Limits of a 40 H7/g6 hole-shaft fit

Step 1: diameter step

$$D = \sqrt{30 \times 50} = 38.73\ \text{mm}$$

Step 2: tolerance unit

$$i = 0.45\,(38.73)^{1/3} + 0.001(38.73) = 0.45(3.383) + 0.039 = 1.561\ \mu\text{m}$$

• IT7 $= 16i = 24.98\ \mu$m, standard value 25 $\mu$m
• IT6 $= 10i = 15.6\ \mu$m, standard value 16 $\mu$m

Step 3: fundamental deviation of shaft g

$$es = -2.5\,D^{0.34} = -2.5(38.73)^{0.34} = -2.5(3.468) = -8.67\ \mu\text{m} \approx -9\ \mu\text{m}$$

Step 4: limits

Hole 40 H7 (lower deviation zero):

• Lower limit $= 40.000$ mm
• Upper limit $= 40.000 + 0.025 = 40.025$ mm
• Tolerance $= 0.025$ mm

Shaft 40 g6 (upper deviation $-0.009$ mm):

• Upper limit $= 40 - 0.009 = 39.991$ mm
• Lower limit $= 39.991 - 0.016 = 39.975$ mm
• Tolerance $= 0.016$ mm

Step 5: clearances

$$C_{max} = \text{hole max} - \text{shaft min} = 40.025 - 39.975 = 0.050\ \text{mm}$$ $$C_{min} = \text{hole min} - \text{shaft max} = 40.000 - 39.991 = 0.009\ \text{mm}$$

Check: $C_{max} - C_{min} = 0.041 = 0.025 + 0.016$.

Both clearances are positive, so this is a clearance fit (close running or sliding fit).

Answer: hole 40.000 to 40.025 mm; shaft 39.975 to 39.991 mm; $C_{max}$ = 0.050 mm, $C_{min}$ = 0.009 mm; clearance fit.