Twenty original exam-style FE Civil questions with fully worked, step-by-step solutions — free. Mathematics, statics, mechanics of materials, dynamics, fluid mechanics, water resources, geotechnical, transportation, engineering economics and construction. Attempt it in about 60 minutes at exam pacing (~3 minutes per question), score yourself, then study every solution before you decide on the full exam.
The 20 questions
Work each question before opening its solution. The exam tip under each solution names the trap the question was built around.
Question 1 — Mathematics
Find the area of the region bounded by the curves y = x² and y = 4.
- 5.33
- 10.67
- 16.0
- 21.33
Answer: B) 10.67 — show the worked solution
The curves intersect where x² = 4, i.e. x = ±2. Area = ∫₋₂² (4 − x²) dx = [4x − x³/3]₋₂² = 32/3 ≈ 10.67.
Why the others are wrong. A) 5.33 — integrated from 0 to 2 only (half the symmetric region). C) 16.0 — the 4×4 rectangle with nothing subtracted. D) 21.33 — the correct area doubled.
Exam tip: sketch first. Symmetry about the y-axis means you can integrate 0→2 and double — but you must remember the doubling step.
Question 2 — Statics
A simply supported beam spans 20 ft. It carries a triangular distributed load that is zero at the left support and reaches a maximum intensity of 400 lb/ft at the right support. The left reaction is most nearly:
- 1.00 kips
- 1.33 kips
- 2.00 kips
- 2.67 kips
Answer: B) 1.33 kips — show the worked solution
Total load W = ½ × 400 × 20 = 4,000 lb. For a triangular distribution the resultant acts one-third of the span from the maximum-intensity end: 20/3 = 6.67 ft left of the right support. Summing moments about the right support: R_left × 20 = 4,000 × 6.67, so R_left = 1,333 lb ≈ 1.33 kips.
Why the others are wrong. C) 2.00 kips — the resultant was placed at midspan (treating the load as uniform). D) 2.67 kips — that is the right reaction (resultant measured from the wrong end). A) 1.00 kip — no consistent method produces this.
Exam tip: for any distributed load, write down the total (area under the curve) and where it acts (centroid) before any arithmetic. Most statics misses are centroid misses.
Question 3 — Mechanics of Materials
A steel rod 10 ft long with a 1.0-in diameter carries an axial tensile load of 20 kips. With E = 29,000 ksi, the elongation is most nearly:
- 0.0105 in
- 0.105 in
- 0.210 in
- 1.05 in
Answer: B) 0.105 in — show the worked solution
δ = PL/AE. A = π/4 × (1.0)² = 0.785 in²; L = 120 in. δ = (20 × 120)/(0.785 × 29,000) = 0.105 in.
Why the others are wrong. A) 0.0105 in — decimal slip (÷10). C) 0.210 in — radius/diameter confusion. D) 1.05 in — decimal slip (×10).
Exam tip: convert length to inches first when E is in ksi. Mixed ft/in units are the single most common axial-deformation error.
Question 4 — Dynamics
A 50-lb block starts from rest and slides 10 ft down a 30° incline. The coefficient of kinetic friction is μk = 0.20. The velocity at the bottom is most nearly:
- 2.56 ft/s
- 14.5 ft/s
- 17.9 ft/s
- 20.8 ft/s
Answer: B) 14.5 ft/s — show the worked solution
Work–energy: T₁ + ΣU = T₂. Driving work = 50 × 10 × sin30° = 250 ft-lb. Friction work = 0.20 × 50 × cos30° × 10 = 86.6 ft-lb. Net = 163.4 ft-lb = ½(W/g)v², so v = √(2 × 163.4 × 32.2/50) = 14.5 ft/s.
Why the others are wrong. C) 17.9 ft/s — friction ignored. D) 20.8 ft/s — friction added instead of subtracted. A) 2.56 ft/s — weight used as mass (the W/g division skipped).
Exam tip: decide the sign of each work term before computing its magnitude: gravity helps here, friction opposes. Sign first, numbers second.
Question 5 — Fluid Mechanics (Venturi Meter)
A venturi meter is installed in a 12-in-diameter pipe. The throat diameter is 6 in. The piezometric head difference between the upstream section and the throat is 2.8 ft. With a discharge coefficient Cd = 0.98, the flow rate is most nearly:
- 2.58 cfs
- 2.67 cfs
- 2.72 cfs
- 10.7 cfs
Answer: B) 2.67 cfs — show the worked solution
A₁ = 0.785 ft²; A₂ = 0.196 ft². V₂ = √[2gΔh/(1 − (A₂/A₁)²)] = √[2 × 32.2 × 2.8/(1 − 0.0625)] = 13.87 ft/s. Q = Cd·V₂·A₂ = 0.98 × 13.87 × 0.196 = 2.67 cfs.
Why the others are wrong. A) 2.58 cfs — the approach-velocity correction skipped. C) 2.72 cfs — Cd never applied; real meters discharge slightly less than the Bernoulli ideal. D) 10.7 cfs — throat velocity multiplied by the upstream area.
Exam tip: in any meter problem the Cd-applied answer sits a few percent below the ideal — a free sanity check. More fluids practice: Continuity, Energy & Momentum.
Question 6 — Pipe Flow (Darcy–Weisbach)
A 1.5-ft-diameter pipe, 2,400 ft long, carries water at 5.0 ft/s. The Darcy friction factor is f = 0.018. Minor losses: entrance (K = 0.5), two 90° flanged elbows (K = 0.9 each), one fully open gate valve (K = 0.15). The total head loss is most nearly:
- 1.9 ft
- 11.2 ft
- 12.1 ft
- 39.8 ft
Answer: C) 12.1 ft — show the worked solution
V²/2g = 0.388 ft. Friction: hf = 0.018 × 1600 × 0.388 = 11.18 ft. Minors: ΣK = 2.45, hm = 0.95 ft. Total = 12.1 ft.
Why the others are wrong. A) 1.9 ft — diameter used in inches in L/D. B) 11.2 ft — friction only; the minors were dropped (worth ~1 ft — exactly the choice spacing). D) 39.8 ft — g = 9.81 used with feet.
Exam tip: glance at L/D first: in the thousands, friction dominates and minors are a correction. More pipe-flow practice: Pipe Flow.
Question 7 — Pipe Flow (Reservoir Discharge)
A reservoir discharges through a pipe (D = 10 in, L = 1,600 ft, f = 0.020) to a free outlet. The water surface is 40 ft above the outlet. ΣK = 6.5. The discharge is most nearly:
- 3.57 cfs
- 4.13 cfs
- 4.47 cfs
- 10.9 cfs
Answer: B) 4.13 cfs — show the worked solution
H = [f(L/D) + ΣK]·V²/2g. f(L/D) = 38.4; total coefficient = 44.9. V = √[2 × 32.2 × 40/44.9] = 7.57 ft/s. Q = 7.57 × 0.545 = 4.13 cfs.
Why the others are wrong. A) 3.57 cfs — f misread. C) 4.47 cfs — minors ignored (~8% of the flow). D) 10.9 cfs — friction ignored.
Exam tip: compare f(L/D) with ΣK before solving: whichever is bigger owns the answer. More practice: Hazen-Williams & Pipe Hydraulics.
Question 8 — Open-Channel Flow (Manning's)
A trapezoidal channel has a 10-ft bottom width, 2H:1V side slopes, and Manning's n = 0.014. It runs on a slope of 0.0008. At a flow depth of 5 ft, the discharge is most nearly:
- 411 cfs
- 878 cfs
- 637 cfs
- 429 cfs
Answer: C) 637 cfs — show the worked solution
A = (10 + 10) × 5 = 100 ft². P = 10 + 10√5 = 32.36 ft (the free surface is not boundary). R = 3.09 ft. V = (1.486/0.014)(3.09)^⅔(0.0008)^½ = 6.37 ft/s. Q = 637 cfs.
Why the others are wrong. A) 411 cfs — the top width was added to the wetted perimeter. B) 878 cfs — R set equal to depth. D) 429 cfs — the SI constant 1.0 used in a US-units problem.
Exam tip: write the unit system at the top of your scratch work. Half of all Manning's errors are constant errors. Free calculator: Manning's Equation Calculator.
Question 9 — Open-Channel Flow (Critical Depth)
A rectangular channel 12 ft wide carries 500 cfs. The critical depth is most nearly:
- 1.14 ft
- 3.78 ft
- 5.67 ft
- 19.8 ft
Answer: B) 3.78 ft — show the worked solution
q = 500/12 = 41.67 cfs/ft. yc = (q²/g)^⅓ = 3.78 ft. Check: Vc = 11.03 ft/s = √(g·yc) ✓ (Fr = 1 at critical).
Why the others are wrong. A) 1.14 ft — wrong exponent. C) 5.67 ft — the critical specific energy (1.5yc): a correct number for a question nobody asked. D) 19.8 ft — total Q cubed instead of q.
Exam tip: convert to q = Q/b immediately in rectangular-channel problems. More practice: Open-Channel Flow.
Question 10 — Hydrology (SCS Curve Number)
A 960-acre watershed has CN = 78. A storm drops 3.5 in of rain. Using the SCS method (Ia = 0.2S), the total runoff volume is most nearly:
- 12 ac-ft
- 80 ac-ft
- 120 ac-ft
- 155 ac-ft
Answer: C) 120 ac-ft — show the worked solution
S = 1000/78 − 10 = 2.82 in; Ia = 0.564 in (< 3.5 in, so runoff occurs). Q = (3.5 − 0.564)²/(3.5 − 0.564 + 2.82) = 1.50 in. V = 1.50 × 960/12 = 120 ac-ft.
Why the others are wrong. A) 12 ac-ft — decimal slip. B) 80 ac-ft — storage double-counted. D) 155 ac-ft — Ia never subtracted; the most common curve-number error.
Exam tip: check P > Ia before touching the equation. More hydrology practice: Hydrology & Runoff.
Question 11 — Hydrology (Time of Concentration + Rational)
A culvert must pass the peak flow from a 28-acre watershed, C = 0.45. Longest flow path 1,800 ft at 0.020 ft/ft slope. Tc = 0.0078·L^0.77·S^(−0.385) (min, ft). IDF: i = 95/(Tc + 12) (in/hr, min). The design peak flow is most nearly:
- 32 cfs
- 51 cfs
- 63 cfs
- 100 cfs
Answer: B) 51 cfs — show the worked solution
Tc = 0.0078 × 1800^0.77 × 0.020^(−0.385) = 11.3 min. i = 95/(11.3 + 12) = 4.08 in/hr. Q = 0.45 × 4.08 × 28 = 51 cfs.
Why the others are wrong. A) 32 cfs — intensity read at a default 25-min duration instead of the computed Tc. C) 63 cfs — a paved-surface C substituted for the given 0.45. D) 100 cfs — Tc dropped from the IDF denominator.
Exam tip: in any Rational Method problem, ask "where does my intensity come from?" first. More practice: Hydrology & Runoff.
Question 12 — Stormwater (Pipe Sizing)
A storm sewer must carry Q = CiA runoff from 12 acres, C = 0.55, i = 4.2 in/hr. The sewer (n = 0.013) on a 1.0% slope flows full. The smallest standard diameter (24, 27, 30, 36 in) that works is:
- 24 in
- 27 in
- 30 in
- 36 in
Answer: B) 27 in — show the worked solution
Q = 0.55 × 4.2 × 12 = 27.7 cfs. Full-flow capacities: 24 in → 22.6 cfs (18% short ✗); 27 in → 31.0 cfs ✓. Smallest adequate: 27 in.
Why the others are wrong. A) 24 in — "looks close" but is 18% undersized; sewers don't round up capacity. C/D — hydraulically fine but not the minimum.
Exam tip: full-pipe capacity scales as D^(8/3): one size up buys ~35–40% more flow.
Question 13 — Groundwater (Thiem)
A well pumps 400 gpm from a 60-ft-thick confined aquifer. At steady state, drawdown is 7.5 ft at 50 ft from the well and 3.0 ft at 200 ft. Well radius 1.0 ft. The drawdown in the pumping well is most nearly:
- 7.5 ft
- 13.0 ft
- 20.2 ft
- 36.7 ft
Answer: C) 20.2 ft — show the worked solution
T = Q·ln(r₂/r₁)/[2π(s₁−s₂)] = 77,000 × ln(4)/(2π × 4.5) ≈ 3,775 ft²/day. Extrapolate: s = 7.5 + [77,000/(2π × 3,775)]·ln(50/1) = 20.2 ft.
Why the others are wrong. A) 7.5 ft — the nearest observation well's drawdown reported as the well's; the extrapolation is the question. B) 13.0 ft and D) 36.7 ft — log₁₀ mixed into the natural-log form.
Exam tip: pick ln or log₁₀ once and stay in it. More practice: Groundwater Flow.
Question 14 — Water Treatment (Chlorine Feed)
A plant treats 1.6 MGD. Chlorine demand is 2.1 mg/L; a 0.4 mg/L free residual must remain. The feed rate is most nearly:
- 5.3 lb/day
- 28.0 lb/day
- 33.4 lb/day
- 334 lb/day
Answer: C) 33.4 lb/day — show the worked solution
Dose = demand + residual = 2.5 mg/L. Feed = 2.5 × 1.6 × 8.34 = 33.4 lb/day.
Why the others are wrong. A) 5.3 lb/day — only the residual fed. B) 28.0 lb/day — only the demand fed. D) 334 lb/day — factor-of-10 slip on 8.34.
Exam tip: dose × flow × 8.34 = lb/day, and the dose is demand plus residual. More practice: Water Treatment.
Question 15 — Wastewater (F/M Ratio)
An activated-sludge plant treats 2.0 MGD at 220 mg/L influent BOD₅. Basin volume 0.60 MG, MLVSS 1,800 mg/L. The F/M ratio is most nearly:
- 0.31 /day
- 0.41 /day
- 0.49 /day
- 4.1 /day
Answer: B) 0.41 /day — show the worked solution
F/M = (2.0 × 220)/(0.60 × 1,800) = 0.41 /day. Cross-check: 3,670/9,010 lb/day ✓. Conventional range ≈ 0.2–0.5 /day.
Why the others are wrong. A) 0.31 /day — MLSS used instead of MLVSS (only the volatile fraction is active biomass). C) 0.49 /day — volume slip. D) 4.1 /day — decimal slip; no conventional process runs there.
Exam tip: memorise the plausible bands, not just formulas. More practice: Wastewater Treatment.
Question 16 — Geotechnical (Bearing Capacity)
A 6-ft-wide strip footing bears 3 ft deep in sand (c = 0, φ = 30°, γ = 120 pcf). Using Terzaghi's equation with Nc = 37.2, Nq = 22.5, Nγ = 19.7, the ultimate bearing capacity is most nearly:
- 8.1 ksf
- 15.2 ksf
- 22.3 ksf
- 7.1 ksf
Answer: B) 15.2 ksf — show the worked solution
q_ult = γDfNq + 0.5γBNγ = 120 × 3 × 22.5 + 0.5 × 120 × 6 × 19.7 = 8,100 + 7,092 = 15,192 psf ≈ 15.2 ksf.
Why the others are wrong. A) 8.1 ksf — the 0.5γBNγ term dropped. D) 7.1 ksf — the surcharge term dropped. C) 22.3 ksf — the 0.5 factor dropped.
Exam tip: Terzaghi has three terms — name the physical mechanism of each before calculating. Dropped terms are the exam's favourite distractor. More practice: Geotechnical Engineering.
Question 17 — Materials (Sand-Cone Test)
A sand-cone test on compacted fill: 6.96 lb of calibrated sand (87 lb/ft³) fills the hole. Excavated soil weighs 9.85 lb wet; water content 11.5%. Lab maximum dry unit weight 116.5 lb/ft³. Percent compaction is most nearly:
- 93.5%
- 94.8%
- 105.7%
- 110.4%
Answer: B) 94.8% — show the worked solution
V = 6.96/87 = 0.0800 ft³. γ_wet = 123.1 lb/ft³. γd = 123.1/1.115 = 110.4 lb/ft³. Percent compaction = 110.4/116.5 = 94.8% — against a 95% spec this lift fails by 0.2 points.
Why the others are wrong. A) 93.5% — moisture correction as ×(1−w) instead of ÷(1+w). C) 105.7% — no moisture correction: wet field value over dry lab maximum, "passing" a failing lift. D) 110.4% — the dry unit weight reported as a percentage.
Exam tip: only the dry field value is ever compared to the Proctor maximum. More practice: Materials.
Question 18 — Transportation (Stopping Sight Distance)
For a design speed of 60 mph, perception–reaction time 2.5 s, friction factor 0.35, on a level grade, the stopping sight distance is most nearly:
- 343 ft
- 563 ft
- 221 ft
- 493 ft
Answer: B) 563 ft — show the worked solution
SSD = 1.47Vt + V²/[30(f ± G)] = 1.47 × 60 × 2.5 + 60²/(30 × 0.35) = 220.5 + 342.9 = 563 ft.
Why the others are wrong. A) 343 ft — braking distance only. C) 221 ft — reaction distance only. D) 493 ft — the 1.47 mph→ft/s conversion skipped.
Exam tip: SSD is always two terms — say "reaction plus braking" aloud before calculating. More practice: Transportation Engineering.
Question 19 — Engineering Economics (EUAC)
Two pump stations, 15-year life, 6% interest. A: first cost $85,000, O&M $6,500/yr, salvage $5,000. B: first cost $120,000, O&M $3,800/yr, salvage $8,000. The preferred alternative and its EUAC are most nearly:
- Alternative A, $15,040/yr
- Alternative B, $15,810/yr
- Alternative A, $15,250/yr
- Alternative B, $16,160/yr
Answer: A) Alternative A, $15,040/yr — show the worked solution
(A/P,6%,15) = 0.10296; (A/F,6%,15) = 0.04296. EUAC_A = 85,000(0.10296) + 6,500 − 5,000(0.04296) = $15,037/yr. EUAC_B = $15,812/yr. Minimum wins: A.
Why the others are wrong. B) — B's EUAC computed correctly but it is the higher one. C) — right alternative, salvage ignored. D) — both errors stacked.
Exam tip: template every time: first cost × (A/P) + annual − salvage × (A/F), and write "MIN wins" first. More practice: Engineering Economics.
Question 20 — Construction (Earthwork)
Cut areas at 100-ft stations: 240 ft² (Sta 10+00), 360 ft² (Sta 11+00), 180 ft² (Sta 12+00). Shrinkage factor 0.90. By average end area, the compacted fill volume is most nearly:
- 1,730 yd³
- 1,900 yd³
- 2,110 yd³
- 2,320 yd³
Answer: B) 1,900 yd³ — show the worked solution
Segment volumes: 100(240+360)/2 = 30,000 ft³; 100(360+180)/2 = 27,000 ft³. Total in place = 57,000 ft³ = 2,111 yd³. Compacted: 2,111 × 0.90 = 1,900 yd³.
Why the others are wrong. A) 1,730 yd³ — all three areas averaged first (the middle station serves both segments). C) 2,110 yd³ — shrinkage never applied. D) 2,320 yd³ — swell (×1.10) instead of shrinkage; same factor, wrong direction.
Exam tip: label every earthwork quantity bank, loose, or compacted the moment you compute it. More practice: Construction Engineering.
Answer key: 1-B · 2-B · 3-B · 4-B · 5-B · 6-C · 7-B · 8-C · 9-B · 10-C · 11-B · 12-B · 13-C · 14-C · 15-B · 16-B · 17-B · 18-B · 19-A · 20-B
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