to Solutions Exercises for Chapter 16
16.1. An assembly consists of three stacked parts. Due to manufacturing imperfections the parts are not identical, and their thicknesses follow a normal distribution with a range of natural variation of .006 inches (which represents six standard deviations). What tolerance range (i.e., 3s limits ) could you hope to hold on the thickness of the stacked assembly? 6s = 0.012 s = .002 s 2Thickness = s 2 + s 2 + s 2 = .000012 s Thickness =
.000012 = .003464
3 s Thickness = 0.0104 16.2 A finish on a metal consist of 3 coats. A primer coat, a finish coat and a clear coat. The mean and standard deviation of the thicknesses of the three coats are P = 0.0025, s P = 0.0013, F = 0.0021, s F = 0.0011, C = 0.0017, s C = 0.0008. What is the mean and standard deviation of the finish?
Thickness = P + F + C = .0025 + .0021 + .0017 Thickness = .0063
2 2 2 2 Thickness = P + F + C
= 0013) (. 2 + (.0011) 2 + (. 0008) 2 = .00000354 Thickness = 0.00188
16.3 Finish part dimensions are measured on a micrometer, and the variability in measurements can be assumed to be a sum of the part to part variance, the variance due to gage, and finally the variance of repeat measurements. Symbolically this is expressed as:
2 2 2 2 T = P + G + R
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If the three variance components are respectively: 2.73, 0.034, and 0.132.
a) Calculate the total variance in measured part dimensions
2 2 T = 2 + G + 2 = 2.73 + 0 .034 + 0 .132 P R
= 2.896
b) Calculate the percentage or proportion of the total variation in measures that is actually due to part to part differences in dimensions.
2 P 2.73 100 2 = 100 = 94 .3% 2.896 T
c) Would micrometer measurements taken in this way be accurate enough to use in a control charts? Why or why not? Yes, because actual part to part variability is > 90% of total.
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