SME
Engineering Graphics & CAD · Problem type
Virtual condition: do the parts fit?
Use this for “A hole is Ø10 at MMC with position Ø0.1; the pin is Ø9.8 at MMC with position Ø0.05. Will they always assemble?”
This is an example. Type your own values to start a new problem.
VCₕ (mm)
=
MMCₕ (mm)
−
tₕ (mm)
VCₛ (mm)
=
MMCₛ (mm)
+
tₛ (mm)
c (mm)
=
VCₕ (mm)
−
VCₛ (mm)
fit
Do they fit?
worked out (not typed)
Formulas
VCₕ = MMCₕ − tₕ
9.9 = 10 − 0.1
VCₛ = MMCₛ + tₛ
9.85 = 9.8 + 0.05
c
=
VCₕ
−
VCₛ
0.05 = 9.9 − 9.85
fit: always fit when VCₛ ≤ VCₕ
9.85 ≤ 9.9, so always fit
Assumptions
•
Both position tolerances are stated at MMC, so each part’s worst case is its virtual condition (VC).
•
A hole’s VC is its MMC size less its tolerance; a shaft’s is its MMC size plus its tolerance.
•
The parts always fit when the shaft’s VC is at most the hole’s; the gap between them is the least clearance.
Step-by-step
WE KNOW
MMCₕ = 10 mm, tₕ = 0.1 mm, MMCₛ = 9.8 mm, tₛ = 0.05 mm
FIND
hole virtual condition (VCₕ), shaft virtual condition (VCₛ), least clearance (c), do they fit? (fit)
Step 1 · Find hole virtual condition (VCₕ)
Use VCₕ = MMCₕ − tₕ
At MMC the hole is smallest, and its position may wander by the tolerance: the pin must clear a hole that much smaller.
VCₕ = 10 − 0.1
VCₕ = 9.9 mm
Step 2 · Find shaft virtual condition (VCₛ)
Use VCₛ = MMCₛ + tₛ
At MMC the shaft is largest, and its position may wander by the tolerance: it sweeps a circle that much larger.
VCₛ = 9.8 + 0.05
VCₛ = 9.85 mm
Step 3 · Find least clearance (c)
Use c = VCₕ − VCₛ
The least clearance is the room left between the two worst cases; below 0 they can collide.
c
=
9.9
−
9.85
c
=
0.05
mm
Step 4 · Find do they fit? (fit)
Use fit: always fit when VCₛ ≤ VCₕ
The shaft’s worst case is no bigger than the hole’s, so any good parts go together.
9.85 ≤ 9.9: always fit
fit = the shaft’s VC against the hole’s
fit = always fit
Check
Put the numbers back in. Both sides should match.
9.9 = 10 − 0.1 ✓
9.85 = 9.8 + 0.05 ✓
0.05 = 9.9 − 9.85 ✓
9.85 ≤ 9.9, so always fit ✓
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