Preparing for the FE Civil exam requires consistent practice with realistic problems. The NCEES FE Civil exam covers 18 topic areas and tests your ability to apply fundamental engineering concepts under timed conditions. The best way to build confidence and identify weak areas is to work through practice problems that mirror the actual exam format.
Below, we have compiled a set of free practice problems spanning five core FE Civil topics: Mathematics, Statics, Fluid Mechanics, Structural Engineering, and Geotechnical Engineering. Each problem includes four multiple-choice options and a detailed step-by-step solution you can reveal when you are ready.
How to Use This Resource: Try solving each problem on your own before clicking "Show Solution." Use the NCEES FE Reference Handbook as you would on exam day. Time yourself -- aim for about 3 minutes per problem to simulate real exam pacing.
Why Practice Problems Matter
Research consistently shows that active recall -- the process of retrieving information from memory -- is one of the most effective study strategies. Simply re-reading notes or watching videos is passive learning. Working through problems forces your brain to actively engage with the material, strengthening neural pathways and improving long-term retention.
Identify Weak Areas
Practice problems reveal exactly which topics need more study time, so you can prioritize efficiently.
Build Exam Pacing
The FE exam gives you about 3 minutes per question. Regular practice builds the speed you need.
Learn the Reference Handbook
Practice problems teach you where to find formulas and tables in the FE Reference Handbook quickly.
Boost Confidence
The more problems you solve correctly, the more confident you will feel walking into the testing center.
Mathematics (2 Problems)
The Mathematics and Statistics section accounts for approximately 7-11% of the FE Civil exam. Topics include analytic geometry, calculus, linear algebra, and probability and statistics.
Problem 1 -- Definite Integration
Evaluate the definite integral:
$$\int_0^2 (3x^2 + 2x - 1)\,dx$$
A) 8
B) 10
C) 12
D) 14
Answer: B) 10
Integrate term by term:
$$\int(3x^2 + 2x - 1)\,dx = x^3 + x^2 - x + C$$
Evaluate from 0 to 2:
\(F(2) = 2^3 + 2^2 - 2 = 8 + 4 - 2 = 10\)
\(F(0) = 0^3 + 0^2 - 0 = 0\)
\(F(2) - F(0) = 10 - 0 = \textbf{10}\)
Problem 2 -- Standard Deviation
A set of five concrete cylinder compressive strength test results are: 28, 32, 30, 34, and 26 MPa. What is the sample standard deviation of these results?
A) 2.4 MPa
B) 3.2 MPa
C) 4.0 MPa
D) 4.5 MPa
Answer: B) 3.2 MPa
Step 1: Calculate the mean:
$$\bar{x} = \frac{28 + 32 + 30 + 34 + 26}{5} = \frac{150}{5} = 30 \text{ MPa}$$
Step 2: Calculate the squared deviations:
\((28-30)^2 = 4\), \((32-30)^2 = 4\), \((30-30)^2 = 0\), \((34-30)^2 = 16\), \((26-30)^2 = 16\)
Step 3: Sum of squared deviations \(= 4 + 4 + 0 + 16 + 16 = 40\)
Step 4: Sample variance:
$$s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1} = \frac{40}{5-1} = \frac{40}{4} = 10$$
Step 5: Sample standard deviation:
$$s = \sqrt{10} = 3.16 \approx \textbf{3.2 MPa}$$
Statics (2 Problems)
Statics is one of the most heavily weighted topics on the FE Civil exam (approximately 7-11%). It covers resultants of force systems, equilibrium of rigid bodies, frames, trusses, and centroids.
Problem 3 -- Simple Beam Reactions
A simply supported beam has a span of 10 m. A concentrated load of 20 kN is applied 4 m from the left support (A). What is the vertical reaction at support B (right end)?
A) 8 kN
B) 10 kN
C) 12 kN
D) 14 kN
Answer: A) 8 kN
Take moments about point A \((\sum M_A = 0)\):
$$R_B \times 10 - 20 \times 4 = 0$$
$$R_B = \frac{20 \times 4}{10} = \frac{80}{10} = \textbf{8 kN}$$
Verification: \(R_A = 20 - 8 = 12\) kN. Check \(\sum M_B = 0\): \(12(10) - 20(6) = 120 - 120 = 0\). Confirmed.
Problem 4 -- Truss Member Force (Method of Joints)
A simple triangular truss has a pin at A (left) and a roller at C (right). The horizontal distance from A to C is 6 m. Point B is directly above the midpoint of AC at a height of 4 m. A vertical downward load of 12 kN is applied at joint B. What is the force in member AB?
A) 6.0 kN (Tension)
B) 7.5 kN (Compression)
C) 7.5 kN (Tension)
D) 10.0 kN (Compression)
Answer: B) 7.5 kN (Compression)
Step 1: By symmetry, \(R_A = R_C = \frac{12}{2} = 6\) kN (upward).
Step 2: Member AB goes from A(0,0) to B(3,4).
$$L_{AB} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \text{ m}$$
Direction cosines: \(\cos\theta = \frac{3}{5}\), \(\sin\theta = \frac{4}{5}\)
Step 3: At joint A, assuming all members in tension (forces directed away from joint):
$$\sum F_y = 0: \quad 6 + F_{AB}\left(\frac{4}{5}\right) = 0 \implies F_{AB} = -7.5 \text{ kN}$$
The negative sign means the assumed tension direction was wrong -- the member is in compression.
\(|F_{AB}| = \textbf{7.5 kN (Compression)}\)
Fluid Mechanics (2 Problems)
Fluid Mechanics represents approximately 7-11% of the FE Civil exam, covering fluid properties, hydrostatics, Bernoulli's equation, pipe flow, and open channel flow.
Problem 5 -- Hydrostatic Force on a Submerged Gate
A rectangular gate is 2 m wide and 3 m tall, submerged vertically in water with its top edge at the water surface. What is the total hydrostatic force on the gate? (Use \(\gamma_{water} = 9.81\) kN/m\(^3\))
A) 58.9 kN
B) 73.6 kN
C) 88.3 kN
D) 117.7 kN
Answer: C) 88.3 kN
The hydrostatic force on a submerged plane surface is:
$$F = \gamma \cdot \bar{h} \cdot A$$
where \(\bar{h}\) is the depth to the centroid of the gate.
The gate is 3 m tall with its top at the surface, so the centroid is at:
$$\bar{h} = \frac{3}{2} = 1.5 \text{ m below the surface}$$
Area: \(A = 2 \times 3 = 6 \text{ m}^2\)
$$F = 9.81 \times 1.5 \times 6 = \textbf{88.29} \approx \textbf{88.3 kN}$$
Problem 6 -- Bernoulli's Equation (Pipe Flow)
Water flows through a horizontal pipe that narrows from a diameter of 200 mm to 100 mm. The velocity in the larger section is 1.5 m/s and the pressure is 200 kPa. What is most nearly the pressure in the narrower section? (Use \(\rho = 1000\) kg/m\(^3\); assume no losses.)
A) 173 kPa
B) 183 kPa
C) 193 kPa
D) 200 kPa
Answer: B) 183 kPa
Step 1: Find velocity in the narrow section using continuity \((A_1 V_1 = A_2 V_2)\):
The diameter ratio is \(\frac{200}{100} = 2\), so the area ratio is \(2^2 = 4\).
$$V_2 = V_1 \times \frac{A_1}{A_2} = 1.5 \times 4 = 6.0 \text{ m/s}$$
Step 2: Apply Bernoulli's equation (horizontal pipe, \(z_1 = z_2\)):
$$P_1 + \tfrac{1}{2}\rho V_1^2 = P_2 + \tfrac{1}{2}\rho V_2^2$$
$$200{,}000 + \tfrac{1}{2}(1000)(1.5)^2 = P_2 + \tfrac{1}{2}(1000)(6.0)^2$$
$$200{,}000 + 1{,}125 = P_2 + 18{,}000$$
$$P_2 = 201{,}125 - 18{,}000 = 183{,}125 \text{ Pa} \approx \textbf{183 kPa}$$
Structural Engineering (2 Problems)
Structural Engineering topics cover approximately 7-11% of the FE Civil exam, including analysis of beams, columns, frames, load combinations, and design concepts.
Problem 7 -- Maximum Bending Moment (UDL)
A simply supported beam with a span of 8 m carries a uniformly distributed load (UDL) of 5 kN/m over its entire length. What is the maximum bending moment?
A) 20 kN·m
B) 30 kN·m
C) 40 kN·m
D) 50 kN·m
Answer: C) 40 kN·m
For a simply supported beam with a UDL, the maximum bending moment occurs at midspan:
$$M_{max} = \frac{wL^2}{8}$$
Where \(w = 5\) kN/m and \(L = 8\) m:
$$M_{max} = \frac{5 \times 8^2}{8} = \frac{5 \times 64}{8} = \frac{320}{8} = \textbf{40 kN}{\cdot}\textbf{m}$$
Problem 8 -- Euler Buckling Load
A steel column is 4 m long with both ends pinned. The column has a moment of inertia \(I = 5.0 \times 10^6\) mm\(^4\) and modulus of elasticity \(E = 200\) GPa. What is most nearly the critical (Euler) buckling load?
A) 412 kN
B) 514 kN
C) 617 kN
D) 720 kN
Answer: C) 617 kN
The Euler critical buckling load for a pin-pin column (\(K = 1.0\)) is:
$$P_{cr} = \frac{\pi^2 EI}{(KL)^2}$$
Convert units: \(E = 200{,}000\) N/mm\(^2\), \(I = 5.0 \times 10^6\) mm\(^4\), \(KL = 4{,}000\) mm
$$P_{cr} = \frac{\pi^2 \times 200{,}000 \times 5{,}000{,}000}{(4{,}000)^2}$$
$$P_{cr} = \frac{9.8696 \times 10^{12}}{16 \times 10^6} = 616{,}850 \text{ N} \approx \textbf{617 kN}$$
Geotechnical Engineering (2 Problems)
Geotechnical Engineering accounts for approximately 7-11% of the FE Civil exam, covering soil classification, effective stress, consolidation, shear strength, and bearing capacity.
Problem 9 -- Effective Stress Calculation
A soil profile consists of 3 m of dry sand (unit weight \(\gamma_d = 17\) kN/m\(^3\)) overlying 5 m of saturated clay (saturated unit weight \(\gamma_{sat} = 20\) kN/m\(^3\)). The groundwater table is at the interface between the sand and clay layers. What is the effective vertical stress at the bottom of the clay layer? (Use \(\gamma_w = 9.81\) kN/m\(^3\))
A) 91.0 kPa
B) 101.9 kPa
C) 109.1 kPa
D) 151.0 kPa
Answer: B) 101.9 kPa
Step 1: Total vertical stress at the bottom of the clay:
$$\sigma_{total} = \gamma_d \times H_{sand} + \gamma_{sat} \times H_{clay}$$
$$\sigma_{total} = 17 \times 3 + 20 \times 5 = 51 + 100 = 151 \text{ kPa}$$
Step 2: Pore water pressure at the bottom of the clay:
$$u = \gamma_w \times H_w = 9.81 \times 5 = 49.05 \text{ kPa}$$
(The water table is at the top of the clay, so the water column height equals the clay thickness.)
Step 3: Effective stress:
$$\sigma' = \sigma_{total} - u = 151 - 49.05 = \textbf{101.95} \approx \textbf{101.9 kPa}$$
Problem 10 -- Soil Classification (USCS)
A soil sample has the following properties: 58% passes the No. 200 sieve, liquid limit (LL) = 45, and plastic limit (PL) = 22. How would this soil be classified under the Unified Soil Classification System (USCS)?
A) CL
B) CH
C) ML
D) MH
Answer: A) CL
Step 1: Since more than 50% passes the No. 200 sieve (58%), the soil is fine-grained.
Step 2: Calculate the Plasticity Index (PI):
$$PI = LL - PL = 45 - 22 = 23$$
Step 3: Use the Plasticity Chart:
\(LL = 45\) (less than 50 → "L" suffix for low plasticity).
The A-line equation:
$$PI = 0.73 \times (LL - 20) = 0.73 \times 25 = 18.25$$
Since \(PI = 23 > 18.25\), the point plots above the A-line, indicating Clay (C).
Classification: CL (Low-plasticity clay)
Bonus Problems
Here are two additional problems to further test your preparation across different FE Civil topics.
Problem 11 -- Engineering Economics (Time Value of Money)
An engineer invests $10,000 in an account that earns 6% annual interest compounded annually. What is the value of the investment after 5 years?
A) $12,625
B) $13,382
C) $13,971
D) $14,185
Answer: B) $13,382
Use the future value formula for compound interest:
$$F = P(1 + i)^n$$
Where \(P = \$10{,}000\), \(i = 0.06\), \(n = 5\):
$$F = 10{,}000 \times (1.06)^5$$
$$(1.06)^5 = 1.33823$$
$$F = 10{,}000 \times 1.33823 = \textbf{\$13{,}382}$$
Problem 12 -- Mechanics of Materials (Normal Stress)
A steel rod with a circular cross-section has a diameter of 25 mm and is subjected to an axial tensile load of 50 kN. What is most nearly the normal stress in the rod?
A) 81.5 MPa
B) 91.2 MPa
C) 101.9 MPa
D) 112.6 MPa
Answer: C) 101.9 MPa
Normal stress is defined as:
$$\sigma = \frac{P}{A}$$
Calculate the cross-sectional area:
$$A = \frac{\pi d^2}{4} = \frac{\pi (25)^2}{4} = \frac{\pi (625)}{4} = 490.87 \text{ mm}^2$$
Calculate stress:
$$\sigma = \frac{50{,}000}{490.87} = \textbf{101.9 MPa}$$
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Start Your Free TrialTips for Maximizing Your Practice
- Simulate Real Exam Conditions: Practice with a timer and only use the NCEES FE Reference Handbook. No textbooks, no notes, no internet searches.
- Review Every Solution: Even if you got the answer right, read the solution to make sure you solved it the most efficient way. The FE exam is a race against the clock.
- Track Your Performance: Keep a log of which topics and question types you are getting wrong. Focus your study time on those weak areas.
- Practice in Batches: Do 20-30 problems in a sitting to build endurance. The actual exam is 5 hours and 20 minutes with 110 questions.
- Do Not Memorize -- Understand: The FE exam tests your ability to apply concepts, not memorize formulas. Focus on understanding the underlying principles.