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Probabilistic Design and Reliability | Module 6 of 10

Lifetime Reliability and Failure Data

Lifetime data are not just failed parts in a spreadsheet. Surviving parts, censored tests, hazard behavior, and mechanism knowledge all shape a defensible reliability statement.

Readiness check

Tick only what you can do closed-notes before using this module for design decisions.

Read a survival probability as the fraction still functioning.
Explain why a test can end before a part fails.
Use natural logarithms in a formula.
Recognize a right-censored observation.
Separate observed sample behavior from population behavior.

0 or 1 weak itemsContinue with this module.

2 weak itemsReview data handling in Programming, Module 10.

3 or more weak itemsReview statistics and measurement in Measurements, Module 4.

The core idea

Lifetime reliability describes the probability an item survives beyond time t. Hazard describes the instantaneous failure tendency among items that have survived to t.

R(t) = P[T > t] = 1 - F(t)exponential: R(t) = exp(-lambda t)Weibull: R(t) = exp(-(t/eta)^beta)

Lifetime reliability uses time-to-failure as the random variable. The reliability function R(t) is survival probability. The cumulative failure distribution F(t) is one minus survival. The probability density describes where failures occur in time. The hazard function is different: it is the failure rate among items that have already survived to that time. Exponential lifetimes have constant hazard, useful for some random electronic failures but often too simple for mechanical wear. Weibull models are useful because beta can describe decreasing, constant, or increasing hazard, but a fitted Weibull curve is not proof of a failure mechanism. Lognormal models are useful for multiplicative lifetime scatter. Censoring matters because parts that have not failed by the test end still carry information: they survived at least that long.

The skill works when: the failure event, units, assumptions, distributions, and checks are visible.

The skill breaks down when: a probability is reported without the mechanical model and evidence boundary.

The concept. Lifetime reliability describes the probability an item survives beyond time t. Hazard describes the instantaneous failure tendency among items that have survived to t.

The skills, taught in order

Define the time origin and mission

A lifetime statement needs start time, duty cycle, environment, and what counts as failure.

Use survival and hazard correctly

R(t), F(t), density, and hazard answer different questions.

Respect censoring

Right-censored data are not failed at the censor time; they survived at least that long.

Choose a lifetime family cautiously

Exponential, Weibull, and lognormal models encode different hazard behavior.

Report uncertainty in fitted models

Small samples do not support precise long-life guarantees.

Engineering connection: use these skills to turn variability into a design decision, not just a plot.

Worked example 1: Weibull reliability for bearing life

A fitted Weibull lifetime model for a teaching data set has beta = 2.0 and eta = 1000 hours. Estimate reliability at 500 hours and 1000 hours.

ProblemA fitted Weibull lifetime model for a teaching data set has beta = 2.0 and eta = 1000 hours. Estimate reliability at 500 hours and 1000 hours.
AssumptionsThe fitted parameters are accepted for this exercise. Failure mode and use condition are consistent.
ModelR(t) = exp(-(t/eta)^beta)
SolveAt 500 hours, R = exp(-(0.5)^2) = exp(-0.25) = 0.7788. At 1000 hours, R = exp(-1) = 0.3679.
CheckFor beta = 2, hazard increases with time, so reliability drops more quickly as the component ages.
ConclusionReliability is about 0.779 at 500 hours and 0.368 at 1000 hours.

Result. Reliability is about 0.779 at 500 hours and 0.368 at 1000 hours.

Worked example 2: right-censored test observations

Five springs are tested to 100000 cycles. Failures occur at 42000, 61000, and 77000 cycles. Two springs have not failed when the test stops at 100000 cycles. Identify the observed data.

ProblemFive springs are tested to 100000 cycles. Failures occur at 42000, 61000, and 77000 cycles. Two springs have not failed when the test stops at 100000 cycles. Identify the observed data.
AssumptionsAll springs were under the same test condition.
Modelfailure times: observed failurescensored times: survived at least the censor time
SolveThe failure observations are 42000, 61000, and 77000 cycles. The two nonfailed springs are right-censored at 100000 cycles, not failures at 100000 cycles.
CheckTreating censored springs as failures at 100000 cycles understates their actual unknown lifetime and biases the model.
ConclusionThe data set contains three failures and two right-censored observations.

Result. The data set contains three failures and two right-censored observations.

Python activity

Engineering question: Compute a simple Weibull survival curve for a synthetic lifetime model.

Assumptions and units: Use the units shown in the code comments. Keep the random seed fixed while learning so the result is reproducible.

from math import exp

beta = 2.0
eta = 1000.0
times = [250, 500, 750, 1000, 1250]

for t in times:
    reliability = exp(-((t / eta) ** beta))
    hazard = (beta / eta) * ((t / eta) ** (beta - 1.0))
    r = reliability
    h = hazard
    print(int(t), "hours: R =", round(r, 4), "hazard =", round(h, 6))

print("Note: beta > 1 gives increasing hazard in this model.")

Numerical checks: Reliability must decrease as time increases. With beta = 2, hazard should increase with time.

Limitations: The activity uses a synthetic model. Real fits require failure-mode consistency, censoring treatment, and uncertainty intervals.

Misconceptions and diagnostics

Mistake	Diagnostic question	Correction
The Weibull distribution is universally correct for lifetimes.	Ask: what physical mechanism and data support the chosen family?	Correct the assumption, then rerun the calculation or rewrite the report.
Censored parts failed at the test end.	Ask: did the part fail, or did observation stop while it was still working?	Correct the assumption, then rerun the calculation or rewrite the report.
A bathtub curve explains every mechanical component.	Ask: what actual hazard evidence supports that shape?	Correct the assumption, then rerun the calculation or rewrite the report.
Observed samples are the true population behavior.	Ask: how much uncertainty remains from sample size and test conditions?	Correct the assumption, then rerun the calculation or rewrite the report.

Practice ladder

Recognize

Identify which observations are right-censored in a test stopped at 200 hours with some parts still running.

Solution guidance

Write the governing equation, state units and assumptions, then compare the answer with the relevant limit state or design decision.

Calculate

For an exponential model with lambda = 0.001 per hour, compute R(1000).

Solution guidance

Write the governing equation, state units and assumptions, then compare the answer with the relevant limit state or design decision.

Diagnose

A report treats all suspended tests as failures. Explain the bias.

Solution guidance

Write the governing equation, state units and assumptions, then compare the answer with the relevant limit state or design decision.

Design

Write a data-collection plan for a bearing life test that records censoring correctly.

Solution guidance

Write the governing equation, state units and assumptions, then compare the answer with the relevant limit state or design decision.

Working with AI, and proving it yourself

Useful AI support

Ask for alternative explanations of terminology, candidate code tests, or a checklist of assumptions to verify.

Ask it to challenge a derivation, then prove the result with units, limiting cases, and an independent calculation.

Do not let AI invent material distributions, fabricate failure data, claim standards compliance, or make final safety decisions.

Portfolio evidence

Create one short reliability note for this module. It must include the engineering question, assumptions, input definitions, equations, calculation, checks, interpretation, and limitations.

Verification required: dimensional analysis, a benchmark case, convergence or sample-size check, and a plain-language residual-risk statement.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

What is R(t)?

The probability lifetime T exceeds time t.

What is F(t)?

The cumulative failure probability by time t, equal to 1 - R(t).

What does exponential lifetime assume about hazard?

Constant hazard.

What does Weibull beta > 1 indicate?

Increasing hazard in the Weibull model.

What is right censoring?

Observation stops before failure, so the item is known to survive at least to the censor time.

TodayFinish the review and first two practice levels.

+1 dayRebuild one worked example from a blank page.

+3 daysChange one assumption and predict the effect.

+7 daysConnect this module to the capstone bracket.

+30 daysReuse the method in a new mechanical component.

References and source discipline

This course synthesizes original MechCompass explanations from established reliability engineering sources. Confirm current editions and standards before using them for formal design approval.

Topic in this module	Reference direction
Reliability functions and lifetime distributions	Elsayed, Reliability Engineering, 3rd edition
Censoring and reliability data analysis	NIST/SEMATECH Engineering Statistics Handbook
Mechanical reliability life models	Le, Reliability-Based Mechanical Design, Volume 2