Question 24.7: An assembly of three components A, B and C is shown in Fig. ...

Step I Population of component-A (A)
For the population of component A,

\mu_{A}=40.00 mm \quad \hat{\sigma}_{A}=\frac{0.09}{3}=0.03 mm          (a).

Step II Population of component–B (B)
For the population of the component B,

\mu_{B}=60.00 mm \quad \hat{\sigma}_{B}=\frac{0.09}{3}=0.03 mm .        (b).

Step III Population consisting of addition of populations A and B (X)
A third population consisting of the addition of the populations A and B is formed and denoted by the letter X. Therefore,

\mu_{X}=\mu_{A}+\mu_{B}=40+60=100 mm .

\hat{\sigma}_{X}=\sqrt{\left(\hat{\sigma}_{A}\right)^{2}+\left(\hat{\sigma}_{B}\right)^{2}}=\sqrt{(0.03)^{2}+(0.03)^{2}} .

= 0.0424 mm                (c)
Step IV Population of component-C (C)
For the population of the component C,

\mu_{C}=100.09 mm \quad \hat{\sigma}_{c}=\frac{0.09}{3}=0.03 mm              (d).

Step V Population of interference (I)
The population for interference is denoted by the letter I. It is obtained by subtracting the population C from the population X.

\mu_{I}=\mu_{X}-\mu_{C}=100.00-100.09=-0.09 mm .

\hat{\sigma}_{I}=\sqrt{\left(\hat{\sigma}_{X}\right)^{2}+\left(\hat{\sigma}_{C}\right)^{2}}=\sqrt{(0.0424)^{2}+(0.03)^{2}} .

= 0.052 mm
Step VI Percentage of assemblies with interference
When interference is zero,

I=0 \text { and } Z=\frac{I-\mu_{I}}{\hat{\sigma}_{I}}=\frac{0-(-0.09)}{0.052}=+1.73 .

When Z is less than 1.73, I is negative, giving a clearance fi t. The interference fi t is given by the area below the normal curve from Z = + 1.73 to Z = ∞. From Table 24.6, the area below the normal curve from Z = 0 to Z = + 1.73 is 0.4582. Therefore, the percentage of interference assemblies is (0.5 – 0.4582) × 100 or 4.18%.

Table 24.6 Areas under normal curve from 0 to Z