Acceptance of a 25 H7/g6 Shaft and Bore
An assembly drawing specifies a nominal Ø25 mm shaft and bearing bore with the fit designation 25 H7/g6. The workshop table gives hole H7 = 25.000/25.021 mm and shaft g6 = 24.980/24.993 mm. The drawing notes Ra 1.6 µm on the shaft journal, Ra 3.2 µm in the bore, and a general note ‘surface finish unless otherwise stated: Ra 12.5 µm’. During inspection, the shaft measures 24.995 mm and the bore measures 25.018 mm. The inspector also finds that the surface-finish symbol on the shaft has been omitted because ‘the shaft is bought out and the supplier will know the fit from 25 g6’. The store has accepted the parts because the measured clearance is positive. The design office is reviewing the inspection report before batch acceptance.
1. Calculate the maximum and minimum clearance for 25 H7/g6, classify the fit, and decide whether the measured shaft and bore should be accepted. Show the limit calculations.
Read the worked answer
Hole H7: lower limit 25.000 mm, upper limit 25.021 mm. Shaft g6: lower limit 24.980 mm, upper limit 24.993 mm.
Maximum clearance = hole max − shaft min = 25.021 − 24.980 = 0.041 mm (41 µm). Minimum clearance = hole min − shaft max = 25.000 − 24.993 = 0.007 mm (7 µm).
Since the minimum clearance is positive, the fit is always clearance; H7/g6 is a clearance fit (close running).
Measured shaft 24.995 mm is greater than the upper limit 24.993 mm, so it is nonconforming and should be rejected. Measured bore 25.018 mm lies between 25.000 and 25.021 mm, so it is accepted. The assembly clearance would be 25.018 − 24.995 = 0.023 mm, which is inside 0.007–0.041 mm, but that does not make the shaft acceptable: acceptance is by individual limits, not by the resulting clearance alone.
2. A student reads g6 as +0.007/+0.020 and concludes the fit is an interference fit. Diagnose the error and state the correct classification.
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The student has reversed the sign and the order of the shaft deviations. For a shaft with fundamental deviation g, the upper deviation is negative and the lower deviation is more negative; the given g6 limits are 24.980/24.993 mm, i.e. −0.020/−0.007 mm from nominal.
If one incorrectly uses +0.007/+0.020 mm, the shaft becomes 25.007–25.020 mm. Then minimum clearance = 25.000 − 25.020 = −0.020 mm (interference) and maximum clearance = 25.021 − 25.007 = +0.014 mm, which would look like a transition fit. The correct reading gives 0.007–0.041 mm clearance, always positive, so it is a clearance fit. The sign of the fundamental deviation and the order (upper/lower) must be checked before classifying.
3. The shaft's surface-finish symbol has been omitted on the grounds that the shaft is bought out. Is this acceptable? Explain how the Ra values should be applied.
Read the worked answer
No. A purchased part may be controlled by a supplier drawing, but the assembly drawing must still identify the required fit and finish, or explicitly reference the supplier's part number and specification. The note ‘25 g6’ controls size and fit, not surface texture.
The shaft journal requires Ra 1.6 µm, which is much finer than the general Ra 12.5 µm note; the bore requires Ra 3.2 µm. Each functionally different surface must carry its symbol or be covered by a clear general note; the general note cannot be used to infer a finer requirement. Omitting the symbol risks the supplier delivering a 12.5 µm finish, which may fail lubrication or fretting requirements. The correct action is to add the Ra 1.6 symbol on the shaft journal (or a note on the purchase specification) and keep Ra 3.2 on the bore.