Learn: Miner's cumulative damage — three blocks
Miner’s rule estimates cumulative fatigue damage by adding the fractions of life consumed at several stress levels. This lesson uses three blocks of cycles and supplied fatigue lives for those blocks.
What the formula is saying
For each block divide applied cycles nᵢ by the reference life Nᵢ at that block’s stress level. Add n₁/N₁, n₂/N₂ and n₃/N₃. Each fraction is a contribution to the linear damage index D.
Read the symbols in plain language
- n₁
- Applied cycles block 1
Whole number of cycles already applied in this stress-range block; zero is permitted.
cyclesUse cycles as the base unit shown here. Every n and N is a cycle count; each ratio and their sum are dimensionless. Applied counts may be zero but reference lives must be strictly positive. Enter whole cycle counts for these study blocks.
- N₁
- Cycles to failure block 1
Positive whole-number fatigue life from a compatible S–N curve at this block’s stress range and mean-stress conditions.
cyclesUse cycles as the base unit shown here. Every n and N is a cycle count; each ratio and their sum are dimensionless. Applied counts may be zero but reference lives must be strictly positive. Enter whole cycle counts for these study blocks.
- n₂
- Applied cycles block 2
Whole number of cycles already applied in this stress-range block; zero is permitted.
cyclesUse cycles as the base unit shown here. Every n and N is a cycle count; each ratio and their sum are dimensionless. Applied counts may be zero but reference lives must be strictly positive. Enter whole cycle counts for these study blocks.
- N₂
- Cycles to failure block 2
Positive whole-number fatigue life from a compatible S–N curve at this block’s stress range and mean-stress conditions.
cyclesUse cycles as the base unit shown here. Every n and N is a cycle count; each ratio and their sum are dimensionless. Applied counts may be zero but reference lives must be strictly positive. Enter whole cycle counts for these study blocks.
- n₃
- Applied cycles block 3
Whole number of cycles already applied in this stress-range block; zero is permitted.
cyclesUse cycles as the base unit shown here. Every n and N is a cycle count; each ratio and their sum are dimensionless. Applied counts may be zero but reference lives must be strictly positive. Enter whole cycle counts for these study blocks.
- N₃
- Cycles to failure block 3
Positive whole-number fatigue life from a compatible S–N curve at this block’s stress range and mean-stress conditions.
cyclesUse cycles as the base unit shown here. Every n and N is a cycle count; each ratio and their sum are dimensionless. Applied counts may be zero but reference lives must be strictly positive. Enter whole cycle counts for these study blocks.
- D
- Result to find
Miner's cumulative damage — three blocks. Miner accumulation adds compatible life fractions without applying an extra percentage conversion.
ratio / no unit
Sort out the units first
Every n and N is a cycle count; each ratio and their sum are dimensionless. Applied counts may be zero but reference lives must be strictly positive. Enter whole cycle counts for these study blocks.
Assumptions before calculating
Assume compatible S–N data and stress definitions for every block, with mean-stress and detail effects already treated appropriately. The linear accumulation assumption ignores the order in which blocks are applied.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find D and explain the result in the stated output unit.
- n₁ · Applied cycles block 1
- 10000 cycles
- N₁ · Cycles to failure block 1
- 100000 cycles
- n₂ · Applied cycles block 2
- 5000 cycles
- N₂ · Cycles to failure block 2
- 50000 cycles
- n₃ · Applied cycles block 3
- 1000 cycles
- N₃ · Cycles to failure block 3
- 20000 cycles
Find the first block’s consumed life fraction
Divide applied cycles by fatigue life at the first block stress level.
(10000) ÷ (100000) = 0.1Find the second block’s consumed life fraction
Use the second block reference life rather than reusing the first denominator.
(5000) ÷ (50000) = 0.1Find the third block’s consumed life fraction
The third stress level has its own applied count and compatible reference life.
(1000) ÷ (20000) = 0.05Add the linear damage contributions
Miner accumulation adds compatible life fractions without applying an extra percentage conversion.
(0.1) + (0.1) + (0.05) = 0.25
Does this worked answer make sense?
Zero applied cycles in every block give D = 0. Doubling all applied counts doubles D for unchanged reference lives. The three displayed damage fractions should add exactly to the reported index within rounding.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- n₁ · Applied cycles block 1
- 20000 cycles
- N₁ · Cycles to failure block 1
- 100000 cycles
- n₂ · Applied cycles block 2
- 10000 cycles
- N₂ · Cycles to failure block 2
- 50000 cycles
- n₃ · Applied cycles block 3
- 3000 cycles
- N₃ · Cycles to failure block 3
- 20000 cycles
Find the first block’s consumed life fraction
Divide applied cycles by fatigue life at the first block stress level.
(20000) ÷ (100000) = 0.2Find the second block’s consumed life fraction
Use the second block reference life rather than reusing the first denominator.
(10000) ÷ (50000) = 0.2Find the third block’s consumed life fraction
The third stress level has its own applied count and compatible reference life.
(3000) ÷ (20000) = 0.15Add the linear damage contributions
Miner accumulation adds compatible life fractions without applying an extra percentage conversion.
(0.2) + (0.2) + (0.15) = 0.55
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- n₁ · Applied cycles block 1
- 15000 cycles
- N₁ · Cycles to failure block 1
- 75000 cycles
- n₂ · Applied cycles block 2
- 12000 cycles
- N₂ · Cycles to failure block 2
- 60000 cycles
- n₃ · Applied cycles block 3
- 2000 cycles
- N₃ · Cycles to failure block 3
- 10000 cycles
Find: Learn: Miner's cumulative damage — three blocks
A hint, not the answer
For each block divide applied cycles nᵢ by the reference life Nᵢ at that block’s stress level. Add n₁/N₁, n₂/N₂ and n₃/N₃. Each fraction is a contribution to the linear damage index D.
Every n and N is a cycle count; each ratio and their sum are dimensionless. Applied counts may be zero but reference lives must be strictly positive. Enter whole cycle counts for these study blocks.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the first block’s consumed life fraction
Divide applied cycles by fatigue life at the first block stress level.
(15000) ÷ (75000) = 0.2Find the second block’s consumed life fraction
Use the second block reference life rather than reusing the first denominator.
(12000) ÷ (60000) = 0.2Find the third block’s consumed life fraction
The third stress level has its own applied count and compatible reference life.
(2000) ÷ (10000) = 0.2Add the linear damage contributions
Miner accumulation adds compatible life fractions without applying an extra percentage conversion.
(0.2) + (0.2) + (0.2) = 0.6
Avoid the common trap
Do not add applied cycles and divide by an arbitrary average N. Each block must use the life corresponding to its own stress level, and the result must not be silently capped at 1.
When this method applies — and when it does not
D near 1 is the nominal exhaustion criterion of this idealized rule, not a guaranteed failure time or safety boundary. Load sequence, interaction, endurance limits, environment and statistical scatter can make actual behavior differ substantially.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Miner's cumulative damage — three blocks. Stress cycles, ranges and life under repeated loading. Linear cumulative damage is an idealization and does not account for load-sequence effects.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.