--- title: "Assembly constraints and tolerance fit calculation" subject: "Computer Aided Design" chapter: 5 key: "semester_4/computer_aided_design:assembly-constraints-tolerance-fit" source: https://www.bctengineer.com/q/semester_4/computer_aided_design/assembly-constraints-tolerance-fit site: BCT Engineering --- # Assembly constraints and tolerance fit calculation Computer Aided Design · Chapter 5: CAD Modeling Techniques - Asked in: year not given, 8 marks - Video: [Tolerance, Limits and Fits with example](https://www.youtube.com/watch?v=fcSFVnSw6NM) — CAD CAM MANUFACTURING INDIA - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/computer_aided_design/assembly-constraints-tolerance-fit - Whole chapter: https://www.bctengineer.com/md/semester_4/computer_aided_design/chapter-5 ## Question (a) Explain the common assembly constraints (mates) used in a CAD assembly. (b) A hole is dimensioned 40 +0.025/0 mm and the mating shaft 40 -0.009/-0.025 mm. Calculate the limits of size, tolerances, maximum and minimum clearance, and state the type of fit. ## Short answer Hole limits are 40.000 to 40.025 mm with tolerance 0.025 mm; shaft limits are 39.975 to 39.991 mm with tolerance 0.016 mm. Subtracting shaft from hole gives Cmax=0.050 mm and Cmin=0.009 mm. Since minimum clearance is positive, the fit is a clearance fit, exactly H7/g6. ## Answer #### (a) Assembly constraints (mates) An assembly is made by placing parts and removing their degrees of freedom (a free part has six: three translations and three rotations) using constraints. | Constraint | Meaning | Degrees of freedom removed | |---|---|---| | Coincident / mate | Two planes, edges or points touch | Up to 3 for a face-to-face mate | | Concentric / insert | Axes of two cylinders coincide | 4 (shaft in hole, free to slide and spin) | | Parallel | Two faces or axes parallel | 2 | | Perpendicular | Faces or axes at 90 deg | 1 | | Tangent | Surface touches another surface | 1 | | Distance / offset | Fixed gap between faces | Up to 3 for parallel faces (like a mate with a gap) | | Angle | Fixed angle | 1 | | Fixed / ground | Part cannot move | 6 | A fully constrained assembly may be built bottom-up or top-down. Interference and clearance checks confirm correct assembly. #### (b) Limits and fit **Hole:** - Upper limit $=40+0.025=40.025$ mm; lower limit $=40.000$ mm. - Tolerance $=40.025-40.000=0.025$ mm. **Shaft:** - Upper limit $=40-0.009=39.991$ mm; lower limit $=40-0.025=39.975$ mm. - Tolerance $=39.991-39.975=0.016$ mm. **Clearances:** $$ \begin{aligned} C_{max}&=\text{hole}_{max}-\text{shaft}_{min}=40.025-39.975=0.050\ \text{mm} \\ C_{min}&=\text{hole}_{min}-\text{shaft}_{max}=40.000-39.991=0.009\ \text{mm} \end{aligned} $$ Fit tolerance $=C_{max}-C_{min}=0.041$ mm $=0.025+0.016$. Since $C_{min}>0$, the shaft is always smaller than the hole, so it is a **clearance fit** (a sliding/running fit; these limits are exactly the ISO fit H7/g6 for 40 mm). **Answer: hole 40.000 to 40.025, shaft 39.975 to 39.991; C_max = 0.050 mm, C_min = 0.009 mm; clearance fit.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How many degrees of freedom does a free part have, and what do assembly constraints do? A. Three; constraints add motion B. Six (3 translations and 3 rotations); constraints remove them C. Four; constraints rotate the part D. Six; constraints scale the part Answer: B. Mates progressively remove the six degrees of freedom of a part. (Key point: Degrees of freedom) 2. A concentric (insert) constraint between a shaft and a hole removes how many degrees of freedom, leaving what? A. 6; nothing B. 4; the shaft can still slide and spin C. 1; the shaft can move in any direction D. 2; the shaft can only rotate Answer: B. Coinciding the axes removes 4 DOF, leaving sliding along and rotation about the axis. (Key point: Concentric constraint) 3. Hole 40 +0.025/0 and shaft 40 -0.009/-0.025. What are the hole and shaft tolerances? A. Hole 0.025 mm, shaft 0.016 mm B. Hole 0.025 mm, shaft 0.034 mm C. Hole 0.050 mm, shaft 0.009 mm D. Hole 0.016 mm, shaft 0.025 mm Answer: A. Hole 40.025-40.000 = 0.025; shaft 39.991-39.975 = 0.016. (Key point: Limits and tolerances) 4. How are C_max, C_min found and what is the fit? A. C_max = 0.050 (hole max - shaft min), C_min = 0.009 (hole min - shaft max); clearance fit B. C_max = 0.050, C_min = -0.009; transition fit C. C_max = 0.034, C_min = 0.009; interference fit D. C_max = 0.041, C_min = 0; transition fit Answer: A. 40.025-39.975 = 0.050 and 40.000-39.991 = 0.009. Since C_min is positive, the shaft is always smaller: clearance fit. (Key point: Clearances and fit type) ## Related in Computer Aided Design - [CAD software modules: part, assembly, drafting](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Acad-software-modules) — Computer Aided Design, chapter 1 - [Classification of CAD surface entities](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Asurface-entities-classification) — Computer Aided Design, chapter 4 - [STL file: facet normals and file size](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Astl-format-numerical) — Computer Aided Design, chapter 6 ## Related in other subjects - [Limits of a 40 H7/g6 hole-shaft fit](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Afit-40-h7-g6-calculation) — Metrology, chapter 9 - [Limits, fits and tolerance calculation](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Alimits-fits-tolerance-numerical) — Manufacturing and Production Processes, chapter 3 - [Classify fits from given limits of size](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Aclassify-fits-from-limits) — Metrology, chapter 9 - [Basic terms of limits, fits and tolerances](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Alimits-fits-terminology) — Metrology, chapter 9