Question 24.5: The recommended class of transition fit between the recess a...

Given Class of fi t = 60H6-j5
There are two populations—population of recess dimension denoted by the letter R and that of spigot dimension denoted by the letter S.
Step I Population of recess (R)
The limiting dimensions for recess 60H6 are (Table 3.2)

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

\frac{60.019}{60.000} mm \quad \text { or } \quad 60.0095 \pm 0.0095 mm .

The design tolerance and natural tolerance are equal. Therefore,

\mu_{R}=60.0095 mm .

\text { and } \quad \hat{\sigma}_{R}=\frac{0.0095}{3}=0.00317 mm .

Step II Population of spigot (S)
The limiting dimensions for spigot 60j5 are [Table 3.3(b)]

Table 3.3b Tolerances for shafts of sizes up to 100 mm (from j to s)

\frac{60.006}{59.993} mm \quad \text { or } \quad 59.9995 \pm 0.0065 mm .

Therefore,

\mu_{S}=59.9995 \text { and } \hat{\sigma}_{S}=\frac{0.0065}{3}=0.00217 mm .

Step III Population of interference (I)
The population for interference is denoted by the letter I. It is obtained by subtracting the population of recess from the population of the spigot. From Eq. 24.12,

\mu_{I}=\mu_{S}-\mu_{R}                 (24.12).

\mu_{I}=\mu_{S}-\mu_{R}=59.9995-60.0095=-0.01 mm .

From Eq. 24.14,

\hat{\sigma}_{I}=\sqrt{\left(\hat{\sigma}_{R}\right)^{2}+\left(\hat{\sigma}_{S}\right)^{2}}                 (24.14).

\hat{\sigma}_{I}=\sqrt{\left(\hat{\sigma}_{R}\right)^{2}+\left(\hat{\sigma}_{S}\right)^{2}} .

=\sqrt{(0.00317)^{2}+(0.00217)^{2}} .

= 0.00384 mm.

Step IV Probability of interference fi t
When the interference is zero,

I=0 \quad \text { and } \quad Z=\frac{I-\mu_{I}}{\hat{\sigma}_{I}}=\frac{0-(-0.01)}{0.00384}=+2.6 .

As shown in Fig. 24.14, the interference will occur only when Z > 2.60. When Z is less than 2.60,

the value of I will be negative, giving a clearance fit. The probability of interference fi t is therefore given by the area below the normal curve from Z = 2.60 to Z = +∞. From Table 24.6, the area below the normal curve from Z = 0 to Z = 2.60 is 0.4953.
Therefore, the probability of interference fit is (0.5 – 0.4953) × 100 or 0.47%.

Table 24.6 Areas under normal curve from 0 to Z