.

a) 125 kips (tension); 0.125 in

b) 125 kips (compression); 0.125 in

c) 175 kips (tension); 0.175 in

d) 175 kips (compression); 0.175 in



2. Which of the following load is not considered as lateral load:

a) Earthquake

b) Wind

c) Rain

d) Soil pressure



3. A simply supported beam is given a uniformly distributed dead

load 3 kips/ft and a concentrated live load 10 kips in the mid-

span of AB. Determine the ultimate shear force of this beam

(use load combination – strength design on ASCE 7).

a) 32 kips

b) 42 kips

c) 52 kips

d) 62 kips



4. Solve for maximum deflection of this cantilever beam. Given

that the modulus of elasticity is 3,500 ksi and the cross-

sectional area is 8” x 12”.

a) 0.62 in

b) 0.73 in

c) 0.82 in

d) 0.93 in



5. From the figure shown below, find the normal stress of member

AB.

a) 0.4 psi (tension)

b) 0.5 psi (tension)

c) 0.6 psi (compression)

d) 0.7 psi (compression)



6. Which of the following statements is true?

a) Maximum bending moment occurred at a point where maximum

shear force occurred.

b) Maximum bending moment occurred at a point where zero shear

force occurred.

c) Maximum bending moment occurred at a point where any

supports (hinge, fixed, roller) are located.

d) Maximum bending moment always occurred at the mid-span of a

beam.



7. Determine which of the following structures are considered as

statically indeterminate structure.

a) 1, 2, 3, 4

b) 1, 3, 4, 5

c) 1, 2, 3, 5

d) 2, 3, 4, 5

1 2

3 4 5



8. Solve for the influence line of vertical support reaction at point B

(VB) for the structure below.

a)

b)

c)

d)



9. Determine the suitable location for distributed load to obtain

maximum positive bending moment at point I. Note: use influence line

to solve this problem.

a) Segment A-B

b) Segment B-C

c) Segment C-D

d) Segment A-D



10. A retaining wall is built with section shown in the figure below.

Calculate the total soil pressure against the retaining wall at depth of

15 ft.

a) 345 psf

b) 445 psf

c) 545 psf

d) 645 psf



11. A structure below is in equilibrium condition. The length of

member BE is 36 ft, while the length of member DF is 48 ft. Member

BE and DF have the same material properties and same cross-

sectional area. Determine the internal force in member DF shown in

the following figure. Note that the member AD is very stiff, and the

compatibility requirement must be satisfied.

a) 22500 lb

b) 27500 lb

c) 32500 lb

d) 37500 lb



12. A 6” x 10” beam has an elastic modulus of 3,500 ksi. Calculate

the maximum deflection of this simply supported beam.

a) 0.7 in

b) 0.8 in

c) 0.9 in

d) 1 in



13. A statically determinate structure with additional hinge at point C

is shown in the figure below. Determine the vertical reaction at point

B.

a) 36.67 kips

b) 46.67 kips

c) 56.67 kips

d) 66.67 kips



14. A portal frame is illustrated in the figure below. Which of the

following segments are having no shear force?

a) AC and DE

b) CD and BF

c) EF and FG

d) All options are false



15. According to ACI 318-14, the design of reinforced concrete uses

LRFD concept, which has load factor and resistance factor. Determine

the resistance factor of strength design for shear, tension,

compression without spiral reinforcement, torsion. Note: write them by

order.

a) 0.75; 0.9; 0.65; 0.75

b) 0.65; 0.75; 0.9; 0.75

c) 0.9; 0.65; 0.75; 0.75

d) 0.75; 0.9; 0.75; 0.65



16. A reinforced concrete beam with single reinforcement has a cross-

sectional view as follows.

The compressive strength of concrete is 4000 psi and the steel

reinforcement uses Grade 60. Determine the design moment strength

of this beam section.

a) 80 kips-ft

b) 100 kips-ft

c) 120 kips-ft

d) 140 kips-ft



17. A steel profile ASTM A992 W18x50 is placed to sustain a

concentrated load and distributed load as shown in the figure below.

Note that the applied loading has been factored.

Determine if the steel profile is compact or non-compact, then judge if

the steel profile is enough to sustain the loading (use LRFD design

based on AISC). Check the compactness only, by assuming the beam

is laterally supported.

a) Compact section, sufficient

b) Compact section, insufficient

c) Non-compact section, sufficient

d) Non-compact section, insufficient



18. A customized steel section is formed by welding plates (PL8x½)

to the top flange and bottom flange of steel profile W12x96. Determine

the increasing inertia moment about its strong and weak axis for this

customized steel section.

a) 21% and 30%

b) 21% and 41%

c) 30% and 15%

d) 41% and 15%



19. Design the required spacing of #3 stirrups for this reinforced

concrete beam (8”x16”) displayed in the figure below.

The compressive strength of concrete is 3000 psi and the steel

reinforcement uses Grade 60.

a) 7 in

b) 9 in

c) 11 in

d) 13 in



20. The elastic modulus of concrete depends on ….

a) Section width

b) Section height

c) Effective depth

d) Compressive strength



21. A HSS column 12” x 12” x 3/8” is designed to resist axial

compression force with no eccentricity. The HSS section uses A36

material based on ASTM specification. The length of column is 15 ft,

assume that the effective length factor is 1. Determine the design

compression load for this column based on LRFD method.

a) 400 kips

b) 480 kips

c) 500 kips

d) 580 kips



22. A section of slab and beam is shown in the following figure. The

slab and beams are cast monolithically, so that the concrete beam can

be considered as a T-beam section. The length of the beam is 13 ft.

Determine the top flange effective width of this interior T-beam

according to ACI 318.

a) 8 in

b) 14 in

c) 24 in

d) 47 in



23. A 12 in x ½ in of thick plate shown in the figure below is

subjected to tensile loading. Calculate the net area due to 5 x ¾” bolt

(bolt hole 13/16”) attached in the thick plate according to AISC.

a) 4.675 in2

b) 4.775 in2

c) 4.875 in2

d) 4.975 in2



24. A corbel structure is designed to support an ultimate vertical load

of 55 kips at distance of 6 in from the face of column (see figure

below). The compressive strength of normal-weight concrete is 4000

psi and the steel reinforcement uses Grade 60. Calculate the required

main steel reinforcement area in the corbel based on ACI 318.

a) 0.83 in2

b) 0.87 in2

c) 0.92 in2

d) 0.98 in2



25. Which of the following is the adjustment factor for glulam

application only using the ASD method according to NDS?

a) Temperature factor

b) Load duration factor

c) Wet service factor

d) Beam stability factor



26. Determine the minimum steel reinforcement applied to a one-way

slab with a thickness of 3.5 in. The steel reinforcement used is Grade

60.

a) #3 – 10”

b) #3 – 12”

c) #3 – 17”

d) #3 – 18”



27. A W-shape column (W14x109 from ASTM A992) is subjected to

ultimate axial force of 450 kips, ultimate bending moment X-axis of

300 kips-ft, and ultimate bending moment Y-axis of 100 kips-ft. The

length of the column is 15 ft. Assume that the column has a pinned-

pinned connection at the ends. By using the LRFD method, calculate

the interaction ratio and determine if the section is sufficient enough to

sustain the loading or not.

a) Not enough information

b) 0.94; sufficient

c) 0.98; sufficient

d) 1.01; not sufficient



28. A simply supported beam is made of Douglas Fir-Larch No. 1 and

has a cross section of 4” x 10”. The applied distributed load is a dead

load. Determine the allowable bending stress according to NDS (use

the ASD method) and judge if the beam is sufficient enough or not.

a) 880; not sufficient

b) 990; not sufficient

c) 1038; sufficient

d) 1138; sufficient



29. A retaining wall made of reinforced concrete is subjected to soil

pressure. The height of the wall is 15 ft, while the soil pressure is

equivalent to a 45 lb/ft3. The compressive strength of concrete is 5000

psi and the steel reinforcement used is Grade 60. Calculate the steel

reinforcement area attached in the wall if the steel percentage is

determined as 1%.

a) 0.9 in2

b) 1.0 in2

c) 1.1 in2

d) 1.2 in2



30. Which of the following statement is false?

a) Concrete and steel are isotropic, while wood is orthotropic.

b) Concrete is good at resisting tension forces, while steel is good at

resisting compression forces.

c) ACI (concrete), AISC (steel), and NDS (wood) adopt the LRFD

method.

d) Wood has many design values based on its species and location.



31. Determine the minimum size of fillet welds if we weld plate 10” x

½” and plate 10” x 3/8” together.

a) 1/8”

b) 3/16”

c) 1/4”

d) 5/16”



32. Seismic resistant design adopts the concept of capacity design,

which requires the column strength should be larger than the beam

strength (strong column weak beam). Therefore, the sway mechanism

(see figure below) will be considered as the expected failure

mechanism.

According to the ACI 318, please specify the minimum multiplication

factor of beam strength to be the minimum column strength to ensure

the expected failure mechanism.

a) 1.0

b) 1.1

c) 1.2

d) 1.3



33. Which of the following statements is not the reason of prestress

loss in prestressed concrete?

a) Imperfection of prestressed steel stress

b) Creep of concrete

c) Relaxation of prestressed steel stress

d) Shrinkage of concrete



34. Some exterior reinforced concrete columns in a building are

designed as edge columns due to their positions. The typical edge

column of 10” x 10” (hatched area) and the pile cap of a group pile is

displayed in the following figure.

If the edge column transfers 500 kips, determine the maximum axial

compression sustained by one pile.

a) 125 kips

b) 260 kips

c) 340 kips

d) 425 kips



35. A 15-ft simply supported rafter is made of Douglas Fir-Larch No. 2

with a cross section of 3” x 8”. Determine the allowable distributed

snow load which can be sustained by the rafter (ignore the self-weight

of rafter) according to NDS (use ASD method).

a) 0.1 kips/ft

b) 0.2 kips/ft

c) 0.8 kips/ft

d) 0.9 kips/ft



36. A combined footing displayed in the following figure is subjected

to column force of 500 kips each.

Calculate the maximum bearing pressure due to this load. Note that

the self-weight of combined footing (normal-weight 0.15 kips/ft3)

should be included too.

a) 5 kips/ft2

b) 8 kips/ft2

c) 11 kips/ft2

d) 14 kips/ft2



37. According to OSHA, what is the height limit from an unprotected

edge working condition for employees to do connecting activities such

as welding, bolting, cutting, bracing, etc., where fall protection is

required?

a) 5 ft

b) 6 ft

c) 7 ft

d) 8 ft



38. In determining the seismic design category, an engineer should

classify the site class of its structure based on the site soil properties.

According to the ASCE, which site class should be used if the soil

properties are unknown?

a) Site class A

b) Site class D

c) Site class E

d) Site class F



39. A masonry structure is designed using the ASD method. Based on

the IBC, determine most nearly the minimum length of lap splice

(reinforcement splice) if the 3/8” rebar and Grade 40 steel is used in

the masonry structure. Assume that the structure is designed until the

rebar reaches its yield stress.

a) 12 in

b) 15 in

c) 25 in

d) 30 in



40. A simply supported prestressed beam (PCI standard rectangular

beam 16RB24) is designed by using PCI guideline. Assume that the

effective depth is 21 in. The compressive strength of normal-weight

concrete is 6000 psi. Determine the required prestressing steel area if

the prestressing steel uses low-relaxation strand with 270 ksi.

a) 1.0 in2

b) 1.5 in2

c) 2.0 in2

d) 2.5 in2



PROBLEM SOLUTIONS



PROBLEM 1 SOLUTION:

AXIAL FORCE

Based on structure’s equilibrium, we can obtain the vertical reaction at

point B by using the following equation.

( )

=

− =

=

0

300 15 45 0

100 kips

A

B

B

M

V

V

Draw free-body diagram at point B and check the equilibrium of Y

direction at point B:

( )

=

− =

− =

=

0

40

5

4100 0

5

125 kips compression

y

B BD

BD

BD

F

V F

F

F

After obtaining the axial force in the member BD, calculate the

elongation of member BD by using the following equation. Note that 1

ft equals to 12 in.

( )

=

=

=

125 25 12

30000 10

0.125 in

BD BDBD

BD

BD

F L

EA

(Answer B)



PROBLEM 2 SOLUTION:

LOADS

Lateral load is a type of loading which is applied horizontally to the

structural system. This kind of load causes lateral deformation of

structure (horizontal deformation).

Rain load is determined as the applied loading due to some amount of

water accumulated on a roof from blockage of the primary drainage

system. Therefore, rain load is considered as one of gravity loads.

(Answer C)



PROBLEM 3 SOLUTION:

SHEAR FORCE

Based on ASCE 7, the load combination – strength design involving

dead load (D) and live load (L) shall be designed by using:

1.2D + 1.6L

Based on structure’s equilibrium, we can obtain the vertical reaction at

point B by using the following equation.

( )

=

+ − =

=

0

1.6 12 1.2 36 18 24 0

105.2 kips

A

L D B

B

M

P w V

V

Then, the vertical reaction at point A can be obtained by considering the

equilibrium of vertical forces.

( )

=

+ = +

+ = +

=

0

1.2 36 1.6

105.2 1.2 3 36 1.6 10

40.4 kips

Y

A B D L

A

A

F

V V w P

V

V

Now we can solve the ultimate shear force ( uV ) by calculating at each

point.

Point A :

= = 40.4 kipsu A

V V

Point D :

= − = −1.2 w 12 2.8 kipsu A D

V V (right before the concentrated live

load)



= − − = −1.2 w 12 1.6 18.8 kipsu A D L

V V P (right after the

concentrated live load)

Point B :

= − − = −1.2 w 24 1.6 62 kipsu A D L

V V P (right before vertical

support)

= − − + =1.2 w 24 1.6 43.2 kipsu A D L B

V V P V (right after vertical

support)

Point C :

= 0 (free-end)u

V

From the calculations above, we can determine that the ultimate shear

force is occurred at point B right before vertical support (62 kips).

(Answer D)



PROBLEM 4 SOLUTION:

DEFLECTION

In the case of cantilever beam, the maximum deflection is occurred in

the free-end point. Therefore, we should calculate the deflection in

point B.

The maximum deflection at the free-end can be calculated as follows.

( )

( )

=

=

=

4

4

3

8

5 4 12

18 3500 8 12

12

0.82 in

B

B

B

wL

EI

(Answer C)



PROBLEM 5 SOLUTION:

NORMAL STRESS

Before we calculate the normal stress AB , we need to obtain the

force of element AB ( ABF ) first.

From the free-body diagram, we can find = − + = −80 30 50 lbAB

F .

=

=

=

50

6 12

0.7 psi (compression)

AB

AB

AB

AB

AB

F

A

(Answer D)



PROBLEM 6 SOLUTION:

BENDING MOMENT

By understanding the relationship between the shear force diagram

and the bending moment diagram, it is clear that the bending moment

diagram represents the area under the shear force diagram. The

relationship between shear force (Q) and bending moment (M) is

explained in this equation:

dMQ

dx=

Where x represents a segment of structure.

Therefore, through the mathematical way, we can get the result:

At every point the shear force equals to zero, the bending moment will

have inflection point (either maximum or minimum).

(Answer B)



PROBLEM 7 SOLUTION:

BASIC CONCEPTS OF STRUCTURES

A structure is called as statically determinate structure if the equilibrium

equation can solve all the unknown internal forces of the structure.

Otherwise, the structure is considered as a statically indeterminate

structure when the equilibrium equations are not sufficient to solve the

unknown internal forces of structure. This will need the compatibility

and constitutive equation to solve the indeterminate structure problem.

Hint: Use (m + r – 2j) for truss and (3m + r – 3j) for frame structures.

m = # of members; r = # of reactions; j = # of joints/nodes

Degree = m + r – 2j = 16 + 3 – 2(8) = 3 Degree = 3m + r – 3j = 3(1) + 6 - 3(2) =3

Degree = 3(3) + 4 - 3(4) =1 Degree = 3(2) + 3 - 3(3) =0 Degree = 3(1) + 4 - 3(2) =1

(Answer C)

1 2

3 4 5



PROBLEM 8 SOLUTION:

INFLUENCE LINE - LOADS

When we solve for influence line, we should assume that unit load of 1

is moving along the structure.

The easiest way is do the influence line by logic. If we put a unit load

of 1 at point A, the vertical reaction at point B will be zero. Otherwise,

if we put a unit load of 1 at point B, the vertical reaction at point B will

be 1.

(Answer D)



PROBLEM 9 SOLUTION:

INFLUENCE LINE - LOADS

In order to obtain the maximum positive bending moment at point I,

we need to find the influence line for bending moment at point I.

Influence line for VB:

Influence line for VC:

Influence line for MI :

By looking at the influence line for MI, it can be seen that the

maximum positive bending moment will be reached if the distributed

load is placed at segment B-C.

(Answer B)



PROBLEM 10 SOLUTION:

EARTH PRESSURE LOAD

Based on Rankine theory,

2 2 33tan 45 tan 45 0.2948

2 2a

K

= − = − =

Active earth pressure at depth = 15 ft:

0.2948 100 15 442.2 psf 445 psfa a

K z = = =

(Answer B)




COMPATIBILITY

Compatibility equation:

3 6

2

2

4836

2

2

3

B D

DB

BE BE DF DF

DFBE

BE DF

l l

ll

P l P l

EA EA

PP

P P

=

=

=

=

=

Equilibrium equation:

0

30000 6 3 6 0

C

BE DF

M

P P

=

− + + =

Substitute PBE from compatibility equation into the equilibrium equation.

230000 6 3 6 0

3

8 180000

22500 lb

DF DF

DF

DF

P P

P

P

− + + =

=

=


(Answer A)



SUPERPOSITION METHOD - DEFLECTION

We can solve this problem by separating the loading, and then

calculate the deflection of each loading case, and finally we look at the

superposition of both deflections.

1. The concentrated load of 30 kips

( )3

3

13

30 10 120.62 in

14848 3500 6 10

12

PL

EI

= = =

2. The distributed load of 2 kips/ft

( )

= = =

4

4

23

25 10 12

5 12 0.26 in1384

384 3500 6 1012

wL

EI

Total deflection is the summation of both deflections.

= + = 1 2

0.88 in 0.9 in

(Answer C)




SUPPORT REACTION

Structure

Equilibrium condition:

0

80 (Eq.1)

V

A B D

F

V V V

=

+ + =

0

0 (Eq.2)

H

A

F

H

=

=

0

16 8 30 4 50 12 0

2 90 (Eq.3)

A

D B

B D

M

V V

V V

=

+ − − =

+ =

Hinge condition:

0 (right side)

6 50 2 0

16.67 kips

C

D

D

M

V

V

=

− =

=

Substitute VD to Eq.3 to obtain VB :

2 90 (Eq.3)

2 16.67 90

56.67 kips

B D

B

B

V V

V

V

+ =

+ =

=

(Answer C)




SHEAR FORCE DIAGRAM

Solve the free-body diagram, and then we can clearly see which

segment having no shear force.

(Answer B)




Strength resistance factor (ACI 318-14):

Shear → 0.75 =

Tension → 0.9 =

Compression without spiral reinforcement → 0.65 =

Torsion → 0.75 =

(Answer A)




CONCRETE MOMENT DESIGN

Given values in the problem:

'

2

4000 psi

60000 psi

3#6 3 0.44 1.32 in

20 2.5 17.5 in

c

y

s

f

f

A

d

=

=

= = =

= − =

Solve for under-reinforced condition:

'

0

0.85

0.85 4000 12 1.32 60000

1.94 in

c s y

H

C T

f ab A f

a

a

=

=

=

=

=

Design moment strength:

( )

( )

10.92

10.9 1.32 60000 17.5 1.942

1178258.4 lb-in=98.2 kips-ft

n s y

n

n

M A f d a

M

M

= −

= −

=

(Answer B)




STEEL DESIGN

The ultimate moment for design is:

2 230 20 3 20300 kips-ft

4 8 4 8u u

u

P L w LM

= + = + =

Check the compactness of the section through equations or tables.

Based on the steel profile table, this section is compact.

Given values in the question:

50 ksiASTM A992

65 ksi

y

u

f

f

=

=

3101 in W18x50 propertiesx

Z = →

Design moment strength:

0.9 0.9 101 50 4545 kips-in=378 kips-ftn x y

M Z f = = =

(sufficient)n u

M M

(Answer A)

Note:

In the case of no lateral support, the unbraced length of the beam must

be checked, and the lateral torsional buckling must be considered in the

nominal moment strength calculation. The easiest way to check this is

by checking AISC Table 3-10 (Available Moment Strength vs Unbraced

Length).




BUILT-UP STEEL SECTION

According to the steel section table, the properties of a W12x96 are

given:

2

4

4

28.2 in

12.7 in

833 in

270 in

x

y

A

d

I

I

=

=

=

=

Customized steel section:

3 2 41833 2 8 0.5 8 0.5 6.605 1182 in

12x

I

= + + =

3 41270 2 8 0.5 312 in

12y

I

= + =

The moment of inertia about the strong axis increased 41%, while the

moment of inertia about the weak axis increased 15%.

(Answer D)




STIRRUPS DESIGN

Solve the structural analysis first, and then make the shear force

diagram.

Based on the shear force diagram, the design shear force is 25 kips.

Shear strength of concrete:

( )'2 2 1 3000 8 0.9 16 12.6 kipsc c w

V f b d= = =

Required shear strength of stirrups:

25

12.6 20.73 kips0.75

us c

VV V

= − = − =

For the concrete of 3000 psi, we can obtain the constant based on

ACI Sec. 11.4.6.3.

'0.75 41.1 psi

50 psi

cf

=



50 psi =

The required spacing of stirrups:

0.9 167.2 in

2 2

0.22 6000033 in

50 8

0.22 60000 0.9 169.1 in

20.73 1000

v y

w

v y

s

d

A fs

b

A f d

V

= =

= =

= =

Use stirrups #3 with spacing of 7 in (from center to center).

(Answer A)




CONCRETE PROPERTIES

Elastic modulus of concrete is related to the compressive strength of

concrete ( '

cf ) and the normal-weight/density of concrete ( c

w ). It can be

clearly seen in the equation.

1.5 '33 (psi)c c c

E w f= or '57000 (psi)

c cE f=

(Answer D)




STEEL COMPRESSION MEMBER

Find the section properties of HSS 12” x 12” x 3/8” from steel profile

table.

2

4.73 in

16 ing

r

A

=

=

Check slenderness ratio:

( )1 15 12 360.43 1.5

4.73 29000

y

c c

FKL

r E

= = = →

Calculate the critical stress:

2 20.430.658 0.658 36 33.32 ksic

cr yF F

= = =

Design compression load:

0.9 0.9 33.32 16 479.8 kips 480 kipsn cr g

P F A = = =

(Answer B)




T-BEAM

ACI 318-2014 Sec. 6.3.2 limits the effective flange width of T-beam

section as follows:

+ = + =

+ = + =

+ = + =

,int

13 128 47 in

4 4

min 16 8 16 4 72 in

8 144 152 in

w

e w s

w

Lb

b b h

b s

The effective width of this T-beam is 47 in.

(Answer D)




BOLTS

The net hole diameter:

1 13 1 7 in= 0.875 in

16 16 16 8net h

d d= + = + =

Equation needed to calculate net width due to bolt holes:

2

4n net

sw w d

g= − +

Pattern 1 (A-B-E-F): 12 2 0.875 10.3 inn

w = − =

Pattern 2 (A-B-C-D-E-F): 2 22.5 2.5

12 4 0.875 9.75 in4 2.5 4 2.5

nw = − + + =

2.5

2.5

32

2.5

2

PP

A

B

C

D

E

FG

2.5

2.5

32

2.5

2

PP

A

B

C

D

E

FG



Pattern 3 (A-B-C-D-G): 22.5

12 3 0.875 10 in4 2.5

nw = − + =

Pattern 4 (A-B-D-E-F):2 22.5 2.5

12 3 0.875 10.28 in4 5.5 4 2.5

nw = − + + =

The net width is taken as the minimum of these 4 possible patterns, so

that the net width is 9.75 in.

29.75 0.5 4.875 inn n

A w t= = =

(Answer C)

2.5

2.5

32

2.5

2

PP

A

B

C

D

E

FG

2.5

2.5

32

2.5

2

PP

A

B

C

D

E

FG




CORBEL DESIGN

The required steel area for shear force:

2550.87 in

0.75 60 1.4u

vf

y

VA

f = = =

Based on ACI 318, the corbel must be designed for a tension force of at

least 0.2u

V .

0.2 0.2 55 11 kipsuc u

N V= = =

Therefore, the required steel area for tension force:

2110.25 in

0.75 60uc

n

y

NA

f= = =

The corbel is also designed to resist bending moment.

( ) ( )55 6 11 14 12 352 kips-inu u uc

M V e N h d= + − = + − =

Therefore, the required steel area for bending moment:

23520.73 in

0.9 0.75 60 0.9 12u

f

y

MA

f d= = =

From those 3 factors affecting required steel area, we can calculate the

total required main reinforcement area as follows.



2

2

'2

0.73 0.25 0.98 in

2 2 0.87max 0.25 0.83 in

3 3

0.04 0.04 4 12 140.448 in

60

f n

vfs n

c

y

A A

AA A

f bd

f

+ = + =

= + = + =

= =

The required main reinforcement area is 0.98 in2.

(Answer D)




GLULAM

Based on NDS Table 5.3.1 (the adjustment factor of glulam), the

following are the adjustment factors (ASD design):

1. Load duration factor

2. Wet service factor

3. Temperature factor

4. Beam stability factor

5. Volume factor

6. Flat use factor

7. Curvature factor

8. Stress interaction factor

9. Shear reduction factor

10. Column stability factor

The load duration factor is the only adjustment factor for glulam

following the ASD method.

(Answer B)




CONCRETE SLAB REINFORCEMENT

The minimum reinforcement of a one-way slab:

Assume that we calculate for a slab width of 1 ft.

( ) 2

,min0.0018 0.0018 1 12 3.5 0.0756 in /ft

sA bh= = =

The required spacing of steel reinforcement (use rebar #3):

20.25 0.37512 17.5 in

0.0756s

= =

Check the maximum spacing for slab reinforcement based on ACI

(7.6.5):

= ==

3 5 3.5 10.5 inmin

18 in

hs

From the calculations above, the steel reinforcement use #3-10”.

(Answer A)




COMBINED AXIAL AND MOMENT FOR STEEL DESIGN

Given value in the questions:

50 ksiASTM A992

65 ksi

y

u

f

f

=

=

Check the manual table for available strength of axial and moment:

1220 kipsc c n

P P= = (Manual Tables, Table 4-1)

= = 705 kips-ftcx nx

M M (Manual Tables, Table 3-10)

= = 348 kips-ftcy ny

M M (Manual Tables, Table 3-4)

Check interaction ratio:

4500.37 0.2

1220u

c n

P

P= =

+ +

+ +

81

9

8 300 1000.37 1

9 705 348

1.0036 1 (NOT OK)

ryr rx

c cx cy

MP M

P M M

The steel interaction ratio for the combined axial and moment is 1.0036,

which means that the column is not sufficient enough to resist the

combined load.

(Answer D)




WOOD BEAM DESIGN

Maximum actual bending stress due to distributed load:

2 2

max

3.75 1267.5 kips-ft

8 8

wLM

= = =

2 233.5 9.25

50 in6 6

x

bhS

= = =

max 67500 1216200 psi

50b

x

Mf

S

= = =

Allowable bending stress based on the wood section:

1000 psi (NDS Supplement Table 4A)b

F =

Adjustment factors:

0.9 (permanent-dead load)D

C =

1 (moisture content < 19%)M

C =

1t

C =

1L

C =

1.1F

C =

1r

C =

' 1000 0.9 1 1 1 1.1 1 990 psibx b D M t L F r

F F C C C C C C= = =

Actual stress > allowable stress (not sufficient)

(Answer B)




RETAINING WALL DESIGN

Assume that the width design is 1 ft.

Design lateral load at the base:

1.7 1.6 45 15 1 1080 lb/ft (triangular load)u a

w hb= = =

Design bending moment at the base:

1 1 151080 15 40500 lb-ft

2 3 2 3u u

hM w h

= = =

Effective depth of wall:

2

'

2

40500 12 0.9 1 0.59

60000486000 0.01 60000 12 1 0.59 0.01

5000

8.52 in

u n

y

y

c

M M

ff bd

f

d

d

=

= −

= −

=

Steel reinforcement area:

2in0.01 12 8.52 1.02

fts

A bd= = =

(Answer C)




BASIC CONCEPT OF CONSTRUCTION MATERIAL

The wrong option is - ‘concrete is good to resist the tension force,

while steel is good to resist the compression force’.

Concrete is very good at its compressive strength, while steel is very

good at its yield tensile strength. Therefore, reinforced concrete is

made to ‘replace’ the weakness of concrete in resisting tension; a

composite structure is made to ‘replace’ the weakness of steel

(buckling) in resisting compression.

(Answer B)




WELD CONNECTION

Based on Table J2.4 AISC, the minimum size of fillet welds for material

thickness (the thinner between two plates) between ¼” and ½” is 3/16”.

(Answer B)




SEISMIC RESISTANT DESIGN

To satisfy a strong column weak beam, ACI 318 states that the flexural

strength of the column shall satisfy this equation:

( )6 5nc nb

M M

Where nbM is the sum of the nominal flexural strength of the beams

framing into the joint and ncM is the sum of the nominal flexural

strength of the columns framing into the joint.

It can be seen that the minimum multiplication factor is 6/5.

(Answer C)




PRESTRESSED CONCRETE

ACI 318 specifies the sources of loss of prestressed as follows.

1. Prestressing steel seating at transfer

2. Elastic shortening of concrete

3. Creep of concrete

4. Shrinkage of concrete

5. Relaxation of prestressing steel stress

6. Friction loss due to intended or unintended curvature in post-

tensioning tendons

Therefore, option A is not the reason of loss of prestressed in

prestressed concrete.

(Answer A)




PILE FOUNDATION

The maximum axial compression occurs in the pile where the location is

the farthest from the column centroid. Therefore, we also need to

include the effect of eccentricity in calculating the axial compression.

Effect of eccentricity (from column centroid to pile cap centroid):

5500 3 1292 kips-ft

12

5500 3 1292 kips-ft

12

x

y

M

M

= − =

= − =

The maximum axial compression in the pile:

( )

max 2 2

max 2 2

max

500 1292 1.5 1292 1.5

4 4 1.5 1.5

340 kips

x x y y

i i

M e M ePP

n x y

P

P

+= +

+

+ = +

+

=

(Answer C)




WOOD DESIGN

Allowable bending stress based on the wood section:

900 psi (NDS Supplement Table 4A)b

F =

Adjustment factors:

1.15 (two months-snow load)D

C =

1 (moisture content < 19%)M

C =

1t

C =

1L

C =

1.2F

C =

1r

C =

' 900 1.15 1 1 1 1.1 1 1242 psibx b D M t L F r

F F C C C C C C= = =

Maximum actual bending stress due to distributed load: 2 2

32.5 7.2521.9 in

6 6x

bhS

= = =

( )2

max

115 12

121000

8

15411 psi21.9

b

x

w

Mf w

S

= = =

Allowable snow load: '

1242 15411

0.0806 kips/ft

bx bF f

w

w

=

=

(Answer A)




COMBINED FOOTING

Total forces sustained by the footing:

500 500 0.15 24 5 2.5 1045 kipsP = + + =

Effect of eccentricity (from column centroid to footing centroid):

( )

( )1

2

500 12 3 4500 kips-ft

500 12 5 3500 kips-ft

M

M

= − =

= − =

( )1 2

1 2

4500 35001 ft

500 500

M Me

P P

+ −+= = =

+ +

The maximum bearing pressure in the footing:

max

max 2

2 2

max

1045 1045 1

5 24 5 24

6

10.9 kips/ft 11 kips/ft

P P ef

A S

f

f

= +

= +

=

(Answer C)




OSHA REGULATION

Per OSHA, anything 6 ft or more should have fall protection when they

are doing a connecting activity, such as welding, bolting, cutting, etc.

(Answer B)




SITE CLASS CATEGORY

Site F is used for soils vulnerable to potential failure or collapse under

seismic loading, such as liquefiable soils, quick and highly sensitive

clays, and collapsible weakly cemented soils.

Site Class E is used for a site does not qualify under the criteria for Site

Class F and there is a total thickness of soft clay greater than 10 ft.

Site Class D shall be used where the soil properties are not known in

sufficient detail to determine the site class, unless the authority having

jurisdiction or geotechnical data determines Site Class E or F soils are

present at the site.

Site Class A is used for hard rock conditions with a depth of 100 ft.

Therefore, the correct answer is site class D.

(Answer B)




REINFORCEMENT SPLICE

The minimum length of lap splices for reinforcing bars in tension or

compression:

30.002 0.002 20000 15 in

8

max 12 in

340 40 15 in

8

b s

d

b

d f

l

d

= =

=

= =

Considering that the structure is designed to reach yield stress, which is

more than 80% of the allowable tensile stress, the lap splice should be

increased by 50%.

Therefore, the reinforcement splice in this case should have a minimum

length of 23 in. and the answer is most nearly:

(Answer C)




PRESTRESSED BEAM DESIGN

Maximum bending moment:

2 21 112 20 600 kips-ft

8 8u u

M w L= = =

Find the required '

uK :

' 2

' 2

'

12000

16 21600

12000

1020

u n

u p

u

u

u

M M

K bdM

K

K

=

=

=

=

Determine pu

by using PCI Fig. 4.10.2 (the values of '

uK ):

'

'

'

0.20 10141020 0.202 (interpolation)

0.21 1052

pu u

u pu

pu u

KK

K

= → = = → =

= → =

The required prestressing steel area:

'

20.202 16 21 61.51 in

270

u p c

ps

pu

bd fA

f

= = =

(Answer B)



SCORE SHEET

Correct Answers: ___________

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

Some things to note:

If you can absolutely crush the morning exam then the depth section

will be much easier on yourself. You should still shoot for a high score

here, but just know if you can get in the upper 90% in the morning

then you just need to score more than 20 correct in the afternoon.

Imagine if you got all 40 correct in the morning – you’d only need to

get 16 right on your depth exam (56/80=70%)! A total of 70% is

about the passing score for the PE. What’s interesting though, is that

the depth section will dominate your study time. So, put in the time

and practice and you will get there. Keep at it!



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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