Weibull Reliability Calculator

A two-parameter Weibull model with known fitted parameters.

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1 Inputs

Minimum: 0

Minimum: 0.05

Minimum: 0.1

2 Results

Reliability R(t)
46.73%
Prob. of failure F(t)
53.27%
Hazard rate h(t)
1.141 /1000h
B10 life
268 h
Failure pattern
β>1 wear-out
R(t) = exp[−(t/η)^β]; B10 = η × (−ln 0.9)^(1/β)
A two-parameter Weibull model with known fitted parameters. β describes the hazard trend, not the failure cause. R(t) is a probability, not a maintenance interval; validate the fit and censored observations first.

Worked example

Example values are illustrative. Replace them with your measured data.

InputValue
Time / cycles of interest (t)1000 hours
Shape parameter β1.5 —
Scale / characteristic life η1200 hours

Reliability R(t): 46.73%

References

Related tools

About this calculator

How it works

With shape β, scale η and time t, the tool computes reliability R(t) = exp(−(t/η)^β), probability of failure F(t) = 1 − R(t), hazard rate h(t) = (β/η)(t/η)^(β−1) (shown per 1,000 h), and B10 life = η × (−ln 0.9)^(1/β), the time by which 10% are expected to fail.

Worked example (illustrative inputs)

β = 1.5, η = 1,200 h, t = 1,000 h. R(1,000) = exp(−0.833^1.5) ≈ 46.7%, so F ≈ 53.3%. The hazard rate is about 1.141 per 1,000 h. B10 ≈ 268 h.

How to read the result

β tells you the failure pattern. Below 1 means early-life failures, which point to commissioning, burn-in and installation quality rather than time-based PM. About 1 (within 0.1) means random failures, where condition monitoring beats scheduled replacement. Above 1 means wear-out, where replacement based on time or usage at or before B10 is effective. The example (β = 1.5) is wear-out.

When to use it

Use it once you have β and η fitted from your own life or failure data for a component, to set replacement intervals or check survival at a planned run time.

À propos de ce calculateur

Traduction (brouillon) — à relire.

Fonctionnement

Avec la forme β, l'échelle η et le temps t, l'outil calcule la fiabilité R(t) = exp(−(t/η)^β), la probabilité de défaillance F(t) = 1 − R(t), le taux de défaillance h(t) = (β/η)(t/η)^(β−1) (pour 1 000 h) et la durée B10 = η × (−ln 0,9)^(1/β), temps auquel 10 % sont censés avoir défailli.

Exemple chiffré (valeurs d'illustration)

β = 1,5, η = 1 200 h, t = 1 000 h. R(1 000) = exp(−0,833^1,5) ≈ 46,7 %, donc F ≈ 53,3 %. Le taux de défaillance est d'environ 1,141 pour 1 000 h. B10 ≈ 268 h.

Lire le résultat

β indique le type de défaillance. En dessous de 1, ce sont des défaillances de jeunesse : la mise en service, le rodage et la qualité d'installation comptent plus qu'une maintenance périodique. Autour de 1 (à 0,1 près), les défaillances sont aléatoires et la surveillance d'état vaut mieux que le remplacement planifié. Au-dessus de 1, c'est de l'usure : un remplacement à temps ou à l'usage, au plus tard à B10, est efficace. L'exemple (β = 1,5) relève de l'usure.

Quand l'utiliser

À utiliser une fois β et η ajustés sur vos propres données de durée de vie ou de pannes d'un composant, pour fixer des intervalles de remplacement ou vérifier la survie à une durée donnée.

حول هذه الحاسبة

ترجمة (مسودة) — للمراجعة.

طريقة الحساب

باستخدام معامل الشكل β ومعامل المقياس η والزمن t، تحسب الأداة الموثوقية R(t) = exp(−(t/η)^β)، واحتمال العطل F(t) = 1 − R(t)، ومعدل الخطر h(t) = (β/η)(t/η)^(β−1) (لكل 1000 ساعة)، وعمر B10 = η × (−ln 0.9)^(1/β)، وهو الزمن الذي يُتوقع أن يتعطل عنده 10%.

مثال محسوب (قيم توضيحية)

β = 1.5، η = 1200 ساعة، t = 1000 ساعة. R(1000) = exp(−0.833^1.5) ≈ 46.7%، إذن F ≈ 53.3%. معدل الخطر نحو 1.141 لكل 1000 ساعة. B10 ≈ 268 ساعة.

قراءة النتيجة

تكشف β نمط العطل. أقل من 1 يعني أعطال بداية العمر، وعندها تهم جودة التشغيل الأولي والتركيب أكثر من الصيانة الدورية. نحو 1 (بفارق 0.1) يعني أعطالاً عشوائية، ومراقبة الحالة أفضل فيها من الاستبدال المجدول. أكثر من 1 يعني تآكلاً، والاستبدال حسب الزمن أو الاستخدام عند B10 أو قبله فعّال. المثال (β = 1.5) حالة تآكل.

متى تستخدمه

استخدمها بعد تقدير β وη من بيانات العمر أو الأعطال الخاصة بك لمكوّن ما، لتحديد فترات الاستبدال أو التحقق من البقاء حتى زمن تشغيل مخطط.

In one line

A two-parameter Weibull model with known fitted parameters. β describes the hazard trend, not the failure cause. R(t) is a probability, not a maintenance interval; validate the fit and censored observations first.

Worked example

EXAMPLE

Illustrative values (the calculator defaults), not data from a real site.

Inputs
Time / cycles of interest (t)1000 hours
Shape parameter β1.5 —
Scale / characteristic life η1200 hours
Results
Reliability R(t)46.73%
Prob. of failure F(t)53.27%
Hazard rate h(t)1.141 /1000h
B10 life268 h
Failure patternβ>1 wear-out

Frequently asked questions

What does the shape parameter β actually tell me?

β below 1 signals infant-mortality failures with a decreasing failure rate, β near 1 signals random, constant-rate failures, and β above 1 signals wear-out with an increasing failure rate — think of these as the three regions of the classic bathtub curve. It's the single most useful number for deciding whether preventive replacement makes sense for that failure mode.

What is B10 life and why is it used instead of mean life?

B10 is the time by which 10% of the population is expected to have failed, meaning 90% survive. It's more conservative than mean life, which is why it's a common design and bearing-life specification — it flags meaningful failure risk well before half the population would have failed.

Where do I get β and η for my own equipment?

They come from fitting a Weibull distribution to your actual failure or life data, meaning time-to-failure records for a population of similar components, typically using probability-plotting or maximum-likelihood software — this calculator assumes you already have those two parameters rather than deriving them from raw data.

How is this different from just using MTBF?

MTBF and the simple exponential reliability model implicitly assume a constant failure rate, β equal to 1. The Weibull model is more general — it can represent early-life and wear-out patterns that a single MTBF number hides, which is why it's preferred for components with known wear characteristics like bearings.

How much failure data do I need before I can trust a fitted β and η?

More data points give a more reliable fit; analyses based on only a handful of failures can produce a shape parameter with wide uncertainty. Treat results from small samples as a rough directional signal and combine them with engineering judgment and manufacturer data rather than as a firm number to act on alone.

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