PE Mechanical Machine Design and Materials Exam Prep
Start your PE Mechanical: Machine Design and Materials prep with this original, unofficial problem written to the NCEES specification effective October 2025. All 36 problems, worked solutions, and a 12-week study plan are free on this page.
Problem 1 · Static failure under bending plus torsion
MDM-MOM-01 · Mechanics of Materials · Select one · SI units
A point on the surface of a ductile steel shaft sees a bending stress σ = 120 MPa and a torsional shear stress τ = 60 MPa. All other stress components are zero. The yield strength is Sy = 350 MPa. Using the distortion-energy (von Mises) theory, what is the factor of safety against yielding?
Record your responses, then compare them with the worked solutions. This page does not automatically grade or save your answers.
Show answer and solution
Answer: B. 2.20.
For plane stress with σy = 0, the von Mises effective stress is σ′ = √(σ² + 3τ²) = √(120² + 3 × 60²) = √25,200 = 158.7 MPa. The factor of safety is n = Sy/σ′ = 350/158.7 = 2.20.
- A (2.06) is the maximum-shear-stress (Tresca) answer: τmax = √((120/2)² + 60²) = 84.9 MPa, so n = 350/(2 × 84.9). Tresca is a real failure theory, just a more conservative one. The question asked for distortion energy.
- C (2.61) uses √(σ² + τ²). The shear term needs the factor of 3.
- D (2.92) ignores the torsion entirely (350/120).
Read the theory the problem names before you reach for a formula. Both theories are in your toolkit, and they give different numbers.
Try a change: Make the shear stress 90 MPa and keep everything else. σ′ = √(120² + 3 × 90²) = 194.7 MPa, so n drops to 1.80.
Source: Stanford ME111, Lecture 21 (static failure), slide 5: plane-stress von Mises form; Roylance, MIT 3.11, Yield and Plastic Flow, p. 2 (Tresca) and p. 5 (von Mises). Topic: NCEES MDM specification, Oct 2025, 2G Static Failure; 2F Combined Loading.
By Castleport Test Prep Editorial Team · Exam facts last verified October 8, 2026
Topic weights · All 36 problems · Worked shaft example · What changed in October 2025 · 12-week plan
The exam at a glance
- 80 questions, computer-based, offered year-round at Pearson test centers
- 8 hours of exam time inside a 9-hour appointment (2-minute nondisclosure agreement, 8-minute tutorial, 50-minute scheduled break)
- $400 exam fee paid to NCEES; your licensing board may charge its own application fee
- One on-screen reference: the NCEES PE Mechanical Reference Handbook. No design standards are supplied for PE Mechanical, and you can't bring your own references.
- Both SI and U.S. customary units
- Traditional multiple choice plus alternative items (multiple correct, point-and-click, drag-and-drop, fill-in-the-blank), all scored right or wrong
- Pass/fail results, typically 7–10 days after the exam
- First-time pass rate: 61% of 401 examinees (January–June 2026); repeat takers: 39% of 140
Sources: NCEES PE Mechanical page; MDM specification, p. 1; NCEES Examinee Guide, May 2026, pp. 3, 11, 14, and 16.
What to study first: the 6 knowledge areas by weight
Mechanics of Materials and Mechanical Components and Assemblies carry the two biggest ranges. Together they account for 33 to 50 of the 80 questions, so that's where most of your hours should go.
| Rank | Knowledge area | Questions | What it covers | Practice here |
|---|---|---|---|---|
| 1 | Mechanics of Materials | 17–26 | Axial loading; shear and transverse loading; bending; buckling; torsion; combined loading; static failure; fatigue failure; thermal and interference stresses | Problems 1 and 8–15, plus the worked shaft example |
| 2 | Mechanical Components and Assemblies | 16–24 | Pressurized vessels and piping; hydraulic and pneumatic components; beams, trusses, and frames; springs; vibrating systems; basic machines and mechanisms; basic mechatronics (electromechanical interfaces, sensors, basic circuits, basic controls) | Problems 26–32 |
| 3 | Basic Engineering Practice | 11–17 | Engineering terms, symbols, and drawings; project management and economic analysis; design methodology (design requirements, risk assessment, verification and validation); physical properties of materials | Problems 2–7 |
| 4 (tie) | Mechanical Attachments | 9–14 | Bonds (welds, brazing, chemical bonds, adhesives); non-threaded fasteners (lugs, shackles, retaining rings, pins, anchors, rivets); threaded fasteners (screws, bolts, studs, anchors) | Problems 16–20 |
| 4 (tie) | Power Transmission | 9–14 | Gears and gear trains; bearings; belts, chains, clutches, brakes, and power screws; shafts and keys; motors and engines | Problems 21–25 |
| 6 | Supportive Knowledge | 8–12 | Manufacturing methods (material removal, heat treatment, assembly, additive manufacturing, forming, surface treatment); fits and tolerances; codes and standards; computational methods (FEA, CAE, numerical methods); instrumentation, testing, inspection, and quality; chemical processes (corrosion, oxidation, embrittlement) | Problems 33–36 |
The question ranges and topic lists come from NCEES. The ranking is ours, and it says nothing about which areas are hardest. Three things to keep in mind:
- The ranges overlap. Their minimums add to 70 and their maximums to 107, but every exam has exactly 80 questions, so you can't work out an exact count for any one area.
- NCEES's lists aren't exhaustive. The specification says the listed knowledge areas are examples, not complete or exclusive categories. Don't skip a related topic because its exact name isn't on the list.
- "Codes and Standards" is on the topic list, but no standards documents are supplied for PE Mechanical. Expect questions you can answer from engineering knowledge and the handbook, not lookups in a code book.
Sources: MDM specification, pp. 1–2; Examinee Guide, p. 16.
36 original practice problems with worked solutions
These are original Castleport problems written to the October 2025 Machine Design and Materials topic list. They are not NCEES questions, and this set is not a full-length exam. It doesn't cover every listed subtopic: there's nothing here yet on trusses and frames, clutches and brakes, chains, interference fits, or heat treatment, so give those their own study time.
Most problems are standard multiple choice. A few use formats like the real exam's alternative items: select all that apply, enter a number, and matching. The real exam also uses point-and-click items, which a text page can't reproduce. On the actual exam, every question is scored right or wrong, with no partial credit (Examinee Guide, p. 11).
Work each problem on paper before you open the solution. When you miss one, write down the first step that went wrong. The error log shows what to do with it.
Basic Engineering Practice (6 problems)
Problem 2 · Geometric tolerance categories
MDM-BEP-01 · Basic Engineering Practice · Match each item
Match each geometric characteristic on a drawing to its tolerance category.
| Item | Your match |
|---|---|
| 1. Flatness | ______ |
| 2. Perpendicularity | ______ |
| 3. Position | ______ |
| 4. Total runout | ______ |
Choices: Form · Orientation · Location · Runout
Show answer and solution
Flatness → form. Perpendicularity → orientation. Position → location. Total runout → runout.
- Form controls the shape of a single feature with no datum needed: flatness, straightness, circularity, cylindricity.
- Orientation controls a feature's angle relative to a datum: perpendicularity, parallelism, angularity.
- Location controls where a feature sits: position is the everyday example.
- Runout controls a surface as the part rotates about a datum axis, circular or total.
Profile (of a line or surface) is the fifth category. A quick way to sort a symbol: ask whether it needs a datum (form doesn't) and whether it's about angle, place, or rotation.
Source: Lin and Verma, ASEE Annual Conference 2007 (AC 2007-337), Table 1: geometric tolerance categories. Topic: NCEES MDM specification, Oct 2025, 1A Engineering Terms, Symbols, and Drawings.
Problem 3 · Compare present costs
MDM-BEP-02 · Basic Engineering Practice · Select one
Two machines meet the same production requirement and will be used for two years.
| Payment | Machine A | Machine B |
|---|---|---|
| Purchase now | $12,000 | $14,000 |
| Maintenance, end of year 1 | $2,000 | $500 |
| Maintenance, end of year 2 | $2,000 | $500 |
Use a 10% annual discount rate. Neither machine has any value at the end of year 2. Ignore taxes and every other cost. Which machine has the lower present cost, and by about how much?
Show answer and solution
Answer: C. Machine B, by $603.
Bring each future payment back to today with P = F/(1 + i)ⁿ.
- Machine A: 12,000 + 2,000/1.10 + 2,000/1.10² = $15,471.07
- Machine B: 14,000 + 500/1.10 + 500/1.10² = $14,867.77
- Difference: $603.31 in B's favor.
A compares purchase prices only. B has the right size but the wrong winner. D adds the payments without discounting (A totals $16,000, B totals $15,000). Put every cash flow at its actual date before you compare.
Source: NISTIR 89-4203, Discount Factor Tables for Life-Cycle Cost Analyses (1989), p. 1, Eqs. 2 and 4. Topic: NCEES MDM specification, Oct 2025, 1B Project Management and Economic Analysis.
Problem 4 · Net present worth of an upgrade
MDM-BEP-03 · Basic Engineering Practice · Select one
An equipment upgrade costs $50,000 today and saves $8,000 at the end of each year for 10 years. It has no salvage value. At an interest rate of 7% per year, what is the upgrade's net present worth?
Show answer and solution
Answer: B. +$6,190.
The uniform present-value factor is (P/A, 7%, 10) = [(1.07)¹⁰ − 1]/[0.07 × (1.07)¹⁰] = 7.0236. Present worth of the savings: 8,000 × 7.0236 = $56,189. Net present worth: 56,189 − 50,000 = +$6,189, about +$6,190. A positive value means the upgrade beats the 7% rate.
- A subtracts $50,000 from the undiscounted $80,000 of savings.
- C discounts the whole $80,000 as if it arrived in one lump at year 10 ($40,668), then subtracts the cost.
- D forgets to subtract the $50,000 first cost.
Source: NISTIR 89-4203, Discount Factor Tables for Life-Cycle Cost Analyses (1989), p. 1, Eqs. 2 and 4. Topic: NCEES MDM specification, Oct 2025, 1B Project Management and Economic Analysis.
Problem 5 · Risk priority number
MDM-BEP-04 · Basic Engineering Practice · Select one
Your team rates failure modes on 1–5 scales, where 5 is worst for severity and likelihood, and 5 means the failure is least likely to be detected. A seal leak is rated severity 4, likelihood 3, and detection 2. What is its risk priority number (RPN)?
Show answer and solution
Answer: C. 24.
RPN is the product of the three ratings: 4 × 3 × 2 = 24.
- A (9) adds the ratings.
- B (12) leaves out detection.
- D (60) uses the worst detection rating (5) instead of the rated 2.
One caution worth carrying into the exam: a high-severity failure mode can end up with a modest RPN if it's rare and easy to detect. That's a known blind spot of RPN, so don't rank risks by the product alone. Scales also vary: some organizations use 1–5, others 1–10, so use whatever scale the problem gives you.
Source: Lindsey, NASA GSFC, Using FMECA as a Risk Assessment and Communication Tool, §2.2 and Table 5. Topic: NCEES MDM specification, Oct 2025, 1C Design Methodology (risk assessment).
Problem 6 · Verification or validation?
MDM-BEP-05 · Basic Engineering Practice · Select one
A team is qualifying a new welding fixture. Which activity is validation rather than verification?
Show answer and solution
Answer: C. Operators use the prototype fixture in the real production cell to confirm it does the job they need.
C. Verification asks whether the product was built right: does it meet its stated requirements? Validation asks whether the right product was built: does it satisfy the users' needs in its intended operating environment? Only C tests the fixture in its real use with its real users.
A, B, and D all check the product against written requirements (a specification limit, a drawing, and the "shall" statements), so they're verification. A product can pass every verification check and still fail validation if the requirements missed what users actually needed.
Source: NASA Systems Engineering Handbook, §5.0, §5.3 (verification) and §5.4 (validation). Topic: NCEES MDM specification, Oct 2025, 1C Design Methodology (verification and validation).
Problem 7 · Stiffness per unit weight
MDM-BEP-06 · Basic Engineering Practice · Select one · SI units
You're choosing a material for a lightweight axial tension link where stiffness per unit mass (E/ρ) matters most. Use these values:
| Material | E (GPa) | ρ (kg/m³) |
|---|---|---|
| Steel | 200 | 7,850 |
| Aluminum alloy | 69 | 2,700 |
| Titanium alloy | 114 | 4,430 |
Which statement is correct?
Show answer and solution
Answer: D. All three are within about 1% of each other, so the material choice barely changes axial stiffness per unit mass.
Divide E by ρ for each:
- Steel: 200 × 10⁹/7,850 = 25.5 × 10⁶ N·m/kg
- Aluminum: 69 × 10⁹/2,700 = 25.6 × 10⁶ N·m/kg
- Titanium: 114 × 10⁹/4,430 = 25.7 × 10⁶ N·m/kg
The answer is D. Aluminum is about a third as stiff as steel, but it's also about a third as dense, so the ratios nearly cancel. A and B compare density alone or modulus alone. C has no support in the numbers.
The lesson: for a link loaded in axial tension, switching among these three alloys doesn't buy stiffness per kilogram. If one of them wins, it wins on some other requirement, such as strength, cost, corrosion, or the section shape you can fit.
Source: OpenStax, University Physics Vol. 1, §12.3: stress, strain, and Young's modulus. Topic: NCEES MDM specification, Oct 2025, 1D Physical Properties of Materials.
Mechanics of Materials (9 problems)
Problem 1 (static failure under bending plus torsion) is at the top of this page. It's the ninth Mechanics of Materials problem.
Problem 8 · Elongation of a rod
MDM-MOM-02 · Mechanics of Materials · Select one · U.S. customary units
A steel rod 0.75 in in diameter and 60 in long carries an axial tensile load of 10 kip. Use E = 29,000 ksi. How much does the rod stretch?
Show answer and solution
Answer: C. 0.0468 in.
Area: A = π(0.75)²/4 = 0.4418 in². Elongation: δ = PL/(AE) = (10 × 60)/(0.4418 × 29,000) = 0.0468 in.
- A uses πd² for the area (forgot the /4), which makes the area four times too big.
- B uses d² as the area, as if the rod were square.
- D uses half the correct area.
Keep units consistent: kip, in, and ksi work together directly.
Source: OpenStax, University Physics Vol. 1, §12.3: stress, strain, and Young's modulus. Topic: NCEES MDM specification, Oct 2025, 2A Axial Loading.
Problem 9 · Maximum transverse shear in a rectangular beam
MDM-MOM-03 · Mechanics of Materials · Select one · SI units
A solid rectangular beam is 40 mm wide and 100 mm deep. At one section it carries a transverse shear force V = 12 kN. What is the maximum transverse shear stress at that section?
Show answer and solution
Answer: C. 4.5 MPa.
For a rectangular section, shear stress peaks at the neutral axis at 1.5 times the average: τmax = 1.5V/A = 1.5 × 12,000/(40 × 100) = 4.5 MPa.
- B (3.0) is the average V/A. Shear isn't uniform across the depth: it's zero at the top and bottom and highest at mid-depth.
- D (6.0) uses a factor of 2.
- A (1.5) is half the average.
The 1.5 factor applies to rectangles only. Other shapes have different factors, so check the section before using it.
Source: Roylance, MIT 3.11, Stresses in Beams, p. 10 (maximum shear in a rectangular section). Topic: NCEES MDM specification, Oct 2025, 2B Shear and Transverse Loading.
Problem 10 · Bending stress in a cantilever
MDM-MOM-04 · Mechanics of Materials · Select one · U.S. customary units
A cantilever bar 24 in long carries a 500 lb load at its free end. The bar is a solid rectangle 1 in wide and 2 in deep, and the load acts parallel to the 2-in side. What is the maximum bending stress?
Show answer and solution
Answer: C. 18,000 psi.
The maximum moment in a cantilever is at the fixed end: M = PL = 500 × 24 = 12,000 in-lb. For bending about the strong axis, I = bh³/12 = 1 × 2³/12 = 0.667 in⁴ and c = 1 in, so σ = Mc/I = 12,000 × 1/0.667 = 18,000 psi. (Equivalently, S = bh²/6 = 0.667 in³.)
- D (36,000) bends the bar about its weak axis (S = hb²/6 = 0.333 in³).
- A (4,500) uses PL/4, the midspan moment of a simply supported beam.
- B (9,000) uses PL/2.
Sketch which way the load pushes before you pick b and h. Swapping them doubles the answer here.
Source: Roylance, MIT 3.11, Stresses in Beams, p. 4, Eqs. 5 and 7; Stanford ME111, Lecture 5, slides 5 and 9–10 (σ = Mc/I; I = πD⁴/64; J = πD⁴/32). Topic: NCEES MDM specification, Oct 2025, 2C Bending.
Problem 11 · Euler buckling load
MDM-MOM-05 · Mechanics of Materials · Select one · SI units
A solid round steel column is 30 mm in diameter and 1.2 m long, pinned at both ends. E = 200 GPa and Sy = 250 MPa. What is the Euler critical buckling load?
Show answer and solution
Answer: B. 54.5 kN.
I = πd⁴/64 = π(30)⁴/64 = 39,761 mm⁴. For pinned–pinned ends, the effective length is L. Pcr = π²EI/L² = π² × 200,000 × 39,761/(1,200)² = 54,500 N = 54.5 kN.
Check that Euler applies. The radius of gyration is r = d/4 = 7.5 mm, so the slenderness ratio is L/r = 160. The transition slenderness is √(2π²E/Sy) = √(2π² × 200,000/250) = 126. Because 160 > 126, this is a long column and Euler is the right model. Below the transition, an intermediate-column (Johnson) equation would apply instead.
- A uses an effective length of 2L (fixed–free).
- D uses 0.5L (fixed–fixed).
- C uses about 0.7L (fixed–pinned).
Source: Stanford ME111, Lecture 13 (columns), §§13.3–13.5; Stanford ME111, Lecture 5, slides 5 and 9–10 (σ = Mc/I; I = πD⁴/64; J = πD⁴/32). Topic: NCEES MDM specification, Oct 2025, 2D Buckling.
Problem 12 · Torsional shear stress in a solid shaft
MDM-MOM-06 · Mechanics of Materials · Select one · U.S. customary units
A solid steel shaft 1.25 in in diameter transmits a torque of 4,000 in-lb. What is the maximum torsional shear stress?
Show answer and solution
Answer: B. 10,430 psi.
τ = Tc/J with c = d/2 and J = πd⁴/32, which simplifies to τ = 16T/(πd³) = 16 × 4,000/(π × 1.25³) = 10,430 psi.
- C doubles the right answer, typically by using I (πd⁴/64) in place of J.
- A halves it.
- D treats the 0.625-in radius as the diameter, which multiplies the stress by 8.
Try a change: with a shaft 30 in long and G = 11,500 ksi, the twist is θ = TL/(JG) = 4,000 × 30/(0.2397 × 11,500,000) = 0.0435 rad, about 2.5°.
Source: Roylance, MIT 3.11, Torsion, p. 8, Eqs. 12–14; Stanford ME111, Lecture 5, slides 5 and 9–10 (σ = Mc/I; I = πD⁴/64; J = πD⁴/32). Topic: NCEES MDM specification, Oct 2025, 2E Torsion.
Problem 13 · Principal stress from Mohr's circle
MDM-MOM-07 · Mechanics of Materials · Select one · SI units
At a point in plane stress, σx = 80 MPa, σy = −20 MPa, and τxy = 30 MPa. What is the maximum principal stress?
Show answer and solution
Answer: C. 88.3 MPa.
The center of Mohr's circle is (σx + σy)/2 = 30 MPa. The radius is √[((σx − σy)/2)² + τxy²] = √(50² + 30²) = 58.3 MPa. So σ1 = 30 + 58.3 = 88.3 MPa and σ2 = 30 − 58.3 = −28.3 MPa. The principal plane is rotated θp = ½ tan⁻¹[2τxy/(σx − σy)] = 15.5° from x.
- A is the circle's center.
- B is the radius, which equals the maximum in-plane shear stress, not a principal stress.
- D adds σx and τxy directly.
Source: Roylance, MIT 3.11, Transformation of Stresses, pp. 4–6, Eqs. 11–12. Topic: NCEES MDM specification, Oct 2025, 2F Combined Loading.
Problem 14 · Fatigue safety factor from a stress cycle
MDM-MOM-08 · Mechanics of Materials · Select one · SI units
A part sees a repeated uniaxial stress cycle from 40 MPa to 200 MPa (both tensile). The fully corrected endurance limit is Se = 200 MPa, Sut = 600 MPa, and Sy = 450 MPa. Assume the mean and alternating stresses grow in proportion. What is the fatigue factor of safety using the Goodman line?
Show answer and solution
Answer: C. 1.67.
First split the cycle: σa = (200 − 40)/2 = 80 MPa and σm = (200 + 40)/2 = 120 MPa. The Goodman line gives 1/n = σa/Se + σm/Sut = 80/200 + 120/600 = 0.60, so n = 1.67.
- A uses the full 160 MPa range as the amplitude. Amplitude is half the range.
- B uses Sy in place of Sut. That's the Soderberg line, a more conservative criterion than the one asked for.
- D ignores the mean stress (Se/σa). Tensile mean stress shortens fatigue life, so leaving it out is unconservative.
Don't stop at fatigue: also check first-cycle yielding. The peak stress is 200 MPa, so n = 450/200 = 2.25. Fatigue governs here.
Source: Stanford ME111, Lecture 26 (fatigue), §§26.4–26.5 and §26.11. Topic: NCEES MDM specification, Oct 2025, 2H Fatigue Failure.
Problem 15 · Thermal stress with a gap
MDM-MOM-09 · Mechanics of Materials · Enter a number · U.S. customary units
A steel bar 40 in long sits between two rigid walls with a 0.010-in gap at one end. The bar is heated 100°F. Use α = 6.5 × 10⁻⁶ /°F and E = 29,000 ksi. Enter the compressive stress in the bar, in ksi, to one decimal place.
Your answer: ______ ksi
Show answer and solution
11.6 ksi. Free expansion would be δ = αΔTL = 6.5 × 10⁻⁶ × 100 × 40 = 0.026 in. The first 0.010 in just closes the gap. The walls block the remaining 0.016 in, so σ = E × (0.016/40) = 29,000 × 0.0004 = 11.6 ksi in compression.
Two common wrong answers: 18.9 ksi (EαΔT, ignoring the gap) and 0 (assuming the gap prevents all stress). Always compare the free expansion with the gap first.
Source: OpenStax, University Physics Vol. 2, §1.3: Eq. 1.2 and Thermal Stress, Example 1.4. Topic: NCEES MDM specification, Oct 2025, 2I Thermal Stresses and Interference Stresses.
Mechanical Attachments (5 problems)
Problem 16 · Shear stress in fillet welds
MDM-ATT-01 · Mechanical Attachments · Select one · SI units
A plate is joined by two parallel fillet welds, each 100 mm long with an 8-mm leg. The 60-kN load acts parallel to the welds. Taking the throat as 0.707 × leg, what is the average shear stress on the weld throats?
Show answer and solution
Answer: C. 53.0 MPa.
Throat: t = 0.707 × 8 = 5.66 mm. Total throat area: 5.66 × 100 × 2 welds = 1,131 mm². τ = 60,000/1,131 = 53.0 MPa.
- B uses the 8-mm leg instead of the throat. The weld fails through its smallest section, the throat.
- D counts only one weld.
- A counts the weld length twice (400 mm).
Source: NPTEL (IIT), Design of Welded Joints, Lecture 25, §§25.1–25.2. Topic: NCEES MDM specification, Oct 2025, 3A Bonds (welds).
Problem 17 · Tightening torque for a target preload
MDM-ATT-02 · Mechanical Attachments · Select one · U.S. customary units
A ½-in bolt needs a preload of 8,000 lb. Use T = KFd with a torque coefficient K = 0.20. What tightening torque is required?
Show answer and solution
Answer: B. 66.7 ft-lb.
T = KFd = 0.20 × 8,000 × 0.5 = 800 in-lb = 66.7 ft-lb.
- D leaves the answer in in-lb but labels it ft-lb.
- C uses d = 1 in.
- A uses K = 0.15.
NASA's fastener manual notes that 0.2 is the commonly assumed K, that a value near 0.15 is often more realistic for steel on steel, and that K shouldn't be used blindly. On an exam, use the K you're given.
Source: NASA RP-1228, Fastener Design Manual (1990), printed p. 16, Alternative Torque Formula. Topic: NCEES MDM specification, Oct 2025, 3C Threaded Fasteners.
Problem 18 · Pin in double shear
MDM-ATT-03 · Mechanical Attachments · Select one · U.S. customary units
A solid pin 0.500 in in diameter joins a center lug between two outer plates (a clevis). The lug pulls with 6.00 kip, and each outer plate carries half. Calculate average direct shear only. What is the average shear stress in the pin?
Show answer and solution
Answer: B. 15.3 ksi.
Pin area: A = π(0.500)²/4 = 0.1963 in². The pin is cut at two planes, one on each side of the lug, so each plane carries 3.00 kip. τ = 3.00/0.1963 = 15.3 ksi.
- C puts the whole 6 kip on one plane (single shear).
- A assumes four shear planes.
- D doubles the single-shear answer.
Sketch the cuts and count them. Lug bearing stress, pin bending, and net-section tension in the plates are separate checks.
Source: University of Illinois Mechanics Reference, Stress: Average Shear Stress and Units. Topic: NCEES MDM specification, Oct 2025, 3B Non-threaded Fasteners (pins).
Problem 19 · Eccentrically loaded bolt group
MDM-ATT-04 · Mechanical Attachments · Select one · SI units
Four identical bolts sit at the corners of a 100 mm × 100 mm square. A 20-kN vertical load acts 250 mm horizontally from the group's centroid, in the plane of the joint. What is the resultant shear force on the most heavily loaded bolt?
Show answer and solution
Answer: C. 21.5 kN.
Split the load into a direct shear plus a moment about the centroid.
- Direct shear per bolt: 20/4 = 5.0 kN, acting down.
- Moment: M = 20 × 250 = 5,000 kN·mm.
- Each bolt is r = √(50² + 50²) = 70.7 mm from the centroid, so Σr² = 4 × 5,000 = 20,000 mm².
- Moment shear per bolt: Mr/Σr² = 5,000 × 70.7/20,000 = 17.7 kN, perpendicular to each bolt's radius (at 45° here).
- On the worst bolt, the moment shear's vertical component (12.5 kN) adds to the 5.0 kN direct shear, and its horizontal component is 12.5 kN. Resultant: √(12.5² + 17.5²) = 21.5 kN.
- D adds 5.0 and 17.7 as scalars. They aren't parallel, so add them as vectors.
- B keeps only the moment part; A keeps only the direct part.
Source: NASA RP-1228, Fastener Design Manual (1990), printed p. 17, Finding Shear Loads on Fastener Group. Topic: NCEES MDM specification, Oct 2025, 3C Threaded Fasteners (bolt groups).
Problem 20 · Average shear in an adhesive lap joint
MDM-ATT-05 · Mechanical Attachments · Enter a number · SI units
A single-lap adhesive joint has an overlap 25 mm long and 40 mm wide. It carries a 5.0-kN tensile load. Enter the average shear stress in the adhesive, in MPa, to one decimal place.
Your answer: ______ MPa
Show answer and solution
5.0 MPa. τavg = P/(bL) = 5,000/(40 × 25) = 5.0 MPa.
That's an average, and the real distribution isn't uniform. In a single-lap joint the shear peaks near the ends of the overlap, and the middle carries much less. NASA's single-lap joint report calls the average shear stress an unsatisfactory basis for design, so treat this number as a first check, not a strength verdict.
Source: Hart-Smith, Adhesive-Bonded Single-Lap Joints, NASA CR-112236 (1973), introduction and Eq. 29. Topic: NCEES MDM specification, Oct 2025, 3A Bonds (adhesives).
Power Transmission (5 problems)
Problem 21 · Output torque of a compound gear train
MDM-PT-01 · Power Transmission · Select one · SI units
A motor delivers 10 kW at 1,800 rpm to a 20-tooth pinion that drives a 60-tooth gear. An 18-tooth gear on the same shaft as the 60-tooth gear drives a 54-tooth output gear. Ignore losses. What is the output torque?
Show answer and solution
Answer: C. 477 N·m.
Speed ratio: (60/20) × (54/18) = 3 × 3 = 9, so the output turns at 1,800/9 = 200 rpm. Convert to rad/s: ω = 200 × 2π/60 = 20.94 rad/s. With no losses, power in equals power out, so T = P/ω = 10,000/20.94 = 477 N·m. (Check: input torque 53.1 N·m × 9 = 477 N·m.)
- A is the input torque.
- B applies only the first stage (×3).
- D squares the overall ratio (×81).
P = Tω needs ω in rad/s. Using rpm directly is a classic slip.
Source: KHK USA, How does shaft alignment and speed ratio affect my design? (2020), parallel-axes section; MIT 2.72 Elements of Mechanical Design, Lecture 12, slide 34 (train ratio); MIT 2.004, Lecture 14, p. 14-2 (P = TΩ). Topic: NCEES MDM specification, Oct 2025, 4A Gears and Gear Trains; 4E Motors.
Problem 22 · Ball bearing rating life in hours
MDM-PT-02 · Power Transmission · Select one · SI units
A ball bearing has a basic dynamic load rating C = 30 kN. It carries a constant equivalent load P = 6 kN at 1,200 rpm. What is its basic rating life L10 in hours?
Show answer and solution
Answer: B. 1,736 h.
For ball bearings, L10 = (C/P)³ million revolutions = 5³ = 125 million rev. In hours: L10h = 125 × 10⁶/(60 × 1,200) = 1,736 h.
- C uses the roller-bearing exponent 10/3.
- A uses an exponent of 2.
- D forgets the 60 min/h conversion, so it's really minutes.
L10 is the life that 90% of a large group of identical bearings reach or exceed under these conditions. It isn't a promise about any single bearing.
Source: NTN, Large Bearings catalog engineering section, §1.2, Eq. 1.1 and Table 1.1; Zaretsky et al., NASA/TM-2003-212186, p. 3 Eq. 2 and p. 17. Topic: NCEES MDM specification, Oct 2025, 4B Bearings.
Problem 23 · Net belt pull
MDM-PT-03 · Power Transmission · Select one · U.S. customary units
A flat-belt drive transmits 10 hp. The driving pulley is 8 in in diameter and turns at 1,750 rpm. Ignore centrifugal effects. What is the net belt pull, F1 − F2?
Show answer and solution
Answer: B. 90.0 lb.
Belt speed: v = πdn = π × (8/12) ft × 1,750 rpm = 3,665 ft/min. Power = (F1 − F2)v, and 1 hp = 33,000 ft·lb/min, so F1 − F2 = 10 × 33,000/3,665 = 90.0 lb.
- C uses the radius instead of the diameter in v.
- A halves the answer.
- D is the pulley torque in in-lb (90 × 4 in), not a force.
Source: NPTEL (IIT Kharagpur), Introduction to Belt Drives, §13.1.9, Eq. 13.1.13. Topic: NCEES MDM specification, Oct 2025, 4C Belts, Chains, Clutches, Brakes, and Power Screws.
Problem 24 · Torque to raise a load with a power screw
MDM-PT-04 · Power Transmission · Select one · SI units
A square-thread power screw has a mean diameter of 25 mm and a lead of 5 mm. The thread friction coefficient is 0.10, and collar friction is negligible. What torque raises a 10-kN load?
Show answer and solution
Answer: C. 20.6 N·m.
For a square thread, the raising torque is T = (F·dm/2) × (l + πf·dm)/(π·dm − f·l) = (10,000 × 25/2) × (5 + π × 0.10 × 25)/(π × 25 − 0.10 × 5) = 125,000 × 12.85/78.04 = 20,590 N·mm = 20.6 N·m.
- A is the frictionless torque, Fl/(2π).
- B halves the right answer; D doubles it.
Two quick follow-ups from the same numbers. Efficiency = Fl/(2πT) = 7.96/20.6 = 39%. The screw is self-locking because f > tan λ (equivalently, πf·dm = 7.85 mm > l = 5 mm), so the load won't drive the screw down by itself.
Source: Hijazi, Hashemite University course notes, Screws and Fasteners (Ch. 8), pp. 3–4. Topic: NCEES MDM specification, Oct 2025, 4C Belts, Chains, Clutches, Brakes, and Power Screws.
Problem 25 · Shear stress in a key
MDM-PT-05 · Power Transmission · Select one · U.S. customary units
A ¼ in × ¼ in square key, 1.5 in long, connects a hub to a 1.25-in shaft carrying a torque of 2,000 in-lb. What is the average shear stress in the key?
Show answer and solution
Answer: C. 8,530 psi.
The key force acts at the shaft surface: F = T/(d/2) = 2,000/0.625 = 3,200 lb. It shears the key across its width × length: A = 0.25 × 1.5 = 0.375 in². τ = 3,200/0.375 = 8,530 psi.
- B divides the torque by the full diameter.
- D is the bearing (crushing) stress on half the key height, a separate check that's often the one that governs.
- A is the force, not a stress.
Source: Durali, Sharif University, Shafting Lecture 19, slide 17 (key shear). Topic: NCEES MDM specification, Oct 2025, 4D Shafts and Keys.
Mechanical Components and Assemblies (7 problems)
Problem 26 · Hoop stress in a thin-walled vessel
MDM-MCA-01 · Mechanical Components and Assemblies · Select one · U.S. customary units
A closed cylindrical tank has an inside diameter of 48 in and a 0.375-in wall. The internal pressure is 250 psi. What is the hoop (circumferential) stress?
Show answer and solution
Answer: B. 16,000 psi.
First check that thin-wall theory applies: r/t = 24/0.375 = 64, well above 10. Hoop stress: σθ = pr/t = 250 × 24/0.375 = 16,000 psi.
- A is the axial (longitudinal) stress, pr/(2t). Hoop stress is twice the axial stress, which is why long seams are the critical ones.
- C uses the diameter where the radius belongs.
- D is r/t, the thin-wall check, not a stress.
Source: Stanford ME111, Lecture 10 (pressure vessels), slides 3–4; Roylance, MIT 3.11, Pressure Vessels, p. 3, Eqs. 2–3. Topic: NCEES MDM specification, Oct 2025, 5A Pressurized Vessels and Piping.
Problem 27 · Retract force of a hydraulic cylinder
MDM-MCA-02 · Mechanical Components and Assemblies · Select one · SI units
A double-acting hydraulic cylinder has an 80-mm bore and a 40-mm rod. Supply pressure is 15 MPa. Ignoring friction and back pressure, what force does the cylinder produce while retracting?
Show answer and solution
Answer: B. 56.5 kN.
Retracting, the fluid pushes on the annulus: the bore area minus the rod area. A = π(80² − 40²)/4 = 3,770 mm². F = pA = 15 × 3,770 = 56,550 N = 56.5 kN.
- C is the extend force, using the full bore area (5,027 mm²).
- A uses the rod area alone.
- D adds the rod area to the bore area.
Same pressure, smaller area: a double-acting cylinder always retracts with less force than it extends.
Source: OpenStax, University Physics Vol. 1, §14.3: Pascal's principle, Eq. 14.12. Topic: NCEES MDM specification, Oct 2025, 5B Hydraulic and Pneumatic Components.
Problem 28 · Springs in series
MDM-MCA-03 · Mechanical Components and Assemblies · Select one · U.S. customary units
Two springs with rates 120 lb/in and 180 lb/in are connected end to end (in series) and carry the same load. What is the equivalent spring rate?
Show answer and solution
Answer: A. 72 lb/in.
In series, both springs carry the same force and their deflections add, so 1/k = 1/k1 + 1/k2. k = (120 × 180)/(120 + 180) = 72 lb/in.
- C is the parallel combination (k1 + k2), where both springs deflect the same amount and their forces add.
- B averages the rates.
- D multiplies them without dividing by the sum.
A quick sense check: a series combination is always softer than the softer spring.
Source: University of Alberta, Advanced Dynamics and Vibrations: Springs in Series and Springs in Parallel. Topic: NCEES MDM specification, Oct 2025, 5D Springs.
Problem 29 · Rate of a helical compression spring
MDM-MCA-04 · Mechanical Components and Assemblies · Select one · SI units
A helical compression spring has a 5-mm wire diameter, a 40-mm mean coil diameter, and 10 active coils. The shear modulus is G = 79.3 GPa. What is the spring rate?
Show answer and solution
Answer: C. 9.68 N/mm.
k = Gd⁴/(8D³N) = 79,300 × 5⁴/(8 × 40³ × 10) = 49,562,500/5,120,000 = 9.68 N/mm.
- D uses 20 mm (the mean radius) where the mean diameter belongs, which multiplies k by 8.
- B uses 16 instead of 8 in the denominator.
- A uses d³ instead of d⁴.
D is the mean coil diameter, not the outside diameter, and N counts only the active coils.
Source: NPTEL (IIT Kharagpur), Design of Springs, §7.1.5.4, Eq. 7.1.10. Topic: NCEES MDM specification, Oct 2025, 5D Springs.
Problem 30 · Natural frequency of a machine on mounts
MDM-MCA-05 · Mechanical Components and Assemblies · Enter a number · SI units
A 50-kg machine rests on isolation mounts with a combined stiffness of 200 kN/m. Treat it as an undamped single-degree-of-freedom system. Enter the natural frequency in hertz, to one decimal place.
Your answer: ______ Hz
Show answer and solution
10.1 Hz. ωn = √(k/m) = √(200,000/50) = 63.2 rad/s, and fn = ωn/(2π) = 10.07 Hz.
A common wrong entry is 63.2, which reports rad/s as Hz. Another is 0.3, which leaves k in kN/m.
Why it matters: 10.07 Hz is about 604 rpm. A motor running near 600 rpm on these mounts would sit almost on resonance, which is the opposite of isolation.
Source: OpenStax, University Physics Vol. 1, §15.1, Eqs. 15.9 and 15.11. Topic: NCEES MDM specification, Oct 2025, 5E Vibrating Systems.
Problem 31 · Grashof condition for a four-bar linkage
MDM-MCA-06 · Mechanical Components and Assemblies · Select one · SI units
A four-bar linkage has a fixed ground link of 100 mm, an input crank of 40 mm pinned to the ground, a coupler of 90 mm, and an output link of 70 mm pinned to the ground. What kind of linkage is it?
Show answer and solution
Answer: A. Grashof crank-rocker: the 40-mm crank can rotate fully.
Grashof's condition compares s + l with p + q, where s is the shortest link and l the longest. Here s + l = 40 + 100 = 140, and p + q = 90 + 70 = 160. Since 140 < 160, it's a Grashof linkage, so at least one link can rotate fully. The shortest link (40 mm) is a crank pinned to the ground, so this is a crank-rocker.
- B would need the shortest link to be the ground link.
- C applies when s + l > p + q.
- D applies when s + l = p + q exactly.
Source: UC3M OpenCourseWare, Machine Theory, Unit 2: Introduction to Kinematics (Grashof's law). Topic: NCEES MDM specification, Oct 2025, 5F Basic Machines and Mechanisms.
Problem 32 · Quarter-bridge strain gauge output
MDM-MCA-07 · Mechanical Components and Assemblies · Select one · SI units
A strain gauge with a gauge factor of 2.0 is wired in an initially balanced quarter Wheatstone bridge with 5.0-V excitation. The gauge sees 500 microstrain. Using the small-strain approximation, what is the bridge output voltage?
Show answer and solution
Answer: B. 1.25 mV.
For a quarter bridge, Vout ≈ Vex × GF × ε/4 = 5.0 × 2.0 × 500 × 10⁻⁶/4 = 1.25 mV.
- D drops the ¼.
- C uses ½, as if two active gauges shared the strain (a half bridge).
- A halves the correct answer.
Signals this small are why strain-gauge circuits need amplification and careful noise control.
Source: Cimbala, Penn State ME 345, Strain Gages, p. 4 (quarter-bridge output). Topic: NCEES MDM specification, Oct 2025, 5G Basic Mechatronics (sensors, basic circuits).
Supportive Knowledge (4 problems)
Problem 33 · Clearance limits
MDM-SK-01 · Supportive Knowledge · Select all that apply · SI units
A hole is 30.00 mm +0.03/−0.00. Its shaft is 29.97 ± 0.01 mm. Clearance is hole diameter minus shaft diameter; ignore form, temperature, and load effects. Which statements are correct? Select all that apply.
Show answer and solution
Answer: A and C.
A and C. Write all four limits first. Hole: 30.00 to 30.03 mm. Shaft: 29.96 to 29.98 mm.
- Minimum clearance = smallest hole − largest shaft = 30.00 − 29.98 = 0.020 mm. (A is correct.)
- Maximum clearance = largest hole − smallest shaft = 30.03 − 29.96 = 0.070 mm. (C is correct.)
- B is the nominal clearance (30.00 − 29.97), not a limit.
- D is false: even the smallest hole is bigger than the largest shaft, so this is a clearance fit.
On the real exam, a multiple-correct question earns credit only if every choice is right.
Source: City, University of London, ME1110 Lecture 7 (limits and fits), slides 13 and 19. Topic: NCEES MDM specification, Oct 2025, 6B Fits and Tolerances.
Problem 34 · Galvanic corrosion and hydrogen embrittlement
MDM-SK-02 · Supportive Knowledge · Select all that apply
Which statements are correct? Select all that apply.
Show answer and solution
Answer: A, B and C.
A, B, and C.
- A: The more active metal becomes the anode and corrodes; the farther apart two metals sit in the galvanic series, the stronger the effect.
- B: A small anode area against a large cathode concentrates the attack, so it's rapid and severe. Designers want the anode to be the larger area.
- C: Plating releases hydrogen, and embrittlement problems grow with fastener strength. That's why high-strength plated parts get a post-plating bake.
- D is false. Course guidance is to coat generally the cathode. A coated anode with a scratch in it becomes a tiny anode facing a huge cathode, which is the bad case from B.
Source: NASA RP-1228, Fastener Design Manual (1990), printed pp. 5–6: Galvanic Corrosion and Hydrogen Embrittlement; FAA AC 43-4B, Corrosion Control for Aircraft, ¶2.5.3; U.S. Naval Academy EN380, Ch. 5 Corrosion Types, §5.2, pp. 5-3 to 5-4. Topic: NCEES MDM specification, Oct 2025, 6F Chemical Processes (corrosion, embrittlement).
Problem 35 · Mesh convergence in FEA
MDM-SK-03 · Supportive Knowledge · Select all that apply
You're checking a finite element model of a bracket. Which statements are correct? Select all that apply.
Show answer and solution
Answer: A, B and D.
A, B, and D.
- A is the working definition of convergence.
- B: A perfectly sharp corner creates a stress singularity, so refining the mesh there won't give a converged peak stress. Model the real fillet radius, or judge the result away from the corner.
- D: Convergence studies refine where the stress changes quickly.
- C is false. Convergence only shows the mesh isn't the problem. Wrong loads, boundary conditions, or material data still give a wrong answer, which is why an independent hand check is worth doing.
Source: Getting Started with Abaqus/CAE, §4.4 Mesh convergence (MIT-hosted copy). Topic: NCEES MDM specification, Oct 2025, 6D Computational Methods.
Problem 36 · Choosing a nondestructive test
MDM-SK-04 · Supportive Knowledge · Match each item
Match each nondestructive test method to what it can find.
| Item | Your match |
|---|---|
| 1. Liquid penetrant | ______ |
| 2. Magnetic particle | ______ |
| 3. Ultrasonic | ______ |
| 4. Radiography | ______ |
Choices: Only flaws open to the surface, in non-porous materials (magnetic or not) · Surface and near-surface flaws, in ferromagnetic materials only · Internal (volumetric) flaws, using sound waves · Internal structure, recorded as an image; cracks show best when aligned with the beam
Show answer and solution
- Liquid penetrant: only discontinuities open to the surface. Works on any non-porous material.
- Magnetic particle: surface and some near-surface flaws, and it works only on ferromagnetic material. It's no use on aluminum or austenitic stainless.
- Ultrasonic: evaluates the internal (volumetric) condition, including subsurface flaws.
- Radiography: shows internal structure. A crack is easiest to detect when it lines up with the beam.
Pick the method by asking two questions: is the flaw likely to be at the surface or inside, and is the material ferromagnetic?
Source: Wisconsin DOT Structure Inspection Manual, Part 5, Ch. 4 (penetrant), §5.4.1; Wisconsin DOT Structure Inspection Manual, Part 5, Ch. 7 (magnetic particle), §§5.7.2–5.7.3; Wisconsin DOT Structure Inspection Manual, Part 5, Ch. 5 (ultrasonic), §5.5.1; Wisconsin DOT Structure Inspection Manual, Part 5, Ch. 9 (radiography), §§5.9.1–5.9.2. Topic: NCEES MDM specification, Oct 2025, 6E Instrumentation, Testing, Inspection, and Quality.
How to read your results
Treat your score as feedback on these 36 problems, not as an NCEES score or a prediction of passing. Four to nine problems per area is enough to show which methods you can't yet explain, but not enough to measure an area precisely. NCEES doesn't publish its passing score. It converts your number correct into a scaled score that adjusts for differences between exam forms, then reports pass or fail (Examinee Guide, p. 14). No practice percentage converts into an exam result.
Worked example: one shaft, two checks
This original, unofficial example ties several Mechanics of Materials topics together and shows why passing a yield check doesn't mean a shaft is safe in fatigue. The material properties and required factors are inputs we chose for the exercise, not design rules.
The setup. A solid steel shaft 25 mm in diameter turns in two simple bearings, A and B, 600 mm apart. A transverse force of 1,200 N acts 250 mm from A and stays fixed in direction while the shaft rotates (think of a belt pull). The shaft also carries a steady torque of 120 N·m at the critical section. Ignore shaft weight, axial load, transverse shear, stress concentrations, and dynamic effects.
Use Sy = 300 MPa, Sut = 550 MPa, and a fully corrected endurance limit Se = 140 MPa. The design calls for a yield factor of at least 2.0 and a fatigue factor of at least 1.5. For the fatigue check, use von Mises effective alternating and mean stresses on the Goodman line.
Try it yourself before reading on. The steps below go in order.
Step 1: Reactions and maximum moment
Take moments about A: RB × 600 = 1,200 × 250, so RB = 500 N. Then RA = 1,200 − 500 = 700 N.
The bending moment peaks under the load: M = 700 × 250 = 175,000 N·mm = 175 N·m. Check from the other side: 500 × 350 = 175,000 N·mm. ✓
Step 2: Surface stresses
For a solid circular shaft, σ = 32M/(πd³) and τ = 16T/(πd³). Convert the torque to N·mm first so the units match: 120 N·m = 120,000 N·mm.
- Bending: σ = 32 × 175,000/(π × 25³) = 114.1 MPa
- Torsion: τ = 16 × 120,000/(π × 25³) = 39.1 MPa
N/mm² is the same as MPa, so no further conversion is needed.
Step 3: Yielding
σ′ = √(σ² + 3τ²) = √(114.1² + 3 × 39.1²) = 132.7 MPa, so ny = 300/132.7 = 2.26. That meets the 2.0 requirement. But this check hasn't looked at fatigue yet.
Step 4: Follow one point as the shaft turns
Here's the step people skip. The force is fixed in space, but the shaft rotates under it. A point on the surface passes through the tension side, then the compression side, every revolution. So its bending stress swings from +114.1 to −114.1 MPa: fully reversed. The torque is steady, so its shear stress stays at 39.1 MPa.
That gives σa = 114.1 MPa, σm = 0, τa = 0, and τm = 39.1 MPa. Combining with von Mises:
- σ′a = 114.1 MPa
- σ′m = √3 × 39.1 = 67.7 MPa
A "steady" load doesn't mean steady stress at a point on a rotating part.
Step 5: Fatigue
1/nG = σ′a/Se + σ′m/Sut = 114.1/140 + 67.7/550 = 0.938, so nG = 1.07. That falls short of the required 1.5, even though the yield check passed comfortably.
What a bigger shaft does
Holding the loads and material values fixed:
| Diameter | Bending stress | Torsional shear | von Mises stress | Yield factor | Fatigue factor |
|---|---|---|---|---|---|
| 25 mm | 114.1 MPa | 39.1 MPa | 132.7 MPa | 2.26 | 1.07 |
| 30 mm | 66.0 MPa | 22.6 MPa | 76.8 MPa | 3.91 | 1.84 |
The 30-mm shaft meets both requirements in this simplified model. A real redesign would also revisit the size-dependent endurance corrections, plus keyways, shoulders, deflection, fits, and critical speed, none of which this example includes.
If you got a different answer, find the first step where you went wrong: reactions, units, combining stresses, or the stress history at the rotating point. Log it.
Sources: Roylance, MIT 3.11, Statics, p. 2, Eqs. 1–2; Stanford ME111, Lecture 5, slides 5 and 9–10; Stanford ME111, Lecture 21, slide 5; Stanford ME111, Lecture 26, §26.4, §26.11, and §26.13. The geometry and numbers are original.
What changed in October 2025
The October 2025 specification reorganized the exam into six knowledge areas. The engineering concepts carry over, but the topic map and question ranges are new, so older topic lists and practice exams don't match the current exam.
| October 2025 area (questions) | Closest April 2020 area(s) (questions) | What's visibly different |
|---|---|---|
| Basic Engineering Practice (11–17) | Basic Engineering Practice (7–11), plus physical properties from Material Properties (7–11) | Physical properties of materials now sit here. QA/QC is no longer listed in this area; testing, inspection, and quality now sit under Supportive Knowledge. |
| Mechanics of Materials (17–26) | Strength of Materials (9–14) | Wider range. Adds combined loading, static failure, and thermal and interference stresses as named topics. |
| Mechanical Attachments (9–14) | Joints and Fasteners (10–16) | Non-threaded fasteners (lugs, shackles, retaining rings, pins, anchors, rivets) are named separately. |
| Power Transmission (9–14) | Part of Mechanical Components (16–25) | Gears, bearings, belts and chains, clutches and brakes, power screws, shafts and keys, and motors are now their own area. |
| Mechanical Components and Assemblies (16–24) | The rest of Mechanical Components, plus Vibration (3–5) and Engineering Science and Mechanics (9–14) | Beams, trusses, and frames are now named. Vibration is folded in. Statics, kinematics, and dynamics are no longer a standalone area. |
| Supportive Knowledge (8–12) | Supportive Knowledge (9–14), plus the chemical side of Material Properties | Adds additive manufacturing, forming, and surface treatment. Corrosion, oxidation, and embrittlement are named. |
Both columns of ranges come straight from NCEES. The pairing between them is our reading, since NCEES doesn't publish a crosswalk.
Can you still use your older materials?
- Concept chapters on stress, fatigue, gears, bearings, and so on: yes, still useful.
- Topic lists, weightings, and study plans built on the old areas: replace them with the October 2025 specification.
- Older prep materials: check their topic map against the October 2025 specification. Material organized around the April 2020 Principles/Applications structure can still contain useful engineering practice, but its category labels and weightings are no longer current.
- Anything that tells you to bring reference books, or describes a morning breadth and afternoon depth session: that's from the old paper-and-pencil format. The computer-based exam has been closed book with an electronic reference since at least the April 2020 specification.
Sources: MDM specification, October 2025; MDM specification, April 2020.
Practice with the only reference you'll have
On exam day you get one reference on screen: the NCEES PE Mechanical Reference Handbook, as a searchable PDF. Log in to MyNCEES to download the current version now, and use it for every practice session so it feels familiar by test day.
Two details from the Examinee Guide (p. 10) change how you should practice:
- You search with a search box on the left side of the reference viewer. Ctrl+F doesn't work. Practice finding things by keyword and by the handbook's own section structure, not by a desktop shortcut.
- NCEES posts short video tutorials on using the on-screen references.
A five-minute lookup drill:
- Pick a problem from this page.
- Name the relationship you need in plain words, such as "spring rate" or "bearing life."
- Search the handbook for that phrase.
- Read the variable definitions and units next to the equation, plus any conditions on when it applies.
- Solve, then write down the search words that worked.
Keep a lookup log so slow searches get repeated until they're fast:
| Topic | Search term that worked | Where you found it | Time to find | What slowed you down |
|---|---|---|---|---|
| (example) Problem 1, von Mises | "distortion energy" | Your handbook's failure-theory section | 2 min | Searched "von Mises" first |
Because nothing beyond the handbook is supplied for PE Mechanical, anything not in it has to come from your own understanding.
Your 12-week study plan
Assumptions: about 10 focused hours a week (120 hours total) alongside a full-time job. This is our editorial plan, not an NCEES requirement, and no number of hours guarantees a pass. It covers every knowledge area and gives the most time to the two largest.
| Week | Focus | Do this | Finish the week with |
|---|---|---|---|
| 1 | Setup and baseline | Confirm your test date and download the handbook from MyNCEES. Work all 36 problems untimed before opening any solution, then check them and log every miss and lucky guess. | A completed error log and a list of your weakest areas |
| 2 | Mechanics of Materials I | Axial, shear, bending, torsion, and buckling. Rework Problems 8–12 and Steps 1–2 of the shaft example. Find each formula in the handbook. | Every formula located, with the search term logged |
| 3 | Mechanics of Materials II | Combined loading, static failure theories, fatigue, and thermal and interference stresses. Rework Problems 1 and 13–15 and Steps 3–5 of the shaft example. | A one-page note on which failure theory to use when, and why |
| 4 | Components and Assemblies I | Pressure vessels and piping, hydraulic and pneumatic components, springs. Rework Problems 26–29. | A worked set with units labeled on every line |
| 5 | Components and Assemblies II | Beams, trusses, and frames; vibrating systems; mechanisms; basic mechatronics. Rework Problems 30–32, and find fresh truss and frame problems. | Free-body sketches and a resonance check habit |
| 6 | Basic Engineering Practice | Drawings and symbols, engineering economics, design methodology and risk, material properties. Rework Problems 2–7. | Interest-factor problems solved from the handbook alone |
| 7 | Mechanical Attachments | Welds, bolts (preload, torque, eccentric groups), pins, rivets, adhesives. Rework Problems 16–20. | The bolt-group method written out once from memory |
| 8 | Power Transmission | Gear trains, bearings, belts and chains, clutches and brakes, power screws, shafts and keys, motors. Rework Problems 21–25. | Power–torque–speed conversions done in both unit systems |
| 9 | Supportive Knowledge, then your weakest area | Manufacturing, fits and tolerances, FEA, inspection, corrosion. Rework Problems 33–36. Spend about 4 hours on whichever area your log shows is weakest. | Every area has fresh problems solved without hints |
| 10 | Timed mixed blocks | Work Blocks A and B below, 108 minutes each, solutions closed until each block is done. | A record of time used, flagged questions, and slow lookups |
| 11 | Full rehearsal | Build an 80-question mixed rehearsal from practice material you are allowed to use, then work it under exam-like conditions: 8 hours in two halves, handbook and approved calculator only. If you do not have 80 fresh questions, use the largest fresh mixed set you can assemble and keep the same pacing discipline. | Results logged by area and by error type |
| 12 | Repair and rest | Rework every logged miss and repeat your five slowest lookups. Confirm logistics: ID name match, appointment email, calculator. Nothing new the day before. | A short repair list and a checked exam-day plan |
Week 10 mixed blocks. Each block has 18 problems, giving a time budget of 18 × 6 = 108 minutes. You'll have seen these problems before, so this is pacing and method practice, not a fresh exam.
| Block | Problem numbers, in order | Time budget |
|---|---|---|
| A | 1, 2, 8, 16, 21, 26, 33, 3, 9, 17, 22, 27, 34, 4, 10, 18, 28, 31 | 108 minutes |
| B | 5, 11, 19, 23, 29, 35, 6, 12, 20, 24, 30, 36, 7, 13, 14, 15, 25, 32 | 108 minutes |
Adjusting the plan:
- Six hours a week: stretch the calendar rather than cutting areas. The same 120 hours takes about 20 weeks.
- Fifteen hours a week: about 8 weeks. Keep the task order, the review time, and the full rehearsal.
- Only six weeks: combine weeks in pairs (1 and 2, 3 and 4, and so on). That's about 20 hours a week; if that isn't realistic, consider a later test date.
- Already strong in an area: keep a short review there and move the saved hours to your weakest area.
- Retaking the exam: start from your NCEES diagnostic report. Weigh both how far below passing examinees you were in each area and how many questions that area carries, then confirm each fix on problems you haven't seen (Examinee Guide, pp. 14 and 20–21).
Turn each miss into your next study task
| What went wrong | Next step | How you know it's fixed |
|---|---|---|
| Picked the wrong method or theory | Write one sentence on what the problem asks and which model fits, then rework it with one input changed | You can say why the wrong method doesn't apply |
| Wrong end condition, support, or load path | Redraw the free-body diagram and label every support and load | A second method (another moment point, a sum of forces) agrees |
| Unit slip (N·m vs N·mm, rpm vs rad/s, kN vs N) | Write the units on every line until they cancel to the answer's units | Units check without looking back |
| Radius vs diameter, or I vs J | List each section property and which dimension it uses before plugging in | Same problem, fresh numbers, right the first time |
| Added vectors as scalars (bolt groups, Mohr's circle) | Sketch the components before combining | The resultant direction makes physical sense |
| Slow handbook lookup | Repeat the five-minute drill on that topic | The same lookup takes less time next round |
Error log fields: problem or topic · your answer · first wrong step · error type (from the table) · corrected method · handbook section · result with one input changed · date to review again.
Log lucky guesses too. A right answer you couldn't explain is a miss waiting to happen.
Exam-day rules that change how you practice
- About 6 minutes per question on average. That's 480 minutes ÷ 80 questions. It's a budgeting guide, not a per-question limit.
- Two halves, one clock. After roughly half the questions, you review and submit them, and you can't go back. The halves aren't timed separately, so time you save early carries over. Clear your flagged questions before you submit.
- Breaks. The optional 50-minute break comes after you submit the first half. Unused break time doesn't extend your exam, and unscheduled breaks come out of your exam time.
- No penalty for wrong answers, so answer everything. Some questions are unscored pretest items, and you can't tell which ones.
- One approved calculator. For 2026 exams: any Casio fx-115 or fx-991 model, the HP 33s or HP 35s, or any TI-30X or TI-36X model. An on-screen TI-30XS is also available. NCEES reviews the list every year, so recheck it if your test date is in a later year.
- Scratch work happens in two reusable booklets with three markers. Do some practice on a whiteboard so it isn't new on exam day.
- Arrive 30 minutes early, and make sure the first and last names on your ID match your appointment confirmation.
Sources: Examinee Guide, May 2026, pp. 8–12 and 14; NCEES exams page, calculator policy.
Quick answers
Which PE Mechanical exam should I take?
NCEES offers three separate PE Mechanical exams: HVAC and Refrigeration, Machine Design and Materials, and Thermal and Fluid Systems. Open all three specifications on the NCEES Mechanical page and pick the one whose topics best match the work you actually do.
How hard is it, and what's the pass rate?
For January–June 2026, 61% of 401 first-time examinees passed Machine Design and Materials, and 39% of 140 repeat examinees passed (NCEES). For comparison, first-time rates were 71% for HVAC and Refrigeration and 70% for Thermal and Fluid Systems. These rates describe groups of test takers, not your personal odds. Repeat takers pass at a much lower rate, so prepare fully for your first attempt.
What score do I need to pass?
NCEES doesn't publish a passing score. Results are reported as pass or fail, and anyone quoting a fixed percentage isn't working from an official number. See how NCEES exam results work.
How do I register?
Check your licensing board's approval process first, then register and pay the $400 fee through MyNCEES. Some boards require their own application and fee. See NCEES exam registration steps.
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. See NCEES retake rules.
Can I get testing accommodations?
Yes, but you need to request them during registration. See how to request NCEES exam accommodations.
Do I need four years of experience to sit for it?
NCEES designs the PE for engineers with at least four years of post-college experience, but your licensing board decides when you can sit. Find yours in the NCEES board directory.
Does passing give me a license?
No. Your state or territorial licensing board issues the license, and passing the PE is one of its requirements.
Haven't passed the FE yet?
Start with FE Mechanical exam prep. For an overview of every PE discipline, see PE exam prep.
Sources
Exam facts on this page were last verified October 8, 2026, against the NCEES sources below. We checked the topic specification and ranges, exam format and timing, the fee, reference rules, scoring, retake limits, the 2026 calculator list, and pass rates. We solved and recalculated every practice problem and the worked example, and checked each method against the teaching source cited beside it. The problems have not been reviewed by NCEES or a licensed engineer.
NCEES
- PE Mechanical exam page (format, fee, references, pass rates)
- PE Mechanical: Machine Design and Materials specification, effective October 2025
- PE Mechanical: Machine Design and Materials specification, effective April 2020 (for comparison only)
- NCEES Examinee Guide, May 2026
- NCEES exams page, including the calculator policy
Teaching sources for the problems and worked example
- Stanford ME111, Lecture 21 (static failure), slide 5: plane-stress von Mises form
- Roylance, MIT 3.11, Yield and Plastic Flow, p. 2 (Tresca) and p. 5 (von Mises)
- Lin and Verma, ASEE Annual Conference 2007 (AC 2007-337), Table 1: geometric tolerance categories
- NISTIR 89-4203, Discount Factor Tables for Life-Cycle Cost Analyses (1989), p. 1, Eqs. 2 and 4
- Lindsey, NASA GSFC, Using FMECA as a Risk Assessment and Communication Tool, §2.2 and Table 5
- NASA Systems Engineering Handbook, §5.0, §5.3 (verification) and §5.4 (validation)
- OpenStax, University Physics Vol. 1, §12.3: stress, strain, and Young's modulus
- Roylance, MIT 3.11, Stresses in Beams, p. 4 (Eqs. 5 and 7) and p. 10 (rectangular-section shear)
- Stanford ME111, Lecture 5, slides 5 and 9–10 (σ = Mc/I; I = πD⁴/64; J = πD⁴/32)
- Stanford ME111, Lecture 13 (columns), §§13.3–13.5
- Roylance, MIT 3.11, Torsion, p. 8, Eqs. 12–14
- Roylance, MIT 3.11, Transformation of Stresses, pp. 4–6, Eqs. 11–12
- Stanford ME111, Lecture 26 (fatigue), §§26.4–26.5, §26.11, and §26.13
- OpenStax, University Physics Vol. 2, §1.3: Eq. 1.2 and Thermal Stress, Example 1.4
- NPTEL (IIT), Design of Welded Joints, Lecture 25, §§25.1–25.2
- NASA RP-1228, Fastener Design Manual (1990), printed pp. 5–6 (corrosion, hydrogen embrittlement), p. 16 (torque), and p. 17 (fastener groups)
- University of Illinois Mechanics Reference, Stress: Average Shear Stress and Units
- Hart-Smith, Adhesive-Bonded Single-Lap Joints, NASA CR-112236 (1973), introduction and Eq. 29
- KHK USA, How does shaft alignment and speed ratio affect my design? (2020), parallel-axes section
- MIT 2.72 Elements of Mechanical Design, Lecture 12, slide 34 (train ratio)
- MIT 2.004, Lecture 14, p. 14-2 (P = TΩ)
- NTN, Large Bearings catalog engineering section, §1.2, Eq. 1.1 and Table 1.1
- Zaretsky et al., NASA/TM-2003-212186, p. 3 Eq. 2 and p. 17
- NPTEL (IIT Kharagpur), Introduction to Belt Drives, §13.1.9, Eq. 13.1.13
- Hijazi, Hashemite University course notes, Screws and Fasteners (Ch. 8), pp. 3–4
- Durali, Sharif University, Shafting Lecture 19, slide 17 (key shear)
- Stanford ME111, Lecture 10 (pressure vessels), slides 3–4
- Roylance, MIT 3.11, Pressure Vessels, p. 3, Eqs. 2–3
- OpenStax, University Physics Vol. 1, §14.3: Pascal's principle, Eq. 14.12
- University of Alberta, Advanced Dynamics and Vibrations: Springs in Series and Springs in Parallel
- NPTEL (IIT Kharagpur), Design of Springs, §7.1.5.4, Eq. 7.1.10
- OpenStax, University Physics Vol. 1, §15.1, Eqs. 15.9 and 15.11
- UC3M OpenCourseWare, Machine Theory, Unit 2: Introduction to Kinematics (Grashof's law)
- Cimbala, Penn State ME 345, Strain Gages, p. 4 (quarter-bridge output)
- City, University of London, ME1110 Lecture 7 (limits and fits), slides 13 and 19
- FAA AC 43-4B, Corrosion Control for Aircraft, ¶2.5.3
- U.S. Naval Academy EN380, Ch. 5 Corrosion Types, §5.2, pp. 5-3 to 5-4
- Getting Started with Abaqus/CAE, §4.4 Mesh convergence (MIT-hosted copy)
- Wisconsin DOT Structure Inspection Manual, Part 5, Ch. 4 (penetrant), §5.4.1
- Wisconsin DOT Structure Inspection Manual, Part 5, Ch. 7 (magnetic particle), §§5.7.2–5.7.3
- Wisconsin DOT Structure Inspection Manual, Part 5, Ch. 5 (ultrasonic), §5.5.1
- Wisconsin DOT Structure Inspection Manual, Part 5, Ch. 9 (radiography), §§5.9.1–5.9.2
- Roylance, MIT 3.11, Statics, p. 2, Eqs. 1–2, and p. 5, Example 1
These teaching sources explain the methods. They are not references supplied on the exam.
Castleport Test Prep Editorial Team. This guide was developed with AI-assisted research and editing. Source checking is not professional engineering review. See how Castleport uses sources and AI assistance.
Castleport Test Prep is an independent exam prep publisher. It is not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES). Exam and credential names identify their subjects; trademarks belong to their respective owners. The practice problems are original and unofficial, and nothing on this page guarantees a passing result or a license.