| Nominal Size Range | Permissible Deviation (± mm) | |--------------------|-------------------------------| | 0.5 – 3 | ±0.2 | | >3 – 6 | ±0.5 |
For Class K, the general tolerance for circular run-out is fixed at 0.2 mm.
Duration: 90 minutes
Total marks: 100
Instructions:
Section A — Short answer and definitions (20 marks) general tolerance iso 2768-mk
Section B — Table interpretation and application (20 marks) (Use the ISO 2768‑m and ISO 2768‑k tables below — simplified values given for the exam.)
Simplified linear tolerance table (mm):
Simplified angular tolerance:
Use these values for calculations.
Section C — Drawing interpretation and correction (20 marks) 10. (8) A drawing note reads “Tolerances: ISO 2768‑m”. The drawing also shows a critical hole Ø12 H7 (H7 tolerance explicitly shown). Explain which tolerance controls the hole and why. If the hole callout is Ø12 H7 with no additional notes, give the rationale whether ISO 2768 affects it. 11. (6) On an assembly drawing, a set of mating parts are dimensioned: shaft nominal Ø20 (no tolerance), mating bore nominal Ø20 (no tolerance); note reads ISO 2768‑mk. Is this acceptable for precision fit? Explain what the drawing should show to ensure a clearance fit of 0.02–0.05 mm. 12. (6) A supplier manufactured a part to k class tolerances though the drawing specified m class; the part’s critical dimension of 25 mm is out of tolerance for m but within k. Explain the likely consequences for assembly and recommended actions (3 points).
Section D — Problem solving & design considerations (40 marks) 13. (10) You are designing a bracket with multiple features. Explain, with brief justification, which features you would: a) apply ISO 2768‑m to (3 examples), b) require specific tighter tolerances (3 examples), and c) select ISO 2768‑k for (2 examples). 14. (8) Calculate cumulative tolerance stack-up for three aligned features in series: A, B, and C, nominal lengths 15 mm, 25 mm, and 40 mm respectively, all unspecified on the drawing and ISO 2768‑m applies. Use the simplified table above to compute worst‑case total length tolerance and resulting possible total length range. 15. (8) For the same features as Q14 but B is specified with a tighter machining tolerance of ±0.05 mm (explicit), while A and C remain under ISO 2768‑m, compute the worst‑case total length range. 16. (6) Explain how note “ISO 2768‑m unless otherwise specified” can reduce drawing clutter but also identify two risks associated with relying on general tolerances. 17. (8) A customer requires interchangeable parts with consistent function across suppliers. Propose a concise set of drawing practices (6 actionable items) to ensure interchangeability while using ISO 2768‑m where appropriate.
Extra credit (up to 5 marks)
Answer key (concise) — for examiner use only | Nominal Size Range | Permissible Deviation (±
Section A
Section B (using table) 6. Ø10 mm: a) m: ±0.15 → range 9.85–10.15 mm b) k: ±0.5 → range 9.5–10.5 mm 7. Length 45 mm: a) m: for 30–120 → ±0.2 → 44.8–45.2 mm b) k: for 30–120 → ±0.8 → 44.2–45.8 mm 8. Ø2.5 mm: a) m (≤3) ±0.1 → 2.4–2.6 mm b) k (≤3) ±0.3 → 2.2–2.8 mm 9. Angle 60°: m: ±1.0° → 59.0°–61.0° k: ±3.0° → 57.0°–63.0° If groove requires ±0.05 mm explicit tolerance, that explicit tolerance overrides ISO 2768 for that feature and ISO 2768 does not apply to that groove.
Section C 10. Ø12 H7 explicit callout controls; ISO 2768 does not override an explicit tolerance. H7 defines specific limits (hole basis tolerance); general tolerance ignored for that dimension. 11. Not acceptable for precision fit; drawing must specify tolerances (e.g., shaft Ø19.98–20.00 and bore Ø20.02–20.05) or use fit designation (e.g., H7/g6) to guarantee 0.02–0.05 mm clearance. 12. Consequences: possible assembly interference or functional failure; actions: reject/ rework part or negotiate acceptable nonconformance and update drawing tolerance notes; implement supplier corrective action.
Section D 13. Examples: a) Apply ISO 2768‑m to: nonfunctional external profile, noncritical hole pattern spacing, general bracket thickness. b) Tight tolerances for: mating surfaces/precision bores, shaft journal diameters, location/position of dowel holes. c) Apply ISO 2768‑k to: large noncritical cast features, rough-cut blanks. (Brief justification: cost vs function tradeoff.) 14. A±0.15, B±0.15, C±0.2 → worst‑case total tolerance = ±(0.15+0.15+0.2)=±0.5 mm. Nominal total = 80 mm → range 79.5–80.5 mm. 15. B ±0.05; total ±(0.15+0.05+0.2)=±0.4 mm → range 79.6–80.4 mm. 16. Benefits: reduces clutter, consistent defaults. Risks: designer assumptions hide critical tolerances; possible misinterpretation by manufacturers leading to nonconforming parts. 17. Six practices: Section A — Short answer and definitions (20 marks)
Extra credit
End of examination.