Weibull Reliability: Survival R(t), Hazard Rate & B10 Bearing Life

Use a fitted lifetime model

This example assumes a two-parameter Weibull lifetime distribution with location fixed at zero. Enter positive scale η in B2, positive shape β in B3 and nonnegative age t in B4. Time and scale use the same units, here hours.

The numbers are illustrative, not an estimate from a tested bearing population. This worksheet does not fit parameters, handle censored observations or calculate confidence intervals. Those steps matter before using a lifetime model for reliability decisions.

Survival and cumulative failures

B9 returns R(t)=EXP(−(t/η)^β), the modeled probability of surviving beyond t. B10 returns F(t)=1−R(t), the cumulative failure probability. At t=0 these are one and zero.

At t=η, R=EXP(−1)≈0.367879, regardless of positive β; η is not the time at which every unit fails. For η=10,000, β=2.5 and t=4,000, R≈0.9037587.

=IF(B6<>"OK",B6,IF(B7<>"OK",B7,EXP(-((B4/B2)^B3))))

Hazard is a rate, not a probability

For positive time, B11 evaluates h(t)=(β/η)(t/η)^(β−1), in inverse hours. It describes the instantaneous conditional failure rate among survivors. It is different from the cumulative probability B10 and can exceed one per chosen time unit.

At zero time the right-hand limit is zero for β>1, 1/η for β=1, and unbounded for 0<β<1. The worksheet explicitly reports the unbounded case rather than dividing by zero.

A fixed two-parameter Weibull has decreasing, constant or increasing hazard according to β. It does not trace all three stages of a bathtub-shaped hazard curve within one fitted distribution.

=IF(B6<>"OK",B6,IF(B7<>"OK",B7,IF(B4=0,IF(B3<1,"Unbounded as t approaches 0",IF(B3=1,1/B2,0)),B3/B2*(B4/B2)^(B3-1))))

B10 is the tenth failure percentile

B12 solves R(t)=0.90: B10=η(−LN(0.90))^(1/β). With the example parameters it is approximately 4,065.10 hours. This is a population percentile under the model, not a guaranteed minimum life for an individual unit.

Although bearing L10 also uses a 90% reliability convention, this generic quantile calculation is not an ISO bearing load-life rating calculation or certification. It includes no bearing load, dynamic capacity or operating-condition adjustment.

=IF(B6<>"OK",B6,B2*(-LN(0.9))^(1/B3))

Try shape and scale changes

Set β to 1: hazard becomes constant at 0.0001 per hour for η=10,000. Set β to 0.5 and inspect positive times, then zero time, to see the decreasing-hazard boundary. Set β to 2 to see increasing hazard.

Double η and t together: survival stays the same, B10 doubles and hazard halves. Invalid time blocks time-dependent outputs but leaves B10 available. Invalid scale or shape blocks all outputs.

The cells retain unrounded results. Verification uses rounded comparisons, while the worksheet can display additional digits. Very small tail probabilities can underflow to zero in floating-point arithmetic; that is not evidence of an absolute physical cutoff.

Further reading

• NIST/SEMATECH e-Handbook — Weibull distribution