PE Civil Structural Exam Prep
Free prep for the NCEES PE Civil: Structural exam: 24 original practice problems with worked answers, the code editions for your test date, and a 12-week study plan.
These are original practice problems written by Castleport Test Prep, not official NCEES questions.
Practice problems
1. Floor load to a beam
Analysis of Structures–Loads and Load Applications · Choose one · PCS-01
A floor carries a dead load D = 80 psf and a live load L = 50 psf (no live-load reduction). Interior beams are spaced 8 ft on center and span 30 ft. Use one-way tributary loading; D includes all dead load, including beam self-weight. Using the combination 1.2D + 1.6L, what factored line load reaches one interior beam?
Show answer and explanation
Answer: C. 1.41 kip/ft
Tributary width = 8 ft (half the spacing on each side). Dead line load = 80 × 8 = 640 lb/ft = 0.64 kip/ft. Live line load = 50 × 8 = 400 lb/ft = 0.40 kip/ft. Factored: 1.2(0.64) + 1.6(0.40) = 0.768 + 0.640 = 1.408 ≈ 1.41 kip/ft.
Why the other answers fail: A is 1.4D alone (0.896). B is the unfactored D + L (1.04). D puts the 1.6 factor on both loads (1.6 × 1.04 = 1.664). The span length doesn't change this line load; it enters later calculations such as reactions or moments.
If you missed it: Write the units at every step: psf → kip/ft → kips or kip-ft.
Units: 1 kip = 1,000 lb; ksi = kips per square inch; ksf = kips per square foot; MPa = N/mm².
2. Which combination governs?
Analysis of Structures–Loads and Load Applications · Choose one · PCS-02
A column carries D = 20 kips and L = 2 kips. Checking only the two combinations 1.4D and 1.2D + 1.6L, what is the governing factored axial load?
Show answer and explanation
Answer: C. 28.0 kips
1.4D = 1.4(20) = 28.0 kips. 1.2D + 1.6L = 24.0 + 3.2 = 27.2 kips. The larger value governs: 28.0 kips.
Why the other answers fail: B is the 1.2D + 1.6L result, which is smaller here. A is 1.2D alone. D is what 1.2D + 1.6L would give if L were 4 kips. The lesson: 1.4D beats 1.2D + 1.6L whenever L is less than D/8 (from 0.2D > 1.6L). Light live loads on heavy members are where the dead-only check bites.
If you missed it: Check every combination your problem gives you before picking one. Don't assume the one with live load always governs.
Basis: Calculation from the values given in the problem; the method is shown above.
3. Wall pressure with a surcharge
Analysis of Structures–Loads and Load Applications · Enter a number · PCS-03
A 12-ft-tall cantilever retaining wall has a vertical, smooth back face and holds dry, homogeneous, cohesionless, level backfill with unit weight γ = 120 pcf. Assume enough wall movement to develop active pressure and no water pressure. Use an active pressure coefficient Ka = 1/3. A uniform surcharge q = 250 psf extends across the entire backfill surface. What is the total active lateral force per foot of wall, in lb/ft?
Answer in: lb/ft Round to the nearest whole lb/ft.
Show answer and explanation
Answer: 3,880 lb/ft
Soil (triangle): ½ Ka γ H² = ½ (1/3)(120)(12²) = 2,880 lb/ft, acting H/3 = 4.0 ft above the base. Surcharge (rectangle): Ka q H = (1/3)(250)(12) = 1,000 lb/ft, acting H/2 = 6.0 ft above the base. Total = 3,880 lb/ft. Its combined line of action is (2,880 × 4 + 1,000 × 6)/3,880 ≈ 4.52 ft above the base.
Why the other answers fail: 2,880 forgets the surcharge. 3,380 treats the surcharge as a triangle. 6,760 drops the ½ on the soil triangle. 11,640 leaves out Ka entirely. Sketch both pressure shapes before you add them.
If you missed it: For this backfill and surcharge model, draw the pressure diagram first: triangle for soil, rectangle for the uniform surcharge. Then find each resultant and where it acts.
Basis: Caltrans, Trenching and Shoring Manual, Chapter 4 (§4-3; pp. 4-5–4-6, 4-9–4-10); FHWA, Rockery Design and Construction Guidelines, Chapter 4 (Sliding Resistance, before Fig. 32); Roylance, Statics of Bending: Shear and Bending Moment Diagrams, MIT OCW 3.11 (pp. 2–4)
4. Trace the load path
Analysis of Structures–Loads and Load Applications · Put in order · PCS-04
In this framing arrangement, the girder is supported by columns on spread footings, and the roof deck is supported by joists that frame into the girder. The footings bear on soil. Put this gravity load path in order, from where the load starts to where it ends.
Items to order: column · spread footing · roof deck · girder · soil · joists
Show answer and explanation
Answer: roof deck → joists → girder → column → spread footing → soil
Each element carries its load to whatever supports it.
Why the other answers fail: The common slip is putting the girder before the joists. In this framing arrangement, joists frame into girders, so they come first. Here, the column transfers its load to the spread footing below it.
If you missed it: For each framing plan you study, trace one load to the soil and name every element it passes through.
5. Support reaction
Analysis of Structures–Forces and Load Effects · Choose one · PCS-05
A simply supported beam has a pin at A and a roller at B, 20 ft apart. An 18-kip downward point load acts 8 ft from A. Neglect self-weight. What is the upward reaction at B?
Show answer and explanation
Answer: A. 7.2 kips
Sum moments about A: R_B(20) = 18(8), so R_B = 7.2 kips. Check: R_A = 18 − 7.2 = 10.8 kips. The support closer to the load takes more.
Why the other answers fail: B assumes the load is at midspan. C is the reaction at A. D gives the whole load to one support.
If you missed it: Draw the free-body diagram of the whole beam before solving, and circle which reaction the question asks for.
Basis: Roylance, Statics of Bending: Shear and Bending Moment Diagrams, MIT OCW 3.11 (pp. 2–4); AWC Design Aid No. 6, Beam Design Formulas (Fig. 8, p. 7)
6. Moment under a point load
Analysis of Structures–Forces and Load Effects · Choose one · PCS-06
A simply supported beam has a pin at A and a roller at B, 20 ft apart. An 18-kip downward point load acts 8 ft from A. Neglect self-weight. What is the maximum bending moment?
Show answer and explanation
Answer: B. 86.4 kip-ft
Maximum moment is under the load: M = R_A × 8 = 10.8 × 8 = 86.4 kip-ft. Same answer from Pab/L = 18(8)(12)/20 = 86.4 kip-ft.
Why the other answers fail: A multiplies the wrong reaction (R_B = 7.2) by 8 ft. C is PL/4, which only works for a load at midspan. D is P × a, which ignores that the reaction at A is less than P.
If you missed it: When a load isn't at midspan, compute the moment from a reaction times its distance. Don't reach for PL/4 by habit.
Basis: Roylance, Statics of Bending: Shear and Bending Moment Diagrams, MIT OCW 3.11 (pp. 2–4); AWC Design Aid No. 6, Beam Design Formulas (Fig. 8, p. 7)
7. Cantilever tip deflection
Analysis of Structures–Forces and Load Effects · Enter a number · PCS-07
A steel cantilever is 10 ft long with a transverse 5-kip load at the free end. E = 29,000 ksi and I = 500 in⁴. Assume a prismatic, linear-elastic beam with negligible shear deformation. What is the magnitude of the tip deflection, in inches? (Ignore self-weight.)
Answer in: in. Round to the nearest 0.001 in.
Show answer and explanation
Answer: 0.199 in.
Δ = PL³/(3EI). Convert L to inches: 120 in. Δ = 5(120³)/(3 × 29,000 × 500) = 8,640,000/43,500,000 = 0.199 in.
Why the other answers fail: Leaving L in feet gives about 0.000115 in. (1/1,728 of the right value, since L is cubed). Using the simple-beam midspan formula PL³/(48EI) gives about 0.012 in.
If you missed it: With these E and I units, convert every length to inches first. A cubed length magnifies any unit error.
Basis: AWC Design Aid No. 6, Beam Design Formulas (Fig. 13, p. 10)
8. Flagpole buckling
Analysis of Structures–Forces and Load Effects · Choose one · PCS-08
A 15-ft steel post is fixed at the base and free at the top (theoretical effective length factor K = 2.0). E = 29,000 ksi and I = 50 in⁴ about the buckling axis. What is the elastic (Euler) buckling load?
Show answer and explanation
Answer: B. 110 kips
Pcr = π²EI/(KL)². KL = 2.0 × 180 in = 360 in. Pcr = π²(29,000)(50)/360² = 14,311,000/129,600 ≈ 110 kips.
Why the other answers fail: A uses K = 1.0 (pinned at both ends). C uses K = 0.5 (fixed at both ends). D halves the pinned-pinned load instead of squaring K. Because KL is squared, doubling K cuts the load to one-quarter. Code column design adds strength limits and inelastic behavior, so Euler is a check on behavior, not a design strength.
If you missed it: Before any column problem, sketch the end conditions and write K beside them.
Basis: MIT OCW 4.440 Basic Structural Design, exam equation sheet (Euler buckling, K factors)
9. Axial load plus bending
Analysis of Structures–Forces and Load Effects · Choose one · PCS-09
A member carries an axial compression P = 40 kips and a bending moment M = 20 kip-ft. Assume linear-elastic behavior and bending about a centroidal principal axis. A = 10 in² and S = 40 in³ at both opposite extreme fibers. Treating compression as positive, what are the extreme-fiber stresses?
Show answer and explanation
Answer: A. 10 ksi compression on one face; 2 ksi tension on the other
Axial stress P/A = 40/10 = 4 ksi (compression). Bending stress M/S = (20 × 12)/40 = 6 ksi. Combine: 4 + 6 = 10 ksi compression; 4 − 6 = −2 ksi, which is 2 ksi tension.
Why the other answers fail: B drops the sign change on the second face. C ignores the axial load. D ignores the moment.
If you missed it: Superimpose stresses face by face and keep track of sign. A negative result is the answer telling you the face is in tension.
Basis: Roylance, Stresses in Beams, MIT OCW 3.11 (Eq. 7, p. 4); Roylance, Introduction to Elasticity, MIT OCW 3.11 (Eq. 2, p. 4)
10. Truss apex joint
Analysis of Structures–Forces and Load Effects · Enter a number · PCS-10
At the apex joint of a symmetric truss, a 10-kip load acts straight down. Only two ideal pin-connected, two-force members meet at the joint, each sloping down at 30° below horizontal on either side. What is the force in each member, in kips? (Enter the magnitude.)
Answer in: kips Round to the nearest 0.1 kip.
Show answer and explanation
Answer: 10 kips
Vertical equilibrium at the joint: 2F sin 30° = 10, so F = 10/(2 × 0.5) = 10 kips. Both members push up on the joint, so they're in compression.
Why the other answers fail: Using cos 30° gives 5.77 kips. Splitting the load in half without resolving the angle gives 5 kips. For the same load, flattening the members increases their axial force.
If you missed it: At every truss joint, write ΣFx = 0 and ΣFy = 0 with the angles drawn. Check which trig function goes with the vertical component.
11. Fall protection systems
Temporary Structures and Other Topics · Select all that apply · PCS-11
Under 29 CFR 1926.501(b)(1), a worker is on a walking/working surface with an unprotected side 6 ft or more above a lower level. Which systems does that paragraph list for protecting the worker from falling? Select all that apply.
Show answer and explanation
Answer: A, B, C
Paragraph (b)(1) names three: guardrail systems, safety net systems, or personal fall arrest systems. Credit requires all three and nothing else, like the all-or-nothing scoring NCEES uses for its alternative item types.
Why the other answers fail: D: a warning line is not one of the three systems listed in (b)(1). E: hard hats protect against falling objects, not falls.
If you missed it: When a question cites a paragraph, read that paragraph's list exactly. Don't add options from memory of other site rules.
Basis: 29 CFR 1926.501(b)(1), OSHA; NCEES, Computer-based testing: alternative item types
12. Guardrail height
Temporary Structures and Other Topics · Enter a number · PCS-12
Under 29 CFR 1926.502(b)(1), what is the nominal top-edge height of a guardrail top rail above the walking/working level, in inches?
Answer in: in. Enter the nominal height, not either end of the tolerance band.
Show answer and explanation
Answer: 42 in.
42 inches, plus or minus 3 inches. The same section requires the top rail to withstand at least 200 lb without failure, applied within 2 inches of the top edge at any point along it, outward or downward. Midrails must withstand at least 150 lb outward or downward at any point along the member.
Why the other answers fail: 36 in. is below the standard 39–45-in. band. 48 in. is not the nominal 42-in. answer, but heights above 45 in. are permitted when conditions warrant and all other criteria are met. When employees use stilts, the top edge must be raised by the stilt height.
If you missed it: Pair each dimension you memorize with its load requirement.
Basis: 29 CFR 1926.502(b), OSHA
13. Stripping slab forms
Temporary Structures and Other Topics · Choose one · PCS-13
Under 29 CFR 1926.703(e)(1), when may forms and shores for an elevated concrete slab be removed?
Show answer and explanation
Answer: B. When the employer determines the concrete has gained enough strength to support its own weight and superimposed loads
OSHA sets a strength condition, not a calendar: forms and shores stay until the employer determines the concrete can carry its weight and superimposed loads. The plans and specifications must state removal conditions that have been followed, or an appropriate ASTM compressive-strength test must show adequate strength. Paragraph (e)(2) adds that reshoring stays until the concrete can carry its weight and all loads in place.
Why the other answers fail: A substitutes a calendar age for the regulation's strength condition. C confuses finishing with strength. D ignores the strength condition entirely.
If you missed it: For temporary works, ask what condition triggers each step: installation, loading, or removal.
Basis: 29 CFR 1926.703(e), OSHA
14. Plumbing-up equipment
Temporary Structures and Other Topics · Select all that apply · PCS-14
Plumbing-up equipment is being used to keep a steel frame stable during erection. Under 29 CFR 1926.754(d), which TWO statements are correct? Select two.
Show answer and explanation
Answer: A, B
Paragraph (d)(2) requires the equipment to be in place before the structure is loaded with construction material. Paragraph (d)(3) allows removal only with a competent person's approval.
Why the other answers fail: C invents an automatic removal trigger the rule doesn't contain. D reverses the sequence (d)(2) requires.
If you missed it: Separate the rule for putting a stability measure in from the rule for taking it out.
Basis: 29 CFR 1926.754(d), OSHA
15. Dry unit weight of soil
Design and Details of Structures–Materials and Material Properties · Enter a number · PCS-15
A soil has specific gravity of solids Gs = 2.70 and void ratio e = 0.65. Use γw = 62.4 pcf. What is its dry unit weight, in pcf?
Answer in: pcf Round to the nearest 0.1 pcf.
Show answer and explanation
Answer: 102.1 pcf
Take a sample with 1 ft³ of solids. Then the void volume is Vv = eVs = 0.65 ft³, total volume = 1.65 ft³, and the solids weigh Gs γw = 2.70 × 62.4 = 168.5 lb. Dry unit weight = 168.5/1.65 = 102.1 pcf. That's the relationship γd = Gsγw/(1 + e).
Why the other answers fail: Dividing by (1 − e) gives about 481 pcf, which is impossible for this soil because it exceeds the 168.5-pcf unit weight of the solids themselves. Forgetting γw leaves a dimensionless 1.64.
If you missed it: When you forget a phase formula, rebuild it from a block diagram with Vs = 1.
Basis: FHWA NHI-05-037, Geotechnical Aspects of Pavements (§5.3.1, Tables 5-7 and 5-8)
16. Elongation in SI units
Design and Details of Structures–Materials and Material Properties · Enter a number · PCS-16
A 3.0-m steel tie carries a uniform tensile stress of 100 MPa. Use E = 200,000 MPa and assume linear-elastic behavior. What is the elongation, in millimeters?
Answer in: mm Round to the nearest 0.1 mm.
Show answer and explanation
Answer: 1.5 mm
Strain ε = σ/E = 100/200,000 = 0.0005 (0.05%). Elongation δ = εL = 0.0005 × 3,000 mm = 1.5 mm.
Why the other answers fail: 0.0015 is the elongation in meters entered as millimeters. 0.0005 is the strain, not the elongation. E is stiffness, not strength, so it never tells you whether the bar yields.
If you missed it: Label each material number before using it: stiffness (E), strength (Fy, f′c), or weight (γ). The exam uses both SI and U.S. units.
Basis: Roylance, Introduction to Elasticity, MIT OCW 3.11 (Eqs. 4, 6, 8, pp. 5–6)
17. Wood load duration factor
Design and Details of Structures–Materials and Material Properties · Choose one · PCS-17
Using NDS 2018, a sawn-lumber beam is designed by allowable stress design (ASD). Reference bending design value Fb = 900 psi. All adjustment factors are 1.0 except the load duration factor C_D. The governing combination is dead load plus snow load. Given C_D values: permanent 0.9; ten-year (occupancy live) 1.0; two-month (snow) 1.15; ten-minute (wind/earthquake) 1.6. What is the adjusted bending design value F′b?
Show answer and explanation
Answer: C. 1,035 psi
NDS 2018 applies the C_D of the shortest-duration load in the combination. In D + S, snow (two months) is shorter than dead (permanent), so C_D = 1.15. F′b = 900 × 1.15 = 1,035 psi.
Why the other answers fail: A uses the dead-load factor 0.9 (that's for a dead-load-only check). B skips C_D. D uses the wind factor, but there's no wind in this combination.
If you missed it: For wood problems, write down which loads are in the combination before picking C_D. NCEES specifies ASD for wood design on this exam.
Basis: AWC, NDS 2018 with Commentary, Chapter 2 (§2.3.2, Table 2.3.2, pp. 10–11); NCEES PE Civil–Structural exam specifications and design standards (exams before April 2027); NCEES PE Civil–Structural exam specifications and design standards (beginning April 2027)
18. Brace points and inflection points
Design and Details of Structures–Component Design and Detailing · Choose one · PCS-18
A 24-ft steel beam has effective lateral and torsional restraint only at its two ends. Its bending moment changes sign at midspan, where no physical brace is installed. What is the distance between its effective brace points?
Show answer and explanation
Answer: D. 24 ft
A point of zero moment (an inflection point) doesn't hold the beam against sideways movement or twist. Research on beam bracing states plainly that it's incorrect to treat an inflection point as a brace point. The only real restraints are the two ends, 24 ft apart.
Why the other answers fail: A assumes continuous bracing. B invents quarter-point braces. C treats the inflection point as a brace, which is the exact mistake this question targets.
If you missed it: Mark physical braces on the framing sketch separately from the moment diagram before you pick an unbraced length.
Basis: Yura, Fundamentals of Beam Bracing, AISC Engineering Journal 38(1), 2001 (p. 15)
19. Concrete beam flexural strength
Design and Details of Structures–Component Design and Detailing · Choose one · PCS-19
A nonprestressed, singly reinforced rectangular concrete beam with one layer of tension reinforcement and no axial load has b = 12 in., d = 20 in., As = 3.00 in², fy = 60 ksi, and f′c = 4 ksi. Use an equivalent rectangular stress block with a = Asfy/(0.85f′c b), β1 = 0.85, c = a/β1, and εt = 0.003(d − c)/c. For this ACI 318-14 exercise, use φ = 0.90 if εt ≥ 0.005. What is the design flexural strength φMn?
Show answer and explanation
Answer: B. 240 kip-ft
a = (3.00 × 60)/(0.85 × 4 × 12) = 4.41 in. c = 4.41/0.85 = 5.19 in. εt = 0.003(20 − 5.19)/5.19 = 0.0086 ≥ 0.005, so φ = 0.90. Using unrounded a, Mn = Asfy(d − a/2) ≈ 3,202.94 kip-in = 266.9 kip-ft. φMn = 0.90 × 266.9 = 240 kip-ft.
Why the other answers fail: A is Mn without φ. C uses the full depth d as the lever arm and skips φ: 3.00 × 60 × 20 = 3,600 kip-in = 300 kip-ft. D applies φ twice. On the exam you'd pull β1, the strain limit, and φ from the ACI 318 edition for your test date; here they're given so the arithmetic is the skill.
If you missed it: After any concrete flexure answer, check εt. Use the strain limits and beam ductility requirements in the ACI 318 edition for your test date. ACI 318-19 uses εt ≥ εty + 0.003 for tension control, where εty is the reinforcement yield strain, and requires this nonprestressed beam to be tension-controlled; reducing φ alone does not make it acceptable.
Basis: STRUCTURE, Flexural Design of Reinforced Concrete Beam Sections (Section Analysis Basics; Eq. 7; ACI 318-19); ACI University, ACI 318-19 Changes: High-Strength (video summary)
20. Steel plastic moment
Design and Details of Structures–Component Design and Detailing · Enter a number · PCS-20
A compact, continuously braced steel W-shape bends about its strong axis. Fy = 50 ksi, Zx = 100 in³, and φb = 0.90. Assuming nominal flexural strength Mn equals the plastic moment, what is φMn in kip-ft?
Answer in: kip-ft Round to the nearest whole kip-ft.
Show answer and explanation
Answer: 375 kip-ft
Mp = Fy Zx = 50 × 100 = 5,000 kip-in = 416.7 kip-ft. φMn = 0.90 × 416.7 = 375 kip-ft.
Why the other answers fail: 416.7 forgets φ. 4,500 is the right answer left in kip-in. Using the elastic section modulus Sx instead of Zx gives the yield moment, not the plastic moment.
If you missed it: Know which section property goes with which limit state: Z for plastic moment, S for first yield.
Basis: MIT OCW 1.051, Structural Engineering Design, Quiz 2 Solutions (Question 4(b), p. 5)
21. Eccentric footing pressure
Design and Details of Structures–Component Design and Detailing · Choose one · PCS-21
An 8 ft × 8 ft spread footing carries a vertical load P = 200 kips and a moment M = 80 kip-ft about a centroidal axis parallel to one side (all loads included). Assume a rigid footing with linear soil pressure. What is the maximum soil pressure?
Show answer and explanation
Answer: B. 4.06 ksf
e = M/P = 80/200 = 0.40 ft. B/6 = 8/6 = 1.33 ft, so e < B/6 and the whole footing stays in contact. q = (P/A)(1 ± 6e/B) = (200/64)(1 ± 6 × 0.40/8) = 3.125(1 ± 0.30). q max = 4.06 ksf; q min = 2.19 ksf.
Why the other answers fail: A is the average pressure (ignores the moment). C is the minimum. D is what you get if you double the eccentricity (6e/B = 0.60). If e had exceeded B/6, this formula would no longer apply because part of the footing would lift off.
If you missed it: Compute e and compare it with B/6 before you use the trapezoidal pressure formula.
Basis: FHWA-HRT-14-094, Eccentricity, Bearing Pressure Diagram, and Footing Liftoff
22. Overturning check
Design and Details of Structures–Component Design and Detailing · Enter a number · PCS-22
Per foot of a cantilever retaining wall, the resisting moment about the toe is 120 kip-ft. The active lateral resultant is 6 kips acting 5 ft above the base. What is the factor of safety against overturning (resisting moment ÷ overturning moment)?
Answer in: (ratio) Round to the nearest 0.1.
Show answer and explanation
Answer: 4.0
Overturning moment = 6 × 5 = 30 kip-ft. FS = 120/30 = 4.0.
Why the other answers fail: Using the wall height instead of the resultant's height inflates the overturning moment. Dividing the other way (30/120 = 0.25) inverts the ratio. The minimum FS a design must meet comes from the governing code or project criteria, not from this problem.
If you missed it: Find where each lateral resultant acts (see problem 3) before taking moments about the toe.
Basis: Roylance, Statics of Bending: Shear and Bending Moment Diagrams, MIT OCW 3.11 (pp. 2–4)
23. Pile group capacity
Design and Details of Structures–Component Design and Detailing · Choose one · PCS-23
A pile group has 9 piles. Each pile has an ultimate axial compression capacity of 100 kips. Use a group efficiency of 0.80 and a factor of safety of 2.5 applied to the ultimate group capacity. Use Qug = ηg nQu for this capacity calculation; assume settlement and other limit states do not govern. What is the allowable group capacity?
Show answer and explanation
Answer: A. 288 kips
Ultimate group capacity = 9 × 100 × 0.80 = 720 kips. Allowable = 720/2.5 = 288 kips.
Why the other answers fail: B skips the efficiency (900/2.5). C is the ultimate group capacity, before the factor of safety. D applies the efficiency twice.
If you missed it: For this model, apply the group efficiency to the sum of individual ultimate capacities, then divide by the factor of safety.
Basis: FHWA NHI-06-089, Soils and Foundations, Volume II (Eqs. 9-3 and 9-17; chapter copy)
24. How many bolts?
Design and Details of Structures–Component Design and Detailing · Choose one · PCS-24
A concentrically loaded group of identical bolts shares a factored shear Pu = 150 kips equally, with no bolt tension. Bolt shear governs, with an available strength φRn = 17.9 kips per bolt (bearing and tearout checked separately and not governing). What is the minimum number of bolts?
Show answer and explanation
Answer: B. 9
150/17.9 = 8.38. You can't use part of a bolt, so round up: 9 bolts (9 × 17.9 = 161 kips ≥ 150 kips).
Why the other answers fail: A rounds down, which leaves the connection short (8 × 17.9 = 143 kips). C is more than the minimum. D doesn't follow from the numbers.
If you missed it: For this equal-sharing capacity check, round the fractional bolt count up and check the total against the demand.
Score yourself, then log your misses
Count your correct answers out of 24. That number tells you how you did on these 24 problems. It isn't an NCEES score, it isn't scaled, and it doesn't predict a pass. Three to seven problems per area is too few to rank your areas precisely, so treat a miss as a lead to follow, not a verdict.
The problems span all five exam areas. This is a starter set, not a full-length exam or complete coverage of every topic in the specification.
What does help: figuring out why you missed each one. For every miss, write down the first step that went wrong and sort it into one of these rows.
| What went wrong | What to write down | What to do next |
|---|---|---|
| Wrong physical model | The support, load path, restraint, or contact condition you assumed | Redraw the system and state each assumption before calculating again |
| Units | The first line where the units stopped agreeing | Rework it with units written on every line |
| Wrong property or factor | The value you used and the one you needed | Label each number as stiffness, strength, weight, or a factor before using it |
| Couldn't find the rule | Document, edition, topic, and the search words you tried | Do the reference drill below and write down the locator that worked |
| Arithmetic or calculator entry | The first wrong operation | Redo it once with a rough magnitude check |
| Guessed right | The part you couldn't explain | Treat it as a miss and solve a fresh problem on the same idea later |
Keep a running log with these columns: date, problem or topic, first wrong step, error type, correct method or reference, next practice task, date to revisit.
Which code editions apply to your exam date
NCEES supplies the design standards on-screen during the exam, and it has posted two sets for PE Civil: Structural. If your appointment is in April 2027 or later, study the April 2027 set. Earlier appointments use the current set. The five topic areas and question ranges are the same in both documents; only the standards changed.
| Supplied standard | Exams before April 2027 | Exams beginning April 2027 |
|---|---|---|
| AASHTO LRFD Bridge Design Specifications | 8th edition, 2017 (May 2018 errata noted) | 10th edition, 2024 |
| International Building Code (IBC) | 2018, without supplements | 2024, without supplements |
| ASCE 7, Minimum Design Loads for Buildings and Other Structures | ASCE 7-16 | ASCE 7-22 |
| ACI 318, Building Code Requirements for Structural Concrete and Commentary | 2014 | 2019 (2022) |
| AISC Steel Construction Manual | 15th edition | 16th edition |
| AWC National Design Specification (NDS) for Wood Construction with Commentary | 2018 | 2024 |
| NDS Supplement, Design Values for Wood Construction | 2018 | 2024 |
| AWC Special Design Provisions for Wind and Seismic (SDPWS) with Commentary | 2015 | 2021 |
| 29 CFR (OSHA), listed sections only | July 2020 | 2024 |
| PCI Design Handbook: Precast and Prestressed Concrete | 7th edition, 2010 | 8th edition, 2017 |
| TMS 402/602, Building Code Requirements and Specification for Masonry Structures | 2016 | 2022 |
Sources: NCEES standards before April 2027 and NCEES standards beginning April 2027, design standards pages.
A few details that matter:
- Wood is ASD only. Both lists state that you'll use allowable stress design for wood, not LRFD (load and resistance factor design).
- OSHA coverage is narrow. Both lists name the same sections: 29 CFR 1910.28–.30, 1910.66–.68, and 1910.140; and 29 CFR 1926.104, Subpart L Appendix A, 1926.500–.503, 1926.703–.706, 1926.752, and 1926.754–.758. The lists also include Appendices A–D to 1910.66, Appendices B–D to Subpart M, and the Appendix A listed with Subpart Q.
- The edition is the exam's, not your office's. If your firm designs to a newer code than your exam lists, study the exam's edition. Requirements and numbering can differ between editions.
- You'll need your own access to study. NCEES provides the standards as PDFs during the exam only. It doesn't sell or hand out copies for study, so get the correct editions from the publishers or through your employer or library. The PE Civil Reference Handbook is available through your MyNCEES account.
Pick a test date you can actually prepare for, then study that date's set. Rushing the exam to beat the April 2027 switch only makes sense if you'd be ready anyway.
Make sure this is your exam
This page covers PE Civil: Structural, one of five PE Civil modules: 80 questions, one appointment. NCEES also offers a separate PE Structural exam with vertical and lateral breadth sections and buildings or bridges depth sections, each with its own fee. Not every licensing board accepts it as its PE exam. If your registration says PE Structural, use the NCEES PE Structural page instead. Not sure which PE discipline fits you? Start with our PE exam prep guide.
What the exam covers
NCEES publishes five knowledge areas with a range of questions for each. The ranges tell you roughly how many questions come from an area. They don't say how hard the area is or how long you'll need on it.
| Knowledge area (NCEES label) | Questions | Topics listed by NCEES |
|---|---|---|
| Analysis of Structures–Loads and Load Applications | 12–18 | Dead, live, construction, wind, seismic, moving (vehicular, cranes), snow/rain/ice, impact, earth pressure and surcharge loads; tributary areas and load paths; load combinations |
| Analysis of Structures–Forces and Load Effects | 17–26 | Shear and moment diagrams; axial; shear; flexure; combined stresses; deflection; special topics (torsion, buckling, fatigue, progressive collapse, thermal deformation, bearing) |
| Temporary Structures and Other Topics | 5–8 | Special inspections; submittals; formwork, falsework, scaffolding, shoring and reshoring, bracing, anchorage; impact of construction on adjacent facilities; safety (construction, roadside, work zone) |
| Design and Details of Structures–Materials and Material Properties | 10–15 | Soil classification and properties (strength, permeability, compressibility, phase relationships); concrete (plain, reinforced, cast-in-place, precast, pre- and post-tensioned); steel (structural, cold-formed); timber; masonry (brick veneer, CMU); material test methods and specification conformance |
| Design and Details of Structures–Component Design and Detailing | 26–39 | Horizontal members (beams, slabs, diaphragms, struts); vertical members (columns, bearing walls, shear walls); systems (trusses, braces, frames, composite construction); connections (bolted, welded, bearing, embedded, anchored, post-installed anchors); shallow foundations; deep foundations; retaining walls |
Source: NCEES PE Civil–Structural exam specifications, pp. 1–2 (identical in the pre-April 2027 document).
How to use each area:
- Loads. Get the load to the member before you design anything. Problems 1–4 drill tributary widths, combinations, and lateral earth pressure. Then practice finding the applicable building-load provisions in the edition of ASCE 7 for your date, and bridge loads in AASHTO. For earth pressure, identify the soil, water, surcharge, and wall-restraint model before calculating.
- Forces and load effects. This is the engine behind most of the exam. Draw the free-body diagram, solve equilibrium, then pick a formula. Problems 5–10 cover reactions, moment, deflection, buckling, combined stress, and truss joints.
- Temporary structures. Small in count, but the rules are specific and checkable. Read the OSHA sections in the table above and know what triggers each step: installing, loading, removing. Problems 11–14 practice this.
- Materials. Know what each property tells you (stiffness, strength, weight) and which factors adjust it. Problems 15–17 cover soil phase relationships, steel elongation in SI units, and the wood load duration factor.
- Component design. The biggest area by question range. Plan to work full design problems in each material you don't use at work. Problems 18–24 cover bracing, concrete and steel flexure, footings, retaining walls, piles, and bolts.
Reference drill: which standard do you open first?
On exam day you search on-screen PDFs, and NCEES provides each standard as individual chapters, with only one chapter open and searched at a time. You'll use the search box beside the reference; Ctrl+F doesn't work (NCEES specifications, design standards page; Examinee Guide, p. 10). Practice identifying the document and chapter before typing a search.
Cover the right column and name the standard for each task.
| Task | Open first |
|---|---|
| Live-load design of a highway bridge girder | AASHTO LRFD Bridge Design Specifications |
| Wind and snow loads on a building | ASCE 7 |
| Flexural strength of a reinforced concrete beam | ACI 318 |
| Bending check on a sawn-lumber header (ASD) | AWC NDS, with design values from the NDS Supplement |
| Design data for a precast double tee | PCI Design Handbook |
| Strength of a CMU shear wall | TMS 402/602 |
| Plumbing-up equipment during steel erection | 29 CFR 1926 Subpart R (1926.754) |
| Bolt shear strength in a steel connection | AISC Steel Construction Manual |
Then make it real: pick one row per study session, open that standard in the edition for your date, find the governing section, and write down the edition, section, the conditions that apply, and the search words that got you there. Search by the words in a section title, not only by section numbers, since numbering can differ between editions.
Your 12-week PE Civil Structural study plan
This plan assumes about 10 hours a week (for example, four 75-minute weeknight sessions, a 3-hour weekend problem block, and a 2-hour review block), or about 120 hours in all. That's our editorial example, not an NCEES requirement or a promise of readiness. Stretch any week where your error log keeps filling up.
| Week | Focus | What to do | What you should have at the end |
|---|---|---|---|
| 1 | Setup and baseline | Work all 24 problems above, untimed. Confirm your test date and code set. Get the PE Civil Reference Handbook from MyNCEES and access to the right editions. Start your error log. | A dated code list, 24 attempts, and a list of weak topics |
| 2 | Analysis basics | Reactions, shear and moment diagrams, axial and flexural stress, combined stress, deflection, truss joints | 30+ solved problems with free-body diagrams |
| 3 | Analysis special topics | Buckling and effective length, torsion, fatigue, thermal effects, bearing, progressive collapse concepts | Notes on when each applies and where it lives in the standards |
| 4 | Gravity loads and combinations | Dead and live loads, tributary areas, load paths, load combinations, moving and impact loads (AASHTO for bridges) | A load-calculation sheet with a section reference for each code-based value |
| 5 | Lateral and environmental loads | Wind, seismic, snow, rain and ice in ASCE 7 for your date; earth pressure and surcharge using the applicable soil, water and wall-restraint model | Worked examples of each load type, start to finish |
| 6 | Steel | Tension, compression, and flexural members; bracing; bolted and welded connections in the AISC Manual | For each problem: demand, checks, AISC section, result |
| 7 | Concrete | Beams, slabs, columns, shear, development and anchorage in ACI 318; precast and prestressed basics in the PCI Handbook | A concrete checklist and worked problems |
| 8 | Wood and masonry | Wood members and connections by ASD in the NDS and SDPWS; masonry walls and members in TMS 402/602 | Separate worksheets naming the method and section for each |
| 9 | Soils and foundations | Soil properties and phase relationships, spread and combined footings, piles and drilled shafts, retaining walls | Pressure and load diagrams for each problem, with contact, bearing, and stability checks kept separate |
| 10 | Temporary structures and site safety | Formwork and shoring, steel erection, scaffolding, fall protection in the listed OSHA sections; special inspections and submittals; effects on adjacent facilities | A checklist of trigger conditions (install, load, remove) with CFR locators |
| 11 | Full timed practice | Take an 80-question practice exam under exam rules: 8 hours, one approved calculator, only on-screen PDFs, and a break after you submit the first half. This page supplies 24 starter problems; use a separate, unfamiliar 80-question set matched to your test-date standards for a full simulation. Without one, time shorter sets and review them, but don't treat that drill as a full exam. | Results by area and a timing record: reading, setup, lookup, arithmetic |
| 12 | Repair and logistics | Rework every miss from week 11 with fresh problems on the same ideas. Repeat the reference drill. Check your appointment, ID, and calculator model. | A short list of what's left, and a calm last few days |
Less time, or more?
With 8 weeks, merge weeks 2–3, 4–5, 7–8, and 9–10, and keep the full practice exam. Keeping all 120 study hours means about 15 hours a week. With 16 weeks, give the extra time to component design (weeks 6–9), since that area carries the most questions (26–39), and add a second timed exam. Either way, let your error log decide where extra hours go, not the calendar.
If you're retaking
NCEES sends a diagnostic report showing how you did in each major topic area if you don't pass. Use it to reorder weeks 2–10 so your weakest areas come first, then follow the same plan. Don't turn the report's bars into a guessed raw score; NCEES doesn't publish the passing score. See understanding your NCEES results and the NCEES retake policy.
Exam-day rules that change how you study
- 80 questions, 8 hours of testing. The 9-hour appointment also covers a 2-minute nondisclosure agreement, an 8-minute tutorial, and a 50-minute scheduled break. That's six minutes per question on average, not a per-question limit. (NCEES PE Civil page)
- The first half locks. After you see about half the questions, you're prompted to review and submit them, and you can't go back afterward. Finish your review before the break. The two sections share the eight-hour testing allowance; they are not separately timed. (Examinee Guide, p. 11)
- References are on-screen only. You get the PE Civil Reference Handbook and the listed standards as searchable PDFs. Personal books and notes aren't allowed. Practice with PDFs, not tabbed paper copies, in your last few weeks. (NCEES PE Civil page)
- One approved calculator. For 2026 exams, NCEES approves Casio fx-115 and fx-991 models, HP 33s and 35s, and TI-30X and TI-36X models. An on-screen TI-30XS is also available. If you test in 2027, check the list for that year. (NCEES 2026 calculator memo, p. 1; Examinee Guide, p. 8)
- Scratch work goes on reusable booklets. The test center provides two reusable booklets and three markers. (Examinee Guide, p. 9)
- No partial credit, no penalty for wrong answers. Alternative item types (select all that apply, point and click, drag and drop, fill in the blank) are scored right or wrong. Wrong answers aren't deducted, so answer every question. (NCEES CBT page; Examinee Guide, p. 14)
- Some questions don't count. Every exam includes a limited number of unscored pretest items that don't affect your result. (Examinee Guide, p. 11)
- Pass or fail, on a scaled score. NCEES doesn't publish a passing score because it varies slightly with difficulty. Results usually arrive 7–10 days after the exam. (Examinee Guide, p. 14; NCEES PE Civil page)
Quick answers
What's the pass rate? In NCEES's January–June 2026 data (updated July 2026), 62% of 1,624 first-time takers passed PE Civil: Structural, and 43% of 847 repeat takers passed. These figures cover examinees testing under an NCEES member board's jurisdiction. NCEES updates these figures each January and July. (NCEES PE Civil page, Scoring & Pass Rates)
How much does it cost? The exam fee is $400, paid to NCEES. Your licensing board may also charge its own application fee. See our guide to NCEES registration and scheduling. (NCEES PE Civil page)
Do I need board approval before I register? It depends on the board. Some require a separate application before you can book a seat. NCEES says the exam is designed for engineers with at least four years of post-college experience in their discipline, but your board sets the actual eligibility rules. Find yours in the NCEES licensing board directory.
Does passing make me licensed? No. Passing is one requirement. Your state or territorial board grants the license. (NCEES licensing board directory)
How often can I retake it? NCEES allows one attempt per testing window and no more than three in any 12-month period. The windows are January–March, April–June, July–September, and October–December. Your board may be stricter. (Examinee Guide, p. 5)
Can I get testing accommodations? Yes, if you meet NCEES's eligibility criteria and document your request. Request accommodations during registration and obtain approval before scheduling. (NCEES Examinee Guide, pp. 3, 5) See our guide to NCEES exam accommodations.
Are older PE review materials still useful? For mechanics, load paths, and problem setup, often yes. For anything that depends on a code value or provision, check that the material's edition matches your exam's code set in the table above.
Sources and verification
- NCEES, PE Civil exam page: format, fee, references, results timing, pass rates
- NCEES PE Civil–Structural specifications and design standards, exams before April 2027
- NCEES PE Civil–Structural specifications and design standards, beginning April 2027
- NCEES Examinee Guide, May 2026: retakes (p. 5), calculator (p. 8), test-center supplies (p. 9), reference search (p. 10), sections and pretest items (p. 11), scoring (p. 14)
- NCEES, Computer-based testing: alternative item types
- NCEES, 2026 approved calculators, p. 1, memo dated October 20, 2025
- NCEES, PE Structural exam page
- OSHA: 29 CFR 1926.501, 1926.502, 1926.703, 1926.754
- AWC, NDS 2018 with Commentary, Chapter 2, §2.3.2 and Table 2.3.2
- AWC Design Aid No. 6, Beam Design Formulas with Shear and Moment Diagrams
- David Roylance, MIT OpenCourseWare 3.11: Statics of Bending, Stresses in Beams, Introduction to Elasticity
- MIT OpenCourseWare 4.440, Basic Structural Design, equation sheet: Euler buckling and effective length factors
- Joseph A. Yura, "Fundamentals of Beam Bracing," AISC Engineering Journal, 2001, p. 15
- FHWA-HRT-14-094, Eccentricity, Bearing Pressure Diagram, and Footing Liftoff
- Quimby, Tributary Areas for Gravity Loads (§TA.2)
- Caltrans, Trenching and Shoring Manual, Chapter 4 (§4-3; pp. 4-5–4-6, 4-9–4-10)
- FHWA, Rockery Design and Construction Guidelines, Chapter 4 (Sliding Resistance, before Fig. 32)
- Roylance, Trusses, MIT OCW 3.11 (pp. 2–4)
- FHWA NHI-05-037, Geotechnical Aspects of Pavements (§5.3.1, Tables 5-7 and 5-8)
- STRUCTURE, Flexural Design of Reinforced Concrete Beam Sections (Section Analysis Basics; Eq. 7; ACI 318-19)
- ACI University, ACI 318-19 Changes: High-Strength (video summary)
- MIT OCW 1.051, Structural Engineering Design, Quiz 2 Solutions (Question 4(b), p. 5)
- FHWA NHI-06-089, Soils and Foundations, Volume II (Eqs. 9-3 and 9-17; chapter copy)
- Bendixen, Plastic Design of Eccentrically Loaded Fasteners, AISC Engineering Journal (Eq. 1(b), p. 125)
Castleport Test Prep Editorial Team · Last verified October 7, 2026 (NCEES exam page, both PE Civil–Structural specification documents, the May 2026 Examinee Guide, calculator policy, pass rates, the cited OSHA sections, and the technical sources behind each practice problem)
This guide was developed with AI-assisted research and editing. Every practice answer was solved and checked against the cited sources and the values given in each problem. Source checking is not a professional engineering review. See how Castleport uses sources and AI assistance.
Castleport Test Prep is an independent exam prep publisher, not affiliated with, endorsed by, or approved by NCEES. Exam and credential names identify the subjects discussed; trademarks belong to their respective owners. The practice problems are original and unofficial, and no score on them predicts an exam result or licensure.