PE Civil Structural Exam: 10 Practice Problems with Detailed Solutions

The PE Civil Structural depth section tests your ability to apply design codes—ACI 318, AISC 360, NDS, ASCE 7, and more—to realistic structural engineering problems. Success requires not just knowing the formulas but understanding when and how to apply them under time pressure.

Below you will find 10 practice problems spanning reinforced concrete, structural steel, wood design, foundations, seismic design, and load combinations. Each includes four answer choices and a detailed, step-by-step solution that walks through the governing code provisions.

Tip: Work each problem on your own before checking the solution. On the PE exam, you have roughly 6 minutes per problem—practice building that pace now.

Problem 1: Reinforced Concrete Beam — Flexural Capacity

A singly reinforced rectangular concrete beam has a width b = 14 in., an effective depth d = 21.5 in., and tensile reinforcement A_s = 4.00 in.². The concrete compressive strength is f′_c = 4,000 psi and the steel yield strength is f_y = 60,000 psi. Determine the design flexural strength φM_n of the beam per ACI 318.

A) 290 kip-ft
B) 341 kip-ft
C) 380 kip-ft
D) 410 kip-ft

Show Solution

Answer: B

We use the Whitney stress block approach from ACI 318 to find the nominal moment capacity.

Step 1: Find the depth of the equivalent stress block, a.

a = A_s × f_y / (0.85 × f′_c × b)

a = 4.00 × 60 / (0.85 × 4 × 14)

a = 240 / 47.6 = 5.04 in.

Step 2: Verify that the section is tension-controlled.

For f′_c = 4,000 psi, β₁ = 0.85. The neutral axis depth c = a / β₁ = 5.04 / 0.85 = 5.93 in.

The net tensile strain in the steel: ε_t = 0.003 × (d − c) / c = 0.003 × (21.5 − 5.93) / 5.93 = 0.003 × 2.625 = 0.00787.

Since ε_t = 0.00787 > 0.005, the section is tension-controlled and φ = 0.90.

Step 3: Compute the design flexural strength.

φM_n = φ × A_s × f_y × (d − a/2)

φM_n = 0.90 × 4.00 × 60 × (21.5 − 5.04/2)

φM_n = 0.90 × 240 × (21.5 − 2.52)

φM_n = 216 × 18.98 = 4,100 kip-in.

φM_n = 4,100 / 12 = 341 kip-ft

Problem 2: Structural Steel Beam — Flexural Strength

A W21×50 steel beam (Z_x = 110 in.³) is used as a simply supported floor beam with full lateral bracing along its compression flange. The steel has F_y = 50 ksi. Determine the design flexural strength φM_n per AISC 360.

A) 344 kip-ft
B) 413 kip-ft
C) 458 kip-ft
D) 550 kip-ft

Show Solution

Answer: B

With full lateral bracing, the unbraced length L_b = 0, which is less than L_p. The beam reaches its full plastic moment capacity—no lateral-torsional buckling reduction is needed.

Step 1: Check the limit state.

Since L_b ≤ L_p, the nominal moment is governed by the plastic moment: M_n = M_p = F_y × Z_x.

Step 2: Compute the design flexural strength.

φM_n = φ_b × F_y × Z_x

φM_n = 0.90 × 50 × 110

φM_n = 4,950 kip-in.

φM_n = 4,950 / 12 = 412.5 kip-ft ≈ 413 kip-ft

If the beam were unbraced over a longer length (L_b > L_p), you would need to check for lateral-torsional buckling per AISC 360 Chapter F, which reduces the available strength.

Problem 3: Steel Column — Axial Compression Capacity

A W10×49 steel column (A_g = 14.4 in.², r_y = 2.54 in.) has an unbraced length of 15 ft with pinned ends (K = 1.0) about the weak axis. F_y = 50 ksi and E = 29,000 ksi. Determine the design axial compressive strength φP_n per AISC 360 Chapter E.

A) 339 kips
B) 449 kips
C) 504 kips
D) 648 kips

Show Solution

Answer: B

We follow AISC 360 Chapter E to determine the critical stress F_cr and the design compressive strength.

Step 1: Compute the slenderness ratio.

KL/r = 1.0 × (15 × 12) / 2.54 = 180 / 2.54 = 70.9

Step 2: Determine the buckling regime.

The slenderness limit is 4.71√(E/F_y) = 4.71√(29,000/50) = 4.71 × 24.08 = 113.4.

Since KL/r = 70.9 < 113.4, the column buckles inelastically.

Step 3: Compute the elastic critical stress F_e.

F_e = π²E / (KL/r)² = π² × 29,000 / (70.9)² = 286,088 / 5,027 = 56.9 ksi

Step 4: Compute the critical stress F_cr.

F_cr = 0.658^(F_y/F_e) × F_y

F_y/F_e = 50 / 56.9 = 0.879

F_cr = 0.658^0.879 × 50

To evaluate 0.658^0.879: ln(0.658) = −0.419, so 0.658^0.879 = e^{−0.419 × 0.879} = e^−0.368 = 0.692.

F_cr = 0.692 × 50 = 34.6 ksi

Step 5: Compute the design compressive strength.

φP_n = φ_c × F_cr × A_g = 0.90 × 34.6 × 14.4 = 449 kips

Problem 4: ASCE 7 Load Combinations — Governing Factored Load

A roof beam supports the following service-level loads: dead load D = 30 kips, floor live load L = 50 kips, snow load S = 20 kips, and wind load W = 25 kips (downward component). Determine the governing factored load using ASCE 7 LRFD load combinations.

A) 118 kips
B) 121 kips
C) 126 kips
D) 142 kips

Show Solution

Answer: C

We evaluate the relevant ASCE 7 LRFD load combinations and select the one producing the largest total factored load.

Combination 1: 1.4D = 1.4(30) = 42 kips

Combination 2: 1.2D + 1.6L + 0.5S = 1.2(30) + 1.6(50) + 0.5(20) = 36 + 80 + 10 = 126 kips

Combination 3: 1.2D + 1.6S + 1.0L = 1.2(30) + 1.6(20) + 1.0(50) = 36 + 32 + 50 = 118 kips

Combination 4: 1.2D + 1.0W + 1.0L + 0.5S = 1.2(30) + 1.0(25) + 1.0(50) + 0.5(20) = 36 + 25 + 50 + 10 = 121 kips

Combination 5: 0.9D + 1.0W = 0.9(30) + 1.0(25) = 27 + 25 = 52 kips

Combination 2 governs with a factored load of 126 kips.

Note: On the PE exam, always evaluate all relevant combinations. Combination 2 (1.2D + 1.6L + 0.5S) frequently governs for members carrying significant live load.

Problem 5: Reinforced Concrete — Development Length of Reinforcing Bar

Determine the tension development length l_d for a #8 bottom bar (d_b = 1.0 in.) in normal-weight concrete with f′_c = 4,000 psi and f_y = 60,000 psi. The bar has clear spacing ≥ d_b and clear cover ≥ d_b, and is uncoated. Use the ACI 318 simplified equation for bars #7 and larger.

A) 30 in.
B) 38 in.
C) 48 in.
D) 54 in.

Show Solution

Answer: B

ACI 318 provides a simplified equation for tension development length. For #7 bars and larger with clear spacing ≥ d_b and clear cover ≥ d_b:

l_d = (f_y × ψ_t × ψ_e) / (25 × λ × √f′_c) × d_b

Step 1: Identify the modification factors.

ψ_t = 1.0 (bottom bar, not a top bar with > 12 in. of concrete below)

ψ_e = 1.0 (uncoated reinforcement)

λ = 1.0 (normal-weight concrete)

Step 2: Compute the development length.

l_d = (60,000 × 1.0 × 1.0) / (25 × 1.0 × √4,000) × 1.0

l_d = 60,000 / (25 × 63.25) × 1.0

l_d = 60,000 / 1,581 = 37.9 in.

l_d ≈ 38 in.

If the bar had been a top bar (ψ_t = 1.3) or epoxy-coated (ψ_e = 1.5), the required development length would increase significantly.

Problem 6: Wood Beam Design — Adjusted Bending Stress

A No. 2 Douglas Fir-Larch 2×12 joist (actual dimensions 1.5 in. × 11.25 in.) is used as a repetitive floor joist with normal load duration in dry service conditions. The NDS reference design value for bending is F_b = 900 psi. The applicable adjustment factors are: C_D = 1.0 (normal duration), C_M = 1.0 (dry service), C_L = 1.0 (laterally braced), C_F = 1.0 (2×12), and C_r = 1.15 (repetitive member). The joist carries a uniform load producing a maximum moment of 3.0 kip-ft. Does the joist have adequate bending capacity?

A) Yes — f_b = 950 psi < F_b′ = 1,035 psi
B) No — f_b = 1,138 psi > F_b′ = 1,035 psi
C) Yes — f_b = 1,138 psi < F_b′ = 1,200 psi
D) No — f_b = 1,422 psi > F_b′ = 1,035 psi

Show Solution

Answer: B

Per NDS, the adjusted design value F_b′ is found by multiplying the reference value by all applicable adjustment factors.

Step 1: Compute the adjusted bending design value.

F_b′ = F_b × C_D × C_M × C_L × C_F × C_r

F_b′ = 900 × 1.0 × 1.0 × 1.0 × 1.0 × 1.15 = 1,035 psi

Step 2: Compute the section modulus.

S_x = b × h² / 6 = 1.5 × (11.25)² / 6 = 1.5 × 126.56 / 6 = 31.64 in.³

Step 3: Compute the actual bending stress.

M = 3.0 kip-ft = 3,000 × 12 = 36,000 lb-in.

f_b = M / S_x = 36,000 / 31.64 = 1,138 psi

Step 4: Compare actual stress to adjusted allowable stress.

f_b = 1,138 psi > F_b′ = 1,035 psi — the joist is overstressed.

The joist does not have adequate bending capacity. Option C is a common trap—it incorrectly uses a higher F_b′ value by misapplying the adjustment factors (for example, using C_D = 1.15 for snow duration when the problem specifies normal duration). Option A uses an incorrect section modulus. Always compute S from the actual dressed dimensions and apply only the adjustment factors given in the problem.

The designer should select a larger section (e.g., 2×14 or 3×12) or reduce the span to bring the actual stress below the adjusted allowable.

Problem 7: Shallow Foundation — Terzaghi Bearing Capacity

A continuous (strip) footing is 4 ft wide and founded at a depth D_f = 3 ft in a cohesive soil with c = 500 psf, φ = 20°, and unit weight γ = 120 pcf. Using Terzaghi’s bearing capacity equation with bearing capacity factors N_c = 17.7, N_q = 7.4, and N_γ = 5.0, determine the allowable bearing capacity q_a using a factor of safety of 3.

A) 2,950 psf
B) 3,600 psf
C) 4,240 psf
D) 5,100 psf

Show Solution

Answer: C

Terzaghi’s bearing capacity equation for a continuous (strip) footing is:

q_u = c × N_c + q × N_q + 0.5 × γ × B × N_γ

where q = γ × D_f is the overburden pressure at the footing base.

Step 1: Compute the overburden pressure.

q = γ × D_f = 120 × 3 = 360 psf

Step 2: Compute the ultimate bearing capacity.

q_u = 500 × 17.7 + 360 × 7.4 + 0.5 × 120 × 4 × 5.0

q_u = 8,850 + 2,664 + 1,200

q_u = 12,714 psf

Step 3: Apply the factor of safety.

q_a = q_u / FS = 12,714 / 3 = 4,238 psf ≈ 4,240 psf

The three terms in Terzaghi’s equation represent: cohesion resistance (8,850 psf), surcharge/overburden resistance (2,664 psf), and self-weight resistance (1,200 psf). For this soil, the cohesion term dominates.

Problem 8: Retaining Wall — Factor of Safety Against Overturning

A concrete cantilever retaining wall retains soil with a height H = 12 ft behind the wall. The retained soil has γ = 120 pcf and φ = 30°. The wall has the following resisting forces and moment arms about the toe: wall self-weight W₁ = 3,600 lb/ft acting at 2.5 ft from the toe, and soil above the heel W₂ = 5,280 lb/ft acting at 5.0 ft from the toe. Determine the factor of safety against overturning about the toe.

A) 2.0
B) 2.5
C) 3.1
D) 3.8

Show Solution

Answer: C

The factor of safety against overturning is the ratio of the resisting moment to the overturning moment, both taken about the toe of the wall.

Step 1: Compute the active earth pressure coefficient.

K_a = tan²(45° − φ/2) = tan²(45° − 15°) = tan²(30°) = (0.5774)² = 0.333

Step 2: Compute the active earth pressure resultant.

P_a = 0.5 × K_a × γ × H² = 0.5 × 0.333 × 120 × (12)²

P_a = 0.5 × 0.333 × 120 × 144 = 2,880 lb/ft

Step 3: Compute the overturning moment about the toe.

The resultant P_a acts at H/3 = 12/3 = 4.0 ft above the base.

M_O = P_a × H/3 = 2,880 × 4.0 = 11,520 lb-ft/ft

Step 4: Compute the resisting moment about the toe.

M_R = W₁ × x₁ + W₂ × x₂

M_R = 3,600 × 2.5 + 5,280 × 5.0

M_R = 9,000 + 26,400 = 35,400 lb-ft/ft

Step 5: Compute the factor of safety.

FS_OT = M_R / M_O = 35,400 / 11,520 = 3.07 ≈ 3.1

The minimum required factor of safety against overturning is typically 2.0. With FS = 3.1, this wall is adequate.

Problem 9: Seismic Design — Equivalent Lateral Force Base Shear

A 5-story special moment frame building has the following seismic design parameters per ASCE 7: S_DS = 1.0g, S_D1 = 0.60g, R = 8, I_e = 1.0, and the fundamental period T = 0.50 s. The effective seismic weight is W = 5,000 kips. Determine the seismic base shear V using the equivalent lateral force procedure.

A) 220 kips
B) 440 kips
C) 625 kips
D) 750 kips

Show Solution

Answer: C

Per ASCE 7, the seismic response coefficient C_s is computed and then checked against upper and lower bounds.

Step 1: Compute the initial seismic response coefficient.

C_s = S_DS / (R / I_e) = 1.0 / (8 / 1.0) = 1.0 / 8 = 0.125

Step 2: Check the upper bound.

C_s need not exceed S_D1 / [T × (R / I_e)] = 0.60 / (0.50 × 8) = 0.60 / 4.0 = 0.150

Since 0.125 < 0.150, the initial value governs (no reduction from the upper bound).

Step 3: Check the lower bound.

C_s shall not be less than 0.044 × S_DS × I_e = 0.044 × 1.0 × 1.0 = 0.044

Since 0.125 > 0.044, the lower bound does not control.

Step 4: Compute the base shear.

V = C_s × W = 0.125 × 5,000 = 625 kips

The high R factor of 8 reflects the ductility of the special moment frame system, which significantly reduces the seismic design forces compared to less ductile systems.

Problem 10: Steel Bolt Connection — Shear Capacity

A bolted lap splice connection uses six 3/4-in. diameter A325-N bolts in single shear. The nominal shear stress for A325-N bolts (threads included in the shear plane) is F_nv = 54 ksi. Determine the total design shear strength φR_n of the bolt group per AISC 360.

A) 80.6 kips
B) 95.8 kips
C) 107.3 kips
D) 143.1 kips

Show Solution

Answer: C

Per AISC 360 Section J3.6, the design shear strength of a bolt is φR_n = φ × F_nv × A_b, where φ = 0.75 for bolt shear.

Step 1: Compute the nominal bolt area.

A_b = π × d² / 4 = π × (0.75)² / 4 = π × 0.5625 / 4 = 0.4418 in.²

Step 2: Compute the design shear strength per bolt.

φr_n = φ × F_nv × A_b = 0.75 × 54 × 0.4418 = 17.89 kips/bolt

Step 3: Compute the total design shear strength for the bolt group.

φR_n = n × φr_n = 6 × 17.89 = 107.3 kips

If the bolts were A325-X (threads excluded from the shear plane), the nominal shear stress would increase to F_nv = 68 ksi, giving a 26% higher capacity. Always verify whether threads are included or excluded in the shear plane.

Key Takeaways for the PE Civil Structural Exam

These ten problems illustrate the range of design codes and analytical methods you must be comfortable with on exam day. Here are a few tips to carry forward:

Know your code provisions cold. The PE exam is open-book, but you cannot afford to search for basic equations. Memorize the Whitney stress block formula, AISC flexure provisions, Terzaghi’s equation, and the ASCE 7 load combinations.
Verify limit states. Problems 1 and 3 demonstrate that you must check whether a section is tension-controlled or which buckling regime applies before plugging into a formula.
Watch your units. Many exam errors come from mixing kips and pounds, or inches and feet. Convert early and label every intermediate result.
Practice the full workflow. On the real exam, you will not be told which code section to use. Practice identifying the correct approach from the problem statement alone.

Ready for More Practice?

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