FE Other Disciplines Practice Problems

Worked FE Other Disciplines 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

8–12 questions on the exam · Topic guide

Problem 1 · Differential equations

The Laplace transform of f(t) = 3e^(−2t) sin 4t is:

  1. A.12/[(s + 2)² + 16]
  2. B.12/[(s² + 16)(s + 2)]
  3. C.12/[(s − 2)² + 16]
  4. D.3(s + 2)/[(s + 2)² + 16]
Show the answer and solution

Answer: A: 12/[(s + 2)² + 16]

L{sin bt} = b/(s² + b²), and multiplying by e^(−at) shifts s to s + a. With b = 4 and a = 2: L{3e^(−2t) sin 4t} = 3(4)/[(s + 2)² + 16] = 12/[(s + 2)² + 16]. The transform of a product is not the product of the transforms, so the last option is wrong.

Differential equations · Reference Handbook: Mathematics

Problem 2 · Analytic geometry

Two straight pipe runs cross in plan. One follows y = 2x + 3 and the other follows y = 0.25x − 1 (x and y in meters). The acute angle between the two runs is most nearly:

  1. A.49.4°
  2. B.60.3°
  3. C.74.1°
  4. D.130.6°
Show the answer and solution

Answer: A: 49.4°

tan θ = |(m₁ − m₂)/(1 + m₁m₂)| = |(2 − 0.25)/(1 + 0.5)| = 1.1667, so θ = 49.4°. Omitting the denominator gives arctan 1.75 = 60.3°, and 130.6° is the obtuse angle, not the acute one.

Analytic geometry · Reference Handbook: Mathematics

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2. Probability and Statistics

6–9 questions on the exam · Topic guide

Problem 3 · Sample distributions and sizes

A lab wants to estimate the mean strength of a material to within ±2 MPa at 95% confidence. The standard deviation is known to be 6 MPa, and z(0.025) = 1.96. The minimum number of specimens required is:

  1. A.6
  2. B.25
  3. C.35
  4. D.139
Show the answer and solution

Answer: C: 35

n = (zσ/E)² = (1.96 × 6/2)² = 5.88² = 34.6. The sample size always rounds up, so n = 35. Rounding down to 34 would give a margin slightly larger than 2 MPa.

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

Problem 4 · Statistics

Daily water demand in a town is normally distributed with a mean of 18.0 MGD and a standard deviation of 2.4 MGD. For the standard normal distribution, Φ(1.645) = 0.950. The demand that is exceeded on only 5% of days is most nearly:

  1. A.14.1 MGD
  2. B.21.1 MGD
  3. C.21.9 MGD
  4. D.22.7 MGD
Show the answer and solution

Answer: C: 21.9 MGD

The 95th percentile is x = μ + zσ = 18.0 + 1.645(2.4) = 21.9 MGD. Using 1.96 gives 22.7 MGD, but 1.96 leaves 2.5% in each tail, not 5% in the upper tail.

Statistics · Reference Handbook: Engineering Probability and Statistics

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3. Chemistry

5–8 questions on the exam · Topic guide

Problem 5 · Chemical reactions

Octane (C8H18, M = 114.23 g/mol) burns completely: 2C8H18 + 25O2 → 16CO2 + 18H2O. Take M(CO2) = 44.01 g/mol and M(H2O) = 18.02 g/mol. The mass of CO2 produced per kilogram of octane burned is most nearly:

  1. A.0.841 kg
  2. B.1.42 kg
  3. C.3.08 kg
  4. D.8.00 kg
Show the answer and solution

Answer: C: 3.08 kg

Each mole of octane gives 8 mol CO2. Mass CO2 per kg fuel = 8(44.01)/114.23 = 3.08 kg. The product is heavier than the fuel because it picks up oxygen from the air.

Chemical reactions · Reference Handbook: Chemistry and Biology

Problem 6 · Chemical reactions

For the gas-phase reaction N2O4 ⇌ 2NO2 at a certain temperature, the equilibrium concentrations are [NO2] = 0.030 mol/L and [N2O4] = 0.0045 mol/L. The equilibrium constant K_c is most nearly:

  1. A.0.150
  2. B.0.200
  3. C.5.00
  4. D.6.67
Show the answer and solution

Answer: B: 0.200

K_c = [NO2]²/[N2O4] = (0.030)²/0.0045 = 0.00090/0.0045 = 0.200. Each concentration is raised to its stoichiometric coefficient.

Chemical reactions · Reference Handbook: Chemistry and Biology

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4. Instrumentation and Controls

4–6 questions on the exam · Topic guide

Problem 7 · Data acquisition

A pressure transmitter spans 0 to 600 kPa gauge with a 4 to 20 mA output. The logger reads 13.6 mA, and local atmospheric pressure is 101.3 kPa. The absolute pressure is most nearly:

  1. A.360 kPa
  2. B.389 kPa
  3. C.461 kPa
  4. D.509 kPa
Show the answer and solution

Answer: C: 461 kPa

The signal is (13.6 − 4)/(20 − 4) = 0.600 of span, so the gauge pressure is 0.600(600) = 360 kPa. Absolute pressure = 360 + 101.3 = 461 kPa. The 4 mA live zero lets the logger tell a zero reading from a broken loop.

Data acquisition · Reference Handbook: Instrumentation, Measurement, and Control

Problem 8 · Sensors and transducers

A strain-gauge load cell has a capacity of 500 kg and a rated output of 2.0 mV/V at full capacity. It is excited at 10.0 V and reads 12.6 mV. Assuming a linear response, the load on the cell is most nearly:

  1. A.63.0 kg
  2. B.315 kg
  3. C.794 kg
  4. D.3,150 kg
Show the answer and solution

Answer: B: 315 kg

Full-scale output = 2.0 mV/V × 10.0 V = 20.0 mV. Load = (12.6/20.0)(500) = 315 kg. Rated output in mV/V must be multiplied by the excitation voltage before comparing it with the reading.

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

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5. Engineering Ethics and Societal Impacts

5–8 questions on the exam · Topic guide

Problem 9 · Societal impacts

Select all that apply.

An engineer is preparing a cradle-to-gate life-cycle assessment of an aluminum bicycle frame. Which stages fall inside the system boundary? Select all that apply.

  1. A.Bauxite mining and alumina production
  2. B.Smelting and alloying the aluminum
  3. C.Tube forming and welding the frame at the factory
  4. D.Riding the bicycle during its service life
  5. E.Recycling or landfilling the frame at end of life
Show the answer and solution

Answer: A: Bauxite mining and alumina production; B: Smelting and alloying the aluminum; C: Tube forming and welding the frame at the factory

A cradle-to-gate assessment covers raw material extraction through manufacture, ending when the product leaves the factory gate. Use and end-of-life stages are added only in a cradle-to-grave assessment.

Societal impacts · Reference Handbook: Ethics and Professional Practice

Problem 10 · Codes of ethics

Under the NCEES Model Law, the engineer in 'responsible charge' of engineering work is the licensee who:

  1. A.Signs the services contract on behalf of the firm
  2. B.Exercises direct control and personal supervision of the work and makes the independent professional decisions
  3. C.Reviews the drawings once they are complete, without having directed the work
  4. D.Holds the highest management position in the firm, whether or not licensed
Show the answer and solution

Answer: B: Exercises direct control and personal supervision of the work and makes the independent professional decisions

Responsible charge means direct control and personal supervision of engineering work, with the licensee making the independent professional judgments. Only work done under a licensee's responsible charge may carry that licensee's seal.

Codes of ethics · Reference Handbook: Ethics and Professional Practice

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6. Safety, Health, and Environment

6–9 questions on the exam · Topic guide

Problem 11 · Hazard communications

An NFPA 704 placard on a storage tank shows the number 4 in the red diamond. This indicates:

  1. A.Severe health hazard
  2. B.Severe flammability hazard: the material vaporizes readily and burns readily at normal conditions
  3. C.Severe instability (reactivity) hazard
  4. D.Minimal flammability hazard
Show the answer and solution

Answer: B: Severe flammability hazard: the material vaporizes readily and burns readily at normal conditions

In the NFPA 704 diamond, blue is health, red is flammability, yellow is instability, and white is special hazards. Ratings run from 0 to 4, and 4 is the most severe.

Hazard communications · Reference Handbook: Safety

Problem 12 · Confined space entry and ventilation rates

Before entry into a permit-required confined space, the atmosphere is tested with a multigas meter. In what order should the hazards be tested?

  1. A.Any order, because all readings appear at the same time
  2. B.Oxygen first, then flammable gases and vapors, then toxic gases
  3. C.Flammable gases first, then toxic gases, then oxygen
  4. D.Toxic gases first, then oxygen, then flammable gases
Show the answer and solution

Answer: B: Oxygen first, then flammable gases and vapors, then toxic gases

Oxygen is tested first because many combustible-gas sensors need adequate oxygen to read correctly and because oxygen deficiency is immediately dangerous. Flammable gases come next because of the fire and explosion hazard, and toxic gases last.

Confined space entry and ventilation rates · Reference Handbook: Safety

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7. Engineering Economics

6–9 questions on the exam · Topic guide

Problem 13 · Project selection

A $120,000 CNC router is depreciated by MACRS as 5-year property. The MACRS rates for years 1 through 6 are 20.00%, 32.00%, 19.20%, 11.52%, 11.52%, and 5.76%. The book value at the end of year 3 is most nearly:

  1. A.$20,736
  2. B.$23,040
  3. C.$34,560
  4. D.$48,000
Show the answer and solution

Answer: C: $34,560

Accumulated depreciation through year 3 = 120,000(0.20 + 0.32 + 0.192) = 120,000(0.712) = $85,440. Book value = 120,000 − 85,440 = $34,560. MACRS ignores salvage value, and its percentages apply to the original cost basis each year.

Project selection · Reference Handbook: Engineering Economics

Problem 14 · Time value of money

A supplier charges 1.5% per month on overdue balances, compounded monthly. What is the effective annual interest rate? Enter the rate in percent to two decimal places.

Fill in the blank: answer in %.

Show the answer and solution

Answer: 19.56 % (accepted 19.51 to 19.61)

i_eff = (1 + 0.015)^12 − 1 = 1.19562 − 1 = 0.1956, or 19.56%. The 18% figure is the nominal annual rate, which ignores the monthly compounding.

Time value of money · Reference Handbook: Engineering Economics

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8. Statics

9–14 questions on the exam · Topic guide

Problem 15 · Centroid of area

An area consists of a rectangle 400 mm wide and 200 mm tall with a semicircle of radius 100 mm attached to the middle of its top edge (flat side down, curved side up). The distance from the bottom of the rectangle to the centroid of the composite area is most nearly:

  1. A.90.6 mm
  2. B.100.0 mm
  3. C.123.4 mm
  4. D.140.2 mm
Show the answer and solution

Answer: C: 123.4 mm

Rectangle: A = 80,000 mm², ȳ = 100 mm. Semicircle: A = π(100)²/2 = 15,708 mm², ȳ = 200 + 4(100)/(3π) = 242.4 mm. ȳ = (80,000·100 + 15,708·242.4)/95,708 = 123.4 mm.

Centroid of area · Reference Handbook: Statics

Problem 16 · Equivalent force systems

An 800 N force acts at point P, located 1.2 m to the right of and 0.5 m above point O. The force is directed downward and to the right at 60° below the horizontal. The force is replaced by an equivalent force-couple system at O. The magnitude of the couple is most nearly:

  1. A.631 N·m
  2. B.831 N·m
  3. C.960 N·m
  4. D.1,031 N·m
Show the answer and solution

Answer: D: 1,031 N·m

Fx = 800 cos 60° = 400 N (to the right) and Fy = −800 sin 60° = −692.8 N. M_O = xFy − yFx = 1.2(−692.8) − 0.5(400) = −831.4 − 200 = −1,031 N·m (clockwise). Both components turn the body clockwise about O, so their moments add.

Equivalent force systems · Reference Handbook: Statics

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9. Dynamics

9–14 questions on the exam · Topic guide

Problem 17 · Kinematics of particles

A ball is launched from level ground at 20 m/s and 35° above the horizontal. Neglect air resistance and take g = 9.81 m/s². Its horizontal range on the level ground is most nearly:

  1. A.13.9 m
  2. B.19.2 m
  3. C.23.4 m
  4. D.38.3 m
Show the answer and solution

Answer: D: 38.3 m

R = v₀² sin 2θ/g = (20)² sin 70°/9.81 = 400(0.9397)/9.81 = 38.3 m. Equivalently, flight time = 2(20 sin 35°)/9.81 = 2.339 s and R = 20 cos 35°(2.339) = 38.3 m.

Kinematics of particles · Reference Handbook: Dynamics

Problem 18 · Kinematics of mechanisms

In a Scotch yoke mechanism, a crank of 0.10 m radius turns at a constant 600 rpm and drives a slotted yoke in simple harmonic motion, x = r cos ωt. The maximum acceleration of the yoke is most nearly:

  1. A.6.28 m/s²
  2. B.10.0 m/s²
  3. C.62.8 m/s²
  4. D.395 m/s²
Show the answer and solution

Answer: D: 395 m/s²

ω = 600(2π/60) = 62.83 rad/s. For x = r cos ωt, a_max = rω² = 0.10(62.83)² = 395 m/s².

Kinematics of mechanisms · Reference Handbook: Dynamics

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10. Strength of Materials

9–14 questions on the exam · Topic guide

Problem 19 · Stress transformations and Mohr's circle

A point on a machine bracket is in plane stress with σx = 80 MPa (tension), σy = −20 MPa (compression) and τxy = 40 MPa. The maximum principal stress is most nearly:

  1. A.30.0 MPa
  2. B.64.0 MPa
  3. C.80.0 MPa
  4. D.94.0 MPa
Show the answer and solution

Answer: D: 94.0 MPa

Center C = (80 + (−20))/2 = 30 MPa; radius R = √[((80 − (−20))/2)² + 40²] = √(50² + 40²) = 64.03 MPa. σ1 = C + R = 94.0 MPa (and σ2 = -34.0 MPa). The shear stress always makes σ1 exceed the larger of σx and σy.

Stress transformations and Mohr's circle · Reference Handbook: Mechanics of Materials

Problem 20 · Stress and strain caused by axial loads

A stepped bar has a 4 ft steel segment (A = 0.50 in², E = 29,000 ksi) joined end to end with a 3 ft aluminum segment (A = 1.00 in², E = 10,000 ksi). One end is fixed and a 10 kip axial tension is applied at the other end. The total elongation is most nearly:

  1. A.0.0331 in.
  2. B.0.0360 in.
  3. C.0.0455 in.
  4. D.0.0691 in.
Show the answer and solution

Answer: D: 0.0691 in.

Both segments carry the full 10 kips in series. Steel: δ = PL/(AE) = 10(48)/[0.50(29,000)] = 0.0331 in. Aluminum: δ = 10(36)/[1.00(10,000)] = 0.0360 in. Total = 0.0691 in. Using E = 29,000 ksi for the aluminum understates its stretch by a factor of 2.9 and gives 0.0455 in.

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

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11. Materials

6–9 questions on the exam · Topic guide

Problem 21 · Properties (e.g., chemical, electrical, mechanical, physical, thermal)

An aluminum ring with a bore of 50.00 mm at 20°C must be heated so it just slides over a steel shaft that measures 50.05 mm at 20°C; the shaft stays at 20°C. The coefficient of thermal expansion is 23.0 × 10⁻⁶ /°C for the aluminum and 12.0 × 10⁻⁶ /°C for the steel. The minimum temperature rise of the ring is most nearly:

  1. A.21.7 °C
  2. B.43.5 °C
  3. C.83.3 °C
  4. D.90.9 °C
Show the answer and solution

Answer: B: 43.5 °C

The bore must grow by 0.05 mm: ΔD = αDΔT, so ΔT = 0.05/[(23.0 × 10⁻⁶)(50.00)] = 43.5°C. Only the ring is heated, so the steel coefficient and the coefficient difference do not apply.

Properties (e.g., chemical, electrical, mechanical, physical, thermal) · Reference Handbook: Materials Science/Structure of Matter

Problem 22 · Engineered materials (e.g., composites, polymers)

A glass-fiber/polyester composite contains 40% glass fibers by mass. The glass density is 2.54 g/cm³ and the cured polyester density is 1.38 g/cm³. Assuming no voids, the composite density is most nearly:

  1. A.1.69 g/cm³
  2. B.1.84 g/cm³
  3. C.1.96 g/cm³
  4. D.2.08 g/cm³
Show the answer and solution

Answer: A: 1.69 g/cm³

Work per gram of composite: volume = 0.40/2.54 + 0.60/1.38 = 0.1575 + 0.4348 = 0.5923 cm³, so ρ = 1/0.5923 = 1.69 g/cm³. The linear rule of mixtures for density needs volume fractions; applying it to mass fractions gives 1.84 g/cm³.

Engineered materials (e.g., composites, polymers) · Reference Handbook: Materials Science/Structure of Matter

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12. Fluid Mechanics

12–18 questions on the exam · Topic guide

Problem 23 · 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

Problem 24 · Pipe and duct flow and friction losses

Water flows at 0.15 m³/s through 850 m of 300 mm diameter pipe. The Darcy friction factor is 0.018. Take g = 9.81 m/s². Enter the friction head loss in meters, to one decimal place.

Fill in the blank: answer in m.

Show the answer and solution

Answer: 11.7 m (accepted 11.5 to 11.9)

V = Q/A = 0.15/[(π/4)(0.30)²] = 2.122 m/s. h_f = f(L/D)(V²/2g) = 0.018(850/0.30)(2.122²/19.62) = 11.7 m.

Pipe and duct flow and friction losses · Reference Handbook: Fluid Mechanics

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13. Basic Electrical Engineering

6–9 questions on the exam · Topic guide

Problem 25 · DC circuit analysis

Node A connects through a 2 Ω resistor to the positive terminal of a 12 V source, through a 5 Ω resistor to the positive terminal of a 24 V source, and through a 10 Ω resistor to the reference node. The negative terminals of both sources are at the reference node. The voltage of node A relative to the reference is most nearly:

  1. A.1.50 V
  2. B.13.5 V
  3. C.15.4 V
  4. D.18.0 V
Show the answer and solution

Answer: B: 13.5 V

KCL at A (currents leaving): (V − 12)/2 + (V − 24)/5 + V/10 = 0. Multiplying by 10: 5V − 60 + 2V − 48 + V = 0, so 8V = 108 and V = 13.5 V. Dropping the 10 Ω branch gives 7V = 108, or 15.4 V, which ignores the current drawn to the reference node.

DC circuit analysis · Reference Handbook: Electrical and Computer Engineering

Problem 26 · 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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14. Thermodynamics and Heat Transfer

9–14 questions on the exam · Topic guide

Problem 27 · Conduction

A house wall consists of 13 mm gypsum board (k = 0.17 W/(m·K)), 90 mm of insulation (k = 0.040 W/(m·K)), and 20 mm wood siding (k = 0.12 W/(m·K)). The inside film coefficient is 8 W/(m²·K), and the outside film coefficient is 25 W/(m²·K). The indoor air is at 21°C and the outdoor air at −10°C. The steady heat flux through the wall is most nearly:

  1. A.4.14 W/m²
  2. B.11.7 W/m²
  3. C.12.4 W/m²
  4. D.13.8 W/m²
Show the answer and solution

Answer: B: 11.7 W/m²

R″ = 1/8 + 0.013/0.17 + 0.090/0.040 + 0.020/0.12 + 1/25 = 0.125 + 0.076 + 2.250 + 0.167 + 0.040 = 2.658 m²·K/W. q″ = ΔT/R″ = 31/2.658 = 11.7 W/m². The resistances are in series, so they add, and the film resistances are part of the total.

Conduction · Reference Handbook: Heat Transfer

Problem 28 · Processes

Gas in a piston-cylinder device expands polytropically with n = 1.3 from 1.0 MPa and 0.050 m³ to a final volume of 0.20 m³. What is the boundary work done by the gas? Enter the value in kJ to one decimal place.

Fill in the blank: answer in kJ.

Show the answer and solution

Answer: 56.7 kJ (accepted 55.9 to 57.5)

P2 = P1(V1/V2)^n = 1,000(0.25)^1.3 = 164.9 kPa. W = (P2V2 − P1V1)/(1 − n) = (164.9·0.20 − 1,000·0.050)/(1 − 1.3) = (32.98 − 50.0)/(−0.3) = 56.7 kJ.

Processes · Reference Handbook: Thermodynamics

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Common questions

Are these FE Other Disciplines practice problems like the real exam?

They are original problems written to the NCEES FE Other Disciplines 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 Other Disciplines 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).