FE Civil Practice Problems

Worked FE Civil practice problems, one or more from each knowledge area of the NCEES specifications. Try each problem, then open the solution. When you are ready, take a free 10-question drill on any topic or the full 110-question practice exam.

110 questions on the real 5 hours 20 minutes clock, weighted to the NCEES specifications, with a score by knowledge area and a worked solution for every answer.

Or start with a free 10-question drill

1. Mathematics and Statistics

8–12 questions on the exam · Topic guide

Problem 1 · Analytic geometry

The boundary of an elliptical culvert opening is 9x² + 25y² = 225, with x and y in feet measured from the center. The distance between the two foci of the ellipse is most nearly:

  1. A.4 ft
  2. B.6 ft
  3. C.8 ft
  4. D.10 ft
Show the answer and solution

Answer: C: 8 ft

Dividing by 225 gives x²/25 + y²/9 = 1, so a = 5 ft and b = 3 ft. Then c = √(a² − b²) = √16 = 4 ft, and the foci are 2c = 8 ft apart. The value 4 ft is only the center-to-focus distance.

Analytic geometry · Reference Handbook: Mathematics

Problem 2 · Statistics

Daily water demand in a town is normally distributed with a mean of 18.0 MGD and a standard deviation of 2.4 MGD. For the standard normal distribution, Φ(1.645) = 0.950. The demand that is exceeded on only 5% of days is most nearly:

  1. A.14.1 MGD
  2. B.21.1 MGD
  3. C.21.9 MGD
  4. D.22.7 MGD
Show the answer and solution

Answer: C: 21.9 MGD

The 95th percentile is x = μ + zσ = 18.0 + 1.645(2.4) = 21.9 MGD. Using 1.96 gives 22.7 MGD, but 1.96 leaves 2.5% in each tail, not 5% in the upper tail.

Statistics · Reference Handbook: Engineering Probability and Statistics

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2. Ethics and Professional Practice

4–6 questions on the exam · Topic guide

Problem 3 · Licensure

A PE licensed in one state is invited to design a project located in another state where she holds no license. Before offering or performing engineering services there, she generally must:

  1. A.Notify NCEES, which authorizes practice in every state
  2. B.Retake both the FE and PE exams in the second state
  3. C.Obtain a license from that state's board, typically by comity (endorsement) based on her existing credentials
  4. D.Do nothing further, because a PE license issued in any state is valid nationwide
Show the answer and solution

Answer: C: Obtain a license from that state's board, typically by comity (endorsement) based on her existing credentials

Engineering licenses are granted by individual state and territorial boards. A licensee who wants to practice in another jurisdiction applies there, usually by comity, in which the board recognizes her education, exams, and experience without requiring the exams again.

Licensure · Reference Handbook: Ethics and Professional Practice

Problem 4 · Public protection issues

A state engineering licensing board receives a complaint that a licensee sealed plans for a type of structure outside her competence and part of the structure later failed. The primary purpose of the board's action on this complaint is to:

  1. A.Recover monetary damages for the owner of the failed structure
  2. B.Protect the public by investigating and, if warranted, disciplining the licensee, for example by reprimand, suspension, or revocation
  3. C.Protect the fees and reputation of licensed engineers in the state
  4. D.Prosecute the licensee and impose a prison sentence
Show the answer and solution

Answer: B: Protect the public by investigating and, if warranted, disciplining the licensee, for example by reprimand, suspension, or revocation

Licensing boards exist to protect the public by setting qualifications, investigating complaints, and disciplining licensees who violate the law or rules. The owner's financial recovery and any criminal case proceed separately in the courts.

Public protection issues · Reference Handbook: Ethics and Professional Practice

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3. Engineering Economics

5–8 questions on the exam · Topic guide

Problem 5 · Time value of money

A bond with a face value of $1,000 pays a 5% annual coupon ($50 at the end of each year) and matures in 8 years, when the $1,000 face value is repaid. An investor buys it today for $920. What rate of return (yield to maturity) does the investor earn? Enter the rate in percent to one decimal place.

Fill in the blank: answer in %.

Show the answer and solution

Answer: 6.3 % (accepted 6.2 to 6.4)

Solve 920 = 50(P/A, i, 8) + 1,000(P/F, i, 8). At 6%: 50(6.2098) + 1,000(0.6274) = 937.9. At 7%: 50(5.9713) + 1,000(0.5820) = 880.6. Interpolating gives i ≈ 6.3%. The yield exceeds the coupon rate because the bond was bought below face value.

Time value of money · Reference Handbook: Engineering Economics

Problem 6 · Time value of money

A $30,000 vehicle loan carries a nominal interest rate of 6% per year compounded monthly and is repaid in 60 equal end-of-month payments. The monthly payment is most nearly:

  1. A.$500
  2. B.$580
  3. C.$650
  4. D.$1,856
Show the answer and solution

Answer: B: $580

The monthly rate is 6%/12 = 0.5% and n = 60. A = 30,000(A/P, 0.5%, 60) = 30,000[0.005(1.005)^60/((1.005)^60 − 1)] = 30,000(0.019333) = $580. Using 6% per month greatly overstates the interest.

Time value of money · Reference Handbook: Engineering Economics

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4. Statics

8–12 questions on the exam · Topic guide

Problem 7 · Area moments of inertia

A rectangle is 4 in wide and 10 in tall. Find its moment of inertia about a horizontal axis located 2 in below its bottom edge. Enter the value in in⁴ to the nearest whole number.

Fill in the blank: answer in in⁴.

Show the answer and solution

Answer: 2293 in⁴ (accepted 2268 to 2318)

I_c = 4(10)³/12 = 333.3 in⁴ and A = 40 in². The centroid is 5 + 2 = 7 in above the axis, so I = 333.3 + 40(7)² = 333.3 + 1,960 = 2,293 in⁴.

Area moments of inertia · Reference Handbook: Statics

Problem 8 · Equivalent force systems

An 800 N force acts at point P, located 1.2 m to the right of and 0.5 m above point O. The force is directed downward and to the right at 60° below the horizontal. The force is replaced by an equivalent force-couple system at O. The magnitude of the couple is most nearly:

  1. A.631 N·m
  2. B.831 N·m
  3. C.960 N·m
  4. D.1,031 N·m
Show the answer and solution

Answer: D: 1,031 N·m

Fx = 800 cos 60° = 400 N (to the right) and Fy = −800 sin 60° = −692.8 N. M_O = xFy − yFx = 1.2(−692.8) − 0.5(400) = −831.4 − 200 = −1,031 N·m (clockwise). Both components turn the body clockwise about O, so their moments add.

Equivalent force systems · Reference Handbook: Statics

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5. Dynamics

4–6 questions on the exam · Topic guide

Problem 9 · Newton's second law for particles

A 900 kg elevator car accelerates upward at 1.5 m/s². Take g = 9.81 m/s². The tension in the hoisting cable is most nearly:

  1. A.1,350 N
  2. B.7,480 N
  3. C.8,830 N
  4. D.10,200 N
Show the answer and solution

Answer: D: 10,200 N

ΣF = ma upward: T − mg = ma, so T = m(g + a) = 900(9.81 + 1.5) = 10,179 N ≈ 10,200 N.

Newton's second law for particles · Reference Handbook: Dynamics

Problem 10 · Work-energy of particles

A hoist lifts an 800 kg load at a constant 0.50 m/s. The motor-and-drive efficiency is 85%. Take g = 9.81 m/s² and 1 hp = 745.7 W. The electrical power input to the hoist is most nearly:

  1. A.4.47 hp
  2. B.5.26 hp
  3. C.6.19 hp
  4. D.8.39 hp
Show the answer and solution

Answer: C: 6.19 hp

Output power = mgv = 800(9.81)(0.50) = 3,924 W. Input = 3,924/0.85 = 4,616 W = 4,616/745.7 = 6.19 hp.

Work-energy of particles · Reference Handbook: Dynamics

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6. Mechanics of Materials

7–11 questions on the exam · Topic guide

Problem 11 · Stress transformations and Mohr's circle

The outer surface of a pressurized housing is in plane stress (σz = 0) with σx = 12 ksi, σy = 4 ksi and τxy = 3 ksi. The absolute maximum shear stress at the point, considering all three dimensions, is most nearly:

  1. A.1.50 ksi
  2. B.4.00 ksi
  3. C.5.00 ksi
  4. D.6.50 ksi
Show the answer and solution

Answer: D: 6.50 ksi

C = 8 ksi, R = √(4² + 3²) = 5 ksi, so σ1 = 13 ksi, σ2 = 3 ksi, σ3 = 0. Because both in-plane principal stresses are tensile, the largest Mohr's circle spans σ1 to σ3 = 0: τ_abs,max = (13 − 0)/2 = 6.5 ksi. The in-plane value of 5.0 ksi understates it.

Stress transformations and Mohr's circle · Reference Handbook: Mechanics of Materials

Problem 12 · Material failure

A steel tie rod 20 mm in diameter is made of steel with a yield strength of 250 MPa. The design requires a factor of safety of 2.5 against yielding. The allowable axial tensile load is most nearly:

  1. A.31.4 kN
  2. B.78.5 kN
  3. C.126 kN
  4. D.196 kN
Show the answer and solution

Answer: A: 31.4 kN

A = π(20)²/4 = 314.2 mm². σ_allow = Sy/FS = 250/2.5 = 100 MPa. P_allow = σ_allow·A = 100(314.2) = 31416 N = 31.4 kN. The yield load without a factor of safety is 78.5 kN.

Material failure · Reference Handbook: Mechanics of Materials

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7. Materials

5–8 questions on the exam · Topic guide

Problem 13 · Test methods and specifications of metals, concrete, aggregates, asphalt, and wood

A coarse aggregate specific gravity test gives an oven-dry mass of 2,900 g, a saturated-surface-dry (SSD) mass in air of 3,000 g, and an SSD mass submerged in water of 1,850 g. The bulk specific gravity on an SSD basis is most nearly:

  1. A.2.52
  2. B.2.61
  3. C.2.76
  4. D.2.86
Show the answer and solution

Answer: B: 2.61

Bulk SSD specific gravity = B/(B − C) = 3000/(3000 − 1850) = 3000/1150 = 2.609. For comparison, bulk oven-dry = A/(B − C) = 2.522 and apparent = A/(A − C) = 2.762; absorption = (B − A)/A = 3.45%.

Test methods and specifications of metals, concrete, aggregates, asphalt, and wood · Reference Handbook: Civil Engineering

Problem 14 · Physical and mechanical properties of metals, concrete, aggregates, asphalt, and wood

For a pour into heavily congested reinforcement, a supplier must raise the slump of an approved concrete mix from 3 in. to 7 in. without reducing its 28-day compressive strength. The appropriate change is to:

  1. A.Double the dose of air-entraining admixture to raise the air content
  2. B.Add a high-range water-reducing admixture (superplasticizer) at the same water-cement ratio
  3. C.Increase the proportion of coarse aggregate in the mix
  4. D.Add water at the truck until the target slump is reached
Show the answer and solution

Answer: B: Add a high-range water-reducing admixture (superplasticizer) at the same water-cement ratio

Compressive strength is governed mainly by the water-cement ratio. A superplasticizer disperses the cement particles and greatly increases slump without additional water, so the w/c ratio and the strength are unchanged. Retempering with water is the field shortcut that specifications prohibit beyond the design water content because it weakens the concrete.

Physical and mechanical properties of metals, concrete, aggregates, asphalt, and wood · Reference Handbook: Civil Engineering

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8. Fluid Mechanics

6–9 questions on the exam · Topic guide

Problem 15 · Pumps

A pump draws 40°C water from an open tank at atmospheric pressure (101.3 kPa). The tank's water surface is 3.0 m below the pump centerline, and the suction-line head loss is 1.2 m. At 40°C the vapor pressure is 7.38 kPa and γ = 9.73 kN/m³. Neglect the velocity head at the inlet. The net positive suction head available is most nearly:

  1. A.5.45 m
  2. B.6.21 m
  3. C.6.65 m
  4. D.11.5 m
Show the answer and solution

Answer: A: 5.45 m

NPSHA = (p_atm − p_v)/γ − z_lift − h_L = (101.3 − 7.38)/9.73 − 3.0 − 1.2 = 9.65 − 4.2 = 5.45 m. The pump sits above the water, so the 3.0 m is a lift that reduces the available head.

Pumps · Reference Handbook: Fluid Mechanics

Problem 16 · Fluid transport systems

Select all that apply.

Select all statements that are true about operating centrifugal pumps together.

  1. A.Pumps in parallel operate at the same head, and their flows add.
  2. B.Pumps in series carry the same flow, and their heads add.
  3. C.Two identical pumps in parallel always double the flow through a given piping system.
  4. D.The operating point is where the combined pump curve intersects the system curve.
  5. E.Adding a second pump in series lowers the system curve.
Show the answer and solution

Answer: A: Pumps in parallel operate at the same head, and their flows add.; B: Pumps in series carry the same flow, and their heads add.; D: The operating point is where the combined pump curve intersects the system curve.

The combined curve for parallel pumps adds flows at equal head, and for series pumps adds heads at equal flow. The system operates where that combined curve meets the system curve. Because friction head rises roughly with Q², a second parallel pump raises the flow by less than 100%, and the pumps never change the system curve itself.

Fluid transport systems · Reference Handbook: Fluid Mechanics

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9. Surveying

6–9 questions on the exam · Topic guide

Problem 17 · Area computations

A four-sided parcel has corner coordinates (east, north) in feet of A (0, 0), B (420, 80), C (510, 390), and D (130, 450). Take 1 acre = 43,560 ft². The area of the parcel is most nearly:

  1. A.1.41 ac
  2. B.3.46 ac
  3. C.6.93 ac
  4. D.37.3 ac
Show the answer and solution

Answer: B: 3.46 ac

Coordinate (shoelace) method: 2A = Σ(xᵢyᵢ₊₁ − xᵢ₊₁yᵢ) = 0 + 123,000 + 178,800 + 0 = 301,800 ft², so A = 150,900 ft² = 150,900/43,560 = 3.46 ac. Forgetting the ½ doubles the area to 6.93 ac.

Area computations · Reference Handbook: Civil Engineering

Problem 18 · Angles, distances, and trigonometry

To find the distance across a river, a surveyor lays out baseline AB = 250.00 m and sights point C on the far bank. The measured angles are CAB = 62°10' and ABC = 71°35'. The distance AC is most nearly:

  1. A.190 m
  2. B.268 m
  3. C.306 m
  4. D.328 m
Show the answer and solution

Answer: D: 328 m

Angle C = 180° − 62°10' − 71°35' = 46°15'. By the law of sines, AC = AB sin B/sin C = 250.00(sin 71°35')/(sin 46°15') = 250.00(0.9488)/(0.7224) = 328 m. AC is the side opposite angle B, so sin B goes in the numerator.

Angles, distances, and trigonometry · Reference Handbook: Civil Engineering

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10. Water Resources and Environmental Engineering

10–15 questions on the exam · Topic guide

Problem 19 · Water distribution systems

Two pipes connect the same two junctions in a distribution system: pipe 1 is 300 mm in diameter and 800 m long, and pipe 2 is 200 mm in diameter and 400 m long. Both have the same Darcy friction factor, and minor losses are negligible. A total of 0.15 m³/s flows between the junctions. The flow in pipe 2 is most nearly:

  1. A.0.0313 m³/s
  2. B.0.0462 m³/s
  3. C.0.0509 m³/s
  4. D.0.0750 m³/s
Show the answer and solution

Answer: C: 0.0509 m³/s

Parallel pipes have equal head loss: h_f = f(L/D)(V²/2g) ∝ LQ²/D⁵, so Q ∝ √(D⁵/L). Q₁/Q₂ = √[(0.3/0.2)⁵(400/800)] = √3.797 = 1.949. Q₂ = 0.15/(1 + 1.949) = 0.0509 m³/s and Q₁ = 0.0991 m³/s.

Water distribution systems · Reference Handbook: Fluid Mechanics

Problem 20 · Groundwater

A well fully penetrates an unconfined aquifer with hydraulic conductivity K = 30 m/d resting on a horizontal impermeable base. At steady state, the saturated thickness is 18.0 m at an observation well 10 m from the pumped well and 20.0 m at another 80 m away. Use Q = πK(h₂² − h₁²)/ln(r₂/r₁). Enter the pumping rate in m³/d to the nearest 10 m³/d.

Fill in the blank: answer in m³/d.

Show the answer and solution

Answer: 3440 m³/d (accepted 3400 to 3480)

Q = π(30)(20.0² − 18.0²)/ln(80/10) = π(30)(76)/2.079 = 3,440 m³/d. In an unconfined aquifer the flow area shrinks as the water table drops, so the heads enter as squares.

Groundwater · Reference Handbook: Environmental Engineering

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11. Structural Engineering

10–15 questions on the exam · Topic guide

Problem 21 · Analysis of statically determinant beams, columns, trusses, and frames

A symmetric triangular roof truss spans 8 m between a pin support and a roller support. Its apex is 3 m above the bottom chord at midspan, so each sloping member is 5 m long, and the bottom chord is a single member joining the supports. A 60 kN vertical load acts at the apex. The force in each sloping member is most nearly:

  1. A.30.0 kN
  2. B.37.5 kN
  3. C.40.0 kN
  4. D.50.0 kN
Show the answer and solution

Answer: D: 50.0 kN

Each support reaction is 30 kN. At a support joint, the sloping member's vertical component balances the reaction: F(3/5) = 30, so F = 50.0 kN (compression). Its horizontal component, 50(4/5) = 40 kN, is the bottom-chord tension.

Analysis of statically determinant beams, columns, trusses, and frames · Reference Handbook: Civil Engineering

Problem 22 · Elementary statically indeterminate structures

A propped cantilever is fixed at A and supported on a roller at B, with a span of 6 m. A single 80 kN point load acts at midspan. Using standard beam results, enter the roller reaction at B in kN to one decimal place.

Fill in the blank: answer in kN.

Show the answer and solution

Answer: 25 kN (accepted 24.6 to 25.4)

For a propped cantilever with a midspan point load, the roller reaction is 5P/16 = 5(80)/16 = 25.0 kN, the fixed-end reaction is 11P/16 = 55.0 kN and the fixed-end moment is 3PL/16 = 90 kN·m. The fixed support's rotational restraint draws load toward A.

Elementary statically indeterminate structures · Reference Handbook: Civil Engineering

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12. Geotechnical Engineering

10–15 questions on the exam · Topic guide

Problem 23 · Consolidation and differential settlement

A 3.0 m thick overconsolidated clay layer has a recompression index Cr = 0.05, a compression index Cc = 0.30 and e0 = 0.80. At mid-depth the current effective stress is 80 kPa, the preconsolidation stress is 120 kPa, and a building will raise the effective stress to 140 kPa. The consolidation settlement is most nearly:

  1. A.20.3 mm
  2. B.48.1 mm
  3. C.86.7 mm
  4. D.122 mm
Show the answer and solution

Answer: B: 48.1 mm

The final stress exceeds the preconsolidation stress, so recompression applies from 80 to 120 kPa and virgin compression from 120 to 140 kPa: S = [3.0/1.80][0.05 log(120/80) + 0.30 log(140/120)] = 1.667(0.00880 + 0.02008) = 0.0481 m ≈ 48 mm.

Consolidation and differential settlement · Reference Handbook: Civil Engineering

Problem 24 · Bearing capacity

A square footing is founded 1.2 m deep in a saturated clay analyzed undrained (φ = 0, c = 50 kPa, γ = 18 kN/m³). Use Terzaghi's square-footing equation q_ult = 1.3cNc + γDfNq + 0.4γBNγ with Nc = 5.7, Nq = 1.0 and Nγ = 0. Enter the ultimate bearing capacity in kPa to the nearest whole number.

Fill in the blank: answer in kPa.

Show the answer and solution

Answer: 392 kPa (accepted 388 to 396)

q_ult = 1.3(50)(5.7) + 18(1.2)(1.0) + 0 = 370.5 + 21.6 = 392.1 kPa ≈ 392 kPa. With φ = 0, Nγ = 0, so the footing width does not enter the result.

Bearing capacity · Reference Handbook: Civil Engineering

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13. Transportation Engineering

9–14 questions on the exam · Topic guide

Problem 25 · Transportation planning

A proposed 240-unit apartment complex has a weekday trip rate of 9.44 trips per dwelling unit and a PM peak-hour rate of 0.99 trips per dwelling unit, with 63% of PM peak-hour trips entering. The number of entering trips in the PM peak hour is most nearly:

  1. A.150
  2. B.238
  3. C.1,427
  4. D.2,266
Show the answer and solution

Answer: A: 150

PM peak-hour trips = 240(0.99) = 237.6, and entering trips = 0.63(237.6) = 150. The daily rate describes 24-hour volume and does not apply to a peak-hour question.

Transportation planning · Reference Handbook: Civil Engineering

Problem 26 · Transportation planning

A 3.2 mi highway segment with an AADT of 14,500 veh/day recorded 21 crashes in 3 years. The crash rate per million vehicle-miles traveled is most nearly:

  1. A.0.41
  2. B.1.24
  3. C.1.32
  4. D.3.97
Show the answer and solution

Answer: A: 0.41

Vehicle-miles = 14,500(365)(3)(3.2) = 50.8 million. Rate = 21/50.8 = 0.41 crashes per million vehicle-miles. Both the exposure period and the segment length belong in the denominator.

Transportation planning · Reference Handbook: Civil Engineering

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14. Construction Engineering

8–12 questions on the exam · Topic guide

Problem 27 · Construction operations and methods (e.g., safety, equipment, productivity analysis, temporary erosion control)

On a highway construction site, runoff from disturbed slopes concentrates in a newly cut roadside ditch that carries significant flow during storms. The most appropriate temporary practice to slow the water and trap sediment within the ditch itself is:

  1. A.A single row of straw wattles laid along the top of the slope
  2. B.Rock check dams placed at intervals across the ditch
  3. C.Seeding the ditch with no other protection
  4. D.A silt fence installed across the ditch bottom
Show the answer and solution

Answer: B: Rock check dams placed at intervals across the ditch

Check dams of riprap or stone placed across a channel reduce velocity, limit scour and let sediment settle behind each dam. Silt fence and fiber rolls are perimeter and sheet-flow controls placed along the contour, not in channels carrying concentrated flow.

Construction operations and methods (e.g., safety, equipment, productivity analysis, temporary erosion control) · Reference Handbook: Civil Engineering

Problem 28 · Interpretation of engineering drawings

On a floor plan drawn at a scale of 1/4 in. = 1 ft-0 in., a wall measures 6-3/8 in. long with a ruler. The actual length of the wall is most nearly:

  1. A.1.59 ft
  2. B.25.5 ft
  3. C.51.0 ft
  4. D.76.5 ft
Show the answer and solution

Answer: B: 25.5 ft

At 1/4 in. = 1 ft, each inch on the drawing represents 4 ft. 6.375 in. × 4 ft/in. = 25.5 ft (25 ft-6 in.).

Interpretation of engineering drawings · Reference Handbook: Civil Engineering

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Common questions

Are these FE Civil practice problems like the real exam?

They are original problems written to the NCEES FE Civil CBT specifications, at the length of a real FE question: about three minutes each with a calculator and the FE Reference Handbook. Most are multiple choice with four options; some are select-all-that-apply or numeric fill-in questions, the exam’s alternative item types.

How many FE Civil practice problems should I do?

Work through every knowledge area, then sit at least one full 110-question practice exam on the real 5-hour-20-minute clock. The score by knowledge area shows where to spend the rest of your time.

Ready for exam day?

110 questions on the real 5 hours 20 minutes clock, weighted to the NCEES specifications, with a score by knowledge area and a worked solution for every answer.

Or start with a free 10-question drill

Practice problems for other FE exams

Original independent practice problems written to the published NCEES exam specifications. Not affiliated with or endorsed by NCEES (National Council of Examiners for Engineering and Surveying).