Risk and Reliability of Design

Risk is the combination of the probability that a hazard causes harm and the severity of that harm: $\text{Risk} = \text{probability} \times \text{consequence}$. Zero risk is impossible, so designers aim for risk that is as low as reasonably practicable (ALARP).

Risk and society

• People accept voluntary risks (driving, climbing) more easily than involuntary ones (a dam or chemical plant nearby).
• Rare events with large consequences (aircraft crash, bridge collapse) are feared more than common small ones, though total harm may be less.
• Acceptance depends on benefit, familiarity, control and trust. Engineers have an ethical duty to protect public safety, health and welfare before cost.

Regulations and standards

• Regulations/laws are legally binding (safety acts, boiler, pressure vessel and lift rules, building codes).
• Standards and codes (ISO, IEC, ASME, IS, NS) give accepted minimum design rules, factors of safety, test methods and dimensions; following them also helps in liability defence.
• Product liability puts responsibility on the designer and maker for defects and missing warnings.

Risk assessment steps

1. Define the machine limits, use and misuse.
2. Identify hazards (crushing, shearing, entanglement, high pressure, noise).
3. Estimate risk: likelihood (exposure, probability of event, chance to avoid) and severity.
4. Evaluate against criteria: acceptable or not.
5. Reduce risk in order: eliminate by design, use guards and protective devices, then warnings and training, then personal protective equipment.
6. Document, review and repeat after any change.

(a) Exponential law

$R(t) = e^{-\lambda t}$, MTBF $= 1/\lambda$, with $\lambda = 2.5\times10^{-4}\ \text{h}^{-1}$.

Reliability at 400 h

$$R(400) = e^{-2.5\times10^{-4}\times 400} = e^{-0.1} = 0.9048$$

MTBF

$$\text{MTBF} = \frac{1}{2.5\times10^{-4}} = 4000\ \text{h}$$

Time for $R = 0.95$

$$t = -\frac{\ln 0.95}{\lambda} = \frac{0.05129}{2.5\times10^{-4}} = 205.2\ \text{h}$$

Failures among 500 components in 1000 h

$$R(1000) = e^{-0.25} = 0.7788,\quad \text{failures} = 500(1-0.7788) = 110.6 \approx 111$$

(b) System reliability

        +--- B (0.90) ---+
 --A----|                |----D----
 (0.95) +--- C (0.90) ---+   (0.98)

Parallel pair: $R_{BC} = 1 - (1-0.90)(1-0.90) = 1 - 0.01 = 0.99$

Series system: $R_s = R_A \times R_{BC} \times R_D$

$$R_s = 0.95 \times 0.99 \times 0.98 = 0.9217$$

Note that without the redundancy in B and C the reliability would be $0.95\times0.90\times0.98 = 0.838$, so the parallel pair improves the system by about 8 percentage points. The weakest series element (A) now limits reliability.

Answer: (a) $R(400)=0.905$, MTBF $=4000$ h, $t_{0.95}=205$ h, about 111 failures; (b) $R_s = 0.922$.

(a) Stress-strength interference

In probabilistic design, stress $s$ and strength $S$ are random variables with distributions. Failure occurs when stress exceeds strength. The region where the two distribution curves overlap is the interference; its area is related to the probability of failure.

 f
 |   stress     strength
 |    /\          /\
 |   /  \        /  \
 |  /    \__/\__/    \
 +-------------------------> MPa
       overlap = interference

Reliability $R = P(S > s)$. A larger gap between the means or smaller scatter gives higher reliability. The factor of safety based on means alone does not show this.

(b) Numerical

For normal variables the margin $M = S - s$ is also normal:

$$\mu_M = 350 - 250 = 100\ \text{MPa}$$ $$\sigma_M = \sqrt{30^2 + 25^2} = \sqrt{1525} = 39.05\ \text{MPa}$$

Standard normal variate (safety index):

$$z = \frac{\mu_M}{\sigma_M} = \frac{100}{39.05} = 2.56$$

From the normal table $\Phi(2.56) = 0.9948$, so

$$R = 0.9948,\qquad P_f = 1 - R = 0.0052\ (\text{about } 1 \text{ in } 190)$$

Mean strength for $R = 0.999$: $z = 3.09$.

$$\mu_S - 250 = 3.09 \times 39.05 = 120.7 \;\Rightarrow\; \mu_S = 370.7\ \text{MPa}$$

Answer: $R = 0.995$, $P_f = 0.0052$; required mean strength about 371 MPa.

Hazard analysis finds what can go wrong, how likely it is and what it causes, so that design changes can remove or reduce it.

FMEA (bottom-up)

For each component list the failure mode, its effect on the system, its cause and current controls. Rate severity (S), occurrence (O) and detection (D) from 1 to 10 and calculate the risk priority number $RPN = S \times O \times D$. Highest RPN items are fixed first.

FTA (top-down)

Start with an undesired top event and trace the combinations of causes down to basic events using logic gates.

        [ Machine stops ]
              |
            (OR)
         /          \
  [Motor fails]   [Power lost]
                       |
                     (AND)
                   /      \
            [Mains off] [Backup fails]

AND gate: all inputs needed; OR gate: any input is enough. Probabilities of basic events give the top-event probability.

Design for reliability

• Use proven components, adequate factor of safety and derating (use below rated load).
• Redundancy (parallel units, standby) for critical functions.
• Simplify: fewer parts mean fewer failures in a series system.
• Fail-safe and fail-soft design, protective devices, good lubrication and sealing, controlled manufacturing and testing.

Maintenance and repair

Preventive maintenance (inspection, replacement) reduces failure rate during useful life. Availability $= \dfrac{MTBF}{MTBF + MTTR}$, so easy access, modular parts and a short repair time (MTTR) raise availability.