Question 24.6: The recommended class of fit for the journal and the bush of...

Given Class of fi t = 40H6-e7.

X_{1}=0.08 mm \quad X_{2}=0.06 mm .

There are two populations—population of the bushes denoted by the letter B and that of the
journals denoted by the letter J.
Step I Population of bushes (B)
The limiting dimensions for the bush 40H6 are (from Table 3.2).

Table 3.2 Tolerances for holes of sizes up to 100 mm (H5 to H11)

\frac{40.016}{40.000} mm \quad \text { or } \quad 40.008 \pm 0.008 mm .

Since design tolerance and natural tolerance are equal,

\mu_{B}=40.008 mm \quad \hat{\sigma}_{B}=\frac{0.008}{3}=0.002667 mm .          (a)

Step II Population of journal
The limiting dimensions for the journal 40e7 are [Table 3.3(a)].

Table 3.3a Tolerances for shafts of sizes up to 100 mm (from d to h)

\frac{39.950}{39.925} mm \quad \text { or } \quad 39.9375 \pm 0.0125 mm .

Therefore,

\mu_{J}=39.9375 mm .

\hat{\sigma}_{J}=\frac{0.0125}{3}=0.004167 mm                 (b).

Step III Population of clearance (C)
The letter C denotes the population for clearance.
It is obtained by subtracting the population of the journal from the population of the bushes.
Therefore,

\mu_{C}=\mu_{B}-\mu_{J}=40.008-39.9375=0.0705 mm .

=\sqrt{(0.002667)^{2}+(0.004167)^{2}} .

= 0.004947 mm
Step IV Percentage of rejected assemblies
For maximum clearance,

X_{1}=0.08 mm .

\text { and } \quad Z_{1}=\frac{X_{1}-\mu_{C}}{\hat{\sigma}_{C}}=\frac{0.08-0.0705}{0.004947}=+1.92 .

For minimum clearance,

X_{2}=0.06 mm .

\text { and } \quad Z_{2}=\frac{X_{2}-\mu_{C}}{\hat{\sigma}_{C}}=\frac{0.06-0.0705}{0.004947}=-2.12 .

From Table 24.6, the areas below the normal curve from Z = 0 to 1.92 and from Z = 0 to 2.12 are 0.4726 and 0.4830 respectively.
Therefore, the percentage of rejected assemblies is [1 – (0.4726 + 0.4830)] × 100 or 4.44%.

Table 24.6 Areas under normal curve from 0 to Z