Safety Margin on Strength, and Specification Limits from a Pilot Run: 22-Mec-B4 Integrated Manufacturing Systems, December 2018, Q4

Question 4: Safety Margin on Strength, and Specification Limits from a Pilot Run (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Part (a): a deterministic applied stress and a strength distribution whose scatter is a fixed fraction of its own mean. Part (b): a pilot-run sample of 20 bores.

Given data
Quantity	Symbol	Value
(a) Maximum applied stress	$S$	40,000 psi
(a) Strength scatter	$\sigma_s$	0.10 $\mu_s$
(a) Safety margins required	SM	3.0, 4.0, 5.0
(b) Sample size	$n$	20
(b) Sample mean bore	$\bar{x}$	9.8576 in
(b) Sample standard deviation	$s$	0.00015 in

Find. (a) the average strength that must be specified at each of the three safety margins; (b)(i) limits containing 99 per cent of production when the sample estimates are taken as exact, and (b)(ii) limits that are 95 per cent certain to contain 99 per cent of production when they are not.

Figure 4.1 — Part (a). Raising the required safety margin pushes the whole strength distribution to the right, but because the scatter is a fixed 10 per cent of the mean, each distribution also gets wider as it moves — which is why the required mean strength grows faster than the safety margin does.

Part (a) — the specification on average strength

Approach. Write the safety margin as the number of strength standard deviations that separate the mean strength from the applied stress, substitute $\sigma_s = 0.10\mu_s$, and solve the resulting linear equation for $\mu_s$.

Part (a) — define the safety margin. With the stress deterministic at $S = 40{,}000$ psi and only the strength random, $$\text{SM} = \frac{\mu_s - S}{\sigma_s}$$ which is exactly the standard normal deviate at which the strength distribution reaches the applied stress; the reliability of the component is therefore $R = \Phi(\text{SM})$.
Substitute the proportional scatter and solve. Putting $\sigma_s = 0.10\mu_s$, $$\text{SM} = \frac{\mu_s - S}{0.10\,\mu_s} \quad\Longrightarrow\quad \mu_s\,(1 - 0.10\,\text{SM}) = S \quad\Longrightarrow\quad \mu_s = \frac{S}{1 - 0.10\,\text{SM}}$$ The denominator is what makes the answer grow faster than linearly: every increase in the required margin also inflates the standard deviation that the margin is measured in.
Evaluate at the three margins. $$\mu_s(\text{SM}=3.0) = \frac{40{,}000}{1 - 0.30} = \frac{40{,}000}{0.70} = 57{,}143 \text{ psi}$$ $$\mu_s(\text{SM}=4.0) = \frac{40{,}000}{0.60} = 66{,}667 \text{ psi}, \qquad \mu_s(\text{SM}=5.0) = \frac{40{,}000}{0.50} = 80{,}000 \text{ psi}$$ $$\boxed{\;\mu_s = 57{,}143 \ /\ 66{,}667 \ /\ 80{,}000 \text{ psi at SM} = 3 / 4 / 5\;}$$ The corresponding strength standard deviations are 5,714, 6,667 and 8,000 psi, and the implied reliabilities are $\Phi(3) = 0.99865$, $\Phi(4) = 0.999968$ and $\Phi(5) = 0.9999997$.

The engineering reading of these three numbers is that moving from a safety margin of 3 to one of 5 costs a 40 per cent increase in specified mean strength — a material change, a heat treatment, or a section increase — while buying a reduction in failure probability from about 1 in 740 to about 1 in 3.5 million. Which of the three is correct is a consequences-of-failure judgement, not a statistical one.

Part (b) — setting specification limits from the pilot run

Approach. Part (i) treats $\bar{x}$ and $s$ as if they were $\mu$ and $\sigma$, so the limits are a straight normal-percentile calculation. Part (ii) admits that they are estimates from only 20 pieces, which calls for a two-sided statistical tolerance interval whose factor is larger than the normal deviate.

Part (b)(i) — treat the estimates as exact. If $\mu = 9.8576$ in and $\sigma = 0.00015$ in were known without error, the central 99 per cent of a normal population lies within $z_{0.995}$ standard deviations of the mean, with $z_{0.995} = 2.576$: $$\bar{x} \pm z_{0.995}\,s = 9.8576 \pm 2.576(0.00015) = 9.8576 \pm 0.000386$$ $$\boxed{\;9.85721 \text{ to } 9.85799 \text{ in}\;}$$
Part (b)(ii) — allow for the sampling error. With only 20 pieces, $\bar{x}$ and $s$ are themselves random, and interval limits computed as if they were not will contain 99 per cent of production well under half the time — about 40 per cent of samples of 20. The correct construction is the two-sided normal tolerance interval $\bar{x} \pm k s$, whose factor $k$ is chosen so that the interval covers a proportion $P$ of the population with confidence $\gamma$. For $n = 20$, $P = 0.99$ and $\gamma = 0.95$ the tabulated factor is $$k = 3.615$$ Howe's closed-form approximation confirms the table value from the underlying chi-square: $$k \approx \sqrt{\frac{(n-1)\left(1 + 1/n\right)z_{0.995}^{2}}{\chi^{2}_{0.05,\,n-1}}} = \sqrt{\frac{19(1.05)(2.5758)^{2}}{10.117}} = 3.617$$
Apply the tolerance factor. $$\bar{x} \pm k\,s = 9.8576 \pm 3.615(0.00015) = 9.8576 \pm 0.000542$$ $$\boxed{\;9.85706 \text{ to } 9.85814 \text{ in}\;}$$ The interval is 1.40 times wider than the one in part (i), and that widening is the entire price of not knowing $\mu$ and $\sigma$. It is also the reason a pilot run of 20 is a thin basis for a production specification: at $n = 100$ the factor falls to about 2.93, and the limits tighten accordingly.

Check: what the specification is for. Part (b) asks for limits that contain production, which is a description of what the pilot process actually makes, not a statement of what the impeller needs functionally. The paper's own phrasing — “all the units functioned properly, and so it was decided to use the data” — licenses exactly this reading. Note in passing that such limits guarantee only conformance to past behaviour: if the pilot process is later improved or degraded, the limits carry no functional authority, and a proper drawing tolerance should still be established from the impeller's fit and performance requirements.

Question 4 — results
Quantity	Value
(a) Required mean strength at SM = 3.0	57,143 psi ($\sigma_s = 5{,}714$ psi)
(a) Required mean strength at SM = 4.0	66,667 psi ($\sigma_s = 6{,}667$ psi)
(a) Required mean strength at SM = 5.0	80,000 psi ($\sigma_s = 8{,}000$ psi)
(a) Implied reliabilities	0.99865, 0.999968, 0.9999997
(b)(i) Normal deviate for 99 per cent	$z_{0.995} = 2.576$
(b)(i) Specification limits	9.85721 to 9.85799 in (± 0.000386 in)
(b)(ii) Tolerance factor, $n=20$, 95 per cent / 99 per cent	$k = 3.615$ (Howe check 3.617)
(b)(ii) Specification limits	9.85706 to 9.85814 in (± 0.000542 in)
(b) Ratio of the two interval widths	1.40

Topic. Stress–strength interference and the safety margin; statistical tolerance intervals as against normal-percentile limits.

Key relations. $\text{SM} = (\mu_s - S)/\sigma_s$ and $R = \Phi(\text{SM})$ for deterministic stress; $\mu_s = S/(1 - c\,\text{SM})$ when $\sigma_s = c\,\mu_s$; $\bar{x} \pm z_{(1+P)/2}\,\sigma$ when the parameters are known; $\bar{x} \pm k\,s$ with $k = k(n, P, \gamma)$ when they are estimated, and Howe's approximation $k \approx \{(n-1)(1 + 1/n)z^2_{(1+P)/2}/\chi^2_{1-\gamma,\,n-1}\}^{1/2}$.

Why this works. The safety margin is the standard normal deviate at which the strength distribution meets the applied stress, so it converts directly into a probability of failure; that is what makes it a more useful design statement than a deterministic factor of safety, which says nothing about scatter. The tolerance interval works because the sampling distributions of $\bar{x}$ and of $s$ are known exactly for a normal population — normal and chi-square respectively — so the extra width needed to absorb their joint variability can be computed once and tabulated as $k$.

Common pitfall. In (a), forgetting that $\sigma_s$ moves with $\mu_s$ and simply writing $\mu_s = S + 3\sigma_s$ with $\sigma_s = 0.10(40{,}000) = 4{,}000$ psi, which gives 52,000 psi instead of 57,143 psi. In (b), confusing the three intervals that share the same arithmetic shape: a confidence interval on the mean uses $s/\sqrt{n}$ and is far narrower; a prediction interval covers one future observation; only the tolerance interval covers a stated proportion of the whole population, and only it answers this question. Using Student's $t$ in place of $k$ in part (ii) is the same mistake in another costume.

Source. C. E. Ebeling, An Introduction to Reliability and Maintainability Engineering, 3rd ed., Ch. 12 (design for reliability: stress–strength interference and the safety margin); D. C. Montgomery, Introduction to Statistical Quality Control, 8th ed., Ch. 4 and Appendix VII (tolerance intervals and their factors); A. J. Duncan, Quality Control and Industrial Statistics, 5th ed., chapter on setting tolerances from sample data.

Question 4 of 7: Safety Margin on Strength, and Specification Limits from a Pilot Run

Question 4: Safety Margin on Strength, and Specification Limits from a Pilot Run (20 marks)

Part (a) — the specification on average strength

Part (b) — setting specification limits from the pilot run