Bearing Cover Spigot Tolerance Stack
A bearing housing is being detailed for a small gearbox. The housing bore is 80 mm diameter and 45 ±0.10 mm deep from the front face to the shoulder. A standard bearing of width 20 ±0.05 mm seats against the shoulder. A cover plate has a spigot that enters the bore; the spigot length is 24 ±0.10 mm from the cover face. A compressed paper gasket, nominal 1.0 mm with a tolerance of ±0.20 mm, is fitted between the cover face and the housing front face. The cover is bolted down until the gasket is compressed, so the cover face does not touch the housing face. The designer must ensure the spigot does not clamp the bearing outer ring before the gasket is fully compressed, and must keep at least 0.20 mm running clearance if possible. The drawing currently uses chain dimensioning from the front-face datum, and the shop has asked whether the bore depth can be inspected reliably.
1. Calculate the nominal and worst-case clearance between the spigot end and the bearing outer ring. State whether interference is possible and show the stack.
Read the worked answer
Let D = bore depth, W = bearing width, L = spigot length, t = gasket thickness. With the cover face held off the housing face by the gasket, the spigot end is at L + t from the housing face. The bearing front face is at D − W. Clearance C = (D − W) − (L + t) = D − W − L − t. Nominal: 45 − 20 − 24 − 1.0 = 0.00 mm. Worst-case minimum: D_min − W_max − L_max − t_max = 44.90 − 20.05 − 24.10 − 1.20 = −0.45 mm. Worst-case maximum: D_max − W_min − L_min − t_min = 45.10 − 19.95 − 23.90 − 0.80 = +0.45 mm. The stack is centred on zero and interference up to 0.45 mm is possible; the spigot can clamp the bearing before the gasket is compressed. The 0.20 mm running clearance is not guaranteed.
2. Recommend a dimensioning or tolerance change that guarantees at least 0.20 mm clearance in the worst case without changing the bearing. Give a specific numeric change and explain why it is preferable to tightening all tolerances.
Read the worked answer
Require L_max ≤ D_min − W_max − t_max − 0.20 = 44.90 − 20.05 − 1.20 − 0.20 = 23.45 mm. With the existing ±0.10 mm spigot tolerance, set the nominal spigot length to 23.35 mm, giving L_max = 23.45 mm and L_min = 23.25 mm. Then worst-case clearance is 0.20 mm at the low end and 1.10 mm at the high end. This is preferable to tightening D, W and t because those tolerances are already at reasonable machining and gasket levels; tightening all of them would increase cost and still leave the nominal stack at zero. Reducing the spigot length is a single functional change and keeps the bearing standard. Alternatively, specify the gasket compressed thickness as 0.8 ±0.1 mm and keep L = 24; but the spigot change is more robust because it does not depend on gasket compression.
3. The shop proposes to inspect the 45 mm bore depth with a depth micrometer from the front face while the housing is clamped on its base. Assess this method and state one alternative that matches the functional datum.
Read the worked answer
A depth micrometer from the front face is functionally correct because the front face is the datum that locates the cover and gasket. However, the reading must be taken to the shoulder, not to a bore step or burr; the shoulder must be flat and clean. Clamping on the base may deflect a thin housing, so inspect in the free or supported state specified, or use a CMM with the same datum. A better shop check is a dial indicator on a setting ring or a depth gauge zeroed on a surface plate with the front face located; this matches the functional stack better than measuring from the outside base. Require the drawing to show the front face as datum A and the bore axis as datum B, and change the 45 mm dimension to a directly toleranced depth from A so the inspector does not have to add chain dimensions.