Question 19.4: A product is made by aligning four components (see Figure 19...
A product is made by aligning four components (see Figure 19.11) whose lengths are x_1, x_2, x_3, and x_4. The overall length is denoted by Ζ. The tolerance limits of the lengths are known to be 8.000 ± 0.030, 10.000 ± 0.040, 15.000 ± 0.050, and 7.000 ± 0.030 mm, respectively. If the lengths of the four components are independently normally distributed, estimate the tolerance limits for the length of the overall assembly.
Assuming ±3σ natural tolerance limits:
\sigma_{1}=0.03 / 3=0.01 mm\sigma_{2}=0.04 / 3=0.0133333 mm
\sigma_{3}=0.05 / 3=0.0166667 mm
\sigma_{4}=0.03 / 3=0.01 mm
\sigma_{z}^{2}=\sigma_{1}^{2}+\sigma_{2}^{2}+\sigma_{3}^{2}+\sigma_{4}^{2}=0.01^{2}+0.013333^{2}+0.0166667^{2}+0.01^{2}=6.555 \times 10^{-4}
\sigma_{z}=0.0256
Overall variability =6 \sigma_{z}.
Tolerance limits =\pm\left(6 \sigma_{z} / 2\right)=\pm 3 \sigma_{z}=\pm 0.077.
The mean length of the overall assembly is given by
\mu_{z}=\mu_{x_{1}}+\mu_{x_{2}}+\mu_{x_{3}}+\mu_{x_{4}}=8+10+15+7=40 mmz=40.000 \pm 0.077 mm
So 99.73% of all items would have an overall dimension within these tolerance limits.
(By comparison, the sure fit or extreme variability tolerance limits would be z=40.000 \pm 0.150 mm, i.e. 100% of all items would be within this tolerance limit.)
The basic normal model indicated a better, or more attractive (certainly for marketing purposes), tolerance band. If the sure-fit model was used and an assembly tolerance of ±0.077 mm was desired, the tolerances on the individual components would need to be 8.000 ± 0.019, 10.000 ± 0.019, 15.000 ± 0.019, and 7.000 ± 0.019 mm. This is a very fine or tight tolerance and would be costly to achieve. As the statistical model takes account of 99.73% of the assemblies, use of the sure-fit approach only gives 0.27% of the assemblies with better quality and would require considerable extra effort and expense.