PE Civil (NCEES Principles and Practice of Engineering) Practice Test

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In the PE Civil (NCEES) guide: 37 topic briefings, each opening with exactly what the exam tests there. Practice here stays free.

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NCEES PE Civil Exam — Exam facts
Administering bodyNational Council of Examiners for Engineering and Surveying (NCEES) — exam delivered by Pearson VUE

Source: NCEES — PE Civil exam

Questions80 questions

Source: NCEES — PE Civil exam

Time limit480 minutes

Source: NCEES — PE Civil exam

Passing scoreNot published by NCEES

What we read and found nothing in: NCEES — PE Civil exam

Fees
  • $400 — Exam fee (NCEES, per attempt)

Source: NCEES — PE Civil exam

Languages offeredNot published by NCEES

What we read and found nothing in: NCEES — PE Civil exam

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Frequently asked questions

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A full bank of original PE Civil (NCEES Principles and Practice of Engineering) practice questions across the official content areas, weighted like the real exam, with explanations. Free, no signup.

What is the PE Civil (NCEES Principles and Practice of Engineering) exam like?+

About 80 questions, 540 minutes. Practice by topic here, then take the full timed mock exam to gauge readiness.

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PrepPass practice is in English, 中文 and Español. The official exam is in English — switch the question language to English any time to rehearse the exact terminology you'll see on test day.

Is there a study guide for the PE Civil (NCEES Principles and Practice of Engineering)?+

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Sample practice questions

A few real questions from this free bank, with full explanations. Use the practice tool above for the whole set.

  1. 1. Structural Engineering

    A simply supported beam of span 8 m carries a uniformly distributed load of 4 kN/m over its full length. What is the vertical reaction at each support?

    • a.64 kN
    • b.32 kN
    • c.16 kN
    • d.8 kN

    Answer: c

    Explanation: The total load is w·L = 4 x 8 = 32 kN. By symmetry each support carries half: 32/2 = 16 kN, which also equals w·L/2. Symmetric loading always splits equally between the two supports.

  2. 2. Structural Engineering

    What is the plastic section modulus Z of a rectangular section 200 mm wide and 300 mm deep about its strong axis?

    • a.3.0x10^6 mm^3
    • b.2.25x10^6 mm^3
    • c.4.5x10^6 mm^3
    • d.9.0x10^6 mm^3

    Answer: c

    Explanation: The plastic section modulus of a rectangle is Z = b·h^2/4 = 200 x 300^2 / 4 = 200 x 90,000 / 4 = 4.5x10^6 mm^3. Z is used for the fully plastic moment M_p = Z·Fy, whereas the elastic S = b·h^2/6 governs first yield.

  3. 3. Structural Engineering

    A steel bar undergoes an axial strain of 5x10^-4. With Poisson's ratio 0.3, what is the magnitude of the lateral (transverse) strain?

    • a.1.5x10^-4
    • b.5x10^-4
    • c.1.67x10^-4
    • d.3x10^-4

    Answer: a

    Explanation: Lateral strain = -nu x axial strain = -0.3 x 5x10^-4 = -1.5x10^-4 (a contraction). Poisson's ratio relates the transverse contraction to the axial extension; for steel it is about 0.3.

  4. 4. Geotechnical Engineering

    A 4 m thick clay layer with double drainage has cv = 2 m^2/yr. What is the time factor after 1 year?

    • a.0.5
    • b.0.25
    • c.1.0
    • d.0.125

    Answer: a

    Explanation: With double drainage the drainage path H_dr is half the layer thickness = 2 m. Time factor Tv = cv·t/H_dr^2 = 2 x 1 / 2^2 = 2/4 = 0.5. Double drainage quarters the time compared with single drainage because the path is halved and it is squared.

  5. 5. Water Resources and Environmental

    A rectangular open channel is 4 m wide and flows at a depth of 1.5 m. What is the hydraulic radius?

    • a.0.857 m
    • b.0.667 m
    • c.4 m
    • d.1.5 m

    Answer: a

    Explanation: R = A/P. Area A = 4 x 1.5 = 6 m^2; wetted perimeter (bed plus two sides) P = 4 + 2(1.5) = 7 m; R = 6/7 = 0.857 m. The hydraulic radius uses the wetted perimeter, not the flow depth or width alone.

  6. 6. Water Resources and Environmental

    Normal depth in an open channel is the depth at which:

    • a.the flow is critical
    • b.the velocity is maximum
    • c.the velocity is zero
    • d.flow is uniform, with gravity balancing friction and the water surface parallel to the bed

    Answer: d

    Explanation: Normal (uniform) depth occurs when the gravitational driving force exactly balances boundary friction, so the depth, velocity, and energy grade stay constant and the water surface parallels the channel bed. Manning's equation solves for this depth.

  7. 7. Transportation Engineering

    Under the Greenshields model with a free-flow speed of 60 mph, at what speed does maximum flow (capacity) occur?

    • a.45 mph
    • b.30 mph
    • c.60 mph
    • d.15 mph

    Answer: b

    Explanation: In the Greenshields model, capacity occurs at the optimum speed v_o = v_f/2 = 60/2 = 30 mph (and density k_j/2). Flow falls off on either side of this point.

  8. 8. Construction Engineering and Management

    An embankment requires 1,800 compacted cubic yards. If the borrow soil has a shrinkage of 10%, what bank volume must be excavated?

    • a.1,800 cubic yards
    • b.2,000 cubic yards
    • c.1,980 cubic yards
    • d.1,620 cubic yards

    Answer: b

    Explanation: Bank volume = compacted volume / (1 - shrinkage) = 1,800 / 0.90 = 2,000 CY. You must excavate more bank material than the finished fill because compaction reduces the volume.

  9. 9. Construction Engineering and Management

    A mass haul (mass diagram) plots what quantity along a roadway alignment?

    • a.cumulative cut minus fill along the alignment
    • b.concrete strength gain
    • c.the crane load chart
    • d.the total accumulated project cost plotted against elapsed calendar time

    Answer: a

    Explanation: The mass diagram plots cumulative net earthwork (cut positive, fill negative) versus station. Its shape guides haul direction, haul distance, and balancing of cut and fill.

  10. 10. Engineering Materials

    An oven-dry aggregate sample of 470.5 g occupies a volume of 180 cubic centimeters. What is its bulk specific gravity? (water = 1.0 g/cc)

    • a.2.61
    • b.2.71
    • c.0.38
    • d.180

    Answer: a

    Explanation: Bulk specific gravity = oven-dry mass / (volume x density of water) = 470.5 / (180 x 1.0) = 2.61. Most natural aggregates have specific gravities around 2.6 to 2.7.

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