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398 Mechanical Engineering Design
EXAMPLE 7–7 Find the shaft and hole dimensions for a loose running fit with a 34-mm basic size.
Solution From Table 7–9, the ISO symbol is 34H11/c11. From Table A–11, we find that toler-
ance grade IT11 is 0.160 mm. The symbol 34H11/c11 therefore says that
D =
d =
0.160 mm. Using Eq. (7–36) for the hole, we get
Answer D max = D +
D = 34 + 0.160 = 34.160 mm
Answer D min = D = 34.000 mm
The shaft is designated as a 34c11 shaft. From Table A–12, the fundamental deviation
is δ F =−0.120 mm. Using Eq. (7–37), we get for the shaft dimensions
Answer d max = d + δ F = 34 + (−0.120) = 33.880 mm
Answer d min = d + δ F −
d = 34 + (−0.120) − 0.160 = 33.720 mm
EXAMPLE 7–8 Find the hole and shaft limits for a medium drive fit using a basic hole size of 2 in.
Solution The symbol for the fit, from Table 7–8, in inch units is (2 in)H7/s6. For the hole, we use
Table A–13 and find the IT7 grade to be
D = 0.0010 in. Thus, from Eq. (7–36),
Answer D max = D +
D = 2 + 0.0010 = 2.0010 in
Answer D min = D = 2.0000 in
The IT6 tolerance for the shaft is
d = 0.0006 in. Also, from Table A–14, the
fundamental deviation is δ F = 0.0017 in. Using Eq. (7–38), we get for the shaft that
Answer d min = d + δ F = 2 + 0.0017 = 2.0017 in
Answer d max = d + δ F +
d = 2 + 0.0017 + 0.0006 = 2.0023 in
Stress and Torque Capacity in Interference Fits
Interference fits between a shaft and its components can sometimes be used effectively
to minimize the need for shoulders and keyways. The stresses due to an interference fit
can be obtained by treating the shaft as a cylinder with a uniform external pressure, and
the hub as a hollow cylinder with a uniform internal pressure. Stress equations for these
situations were developed in Sec. 3–16, and will be converted here from radius terms
into diameter terms to match the terminology of this section.
