B10 Life for Weibull and Lognormal Distributions

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This video from the Institute of Quality and Reliability explains how to determine BX life for products that follow Weibull and log-normal distributions. BX life is defined as the time by which X percent of items are expected to fail. For example, B10 life means that 10% of items are expected to fail, which implies a 90% reliability.

We recommend watching following videos before watching this video for better learning experience:

Weibull Distribution Part-1

B10 Life for Exponential Distribution

Normal Distribution and Z-Score

Lognormal Distribution

Weibull Distribution

The Weibull distribution is used when the failure rate (λ) varies with time. Its reliability function is given by:

$$ R\left(T\right)=e^{-\left(\frac{T}{\eta}\right)^{\beta}} $$

Where:

• T is time.
• η (eta) is the characteristic life or scale parameter.
• β (beta) is the shape parameter.

To calculate B10 life (where R(T)=0.9):

1. Start with the reliability function:

$$ 0.9=e^{-\left(\frac{T}{\eta}\right)^{\beta}} $$

1. Take the natural logarithm of both sides: ln(0.9)=−(T/η)β

$$ \ln(0.9)=-{\left(\frac{T}{\eta}\right)^{\beta}} $$

1. Solve for T: T=η⋅(−ln(0.9))1/β

$$ T=\eta\centerdot\left(-\ln\left(0.9\right)\right)^{\frac{1}{\beta}} $$

Example: An oil seal with a characteristic life (η) of 6,000 hours and a shape parameter (β) of 2.5 has a B10 life calculated as approximately 2,439 hours.

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Log-Normal Distribution

The log-normal distribution is often used for failures caused by fatigue or wear-out. To determine BX life for this distribution, the failure times are transformed into their natural logarithms, which then follow a normal distribution.

The calculation uses the Z-score formula:

Z=(x−μ)/σ

Where:

• x is the natural logarithm of the time (ln(T)) you want to find.
• μ (mu) is the log mean of the distribution.
• σ (sigma) is the log standard deviation of the distribution.

For B10 life, you need the Z-score corresponding to a cumulative failure probability of 0.10 (10%). From a standard normal distribution table, this Z-score is approximately -1.28.

Example: Refrigerator door handles experience fatigue failures following a log-normal distribution with a log mean (μ) of 11 and a log standard deviation (σ) of 0.7. The doors open 40 times per day.

1. Using the Z-score for B10 life (-1.28): −1.28=(ln(T)−11) / 0.7
2. Solve for ln(T): ln(T)=−1.28×0.7+11=10.104
3. Solve for T (Cycles): T=e^10.104≈24,440 cycles.
4. Convert cycles to days (40 cycles/day): 24,440 cycles/40 cycles/day=611 days.

This indicates that 10% of the door handles are expected to fail by 611 days, suggesting a potential need for design improvement.