Page 233 - CITS - ED - Mechanical

ENGINEERING DRAWING - CITS

Fig 5 Fig 6

The limits of the hole are 25.000 and 25.021 mm and the limits of th eshaft are 25.022 and 25.035. The shaft is
always bigger than the hole. This is an interference fit.
Maximum interference: In an interference fit, it is the algebraic difference between the minimum hole and the
maximum shaft. (Fig 7)

Minimum interference: In an interference fit, it is the algebraic difference between the maximum hole and minimum
shaft. (Figs 7 & 8)
In the example shown in figure 6,

the maximum interference is = 25.035 - 25.000
= 0.035.
the minimum interference is = 25.022 - 25.021
= 0.001.
Transition fit: It is a fit which may sometimes provide clearance and sometimes interference. When this class of
fit is represented graphically, the tolerance zones of the hole and shaft will overlap each other. (Fig 8)
Example
Fit 75 H8/j7 (Fig 9)

Fig 7 Fig 8

Fig 9

The limits of the hole are 75.000 and 75.046 mm and those of the shaft are 75.018 and 74.988 mm.
Maximum clearance = 75.046 - 74.988
= 0.058 mm.
If the hole is 75.000 and the shaft 75.018 mm, the shaft is 0.018 mm bigger than the hole. This results in interfer-
ence. This is a transition fit because it can result in a clearance fit or an interference fit.

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CITS :Engineering Drawing (Mechanical) - Exercise 9