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NCEES Engineering Licensure

FE Civil Practice Exam

FE Civil exam practice — the NCEES discipline exam, your first step toward the PE.

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Built for FE Civil candidates

  • NCEES FE Civil topic mix: statics, structural analysis & design, geotechnical, water resources & environmental, transportation, and construction
  • Answerable from the NCEES FE Reference Handbook + first principles
  • Mix of calculation and conceptual items, in SI and US customary units
  • Step-by-step solutions on every question (first 2 free; all with Premium)
  • Follow-up study chat to dig into any concept that trips you up

The FE Civil exam is one of the seven NCEES Fundamentals of Engineering discipline exams — a 110-question, computer-based test, taken near graduation, that is the first step toward becoming a licensed Professional Engineer (PE). Every question is meant to be answered from the FE Reference Handbook and engineering fundamentals, so practice that drills the right Civil topic mix, in the right units, with worked solutions is exactly how you build exam-day speed.

Crameleon builds your practice exam on demand — pick the pathway and module you're studying, and we generate a fresh paper with step-by-step solutions (the first 2 free; all of them with Premium). Every question comes with a follow-up study chat, so you can dig into any concept that trips you up without leaving the page. The generation takes under 2 minutes and your first exam is free.

Sample FE Civil exam questions

Three original questions on FE Civil exam topics, each with a worked solution.

Question 1Multiple choiceGeotechnical Engineering

A level site is underlain by a deep sand deposit. The groundwater table is 3 m below the ground surface. Above the water table the sand has a moist unit weight of 18 kN/m³; below the water table its saturated unit weight is 20 kN/m³. Take the unit weight of water as 9.81 kN/m³ and neglect capillary rise. What is the effective vertical stress at a depth of 7 m below the ground surface?

  1. A.

    40.8 kPa

  2. B.

    65.3 kPa

  3. C.

    94.8 kPa

  4. D.

    134 kPa

Worked solution
Answer:C

Effective stress is the total vertical stress minus the pore water pressure: σ′=σ−u\sigma' = \sigma - u. The soil skeleton carries only the effective stress.

Total vertical stress at 7 m is the weight of everything above, layer by layer, using the moist unit weight above the water table and the saturated unit weight below it:

σ=3(18)+4(20)=54+80=134 kPa\sigma = 3(18) + 4(20) = 54 + 80 = 134\,\text{kPa}

Pore water pressure is hydrostatic below the water table, so it depends only on the depth below the water table (7 − 3 = 4 m), not on the depth below the ground surface:

u=4(9.81)=39.2 kPau = 4(9.81) = 39.2\,\text{kPa}

Effective stress:

σ′=134−39.2=94.8 kPa\sigma' = 134 - 39.2 = 94.8\,\text{kPa}

The same result comes from using the buoyant unit weight for the submerged sand: 3(18)+4(20−9.81)=54+40.8=94.8 kPa3(18) + 4(20 - 9.81) = 54 + 40.8 = 94.8\,\text{kPa}.

Effective vertical stress ≈ 94.8 kPa (option C)

The most tempting wrong answer is D (134 kPa), which is the total stress; it ignores the pore water pressure that supports part of the overburden. B (65.3 kPa) subtracts a full 7 m of water pressure as if the water table were at the ground surface. A (40.8 kPa) counts only the submerged layer and forgets the 3 m of moist sand above the water table.

Question 2Short answerWater Resources and Environmental Engineering

A rectangular concrete channel 3.0 m wide carries water at a uniform-flow depth of 1.2 m. The channel bed slope is 0.002 and Manning's roughness coefficient is n=0.015n = 0.015. Using Manning's equation in SI form, Q=1nAR2/3S1/2Q = \frac{1}{n}AR^{2/3}S^{1/2}, determine the discharge in m³/s.

Worked solution
Answer:

Q ≈ 8.2 m³/s

Manning's equation relates discharge to the channel geometry, roughness and slope for uniform flow. First find the geometric quantities.

Flow area (width times depth):

A=3.0×1.2=3.6 m2A = 3.0 \times 1.2 = 3.6\,\text{m}^2

Wetted perimeter, which is the bed plus the two wetted sides but not the free surface, because the water surface has no friction with a boundary:

P=3.0+2(1.2)=5.4 mP = 3.0 + 2(1.2) = 5.4\,\text{m}

Hydraulic radius:

R=AP=3.65.4=0.667 mR = \frac{A}{P} = \frac{3.6}{5.4} = 0.667\,\text{m}

Substitute into Manning's equation, with R2/3=0.6670.667=0.763R^{2/3} = 0.667^{0.667} = 0.763 and S1/2=0.0020.5=0.0447S^{1/2} = 0.002^{0.5} = 0.0447:

Q=10.015(3.6)(0.763)(0.0447)=8.19 m3/sQ = \frac{1}{0.015}(3.6)(0.763)(0.0447) = 8.19\,\text{m}^3\text{/s}

The mean velocity is V=Q/A=8.19/3.6=2.28V = Q/A = 8.19/3.6 = 2.28 m/s, which is a reasonable velocity for a concrete channel on a mild slope.

Q ≈ 8.2 m³/s

Question 3Worked problemStructural Engineering

A beam ABC is supported by a pin at A and a roller at B and has an overhang beyond B. Taking A as the origin, support B is at x=24 ftx = 24\,\text{ft} and the free end C is at x=30 ftx = 30\,\text{ft}. The beam carries a uniformly distributed load of 2.0 kip/ft downward over its entire 30 ft length. Neglect the beam's self-weight.

(a) Determine the support reactions at A and B.

(b) Determine the bending moment at B.

(c) Determine the maximum positive bending moment in span AB and its location.

(d) If the beam is a steel W-shape with an elastic section modulus of S=60 in3S = 60\,\text{in}^3 about the bending axis, determine the maximum bending stress in the beam.

Worked solution
Answer:

(a) RAR_A = 22.5 kip, RBR_B = 37.5 kip; (b) MBM_B = −36 kip·ft; (c) Mmax⁡M_{\max} = +126.6 kip·ft at x = 11.25 ft; (d) σmax⁡\sigma_{\max} ≈ 25.3 ksi

(a) Reactions. The resultant of the distributed load is 2.0×30=602.0\times 30 = 60 kip acting at the midpoint of the loaded length, x=15x = 15 ft. Take moments about A so that RAR_A drops out:

∑MA=0:RB(24)−60(15)=0⇒RB=37.5 kip\sum M_A = 0: \quad R_B(24) - 60(15) = 0 \quad \Rightarrow \quad R_B = 37.5\,\text{kip}

Vertical equilibrium: RA=60−37.5=22.5R_A = 60 - 37.5 = 22.5 kip. Both act upward.

(b) Moment at B. Cut the beam just at B and consider the 6 ft overhang to the right. The only load on it is 2.0×6=122.0\times 6 = 12 kip acting 3 ft from B, which bends the beam concave downward (hogging), so the moment is negative:

MB=−wa22=−2.0(6)22=−36 kip⋅ftM_B = -\frac{wa^2}{2} = -\frac{2.0(6)^2}{2} = -36\,\text{kip·ft}

(c) Maximum positive moment. The bending moment is a maximum where the shear is zero. In span AB the shear is V(x)=RA−wx=22.5−2.0xV(x) = R_A - wx = 22.5 - 2.0x, which is zero at x=11.25x = 11.25 ft. The moment there is

M=RAx−wx22=22.5(11.25)−2.0(11.25)22=253.1−126.6=126.6 kip⋅ftM = R_A x - \frac{wx^2}{2} = 22.5(11.25) - \frac{2.0(11.25)^2}{2} = 253.1 - 126.6 = 126.6\,\text{kip·ft}

This is positive (sagging). It is also the largest moment anywhere in the beam, because 126.6>36126.6>36.

(d) Bending stress. Flexure gives σ=M/S\sigma = M/S with MM converted to kip·in so the units match the section modulus:

σmax⁡=126.6×1260=25.3 ksi\sigma_{\max} = \frac{126.6 \times 12}{60} = 25.3\,\text{ksi}

RAR_A = 22.5 kip, RBR_B = 37.5 kip; MBM_B = −36 kip·ft; Mmax⁡M_{\max} = +126.6 kip·ft at x = 11.25 ft; σmax⁡\sigma_{\max} ≈ 25.3 ksi

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