SME
Machine Design · Problem type
Miner’s rule for damage
Use this for “A part runs 20,000 cycles where it would last 80,000, then 100,000 where it would last 500,000. How much of its life is used?”
Diagram
D
=
Σnᵢ
Nᵢ
=
20,000
80,000
+
100,000
500,000
=
0.45.
45 % of the life is used; the bar fills at D = 1.
Repeating
the
whole
set:
1
D
=
2.22
times
to
failure.
This is an example. Type your own values to start a new problem.
D
=
n₁
N₁
+
n₂
N₂
Repeats
Sets to failure
calculated
Formulas
D
=
n₁
N₁
+
n₂
N₂
0.45
=
20,000
80,000
+
100,000
500,000
Repeats
=
1
D
2.2222
=
1
0.45
Assumptions
•
The same stress cycle repeats; stresses at the critical point stay under the yield strength.
•
Damage adds in proportion to cycles, whatever the order of the blocks (Miner’s rule).
•
Each life N is read from the S–N line at that block’s stress; failure comes at D = 1.
Step-by-step
WE KNOW
n₁ = 20,000 cycles, N₁ = 80,000 cycles, n₂ = 100,000 cycles, N₂ = 500,000 cycles
FIND
damage used (D), sets to failure (Repeats)
Step 1 · Find damage used (D)
Use D = n₁ ÷ N₁ + n₂ ÷ N₂
Each block uses its cycles’ share of the life at its stress; the shares add.
D
=
20,000
80,000
+
100,000
500,000
D
=
0.25
+
0.2
D
=
0.45
Step 2 · Find sets to failure (Repeats)
Use Repeats = 1 ÷ D
The part fails when the damage reaches 1.
Repeats
=
1
0.45
Repeats = 2.2222
Check
Put the numbers back in. Both sides should match.
0.45
=
20,000
80,000
+
100,000
500,000
✓
2.2222
=
1
0.45
✓
Topic check
See what you know: 5 questions on this topic
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