22-Mec-B5 Product Design and Development · December 2016

Question 6 of 7: Enhancing Reliability and Robustness

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 6: Enhancing Reliability and Robustness (15 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.

A — The process (9 marks)

Reliability and robustness are related but not identical, and the process must respect the difference. Reliability is the probability that the product performs its function for a stated time under stated conditions; robustness is insensitivity of that function to variation it cannot control. Robustness is the cheaper of the two to buy, because it is bought in design rather than in tolerances or in inspection, so the sequence below attacks robustness first.

Step 1 — define the function and the failure. Write the intended function as a measurable response with a target and limits, the duty cycle it must survive, and an unambiguous definition of failure. For the roof-rack clamp the response is the preload it applies to the crossbar, target 2.00 kN with limits of 1.40 and 2.60 kN: below the lower limit the rack creeps under braking, above the upper limit the crossbar is crushed.

Step 2 — enumerate the failure modes. Design FMEA across the function tree, scoring severity, occurrence and detection, with a fault-tree analysis on anything whose severity is a safety issue. A rack releasing at highway speed is severity 10, so it gets a fault tree regardless of its risk priority number.

Step 3 — identify the noise factors. This is the step candidates most often omit and it is where robustness is won or lost. The five classical noises are piece-to-piece variation (crossbar width and wall thickness across vehicle brands), change over time (strap relaxation, thread wear, corrosion), customer usage (a user who tightens by feel), external environment (temperature swing from −30 to +40 degrees Celsius, salt, grit) and system interaction (roof flex).

Step 4 — parameter design. Choose the levels of the control factors — cam profile, lever length, spring rate, jaw pad modulus — that make the response least sensitive to those noises, using a designed experiment with the noises deliberately applied as an outer array and the signal-to-noise ratio as the response. For a nominal-the-best characteristic the ratio is $SN=10\log_{10}(\mu^{2}/\sigma^{2})$, which rewards reducing variation before centring the mean, since the mean can usually be shifted afterwards with a single cheap adjustment.

Step 5 — tolerance design, and only then. Tighten tolerances only on the few characteristics where parameter design has left residual sensitivity, because tolerance is bought with money for ever. Set the targets in capability terms, $C_p=T/6\sigma \ge 1.33$.

Step 6 — design margin against the physics of failure. Derate against the governing mechanism — fatigue, creep, wear, corrosion, electromigration — and place redundancy or a fail-safe where severity demands it; the rack carries an independent secondary strap so that clamp failure is not release.

Step 7 — lock it down. A control plan tied to the FMEA, capability monitoring on the characteristics identified, and a change process that re-opens the FMEA whenever a supplier or a process moves.

Given. Two candidate cam designs from the parameter-design experiment. Design A delivers a mean preload of 2.05 kN with a standard deviation of 0.31 kN; design B delivers 2.00 kN with 0.12 kN. The specification is 2.00 ± 0.60 kN. A field failure caused by preload error costs $95 to put right, and the deviation at which that cost is incurred is 0.90 kN.

Find. The signal-to-noise ratio, process capability and expected quality loss of each design, and hence which to release.

1. Compare the designs on signal-to-noise ratio. For a nominal-the-best response,$$SN_A=10\log_{10}\!\left(\frac{2.05^{2}}{0.31^{2}}\right)=16.41\ \text{dB},\qquad SN_B=10\log_{10}\!\left(\frac{2.00^{2}}{0.12^{2}}\right)=24.44\ \text{dB}$$a gain of 8.03 dB, which on this scale means design B’s variance is roughly one sixth of design A’s.
2. Check both against the specification. With a tolerance width $T=1.20$ kN,$$C_{p,A}=\frac{1.20}{6\times 0.31}=0.645,\qquad C_{p,B}=\frac{1.20}{6\times 0.12}=1.667$$so design A cannot meet the specification at all, while design B has room to spare. Even if design B’s mean drifted to 2.06 kN in production its capability index would still be $C_{pk}=0.54/(3\times 0.12)=1.50$, comfortably above the 1.33 target.
3. Price the residual variation. Taguchi’s quadratic loss function values any departure from target, not merely a departure outside the limits: $L=k\bigl(\sigma^{2}+(\mu-m)^{2}\bigr)$ with $k=A_0/\Delta_0^{2}=95/0.90^{2}=117.28$ dollars per kilonewton squared. Substituting,$$L_A=117.28\,(0.0961+0.0025)=11.56,\qquad L_B=117.28\,(0.0144)=1.69$$in dollars per unit, so$$\boxed{\Delta L=9.88\ \text{per unit},\ \text{about}\ 237{,}000\ \text{per year at 24,000 units}}$$in dollars. Design B is released, and note that the argument never needed a single unit to fall outside the tolerance band to justify itself.

B — Validation during the design process (3 marks)

Validation must be planned as a document, not improvised: a design verification plan and report that lists, for every requirement, the test, the sample size, the acceptance criterion and the responsible engineer. Its backbone has four parts. Confirmation runs re-build the parameter-design optimum with the noises applied and check that the predicted signal-to-noise ratio is actually achieved, because designed experiments occasionally predict an optimum that interactions do not deliver. Highly accelerated life testing steps temperature, vibration and load past the specification until the product breaks, to find the operating and destruct margins and expose failure modes the FMEA missed. Accelerated life testing then demonstrates the reliability target on a correlated duty cycle. Field-correlated validation closes the loop with instrumented vehicles on real roads.

Given. The clamp must demonstrate a reliability of 0.98 at the end of the warranty life, at 95 % confidence, with no failures permitted. The dominant mechanism is fatigue of the cam pivot, for which the Weibull shape parameter is 2.0.

Find. The sample size for a zero-failure demonstration, and the reduction available if the test is run past the warranty life.

1. Size the zero-failure test. If $n$ units survive to the target life, the confidence that reliability is at least $R$ is $C=1-R^{n}$, so$$n=\frac{\ln(1-C)}{\ln R}=\frac{\ln 0.05}{\ln 0.98}=\frac{-2.9957}{-0.020203}=149\ \text{units}$$which is an expensive test for a clamp with a die-cast body.
2. Trade test time against sample size. Testing to $k$ times the required life converts, under a Weibull model, to an equivalent reliability $R^{k^{\beta}}$, so the sample size falls by $k^{\beta}$. Running to 1.5 times the warranty life with $\beta=2$ gives $k^{\beta}=2.25$ and$$\boxed{n=\frac{149}{2.25}=66\ \text{units tested to 1.5 lives}}$$which is the practical reason accelerated methods exist: the same statistical statement for 44 % of the hardware.

C — Measuring success in the long run (3 marks)

Long-run success is measured from the field, not from the laboratory, and the metrics must be leading as well as lagging. The primary lagging measure is the warranty claim rate per thousand units in service; alongside it sit the cost of poor quality, the returned-parts analysis that classifies each failure against the FMEA, and the customer complaint and safety-recall record. The leading measures are process capability trends on the characteristics the control plan identified, supplier incoming quality, and audit-fleet teardown results. The step that turns claims into engineering is fitting a Weibull distribution to the field returns, because the shape parameter says which mechanism is acting: below one, infant mortality and therefore a process or supplier problem; near one, random external events; above one, wear-out and therefore a design-life problem.

Given. In the first full year, 214 warranty claims are received against 24,000 racks in service. The returned-parts analysis fits a Weibull shape parameter of 2.0.

Find. The claim rate per thousand, the characteristic life and the B10 life implied by the field data.

1. Reduce the raw claims to comparable measures. The rate is $214/24{,}000\times 1000=8.92$ claims per thousand units per year, so the one-year field reliability is $R(1)=0.99108$. Inverting the Weibull survival function $R(t)=e^{-(t/\eta)^{\beta}}$ at $t=1$ year,$$\eta=\frac{t}{\bigl(-\ln R\bigr)^{1/\beta}}=\frac{1}{(0.008957)^{1/2}}=10.57\ \text{years}$$and the life by which a tenth of the population has failed is$$\boxed{B_{10}=\eta\,(-\ln 0.9)^{1/\beta}=10.57\times 0.3246=3.43\ \text{years}}$$A shape parameter of 2.0 says this is wear-out, not infant mortality, so the corrective action is a design-life change rather than a supplier audit.