How are maximum and minimum clearance calculated?

Max = hole max - shaft min; min = hole min - shaft max

Max = hole max - shaft max; min = hole min - shaft min

Max = shaft max - hole min; min = shaft min - hole max

For shaft (i) (39.950 to 39.975) with the hole 40.000 to 40.039, the clearances and fit are

0.089 and 0.025 mm; clearance fit

0.064 and 0.025 mm; transition fit

-0.004 and -0.059 mm; interference fit

0.089 and -0.025 mm; transition fit

Practice · 3+5 marks

Limits, fits and tolerance calculation

Short answer

Unilateral tolerance varies only one way from basic size; bilateral varies both sides. Geometric tolerance controls form, orientation, location or run-out, with symbols such as straightness, flatness, parallelism and position. For hole 40+0.039/0: shaft (i) clearance fit, clearances 0.025-0.089 mm; shaft (ii) interference fit, interferences 0.004-0.059 mm; shaft (iii) transition fit, 0.013 mm clearance to 0.042 mm interference.

(a) Differentiate between unilateral and bilateral tolerance, and explain geometric tolerance with any four characteristic symbols. (b) A hole is dimensioned $\phi 40^{+0.039}_{0}$ mm. Three shafts are available: (i) $\phi 40^{-0.025}_{-0.050}$ , (ii) $\phi 40^{+0.059}_{+0.043}$ , (iii) $\phi 40^{+0.042}_{+0.026}$ . For each, find the maximum and minimum clearance (or interference) and state the type of fit.

(a) Unilateral and bilateral tolerance; geometric tolerance

Point	Unilateral	Bilateral
Variation	In one direction only from basic size	On both sides of basic size
Example	$40^{+0.04}_{0}$	$40 \pm 0.02$
Use	Hole basis / shaft basis systems, easy fits	Positioning, centre distances

Geometric tolerance limits the form, orientation, location and run-out of a feature, not just its size. It is given in a feature control frame on the drawing. Characteristic symbols:

Straightness (a straight line) and flatness (a parallelogram) - form.
Parallelism (two parallel lines) and perpendicularity (a right-angle) - orientation.
Concentricity (two circles) and position - location.
Circular run-out (arrow) - run-out.

(b) Fit calculation

Hole: maximum size $= 40.039$ mm, minimum size $= 40.000$ mm.

Maximum clearance $=$ hole max $-$ shaft min; minimum clearance $=$ hole min $-$ shaft max (negative value = interference).

Shaft (i): shaft max $= 39.975$ , min $= 39.950$

C_{max} = 40.039 - 39.950 = 0.089\ \text{mm},\qquad C_{min} = 40.000 - 39.975 = 0.025\ \text{mm}

Both positive, so it is a clearance fit (running fit).

Shaft (ii): shaft max $= 40.059$ , min $= 40.043$

C_{max} = 40.039 - 40.043 = -0.004\ \text{mm},\qquad C_{min} = 40.000 - 40.059 = -0.059\ \text{mm}

Both negative, so maximum interference $= 0.059$ mm and minimum interference $= 0.004$ mm: an interference (press) fit.

Shaft (iii): shaft max $= 40.042$ , min $= 40.026$

C_{max} = 40.039 - 40.026 = 0.013\ \text{mm},\qquad C_{min} = 40.000 - 40.042 = -0.042\ \text{mm}

The result can be clearance or interference, so it is a transition fit.

Shaft	Max clearance (mm)	Min clearance (mm)	Fit
(i)	0.089	0.025	Clearance
(ii)	$-0.004$	$-0.059$	Interference
(iii)	0.013	$-0.042$	Transition

Answer: (i) clearance fit, 0.089 to 0.025 mm; (ii) interference fit, 0.004 to 0.059 mm; (iii) transition fit, 0.013 mm clearance to 0.042 mm interference.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Limits, fits and tolerance calculation

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