1. The Fundamentals of the 2-Parameter Weibull Distribution
The Weibull cumulative distribution function (CDF), representing the probability that a component will fail prior to operating age t, is expressed as:
Reliability R(t) = 1 - F(t) = e^[ - (t / η)^β ]
Hazard Rate h(t) = (β / η) × (t / η)^(β - 1)
Where:
- β (Beta - Shape Parameter): Dimensionless. Controls the slope of the hazard function and identifies the failure mechanism.
- η (Eta - Characteristic Life / Scale Parameter): Measured in time units (hours, cycles, kilometers). The operating age at which 63.2% of the population will have failed, regardless of the β value.
2. Interpreting the Shape Parameter (β) and the Bathtub Curve
β = 1.0 : Constant Failure Rate (CFR) → Random environmental stress (Exponential Distribution)
β > 1.0 : Increasing Failure Rate (IFR) → Mechanical Wear-out, erosion, fatigue, chemical degradation
The value of β directly dictates maintenance strategy:
- If β ≤ 1.0, time-based preventive maintenance (PM) is counterproductive. Replacing parts at fixed calendar intervals actually increases total failures due to infant mortality from human installation errors.
- If β > 1.5 - 2.0, time-based replacement is mathematically justified, because older components have an escalating probability of breakdown.
3. Calculating B10 Life
In mechanical design (such as bearing life per ISO 281), B10 Life represents the point in time where 10% of the population is expected to fail (90% reliability):
A fleet of 5 heavy electric motor bearings operated to failure with the following operating hours recorded: 1,200, 1,750, 2,100, 2,600, and 3,100 hours.
- Step 1: Fit Weibull Parameters
Using median rank regression (Bernard's approximation (i - 0.3)/(N + 0.4)):
Calculated Shape Parameter β = 2.65
Calculated Characteristic Life η = 2,420 operating hours - Step 2: Physical Interpretation
Because β = 2.65 (> 1.0), the bearings exhibit classic mechanical fatigue wear-out. Time-based replacement is viable. - Step 3: Calculate B10 Life
B10 = 2,420 × (0.10536)^(1 / 2.65) = 2,420 × 0.4283 = 1,036 hours.
90% of bearings will survive past 1,036 operating hours without failure.
Frequently Asked Questions
Why does Eta always correspond to 63.2% failure probability?
When t = η, F(η) = 1 - e^(-1) = 1 - (1 / 2.71828) = 1 - 0.3679 = 0.6321, or 63.2%, regardless of the value of β.
How many failure points are needed for an accurate Weibull plot?
While mathematical fitting can technically occur with as few as 3-5 failures, industrial standards (IEC 61649) recommend at least 10 to 20 failure data points for tight 90% confidence bounds.