How to use these PE Exam practice questions
These nine problems practise general engineering calculations and reasoning. They are not an official question set or a blueprint for any one PE discipline. The PE exam is computer-based and timed, and its questions test applied engineering judgment, quantitative problem-solving, and standards-based decisions. The scope and reference materials differ by discipline, so check the NCEES website for your discipline’s specifications and current exam information.
MySummaries’ practice set below uses short, self-contained problems. For each one, choose an answer before opening the explanation. Then check whether you selected the governing model, kept units consistent, stated the necessary assumption, and checked whether the result is reasonable. Those habits matter just as much as arithmetic.
Question 1 tests normal stress and conversion from force per area to megapascals.
A concentric axial tensile load of 120 kN acts on a prismatic bar with a cross-sectional area of 600 mm². Assuming uniform stress, what is the average normal stress?
Use σ = P/A. Convert 120 kN to 120,000 N; 120,000 N / 600 mm² = 200 N/mm² = 200 MPa. The strongest distractor is 240 MPa, which can result from using an incorrect area or arithmetic; the given area is 600 mm², not 500 mm².
The key conversion is 1 N/mm² = 1 MPa. Keep the area in square millimetres and the force in newtons to make that conversion direct.
Question 2 tests electrical power for a resistive load, with voltage and resistance given.
A 12 Ω resistive load is connected across a 240 V supply. Under steady-state conditions, what power does the load dissipate?
For a resistor with voltage specified, P = V²/R = 240²/12 = 4,800 W = 4.8 kW. The strongest distractor is 2.0 kW, which does not follow from either P = V²/R or the corresponding current I = V/R = 20 A and P = VI = 4.8 kW.
Use the form of the power equation that matches the known quantities, then verify it by a second form when practical. Here, the calculated current is 20 A, and 240 V × 20 A confirms the result.
Question 3 tests hydrostatic gauge pressure and the distinction between gauge and absolute pressure.
A point is 8.0 m below the free surface of water. Take the water density as 1,000 kg/m³ and gravitational acceleration as 9.81 m/s². What is the gauge pressure at the point?
Hydrostatic gauge pressure is p = ρgh = 1,000 × 9.81 × 8.0 = 78,480 Pa = 78.5 kPa. The strongest distractor is 179.8 kPa, which adds approximately atmospheric pressure and therefore gives an absolute pressure, not the requested gauge pressure.
Read the requested pressure type before calculating. Adding atmospheric pressure is appropriate only when the problem asks for absolute pressure.
Question 4 tests the maximum bending moment for a simply supported beam with a central point load.
A simply supported beam spans 6 m and carries a single 12 kN point load at midspan. Neglect the beam's self-weight. What is the maximum bending moment?
For a simply supported beam with a central point load, Mmax = PL/4. Thus Mmax = 12 × 6 / 4 = 18 kN·m. The strongest distractor is 36 kN·m, which results from omitting the division by four in the governing expression.
The formula depends on the support and loading arrangement. Do not use this result for a cantilever or an off-centre load without deriving the corresponding reactions and moment diagram.
Question 5 tests free thermal expansion using a stated coefficient and temperature change.
A 20 m member has a coefficient of linear thermal expansion of 12 × 10⁻⁶ /°C. If its temperature rises by 50°C and expansion is unrestrained, what is its change in length?
For free expansion, ΔL = αLΔT = 12 × 10⁻⁶ × 20 × 50 m = 0.012 m = 12 mm. The strongest distractor is 1.2 mm, a factor-of-ten error in the powers of ten or unit conversion.
The stated assumption—unrestrained expansion—is important. If movement is restrained, the problem becomes one of thermal stress and requires a different model.
Question 6 tests pump input power from flow, head, fluid density, and efficiency.
A pump moves water at 0.020 m³/s through a total head of 25 m. Take water density as 1,000 kg/m³, g = 9.81 m/s², and pump efficiency as 70%. What input power is required?
Hydraulic output power is ρgQH = 1,000 × 9.81 × 0.020 × 25 = 4,905 W. Input power is output divided by efficiency: 4,905/0.70 = 7,007 W, or 7.01 kW. The strongest distractor is 4.91 kW, which is the hydraulic output and fails to account for pump losses.
Write down whether the question asks for hydraulic output or equipment input. Dividing by efficiency is required to obtain input power from useful hydraulic output.
Question 7 tests equivalent resistance and current in a parallel circuit.
Two resistors, 6 Ω and 3 Ω, are connected in parallel across a 12 V source. What is the total current supplied by the source?
The parallel equivalent resistance is 1/R = 1/6 + 1/3 = 1/2, so R = 2 Ω. The source current is I = V/R = 12/2 = 6 A. The strongest distractor is 4 A, which can arise from adding the branch currents incorrectly; the actual branch currents are 2 A and 4 A, which sum to 6 A.
A useful check is to calculate each branch current: 12/6 = 2 A and 12/3 = 4 A. Their sum agrees with the equivalent-resistance result.
Question 8 tests axial deformation using Hooke’s law for a linearly elastic member.
A 2.0 m bar is under an axial tensile stress of 120 MPa. Assume linear elastic behaviour and an elastic modulus of 200 GPa. What is the bar's elongation?
Strain is σ/E = 120 MPa / 200,000 MPa = 0.0006. Elongation is strain times length: 0.0006 × 2,000 mm = 1.2 mm. The strongest distractor is 12 mm, a factor-of-ten error; keep stress and modulus in the same units before dividing.
This calculation assumes a uniform bar and a constant elastic modulus over the length. If the problem gives changing area or material properties, use the corresponding deformation relationship rather than treating the whole bar as uniform.
Question 9 tests vector addition and reporting a resultant’s direction.
Two perpendicular forces act at a point: 8 kN east and 6 kN north. What is the magnitude and direction of their resultant, measured from east towards north?
Resolve the perpendicular components: R = √(8² + 6²) = 10 kN. The direction from east is tan⁻¹(6/8) = 36.9° north of east. The strongest distractor is 10 kN at 53.1°; that angle comes from reversing the component ratio and measuring from the north axis instead.
For a final check, the resultant must be larger than either component but smaller than their arithmetic sum. State the reference axis and direction so the angle is unambiguous.
Review the mistakes, not just the score
A correct answer reached through a guessed formula is not yet reliable. For each missed problem, record the governing equation, the assumptions that make it applicable, and the unit conversion or check that would have caught the error. For a discipline-specific problem, also identify the applicable standard or reference and confirm its limits rather than relying on a remembered value.
A review summary can rank recurring weak areas and bring repeated errors back for another attempt. For this set, the useful categories are model selection, unit conversion, assumptions, and result checks.
MySummaries can turn your own PE study notes into a revision board, then generate practice and review activities from that material. Use it to organise discipline-specific equations, limits, and worked examples alongside practice questions. Open MySummaries.