How are Cmax and Cmin found from limits of size?

Cmax = hole max - shaft max; Cmin = hole min - shaft min

Cmax = shaft max - hole min; Cmin = shaft min - hole max

Cmax = hole tolerance + shaft tolerance only

Hole 30.000-30.021, shaft 29.967-29.980. What is the fit?

Clearance fit, 0.020 to 0.054 mm

Transition fit, +0.054 to -0.020 mm

Interference fit, 0.020 to 0.054 mm

Clearance fit, 0.013 to 0.021 mm

Hole 50.000-50.025, shaft 50.002-50.018. What is the fit?

Transition: Cmax = +0.023, Cmin = -0.018 mm

Interference: 0.023 to 0.018 mm

Transition: Cmax = +0.043, Cmin = -0.018 mm

Hole 25.000-25.021, shaft 25.035-25.048. What is the fit and its interference range?

Interference fit, 0.014 to 0.048 mm

Clearance fit, 0.014 to 0.048 mm

Transition fit, +0.014 to -0.048 mm

Interference fit, 0.021 to 0.048 mm

Practice · 6 marks

Classify fits from given limits of size

Short answer

(a) Clearance fit: hole tolerance 0.021 mm, shaft tolerance 0.013 mm; clearance 0.020 to 0.054 mm. (b) Transition fit: hole tolerance 0.025 mm, shaft 0.016 mm; +0.023 to -0.018 mm. (c) Interference fit: hole tolerance 0.021 mm, shaft 0.013 mm; interference 0.014 to 0.048 mm.

The limits of size of three hole-shaft pairs are given below. For each, find the tolerances, the maximum and minimum clearance (or interference), and state the type of fit. (a) Hole 30.000 to 30.021 mm; shaft 29.967 to 29.980 mm. (b) Hole 50.000 to 50.025 mm; shaft 50.002 to 50.018 mm. (c) Hole 25.000 to 25.021 mm; shaft 25.035 to 25.048 mm.

Use: $C_{max} =$ hole max $-$ shaft min; $C_{min} =$ hole min $-$ shaft max. A negative value is interference.

	(a)	(b)	(c)
Hole tolerance	$0.021$	$0.025$	$0.021$
Shaft tolerance	$0.013$	$0.016$	$0.013$
$C_{max}$ = hole max - shaft min	$30.021 - 29.967 = +0.054$	$50.025 - 50.002 = +0.023$	$25.021 - 25.035 = -0.014$
$C_{min}$ = hole min - shaft max	$30.000 - 29.980 = +0.020$	$50.000 - 50.018 = -0.018$	$25.000 - 25.048 = -0.048$
Fit type	Clearance	Transition	Interference

Results

(a) Clearance fit: the clearance always lies between 0.020 mm and 0.054 mm. The shaft is always smaller than the hole, so it is a running fit (it matches 30 H7/f6).
(b) Transition fit: the maximum clearance is 0.023 mm and the maximum interference is 0.018 mm. Depending on the actual sizes, the assembly may have clearance or interference. It is used for accurate location (it matches 50 H7/k6).
(c) Interference fit: the minimum interference is 0.014 mm and the maximum is 0.048 mm. It is always a press fit (it matches 25 H7/s6).

All three pairs are on the hole basis, as the lower limit of the hole equals the basic size in each (lower deviation zero).

Answer: (a) clearance, 0.020 to 0.054 mm; (b) transition, +0.023 to -0.018 mm; (c) interference, 0.014 to 0.048 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Classify fits from given limits of size

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