FE Mechanical Exam Prep: 30 Free Practice Problems and a 10-Week Plan
Start your FE Mechanical exam prep with Problem 1 below.
Practice problems
Original, unofficial practice; these are not NCEES questions or a full-length exam.
Math, statistics, ethics, and economics
Problem 1 · Mathematics
Use one iteration of Newton's method on f(x) = x² − 5, starting from x₀ = 2. What is x₁?
- A. 2.236
- B. 2.25
- C. 2.50
- D. 1.75
Show answer and explanation
Answer: B. f(2) = 2² − 5 = −1, and f′(x) = 2x, so f′(2) = 4. Newton's update is x₁ = x₀ − f(x₀)/f′(x₀) = 2 − (−1)/4 = 2.25.
Common misses: A is the true root, √5 ≈ 2.236. One iteration gets you close, not all the way there. C uses f′(x) = x instead of 2x. D adds f/f′ instead of subtracting it.
FE Mechanical area 1, Mathematics: Numerical methods. Handbook search to try: "Newton's method". Principle: OpenStax Calculus Vol. 1, 4.9 Newton's Method (Eq. 4.8).
The 30 problems sample all 14 subjects on the NCEES FE Mechanical exam, and each one has a worked explanation. Then use the subject map and the 10-week plan to decide what to study next.
Work them with an approved calculator and the FE Reference Handbook.
Problem 2 · Probability and Statistics
Four test points (x, y) are (1, 2), (2, 3), (3, 5), and (4, 6). What is the slope b₁ of the least-squares line y = b₀ + b₁x?
- A. 1.33
- B. 0.50
- C. 1.40
- D. 0.71
Show answer and explanation
Answer: C. The means are x̄ = 2.5 and ȳ = 4. Then Σ(x − x̄)(y − ȳ) = 3 + 0.5 + 0.5 + 3 = 7 and Σ(x − x̄)² = 2.25 + 0.25 + 0.25 + 2.25 = 5. The slope is b₁ = 7 ÷ 5 = 1.40.
Common misses: A is the slope through only the first and last points, (6 − 2) ÷ 3. Least squares uses every point. B is the intercept, b₀ = 4 − 1.40(2.5) = 0.50. D flips the ratio (5 ÷ 7).
FE Mechanical area 2, Probability and Statistics: Regression and curve fitting. Handbook search to try: "least squares". Principle: NIST/SEMATECH e-Handbook, 4.4.3.1 Least Squares.
Problem 3 · Ethics and Professional Practice
Select all that apply. Which statements about U.S. intellectual property protection are correct?
- A. A patent gives its owner the right to exclude others for a limited time in exchange for publicly disclosing the invention.
- B. Trade secret protection depends on reasonable efforts to keep the information secret.
- C. A U.S. utility patent generally expires 20 years from the first nonprovisional application's filing date, subject to adjustments or extensions.
- D. Trade secret protection automatically expires after 20 years.
Show answer and explanation
Answers: A, B, C. A, B, and C are all correct. A patent is a limited-time right to exclude, granted in exchange for full public disclosure. A trade secret stays protected only while the owner takes reasonable steps to keep it secret. Utility patents generally expire 20 years from the first nonprovisional application's filing date, subject to adjustments or extensions. You need all three, and not D, to get this item right.
Common misses: D: trade secrets have no fixed expiration. Protection lasts as long as the information stays secret and the other conditions hold. That's a tradeoff against a patent: no fixed time limit, but continuing trade secret protection depends on secrecy.
FE Mechanical area 3, Ethics and Professional Practice: Intellectual property. Handbook search to try: "intellectual property". Principle: USPTO, Introduction to Intellectual Property (PDF pp. 10, 25: patent rights and disclosure); USPTO, Patent Essentials; USPTO, Trade secret policy.
Problem 4 · Engineering Economics
You invest $1,000 today and receive one payment of $1,210 at the end of year 2. There are no other cash flows. What is the rate of return?
- A. 10.0% per year
- B. 10.5% per year
- C. 21.0% per year
- D. 8.7% per year
Show answer and explanation
Answer: A. The rate of return is the interest rate that makes the present worth of the income equal the cost: 1,000 = 1,210 ÷ (1 + i)². So (1 + i)² = 1.21 and i = 0.10, or 10.0% per year.
Common misses: B treats the $210 gain as simple interest spread over two years (21% ÷ 2). That ignores compounding. C is the total two-year gain, not an annual rate. D divides the $210 gain by $1,210 and then by two years, using the final amount as the base and ignoring compounding.
FE Mechanical area 4, Engineering Economics: Time value of money: rate of return. Handbook search to try: "rate of return". Principle: Penn State EME 460, Rate of Return (ROR) Calculation; Penn State EME 460, Nominal, Period, and Effective Interest Rates (F = P(1 + i)ⁿ).
These four areas appear on every FE exam. For 21 more problems on these shared skills, see our all-discipline FE practice set.
Electricity and magnetism
Problem 5 · Electricity and Magnetism
A single-phase load draws 10 A rms at 120 V rms with a lagging power factor of 0.80. Assume sinusoidal voltage and current. What average (real) power does it use?
- A. 1,200 W
- B. 720 W
- C. 1,500 W
- D. 960 W
Show answer and explanation
Answer: D. Average power is P = V_rms × I_rms × cos φ, and cos φ is the power factor. P = 120 × 10 × 0.80 = 960 W.
Common misses: A is the apparent power, 1,200 VA. It ignores the power factor. B uses sin φ = 0.60 instead of cos φ. C divides by the power factor instead of multiplying. Real power can't be larger than V_rms × I_rms.
FE Mechanical area 5, Electricity and Magnetism: AC circuit analysis. Handbook search to try: "power factor". Principle: OpenStax University Physics Vol. 2, 15.4 Power in an AC Circuit.
Problem 6 · Electricity and Magnetism
An uncharged 100 µF capacitor charges through a 10 kΩ resistor from an ideal 12 V DC source. What is the capacitor voltage 1.0 s after the switch closes?
- A. 12.0 V
- B. 4.41 V
- C. 7.59 V
- D. 6.00 V
Show answer and explanation
Answer: C. The time constant is τ = RC = (10,000 Ω)(100 × 10⁻⁶ F) = 1.0 s. While charging, V = 12(1 − e^(−t/τ)). At t = τ, V = 12(1 − e⁻¹) = 12(0.63212…) = 7.59 V.
Common misses: A assumes the capacitor is fully charged after one time constant. It's about 63% charged. B is 12e⁻¹, the discharging form. D assumes the capacitor is half charged after one time constant.
FE Mechanical area 5, Electricity and Magnetism: DC circuit analysis (RC transient). Handbook search to try: "RC circuit". Principle: OpenStax University Physics Vol. 2, 10.5 RC Circuits (Eqs. 10.8–10.9).
Statics
Problem 7 · Statics
A horizontal beam spans 5.00 m from a pin support at A to a roller support at B. It carries a downward uniform load of 2.00 kN/m along its whole length plus a downward point load of 6.00 kN located 1.50 m from A. Ignore the beam's weight; there are no horizontal loads. What is the vertical reaction at B?
- A. 1.80 kN upward
- B. 6.80 kN upward
- C. 8.00 kN upward
- D. 9.20 kN upward
Show answer and explanation
Answer: B. Replace the uniform load with its resultant: 2.00 × 5.00 = 10.0 kN acting at midspan, 2.50 m from A. Take moments about A so the reaction there drops out: B_y(5.00) − 10.0(2.50) − 6.00(1.50) = 0, so B_y = (25.0 + 9.00) ÷ 5.00 = 6.80 kN. Check with vertical balance: A_y = 16.0 − 6.80 = 9.20 kN, and 9.20 + 6.80 equals the 16.0 kN total load.
Common misses: A includes the point load's moment but leaves out the uniform load. C splits the 16.0 kN total equally, but the off-center point load makes these reactions unequal. D is the reaction at A, not B.
FE Mechanical area 6, Statics: Equilibrium of rigid bodies. Handbook search to try: "equilibrium". Principle: OpenStax University Physics Vol. 1, 12.1 Conditions for Static Equilibrium (Eqs. 12.2, 12.5).
Problem 8 · Statics
A 50 kg crate rests on a level floor. The coefficient of static friction between crate and floor is 0.40. Use g = 9.81 m/s². What horizontal push is needed to start the crate sliding?
- A. 196 N
- B. 491 N
- C. 20 N
- D. 1,226 N
Show answer and explanation
Answer: A. On a level floor with a horizontal push, the normal force equals the weight: N = mg = 50 × 9.81 = 490.5 N. Static friction can resist up to μ_s N = 0.40 × 490.5 = 196 N, so the push has to reach about 196 N.
Common misses: B is the weight (the normal force), not the friction limit. C multiplies μ by the mass and forgets g. D divides the normal force by the friction coefficient instead of multiplying.
FE Mechanical area 6, Statics: Static friction. Handbook search to try: "friction". Principle: OpenStax University Physics Vol. 1, 6.2 Friction (Eq. 6.1).
Problem 9 · Statics
A solid rectangular section is 100 mm wide and 200 mm tall. What is its area moment of inertia about the horizontal axis through its centroid?
- A. 2.67 × 10⁸ mm⁴
- B. 6.67 × 10⁷ mm⁴
- C. 1.67 × 10⁷ mm⁴
- D. 1.33 × 10⁸ mm⁴
Show answer and explanation
Answer: B. For a rectangle about its centroidal axis, I = bh³/12, where h is the dimension perpendicular to that axis (the 200 mm height). I = 100 × 200³ ÷ 12 = 6.67 × 10⁷ mm⁴.
Common misses: A is bh³/3, the moment of inertia about the base, not the centroid. C swaps b and h (100 mm cubed instead of 200 mm). D uses bh³/6.
FE Mechanical area 6, Statics: Centroids and moments of inertia. Handbook search to try: "moment of inertia". Principle: Roylance, Stresses in Beams (MIT 3.11), Eq. 5.
Dynamics, kinematics, and vibrations
Problem 10 · Dynamics, Kinematics, and Vibrations
A vehicle traveling at 25 m/s brakes with a constant deceleration of 5.0 m/s². How far does it travel before it stops? Enter meters to one decimal place.
Enter a number in m.
Show answer and explanation
Answer: 62.5 m. Use v² = v₀² + 2a(x − x₀) with a = −5.0 m/s² and final speed 0: 0 = 25² − 2(5.0)d, so d = 625 ÷ 10 = 62.5 m.
Common misses: 125 m drops the 2. 187.5 m uses the 5 s stopping time but inserts +5.0 m/s² in d = v₀t + ½at². 5 is the stopping time in seconds (25 ÷ 5.0), not a distance.
FE Mechanical area 7, Dynamics, Kinematics, and Vibrations: Kinematics of particles. Handbook search to try: "constant acceleration". Principle: OpenStax University Physics Vol. 1, 3.4 Motion with Constant Acceleration (Eq. 3.14).
Problem 11 · Dynamics, Kinematics, and Vibrations
A 2.0 kg block on a frictionless horizontal surface compresses a spring (k = 800 N/m) by 0.10 m and is released from rest. Assume the spring is ideal and massless, with its other end fixed; the block is not attached to it. What is its speed when it leaves the spring?
- A. 2.0 m/s
- B. 1.41 m/s
- C. 6.32 m/s
- D. 40 m/s
Show answer and explanation
Answer: A. All the spring energy becomes kinetic energy: ½kx² = ½mv². So v = x√(k/m) = 0.10 × √(800 ÷ 2.0) = 0.10 × 20 = 2.0 m/s.
Common misses: B keeps the ½ on only one side of the equation. C uses √(kx/m), mixing up spring force and spring energy. D skips the square root and doesn't square the displacement.
FE Mechanical area 7, Dynamics, Kinematics, and Vibrations: Work-energy of particles. Handbook search to try: "spring energy". Principle: OpenStax University Physics Vol. 1, 15.2 Energy in Simple Harmonic Motion (Eq. 15.12).
Problem 12 · Dynamics, Kinematics, and Vibrations
A 4.0 kg mass on a spring with k = 1,600 N/m vibrates freely with no damping. Treat the spring as ideal and massless, with its other end fixed. What is its natural frequency?
- A. 20 Hz
- B. 0.314 Hz
- C. 400 Hz
- D. 3.18 Hz
Show answer and explanation
Answer: D. The natural angular frequency is ωₙ = √(k/m) = √(1,600 ÷ 4.0) = 20 rad/s. In hertz, f = ωₙ ÷ 2π = 20 ÷ 6.283 = 3.18 Hz.
Common misses: A reports 20 rad/s as if it were hertz. Watch the units on ω. B is the period in seconds (2π ÷ 20), not a frequency. C is k/m without the square root.
FE Mechanical area 7, Dynamics, Kinematics, and Vibrations: Free and forced vibrations. Handbook search to try: "natural frequency". Principle: OpenStax University Physics Vol. 1, 15.1 Simple Harmonic Motion (Eqs. 15.9, 15.11).
Mechanics of materials
Problem 13 · Mechanics of Materials
A steel rod 2.0 m long with a cross-sectional area of 200 mm² carries an axial tensile load of 20 kN. Its modulus of elasticity is 200 GPa. Assume linear elastic behavior. What is the rod's elongation?
- A. 0.10 mm
- B. 1.0 mm
- C. 10 mm
- D. 0.50 mm
Show answer and explanation
Answer: B. δ = PL/(AE). Work in N and mm, using 200 GPa = 200,000 N/mm²: δ = (20,000 N)(2,000 mm) ÷ [(200 mm²)(200,000 N/mm²)] = 1.0 mm.
Common misses: A and C are off by a factor of 10, the usual sign of a unit slip (GPa to N/mm², or m to mm). D doubles the area.
FE Mechanical area 8, Mechanics of Materials: Stress and strain caused by axial loads. Handbook search to try: "axial deformation". Principle: OpenStax University Physics Vol. 1, 12.3 Stress, Strain, and Elastic Modulus (Eq. 12.36).
Problem 14 · Mechanics of Materials
A solid circular steel shaft 40 mm in diameter carries a torque of 500 N·m. Assume linear elastic behavior. What is the maximum shear stress in the shaft?
- A. 39.8 MPa
- B. 79.6 MPa
- C. 318 MPa
- D. 4.97 MPa
Show answer and explanation
Answer: A. The maximum shear stress is at the surface: τ = Tr/J, where J = πr⁴/2 for a solid shaft. With r = 0.020 m, J = 2.513 × 10⁻⁷ m⁴ and τ = (500)(0.020) ÷ (2.513 × 10⁻⁷) = 39.8 × 10⁶ Pa = 39.8 MPa. The shortcut τ = 16T/(πd³) gives the same number.
Common misses: B uses the area moment I = πd⁴/64 in place of the polar moment J (or uses the diameter as the outer distance c). C puts the radius where 16T/(πd³) needs the diameter. D uses 2T/(πd³), the radius formula with the diameter plugged in.
FE Mechanical area 8, Mechanics of Materials: Stress and strain caused by torsional loads. Handbook search to try: "torsion". Principle: Roylance, Shear and Torsion (MIT 3.11), Eqs. 12 and 14.
Problem 15 · Mechanics of Materials
A plane-stress element has σx = 80 MPa and σy = 20 MPa (both tension) and τxy = 40 MPa. What is the maximum in-plane shear stress?
- A. 40 MPa
- B. 30 MPa
- C. 50 MPa
- D. 100 MPa
Show answer and explanation
Answer: C. On Mohr's circle the center is (σx + σy)/2 = 50 MPa and the radius is √[((σx − σy)/2)² + τxy²] = √(30² + 40²) = 50 MPa. The maximum in-plane shear stress equals the radius: 50 MPa. (The principal stresses are 50 ± 50, or 100 MPa and 0.)
Common misses: A is just the applied τxy; rotating the element finds a bigger shear. B is (σx − σy)/2 by itself, one leg of the triangle. D is the largest principal stress, center plus radius, which is a normal stress.
FE Mechanical area 8, Mechanics of Materials: Stress transformations and Mohr's circle. Handbook search to try: "Mohr's circle". Principle: Roylance, Transformation of Stresses and Strains (MIT 3.11), principal stresses and Eq. 12.
Material properties and processing
Problem 16 · Material Properties and Processing
On a stress-strain diagram for a metal, what does the slope of the initial straight-line portion represent?
- A. Yield strength
- B. Ductility
- C. Modulus of elasticity
- D. Ultimate tensile strength
Show answer and explanation
Answer: C. In the linear elastic range, stress = (modulus of elasticity) × strain. So the slope of that straight segment is the modulus of elasticity, also called Young's modulus.
Common misses: A and D are stress values, points read off the curve, not a slope. B describes how much a material stretches plastically before it breaks.
FE Mechanical area 9, Material Properties and Processing: Stress-strain diagrams. Handbook search to try: "stress-strain". Principle: OpenStax University Physics Vol. 1, 12.3 Stress, Strain, and Elastic Modulus (Eq. 12.33); OpenStax University Physics Vol. 1, 12.4 Elasticity and Plasticity (Fig. 12.25).
Problem 17 · Material Properties and Processing
A galvanized (zinc-coated) steel bracket is scratched through to bare steel and left outdoors in wet weather. What happens at the scratch?
- A. The steel corrodes first and protects the zinc.
- B. The zinc corrodes first and protects the exposed steel.
- C. Neither metal corrodes, because the two metals are in contact.
- D. Both corrode at the same rate as unprotected steel.
Show answer and explanation
Answer: B. Zinc oxidizes more readily than iron. In the wet zinc–steel couple, zinc acts as the anode and is consumed, while the exposed steel acts as the cathode and is protected. That's why a scratched galvanized part doesn't rust right away.
Common misses: A reverses the anode and cathode. C ignores the galvanic couple that the contact creates. D treats the zinc as only a paint-like barrier. At the scratch it's the sacrificial protection that matters.
FE Mechanical area 9, Material Properties and Processing: Corrosion mechanisms and control. Handbook search to try: "corrosion". Principle: OpenStax Chemistry 2e, 17.6 Corrosion.
Fluid mechanics
Problem 18 · Fluid Mechanics
What is the gauge pressure 10 m below the free surface of a water tank that is open to the atmosphere? Use ρ = 1,000 kg/m³ and g = 9.81 m/s².
- A. 199 kPa
- B. 9.81 kPa
- C. 98.1 kPa
- D. 981 kPa
Show answer and explanation
Answer: C. Gauge pressure is pressure above atmospheric: p = ρgh = 1,000 × 9.81 × 10 = 98,100 Pa = 98.1 kPa.
Common misses: A adds standard atmospheric pressure (about 101.3 kPa), so it's the absolute pressure. Read the question for "gauge" or "absolute." B and D are off by a factor of 10.
FE Mechanical area 10, Fluid Mechanics: Fluid statics. Handbook search to try: "hydrostatic pressure". Principle: OpenStax University Physics Vol. 1, 14.1 Fluids, Density, and Pressure (Eq. 14.4); OpenStax University Physics Vol. 1, 14.2 Measuring Pressure (Eq. 14.11).
Problem 19 · Fluid Mechanics
Water flows steadily through a horizontal pipe that narrows from 100 mm to 50 mm diameter. The velocity in the larger section is 2.0 m/s. Ignoring losses, what is the pressure drop p₁ − p₂? Use ρ = 1,000 kg/m³.
- A. 30 kPa
- B. 6.0 kPa
- C. 60 kPa
- D. −30 kPa
Show answer and explanation
Answer: A. First, continuity: area goes with diameter squared, so V₂ = V₁(D₁/D₂)² = 2.0 × 4 = 8.0 m/s. Then Bernoulli at the same elevation: p₁ − p₂ = ½ρ(V₂² − V₁²) = 0.5 × 1,000 × (64 − 4) = 30,000 Pa = 30 kPa.
Common misses: B scales the velocity by the diameter ratio (×2) instead of the area ratio (×4). C leaves out the ½. D has the sign backward: pressure drops where the flow speeds up.
FE Mechanical area 10, Fluid Mechanics: Energy, impulse, and momentum. Handbook search to try: "Bernoulli". Principle: OpenStax University Physics Vol. 1, 14.5 Fluid Dynamics (continuity, Eq. 14.14); OpenStax University Physics Vol. 1, 14.6 Bernoulli's Equation.
Problem 20 · Fluid Mechanics
Water (ρ = 1,000 kg/m³, dynamic viscosity μ = 1.0 × 10⁻³ Pa·s) flows at a mean speed of 0.50 m/s in a tube with a 20 mm inside diameter. What are the Reynolds number and flow regime?
- A. 10,000; turbulent
- B. 5,000; turbulent
- C. 10; laminar
- D. 1.0 × 10⁷; turbulent
Show answer and explanation
Answer: A. Re = ρVD/μ = (1,000)(0.50)(0.020) ÷ (1.0 × 10⁻³) = 10,000. That's far above the laminar range (below roughly 2,000), so the flow is turbulent.
Common misses: B uses the radius instead of the diameter. C leaves the density out. D plugs in 1.0 × 10⁻⁶, which is roughly water's kinematic viscosity in m²/s, a different property, in place of the dynamic viscosity μ.
FE Mechanical area 10, Fluid Mechanics: Internal flow. Handbook search to try: "Reynolds number". Principle: OpenStax University Physics Vol. 1, 14.7 Viscosity and Turbulence (Eq. 14.20).
Thermodynamics
Problem 21 · Thermodynamics
A reversible (Carnot) heat engine runs between reservoirs at 327°C and 27°C. What is its thermal efficiency?
- A. 91.7%
- B. 50.0%
- C. 8.3%
- D. 200%
Show answer and explanation
Answer: B. Temperature ratios need absolute temperatures. T_H = 327 + 273.15 = 600.15 K and T_C = 27 + 273.15 = 300.15 K. η = 1 − T_C/T_H = 1 − 300.15/600.15 ≈ 0.499875, or 50.0%.
Common misses: A plugs in Celsius: 1 − 27/327. C is 27/327, also Celsius. D is T_H/T_C, and an efficiency can't exceed 100%.
FE Mechanical area 11, Thermodynamics: Power cycles. Handbook search to try: "Carnot". Principle: OpenStax University Physics Vol. 2, 4.5 The Carnot Cycle (Eq. 4.5); OpenStax University Physics Vol. 2, 1.2 Thermometers and Temperature Scales (Table 1.1).
Problem 22 · Thermodynamics
A refrigerator removes heat from its cold space at 5.0 kW while using 1.25 kW of input power. What is its coefficient of performance (COP)?
- A. 5.0
- B. 0.25
- C. 6.25
- D. 4.0
Show answer and explanation
Answer: D. For a refrigerator, the useful effect is the heat removed: COP_R = Q̇_L/Ẇ = 5.0 ÷ 1.25 = 4.0.
Common misses: A is the heat-pump COP, Q̇_H/Ẇ = 6.25 ÷ 1.25. Same machine, different goal. B flips the ratio. C is the heat rejected to the surroundings, 5.0 + 1.25 = 6.25 kW, which is a rate, not a COP.
FE Mechanical area 11, Thermodynamics: Refrigeration and heat pump cycles. Handbook search to try: "coefficient of performance". Principle: OpenStax University Physics Vol. 2, 4.3 Refrigerators and Heat Pumps (Eqs. 4.3, 4.4).
Problem 23 · Thermodynamics
A turbine runs steadily with one inlet and one outlet. Mass flow is 0.800 kg/s. The fluid enters with a specific enthalpy of 780 kJ/kg and leaves at 530 kJ/kg. The turbine loses heat to the surroundings at 12.0 kW. Kinetic and potential energy changes are negligible. What shaft power does the turbine deliver? Enter kW to the nearest whole number.
Enter a number in kW.
Show answer and explanation
Answer: 188 kW. Steady-flow energy balance with heat in and work out as positive: Q̇ − Ẇ = ṁ(h_out − h_in). Heat leaves, so Q̇ = −12.0 kW: −12.0 − Ẇ = 0.800(530 − 780) = −200 kW, which gives Ẇ = 188 kW. In words: the fluid gives up 200 kW, 12 kW escapes as heat, and 188 kW comes out on the shaft.
Common misses: 200 ignores the heat loss. 212 adds the heat loss instead of subtracting it. 235 is the specific work in kJ/kg (250 − 12/0.800), not power; it was never multiplied by the mass flow.
FE Mechanical area 11, Thermodynamics: Performance of components. Handbook search to try: "steady flow energy". Principle: MIT Unified Engineering Thermodynamics, Chapter 6: steady flow energy equation; turbine.
Heat transfer
Problem 24 · Heat Transfer
A plane wall 0.20 m thick (k = 0.80 W/m·K) has an area of 10 m². Its two surfaces are held 25 K apart at steady state. Assume one-dimensional conduction with no internal heat generation. What is the conduction heat rate through the wall? Enter watts to the nearest whole number.
Enter a number in W.
Show answer and explanation
Answer: 1,000 W. Steady one-dimensional conduction: q = kAΔT/L = (0.80)(10)(25) ÷ 0.20 = 1,000 W.
Common misses: 40 W multiplies by the thickness instead of dividing. 100 W leaves out the area; that's the heat flux in W/m², not the total rate.
FE Mechanical area 12, Heat Transfer: Conduction. Handbook search to try: "conduction". Principle: OpenStax University Physics Vol. 2, 1.6 Mechanisms of Heat Transfer (Eq. 1.9).
Problem 25 · Heat Transfer
A 1.0 m² surface with emissivity 0.90 is at 127°C inside large surroundings at 27°C. What is its net radiation heat loss? Use σ = 5.67 × 10⁻⁸ W/m²·K⁴.
- A. 13.2 W
- B. 994 W
- C. 894 W
- D. 5.10 W
Show answer and explanation
Answer: C. Radiation needs absolute temperatures: 400.15 K and 300.15 K. q = εσA(T_s⁴ − T_sur⁴) = 0.90 × 5.67 × 10⁻⁸ × 1.0 × (400.15⁴ − 300.15⁴) ≈ 894 W.
Common misses: A uses Celsius. B leaves out the emissivity, treating the surface as a perfect blackbody. D raises the temperature difference (100 K) to the fourth power instead of subtracting the two fourth powers.
FE Mechanical area 12, Heat Transfer: Radiation. Handbook search to try: "radiation". Principle: OpenStax University Physics Vol. 2, 1.6 Mechanisms of Heat Transfer (net radiation); OpenStax University Physics Vol. 2, 1.2 Thermometers and Temperature Scales (Table 1.1).
Measurements, instrumentation, and controls
Problem 26 · Measurements, Instrumentation, and Controls
Power is calculated as P = VI from independent measurements V = 12.0 ± 0.1 V and I = 2.00 ± 0.02 A. The ± values are standard uncertainties. Using first-order propagation in quadrature (root-sum-square), what is the combined standard uncertainty in P?
- A. ±0.44 W
- B. ±0.013 W
- C. ±0.12 W
- D. ±0.31 W
Show answer and explanation
Answer: D. For a product of independent measurements, first-order propagation combines the relative standard uncertainties in quadrature: √[(0.1/12.0)² + (0.02/2.00)²] = √(0.0000694 + 0.000100) = 0.0130. P = 24.0 W, so the combined standard uncertainty is 24.0 × 0.0130 ≈ 0.31 W, reported as P ≈ 24.0 ± 0.31 W.
Common misses: A adds the relative uncertainties directly instead of using the root-sum-square combination the question asks for. B is the relative uncertainty (1.3%), never converted back to watts. C adds 0.1 and 0.02 directly even though the input uncertainties have different units.
FE Mechanical area 13, Measurements, Instrumentation, and Controls: Measurement uncertainty. Handbook search to try: "propagation of error". Principle: NIST/SEMATECH e-Handbook, 2.5.5.2 Formulas for functions of two variables; NIST, Combining uncertainty components (first-order propagation and product rule).
Problem 27 · Measurements, Instrumentation, and Controls
A negative-feedback loop has a forward-path gain G = 10 and a feedback-path gain H = 0.40. Both are constant gains. What is the closed-loop gain C/R?
- A. 2.0
- B. 4.0
- C. 10
- D. −3.3
Show answer and explanation
Answer: A. For negative feedback, C/R = G/(1 + GH) = 10 ÷ (1 + 10 × 0.40) = 10 ÷ 5 = 2.0.
Common misses: B is the loop gain GH. C is the forward gain alone, as if the loop were open. D uses the positive-feedback form, G/(1 − GH).
FE Mechanical area 13, Measurements, Instrumentation, and Controls: Control systems (feedback, block diagrams). Handbook search to try: "feedback". Principle: MIT OpenCourseWare 2.004 Systems, Modeling, and Control II, Lecture 11 (negative feedback).
Mechanical design and analysis
Problem 28 · Mechanical Design and Analysis
A point on the surface of a ductile steel shaft is in pure torsional shear with τ = 100 MPa. The yield strength is S_y = 350 MPa. Using the distortion-energy (von Mises) theory, what is the factor of safety against yielding?
- A. 1.75
- B. 2.02
- C. 3.50
- D. 1.17
Show answer and explanation
Answer: B. In pure shear, the von Mises stress is σ′ = √3 τ = 1.732 × 100 = 173.2 MPa. The factor of safety is n = S_y/σ′ = 350 ÷ 173.2 = 2.02.
Common misses: A is the maximum-shear-stress (Tresca) result, S_y/(2τ). It's more conservative, but it's not the theory the question names. C compares τ directly to S_y as if shear and normal stress were interchangeable. D divides by 3τ.
FE Mechanical area 14, Mechanical Design and Analysis: Failure theories and analysis. Handbook search to try: "von Mises". Principle: Stanford ME111, Class 21: Yield Criteria for Ductile Materials (pure shear).
Problem 29 · Mechanical Design and Analysis
A thin-walled cylindrical tank has an inside radius of 500 mm, a wall thickness of 10 mm, and an internal gauge pressure of 2.0 MPa. What is the hoop (circumferential) stress in the wall?
- A. 50 MPa
- B. 200 MPa
- C. 100 MPa
- D. 0.04 MPa
Show answer and explanation
Answer: C. For a thin-walled cylinder, σ_hoop = pr/t = (2.0)(500) ÷ 10 = 100 MPa. Quick check: with r/t = 50, the wall stress should be much larger than the pressure, and it is.
Common misses: A is the axial (longitudinal) stress, pr/(2t), which is half the hoop stress. B uses the diameter in place of the radius. D flips the ratio to pt/r, giving a stress far smaller than the pressure itself.
FE Mechanical area 14, Mechanical Design and Analysis: Pressure vessels and piping. Handbook search to try: "thin-walled pressure vessel". Principle: Roylance, Pressure Vessels (MIT 3.11), Eqs. 2 and 3.
Problem 30 · Mechanical Design and Analysis
A gearbox receives 10.0 kW at its input shaft. Its mechanical efficiency is 90.0% (output power ÷ input power). The output shaft turns steadily at 600 rev/min. What torque does the output shaft carry?
- A. 15.0 N·m
- B. 143 N·m
- C. 159 N·m
- D. 177 N·m
Show answer and explanation
Answer: B. Output power is 0.900 × 10.0 kW = 9,000 W. Convert speed to rad/s: 600 rev/min × 2π rad/rev ÷ 60 s/min = 20π ≈ 62.83 rad/s. Since P = Tω, T = 9,000 ÷ 62.83 = 143 N·m.
Common misses: A divides power by 600 as if rev/min were rad/s. C uses the input power and skips the efficiency. D divides by the efficiency, which would make output power bigger than input.
FE Mechanical area 14, Mechanical Design and Analysis: Power transmission. Handbook search to try: "power transmission". Principle: OpenStax University Physics Vol. 1, 10.8 Work and Power for Rotational Motion (Eq. 10.31).
What your results mean
A count of right answers tells you how you did on these 30 problems. That's all it tells you. NCEES converts the number of scored questions you answer correctly into a scaled score and compares it with a minimum standard that it doesn't publish (NCEES exam scoring). So "22 of 30" can't be turned into "ready" or "not ready," and one to three problems per subject is far too few to grade a subject.
Use the results to order your study instead:
- Missed a problem? Write down why in your error log (see Turn misses into your next session), then move that subject earlier in the plan.
- Right, but it took a while or you guessed? Count it as a miss for planning. A slow or shaky right answer still needs work.
- Right and quick? Good sign for that skill, but each subject covers far more than one problem shows. Keep it on the schedule.
Answer key
| Problem | Knowledge area | Answer |
|---|---|---|
| 1 | Mathematics | B |
| 2 | Probability and Statistics | C |
| 3 | Ethics and Professional Practice | A, B, C |
| 4 | Engineering Economics | A |
| 5 | Electricity and Magnetism | D |
| 6 | Electricity and Magnetism | C |
| 7 | Statics | B |
| 8 | Statics | A |
| 9 | Statics | B |
| 10 | Dynamics, Kinematics, and Vibrations | 62.5 m |
| 11 | Dynamics, Kinematics, and Vibrations | A |
| 12 | Dynamics, Kinematics, and Vibrations | D |
| 13 | Mechanics of Materials | B |
| 14 | Mechanics of Materials | A |
| 15 | Mechanics of Materials | C |
| 16 | Material Properties and Processing | C |
| 17 | Material Properties and Processing | B |
| 18 | Fluid Mechanics | C |
| 19 | Fluid Mechanics | A |
| 20 | Fluid Mechanics | A |
| 21 | Thermodynamics | B |
| 22 | Thermodynamics | D |
| 23 | Thermodynamics | 188 kW |
| 24 | Heat Transfer | 1,000 W |
| 25 | Heat Transfer | C |
| 26 | Measurements, Instrumentation, and Controls | D |
| 27 | Measurements, Instrumentation, and Controls | A |
| 28 | Mechanical Design and Analysis | B |
| 29 | Mechanical Design and Analysis | C |
| 30 | Mechanical Design and Analysis | B |
What's on the FE Mechanical exam
The FE Mechanical exam has 110 questions across 14 knowledge areas. The ranges below are NCEES's published question counts from the FE Mechanical specification (PDF), effective beginning with the July 2020 exams. The last column shows where each area falls in our 10-week plan.
NCEES uses both SI and U.S. Customary units. This starter set's engineering calculations use SI/metric quantities, so include U.S. Customary-unit problems in your broader practice too.
| No. | Knowledge area | Questions | Plan week |
|---|---|---|---|
| 1 | Mathematics | 6–9 | 2 |
| 2 | Probability and Statistics | 4–6 | 2 |
| 3 | Ethics and Professional Practice | 4–6 | 8 |
| 4 | Engineering Economics | 4–6 | 8 |
| 5 | Electricity and Magnetism | 5–8 | 3 |
| 6 | Statics | 9–14 | 2 |
| 7 | Dynamics, Kinematics, and Vibrations | 10–15 | 3 |
| 8 | Mechanics of Materials | 9–14 | 4 |
| 9 | Material Properties and Processing | 7–11 | 4 |
| 10 | Fluid Mechanics | 10–15 | 5 |
| 11 | Thermodynamics | 10–15 | 6 |
| 12 | Heat Transfer | 7–11 | 7 |
| 13 | Measurements, Instrumentation, and Controls | 5–8 | 7 |
| 14 | Mechanical Design and Analysis | 10–15 | 8 |
The ranges tell you how much of the exam a subject takes up. They don't tell you how hard it is, or how much work you need on it. That comes from your own misses. The specification also lists the subtopics under each area (Mechanical Design and Analysis alone has 13, from springs and bearings to GD&T), so open the PDF when you pick practice problems.
Six areas carry the most questions: Dynamics, Kinematics, and Vibrations; Fluid Mechanics; Thermodynamics; and Mechanical Design and Analysis (10–15 each), plus Statics and Mechanics of Materials (9–14 each). Together, their published range endpoints add up to 58–88. These are endpoint sums, not a possible combined question-count range on a 110-question exam.
Should you skip a small area? We wouldn't. Your score counts correct answers, and even the smallest areas hold 4–8 questions each. Spend less time there, not zero.
Your 10-week FE Mechanical study plan
Assumes about 8 hours a week (80 hours in total) for someone who has finished, or nearly finished, a mechanical engineering degree. That's our planning estimate, not an NCEES requirement or a promise that 80 hours is enough. If a week runs long, give it more time instead of skipping what you haven't learned.
| Week | Focus (published question ranges) | How to spend about 8 hours | Done when |
|---|---|---|---|
| 1 | Set up and sample | Work all 30 problems above with your calculator and handbook (about 2 h). Read every explanation and log each miss (3 h). Download the handbook, check your calculator model, and do the five-minute handbook drill (1 h). Skim all 14 areas in the specification and mark the subtopics that look unfamiliar (2 h). | Your error log is started, your weak areas are ranked, and weeks 2–10 are on your calendar. |
| 2 | Statics (9–14), Mathematics (6–9), Probability and Statistics (4–6). Together: 19–29 | Statics 4 h, math 2 h, statistics 1 h, corrections 1 h. | Every subtopic in these areas has been practiced, and misses are logged. |
| 3 | Dynamics, Kinematics, and Vibrations (10–15), Electricity and Magnetism (5–8). Together: 15–23 | Dynamics and vibrations 5 h, electricity and magnetism 2 h, redo week 2 misses 1 h. | Week 2 misses retried without looking at the old work. |
| 4 | Mechanics of Materials (9–14), Material Properties and Processing (7–11). Together: 16–25 | Mechanics of materials 4 h, materials 3 h, mixed set of 10 problems from weeks 2–3 1 h. | Earlier areas still warm. |
| 5 | Fluid Mechanics (10–15) | Fluids 6 h, mixed review 1 h, corrections 1 h. | You can tell gauge from absolute pressure and flow rate from velocity without thinking. |
| 6 | Thermodynamics (10–15) | Thermo 6 h, including table and chart lookups in the handbook (steam tables, psychrometric chart), mixed review 1 h, corrections 1 h. | You can find and read property tables quickly. |
| 7 | Heat Transfer (7–11), Measurements, Instrumentation, and Controls (5–8). Together: 12–19 | Heat transfer 4 h, measurements and controls 3 h, mixed set of 15 problems from weeks 2–6 1 h. | Heat-transfer and controls misses are logged, and the mixed set is corrected. |
| 8 | Mechanical Design and Analysis (10–15), Ethics (4–6), Engineering Economics (4–6). Together: 18–27 | Mechanical design 5 h, economics 1.5 h, ethics 1 h, corrections 0.5 h. | All 14 areas visited. Check every area against your log. |
| 9 | Repair and pacing | Your two biggest gaps 4 h. Two pacing blocks of 30 mixed problems, 90 minutes each, with the handbook PDF and your calculator (3 h). Review misses and slow problems 1 h. | You know your real pace and what slows you down. |
| 10 | Taper and logistics | Light mixed practice 3 h. Redo old misses without solutions 3 h. Light handbook and calculator review 1 h. Check your ID, calculator, appointment time, and route 1 h. No new topics in the last few days. | Logistics checked; you walk in rested. |
Adding the lower and upper endpoints in weeks 2–8 gives 100 and 153, respectively—the same endpoint sums calculated from the specification, not the exam's length. The way we've grouped the areas is our planning aid; it is our study sequence, not NCEES's subject order or the question order on the exam.
Where to find more problems: your old coursework and textbooks, the free teaching sources linked under each problem above, and our all-discipline FE practice set for the shared subjects. Use problems you're allowed to use; never use real or recalled exam questions.
A repeatable study session
For a 60-minute session, try this: 10 minutes redoing an old miss without peeking, 35 minutes on new problems, and 15 minutes checking work and logging what went wrong. If you can't set up a problem at all, shift time toward reading the concept before you practice more.
Before you close out a problem, say three things to yourself: what told you which model to use, what assumptions made it work, and what answer would have been obviously wrong. A right answer you can't explain is a guess that happened to land.
If your situation is different
| Your situation | How to adjust |
|---|---|
| About 4 hours a week | Stretch each week across two. Same work, 20 weeks. |
| About 12 hours a week | Fit weeks 2–8 into five weeks, but only if week 1 shows few big gaps. Faster calendars don't shrink what you need to learn. |
| Several years out of school | Add 2–3 weeks before week 2 to rebuild math, statics, and thermo basics. Ten weeks may be too short; that's fine. |
| Your appointment is close | Do week 1, then spend your remaining time on your biggest logged gaps and mixed practice. If your plan changes, check the rescheduling rules before you move the appointment. |
| Retaking after a fail | NCEES sends a diagnostic report showing your relative performance by knowledge area (NCEES exam scoring). Put your weakest areas first, but still cover all 14. The report doesn't count how many points you missed by. |
Pacing math
The exam gives you 5 hours 20 minutes (320 minutes) for 110 questions. That's about 2 minutes 55 seconds per question on average (320 ÷ 110 ≈ 2.91 minutes). Some questions take a minute; some calculations take four. Practice moving on from a stuck problem, flagging it, and coming back. Do that before you submit the first section, because once you submit it you can't return to those questions.
Use the handbook you'll get on exam day
On exam day, NCEES gives you the current FE Reference Handbook on screen as a searchable PDF. You search it with a search box on the left side of the screen; Ctrl+F doesn't work in the exam software (NCEES Examinee Guide, May 2026, p. 10). You can't bring your own copy or notes.
Download the same handbook free through your MyNCEES account. You may print it for personal study, but not share or post it without written permission from NCEES (NCEES help: exam reference handbooks). Use the version MyNCEES shows for your exam, since NCEES revises it from time to time.
The handbook gives you formulas and tables. It doesn't tell you which one fits a problem. Use this routine on every practice problem:
- Name what's asked for and its units before you search.
- Name the governing idea: equilibrium, energy balance, continuity, torsion, time value of money.
- Search a short concept term, not the problem's wording. Each problem above suggests a term to try. These are starting points we chose, not exact handbook headings.
- Read the variable definitions and conditions before you plug in. Is it gauge or absolute pressure? Polar moment J or area moment I? Kelvin or Celsius?
- Log the search term that worked next to the topic in your error log.
Five-minute drill: open your handbook and find (1) the torsion shear stress formula, (2) the Carnot efficiency, (3) the steam tables, (4) the Reynolds number, and (5) the interest-factor tables. Time yourself in week 1 and again in week 9. At home your PDF viewer's Ctrl+F works, so practice with short search terms; the exam's search box is a different tool.
Turn misses into your next session
After every missed or slow problem, write down what actually stopped you. Then fix that cause, not just that problem.
| Problem and topic | Your answer and time | Cause | Fix and search term that worked | Retry date and result |
|---|---|---|---|---|
| Example: Problem 21, Carnot efficiency | 91.7%, about 2 min | Units: used Celsius in a temperature ratio | Convert both temperatures to kelvin before any ratio. Search term: "Carnot" | Your date and result |
| Your row |
| Cause | What fixes it |
|---|---|
| Concept | Read the linked principle, explain it in your own words, then solve a simpler example. |
| Model or setup | Draw the body or control volume. List what acts on it, or what enters and leaves, before you pick an equation. |
| Units or sign | Convert the inputs on a separate line. Write your sign convention before you start. |
| Algebra or calculator | Redo the exact keystrokes. Check parentheses, scientific notation, and degree/radian mode. Keep full precision until the end. |
| Handbook lookup | Practice finding the topic and reading its variable definitions without solving anything. |
| Time | Work a short timed mixed set. Note where you stall, and practice flagging and moving on. |
FE Mechanical exam facts to check before you book
| Detail | FE exam |
|---|---|
| Questions | 110, including unscored pretest items that are not identified during the exam |
| Exam time | 5 hours 20 minutes, plus a 25-minute scheduled break |
| Full appointment | The May 2026 Examinee Guide lists 5 hours 55 minutes (including a 2-minute nondisclosure agreement and an 8-minute tutorial); the NCEES FE webpage says 6 hours. Plan for about six. |
| Fee | $225, paid to NCEES. Your licensing board may charge its own application fee. |
| Where and when | Computer-based, at NCEES-approved Pearson test centers, year-round |
| Results | Pass or fail, typically within 7–10 days |
Sources: NCEES FE exam page; NCEES Examinee Guide, May 2026, pp. 3, 11, and 16; NCEES exam scoring.
Rules worth knowing before exam day:
- The first half locks. The exam has two sections. After about half the questions, you review and submit them, and you can't go back. The time isn't split between sections, so manage it yourself. The optional scheduled break comes after you submit the first section (Examinee Guide, pp. 11–12).
- Not every question is single-answer. Expect multiple-correct, point-and-click, drag-and-drop, and fill-in-the-blank questions too. None give partial credit (NCEES computer-based testing). The starter set above includes multiple-correct and numeric-entry problems, but it doesn't copy the exam's screens.
- Answer everything. Your score is based on correct answers, and nothing is taken off for wrong ones (NCEES exam scoring).
- Bring one approved calculator. For 2026, NCEES allows Casio fx-115 and fx-991 models, the HP 33s and HP 35s only, and Texas Instruments TI-30X and TI-36X models (NCEES 2026 approved calculator list, p. 1). Check it again for your exam year. An on-screen TI-30XS is also available during the exam (Examinee Guide, p. 8). Practice on the calculator you'll bring.
- Scratch materials are provided: two reusable booklets and three markers (Examinee Guide, p. 9).
- Bring an accepted, current, physical photo ID whose name matches your appointment. Digital and student IDs aren't accepted. Arrive 30 minutes early (Examinee Guide, pp. 8–9).
For the steps around the exam, see our guides to NCEES exam registration, results, the retake policy, and accommodations.
Common questions about FE Mechanical prep
What score do I need to pass? NCEES doesn't publish one. Your correct answers become a scaled score, which is compared with a minimum standard, and you get a pass or fail result (NCEES exam scoring). Anyone quoting an exact passing percentage is guessing.
Should I take FE Mechanical or FE Other Disciplines? They're separate exams with different subject lists. Compare the Mechanical specification and the Other Disciplines specification with your coursework, and pick the one that matches what you studied. Your licensing board decides whether you're eligible, so check its requirements before you register (Examinee Guide, p. 2).
What happens if I fail? You get a diagnostic report by knowledge area. NCEES allows one attempt per testing window (January–March, April–June, July–September, October–December) and no more than three in 12 months; some boards are stricter (Examinee Guide, p. 5). Use the retake row in the plan above.
Does passing make me an EIT or a PE? No. NCEES describes the FE as generally the first step toward a professional engineer license (NCEES FE exam page). Engineer intern (EI) or engineer-in-training (EIT) certification, and every later step, goes through your state board. Find yours in the NCEES licensing board directory.
Which specification should I study from? The FE Mechanical specification linked from the NCEES FE exam page, which took effect with the July 2020 exams. Check that page again before your test date. A year printed on a prep book's cover doesn't change which specification applies.
Sources and verification
Last verified: October 6, 2026. We checked the NCEES FE exam page, the FE Mechanical and Other Disciplines specifications, the May 2026 Examinee Guide, NCEES's scoring, computer-based testing, and handbook-access pages, and the 2026 calculator policy. Each problem's teaching principle was checked against the source linked beneath it, and every numerical answer and numerical wrong option was recalculated. The conceptual options were checked against their cited principles.
AI-assisted tools were used to develop and check this content. This is editorial source checking, not professional engineering review.
Official sources:
- NCEES: FE exam
- NCEES: FE Mechanical CBT exam specifications (PDF), effective July 2020
- NCEES Examinee Guide, May 2026 (PDF) · current guide landing page
- NCEES: Exam scoring
- NCEES: Computer-based testing
- NCEES: 2026 approved calculator list (official memo, p. 1)
- NCEES help: Exam reference handbooks
- NCEES member licensing board directory
Teaching sources for each problem are linked directly beneath it. How we check exam facts: our methodology. Spot an error? Our corrections page explains how to report it.
Written by the Castleport Test Prep Editorial Team.
Castleport Test Prep is an independent exam prep publisher, not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES), Pearson, or any state engineering licensing board. The practice problems on this page are original and unofficial; they are not NCEES exam questions. Exam and credential names identify their subjects; trademarks belong to their respective owners. This resource doesn't guarantee an exam result or determine eligibility or licensure.