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Free FE practice problems with full solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

Here are five free FE practice problems from five areas of the exam: fluid mechanics, statics, water treatment, engineering economics, and ethics. Each one is original, generated by our software from the formulas in the FE Reference Handbook, and checked by code before it was published. Every solution gives the handbook section and page, the equation, the substitution with units, and the mistake behind each wrong option.

How these problems are made

The problem engine draws the numbers for each problem within realistic ranges, solves it, and builds the wrong options from named mistakes such as a unit slip or the wrong formula. Code recomputes every answer before a problem is published, and no problem copies wording from the handbook. That is the same engine that writes the daily problem inside the lab.

The problems

Problem 1 · FE, Pipe flow, Darcy-Weisbach

A 12 in. diameter pipe that is 2,050 ft long carries 1,650 gpm of water. The Darcy friction factor for the pipe is 0.033. The head loss due to friction in the pipe is most nearly:

  • A 23.0 ft
  • B 75.6 ft
  • C 46.1 ft
  • D 4.92 ft

Handbook: Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), FE Reference Handbook 10.6

Show the worked solution
Answer
A (23.0 ft)
Given
Q = 1,650 gpm, D = 12 in., L = 2,050 ft, f = 0.033, g = 32.174 ft/sec² (handbook), 7.481 gal/ft³ (handbook)
Find
head loss due to friction, hf (ft)
Handbook
Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation); Units and Conversion Factors (7.481 gal per ft³), pages 3 and 187
Equation
hf = f(L/D)V²/(2g), with V = Q/A and A = πD²/4
Substitute
  1. Q = (1,650 gal/min)/(7.481 gal/ft³ × 60 sec/min) = 3.676 ft³/sec
  2. D = 12 in./(12 in./ft) = 1.000 ft
  3. A = πD²/4 = π(1.000 ft)²/4 = 0.7854 ft²
  4. V = Q/A = (3.676 ft³/sec)/(0.7854 ft²) = 4.680 ft/sec
  5. hf = f(L/D)V²/(2g) = 0.033 × (2,050 ft/1.000 ft) × (4.680 ft/sec)²/(2 × 32.174 ft/sec²) = 23.0 ft
Result
23.0 ft, 3 significant figures
Check
units (ft/ft)(ft/sec)²/(ft/sec²) = ft; V = 4.68 ft/sec is an ordinary velocity for water in a pipe, and the loss is 1.12 ft per 100 ft of pipe.
Why the others are wrong
  • B: used g = 9.807 m/s² in a USCS problem
  • C: used the radius instead of the diameter in L/D, which doubles the loss
  • D: did not square the velocity

Problem 2 · FE Civil, Truss member force

A three-member plane truss has joints A (0, 0), B (13.0, 6.0) and C (21.0, 0), with coordinates in m. The members are AB, BC and AC. A is a pin support and C is a roller on a horizontal surface. Joint B carries a downward load of 185 kN and a horizontal load of 10 kN in the +x direction. The force in member BC is most nearly:

  • A 154 kN (C)
  • B 196 kN (C)
  • C 117 kN (C)
  • D 113 kN (C)

Handbook: Statics, Plane Truss: Method of Joints, FE Reference Handbook 10.6

Show the worked solution
Answer
B (196 kN in compression)
Given
L = 21.0 m, a = 13.0 m, h = 6.0 m, P = 185 kN, H = 10 kN
Find
force in member BC, magnitude and tension (T) or compression (C)
Handbook
Statics, Plane Truss: Method of Joints, page 98
Equation
Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive
Substitute
  1. Reactions, sum of moments about A = 0: Cy = (P·a + H·h)/L = ((185 kN)(13.0 m) + (10 kN)(6.0 m))/(21.0 m) = 117.4 kN
  2. Sum Fy = 0: Ay = P - Cy = 67.62 kN; sum Fx = 0: Ax = -H = -10 kN
  3. Length BC = √((L - a)² + h²) = √(8.0² + 6.0²) m = 10.00 m
  4. Joint C, sum Fy = 0: Cy + F_BC(h/BC) = 0, so F_BC = -Cy·BC/h = -(117.4 kN)(10.00 m)/(6.0 m) = -195.6 kN
  5. F_BC = -195.6 kN, so BC carries 196 kN in compression (C)
Result
196 kN (C), 3 significant figures
Check
with all three member forces, joint B also balances (sum Fx = 0 and sum Fy = 0 including the load); a positive member force is tension, a negative one compression.
Why the others are wrong
  • A: assumed each support carries P/2, but B is not at midspan
  • C: set the member force equal to the reaction without resolving it along the member
  • D: used the reaction at A in the equilibrium of joint C

Problem 3 · FE Environmental, Clarifier overflow rate and detention time

A circular primary clarifier with a diameter of 23.0 m treats an average flow of 22,500 m³/d. Find the overflow rate (surface loading rate) in m³/(m²·d).

Enter your answer in m³/(m²·d), 3 significant figures.

Handbook: Environmental Engineering, Clarifier, FE Reference Handbook 10.6

Show the worked solution
Answer
54.2 m³/(m²·d)
Given
D = 23.0 m, Q = 22,500 m³/d
Find
overflow rate vo (m³/(m²·d))
Handbook
Environmental Engineering, Clarifier (overflow rate), page 345
Equation
vo = Q/A_surface, with A = πD²/4
Substitute
  1. A = πD²/4 = π(23.0 m)²/4 = 415.5 m²
  2. vo = Q/A = (22,500 m³/d)/(415.5 m²) = 54.2 m³/(m²·d)
Result
54.2 m³/(m²·d), 3 significant figures
Check
units (m³/d)/m² = m³/(m²·d), the same as m/d; the rate does not depend on the depth.
Common wrong answers
  • 311 m³/(m²·d): divided Q by the circumference πD (that is the weir loading idea, per m of weir) instead of the surface area
  • 217 m³/(m²·d): put the radius into πD²/4, so the area is 4 times too small and the rate 4 times too large
  • 13.5 m³/(m²·d): used πD² instead of πD²/4, so the area is 4 times too large and the rate 4 times too small

Problem 4 · FE, Time value of money

A contractor plans to sell a crane for $30,000 at the end of year 23. At a MARR of 4% per year, the present worth of that sale is most nearly:

  • A $15,600
  • B $73,900
  • C $12,700
  • D $12,200

Handbook: Engineering Economics, factor formulas and Factor Tables, FE Reference Handbook 10.6

Show the worked solution
Answer
D ($12,200)
Given
F = $30,000, n = 23 years, i = 4% per year
Find
present worth P of the resale ($)
Handbook
Engineering Economics, factor formulas (P/F) and Factor Table (i = 4%), pages 235 and 240
Equation
P = F(P/F, i, n), with (P/F, i, n) = (1 + i)⁻ⁿ
Substitute
  1. (P/F, 4%, 23) = 1/(1.04)^23 = 0.4057 (factor table, i = 4%)
  2. P = F(P/F, 4%, 23) = ($30,000)(0.4057) = $12,200
Result
$12,200, 3 significant figures
Check
P is less than F = $30,000, as a future amount is always worth less today when i > 0.
Why the others are wrong
  • A: discounted with simple interest, F/(1 + ni), instead of compound (1 + i)⁻ⁿ
  • B: used (F/P), which moves money forward in time, instead of (P/F)
  • C: used n - 1 = 22 years instead of 23

Problem 5 · FE, Ethics scenario

While designing a site grading plan next to a neighboring property, Dana, a licensed engineer, reviews the public plans for a retaining wall designed by another firm and finds a calculation error that could affect public safety.

Which statements are correct under the Model Rules of Professional Conduct? Select all that apply.

  • A Dana should post the error online and criticize the other firm to warn the public.
  • B Dana must not try to damage the other engineer's reputation falsely or maliciously.
  • C Dana should make a reasonable effort to inform the engineer who prepared the plans.
  • D Because the error is in another firm's work, Dana has no responsibility to say anything.

Handbook: Ethics and Professional Practice, Model Rules Section 240.15 (Rules of Professional Conduct), FE Reference Handbook 10.6

Show the worked solution
Answer
B, C
Given
the scenario in the post
Find
every statement that is correct under the Model Rules of Professional Conduct (Section 240.15)
Handbook
Ethics and Professional Practice, Model Rules Section 240.15, Rules of Professional Conduct, pages 4 and 5
Equation
none (conceptual item); each statement is tested against one rule of Section 240.15
Substitute
  1. B: correct. No malicious or false injury to another licensee's reputation or practice. (Model Rules 240.15 C.3, handbook page 5)
  2. C: correct. A licensee informs another licensee whose work is believed to hold a material error affecting public safety, unless the law prohibits it. (Model Rules 240.15 C.4, handbook page 5)
Result
B and C are correct (2 of 4 statements)
Check
every correct statement matches what a rule requires, and every wrong one breaks a rule.
Why the others are wrong
  • A: Indiscriminate public criticism takes the place of informing the engineer. (Model Rules 240.15 C.3, handbook page 5)
  • D: A reasonable effort to inform the other licensee is expected. (Model Rules 240.15 C.4, handbook page 5)

More practice by topic

Each topic page has ten problems with full solutions: FE Pipe Flow Practice Problems (Darcy-Weisbach), FE Truss Practice Problems (Method of Joints), FE Clarifier Practice Problems: Overflow Rate and Detention Time, FE Engineering Economics Practice Problems: Time Value of Money, FE Ethics Practice Problems (Select All That Apply). A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

Are these real FE exam questions?

No. They are original problems written to match the topics in the NCEES exam specifications. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee signs.

Do I need the FE Reference Handbook to solve them?

Yes. Download it free in MyNCEES and keep it open as a PDF, the way you will use it on exam day. Each solution gives the page so you can practice finding it.

How do the multiple-correct questions work?

Select every statement that is correct. The FE scores these items right or wrong, with no partial credit.

Sources

  1. NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
  3. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.