PE Civil Construction Exam Prep
Start your PE Civil Construction exam prep with this original, unofficial problem. All 36 problems, worked solutions, and the 12-week plan are free on this page.
Problem 1 · Construction loads: area load to support reaction
CON-SUP-01 · Select one
An idealized temporary beam is simply supported over a 10-ft span with no overhangs. It carries a 6-ft tributary width of a uniform downward area load of 180 lb/ft² (concrete, forms, workers, and materials). The beam itself weighs 20 lb/ft. There are no other loads. What is the vertical reaction at each end?
- A. 5.40 kip
- B. 11.00 kip
- C. 5.50 kip
- D. 1.10 kip
Show answer and solution
Answer: C. 5.50 kip.
Follow the load path from area load to line load to reactions. Line load from the deck: 180 lb/ft² × 6 ft = 1,080 lb/ft. Add self-weight: 1,080 + 20 = 1,100 lb/ft. Total load on the span: 1,100 × 10 = 11,000 lb. A symmetric uniform load splits evenly, so each end carries wL/2 = 5,500 lb = 5.50 kip. A forgets the beam's own weight. B is the total load on both supports. D is the line load (1.10 kip/ft) mistaken for a force. This reaction is the demand on the support; it says nothing yet about whether the beam, connection, or shore below has the capacity to carry it.
Try a change: Widen the tributary width to 8 ft and keep the 20 lb/ft self-weight. Does everything scale by 8/6? No: only the deck load changes. w = 180(8) + 20 = 1,460 lb/ft, so each reaction = 1,460(10)/2 = 7,300 lb = 7.30 kip.
Source: American Wood Council, Beam Formulas with Shear and Moment Diagrams (Design Aid 6), Figure 1, p. 4
By Castleport Test Prep Editorial Team · Exam facts last verified October 7, 2026
Topic weights · References for my test date · More practice problems · 12-week plan
The exam at a glance
- 80 questions, computer-based, year-round at Pearson VUE test centers
- 8 hours of exam time inside a 9-hour appointment (2-min agreement, 8-min tutorial, 50-min scheduled break)
- $400 exam fee paid to NCEES; your licensing board may charge its own application fee
- References are supplied on screen: the NCEES PE Civil Reference Handbook plus the design standards listed for your test date. You can't bring your own.
- Both SI and U.S. customary units
- Results are pass/fail, typically 7–10 days after the exam
- First-time pass rate: 60% of 1,000 examinees (January–June 2026)
Sources: NCEES PE Civil exam page; NCEES PE Civil: Construction exam specifications.
What to study first: the 11 areas by weight
Use the official topic weights below: temporary works and construction operations carry the most questions. Your test date decides which design-standard editions you get on screen (they change for exams starting April 2027), but the topics and their weights stay the same.
NCEES publishes a question range for each of the 11 knowledge areas. Here they are, heaviest first.
| Rank | Knowledge area | Questions | What it covers | Practice here |
|---|---|---|---|---|
| 1 | Design for Support of Construction Loads | 10–15 | Formwork; falsework and scaffolding; shoring and reshoring; bracing and anchorage; support of excavation; construction loads on permanent structures (erection analysis, equipment loading) | Problems 1–5 |
| 2 | Construction Operations and Methods | 9–14 | Lifting and rigging; cranes (stability, outrigger loads, lifting capacity, crane types); dewatering and pumping; equipment selection and operations; deep foundation installation; excavation and embankment (cut and fill, borrow pit volume, haul distances) | Problems 6–8 |
| 3 | Project Planning and Scheduling | 7–11 | Construction sequencing; activity time analysis; network analysis, CPM, and linear schedules; resource scheduling and leveling; time–cost trade-off | Problems 9–11 |
| 3 | Material, Production, and Execution Quality Control | 7–11 | Test methods and spec conformance; weld and bolt installation; QA/QC; concrete placement; concrete maturity and early strength; compaction of soils, asphalt, and aggregates | Problems 12–15 |
| 3 | Structural Mechanics | 7–11 | Dead, live, construction, wind, and preloading loads; bending, shear, axial, combined stress, and deflection; beams, columns, one-way slabs; foundations, retaining walls, trusses, frames, two-way slabs, slab-on-grade | Problems 16–18 |
| 6 | Soil Mechanics | 6–9 | Lateral earth pressure; consolidation and settlement; effective and total stress; bearing capacity; slope stability | Problems 19–20 |
| 6 | Estimating Quantities and Costs | 6–9 | Quantity take-off; cost estimating; engineering economics (break-even, net present value, life-cycle cost); productivity and earned value | Problems 21–24 |
| 8 | Site Layout and Development | 5–8 | Staking, benchmarks, and elevations; horizontal and vertical curve elements; site investigations (adjacent structures, utilities) | Problems 25–27 |
| 8 | Material Properties | 5–8 | Soil, rock, and aggregates; concrete mix design, admixtures, and proportioning; structural, reinforcing, and prestressing steel; sawn and engineered wood | Problems 28–29 |
| 10 | Hydraulics and Hydrology | 4–6 | Open-channel flow, culverts, storm sewers; rational and SCS/NRCS methods, time of concentration; detention and retention ponds | Problems 30–32 |
| 10 | Health and Safety | 4–6 | OSHA construction regulations and safety management; work zone and public safety, including maintenance of traffic | Problems 33–36 |
The ranges are NCEES's; the ranking is ours, and it says nothing about which topics are hardest. Two takeaways:
- The top two areas alone are 19 to 29 questions. Scheduling, quality control, and structural mechanics each carry a 7–11-question range. That's where most of your hours should go.
- There's no separate breadth section anymore. Since April 2024, PE Civil exams are discipline-specific. Soils, hydraulics, and structural topics still appear, written in a construction context, so don't skip them.
Source: NCEES PE Civil: Construction exam specifications, effective April 2024, pp. 1–3.
Which references you'll get on exam day
NCEES supplies the handbook and a specific list of design standards as searchable PDFs. Your test date decides the list. The topics and question ranges are identical in both NCEES documents; only the references change.
| Reference supplied on screen | Exams before April 2027 | Exams starting April 2027 | What it means for you |
|---|---|---|---|
| NCEES PE Civil Reference Handbook | Current version | Current version | Download it free in MyNCEES and practice with it |
| ACI 347R, Guide to Formwork for Concrete | 2014, listed on its own | Included in the appendix of ACI SP-4 | Same guide; learn where it sits inside SP-4 |
| ACI SP-4, Formwork for Concrete | 8th ed., 2014 | 8th ed., 2014 | No change |
| ACI 318, Building Code Requirements for Structural Concrete | Not listed | 2019 (reapproved 2022), added | A new code to learn to navigate |
| AISC Steel Construction Manual | 15th edition | 16th edition | Table and section locations can move between editions |
| ASCE 37-14, Design Loads on Structures During Construction | 2nd ed., 2015 | 2nd ed., 2015 | No change |
| CMWB, Standard Practice for Bracing Masonry Walls Under Construction | 2012 | Not listed | Not supplied if you test April 2027 or later |
| PCA EB001, Design and Control of Concrete Mixtures | 17th ed., 2021 | 17th ed., 2021 | No change (PCA is now the American Cement Association) |
| MUTCD Part 6, temporary traffic control | 2009 edition, with Revisions 1 and 2 (May 2012) | 11th edition, 2023 | Chapter and table numbers moved; some values changed |
| Code of Federal Regulations, Title 29 | July 2020: Parts 1903 and 1926 | 2024: Parts 1903, 1904, and 1926 | Part 1904 (injury recordkeeping and reporting) is added |
Sources: specifications for exams before April 2027, p. 4; specifications for exams starting April 2027, p. 4.
Why the edition matters, in one example. Both MUTCD editions use the same taper-length formulas (L = WS²/60 at 40 mph or less, L = WS at 45 mph or more). But the 2009 edition's table lists a downstream taper of 100 ft per lane, while the 11th edition lists 50 ft minimum and 100 ft maximum, and the tables themselves moved from 6C-3 and 6C-4 to 6B-3 and 6B-4. (2009 MUTCD, Section 6C.08; 11th edition, Section 6B.08)
When choosing references, match the edition to your test date. If your date could slip into April 2027 or later, plan around the newer list. Dropping a reference from the list doesn't mean its subject leaves the exam; formwork, bracing and anchorage, and traffic control are still covered by the topic list.
Practice finding the rule, not just a word
On the exam you search the supplied PDFs with a search box on the left side of the reference window. Ctrl+F doesn't work (NCEES Examinee Guide, May 2026, p. 10). Design standards are supplied as individual chapters, and only one chapter at a time can be opened and searched. Knowing which document answers which kind of question saves more time than any search trick:
| If the problem is about… | Start in… |
|---|---|
| Formwork pressure, ties, form design | ACI SP-4 (and ACI 347R) |
| Load cases and combinations during construction | ASCE 37-14 |
| Steel member properties and capacities | AISC Steel Construction Manual |
| Concrete materials, mix proportioning, curing | PCA EB001 |
| Concrete design requirements (April 2027 and later) | ACI 318 |
| Work zone signs, tapers, channelizing devices | MUTCD Part 6 |
| Excavations, fall protection, scaffolds, cranes, concrete construction safety | 29 CFR 1926 |
| Equations for soils, hydraulics, surveying, economics, scheduling | NCEES PE Civil Reference Handbook |
A good lookup drill takes five steps: pick the document, use its contents to pick the chapter, search a technical term from the problem, read the definitions and nearby exceptions, and note where you found it. Keep a log so slow lookups get repeated until they're fast:
| Topic | Document and edition | Chapter or section found | Search term that worked | Time to find | What slowed you down |
|---|---|---|---|---|---|
| (example) Trench sloping, Type B soil | 29 CFR 1926, Subpart P | Appendix B, Table B-1 | "maximum allowable slopes" | 3 min | Searched "trench slope" first |
36 original practice problems with worked solutions
These are original Castleport problems written to the NCEES Construction topic list. They are not NCEES questions, and your score on them isn't a prediction of your exam result. This 36-problem set is not a full-length exam and does not exhaust every listed subtopic. Most are standard multiple choice. A few use formats like those on the real exam: select-all-that-apply and enter-a-number. On the actual exam, every question, including those formats, is scored right or wrong with no partial credit (Examinee Guide, p. 11).
Work each problem before reading the answer. When you miss one, write down the first step that went wrong; the error log below shows what to do with it.
Design for Support of Construction Loads (5 problems)
Problem 1, construction loads: area load to support reaction, appears at the start of this guide. Continue here with Problems 2–5.
Problem 2 · Formwork lateral pressure (liquid head)
CON-SUP-02 · Select one
Fresh concrete with a unit weight of 150 lb/ft³ fills a 12-ft-tall wall form. For this calculation, assume a static full-liquid-head pressure distribution. What is the maximum lateral pressure, at the base of the form?
- A. 900 psf
- B. 150 psf
- C. 1,800 psf
- D. 3,600 psf
Show answer and solution
Answer: C. 1,800 psf.
Treating the concrete as a fluid, pressure grows linearly with depth: p = wh = 150 × 12 = 1,800 psf at the base, zero at the top. A is the average pressure over the height (useful for finding the resultant force, not the peak). B mistakes the unit weight (lb/ft³) for a pressure (lb/ft²). D doubles the answer. This calculation uses the stipulated static liquid-head model. OSHA requires formwork to carry the lateral loads it can reasonably be expected to see.
Source: OpenStax, University Physics Volume 1 (2016), Section 14.1: pressure, force and area; static fluid pressure, Equation 14.4; OSHA, 29 CFR 1926.703, Requirements for cast-in-place concrete
Problem 3 · Form tie spacing
CON-SUP-03 · Select one
Form ties have a safe working load of 3,000 lb. The design lateral pressure over a panel is a uniform 1,000 psf. Which tie layout uses the largest tributary area per tie without overloading the ties?
- A. 2.0 ft horizontal × 1.5 ft vertical
- B. 2.0 ft × 2.0 ft
- C. 3.0 ft × 1.5 ft
- D. 1.0 ft × 1.0 ft
Show answer and solution
Answer: A. 2.0 ft horizontal × 1.5 ft vertical.
Each tie can serve 3,000 lb ÷ 1,000 psf = 3.0 ft² of form. A gives exactly 2.0 × 1.5 = 3.0 ft². B serves 4.0 ft² (4,000 lb per tie). C serves 4.5 ft² (4,500 lb). D satisfies the stated tie-force limit but has a smaller tributary area than A. This comparison checks the tie force, not the capacity of the rest of the form system.
Problem 4 · Support of excavation: sloping in Type B soil
CON-SUP-04 · Select one
A 14-ft-deep trench in Type B soil will be sloped without shoring, using OSHA's Appendix B maximum allowable slopes. The ground is level, and a competent person has confirmed that the soil and site conditions permit the tabulated Type B slope, with no distress or surcharge requiring a flatter slope. Both sides are sloped, and the bottom must be 4 ft wide. What is the minimum top width?
- A. 18 ft
- B. 25 ft
- C. 32 ft
- D. 46 ft
Show answer and solution
Answer: C. 32 ft.
Table B-1 sets Type B soil at 1H:1V (45°) for excavations 20 ft deep or less. Each side runs 14 ft horizontally, so the top width is 4 + 2(14) = 32 ft. B uses Type A's ¾H:1V (4 + 2 × 10.5). D uses Type C's 1½H:1V (4 + 2 × 21). A slopes only one side. Past 20 ft deep, sloping or benching must be designed by a registered professional engineer.
Problem 5 · Formwork and shoring requirements
CON-SUP-05 · Select all that apply
Under OSHA's cast-in-place concrete rules (29 CFR 1926.703), which statements are correct? Select all that apply.
- A. Formwork must be able to support, without failure, all vertical and lateral loads that may reasonably be anticipated.
- B. Drawings or plans for the jack layout, formwork (including shoring equipment), working decks, and scaffolds, including all revisions, must be available at the jobsite.
- C. Erected shoring equipment must be inspected immediately before, during, and immediately after concrete placement.
- D. Designing for vertical loads alone is enough when no strong wind is forecast.
- E. An eccentric load on a shore head is acceptable without a special design as long as it is below the shore's axial load rating.
Show answer and solution
Answer: A, B and C.
A matches 1926.703(a)(1). B matches 1926.703(a)(2). C matches 1926.703(b)(3). D is wrong because the rule covers reasonably anticipated lateral loads, with no exception tied to a forecast. E is wrong because 1926.703(b)(7) prohibits eccentric loads on shore heads and similar members unless those members were designed for that loading. A centered axial rating says nothing about an eccentric load.
Source: OSHA, 29 CFR 1926.703, Requirements for cast-in-place concrete
Construction Operations and Methods (3 problems)
Problem 6 · Rigging: sling angle and leg tension
CON-OPS-01 · Select one
A 10,000-lb load with its center of gravity centered between the pick points hangs from a two-leg bridle. Both legs are the same length, and each leg is 60° from horizontal. Assume a static load and negligible sling weight. What is the tension in each leg?
- A. 5,000 lb
- B. 5,774 lb
- C. 7,071 lb
- D. 10,000 lb
Show answer and solution
Answer: B. 5,774 lb.
Each leg's vertical component carries half the load: T sin θ = W/2, so T = 10,000 / (2 × sin 60°) = 10,000 / 1.732 = 5,774 lb. A ignores the sling angle. C is the answer at 45°. D assumes one leg takes the whole load. Flatter angles raise leg tension quickly, which is why low sling angles get so much attention in rigging.
Problem 7 · Equipment production in bank volume
CON-OPS-02 · Select one
A scraper carries 20 loose cubic yards (LCY) per load on a 6.0-minute cycle. Assume a 50-minute productive hour and 25% swell (1 bank cubic yard, BCY, becomes 1.25 LCY). What is the hourly production in BCY?
- A. 133 BCY/hr
- B. 167 BCY/hr
- C. 200 BCY/hr
- D. 107 BCY/hr
Show answer and solution
Answer: A. 133 BCY/hr.
Cycles per hour = 50 ÷ 6.0 ≈ 8.3333. Loose production = (50 ÷ 6.0) × 20 ≈ 166.6667 LCY/hr. Bank production = [(50 ÷ 6.0) × 20] ÷ 1.25 ≈ 133.3333 BCY/hr, or 133 BCY/hr to the nearest whole number. B stops at loose yards. C uses a 60-minute hour and stays in loose yards. D applies an extra 0.8 efficiency factor on top of the 50-minute hour, counting lost time twice. Label every volume as bank, loose, or compacted before you convert.
Source: FHWA Federal Lands, Earthwork Design (updated Jan. 15, 2026): shrink/swell, average end area, haul
Problem 8 · Matching haul units to a loader
CON-OPS-03 · Select one
A loader fills a truck every 3.0 minutes. Each truck's cycle is: 3.0 min loading, a 2-mile haul at 20 mph, 2.0 min dumping, and a 2-mile return at 30 mph. Assume constant cycle times, continuous operation, and no interference between trucks. What is the minimum number of trucks needed to keep the loader working without waiting?
- A. 3
- B. 4
- C. 5
- D. 6
Show answer and solution
Answer: C. 5.
Haul time = 2/20 hr = 6 min. Return time = 2/30 hr = 4 min. Cycle = 3 + 6 + 2 + 4 = 15 min. Trucks needed = 15 ÷ 3 = 5. B leaves out the return trip (11 ÷ 3 = 3.7, rounded to 4). A counts travel time only and rounds down. D supplies the loader but exceeds the minimum; the sixth truck adds waiting rather than increasing loader output. When the division doesn't come out even, rounding up keeps the loader busy and rounding down keeps the trucks busy, so read which one the problem asks for.
Source: worked from first principles using only the inputs given.
Project Planning and Scheduling (3 problems)
Problem 9 · CPM: total float vs. free float
CON-SCHED-01 · Select one
All relationships are finish-to-start with no lag. Use elapsed working days starting at day 0, with no resource constraints. Durations in working days: A = 3 (no predecessor); B = 5 (after A); C = 4 (after A); D = 6 (after B); E = 2 (after C); F = 3 (after D and E). Which statement about activity C is correct?
- A. Total float 5 days, free float 0 days
- B. Total float 5 days, free float 5 days
- C. Total float 0 days; C is critical
- D. Total float 0 days, free float 5 days
Show answer and solution
Answer: A. Total float 5 days, free float 0 days.
Forward pass: A 0–3, B 3–8, C 3–7, D 8–14, E 7–9, F 14–17. The project takes 17 days on the critical path A-B-D-F. Backward pass: F's late start is 14, so E's late finish is 14 and its late start is 12; C's late finish is 12. Total float of C = 12 − 7 = 5 days. Free float of C = E's early start − C's early finish = 7 − 7 = 0. So any slip in C immediately delays E, but the project finish isn't affected until C slips more than 5 days. B gives C the free float that belongs to E (E can slip 5 days before F is affected).
Try a change: Change E's duration from 2 to 7 days. E now runs 7–14, F still starts at 14, and A-C-E-F also totals 17 days. The finish date doesn't move, but there are now two critical paths and C has zero float.
Problem 10 · Time–cost trade-off (crashing)
CON-SCHED-02 · Select one
A project has exactly two start-to-finish paths: A-B-D-F takes 17 working days, and A-C-E-F takes 12 working days. You must finish 2 days earlier. Crash options (cost per day, maximum days): A $800, 1; B $500, 2; D $300, 1; F $1,000, 1. C and E cannot be shortened. What is the minimum added cost?
- A. $600
- B. $800
- C. $1,000
- D. $1,300
Show answer and solution
Answer: B. $800.
Only critical activities shorten the project, and A-C-E-F (12 days) stays shorter than the crashed path, so it never becomes critical here. Take the cheapest day first: D for 1 day ($300). D is then maxed out, so take the next cheapest: B for 1 day ($500). Total = $800. A assumes D can be crashed twice. C would buy two days of B ($1,000), which works but costs more. D crashes D and F ($1,300).
Problem 11 · Effect of a delay on project finish
CON-SCHED-03 · Select one
In an unconstrained finish-to-start CPM schedule, an activity has ES = 10, EF = 15, LS = 12, and LF = 17, all in elapsed working days. It is delayed by 3 days, and nothing else in the schedule changes. How much is project completion delayed?
- A. None
- B. 1 day
- C. 2 days
- D. 3 days
Show answer and solution
Answer: B. 1 day.
Total float = LS − ES = 12 − 10 = 2 days. A 3-day slip uses up the 2 days of float and pushes the finish by 1 day. D treats the activity as critical. A assumes the float covers the whole delay. C confuses the float with the delay.
Material, Production, and Execution Quality Control (4 problems)
Problem 12 · Relative compaction from a field density test
CON-QC-01 · Select one
The standard Proctor maximum dry density for a fill soil is 118.0 pcf. A field test measures a wet (moist) density of 128.5 pcf at 10.5% moisture content by dry soil mass. Relative compaction is field dry density divided by maximum dry density. The specification requires at least 95%. What is the result?
- A. 98.6%, passes
- B. 108.9%, passes
- C. 97.5%, passes
- D. 89.2%, fails
Show answer and solution
Answer: A. 98.6%, passes.
Field dry density = wet density ÷ (1 + w) = 128.5 ÷ 1.105 ≈ 116.2896 pcf. Relative compaction = [128.5 ÷ 1.105] ÷ 118.0 × 100% ≈ 98.5505%, or 98.6%, which meets the stated 95% criterion. B compares wet density to a dry maximum. C multiplies by (1 − w) instead of dividing by (1 + w): right verdict, wrong number, and wrong method. D divides by (1 + w) twice.
Problem 13 · Concrete maturity (Nurse-Saul)
CON-QC-02 · Select one
Sensors in a slab average 20 °C for the first 12 hours and 14 °C for the next 12 hours. Using the Nurse-Saul temperature-time factor with a datum temperature of −10 °C, what is the maturity at 24 hours, and what makes that number usable for estimating in-place strength?
- A. 648 °C·h; a strength–maturity calibration curve developed for this mix
- B. 408 °C·h; a strength–maturity calibration curve developed for this mix
- C. 648 °C·h; one general maturity-versus-strength table that applies to all normal-weight concrete
- D. 34 °C·h; field-cured cylinders only
Show answer and solution
Answer: A. 648 °C·h; a strength–maturity calibration curve developed for this mix.
M = Σ(T − T₀)Δt = (20 − (−10))(12) + (14 − (−10))(12) = 360 + 288 = 648 °C·h. B uses 0 °C as the datum. The maturity number only estimates strength through a calibration curve built with the project's own materials; FHWA notes that a change in the mix design requires a new curve, so there is no one-size table (C). D adds the temperatures without multiplying by time.
Problem 14 · QC vs. QA vs. verification testing
CON-QC-03 · Select one
On a federal-aid highway project, the contractor's own technician runs density tests so the crew can decide whether the rolling pattern needs more passes. Under the federal definitions in 23 CFR 637.203, this testing is:
- A. Quality control
- B. Verification sampling and testing
- C. Independent assurance
- D. Owner acceptance
Show answer and solution
Answer: A. Quality control.
23 CFR 637.203 defines quality control as the contractor/vendor operational techniques and activities performed to fulfill the contract requirements, which describes a contractor adjusting its own process. Verification sampling and testing validates the quality of the product. Independent assurance is an unbiased evaluation of the sampling and testing procedures used in the acceptance program. Owner acceptance is a decision about the work, not the contractor's process testing.
Problem 15 · Cylinder compressive strength
CON-QC-04 · Select one
A concrete cylinder has a measured average diameter of 6.00 in. Its maximum load in a valid compression test is 141,400 lbf. What is its compressive strength, reported to the nearest 10 psi?
- A. 1,250 psi
- B. 5,000 psi
- C. 3,930 psi
- D. 11,250 psi
Show answer and solution
Answer: B. 5,000 psi.
Area = π(6.00)²/4 ≈ 28.2743 in². Strength = 141,400 ÷ [π(6.00)²/4] ≈ 5,001.0 psi, which reports as 5,000 psi. A treats the 6-in. diameter as a radius (area 113.1 in²). C uses d² = 36 in² as the area. D uses the area of a 4-in. cylinder (12.57 in²). One cylinder result describes that specimen; deciding whether a placement is accepted, or whether forms can be stripped, takes the project's acceptance criteria.
Source: Iowa DOT Instructional Memorandum 315, Section II.F, Calculations
Structural Mechanics (3 problems)
Problem 16 · Simple beam under uniform load
CON-STR-01 · Select one
A simply supported 20-ft beam carries a uniform downward construction load of 400 plf over its full span, including its own weight. What are the maximum moment and maximum shear?
- A. 20 kip-ft; 4 kips
- B. 40 kip-ft; 8 kips
- C. 20 kip-ft; 8 kips
- D. 10 kip-ft; 4 kips
Show answer and solution
Answer: A. 20 kip-ft; 4 kips.
M = wL²/8 = 0.4 × 20² / 8 = 20 kip-ft. V = wL/2 = 0.4 × 20 / 2 = 4 kips. B uses wL²/4 and treats the total load as the shear. C has the right moment but the total load as the shear. D uses wL²/16.
Source: American Wood Council, Beam Formulas with Shear and Moment Diagrams (Design Aid 6), Figure 1, p. 4
Problem 17 · Reactions from an off-center point load
CON-STR-02 · Select one
A simply supported beam spans 20 ft between a pin at the left end and a roller at the right end. A 12-kip downward point load acts 5 ft from the left support. Ignore the beam's weight. What is the upward reaction at the right support?
- A. 3 kips
- B. 6 kips
- C. 9 kips
- D. 12 kips
Show answer and solution
Answer: A. 3 kips.
Take moments about the left support: R_right × 20 = 12 × 5, so R_right = 3 kips. Vertical equilibrium gives R_left = 12 − 3 = 9 kips. B splits the load evenly, which only works at midspan. C is the left reaction. D puts the whole load on one support. The support nearer the load always carries more of it.
Source: American Wood Council, Beam Formulas with Shear and Moment Diagrams (Design Aid 6), Figure 8, p. 7
Problem 18 · Axial stress in a timber shore
CON-STR-03 · Enter a number
A shore cut from a nominal 4×4 (actual dressed size 3.5 in. × 3.5 in.) carries a concentric 6,000-lb axial compressive load. Enter the axial compressive stress in psi, rounded to the nearest 10 psi.
Show answer and solution
Answer: 490 psi.
Area = 3.5 × 3.5 = 12.25 in². Stress = 6,000 ÷ 12.25 ≈ 489.80 psi, or 490 psi to the nearest 10 psi. Using the nominal 4 × 4 = 16 in² gives 375 psi, which understates the stress by about a quarter. Always use actual dressed dimensions. Comparing this stress to an allowable value, including column buckling, is the next step in a real check.
Soil Mechanics (2 problems)
Problem 19 · Rankine active earth pressure
CON-SOIL-01 · Select one
A 12-ft cantilever wall with a vertical back face retains level, dry, cohesionless backfill with a unit weight of 120 pcf and a friction angle of 30°. Assume sufficient wall movement to develop active pressure, with no surcharge or seismic load. Ignore wall friction. Using Rankine theory, what is the active resultant per foot of wall, and where does it act?
- A. 2,880 lb/ft at 4 ft above the base
- B. 2,880 lb/ft at 6 ft above the base
- C. 5,760 lb/ft at 4 ft above the base
- D. 8,640 lb/ft at 4 ft above the base
Show answer and solution
Answer: A. 2,880 lb/ft at 4 ft above the base.
Ka = tan²(45° − φ/2) = tan²(30°) = 1/3. Pa = ½ Ka γ H² = ½ × (1/3) × 120 × 12² = 2,880 lb/ft. The pressure diagram is a triangle, so the resultant acts at H/3 = 4 ft above the base. B puts it at mid-height, as if the pressure were uniform. C drops the ½. D uses a coefficient of 1, as if the soil were a fluid of the same weight.
Problem 20 · Total vs. effective stress
CON-SOIL-02 · Enter a number
The groundwater table is 5 ft below level ground. Soil above it weighs 115 pcf; soil below it has a saturated unit weight of 125 pcf. Assume hydrostatic groundwater, no surcharge, and no capillary rise. Use γw = 62.4 pcf. Enter the vertical effective stress at a depth of 15 ft, in psf, rounded to the nearest 10 psf.
Show answer and solution
Answer: 1,200 psf.
Total stress = 5(115) + 10(125) = 575 + 1,250 = 1,825 psf. Pore-water pressure counts only depth below the water table: u = 10 × 62.4 = 624 psf. Effective stress = 1,825 − 624 = 1,201 psf, or 1,200 psf rounded. Subtracting water pressure for the full 15 ft gives 889 psf, incorrectly applying hydrostatic water pressure above the stated water table. Mark the water table on a sketch before you start.
Try a change: Raise the water table to 3 ft below grade, everything else the same. Total stress = 3(115) + 12(125) = 1,845 psf; u = 12 × 62.4 = 748.8 psf; effective stress = 1,096.2 psf. A higher water table lowers effective stress.
Estimating Quantities and Costs (4 problems)
Problem 21 · Average end area volume
CON-EST-01 · Select one
Cut cross-sections are 210 ft² at Sta. 10+00 and 330 ft² at Sta. 11+00, with stations measured in feet. Using the average end area method, what is the cut volume between the two stations?
- A. 540 CY
- B. 1,000 CY
- C. 2,000 CY
- D. 27,000 CY
Show answer and solution
Answer: B. 1,000 CY.
V = L(A₁ + A₂)/2 = 100 × (210 + 330)/2 = 27,000 ft³. Divide by 27 to get 1,000 CY. D forgets to convert cubic feet to cubic yards. C skips the division by 2. A just adds the two areas.
Source: FHWA Federal Lands, Earthwork Design (updated Jan. 15, 2026): shrink/swell, average end area, haul
Problem 22 · Shrink, swell, and truck loads
CON-EST-02 · Select one
An embankment needs 12,000 compacted cubic yards. The borrow material shrinks 15% from bank to compacted (1 BCY becomes 0.85 CCY) and swells 25% from bank to loose (1 BCY becomes 1.25 LCY). Trucks haul 12 LCY per load. How many truckloads are needed?
- A. 1,000
- B. 1,177
- C. 1,250
- D. 1,471
Show answer and solution
Answer: D. 1,471.
Bank yards needed = 12,000 ÷ 0.85 ≈ 14,117.65 BCY. Loose yards hauled = (12,000 ÷ 0.85) × 1.25 ≈ 17,647.06 LCY. Loads = [(12,000 ÷ 0.85) × 1.25] ÷ 12 ≈ 1,470.59, so 1,471 trips are needed, including the final partial load. A ignores both factors. B applies shrinkage only. C applies swell only.
Source: FHWA Federal Lands, Earthwork Design (updated Jan. 15, 2026): shrink/swell, average end area, haul
Problem 23 · Earned value indexes
CON-EST-03 · Select all that apply
At a status date: budget at completion (BAC) $2,000,000; planned value (PV) $800,000; earned value (EV) $720,000; actual cost (AC) $900,000. Select every true statement.
- A. The project is over budget for the work performed so far.
- B. SPI = 0.90.
- C. CPI = 1.25.
- D. If current cost efficiency continues, the estimate at completion is $2,500,000.
- E. Schedule variance is +$80,000.
Show answer and solution
Answer: A, B and D.
CPI = EV ÷ AC = 720,000 ÷ 900,000 = 0.80. Below 1.0 means the work done has cost more than its budgeted value, so A is true and C (the inverse) is false. SPI = EV ÷ PV = 720,000 ÷ 800,000 = 0.90, so B is true. EAC = BAC ÷ CPI = 2,000,000 ÷ 0.80 = $2,500,000, so D is true. Schedule variance = EV − PV = −$80,000 (behind schedule), so E has the wrong sign. Both indexes put earned value on top.
Problem 24 · Equivalent annual cost: buy vs. rent
CON-EST-04 · Select one
A contractor can buy a roller for $150,000 and sell it for $30,000 after 5 years, or rent an equivalent roller for $45,000 per year paid at the end of each year. The purchase is paid now and the resale occurs at the end of year 5. Operating costs are the same either way; ignore any other cost differences. Interest is 8% per year. Which statement is correct?
- A. Buying has an equivalent annual cost of about $32,450, so buying is cheaper
- B. Buying has an equivalent annual cost of about $37,570, so buying is cheaper
- C. Renting is cheaper because 5 × $45,000 is less than the purchase price plus interest
- D. Buying has an equivalent annual cost of about $24,000, so buying is cheaper
Show answer and solution
Answer: A. Buying has an equivalent annual cost of about $32,450, so buying is cheaper.
Capital recovery factor (A/P, 8%, 5) = 0.08(1.08)⁵ / [(1.08)⁵ − 1] ≈ 0.25046. Sinking fund factor (A/F, 8%, 5) = 0.08 / [(1.08)⁵ − 1] ≈ 0.17046. Using unrounded factors, the equivalent annual cost of buying is 150,000(A/P) − 30,000(A/F) ≈ $32,454.77, or about $32,450 to the nearest $10, which is less than $45,000. B ignores the salvage value. C compares undiscounted totals and gets the comparison wrong anyway (5 × $45,000 = $225,000). D is straight-line (150,000 − 30,000) ÷ 5 with no interest.
Site Layout and Development (3 problems)
Problem 25 · Horizontal curve: middle ordinate
CON-SITE-01 · Select one
A circular horizontal curve has a radius of 800 ft and a deflection angle Δ of 40°. What is the middle ordinate?
- A. 51.3 ft
- B. 48.2 ft
- C. 291.2 ft
- D. 558.5 ft
Show answer and solution
Answer: B. 48.2 ft.
M = R[1 − cos(Δ/2)] = 800(1 − cos 20°) ≈ 48.2459 ft, or 48.2 ft. A is the external distance, R[1/cos(Δ/2) − 1], which is easy to mix up with M. C is the tangent length, R tan(Δ/2). D is the curve length, 2πRΔ/360.
Source: Caltrans Bridge Design Details, Attachment E: Horizontal Curve Equations (June 2025)
Problem 26 · Differential leveling to a form
CON-SITE-02 · Enter a number
From a benchmark at elevation 102.45 ft, the backsight reading is 4.32 ft. The foresight reading on the top of a footing form is 1.85 ft. Enter the elevation of the top of the form, in feet, to two decimals.
Show answer and solution
Answer: 104.92 ft.
Height of instrument = 102.45 + 4.32 = 106.77 ft. Top of form = 106.77 − 1.85 = 104.92 ft. If the design top of form is 105.00 ft, the computed difference is 0.08 ft low; whether an adjustment is required depends on the specified elevation tolerance. Adding the foresight instead of subtracting it gives 108.62 ft, a blunder worth catching with a quick sanity check: a smaller rod reading means a higher point.
Problem 27 · Vertical curve elevation
CON-SITE-03 · Select one
An equal-tangent vertical curve starts at a PVC elevation of 100.00 ft. The incoming grade is +2.0%, the outgoing grade is −1.0%, and the curve is 400 ft long. What is the curve elevation 100 ft past the PVC?
- A. 101.63 ft
- B. 102.00 ft
- C. 102.38 ft
- D. 101.25 ft
Show answer and solution
Answer: A. 101.63 ft.
y = PVC elevation + g₁x + [(g₂ − g₁)/(2L)]x² = 100.00 + 0.02(100) + [(−0.01 − 0.02)/(2 × 400)](100)² = 100.00 + 2.00 − 0.375 = 101.625 ft, or 101.63 ft. B is the elevation on the incoming tangent, ignoring the curve. C adds the curve correction instead of subtracting it on this crest curve. D uses (g₂ − g₁)/L without the 2.
Material Properties (2 problems)
Problem 28 · Water–cementitious materials ratio
CON-MAT-01 · Select one
A concrete batch contains 500 lb of portland cement, 100 lb of fly ash, 240 lb of added mixing water, and 30 lb of free water carried in by the aggregates. There are no other water sources or cementitious materials. What is the water–cementitious materials ratio (w/cm)?
- A. 0.40
- B. 0.45
- C. 0.54
- D. 2.22
Show answer and solution
Answer: B. 0.45.
w/cm = (240 + 30) ÷ (500 + 100) = 270 ÷ 600 = 0.45. FHWA defines the ratio as mix water, including free moisture on the aggregates, divided by cement plus supplementary cementitious materials such as fly ash. A leaves out the aggregate free water. C keeps the water but drops fly ash from the denominator. D flips the ratio.
Problem 29 · Batch water correction for aggregate moisture
CON-MAT-02 · Select one
A mix calls for 1,800 lb of coarse aggregate in the saturated-surface-dry (SSD) condition. The stockpile moisture content is 4.0% and the absorption is 1.0%, both based on oven-dry mass. About how much should the batch water be reduced to account for this aggregate's free water?
- A. 18 lb
- B. 54 lb
- C. 72 lb
- D. 90 lb
Show answer and solution
Answer: B. 54 lb.
Oven-dry mass = 1,800 ÷ 1.01 ≈ 1,782.1782 lb. Wet batch weight = (1,800 ÷ 1.01) × 1.04 ≈ 1,853.4653 lb. Free water = (1,800 ÷ 1.01)(0.04 − 0.01) ≈ 53.4653 lb, closest to the listed 54-lb choice. The shortcut 3% × 1,800 lb happens to select the same option, but it applies oven-dry-based percentages to SSD mass; use the oven-dry mass for the calculation. A uses absorption alone. C treats all 4% as free water. D adds moisture and absorption together. This kind of aggregate free water belongs in the w/cm numerator; Problem 28 supplies a different free-water amount.
Source: Caltrans Concrete Technology Manual (June 2013), Chapter 2, p. 2-19: oven-dry, SSD and wet aggregate moisture states; FHWA-04-122 (Feb. 2005), Chapter 5, Section 5.2.2: definition of water-to-cementitious materials ratio
Hydraulics and Hydrology (3 problems)
Problem 30 · Rational method (U.S. units)
CON-HYD-01 · Select one
A 12-acre site has a runoff coefficient C of 0.60. The design rainfall intensity, for a duration equal to the time of concentration, is 3.0 in/hr. Assume the rational method applies. Using the conventional U.S. approximation Q ≈ CiA, what is the peak runoff?
- A. 1.8 cfs
- B. 21.6 cfs
- C. 36.0 cfs
- D. 2.16 cfs
Show answer and solution
Answer: B. 21.6 cfs.
Q ≈ CiA = 0.60 × 3.0 × 12 = 21.6 cfs (with i in in/hr and A in acres, this conventional approximation rounds the unit-conversion factor to 1). A drops the area. C drops the runoff coefficient. D is a decimal slip. Intensity is taken at the time of concentration because that's when the whole area is contributing flow.
Source: TxDOT Hydraulic Design Manual, Chapter 4, Section 12, Rational Method, Equation 4-20
Problem 31 · Composite runoff coefficient
CON-HYD-02 · Enter a number
A drainage area is 4 acres of pavement (C = 0.90) and 8 acres of lawn (C = 0.20). Enter the area-weighted runoff coefficient to two decimals.
Show answer and solution
Answer: 0.43.
C = (4 × 0.90 + 8 × 0.20) ÷ 12 = (3.6 + 1.6) ÷ 12 ≈ 0.4333, or 0.43. A simple average of 0.90 and 0.20 gives 0.55, which ignores that the lawn covers twice as much area.
Source: TxDOT Hydraulic Design Manual, Chapter 4, Section 12, Mixed Land Use, Equation 4-23
Problem 32 · Rational method (SI units)
CON-HYD-03 · Enter a number
For a small catchment where the rational method applies: C = 0.65, i = 72 mm/h, and A = 2.5 ha. In SI units with i in mm/h and A in hectares, Q = CiA/360 gives Q in m³/s. Enter the peak discharge in m³/s, to three decimals.
Show answer and solution
Answer: 0.325 m³/s.
Q = (0.65 × 72 × 2.5) ÷ 360 = 117 ÷ 360 = 0.325 m³/s. The 360 converts mm/h over hectares into m³/s (1 ha = 10,000 m², 1 mm = 0.001 m, 1 h = 3,600 s). The exam uses both SI and U.S. units, so write the unit convention next to any formula before plugging in. The conventional U.S. form rounds its unit-conversion factor to 1; dropping the 360 here would give 117, which is badly wrong.
Source: TxDOT Hydraulic Design Manual, Chapter 4, Section 12, Rational Method, Equation 4-20
Health and Safety (4 problems)
Problem 33 · Fall protection triggers
CON-HS-01 · Select all that apply
For ordinary work on completed supported scaffolds and at unprotected floor edges, apply the height triggers in 29 CFR 1926.451(g)(1) and 1926.501(b)(1). Assume no other hazards or special work conditions trigger protection. Which cases meet those height triggers? Select all that apply.
- A. A worker on a supported scaffold platform 8 ft above the lower level
- B. A worker at an unprotected floor edge 7 ft above the lower level (not on a scaffold)
- C. A worker on a supported scaffold platform 12 ft above the lower level
- D. A worker at an unprotected floor edge 5 ft above the lower level
Show answer and solution
Answer: B and C.
The general rule for walking/working surfaces applies at 6 ft or more above a lower level (1926.501(b)(1)), so B meets that height trigger and D does not. Scaffolds are covered by Subpart L instead, which requires protection more than 10 ft above a lower level (1926.451(g)(1)), so C meets that height trigger and A does not. These answers address the two stated height triggers, not every possible fall hazard or work condition.
Source: OSHA, 29 CFR 1926.501(b)(1), Unprotected sides and edges; OSHA, 29 CFR 1926.451(g)(1), Scaffold fall protection
Problem 34 · Reporting a serious injury to OSHA
CON-HS-02 · Select one
Reference match: Part 1904 is added to the supplied NCEES reference list for exams starting April 2027. This item practices the stated federal reporting rule.
A worker suffers an immediate amputation in a work-related incident at the construction jobsite. The employer learns of the incident and amputation immediately. No one is hospitalized and no one dies. Under federal OSHA's reporting rule, within what time must the employer report the amputation?
- A. 8 hours
- B. 24 hours
- C. 7 days
- D. No report; record it on the OSHA 300 log only
Show answer and solution
Answer: B. 24 hours.
29 CFR 1904.39 requires reporting a fatality within 8 hours and an in-patient hospitalization, amputation, or loss of an eye within 24 hours. Here, the amputation occurs within 24 hours of the work incident and the employer knows immediately, so the 24-hour reporting deadline applies from the amputation. A is the fatality deadline. The 300 log is a separate recordkeeping duty, not a substitute for the report. Part 1904 is named on the 29 CFR reference list for exams starting April 2027.
Source: OSHA, 29 CFR 1904.39(a)(2), (b)(6)–(8): reporting deadline, event window, and employer knowledge
Problem 35 · Work zone merging taper length
CON-HS-03 · Select one
A 12-ft lane is closed on a road with a 45-mph posted speed limit. For this calculation, use W = 12 ft and S = 45 mph in the tabulated MUTCD Part 6 guidance. What minimum merging-taper length does the table recommend?
- A. 270 ft
- B. 405 ft
- C. 540 ft
- D. 1,080 ft
Show answer and solution
Answer: C. 540 ft.
For speeds of 45 mph or more, L = WS = 12 × 45 = 540 ft, and the guidance table gives at least L for a merging taper. B uses WS²/60, the formula for 40 mph or less. A is a shifting taper (0.5L). D doubles it. The taper formulas and the merging-taper rule are the same in the 2009 and 11th editions, although the table numbers changed (Tables 6C-3/6C-4 in 2009, Tables 6B-3/6B-4 in the 11th edition).
Source: FHWA, MUTCD 11th Edition (Dec. 2023), Part 6, Section 6B.08, Tables 6B-3 and 6B-4, p. 775; FHWA, MUTCD 2009 Edition, Section 6C.08, Tables 6C-3 and 6C-4
Problem 36 · Trench egress spacing
CON-HS-04 · Select one
Workers may be anywhere along a straight trench 80 ft long and 4.5 ft deep. Distance x is measured from one end (x = 0 to x = 80 ft). OSHA requires a safe means of egress in trenches 4 ft or more deep so that no more than 25 ft of lateral travel is needed. Assume each ladder is otherwise properly installed. Which ladder arrangement meets the lateral-travel requirement everywhere in the trench?
- A. Ladders at x = 0 and x = 80 ft
- B. One ladder at x = 40 ft
- C. Ladders at x = 20 ft and x = 60 ft
- D. Ladders at x = 0 and x = 40 ft
Show answer and solution
Answer: C. Ladders at x = 20 ft and x = 60 ft.
With ladders at 20 and 60 ft, the farthest any worker can be from a ladder is 20 ft (at either end or at the 40-ft midpoint). A leaves the midpoint 40 ft from a ladder. B leaves both ends 40 ft away. D leaves the far end 40 ft away. Check the worst-case location, not the spacing between ladders. Meeting the egress rule doesn't make the trench safe on its own; cave-in protection is a separate requirement.
Source: OSHA, 29 CFR 1926.651(c)(2), Means of egress from trench excavations
How to read your results
Treat your percentage as feedback on these 36 problems, not an NCEES score or a prediction of passing. Two to five problems per area is enough to show which methods you can't yet explain, but not enough to measure an area precisely. NCEES doesn't publish its passing score, and it converts performance to a scaled score to adjust for differences between exam forms before reporting pass/fail (NCEES exam scoring), so no practice percentage converts into an exam result.
Your 12-week study plan
Example: 12 weeks at 10 hours a week (120 hours). This is our planning example, weighted toward the heaviest NCEES areas. 120 hours isn't a requirement for passing. Keep every area in the plan, then shift hours toward the methods and lookups that give you trouble.
| Week | Focus | Do this | Finish the week with |
|---|---|---|---|
| 1 | Setup and baseline | Confirm your test date and its reference list. Download the handbook from MyNCEES. If you have already studied the opening example, record Problem 1 as review. Work the remaining 35 before revealing their answers, then check them. | A completed error log and a list of the areas where you missed the most |
| 2 | Support of construction loads I | Load paths from area to line to reaction; formwork pressure; tie and sheathing layouts. Rework Problems 1–3. Practice searching ACI SP-4 and ACI 347R. | Two labeled free-body sketches and three lookup-log entries |
| 3 | Support of construction loads II | Falsework, scaffolding, shoring and reshoring, bracing; ASCE 37 load cases; construction loads on permanent structures; support of excavation. Rework Problems 4–5. | A list of what a reaction calculation does not check (capacity, bracing, connections) |
| 4 | Construction operations I | Rigging and sling angles, crane stability and outrigger loads, equipment selection. Rework Problem 6. | A free-body sketch for the sling problem, with its vertical force components checked against the load |
| 5 | Construction operations II | Earthwork: cut and fill, shrink and swell, borrow volume, haul; productivity; dewatering; deep foundations. Rework Problems 7–8 and 21–22. | A volume-conversion chain with bank, loose, and compacted labeled on every line |
| 6 | Scheduling | Forward and backward passes, total vs. free float, crashing, resource leveling, linear schedules. Rework Problems 9–11. | A complete CPM table for a network you draw yourself |
| 7 | Quality control | Compaction testing, maturity, concrete placement, weld and bolt installation, QA vs. QC. Rework Problems 12–15. Practice searching PCA EB001. | For each test result, a note on what it does and doesn't establish |
| 8 | Structural mechanics and soils | Beams, columns, one-way slabs, combined stress, deflection; lateral earth pressure, effective stress, settlement, bearing capacity, slope stability. Rework Problems 16–20. Practice AISC Manual tables in your edition. | Reactions, moments, and earth pressures checked by a second method |
| 9 | Estimating and site layout | Take-off, cost estimating, engineering economics, earned value; leveling, horizontal and vertical curves. Rework Problems 21–27. | A one-page formula sheet with the handbook location of each formula |
| 10 | Materials, hydraulics, safety | Mix design and admixtures, aggregate moisture, steel and wood; rational and NRCS methods, detention; OSHA 1926 (and 1904 if you test April 2027 or later), MUTCD Part 6. Rework Problems 28–36. | Lookup-log entries for every OSHA and MUTCD rule you used |
| 11 | Timed mixed practice | Work Blocks A and B below, allowing 108 minutes per block, with an optional 50-minute break between them. Use your searchable references and keep solutions closed until each block is complete. Spend the rest of the week reviewing misses and slow lookups. | A record of time per block, flagged questions, and slow lookups |
| 12 | Targeted review | Rework every logged miss. Repeat your five slowest lookups. Confirm your appointment, ID, and calculator. | A short final list of methods to review and a checked exam-day plan |
Week 11 mixed blocks: Each block contains 18 problems, giving you a practice budget of 18 × 6 = 108 minutes. These are the same problems you have already studied, so this is repeated method and pacing practice, not a fresh full-length exam or an eight-hour endurance simulation.
| Block | Problem numbers, in order | Time budget |
|---|---|---|
| A | 1, 6, 9, 12, 16, 19, 21, 25, 28, 30, 33, 2, 7, 13, 22, 26, 34, 3 | 108 minutes |
| B | 4, 8, 10, 14, 17, 20, 23, 27, 29, 31, 35, 5, 11, 15, 18, 24, 32, 36 | 108 minutes |
A usable 10-hour week: two 2-hour sessions learning methods and references, two 2-hour sessions solving and checking problems, and one 2-hour mixed session that ends with planning the next week.
Adjusting the plan:
- Fewer hours a week? Stretch the calendar rather than cutting areas. At 6 hours a week, the same 120 hours takes 20 weeks.
- Only 6 weeks? Combine weeks in pairs (1 and 2, 3 and 4, and so on) and keep both timed mixed-practice blocks. That's 20 hours a week; if that's not realistic, consider a later test date.
- Already strong in an area? Keep a short review there and move the saved time to your weakest area.
- Testing April 2027 or later? Spend extra time in Weeks 2, 8, and 10 learning the layout of ACI 318, the AISC 16th edition, the MUTCD 11th edition, and 2024 Title 29.
Turn each miss into your next study task
| What went wrong | Next step | How you know it's fixed |
|---|---|---|
| Picked the wrong method | Write in one sentence what the problem asks for and which model fits, then rework a version with one input changed | You can say why the wrong method doesn't apply |
| Used the wrong reference or missed a condition | Repeat the lookup: document, chapter, section. Read the definitions and exceptions nearby. | You can name the condition that triggers the rule |
| Mixed up units or volume states | Write the conversion chain with units on every line | Units cancel to the one the problem asks for |
| Arithmetic or sign error | Redo it from scratch, then check equilibrium, magnitude, or direction | The check agrees without copying your first attempt |
| Couldn't interpret the result | Write what the number shows and what else a decision would need | You stop treating a reaction, index, or test result as an approval |
| Took too long on a familiar problem | Note where the time went, then practice flagging it and coming back during a timed block | The same lookup or setup takes less time next round |
Error log fields: problem or topic · your answer · first wrong step · error type (from the table above) · corrected method · reference and section · result with one input changed · date to review again.
If your exam is next week
Don't start a new system. Pick the two errors you keep repeating and your two slowest lookups, fix those, do one mixed timed block, and confirm your logistics. If big topic gaps remain, be honest with yourself about them; you can reschedule a Pearson appointment at least 48 hours ahead for a $50 fee paid to Pearson (Examinee Guide, p. 6).
If you're preparing for a retake
If you didn't pass, NCEES sends a diagnostic report showing your relative strength by subject area. It doesn't reveal the passing score or how many questions you were short. Use it alongside your error log to choose your first two focus areas, give each a full week, then rerun the plan. NCEES limits attempts to once per calendar-quarter testing window and no more than three times in 12 months; your board may set stricter limits. See NCEES exam retake rules and how NCEES exam results work.
Exam-day rules that change how you practice
- About 6 minutes per question on average. That's 480 minutes ÷ 80 questions. It's a budgeting guide, not a per-question limit.
- Two halves, one clock. After roughly half the questions, you review and submit them, and you can't go back. The sections aren't timed separately, so time saved in the first half carries over.
- Breaks: the optional 50-minute break comes after you submit the first half. Unused break time doesn't extend your exam time, and unscheduled breaks come out of your exam time.
- No penalty for wrong answers, so answer everything.
- Calculator: one NCEES-approved model. For a 2026 test date, use the 2026 approved-calculator list, p. 1. Recheck NCEES's year-specific policy if your test date changes to a later year.
- Scratch work: you get two reusable booklets and three markers.
- No personal references. Only the supplied handbook and the standards for your test date.
Sources: NCEES Examinee Guide, May 2026, pp. 8–12; NCEES exam scoring.
Quick answers
Do I register with NCEES or my state board?
Check your licensing board's approval process before registering and paying the $400 exam fee through MyNCEES. Some boards require a separate application or fee. See NCEES exam registration steps or find your licensing board.
Can I use older PE Civil study materials with breadth and depth sections?
For concepts, yes. For exam structure, reference editions, and topic weighting, no. Those materials reflect the structure before the April 2024 specifications, when PE Civil stopped using a shared breadth section.
What's the pass rate?
For January–June 2026, 60% of 1,000 first-time examinees passed, and 38% of 417 repeat examinees passed (NCEES). Those rates describe that group of test takers, not your odds.
Does passing the PE exam give me a license?
No. Your state or territorial licensing board issues the license, and passing the PE exam is one of its requirements.
Can I get testing accommodations?
Indicate your need during NCEES exam registration, then follow the NCEES email instructions to continue your request with Pearson. Pearson manages accommodation requests and documentation; you cannot schedule a testing appointment until a decision is made. See how to request NCEES exam accommodations and NCEES's accommodation process.
Haven't passed the FE yet?
Start with FE Civil exam prep.
Sources
Exam facts on this page were last verified October 7, 2026, against the NCEES sources below. The original practice problems were checked against the cited teaching sources and recalculated; they have not been reviewed by NCEES or a licensed engineer.
NCEES
- PE Civil exam page (format, fee, references, pass rates)
- PE Civil: Construction exam specifications, effective April 2024 (topics; design standards for exams before April 2027)
- PE Civil: Construction design standards, effective April 2027
- NCEES Examinee Guide, May 2026
- NCEES exam scoring
Regulations and federal guidance
- OSHA 29 CFR 1926: .451, .501, .651, .703, Subpart P Appendix B
- OSHA, 29 CFR 1904.39
- 23 CFR 637.203
- MUTCD Part 6: 2009 edition and 11th edition
- FHWA: Earthwork Design; Maturity Testing TechBrief FHWA-IF-06-004; FHWA-04-122, Chapter 5; LRFD design example, Design Step P
- GAO: Schedule Assessment Guide, GAO-16-89G; Cost Estimating and Assessment Guide, GAO-20-195G
Technical references
- American Wood Council, Beam Formulas with Shear and Moment Diagrams
- TxDOT Hydraulic Design Manual, Rational Method
- Iowa DOT IM 315, compressive strength calculation
- Caltrans, Horizontal Curve Equations
- Caltrans, Vertical Curve Calculations
- FHWA GEC No. 6, Shallow Foundations, pp. 114–115 (Rankine earth pressure)
- FHWA NHI-05-037, weight-volume relationships and compaction
- NOAA, Geodetic Leveling
- Caltrans, Concrete Construction Materials, p. 2-19 (aggregate moisture states)
- NIST, Life Cycle Costing Manual, pp. 131–132 and 246 (equivalent annual cost)
- OpenStax, University Physics Volume 1: pressure and static fluid head, force equilibrium, and tensile and compressive stress
These teaching sources explain the methods. They are not references supplied on the exam.
Castleport Test Prep is an independent exam prep publisher. It is not affiliated with, endorsed by, or approved by NCEES. Exam and credential names identify their subjects; trademarks belong to their respective owners. The practice problems are original and unofficial, and nothing on this page guarantees a passing result or a license.