Free PE Mechanical Machine Design and Materials Practice Test
40 original, unofficial practice questions written to the current NCEES PE Mechanical: Machine Design and Materials specification (effective October 2025), covering all six knowledge areas. Every question has a worked solution and the reason each wrong answer is wrong; no signup is required.
1. Basic Engineering Practice (6 questions · NCEES range 11–17)
Question 1 · MDM-01
Basic Engineering Practice — 1A Engineering terms, symbols, and drawings
A drawing calls out a fastener thread as 1/2-13 UNC-2A. What do "13" and "A" indicate?
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Answer: A. In a Unified thread callout, 1/2 is the nominal major diameter (in), 13 is threads per inch, UNC is the coarse series, 2 is the class of fit, and the letter tells you which part: A for an external thread (the bolt), B for an internal thread (the nut or tapped hole).
Why the others are wrong: B: B is the letter for an internal thread, not A. C: Unified (inch) callouts give threads per inch, not a metric pitch. D: Length is not part of this callout; "A" is the external-thread designator.
Principle: Koç University, Fasteners notes, "Thread Standards and Definitions" (designation nomenclature); Fastenal Blue Print, Screw Thread Design, thread series and classes of fit (A = external, B = internal). Topic: NCEES MDM specification, 1A Engineering terms, symbols, and drawings.
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This result describes these 40 practice questions. It is not an NCEES score and does not predict whether you will pass.
- Basic Engineering Practice:
- Mechanics of Materials:
- Mechanical Attachments:
- Power Transmission:
- Mechanical Components and Assemblies:
- Supportive Knowledge:
Question 2 · MDM-02
Basic Engineering Practice — 1B Project management and economic analysis
Two presses have 10-year lives. Press 1 costs $50,000, has $8,000/yr operating cost, and a $5,000 salvage value. Press 2 costs $70,000, has $5,000/yr operating cost, and a $10,000 salvage value. At an 8% annual interest rate, what is Press 2's equivalent uniform annual cost (EUAC)?
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Answer: B. EUAC = first cost × (A/P) − salvage × (A/F) + annual cost. (A/P, 8%, 10) = 0.1490 and (A/F, 8%, 10) = 0.0690. Press 2: 70,000(0.1490) − 10,000(0.0690) + 5,000 ≈ $14,740/yr. Press 1 works out to ≈ $15,110/yr, so Press 2 is the lower-cost choice despite its higher price.
Why the others are wrong: A: $11,000 ignores interest: (70,000 − 10,000)/10 + 5,000. C: $13,940 credits the salvage with (A/P) instead of (A/F); salvage arrives at the end, so it's a future amount. D: $15,430 leaves out the salvage value.
Principle: Penn State EME 460, Compound Interest Formulas III, Sec. 6, capital-recovery factor, Eq. 1-6; Penn State EME 460, sinking-fund factor, Sec. 4, sinking-fund deposit factor, Eq. 1-4. Topic: NCEES MDM specification, 1B Project management and economic analysis.
Question 3 · MDM-03
Basic Engineering Practice — 1C Design methodology (risk assessment)
A design FMEA rates four failure modes for Severity (S), Occurrence (O), and Detection (D), each on a 1–10 scale. Seal leak: S 6, O 5, D 4. Bolt fatigue: S 9, O 3, D 5. Sensor drift: S 4, O 7, D 6. Weld crack: S 10, O 2, D 3. Which failure mode has the highest risk priority number (RPN)?
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Answer: D. RPN = S × O × D. Seal leak 120, bolt fatigue 135, sensor drift 168, weld crack 60. Sensor drift ranks highest. Keep the limitation in mind: a severity-10 mode with a low RPN can still deserve action, which is why many teams review high-severity items separately rather than ranking on RPN alone.
Why the others are wrong: A: Seal leak scores 120. B: Bolt fatigue scores 135; high severity alone doesn't make the highest RPN. C: Weld crack has the top severity but the lowest RPN, 60.
Principle: NEON FMEA template (NEON.DOC.000015), "Scoring the RPN Column," p. 5. Topic: NCEES MDM specification, 1C Design methodology (risk assessment).
Question 4 · MDM-04
Basic Engineering Practice — 1D Physical properties of materials
A tension link 500 mm long must have an axial stiffness of 50 kN/mm. If it's made of aluminum (E = 70 GPa, ρ = 2,700 kg/m³), what is the minimum mass of the link?
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Answer: C. Axial stiffness k = AE/L, so A = kL/E = (50,000 N/mm × 500 mm) ÷ 70,000 N/mm² = 357 mm². Mass = ρAL = 2,700 × 357×10⁻⁶ m² × 0.5 m = 0.482 kg. A steel link of equal stiffness (E = 200 GPa, ρ = 7,850 kg/m³) weighs 0.491 kg: for stiffness-limited tension members, E/ρ is nearly the same for the common structural metals, so switching to aluminum barely saves weight.
Why the others are wrong: A: 0.169 kg sizes the aluminum with the steel link's area (125 mm²); that link would be too flexible. B: 0.491 kg is the equal-stiffness steel link. D: 1.40 kg uses steel's density with aluminum's area.
Principle: OpenStax University Physics Vol. 1, 12.3, Eq. 12.36 (Young's modulus) and Eq. 12.42 (shear stress). Topic: NCEES MDM specification, 1D Physical properties of materials.
Question 5 · MDM-05
Basic Engineering Practice — 1C Design methodology (verification and validation)
Select all that apply. For a new assembly fixture, which statements about verification and validation are accurate?
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Answers: A, B, E. Verification asks whether the product meets each stated requirement and uses test, analysis, inspection, or demonstration (A, E). Validation asks whether it fulfills its intended use in its intended environment (B). You need all three, and only those three.
Why the others are wrong: C: NASA treats them as distinct processes with different questions and evidence. D: This reverses them: verification traces to requirements; validation traces to intended use and the stakeholders' concept of operations.
Principle: NASA Systems Engineering Handbook, Sec. 2.4, Sec. 2.4, verification vs. validation. Topic: NCEES MDM specification, 1C Design methodology (verification and validation).
Question 6 · MDM-06
Basic Engineering Practice — 1B Project management and economic analysis
A bracket can be die cast (tooling $24,000/yr, $3.50 per part) or machined (fixture $6,000/yr, $5.75 per part). At what annual quantity are total costs equal, and which process is cheaper at 12,000 parts per year?
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Answer: B. Set total costs equal: 24,000 + 3.50Q = 6,000 + 5.75Q, so Q = 18,000 ÷ 2.25 = 8,000 parts. Above 8,000 parts the lower variable cost wins, so die casting is cheaper at 12,000 (66,000 vs. 75,000).
Why the others are wrong: A: Right quantity, wrong side: below 8,000 machining is cheaper; above it, die casting is. C: 10,667 divides die casting's full fixed cost by the variable-cost difference; use the difference in fixed costs. D: 3,130 divides machining's fixed cost by its unit cost, which isn't a comparison.
Principle: OpenStax Principles of Managerial Accounting, 3.2, Sec. 3.2, fixed and variable costs at break-even. Topic: NCEES MDM specification, 1B Project management and economic analysis.
2. Mechanics of Materials (10 questions · NCEES range 17–26)
Question 7 · MDM-07
Mechanics of Materials — 2A Axial loading (statically indeterminate)
A steel rod (A = 500 mm², E = 200 GPa) sits inside an aluminum tube (A = 1,000 mm², E = 70 GPa). Both are the same length and are compressed together between rigid plates by a 150 kN axial load. Assuming elastic behavior, what is the stress in the steel rod?
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Answer: B. The rigid plates force equal shortening, so each member takes load in proportion to its axial stiffness EA. Steel EA = 1.0×10⁸ N; aluminum EA = 0.7×10⁸ N. Steel carries 150 × 1.0/1.7 = 88.2 kN, so σ = 88235 ÷ 500 = 176 MPa.
Why the others are wrong: A: 100 MPa divides the load by the total area, which assumes equal stress, not equal strain. C: 150 MPa splits the load equally between the members. D: 61.8 MPa is the aluminum's stress (61.8 kN ÷ 1,000 mm²).
Principle: OpenStax University Physics Vol. 1, 12.3, Eq. 12.36 (Young's modulus) and Eq. 12.42 (shear stress). Topic: NCEES MDM specification, 2A Axial loading (statically indeterminate).
Question 8 · MDM-08
Mechanics of Materials — 2B Shear and transverse loading
A rectangular beam 1.5 in wide and 4.0 in deep carries a vertical shear force of 2,700 lbf at a section. What is the maximum transverse shear stress at that section?
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Answer: D. For a rectangle, shear stress is parabolic through the depth and peaks at the neutral axis at τmax = 3V/(2A). A = 1.5 × 4.0 = 6.0 in², so τmax = 1.5 × 2,700 ÷ 6.0 = 675 psi.
Why the others are wrong: A: 450 psi is the average V/A; the peak is 50% higher. B: 900 psi uses 2V/A. C: 1,350 psi uses 3V/A, dropping the 2.
Principle: MIT OCW 3.11, Stresses in Beams, Eqs. 5 and 7 (I = bh³/12, σ = My/I); "Shear stresses," p. 10 (3V/2A, following Eq. 12). Topic: NCEES MDM specification, 2B Shear and transverse loading.
Question 9 · MDM-09
Mechanics of Materials — 2C Bending (deflection)
A steel cantilever (E = 200 GPa) is 1.0 m long with a 30 mm wide × 50 mm deep rectangular section, deep side vertical. A 2.0 kN load acts downward at the free end. Ignoring self-weight, what is the tip deflection?
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Answer: C. I = bh³/12 = 30 × 50³ ÷ 12 = 312,500 mm⁴. With M = Px measured from the tip, Castigliano's theorem gives δ = ∫(Px)(x)dx/EI from 0 to L = PL³/(3EI). δ = 2,000 × 1,000³ ÷ (3 × 200,000 × 312,500) = 10.7 mm.
Why the others are wrong: A: 0.667 mm uses PL³/48EI, the simply supported midspan case. B: 4.00 mm uses an 8EI denominator, which belongs to a cantilever under distributed load (wL⁴/8EI). D: 29.6 mm swaps b and h in I, as if the beam were laid flat.
Principle: MIT OCW 3.11, Stresses in Beams, Eqs. 5 and 7 (I = bh³/12, σ = My/I); "Shear stresses," p. 10 (3V/2A, following Eq. 12); MIT 3.11, Beam Displacements, "Energy method" (Castigliano's theorem), Example 4. Topic: NCEES MDM specification, 2C Bending (deflection).
Question 10 · MDM-10
Mechanics of Materials — 2D Buckling
A solid steel rod 20 mm in diameter and 1.2 m long is fixed at its base and free at its top (a flagpole). E = 200 GPa. What is its Euler critical buckling load?
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Answer: A. I = πd⁴/64 = 7854 mm⁴. Fixed-free gives an effective length of 2L = 2,400 mm. Pcr = π²EI/(Le)² = π² × 200,000 × 7854 ÷ 2,400² = 2692 N ≈ 2.69 kN. Check that Euler applies: r = d/4 = 5 mm, so Le/r = 480, a very slender column.
Why the others are wrong: B: 10.8 kN uses Le = L (pinned-pinned). C: 22.0 kN uses Le = 0.7L (fixed-pinned). D: 43.1 kN uses Le = 0.5L (fixed-fixed).
Principle: Universiti Malaysia Pahang OCW, Columns, "Use of the Euler Formula"; Sec. 9.3, effective lengths. Topic: NCEES MDM specification, 2D Buckling.
Question 11 · MDM-11
Mechanics of Materials — 2E Torsion
A hollow steel shaft (outside diameter 50 mm, inside diameter 40 mm, G = 79.3 GPa) is 1.5 m long and carries a steady torque of 1.2 kN·m. What is the angle of twist?
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Answer: C. J = π(do⁴ − di⁴)/32 = 362,265 mm⁴. θ = TL/(GJ) = 1.2×10⁶ N·mm × 1,500 mm ÷ (79,300 × 362,265) = 0.0627 rad = 3.59°.
Why the others are wrong: A: 1.42° uses the elastic modulus E where the shear modulus G belongs. B: 2.12° uses J for a solid 50 mm shaft. D: 0.0627 is the answer in radians, labeled as degrees.
Principle: MIT OCW 3.11, Torsion, p. 8, Eqs. 12–13 (J, θ = TL/GJ). Topic: NCEES MDM specification, 2E Torsion.
Question 12 · MDM-12
Mechanics of Materials — 2F Combined loading
A short rectangular block, 30 mm wide and 60 mm deep, carries a 40 kN compressive load applied parallel to its axis but offset 10 mm from the centroid along the 60 mm direction. Neglect buckling. What is the maximum compressive stress?
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Answer: A. Superpose axial and bending stress. P/A = 40,000 ÷ 1,800 = 22.2 MPa. M = 40,000 × 10 = 400,000 N·mm; I = 30 × 60³ ÷ 12 = 540,000 mm⁴; Mc/I = 400,000 × 30 ÷ 540,000 = 22.2 MPa. Maximum compression = 44.4 MPa on the side nearer the load; the far side is at 0 because the load sits exactly at the edge of the kern (e = h/6).
Why the others are wrong: B: 0 MPa is the stress on the far face, the minimum, not the maximum. C: 22.2 MPa keeps only one of the two terms (here they happen to be equal). D: 111 MPa swaps b and h in I (I = 135,000 mm⁴).
Principle: OpenStax University Physics Vol. 1, 12.3, Eq. 12.36 (Young's modulus) and Eq. 12.42 (shear stress); MIT OCW 3.11, Stresses in Beams, Eqs. 5 and 7 (I = bh³/12, σ = My/I); "Shear stresses," p. 10 (3V/2A, following Eq. 12). Topic: NCEES MDM specification, 2F Combined loading.
Question 13 · MDM-13
Mechanics of Materials — 2G Static failure
A 40 mm solid shaft of ductile steel (Sy = 350 MPa) carries a bending moment of 600 N·m and a torque of 800 N·m at its critical section. Ignoring stress concentration, what is the factor of safety against yielding by the distortion-energy (von Mises) theory?
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Answer: D. σ = 32M/(πd³) = 95.5 MPa and τ = 16T/(πd³) = 63.7 MPa. With σy = 0, the plane-stress von Mises equation reduces to σ′ = √(σ² + 3τ²) = 145.9 MPa. n = 350 ÷ 145.9 = 2.40.
Why the others are wrong: A: 1.20 compares σ′ with Sy/2, mixing the von Mises stress with a shear-yield limit. B: 2.20 is the maximum-shear-stress (Tresca) result: (Sy/2) ÷ √((σ/2)² + τ²). More conservative, but not what was asked. C: 3.67 ignores the torque.
Principle: IIT Kharagpur, Static Failure Theories, slide 26 of 39, distortion-energy theory (plane stress). Topic: NCEES MDM specification, 2G Static failure.
Question 14 · MDM-14
Mechanics of Materials — 2H Fatigue failure
A steel part sees a fluctuating normal stress with alternating component σa = 120 MPa and mean component σm = 150 MPa. Its fully corrected endurance limit is Se = 210 MPa, Sut = 600 MPa, and Sy = 450 MPa. What is the fatigue factor of safety by the modified Goodman criterion?
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Answer: D. Modified Goodman: σa/Se + σm/Sut = 1/n → 120/210 + 150/600 = 0.571 + 0.250 = 0.821, so n = 1.22. It's above 1 but not by much; check first-cycle yielding separately ((σa + σm) = 270 MPa vs. Sy = 450 MPa is fine here).
Why the others are wrong: A: 0.82 is 1/n, not n. B: 1.11 is the Soderberg result, which uses Sy in place of Sut. C: 1.75 ignores the mean stress (Se/σa).
Principle: IIT Madras ME2200, Fatigue module, "Mean Stress Failure Envelopes" (Modified Goodman, Soderberg). Topic: NCEES MDM specification, 2H Fatigue failure.
Question 15 · MDM-15
Mechanics of Materials — 2H Fatigue failure (notches)
A shaft shoulder has a theoretical stress concentration factor Kt = 2.0 and a notch sensitivity q = 0.80. The nominal alternating bending stress at the shoulder is 60 MPa. What alternating stress should you use in the fatigue check? Enter MPa.
Enter a number in MPa.
Show answer and explanation
Answer: 108 MPa (accept ±0.5 MPa). Kf = 1 + q(Kt − 1) = 1 + 0.80(2.0 − 1) = 1.80. Alternating stress = 1.80 × 60 = 108 MPa.
Common misses: 120 MPa applies the full Kt; real notches are less damaging in fatigue than Kt suggests, which is what q captures. 48 MPa multiplies by q alone.
Principle: WPI ME3320, Lecture 14, slide "Notches and stress concentrations". Topic: NCEES MDM specification, 2H Fatigue failure (notches).
Question 16 · MDM-16
Mechanics of Materials — 2I Thermal stresses and interference stresses
At 20°C a steel bearing has an outside diameter of 80.040 mm and an aluminum housing bore measures 80.000 mm. To install the bearing, only the housing is heated. With α = 23 × 10⁻⁶ /°C for aluminum, what housing temperature gives 0.020 mm of diametral clearance?
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Answer: A. The bore must grow by the 0.040 mm interference plus 0.020 mm clearance = 0.060 mm. ΔD = αDΔT, so ΔT = 0.060 ÷ (23×10⁻⁶ × 80) = 32.6°C, and T = 20 + 32.6 = 52.6°C.
Why the others are wrong: B: 32.6°C is the temperature rise, not the final temperature. C: 41.7°C removes the interference but leaves no assembly clearance. D: 85.2°C uses the radius (40 mm) with a diametral change.
Principle: OpenStax University Physics Vol. 2, 1.3, Eq. 1.2, ΔL = αLΔT. Topic: NCEES MDM specification, 2I Thermal stresses and interference stresses.
3. Mechanical Attachments (5 questions · NCEES range 9–14)
Question 17 · MDM-17
Mechanical Attachments — 3A Bonds (welds)
A plate is joined to a support by two parallel (longitudinal) 1/4 in equal-leg fillet welds, each 4.0 in long. The joint carries 12,000 lbf parallel to the welds. Assuming the load is shared equally, what is the average shear stress on the weld throat?
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Answer: B. Throat t = 0.707 × leg = 0.177 in. Throat area = 2 welds × 4.0 in × 0.177 in = 1.414 in². τ = 12,000 ÷ 1.414 = 8,487 psi.
Why the others are wrong: A: 6,000 psi uses the leg size instead of the throat. C: 12,000 psi uses half the leg as the throat. D: 16,970 psi counts only one weld.
Principle: NPTEL (IIT Roorkee), Welded joints, Secs. 25.1–25.2 (throat = 0.707 × leg). Topic: NCEES MDM specification, 3A Bonds (welds).
Question 18 · MDM-18
Mechanical Attachments — 3C Threaded fasteners (preload)
An M12 bolt has a tensile stress area of 84.3 mm² and a proof strength of 580 MPa. The design calls for a preload equal to 75% of proof load, and the nut factor is K = 0.20 (plain, dry). Using T = K·F·d with d = 12 mm, what tightening torque is required? Enter N·m to the nearest whole number.
Enter a number in N·m.
Show answer and explanation
Answer: 88 N·m (accept ±1 N·m). Proof load = 84.3 × 580 = 48,894 N; preload Fi = 0.75 × 48,894 = 36,670 N. T = 0.20 × 36,670 N × 0.012 m = 88.0 N·m ≈ 88 N·m.
Common misses: 117 N·m tightens to 100% of proof load. 88,000 leaves the diameter in millimeters, giving N·mm. Torque is only an indirect way to set preload: K varies widely with lubrication and finish, so the same torque can give quite different clamp forces.
Principle: Fastenal Torque-Tension chart, formula T = K·D·F and K-value notes. Topic: NCEES MDM specification, 3C Threaded fasteners (preload).
Question 19 · MDM-19
Mechanical Attachments — 3C Threaded fasteners (joint loading)
A bolt is preloaded to Fi = 20 kN in a joint where the member stiffness is three times the bolt stiffness (km = 3kb). An external separating load P = 10 kN is then applied to that bolt's share of the joint. What is the total bolt tension?
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Answer: B. Joint constant C = kb/(kb + km) = 1/4. Before separation the bolt picks up only C·P: Fb = Fi + CP = 20 + 0.25 × 10 = 22.5 kN. The remaining 7.5 kN of P goes into unloading the clamped members.
Why the others are wrong: A: 20.0 kN assumes the preload doesn't change. C: 27.5 kN swaps the stiffnesses (C = 0.75). D: 30.0 kN adds all of P to the bolt, which only happens after the joint separates.
Principle: Çankaya University ME 307, Bolts (2), "Force Analysis of Bolted Joints in Tension"; "Static Loading Requirements"; University of Western Australia, Design notes: Assemblies, Eq. (3a), preloaded joint load sharing. Topic: NCEES MDM specification, 3C Threaded fasteners (joint loading).
Question 20 · MDM-20
Mechanical Attachments — 3B Non-threaded fasteners (pins)
A 12 mm clevis pin connects an eye bar between the two arms of a clevis. The bar pulls with 18 kN. What is the average shear stress in the pin?
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Answer: B. A clevis puts the pin in double shear: the load crosses two shear planes, one at each side of the eye bar. τ = F/(2A) = 18,000 ÷ (2 × π × 12² ÷ 4) = 18,000 ÷ 226.2 = 79.6 MPa.
Why the others are wrong: A: 39.8 MPa counts four shear planes; a single clevis pin has two. C: 62.5 MPa uses d² instead of πd²/4 for the area. D: 159 MPa treats it as single shear.
Principle: OpenStax University Physics Vol. 1, 12.3, Eq. 12.36 (Young's modulus) and Eq. 12.42 (shear stress). Topic: NCEES MDM specification, 3B Non-threaded fasteners (pins).
Question 21 · MDM-21
Mechanical Attachments — 3C Threaded fasteners (joint separation)
Select all that apply. Use the joint from MDM-19: Fi = 20 kN and km = 3kb, so C = 0.25. Which statements are true while the external load stays below the separation load?
Show answer and explanation
Answers: A, B, D. Below separation, Fb = Fi + CP and Fm = Fi − (1 − C)P (A, D). Separation happens when the member force reaches zero: (1 − C)P₀ = Fi, so P₀ = 20 ÷ 0.75 = 26.7 kN (B).
Why the others are wrong: C: C depends only on the stiffness ratio. Higher preload raises the separation load; it doesn't change the bolt's share of the external load.
Principle: Çankaya University ME 307, Bolts (2), "Force Analysis of Bolted Joints in Tension"; "Static Loading Requirements"; University of Western Australia, Design notes: Assemblies, Eq. (3a), preloaded joint load sharing. Topic: NCEES MDM specification, 3C Threaded fasteners (joint separation).
4. Power Transmission (5 questions · NCEES range 9–14)
Question 22 · MDM-22
Power Transmission — 4A Gears and gear trains; 4E Motors
A 3.0 kW motor running at 1,750 rpm drives a compound gear train: a 20-tooth pinion drives a 60-tooth gear, and an 18-tooth pinion on that shaft drives a 54-tooth output gear. Each mesh is 95% efficient. What is the output torque at full motor power?
Show answer and explanation
Answer: D. Motor torque T = P/ω = 3,000 ÷ (1,750 × 2π/60) = 16.37 N·m. Train ratio = (60/20)(54/18) = 9, so output speed is 194 rpm. Torque rises by the ratio and is reduced by each mesh: 16.37 × 9 × 0.95² = 133 N·m.
Why the others are wrong: A: 1.64 N·m divides by the ratio; a speed reducer multiplies torque. B: 140 N·m applies only one mesh efficiency. C: 147 N·m ignores the mesh losses.
Principle: OpenStax University Physics Vol. 1, 10.8, Eq. 10.31, P = τω. Topic: NCEES MDM specification, 4A Gears and gear trains; 4E Motors.
Question 23 · MDM-23
Power Transmission — 4A Gears and gear trains (forces)
A 20° pressure-angle spur pinion with module 3 mm and 24 teeth transmits 100 N·m. What is the radial (separating) force on the pinion?
Show answer and explanation
Answer: C. Pitch diameter d = mz = 3 × 24 = 72 mm. Tangential force Wt = T/(d/2) = 100,000 N·mm ÷ 36 mm = 2,778 N. Radial force Wr = Wt tan 20° = 1,011 N.
Why the others are wrong: A: 506 N uses the full pitch diameter as the lever arm. B: 2,778 N is the tangential force. D: 2,956 N is the total tooth force, Wt/cos 20°.
Principle: KHK, ABCs of Gearing (b), Eq. (2.8), p. 15, d = zm; University of Florida ABE 4171, Shaft Design, slides "Spur Gears" and "Tangential Force" (Wt = T/(D/2), Wr = Wt tan φ). Topic: NCEES MDM specification, 4A Gears and gear trains (forces).
Question 24 · MDM-24
Power Transmission — 4B Bearings
A cylindrical roller bearing has a calculated L10 life of 20,000 h. The equivalent radial load is increased by 50% with speed unchanged. What is the new L10 life?
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Answer: A. L10 = (C/P)^p million revolutions, with p = 10/3 for roller bearings. At constant speed, hours scale the same way: 20,000 ÷ 1.5^(10/3) = 20,000 ÷ 3.86 = 5,177 h. A 50% load increase costs about three-quarters of the life.
Why the others are wrong: B: 5,930 h uses the ball-bearing exponent, 3. C: 8,890 h uses an exponent of 2. D: 13,330 h assumes life is inversely proportional to load.
Principle: NSK, Dynamic Load Ratings and Fatigue Life, "Basic Life Calculation Formula". Topic: NCEES MDM specification, 4B Bearings.
Question 25 · MDM-25
Power Transmission — 4C Belts
A flat belt transmits 5.0 kW at a belt speed of 10 m/s. The coefficient of friction is 0.30 and the wrap angle on the smaller pulley is 160°. Ignoring centrifugal effects and assuming the belt is at the point of slipping, what is the tight-side tension?
Show answer and explanation
Answer: C. Net pull T1 − T2 = P/v = 5,000 ÷ 10 = 500 N. Wrap angle = 160° = 2.793 rad, so T1/T2 = e^(0.30 × 2.793) = 2.311. Then T2 = 500 ÷ (2.311 − 1) = 381 N and T1 = 881 N.
Why the others are wrong: A: 381 N is the slack-side tension. B: 500 N is only the difference in tensions. D: 819 N assumes a 180° wrap.
Principle: IIT Guwahati ME101, Lecture 15 (belt friction), "Belt Friction" slide (tension ratio e^(μβ), β in radians); Glasgow Caledonian Univ., Clutches and Belt Drives summary, Eq. 11B.1, P = (F1 − F2)v. Topic: NCEES MDM specification, 4C Belts.
Question 26 · MDM-26
Power Transmission — 4D Shafts and keys
A 40 mm shaft transmits 300 N·m through a square key 10 mm × 10 mm × 40 mm long. Half the key height bears on the hub. What is the bearing (crushing) stress on the key?
Show answer and explanation
Answer: C. Force at the shaft surface F = T/(d/2) = 300,000 N·mm ÷ 20 mm = 15,000 N. Bearing area = (h/2) × L = 5 × 40 = 200 mm², so σ = 75.0 MPa. (The shear stress, on w × L = 400 mm², is 37.5 MPa.)
Why the others are wrong: A: 18.8 MPa uses the full diameter as the lever arm and the full key height. B: 37.5 MPa is the key's shear stress, or the bearing stress using the full height. D: 150 MPa uses r/2 (10 mm) as the lever arm.
Principle: Al-Mustaqbal University, Keys and Couplings, Sec. 13.9, strength of a sunk key. Topic: NCEES MDM specification, 4D Shafts and keys.
5. Mechanical Components and Assemblies (9 questions · NCEES range 16–24)
Question 27 · MDM-27
Mechanical Components and Assemblies — 5A Pressurized vessels and piping
Using thin-wall membrane theory only (not a code thickness formula), what wall thickness keeps the hoop stress at 20,000 psi in a cylindrical tank with a 48 in inside diameter at 250 psi internal pressure?
Show answer and explanation
Answer: D. Hoop stress σ = pD/(2t) governs because it's twice the axial stress. t = pD/(2σ) = 250 × 48 ÷ 40,000 = 0.300 in. A real design adds the applicable code's formula, joint efficiency, and corrosion allowance.
Why the others are wrong: A: 0.075 in uses the axial formula with the radius. B: 0.150 in uses the axial formula pD/(4t). C: 0.600 in drops the 2 from the hoop formula.
Principle: MIT OCW 3.11, Pressure Vessels, Eqs. 2–3 (axial and hoop stress). Topic: NCEES MDM specification, 5A Pressurized vessels and piping.
Question 28 · MDM-28
Mechanical Components and Assemblies — 5B Hydraulic and pneumatic components
A double-acting hydraulic cylinder has a 4.0 in bore and a 2.0 in rod. At 2,000 psi, what force does it develop while retracting? Ignore friction and back pressure.
Show answer and explanation
Answer: D. Retracting, the pressure acts on the annulus: bore area minus rod area = π/4 × (4² − 2²) = 9.42 in². F = 2,000 × 9.42 = 18,850 lbf.
Why the others are wrong: A: 6,280 lbf uses the rod area alone. B: 24,000 lbf uses (D² − d²) without π/4. C: 25,130 lbf is the extend force on the full bore.
Principle: OpenStax University Physics Vol. 1, 14.3, Eq. 14.12 and Example 14.3 (p = F/A). Topic: NCEES MDM specification, 5B Hydraulic and pneumatic components.
Question 29 · MDM-29
Mechanical Components and Assemblies — 5C Beams, trusses, and frames
A horizontal boom 2.0 m long is pinned to a wall at one end. A cable runs from the boom's free end to a wall anchor 1.5 m directly above the pin. A 10 kN load hangs from the free end. Neglecting the boom's weight, what is the axial compression in the boom?
Show answer and explanation
Answer: B. The cable is the hypotenuse of a 1.5-2.0-2.5 triangle, so its vertical component is 0.6T and horizontal 0.8T. Moments about the pin: 0.6T × 2.0 = 10 × 2.0, so T = 16.7 kN. The boom pushes back against the cable's horizontal pull: 0.8 × 16.7 = 13.3 kN compression.
Why the others are wrong: A: 7.5 kN multiplies the load by 1.5/2.0 instead of dividing. C: 10.0 kN is the cable's vertical component, which balances the load. D: 16.7 kN is the cable tension.
Principle: OpenStax University Physics Vol. 1, 12.1, Eqs. 12.2 and 12.5 (equilibrium conditions). Topic: NCEES MDM specification, 5C Beams, trusses, and frames.
Question 30 · MDM-30
Mechanical Components and Assemblies — 5C Beams, trusses, and frames
A beam is simply supported at A (x = 0) and B (x = 4.0 m) and overhangs to C (x = 5.5 m). It carries a uniform load of 6 kN/m from A to B and a 10 kN point load at C. Ignoring self-weight, what is the largest bending moment magnitude in the beam?
Show answer and explanation
Answer: C. Reactions: ΣMA = 0 gives 4RB = 24(2) + 10(5.5), so RB = 25.75 kN and RA = 8.25 kN. The overhang moment at B is 10 × 1.5 = 15.0 kN·m (hogging). Between the supports, shear is zero at x = 8.25/6 = 1.375 m, where M = 5.67 kN·m (sagging). The largest magnitude is 15.0 kN·m at B.
Why the others are wrong: A: 5.67 kN·m is the largest positive moment, not the largest magnitude. B: 12.0 kN·m uses wL²/8 and ignores the overhang. D: 25.8 is the reaction at B, in kN, not a moment.
Principle: OpenStax University Physics Vol. 1, 12.1, Eqs. 12.2 and 12.5 (equilibrium conditions). Topic: NCEES MDM specification, 5C Beams, trusses, and frames.
Question 31 · MDM-31
Mechanical Components and Assemblies — 5D Springs
A helical compression spring has wire diameter 0.120 in, mean coil diameter 1.00 in, 8 active coils, and G = 11.5 × 10⁶ psi. How far does it deflect under a 50 lbf load? Enter inches to two decimal places.
Enter a number in in.
Show answer and explanation
Answer: 1.34 in (accept ±0.02 in). k = Gd⁴/(8D³Na) = 11.5×10⁶ × 0.120⁴ ÷ (8 × 1.00³ × 8) = 37.3 lbf/in. δ = F/k = 50 ÷ 37.3 = 1.34 in.
Common misses: Using total coils instead of active coils, or the outside diameter instead of the mean diameter, are the usual slips; both change the rate noticeably because D is cubed.
Principle: NPTEL (IIT Kharagpur), Helical springs, Eqs. 7.1.7 and 7.1.10. Topic: NCEES MDM specification, 5D Springs.
Question 32 · MDM-32
Mechanical Components and Assemblies — 5E Vibrating systems
A machine on spring isolators produces a 2,000 N rotating-unbalance force at a frequency three times the mounted natural frequency (r = 3). The damping ratio is 0.10. What force amplitude is transmitted to the floor?
Show answer and explanation
Answer: B. TR = √(1 + (2ζr)²) ÷ √((1 − r²)² + (2ζr)²) = √(1 + 0.36) ÷ √(64 + 0.36) = 0.145. Transmitted force = 2,000 × 0.145 = 291 N. Above r = √2, isolation works, and adding damping actually raises transmission a little.
Why the others are wrong: A: 222 N divides by r², ignoring both the −1 and damping. C: 250 N is the undamped result, 1/(r² − 1). D: 2,000 N assumes the isolators do nothing.
Principle: Arup Strutt help, Vibration isolation, transmissibility formula with damping ratio; Purdue ME, Vibration isolation, isolation only above r = √2. Topic: NCEES MDM specification, 5E Vibrating systems.
Question 33 · MDM-33
Mechanical Components and Assemblies — 5F Basic machines and mechanisms
Select all that apply. A four-bar linkage has a 7 cm ground link, a 2 cm input link pinned to the ground, a 9 cm coupler, and a 6 cm output link. Which statements are true?
Show answer and explanation
Answers: A, B, C. Shortest s = 2, longest l = 9, others p + q = 7 + 6 = 13. s + l = 11 < 13, so it's Grashof (A). The shortest link is adjacent to ground, giving a crank-rocker with the 2 cm link as the crank (B). Grounding the shortest link instead makes a double crank (C).
Why the others are wrong: D: With l = 14, s + l = 16 > 7 + 6 = 13, so no link can fully rotate: non-Grashof.
Principle: Carleton University, Linkage analysis (Ch. 6), Sec. 6.1.2, Grashof condition, Eq. 6.1, pp. 176–177. Topic: NCEES MDM specification, 5F Basic machines and mechanisms.
Question 34 · MDM-34
Mechanical Components and Assemblies — 5F Basic machines and mechanisms
A screw jack with a lead of 6 mm raises a 20 kN load. The overall efficiency of the screw is 35%. What input torque is needed to raise the load at steady speed?
Show answer and explanation
Answer: D. Work balance over one revolution: useful work out = W × lead; work in = T × 2π; efficiency = out ÷ in. So T = W·l/(2πη) = 20,000 × 0.006 ÷ (2π × 0.35) = 54.6 N·m.
Why the others are wrong: A: 6.68 N·m multiplies by the efficiency, which would give out more work than went in. B: 19.1 N·m is the frictionless (100% efficient) torque. C: 343 N·m leaves out the 2π.
Principle: OpenStax College Physics 2e, 7.6, "Efficiency," Eq. 7.68; Çankaya University ME 307, Bolts (1): power screws, power-screw efficiency slide. Topic: NCEES MDM specification, 5F Basic machines and mechanisms.
Question 35 · MDM-35
Mechanical Components and Assemblies — 5G Basic mechatronics (sensors and data acquisition)
A 0–250 bar pressure transducer outputs 0–10 V into a 12-bit analog-to-digital converter with a 0–10 V input range. What is the smallest pressure change the converter can resolve (one LSB)?
Show answer and explanation
Answer: A. LSB = full-scale range ÷ 2^N = 10 V ÷ 4,096 = 2.44 mV. The transducer maps 10 V to 250 bar, so one LSB = 250 ÷ 4,096 = 0.0610 bar. (Some references divide by 2^N − 1; the result rounds to the same value here.) Resolution isn't accuracy: transducer error and noise are usually larger.
Why the others are wrong: B: 0.0153 bar is a 14-bit converter. C: 0.977 bar is an 8-bit converter. D: 2.44 is the LSB in millivolts, not bar.
Principle: Texas Instruments Precision Hub, ADC code to voltage, "Determining LSB size," Eq. 2. Topic: NCEES MDM specification, 5G Basic mechatronics (sensors and data acquisition).
6. Supportive Knowledge (5 questions · NCEES range 8–12)
Question 36 · MDM-36
Supportive Knowledge — 6B Fits and tolerances
Three parts stack in a housing with lengths 20.00 ± 0.05 mm, 35.00 ± 0.10 mm, and 15.00 ± 0.05 mm. Assuming independent, centered, normally distributed variation, what is the statistical (root-sum-square) tolerance on the stack?
Show answer and explanation
Answer: A. RSS = √(0.05² + 0.10² + 0.05²) = √0.015 = ±0.122 mm. The worst-case stack is ±0.200 mm; RSS is tighter because all three parts rarely sit at their limits together.
Why the others are wrong: B: ±0.015 mm is the sum of squares before the square root. C: ±0.067 mm divides the worst case by 3. D: ±0.200 mm is the worst-case (arithmetic) stack.
Principle: NASA Preferred Practice PD-ED-1219, pp. 2 and 6–7, worst-case and RSS tolerance buildup. Topic: NCEES MDM specification, 6B Fits and tolerances.
Question 37 · MDM-37
Supportive Knowledge — 6E Instrumentation, testing, inspection, and quality
A turned diameter is specified 25.000 ± 0.030 mm. The process is stable with mean 25.012 mm and standard deviation 0.004 mm. What is Cpk? Enter to two decimal places.
Enter a number.
Show answer and explanation
Answer: 1.5 (accept ±0.01). USL = 25.030 and LSL = 24.970. Cpk = min[(USL − μ)/(3σ), (μ − LSL)/(3σ)] = min[0.018/0.012, 0.042/0.012] = min[1.50, 3.50] = 1.50.
Common misses: 2.50 is Cp, which ignores that the process is off center. 3.50 uses the farther limit.
Principle: NIST/SEMATECH e-Handbook, 6.1.6, process capability index definitions. Topic: NCEES MDM specification, 6E Instrumentation, testing, inspection, and quality.
Question 38 · MDM-38
Supportive Knowledge — 6F Chemical processes (corrosion and embrittlement)
Select all that apply. In a wet service environment, metal X is anodic to metal Y. Separately, some hardened steel fasteners (above 40 HRC) are zinc electroplated. Which statements are accurate?
Show answer and explanation
Answers: A, B, D. The anodic (less noble) metal corrodes, and a small anode coupled to a large cathode concentrates the corrosion current on the small part (A, B). Hardness increases susceptibility to hydrogen embrittlement, plating can introduce hydrogen, and baking after plating relieves much of it (D).
Why the others are wrong: C: The more noble metal is the cathode and is protected; the anode corrodes. E: Baking reduces the risk; it doesn't remove it entirely.
Principle: NIH ORF Technical Bulletin, Galvanic Corrosion (June 2026), Introduction; area-effect discussion; Fastenal Blue Print, Hydrogen embrittlement, susceptibility, plating, and baking. Topic: NCEES MDM specification, 6F Chemical processes (corrosion and embrittlement).
Question 39 · MDM-39
Supportive Knowledge — 6A Manufacturing methods (surface treatment)
A highly stressed steel spring fails in fatigue from cracks that start at its surface. Which treatment is applied specifically to put the surface into residual compression and improve fatigue life?
Show answer and explanation
Answer: C. Shot peening plastically deforms a thin surface layer, leaving compressive residual stress at and just below the surface. That compression offsets part of the tensile service stress right where fatigue cracks start.
Why the others are wrong: A: Stress relief reduces residual stresses; it doesn't create the surface compression this question is after. B: Plating hardened steel without baking adds hydrogen-embrittlement risk rather than fatigue protection. D: A larger wire lowers the working stress, which can help, but it puts no residual compression into the surface.
Principle: NASA/TM-2001-210843, shot peening, Introduction, pp. 1–2; Fastenal Blue Print, Hydrogen embrittlement, susceptibility, plating, and baking. Topic: NCEES MDM specification, 6A Manufacturing methods (surface treatment).
Question 40 · MDM-40
Supportive Knowledge — 6D Computational methods (numerical methods)
A spring test records force at equal 10 mm intervals: 0, 150, 320, 520, and 760 N at 0, 10, 20, 30, and 40 mm. Using the trapezoidal rule, how much work was done compressing the spring 40 mm?
Show answer and explanation
Answer: A. Work is the area under force vs. displacement. Trapezoidal rule: (Δx/2)[f₀ + 2f₁ + 2f₂ + 2f₃ + f₄] = (10/2)(0 + 300 + 640 + 1,040 + 760) = 13,700 N·mm = 13.7 J.
Why the others are wrong: B: 9.9 J uses left-endpoint rectangles, which underestimates a rising curve. C: 13.6 J is Simpson's rule, a different method (and a fine one, but not what was asked). D: 15.2 J assumes a straight line from 0 to 760 N; this spring stiffens.
Principle: OpenStax Calculus Vol. 2, 3.6, Theorem 3.4, trapezoidal rule. Topic: NCEES MDM specification, 6D Computational methods (numerical methods).
How to work this set
Open the NCEES PE Mechanical Reference Handbook from your MyNCEES account and use your exam calculator, then answer each question before opening its explanation. Units are mixed SI and U.S. Customary, as on the exam. The real PE Mechanical exam gives 8 hours of exam time for 80 questions, an average of 6 minutes per question; at that average pace, these 40 questions take about 4 hours. Work your first pass untimed if you're still learning the methods.
Score your set and find your weak areas
Your score on these 40 questions: ____ / 40. Count a select-all question correct only if you picked every correct option and nothing else, and a number-entry question correct only if it falls within the stated tolerance. That matches NCEES scoring: every question, including alternative item types, is right or wrong, with no partial credit (NCEES computer-based testing).
Mark your misses, and any question you got right by guessing, in this table.
| NCEES knowledge area | Questions on the real exam | Questions here | Your misses |
|---|---|---|---|
| 1. Basic Engineering Practice | 11–17 | MDM-01 to 06 (6) | |
| 2. Mechanics of Materials | 17–26 | MDM-07 to 16 (10) | |
| 3. Mechanical Attachments | 9–14 | MDM-17 to 21 (5) | |
| 4. Power Transmission | 9–14 | MDM-22 to 26 (5) | |
| 5. Mechanical Components and Assemblies | 16–24 | MDM-27 to 35 (9) | |
| 6. Supportive Knowledge | 8–12 | MDM-36 to 40 (5) | |
| Total | 80 | 40 |
The ranges are from the NCEES MDM specification, effective October 2025. We spread our 40 questions roughly in proportion to them. With five to ten questions per area, a tally points you at what to review next; it isn't a measure of how well you know the whole subject.
What your score means. It describes how you did on these 40 questions under the conditions you chose. It doesn't predict your NCEES result. NCEES converts your number of correct answers to a scaled score, compares it with a standard set by subject-matter experts, reports only pass or fail, and doesn't publish the passing score (NCEES Examinee Guide, May 2026, p. 14). Be wary of any "you need 70%" claim.
Turn each miss into a fix
| Question | First wrong step | What I'll change | Reworked a fresh problem correctly? |
|---|---|---|---|
| Example: MDM-24 | Used the ball-bearing exponent for a roller bearing | Check bearing type before choosing p = 3 or 10/3 | Yes / Not yet |
| Your miss: |
Then let the type of mistake decide your next session:
- Most misses in one area: work fresh problems there next, starting with the setups you got wrong.
- You knew the method but couldn't find the equation: spend a short session finding those relationships in the handbook and note the search words that worked. On exam day you search with a search box, and Ctrl+F doesn't work (Examinee Guide, p. 10).
- Setup or unit errors: rework the question from a blank page, writing the unknown, its units, and the governing relationship before any numbers. Then try a fresh variation.
- Misses spread evenly: move to a longer timed mixed set or a full rehearsal.
Getting this same question right a second time mostly tests your memory of the answer, so prove each fix on a fresh problem.
Right but slow? Count it as a miss for planning. At 6 minutes a question, a 15-minute solve costs you another question later.
What this set doesn't copy
The real exam mixes standard multiple choice with alternative item types: multiple correct, point and click, drag and drop, and fill in the blank (Examinee Guide, p. 11). This set uses single-answer, select-all, and number-entry questions only. Our questions follow the current topics but haven't been calibrated against NCEES difficulty, and they aren't NCEES questions.
For a full-length rehearsal, NCEES sells an 80-question PDF practice exam built on the October 2025 specification for $59.95 (NCEES exam prep). It's the only practice exam written by the organization that writes the test. If you use it, check the NCEES errata for that practice exam, posted April 2026, which corrects several of its solutions.
Using an older book or download? Before October 2025, the MDM specification grouped topics under "Principles" and "Applications" with different areas, such as Engineering Science and Mechanics and Material Properties (April 2020 specification). Material written to that version still teaches useful mechanics, but its topic weights and categories no longer match your exam.
PE Mechanical: Machine Design and Materials exam facts
| Detail | What NCEES says |
|---|---|
| Questions | 80 |
| Appointment | 9 hours: 2-minute nondisclosure agreement, 8-minute tutorial, 8-hour exam, 50-minute scheduled break |
| Format | Computer-based at Pearson test centers, year-round |
| Reference | NCEES PE Mechanical Reference Handbook, on screen. No design standards are supplied for PE Mechanical, and no personal references are allowed |
| Units | SI and U.S. Customary |
| Question types | Multiple choice plus alternative item types; no partial credit |
| Fee | $400 per attempt, paid to NCEES; your licensing board may charge its own application fee |
| Results | Usually 7–10 days, pass or fail; a diagnostic report by knowledge area if you don't pass |
| Specification | Effective October 2025 |
Sources: NCEES PE Mechanical page; MDM specification; Examinee Guide, May 2026, pp. 3, 10–11, 14, 16.
What each knowledge area covers
- 1. Basic Engineering Practice (11–17): engineering terms, symbols, and drawings; project management and economic analysis; design methodology (requirements, risk assessment, verification and validation); physical properties of materials.
- 2. Mechanics of Materials (17–26): axial, shear, bending, buckling, torsion, combined loading, static failure, fatigue failure, thermal and interference stresses.
- 3. Mechanical Attachments (9–14): bonds such as welds, brazing, and adhesives; non-threaded fasteners such as pins, rivets, and retaining rings; threaded fasteners.
- 4. Power Transmission (9–14): gears and gear trains; bearings; belts, chains, clutches, brakes, and power screws; shafts and keys; motors and engines.
- 5. Mechanical Components and Assemblies (16–24): pressure vessels and piping; hydraulic and pneumatic components; beams, trusses, and frames; springs; vibrating systems; basic machines and mechanisms; basic mechatronics.
- 6. Supportive Knowledge (8–12): manufacturing methods; fits and tolerances; codes and standards; computational methods; instrumentation, testing, inspection, and quality; chemical processes such as corrosion and embrittlement.
Mechanics of Materials and Mechanical Components and Assemblies together account for 33–50 of the 80 questions, so they deserve the biggest share of your practice time.
Exam-day habits to practice now
- Practice with the handbook you'll get. It's supplied on screen as a searchable PDF, and you can download the current version from your MyNCEES dashboard (Examinee Guide, p. 10; NCEES handbook help).
- Use one approved calculator. You may bring one NCEES-approved model, and an on-screen TI-30XS is available. Check your model against the calculator policy linked from the NCEES exams page (Examinee Guide, p. 8).
- Answer everything. There's no deduction for wrong answers, and unscored pretest questions look like every other question (Examinee Guide, pp. 11 and 14).
- Clear your flags before the midpoint. After about half the questions you review and submit them, and you can't go back (Examinee Guide, p. 11).
- Practice on a whiteboard surface. You get two reusable booklets and three markers, not paper and pencil (Examinee Guide, p. 9).
PE MDM pass rates
| Group (January–June 2026 examinees) | Takers | Pass rate |
|---|---|---|
| First-time takers | 401 | 61% |
| Repeat takers | 140 | 39% |
From the NCEES table updated July 2026, covering examinees under NCEES member boards (NCEES PE Mechanical page). That made MDM's first-time rate the lowest of the three PE Mechanical exams in that period (HVAC and Refrigeration 71%, Thermal and Fluid Systems 70%). Older figures still circulate online; use the NCEES table. These rates describe groups of examinees, not your personal odds.
Common questions
Are these questions as hard as the real exam? We can't promise that. They're written to the current topics with realistic setups, but they haven't been calibrated against NCEES questions. Treat your result as feedback on these 40 questions.
How many do I need right to pass? NCEES doesn't publish a number. The passing score varies slightly with the difficulty of each exam form, and first-time and repeat takers are held to the same standard (Examinee Guide, p. 14).
Do I need to pass the FE first, or have four years of experience? NCEES designs the PE for engineers with at least four years of post-college experience, but your state licensing board decides who may sit (NCEES PE Mechanical page; Examinee Guide, p. 2). Find your board in the NCEES board directory. Passing the exam doesn't by itself grant a license.
How soon can I retake it? Once per calendar-quarter testing window and no more than three times in any 12 months; some boards are stricter (Examinee Guide, p. 5). If you didn't pass, start from your diagnostic report and weigh each area by both your gap and its number of questions.
Next steps
- Build the full plan: PE exam prep: 12-week study plan and error log
- Ready to book? How NCEES exam registration works
- Need accommodations? Requesting NCEES accommodations (you request them during registration)
- After the exam: NCEES exam results · NCEES retake policy
- Still on the FE? Free FE Mechanical practice test
Sources and verification
Last verified October 8, 2026. The current exam format, appointment time, fee, references, scoring, retake rules, specification topics and ranges, and pass rates were rechecked against the NCEES documents below. In this audit, every numerical answer was independently recalculated from the stated inputs, answer-key and JSON synchronization were checked, and the cited teaching references were reviewed for fit to the principle they support. Where a source states a relationship in a different but equivalent form (for example, with radii instead of diameters, or as a load capacity instead of a stress), the explanation shows the rearrangement. This is editorial source checking, not professional engineering review.
Official sources:
- NCEES: PE Mechanical exam, including the pass-rate table updated July 2026
- NCEES PE Mechanical: Machine Design and Materials CBT specification, effective October 2025 (PDF)
- NCEES PE Mechanical: Machine Design and Materials specification, effective April 2020 (PDF, superseded)
- NCEES Examinee Guide, May 2026 (PDF)
- NCEES: Computer-based testing and alternative item types
- NCEES help: Exam reference handbooks
- NCEES PE Mechanical: Machine Design and Materials practice exam
- NCEES errata for the MDM practice exam (PDF, April 2026)
The teaching source for each question is linked in its explanation. How we check exam facts: our methodology. Found an error? Our corrections page explains how to report it.
Written by the Castleport Test Prep Editorial Team. AI-assisted tools were used to research, draft, and check this content.
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 VUE, or any state engineering licensing board. The practice questions 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.