FE Mechanical Practice Problems

Worked FE Mechanical 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.

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1. Mathematics

6–9 questions on the exam · Topic guide

Problem 1 · Single-variable calculus

A landscaped area is bounded by the curves y = x² and y = 4x, with x and y in meters. The area enclosed between the two curves is most nearly:

  1. A.10.7 m²
  2. B.21.3 m²
  3. C.32.0 m²
  4. D.53.3 m²
Show the answer and solution

Answer: A: 10.7 m²

The curves meet at x = 0 and x = 4. Area = ∫₀⁴ (4x − x²) dx = [2x² − x³/3]₀⁴ = 32 − 21.33 = 10.7 m². Adding the two integrals (32 + 21.33 = 53.3 m²) counts the region under the parabola instead of removing it.

Single-variable calculus · Reference Handbook: Mathematics

Problem 2 · Algorithm and logic development

The array [4, 7, 1, 9, 3] goes through exactly one pass of bubble sort in ascending order. The pass compares each adjacent pair from left to right and swaps the pair when the left value is larger. After this single pass, the array is:

  1. A.[1, 3, 4, 7, 9]
  2. B.[1, 4, 7, 3, 9]
  3. C.[4, 1, 7, 3, 9]
  4. D.[1, 4, 3, 7, 9]
Show the answer and solution

Answer: C: [4, 1, 7, 3, 9]

Compare (4, 7): no swap. (7, 1): swap, giving [4, 1, 7, 9, 3]. (7, 9): no swap. (9, 3): swap, giving [4, 1, 7, 3, 9]. One pass moves the largest value to the end, but 1 has moved only one position left, so the array is not yet sorted.

Algorithm and logic development · Reference Handbook: Mathematics

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2. Probability and Statistics

4–6 questions on the exam · Topic guide

Problem 3 · Sample distributions and sizes

The tensile capacity of anchor bolts has a population mean of 120 kN and a standard deviation of 9 kN. A random sample of 9 bolts is tested. For the standard normal distribution, Φ(1.67) = 0.9525. The probability that the sample mean exceeds 125 kN is most nearly:

  1. A.0.0475
  2. B.0.0950
  3. C.0.2877
  4. D.0.9525
Show the answer and solution

Answer: A: 0.0475

The standard error of the mean is σ/√n = 9/√9 = 3 kN, so z = (125 − 120)/3 = 1.67. P(x̄ > 125) = 1 − 0.9525 = 0.0475. Using σ = 9 kN without dividing by √n treats a single bolt instead of the sample mean.

Sample distributions and sizes · Reference Handbook: Engineering Probability and Statistics

Problem 4 · Goodness of fit

A linear regression of pavement roughness against age has a negative slope. The error sum of squares is SSE = 14.2 and the total sum of squares is SST = 182.6. The correlation coefficient r is most nearly:

  1. A.−0.960
  2. B.−0.922
  3. C.0.922
  4. D.0.960
Show the answer and solution

Answer: A: −0.960

R² = 1 − SSE/SST = 1 − 14.2/182.6 = 0.922. |r| = √0.922 = 0.960, and r takes the sign of the slope, so r = −0.960.

Goodness of fit · Reference Handbook: Engineering Probability and Statistics

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3. Ethics and Professional Practice

4–6 questions on the exam · Topic guide

Problem 5 · Public health, safety, and welfare

While inspecting a parking garage for a client's insurance appraisal, an engineer notices severe spalling and exposed, badly corroded reinforcing bars on the balcony slabs of an occupied apartment building next door, which is outside her assignment. What should she do?

  1. A.Mention the observation only in an internal appendix of her client's report
  2. B.Notify the neighboring building's owner and, if the hazard is not promptly addressed, the appropriate public authority
  3. C.Do nothing, because the neighboring building is outside the scope of her contract
  4. D.Give the neighboring owner's contact information to a repair contractor
Show the answer and solution

Answer: B: Notify the neighboring building's owner and, if the hazard is not promptly addressed, the appropriate public authority

The paramount duty to protect public health, safety, and welfare applies even to hazards an engineer observes outside her assignment. She should alert the party able to correct the condition and the responsible authority if needed.

Public health, safety, and welfare · Reference Handbook: Ethics and Professional Practice

Problem 6 · Societal impacts

A design team must choose between two circulating pumps for a 20-year continuous duty. Pump A has the lower purchase price; Pump B costs more but uses about 15% less electricity. Which approach best accounts for the economic and environmental impacts of the choice?

  1. A.Select the pump with the longer manufacturer's warranty
  2. B.Compare life-cycle cost and life-cycle environmental impacts, including energy use over the full 20 years
  3. C.Select the more efficient pump regardless of its cost
  4. D.Select the pump with the lower purchase price, because first cost dominates total cost
Show the answer and solution

Answer: B: Compare life-cycle cost and life-cycle environmental impacts, including energy use over the full 20 years

Sustainability and life-cycle thinking evaluate all costs and impacts over the life of a product, including operation and disposal. For continuously running pumps, energy use over 20 years usually outweighs the purchase price, so a life-cycle comparison is the sound basis for the decision.

Societal impacts · Reference Handbook: Ethics and Professional Practice

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4. Engineering Economics

4–6 questions on the exam · Topic guide

Problem 7 · Time value of money

A packaging machine costs $60,000, has operating and maintenance costs of $8,000 at the end of each year for 6 years, and has a salvage value of $10,000 at the end of year 6. At an interest rate of 8% per year, the present worth of the machine's costs is most nearly:

  1. A.$86,983
  2. B.$90,681
  3. C.$96,983
  4. D.$103,285
Show the answer and solution

Answer: B: $90,681

PW = 60,000 + 8,000(P/A, 8%, 6) − 10,000(P/F, 8%, 6) = 60,000 + 8,000(4.6229) − 10,000(0.6302) = 60,000 + 36,983 − 6,302 = $90,681. Salvage is a receipt at year 6, so it is discounted and subtracted from the costs.

Time value of money · Reference Handbook: Engineering Economics

Problem 8 · Time value of money

An energy-efficiency retrofit costs $10,000 and saves $2,500 at the end of each year for 6 years, with no salvage value. The rate of return on the retrofit is most nearly:

  1. A.8.3%
  2. B.13.0%
  3. C.16.7%
  4. D.25.0%
Show the answer and solution

Answer: B: 13.0%

Find i such that 10,000 = 2,500(P/A, i, 6), so (P/A, i, 6) = 4.000. At 12%, P/A = 4.1114; at 14%, P/A = 3.8887. Interpolating gives i ≈ 13.0%. Simple-return measures ignore the timing of the savings.

Time value of money · Reference Handbook: Engineering Economics

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5. Electricity and Magnetism

5–8 questions on the exam · Topic guide

Problem 9 · AC and DC circuits

A 470 μF capacitor in a power-supply filter is charged to 400 V. The energy stored in the capacitor is most nearly:

  1. A.0.0940 J
  2. B.0.188 J
  3. C.37.6 J
  4. D.75.2 J
Show the answer and solution

Answer: C: 37.6 J

W = CV²/2 = (470 × 10⁻⁶)(400)²/2 = 37.6 J. The charge Q = CV = 0.188 C is a different quantity; energy depends on the square of the voltage.

AC and DC circuits · Reference Handbook: Electrical and Computer Engineering

Problem 10 · AC circuit analysis

A 120 V rms, 60 Hz source supplies a series circuit of a 30 Ω resistor and a 0.10 H inductor. The rms current is most nearly:

  1. A.1.77 A
  2. B.2.49 A
  3. C.3.18 A
  4. D.3.92 A
Show the answer and solution

Answer: B: 2.49 A

X_L = 2πfL = 2π(60)(0.10) = 37.7 Ω. |Z| = √(30² + 37.7²) = 48.2 Ω, so I = 120/48.2 = 2.49 A. Resistance and reactance are 90° apart, so they combine as a right triangle, not by simple addition.

AC circuit analysis · Reference Handbook: Electrical and Computer Engineering

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6. Statics

9–14 questions on the exam · Topic guide

Problem 11 · Static friction

Block A (20 kg) rests on block B (30 kg), which rests on the floor. Block A is tied to the wall by a horizontal cord. The coefficient of static friction is 0.30 at both contact surfaces. Take g = 9.81 m/s². The horizontal force on B needed to start B sliding out from under A is most nearly:

  1. A.88.3 N
  2. B.147 N
  3. C.206 N
  4. D.265 N
Show the answer and solution

Answer: C: 206 N

Friction at the A–B interface = 0.30(20)(9.81) = 58.9 N. The floor's normal force on B carries both blocks: 0.30(50)(9.81) = 147.2 N. Both oppose B's motion, so P = 58.9 + 147.2 = 206 N.

Static friction · Reference Handbook: Statics

Problem 12 · Area moments of inertia

For the T-section with a 200 mm × 20 mm flange on top of a 20 mm × 180 mm web (total depth 200 mm, centroid 142.6 mm above the bottom), the moment of inertia about the horizontal centroidal axis is most nearly:

  1. A.9.85 × 10⁶ mm⁴
  2. B.18.9 × 10⁶ mm⁴
  3. C.28.8 × 10⁶ mm⁴
  4. D.183 × 10⁶ mm⁴
Show the answer and solution

Answer: C: 28.8 × 10⁶ mm⁴

Web: I = 20(180)³/12 + 3,600(142.6 − 90)² = 9.72 × 10⁶ + 9.97 × 10⁶. Flange: I = 200(20)³/12 + 4,000(190 − 142.6)² = 0.13 × 10⁶ + 8.98 × 10⁶. Total ≈ 28.8 × 10⁶ mm⁴.

Area moments of inertia · Reference Handbook: Statics

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7. Dynamics, Kinematics, and Vibrations

10–15 questions on the exam · Topic guide

Problem 13 · Work-energy of particles

A 2.0 kg block slides at 3.0 m/s on a frictionless horizontal surface into a light spring with stiffness 800 N/m. The maximum compression of the spring is most nearly:

  1. A.0.0225 m
  2. B.0.106 m
  3. C.0.150 m
  4. D.0.212 m
Show the answer and solution

Answer: C: 0.150 m

½mv² = ½kx², so x = v√(m/k) = 3.0√(2.0/800) = 3.0(0.05) = 0.150 m.

Work-energy of particles · Reference Handbook: Dynamics

Problem 14 · Free and forced vibrations

A 10 kg mass on a spring of stiffness 4,000 N/m is connected to a viscous damper with a damping coefficient of 80 N·s/m. The damping ratio is most nearly:

  1. A.0.100
  2. B.0.200
  3. C.0.400
  4. D.4.00
Show the answer and solution

Answer: B: 0.200

Critical damping c_c = 2√(km) = 2√(4,000·10) = 400 N·s/m. ζ = c/c_c = 80/400 = 0.200.

Free and forced vibrations · Reference Handbook: Dynamics

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8. Mechanics of Materials

9–14 questions on the exam · Topic guide

Problem 15 · Statically indeterminate systems

A short column is a steel tube (A_s = 3,000 mm², E_s = 200 GPa) completely filled with concrete (A_c = 40,000 mm², E_c = 25 GPa), bonded so both materials shorten equally. A 500 kN axial compressive load is applied through a rigid cap. The compressive stress in the steel is most nearly:

  1. A.1.55 MPa
  2. B.7.81 MPa
  3. C.11.6 MPa
  4. D.62.5 MPa
Show the answer and solution

Answer: D: 62.5 MPa

Compatibility: ε_s = ε_c, so σ_s/E_s = σ_c/E_c. Equilibrium: σ_sA_s + σ_cA_c = P. Then σ_s = PE_s/(A_sE_s + A_cE_c) = 500,000(200,000)/[3000(200,000) + 40,000(25,000)] = 62.5 MPa, and σ_c = 7.81 MPa. Dividing by the total area treats both materials as equally stiff and gives 11.6 MPa.

Statically indeterminate systems · Reference Handbook: Mechanics of Materials

Problem 16 · Stress and strain caused by bending loads

A W12×40 steel beam (S_x = 51.5 in³) is simply supported over a 24 ft span and carries a single 20 kip concentrated load at midspan. Neglecting self-weight, the maximum bending stress is most nearly:

  1. A.2.33 ksi
  2. B.14.0 ksi
  3. C.28.0 ksi
  4. D.55.9 ksi
Show the answer and solution

Answer: C: 28.0 ksi

M = PL/4 = 20(24)/4 = 120 kip·ft = 1440 kip·in. σ = M/S = 1440/51.5 = 28.0 ksi. Forgetting to convert kip·ft to kip·in. gives 2.33 ksi.

Stress and strain caused by bending loads · Reference Handbook: Mechanics of Materials

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9. Material Properties and Processing

7–11 questions on the exam · Topic guide

Problem 17 · Corrosion mechanisms and control

Select all that apply.

A designer must join aluminum panels to a stainless steel frame in a wet outdoor environment. Which of the following measures reduce galvanic corrosion of the joint? Select all that apply.

  1. A.Electrically insulate the two metals from each other with nonconductive washers and sleeves
  2. B.Choose a pair of metals that lie close together in the galvanic series for this environment
  3. C.Make the exposed area of the anodic metal large relative to the cathodic metal
  4. D.Paint only the anodic metal and leave the cathodic metal bare
  5. E.Fasten the panels with small fasteners made of a metal more anodic than both the panels and the frame
Show the answer and solution

Answer: A: Electrically insulate the two metals from each other with nonconductive washers and sleeves; B: Choose a pair of metals that lie close together in the galvanic series for this environment; C: Make the exposed area of the anodic metal large relative to the cathodic metal

Galvanic attack needs electrical contact, an electrolyte and a potential difference. Insulating the metals breaks the circuit, a closely matched pair lowers the driving potential, and a large anode-to-cathode area ratio spreads the corrosion current so the penetration rate is low. If only one metal is coated it should be the cathode, and fasteners should be the more noble component, never a small anode.

Corrosion mechanisms and control · Reference Handbook: Materials Science/Structure of Matter

Problem 18 · Stress-strain diagrams

In a tension test, a round steel specimen with a 12.5 mm diameter and a 50.0 mm gauge length elongates 0.040 mm over the gauge length at a load of 20 kN, well within the linear region. The modulus of elasticity is most nearly:

  1. A.2.04 GPa
  2. B.50.9 GPa
  3. C.160 GPa
  4. D.204 GPa
Show the answer and solution

Answer: D: 204 GPa

A = π(12.5)²/4 = 122.7 mm²; σ = 20,000/122.7 = 163.0 MPa. ε = 0.040/50.0 = 0.00080. E = σ/ε = 163.0/0.00080 = 203,718 MPa ≈ 204 GPa, a typical steel value. Using the strain in percent shrinks the result a hundredfold.

Stress-strain diagrams · Reference Handbook: Materials Science/Structure of Matter

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10. Fluid Mechanics

10–15 questions on the exam · Topic guide

Problem 19 · Pumps

A pump delivers 0.060 m³/s of water against a total dynamic head of 45 m. The pump efficiency is 78%. Take γ = 9.81 kN/m³. Enter the required shaft power in kW, to one decimal place.

Fill in the blank: answer in kW.

Show the answer and solution

Answer: 34 kW (accepted 33.5 to 34.5)

Water power = γQH = 9.81(0.060)(45) = 26.5 kW. Shaft power = water power/η = 26.5/0.78 = 34.0 kW. The motor must supply more than the power delivered to the water, so the efficiency divides.

Pumps · Reference Handbook: Fluid Mechanics

Problem 20 · Compressible flow

Air at a static temperature of 280 K flows at 500 m/s. Take k = 1.4 and R = 287 J/(kg·K). Enter the stagnation temperature in K, to the nearest kelvin.

Fill in the blank: answer in K.

Show the answer and solution

Answer: 404 K (accepted 400 to 408)

c = √(kRT) = √[1.4(287)(280)] = 335.4 m/s, so Ma = 500/335.4 = 1.491. T₀ = T[1 + (k − 1)Ma²/2] = 280[1 + 0.2(2.222)] = 404 K. The same result follows from T₀ = T + V²/(2c_p) with c_p = kR/(k − 1) = 1,004.5 J/(kg·K).

Compressible flow · Reference Handbook: Fluid Mechanics

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11. Thermodynamics

10–15 questions on the exam · Topic guide

Problem 21 · Performance of components

In a counterflow heat exchanger, 2.0 kg/s of oil (c_p = 2.1 kJ/(kg·K)) is cooled from 120°C to 60°C by water (c_p = 4.18 kJ/(kg·K)) that warms from 20°C to 45°C. Neglect heat loss to the surroundings. The required water flow rate is most nearly:

  1. A.1.34 kg/s
  2. B.2.41 kg/s
  3. C.4.80 kg/s
  4. D.9.55 kg/s
Show the answer and solution

Answer: B: 2.41 kg/s

Heat lost by oil = 2.0(2.1)(60) = 252 kW. Water flow = 252/[4.18(45 − 20)] = 2.41 kg/s.

Performance of components · Reference Handbook: Thermodynamics

Problem 22 · Power cycles

An ideal air-standard Diesel cycle has a compression ratio of 18 and a cutoff ratio of 2. Take k = 1.4. The thermal efficiency is most nearly:

  1. A.48.4%
  2. B.58.5%
  3. C.63.2%
  4. D.68.5%
Show the answer and solution

Answer: C: 63.2%

η = 1 − (1/r^(k−1))[(r_c^k − 1)/(k(r_c − 1))] = 1 − (1/18^0.4)[(2^1.4 − 1)/(1.4·1)] = 1 − 0.3146(1.1707) = 63.2%. At the same compression ratio, a Diesel cycle is less efficient than an Otto cycle.

Power cycles · Reference Handbook: Thermodynamics

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12. Heat Transfer

7–11 questions on the exam · Topic guide

Problem 23 · Radiation

An uninsulated pipe 0.10 m in diameter and 2.0 m long has a surface temperature of 450 K and an emissivity of 0.80. It sits in a large room whose walls are at 300 K. Take σ = 5.67 × 10⁻⁸ W/(m²·K⁴). The net radiation heat loss from the pipe is most nearly:

  1. A.14.4 W
  2. B.28.0 W
  3. C.938 W
  4. D.1,172 W
Show the answer and solution

Answer: C: 938 W

A = πDL = 0.628 m². q = εσA(T_s⁴ − T_surr⁴) = 0.80(5.67 × 10⁻⁸)(0.628)(450⁴ − 300⁴) = 0.80(5.67 × 10⁻⁸)(0.628)(3.29 × 10¹⁰) = 938 W. Radiation exchange depends on the difference of the fourth powers of absolute temperatures.

Radiation · Reference Handbook: Heat Transfer

Problem 24 · Conduction

A long aluminum pin fin (k = 200 W/(m·K)) is 5 mm in diameter. Its base is at 120°C, and it is surrounded by air at 20°C with h = 25 W/(m²·K). Treat the fin as infinitely long. The heat transfer from the fin is most nearly:

  1. A.0.154 W
  2. B.3.93 W
  3. C.7.85 W
  4. D.11.1 W
Show the answer and solution

Answer: B: 3.93 W

P = πD = 0.01571 m and A_c = πD²/4 = 1.963 × 10⁻⁵ m². q = √(hPkA_c)·θ_b = √[25(0.01571)(200)(1.963 × 10⁻⁵)](100) = 0.0393(100) = 3.93 W. For an infinitely long fin, all of the heat entering the base is eventually convected away.

Conduction · Reference Handbook: Heat Transfer

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13. Measurements, Instrumentation, and Controls

5–8 questions on the exam · Topic guide

Problem 25 · Control systems

A unity negative-feedback system has the forward-path transfer function G(s) = 40/[s(s + 5)]. The reference input is the ramp r(t) = 2t. The steady-state error is most nearly:

  1. A.0
  2. B.0.125
  3. C.0.222
  4. D.0.250
Show the answer and solution

Answer: D: 0.250

The system is type 1, so K_v = lim s·G(s) = 40/5 = 8 s⁻¹. For a ramp of slope 2, e_ss = 2/K_v = 2/8 = 0.250. The error is finite and constant: the output tracks the ramp with a fixed lag.

Control systems · Reference Handbook: Instrumentation, Measurement, and Control

Problem 26 · Sensors and transducers

An NTC thermistor follows R = R₀ exp[β(1/T − 1/T₀)] with T in kelvins, R₀ = 10.0 kΩ at T₀ = 25°C and β = 3,950 K. Its measured resistance is 3.60 kΩ. Enter the temperature in °C to one decimal place.

Fill in the blank: answer in °C.

Show the answer and solution

Answer: 49.9 °C (accepted 49.4 to 50.4)

1/T = 1/T₀ + ln(R/R₀)/β = 1/298.15 + ln(0.360)/3,950 = 0.0033540 − 0.0002587 = 0.0030953 K⁻¹. T = 323.1 K = 49.9°C. Resistance fell, so the temperature rose, as expected for an NTC device.

Sensors and transducers · Reference Handbook: Instrumentation, Measurement, and Control

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14. Mechanical Design and Analysis

10–15 questions on the exam · Topic guide

Problem 27 · Deformation and stiffness

An aluminum bracket acts as a cantilever 150 mm long with a rectangular section 20 mm wide and 6 mm thick, loaded at its free end so it bends about the axis parallel to the 20 mm width. Take E = 70 GPa. The bracket's stiffness at the free end is most nearly:

  1. A.22.4 N/mm
  2. B.249 N/mm
  3. C.358 N/mm
  4. D.3,360 N/mm
Show the answer and solution

Answer: A: 22.4 N/mm

I = bh³/12 = 20(6)³/12 = 360 mm⁴. For an end-loaded cantilever, δ = PL³/(3EI), so k = 3EI/L³ = 3(70,000)(360)/150³ = 22.4 N/mm. Bending about the thin direction makes the bracket far more flexible than the swapped-axis value.

Deformation and stiffness · Reference Handbook: Mechanics of Materials

Problem 28 · Power transmission

A compound gear train is driven by a 5.0 kW motor at 1,800 rpm. Gear A (20 teeth) on the motor shaft drives gear B (60 teeth). Gear C (15 teeth) is on the same shaft as B and drives gear D (45 teeth) on the output shaft. Neglecting losses, the output torque is most nearly:

  1. A.2.95 N·m
  2. B.26.5 N·m
  3. C.79.6 N·m
  4. D.239 N·m
Show the answer and solution

Answer: D: 239 N·m

Train ratio = (60/20)(45/15) = 9, so n_out = 1,800/9 = 200 rpm. T = P/ω = 5,000/(2π × 200/60) = 239 N·m, which is 9 times the 26.5 N·m input torque.

Power transmission · Reference Handbook: Mechanical Engineering

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Common questions

Are these FE Mechanical practice problems like the real exam?

They are original problems written to the NCEES FE Mechanical 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 Mechanical 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).