N2 Fitting & Machining Theory

44 Module 3 • Limits and fits Example 3.2

A fit between a shaſt and a bush is given as a ratio of 65H9n8. When referring to the relevant tables we determine the following: Basic hole size is 65mm

+ 0.074 Bush – 65H9 = Ø60

–0.000 +0,020

Shaſt – 65n8 = Ø60 –0,066

Determine fromthe information above, the following dimensions: • High limit of the shaſt and bush • Low limit of the shaſt and bush • Tolerance of the shaſt and bush • Minimum allowance between the two mating parts • Maximum allowance between the two mating parts

Solution Shaſt

Low limit = nominal size – low limit = 65 mm – 0,066 mm = 64,934 mm

High limit = nominal size + high limit = 65 mm + 0,020 mm = 65,020 mm

Tolerance = high limit – low limit = 65,020 mm – 64,934 mm = 0,086 mm

Bush

Low limit = nominal size – low limit = 65 mm – 0,000 mm = 65 mm

High limit = nominal size + high limit = 65 mm + 0,074 mm = 65,074 mm

Tolerance = high limit – low limit = 65,074 mm – 65 mm = 0,074 mm

Maximum allowance = maximum limit hole – low limit shaſt = 65,074 – 64,934 = 0,14 mm

Minimum allowance = high limit shaſt – low limit hole = 65,020 – 65 = 0,020 mm

mm mm

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