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Assembly constraints and tolerance fit calculation

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.

(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.

(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.

ConstraintMeaningDegrees of freedom removed
Coincident / mateTwo planes, edges or points touchUp to 3 for a face-to-face mate
Concentric / insertAxes of two cylinders coincide4 (shaft in hole, free to slide and spin)
ParallelTwo faces or axes parallel2
PerpendicularFaces or axes at 90 deg1
TangentSurface touches another surface1
Distance / offsetFixed gap between facesUp to 3 for parallel faces (like a mate with a gap)
AngleFixed angle1
Fixed / groundPart cannot move6

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=40+0.025=40.025 mm; lower limit =40.000=40.000 mm.
  • Tolerance =40.025−40.000=0.025=40.025-40.000=0.025 mm.

Shaft:

  • Upper limit =40−0.009=39.991=40-0.009=39.991 mm; lower limit =40−0.025=39.975=40-0.025=39.975 mm.
  • Tolerance =39.991−39.975=0.016=39.991-39.975=0.016 mm.

Clearances:

Cmax=holemax−shaftmin=40.025−39.975=0.050 mmCmin=holemin−shaftmax=40.000−39.991=0.009 mm\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 =Cmax−Cmin=0.041=C_{max}-C_{min}=0.041 mm =0.025+0.016=0.025+0.016.

Since Cmin>0C_{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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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