PE CIVIL: STRUCTURAL PRACTICE EXAMINATION
STUDY GUIDE | LATEST UPDATE 2026/2027 | ACTUAL
EXAM | PRACTICE QUESTIONS AND ANSWERS |
EXAM REVIEW | 100% CORRECT ANSWERS |
VERIFIED SOLUTIONS
This comprehensive practice examination is designed for candidates preparing for
the NCEES Principles and Practice of Engineering (PE) examination in Civil:
Structural. It delivers a rigorous assessment of structural engineering principles,
design codes, and analysis techniques essential for professional licensure. The 100
questions span the NCEES specification, including structural analysis methods,
design of steel, concrete, wood, and masonry, foundation design, load
determination per ASCE 7, seismic and wind design, and structural detailing. Each
item blends theoretical knowledge with practical, scenario-based problem-solving,
mirroring the depth of the actual exam. Detailed rationales, four to five sentences
each, clarify the correct answer and why alternatives are incorrect. Updated for
the 2026–2027 examination cycle, this resource provides verified solutions and a
publication-quality format, making it an indispensable tool for final review and
confident exam performance.
Table of Contents
I. Structural Analysis Methods
II. Loads and Load Combinations
III. Seismic Design
IV. Wind Design
V. Steel Design (AISC 360)
VI. Concrete Design (ACI 318)
VII. Wood Design (NDS)
VIII. Masonry Design (TMS 402)
IX. Foundations and Retaining Walls
X. Structural Systems and Detailing
, 1. A simply supported steel beam with a span of 30 feet carries a uniform
dead load of 0.5 kips/ft and a uniform live load of 1.2 kips/ft. Using the
ASCE 7 load combination 1.2D + 1.6L, what is the factored design uniform
load?
A) 1.70 kips/ft
B) 2.04 kips/ft
C) 2.52 kips/ft
D) 2.88 kips/ft
Correct Answer: C
The factored uniform load is w_u = 1.2 × 0.5 + 1.6 × 1.2 = 0.6 + 1.92 = 2.52 kips/ft.
Option A simply adds the service loads without factors. Option B incorrectly uses
1.2 × 1.2 + 1.6 × 0.5. Option D multiplies both loads by 1.6. Therefore, the correct
factored load per ASCE 7 is 2.52 kips/ft.
2. A reinforced concrete beam has a width of 14 inches and an effective depth
of 22 inches. The concrete compressive strength is 4,000 psi, and the steel
yield strength is 60,000 psi. The beam is reinforced with 4 #8 bars (A_s =
3.16 in²). Using the simplified rectangular stress block per ACI 318, what is
the nominal flexural strength M_n?
A) 220 ft-kips
B) 260 ft-kips
C) 310 ft-kips
D) 360 ft-kips
Correct Answer: C
The depth of the compression block a = A_s f_y / (0.85 f'_c b) = 3.16 × 60 / (0.85 ×
4 × 14) = 189..6 = 3.98 in. Then M_n = A_s f_y (d – a/2) = 3.16 × 60 × (22 –
1.99) = 189.6 × 20.01 = 3,794 in-kips = 316 ft-kips. Option A uses a different a,
option B underestimates, option D overestimates. Thus, the nominal flexural
strength is approximately 310 ft-kips.
3. A steel column in a braced frame is 16 feet long and pinned at both ends.
The column is a W14×61 with r_y = 2.45 inches and A_g = 17.9 in². The steel
yield strength is 50 ksi. What is the nominal compressive strength P_n per
, AISC 360?
A) 450 kips
B) 580 kips
C) 650 kips
D) 720 kips
Correct Answer: B
The effective length factor K = 1.0. KL/r = 1.0 × 16 × .45 = .45 = 78.4.
The elastic buckling stress F_e = π²E/(KL/r)² = (3.14² × 29,000) / (78.4²) = 286,000 /
6,147 = 46.5 ksi. Since KL/r ≤ 4.71√(E/F_y) ≈ 113, the column is inelastic. F_cr =
0.658^(F_y/F_e) × F_y = 0.658^(50/46.5) × 50 = 0.658^1.075 × 50 = 0.638 × 50 =
31.9 ksi. P_n = F_cr × A_g = 31.9 × 17.9 = 571 kips, rounded to 580 kips. Option A
uses a smaller area, C and D overestimate. Therefore, B is correct.
4. A reinforced concrete cantilever retaining wall is 18 feet tall and retains a
granular backfill with a unit weight of 120 pcf and an angle of internal
friction of 32 degrees. Using Rankine theory, what is the active earth
pressure coefficient K_a?
A) 0.26
B) 0.31
C) 0.36
D) 0.42
Correct Answer: B
Rankine active coefficient K_a = tan²(45° – φ/2). φ = 32°, so φ/2 = 16°. K_a =
tan²(45° – 16°) = tan²(29°) = (0.5543)² = 0.307. Option A uses φ = 35°, C uses φ =
30°, D uses φ = 28°. Therefore, the correct K_a is approximately 0.31.
5. A wood beam is simply supported with a span of 12 feet and carries a
uniform total load of 300 lb/ft. The beam is a 4×10 (actual 3.5 in × 9.25 in)
of Southern Pine No. 2. The reference bending design value F_b is 875 psi.
The size factor C_F is 1.0, the wet service factor C_M is 1.0, and the load
duration factor C_D is 1.0. What is the adjusted bending design value F'_b?
A) 700 psi
B) 875 psi
, C) 1,050 psi
D) 1,200 psi
Correct Answer: B
F'_b = F_b × C_M × C_t × C_L × C_F × C_fu × C_i × C_r × C_D. With all factors except
possibly C_F given as 1.0, and C_F for a 4×10 is approximately 1.0 (size factor for
depth 9.25 in is (12/9.25)^(1/9) ≈ 1.03, but assumed 1.0 here). Thus F'_b = 875 psi.
Option A reduces unnecessarily, C and D increase without justification. Therefore,
B is correct.
6. A steel plate tension member has a gross area of 4.0 in² and a thickness of
0.5 in. It is connected with two 7/8-inch bolts in a single line. The shear lag
factor U is 0.85. What is the effective net area A_e?
A) 2.55 in²
B) 2.85 in²
C) 3.00 in²
D) 3.25 in²
Correct Answer: A
The bolt hole diameter = 7/8 + 1/8 = 1.0 in. Net area A_n = A_g – n × d_h × t = 4.0
– 2 × 1.0 × 0.5 = 3.0 in². Effective net area A_e = A_n × U = 3.0 × 0.85 = 2.55 in².
Option B ignores U, C is A_n, D uses a different gross area. Therefore, the correct
effective net area is 2.55 in².
7. A concrete column is part of a special moment frame in Seismic Design
Category D. According to ACI 318, what is the minimum transverse
reinforcement spacing at the column ends over a length l_o?
A) 4 inches
B) 6 inches
C) 8 inches
D) 12 inches
Correct Answer: A
ACI 318-19 Section 18.7.5.3 requires that the spacing of transverse reinforcement
in the plastic hinge region of special moment frame columns not exceed
STUDY GUIDE | LATEST UPDATE 2026/2027 | ACTUAL
EXAM | PRACTICE QUESTIONS AND ANSWERS |
EXAM REVIEW | 100% CORRECT ANSWERS |
VERIFIED SOLUTIONS
This comprehensive practice examination is designed for candidates preparing for
the NCEES Principles and Practice of Engineering (PE) examination in Civil:
Structural. It delivers a rigorous assessment of structural engineering principles,
design codes, and analysis techniques essential for professional licensure. The 100
questions span the NCEES specification, including structural analysis methods,
design of steel, concrete, wood, and masonry, foundation design, load
determination per ASCE 7, seismic and wind design, and structural detailing. Each
item blends theoretical knowledge with practical, scenario-based problem-solving,
mirroring the depth of the actual exam. Detailed rationales, four to five sentences
each, clarify the correct answer and why alternatives are incorrect. Updated for
the 2026–2027 examination cycle, this resource provides verified solutions and a
publication-quality format, making it an indispensable tool for final review and
confident exam performance.
Table of Contents
I. Structural Analysis Methods
II. Loads and Load Combinations
III. Seismic Design
IV. Wind Design
V. Steel Design (AISC 360)
VI. Concrete Design (ACI 318)
VII. Wood Design (NDS)
VIII. Masonry Design (TMS 402)
IX. Foundations and Retaining Walls
X. Structural Systems and Detailing
, 1. A simply supported steel beam with a span of 30 feet carries a uniform
dead load of 0.5 kips/ft and a uniform live load of 1.2 kips/ft. Using the
ASCE 7 load combination 1.2D + 1.6L, what is the factored design uniform
load?
A) 1.70 kips/ft
B) 2.04 kips/ft
C) 2.52 kips/ft
D) 2.88 kips/ft
Correct Answer: C
The factored uniform load is w_u = 1.2 × 0.5 + 1.6 × 1.2 = 0.6 + 1.92 = 2.52 kips/ft.
Option A simply adds the service loads without factors. Option B incorrectly uses
1.2 × 1.2 + 1.6 × 0.5. Option D multiplies both loads by 1.6. Therefore, the correct
factored load per ASCE 7 is 2.52 kips/ft.
2. A reinforced concrete beam has a width of 14 inches and an effective depth
of 22 inches. The concrete compressive strength is 4,000 psi, and the steel
yield strength is 60,000 psi. The beam is reinforced with 4 #8 bars (A_s =
3.16 in²). Using the simplified rectangular stress block per ACI 318, what is
the nominal flexural strength M_n?
A) 220 ft-kips
B) 260 ft-kips
C) 310 ft-kips
D) 360 ft-kips
Correct Answer: C
The depth of the compression block a = A_s f_y / (0.85 f'_c b) = 3.16 × 60 / (0.85 ×
4 × 14) = 189..6 = 3.98 in. Then M_n = A_s f_y (d – a/2) = 3.16 × 60 × (22 –
1.99) = 189.6 × 20.01 = 3,794 in-kips = 316 ft-kips. Option A uses a different a,
option B underestimates, option D overestimates. Thus, the nominal flexural
strength is approximately 310 ft-kips.
3. A steel column in a braced frame is 16 feet long and pinned at both ends.
The column is a W14×61 with r_y = 2.45 inches and A_g = 17.9 in². The steel
yield strength is 50 ksi. What is the nominal compressive strength P_n per
, AISC 360?
A) 450 kips
B) 580 kips
C) 650 kips
D) 720 kips
Correct Answer: B
The effective length factor K = 1.0. KL/r = 1.0 × 16 × .45 = .45 = 78.4.
The elastic buckling stress F_e = π²E/(KL/r)² = (3.14² × 29,000) / (78.4²) = 286,000 /
6,147 = 46.5 ksi. Since KL/r ≤ 4.71√(E/F_y) ≈ 113, the column is inelastic. F_cr =
0.658^(F_y/F_e) × F_y = 0.658^(50/46.5) × 50 = 0.658^1.075 × 50 = 0.638 × 50 =
31.9 ksi. P_n = F_cr × A_g = 31.9 × 17.9 = 571 kips, rounded to 580 kips. Option A
uses a smaller area, C and D overestimate. Therefore, B is correct.
4. A reinforced concrete cantilever retaining wall is 18 feet tall and retains a
granular backfill with a unit weight of 120 pcf and an angle of internal
friction of 32 degrees. Using Rankine theory, what is the active earth
pressure coefficient K_a?
A) 0.26
B) 0.31
C) 0.36
D) 0.42
Correct Answer: B
Rankine active coefficient K_a = tan²(45° – φ/2). φ = 32°, so φ/2 = 16°. K_a =
tan²(45° – 16°) = tan²(29°) = (0.5543)² = 0.307. Option A uses φ = 35°, C uses φ =
30°, D uses φ = 28°. Therefore, the correct K_a is approximately 0.31.
5. A wood beam is simply supported with a span of 12 feet and carries a
uniform total load of 300 lb/ft. The beam is a 4×10 (actual 3.5 in × 9.25 in)
of Southern Pine No. 2. The reference bending design value F_b is 875 psi.
The size factor C_F is 1.0, the wet service factor C_M is 1.0, and the load
duration factor C_D is 1.0. What is the adjusted bending design value F'_b?
A) 700 psi
B) 875 psi
, C) 1,050 psi
D) 1,200 psi
Correct Answer: B
F'_b = F_b × C_M × C_t × C_L × C_F × C_fu × C_i × C_r × C_D. With all factors except
possibly C_F given as 1.0, and C_F for a 4×10 is approximately 1.0 (size factor for
depth 9.25 in is (12/9.25)^(1/9) ≈ 1.03, but assumed 1.0 here). Thus F'_b = 875 psi.
Option A reduces unnecessarily, C and D increase without justification. Therefore,
B is correct.
6. A steel plate tension member has a gross area of 4.0 in² and a thickness of
0.5 in. It is connected with two 7/8-inch bolts in a single line. The shear lag
factor U is 0.85. What is the effective net area A_e?
A) 2.55 in²
B) 2.85 in²
C) 3.00 in²
D) 3.25 in²
Correct Answer: A
The bolt hole diameter = 7/8 + 1/8 = 1.0 in. Net area A_n = A_g – n × d_h × t = 4.0
– 2 × 1.0 × 0.5 = 3.0 in². Effective net area A_e = A_n × U = 3.0 × 0.85 = 2.55 in².
Option B ignores U, C is A_n, D uses a different gross area. Therefore, the correct
effective net area is 2.55 in².
7. A concrete column is part of a special moment frame in Seismic Design
Category D. According to ACI 318, what is the minimum transverse
reinforcement spacing at the column ends over a length l_o?
A) 4 inches
B) 6 inches
C) 8 inches
D) 12 inches
Correct Answer: A
ACI 318-19 Section 18.7.5.3 requires that the spacing of transverse reinforcement
in the plastic hinge region of special moment frame columns not exceed
