The FE Civil Engineering exam tests your knowledge across a wide range of civil engineering topics—from statics and structural analysis to geotechnical engineering, transportation, hydraulics, and engineering economics. The best way to prepare is to work through realistic problems under exam-like conditions.

Below you will find 10 practice problems that mirror the style, difficulty, and topic distribution you will encounter on the actual NCEES FE Civil exam. Each problem includes four answer choices and a detailed, step-by-step solution. Use these to identify your strengths, target your weaknesses, and build the problem-solving speed you need on test day.

Tip: Try solving each problem on your own before revealing the solution. Time yourself—you should average about 3 minutes per problem on the real exam.

Problem 1: Statics — Reaction Forces on a Simply Supported Beam

A simply supported beam spans 10 m between supports A (pin) and B (roller). A concentrated load of 20 kN acts at a point 4 m from support A. What is the vertical reaction at support B?

  • A) 8 kN
  • B) 12 kN
  • C) 10 kN
  • D) 6 kN
Show Solution

Answer: A

For a simply supported beam, we use static equilibrium to find the reactions.

Step 1: Draw the free-body diagram. Support A is a pin (provides vertical reaction R_A), and support B is a roller (provides vertical reaction R_B). The 20 kN load acts 4 m from A (and therefore 6 m from B).

Step 2: Sum moments about point A to eliminate R_A:

∑M_A = 0: R_B × 10 − 20 × 4 = 0

R_B × 10 = 80

R_B = 80 / 10 = 8 kN

Step 3: Verify with vertical equilibrium: ∑F_y = 0: R_A + R_B = 20, so R_A = 20 − 8 = 12 kN. ✓

The vertical reaction at support B is 8 kN.

Problem 2: Mechanics of Materials — Normal Stress in an Axially Loaded Rod

A solid circular steel rod has a diameter of 25 mm and is subjected to an axial tensile load of 50 kN. What is the normal stress in the rod?

  • A) 63.7 MPa
  • B) 101.9 MPa
  • C) 81.5 MPa
  • D) 127.3 MPa
Show Solution

Answer: B

Normal stress under axial loading is given by σ = P / A, where P is the applied force and A is the cross-sectional area.

Step 1: Calculate the cross-sectional area of the rod:

A = πd² / 4 = π(0.025)² / 4 = π(6.25 × 10&supmin;&sup4;) / 4 = 4.909 × 10&supmin;&sup4; m²

Step 2: Calculate the normal stress:

σ = P / A = 50,000 N / 4.909 × 10&supmin;&sup4; m² = 101.86 × 10&sup6; Pa

σ ≈ 101.9 MPa

This is well below the yield strength of most structural steels (around 250 MPa), so the rod remains in the elastic region.

Problem 3: Fluid Mechanics — Bernoulli’s Equation for Velocity

Water flows through a horizontal pipe that narrows from a diameter of 200 mm (section 1) to 100 mm (section 2). If the pressure at section 1 is 250 kPa and the velocity at section 1 is 2 m/s, what is the velocity at section 2? Assume incompressible, steady flow.

  • A) 4 m/s
  • B) 6 m/s
  • C) 8 m/s
  • D) 10 m/s
Show Solution

Answer: C

For incompressible flow, the continuity equation relates velocities and cross-sectional areas: A_1 × v_1 = A_2 × v_2.

Step 1: Calculate the cross-sectional areas:

A_1 = π(0.200)² / 4 = π(0.04) / 4 = 0.03142 m²

A_2 = π(0.100)² / 4 = π(0.01) / 4 = 0.007854 m²

Step 2: Apply the continuity equation:

v_2 = v_1 × (A_1 / A_2) = 2 × (0.03142 / 0.007854)

Step 3: Simplify the area ratio. Since A is proportional to d², the ratio A_1/A_2 = (d_1/d_2)² = (200/100)² = 4.

v_2 = 2 × 4 = 8 m/s

When the diameter halves, the area decreases by a factor of 4, so the velocity must increase by a factor of 4 to maintain the same volumetric flow rate.

Problem 4: Geotechnical Engineering — Void Ratio from Phase Relationships

A soil sample has a total volume of 0.5 m³ and a volume of solids of 0.3 m³. What is the void ratio of the soil?

  • A) 0.40
  • B) 0.60
  • C) 0.67
  • D) 1.67
Show Solution

Answer: C

The void ratio (e) is defined as the ratio of the volume of voids to the volume of solids: e = V_v / V_s.

Step 1: Determine the volume of voids:

V_v = V_total − V_s = 0.5 − 0.3 = 0.2 m³

Step 2: Calculate the void ratio:

e = V_v / V_s = 0.2 / 0.3 = 0.667

Rounding to two decimal places, e ≈ 0.67.

Note: Do not confuse void ratio with porosity. Porosity is n = V_v / V_total = 0.2 / 0.5 = 0.40, which corresponds to answer choice A. The void ratio uses V_s in the denominator, not V_total.

Problem 5: Structural Analysis — Maximum Moment for a Beam with Uniform Load

A simply supported beam of length L = 8 m carries a uniformly distributed load (UDL) of w = 12 kN/m over its entire span. What is the maximum bending moment in the beam?

  • A) 48 kN·m
  • B) 72 kN·m
  • C) 96 kN·m
  • D) 192 kN·m
Show Solution

Answer: C

For a simply supported beam with a UDL over the entire span, the maximum bending moment occurs at midspan and is given by:

M_max = wL² / 8

Step 1: Substitute the given values:

M_max = 12 × (8)² / 8

Step 2: Compute:

M_max = 12 × 64 / 8 = 768 / 8 = 96 kN·m

The shear diagram for this loading is a straight line from +wL/2 = +48 kN at the left support to −wL/2 = −48 kN at the right support, crossing zero at midspan. The moment diagram is a parabola that peaks at midspan with a value of 96 kN·m.

Problem 6: Transportation Engineering — Stopping Sight Distance

A vehicle is traveling at a design speed of 100 km/h on a level roadway. The driver’s perception-reaction time is 2.5 seconds, and the coefficient of friction between the tires and the pavement is 0.35. What is the minimum stopping sight distance (SSD)? Use g = 9.81 m/s².

  • A) 112 m
  • B) 150 m
  • C) 182 m
  • D) 215 m
Show Solution

Answer: C

Stopping sight distance consists of two parts: the distance traveled during perception-reaction time plus the braking distance.

SSD = v × t + v² / (2gf)

Step 1: Convert speed to m/s:

v = 100 km/h × (1000 m / 3600 s) = 27.78 m/s

Step 2: Calculate the perception-reaction distance:

d_1 = v × t = 27.78 × 2.5 = 69.44 m

Step 3: Calculate the braking distance:

d_2 = v² / (2gf) = (27.78)² / (2 × 9.81 × 0.35)

d_2 = 771.73 / 6.867 = 112.39 m

Step 4: Sum both distances:

SSD = 69.44 + 112.39 = 181.83 m ≈ 182 m

This is the minimum distance a driver needs to see ahead to safely stop. Note that choice A (112 m) is only the braking distance—a common trap. You must add the perception-reaction distance. AASHTO design guidelines are based on this type of calculation.

Problem 7: Engineering Economics — Present Worth Analysis

A city is evaluating a new water treatment system that costs $500,000 to install and will save $85,000 per year in operating costs for 10 years. At the end of 10 years, the system has a salvage value of $50,000. If the discount rate is 6% per year, what is the net present worth (NPW) of this project?

  • A) $153,500
  • B) $155,700
  • C) $157,900
  • D) $152,300
Show Solution

Answer: A

Net present worth is calculated as:

NPW = −Initial Cost + Annual Savings × (P/A, i, n) + Salvage × (P/F, i, n)

Step 1: Compute the P/A factor (present worth of annuity):

(P/A, 6%, 10) = [(1.06)¹° − 1] / [0.06 × (1.06)¹°]

(1.06)¹° = 1.7908

(P/A) = (1.7908 − 1) / (0.06 × 1.7908) = 0.7908 / 0.10745 = 7.3601

Step 2: Compute the P/F factor (present worth of single future amount):

(P/F, 6%, 10) = 1 / (1.06)¹° = 1 / 1.7908 = 0.5584

Step 3: Calculate NPW:

NPW = −$500,000 + $85,000 × 7.3601 + $50,000 × 0.5584

NPW = −$500,000 + $625,609 + $27,920

NPW = $153,529 ≈ $153,500

Since NPW > 0, the project is economically justified at a 6% discount rate. The annual savings more than offset the initial investment over the 10-year life.

Problem 8: Hydraulics — Manning’s Equation for Open Channel Flow

A rectangular open channel is 3 m wide with a flow depth of 1.5 m. The channel has a bed slope of 0.002 and a Manning’s roughness coefficient of n = 0.013. What is the discharge in the channel? Use Manning’s equation: Q = (1/n) × A × R2/3 × S1/2.

  • A) 9.8 m³/s
  • B) 10.5 m³/s
  • C) 12.8 m³/s
  • D) 14.1 m³/s
Show Solution

Answer: C

Manning’s equation for open channel flow (SI units) is:

Q = (1/n) × A × R2/3 × S1/2

Step 1: Calculate the cross-sectional area:

A = width × depth = 3 × 1.5 = 4.5 m²

Step 2: Calculate the wetted perimeter:

P = width + 2 × depth = 3 + 2(1.5) = 6.0 m

Step 3: Calculate the hydraulic radius:

R = A / P = 4.5 / 6.0 = 0.75 m

Step 4: Calculate R2/3:

R2/3 = (0.75)2/3 = 0.8255

Step 5: Calculate S1/2:

S1/2 = (0.002)1/2 = 0.04472

Step 6: Compute the discharge:

Q = (1/0.013) × 4.5 × 0.8255 × 0.04472

Q = 76.923 × 4.5 × 0.8255 × 0.04472

First: A × R2/3 = 4.5 × 0.8255 = 3.7148

Then: 3.7148 × 0.04472 = 0.16609

Finally: Q = 0.16609 / 0.013 = 12.8 m³/s

Manning’s equation is widely used in civil engineering for designing open channels, storm drains, and culverts. The roughness coefficient n depends on the channel material—0.013 is typical for finished concrete.

Problem 9: Environmental Engineering — BOD Removal Efficiency

A wastewater treatment plant receives influent with a BOD5 of 250 mg/L and produces effluent with a BOD5 of 20 mg/L. What is the BOD removal efficiency of the plant?

  • A) 88%
  • B) 90%
  • C) 92%
  • D) 95%
Show Solution

Answer: C

Removal efficiency is calculated as the percentage of the pollutant removed relative to the influent concentration:

Efficiency = [(BOD_in − BOD_out) / BOD_in] × 100%

Step 1: Substitute the given values:

Efficiency = [(250 − 20) / 250] × 100%

Step 2: Compute:

Efficiency = [230 / 250] × 100% = 0.92 × 100% = 92%

A 92% removal efficiency is typical of a well-operating secondary treatment process (e.g., activated sludge). Most regulatory standards require at least 85% BOD removal for secondary treatment.

Problem 10: Surveying — Area by the Coordinate Method

A triangular parcel of land has vertices at the following coordinates: A(0, 0), B(100, 0), and C(60, 80). All units are in meters. What is the area of the parcel using the coordinate method?

  • A) 3,200 m²
  • B) 4,000 m²
  • C) 4,800 m²
  • D) 6,400 m²
Show Solution

Answer: B

The coordinate method (also called the Shoelace formula) computes area from vertex coordinates:

Area = ½ |∑(x_i × y_{i+1} − x_{i+1} × y_i)|

Step 1: List the coordinates in order and repeat the first point at the end:

A(0, 0), B(100, 0), C(60, 80), A(0, 0)

Step 2: Compute the cross-products:

x_A × y_B − x_B × y_A = 0 × 0 − 100 × 0 = 0

x_B × y_C − x_C × y_B = 100 × 80 − 60 × 0 = 8,000

x_C × y_A − x_A × y_C = 60 × 0 − 0 × 80 = 0

Step 3: Sum the cross-products and take half the absolute value:

Area = ½ |0 + 8,000 + 0| = ½ × 8,000 = 4,000 m²

You can verify this with the basic triangle formula: Area = ½ × base × height = ½ × 100 × 80 = 4,000 m². ✓

Tips for Using These Practice Problems

Working through practice problems is one of the highest-value activities in your FE exam preparation, but how you practice matters just as much as how much you practice. Here are some tips to get the most out of these problems:

What Topics Does the FE Civil Exam Cover?

The FE Civil Engineering exam, administered by NCEES, covers the following 16 major topic areas:

The 10 problems above touch on many of these high-weight categories. To fully prepare, you need to practice across all of them—especially statics, mechanics of materials, and geotechnical engineering, which each carry 7–11% of the exam weight and together can account for roughly one-third of all questions.

Next Steps

These 10 problems give you a solid starting point, but the FE Civil exam covers far more ground. To build true exam readiness, you need hundreds of problems across every topic, with instant feedback and detailed explanations for every answer choice.

Disclaimer: This content is for educational purposes only and is not affiliated with, endorsed by, or sponsored by NCEES. “FE” and “Fundamentals of Engineering” are trademarks of NCEES. Always refer to the official NCEES website for the most current exam specifications and policies.