Fatigue — S-N curves, Miner's rule: Worked examples

Before anything, one reminder of what each symbol means in plain words (we never use a symbol we have not earned):

Definition The cast of symbols

$S$ = stress amplitude — how hard one wiggle pulls, in MPa (megapascals). See Stress and Strain.
$N$ = cycles to failure if that amplitude acted all by itself.
$σ_{f}^{'}$ = fatigue strength coefficient — the (extrapolated) stress that breaks the part in a single cycle ( $N = 1$ ); think "the y-intercept of the log–log line".
$b$ = Basquin exponent — how steeply the line droops; a small positive number, ~0.05–0.12.
$n_{i}$ = how many cycles you actually apply at level $i$ .
$D$ = accumulated damage, a dimensionless "fraction of life used". Failure at $D = 1$ .
$σ_{ma x}, σ_{min}$ = the peak and trough stress reached within one cycle (MPa) — the top and bottom of the wiggle.
$σ_{m} = \frac{1}{2} (σ_{ma x} + σ_{min})$ = mean stress — the mid-level the wiggle rocks about; it holds a crack slightly ajar (from the parent note).
$σ_{U T S}$ = ultimate tensile strength — the one-pull breaking stress; the mean-stress rules below de-rate against it.
$S_{e}$ = endurance limit — a stress below which a true-endurance-limit material (plain-carbon / low-alloy steel) lasts effectively forever; its S–N curve goes flat here. Aluminium and titanium do not have one.

The scenario matrix

Everything this topic can throw at you falls into one of these cells. The examples below are labelled with the cell they cover.

Cell	What makes it special	Covered by
A. High stress, low cycles	$S$ near $σ_{f}^{'}$ , so $N$ is small	Ex 1
B. Low stress, huge cycles	$S ≪ σ_{f}^{'}$ , so $N$ explodes	Ex 2
C. Degenerate: $S = 0$	no swing at all → what does Basquin say?	Ex 3
D. Limiting: below / exactly at endurance limit	curve flattens → $N \to \infty$ (steel only); boundary $S = S_{e}$	Ex 4
E. Mixed spectrum, survives	$\sum n_{i} / N_{i} < 1$	Ex 5
F. Mixed spectrum, fails; find where	$\sum$ crosses 1 mid-block	Ex 6
G. Remaining-life / allowable cycles	solve for unknown $n$ given a used-up budget	Ex 7
H. Real-world word problem	launch + on-orbit thermal spectrum	Ex 8
I. Exam twist: mean-stress + "sum the stresses" trap	non-zero mean, Goodman correction, illegal shortcut	Ex 9

Worked Examples

Worked example Example 1 — Cell A: high stress, low cycles

An aluminium fitting has $σ_{f}^{'} = 900$ MPa and $b = 0.10$ . It is cycled at amplitude $S = 450$ MPa (fully reversed, $R = - 1$ ). Find $N$ .

Forecast: $S$ is half of $σ_{f}^{'}$ — quite high. Do you expect thousands, millions, or only a few hundred cycles?

Invert Basquin: $N = (\frac{S}{σ _{f}^{'}})^{- 1/ b}$ . Why this step? We are given $S$ and want $N$ ; Basquin ties them, so solve for the one we lack.
Ratio: $\frac{S}{σ _{f}^{'}} = \frac{450}{900} = 0.5$ . Why? The formula only cares about the ratio of applied stress to the coefficient.
Exponent: $- 1/ b = - 1/0.10 = - 10$ . So $N = 0. 5^{- 10} = 2^{10} = 1024$ . Why? $0. 5^{- 10} = (1/2)^{- 10} = 2^{10}$ — a clean power of two.

Answer: $N \approx 1024$ cycles.

Verify: Plug back: $S = 900 \cdot 102 4^{- 0.10}$ . Since $1024 = 2^{10}$ , $102 4^{- 0.10} = 2^{- 1} = 0.5$ , giving $900 \times 0.5 = 450$ MPa. ✓ Low-life expectation confirmed.

Worked example Example 2 — Cell B: low stress, huge cycles

Same aluminium fitting ( $σ_{f}^{'} = 900$ MPa, $b = 0.10$ ), now cycled gently at $S = 90$ MPa (fully reversed). Find $N$ .

Forecast: We dropped the stress by a factor of 5 versus Example 1. Because the exponent is $- 10$ , the life should not just grow a little — it should explode. How many zeros?

Ratio: $\frac{90}{900} = 0.1$ . Why? Same inversion, new ratio.
Power: $N = 0. 1^{- 10} = 1 0^{10}$ . Why? $0.1 = 1 0^{- 1}$ , and $(1 0^{- 1})^{- 10} = 1 0^{10}$ .

Answer: $N = 1 0^{10}$ cycles.

Verify: $900 \cdot (1 0^{10})^{- 0.10} = 900 \cdot 1 0^{- 1} = 90$ MPa. ✓

Look at the figure below (both Example 1 and Example 2 points are marked on it). The lavender curve is the Basquin line. Notice the butter-coloured double arrow: dropping the linear stress from 450 to 90 MPa (a modest vertical step) slides you an enormous distance to the right along the log- $N$ axis — from the coral dot at $N \approx 1 0^{3}$ to the mint dot at $N = 1 0^{10}$ . That is precisely why $N$ is plotted logarithmically: only a log axis can hold both lives in one picture.

Figure — Fatigue — S-N curves, Miner's rule

Worked example Example 3 — Cell C: degenerate input

$S = 0$ The part is mounted but never loaded — the stress amplitude is exactly $S = 0$ . What does Basquin predict, and is it physical?

Forecast: No wiggle at all. Should the part last forever, or does the formula misbehave?

Rearrange for $N$ : $N = σ_{f}^{'}^{1/ b} \cdot S^{- 1/ b}$ — i.e. $N$ is proportional to $S^{- 1/ b}$ , and $- 1/ b < 0$ . Why this step? We want to see how $N$ behaves as $S \to 0$ , so isolate the $S$ -dependence.
Limit: as $S \to 0^{+}$ , $S^{- 1/ b} \to + \infty$ , so $N \to \infty$ . Why? A negative power of a shrinking number blows up. No swing ⇒ no crack opening ⇒ infinite life.
Guard against nonsense: literally $S = 0$ makes $0^{- 10}$ undefined (division by zero). So we treat $S = 0$ as the limit "infinite life", not by plugging $0$ into the formula. Why? Physics says "no load, no fatigue"; the algebra just needs the limit interpretation, not a raw substitution.

Answer: $N \to \infty$ (unloaded part never fatigues). Do not evaluate $0^{- 1/ b}$ directly.

Verify: try a shrinking sequence $S = 90, 9, 0.9$ MPa with $σ_{f}^{'} = 900, b = 0.10$ : $N = 1 0^{10}, 1 0^{20}, 1 0^{30}$ — growing without bound. ✓

Worked example Example 4 — Cell D: below AND exactly at the endurance limit

A plain-carbon steel bracket has a true endurance limit $S_{e} = 200$ MPa (defined in the cast of symbols above). Consider three orbit stress levels: (a) $S = 150$ MPa, (b) $S = 200$ MPa (exactly $S_{e}$ ), (c) $S = 250$ MPa. Which give infinite life?

Forecast: One is clearly below the flat line, one is clearly above it, and one sits exactly on the corner. What happens at the corner?

Case (a) $S = 150 < S_{e}$ : below the endurance limit ⇒ on the flat branch ⇒ $N \to \infty$ (design-safe, effectively $> 1 0^{7}$ ). Why this step? For steels the sloping Basquin line only lives above $S_{e}$ ; below it the curve is flat, so damage per cycle is essentially zero.
Case (b) $S = 200 = S_{e}$ (the boundary): the endurance limit is defined as the highest stress giving infinite life, so equality is included — at exactly $S_{e}$ the part still survives indefinitely. It is the corner point where the flat line meets the slope; by convention $S = S_{e}$ counts as "infinite life", but with zero margin, so any small overshoot drops you onto the sloping (finite-life) branch. Why? We must resolve the boundary explicitly: " $\leq S_{e}$ survives, $> S_{e}$ is finite". Treating $S = S_{e}$ as finite would wrongly throw away a safe design; treating it as comfortably safe would ignore that there is no cushion.
Case (c) $S = 250 > S_{e}$ : above the limit ⇒ on the sloping Basquin branch ⇒ finite $N$ , computed exactly as in Example 1. Why? Above $S_{e}$ the endurance shortcut no longer applies; you must read the sloped curve.

Answer: (a) infinite, (b) infinite but zero-margin (boundary $S = S_{e}$ included), (c) finite.

Verify (contrast trap): had this been titanium, there is no true endurance limit, so even $150$ MPa gives a finite $N$ — the " $\leq S_{e}$ " rule is a steel-only privilege. ✓ (consistent with the parent note's mistake box)

Worked example Example 5 — Cell E: mixed spectrum that survives

A strut sees two blocks in a mission: $n_{1} = 3 \times 1 0^{4}$ cycles at a level where $N_{1} = 2 \times 1 0^{5}$ , and $n_{2} = 6 \times 1 0^{4}$ cycles where $N_{2} = 3 \times 1 0^{5}$ . Does it survive?

Forecast: Each block uses a slice of the "life budget". Will the two slices together stay under 100 %?

Damage from block 1: $D_{1} = \frac{n _{1}}{N _{1}} = \frac{3 \times 1 0 ^{4}}{2 \times 1 0 ^{5}} = 0.15$ . Why this step? Miner's rule: one cycle at level $i$ spends $1/ N_{i}$ of the budget, so $n_{i}$ cycles spend $n_{i} / N_{i}$ .
Damage from block 2: $D_{2} = \frac{6 \times 1 0 ^{4}}{3 \times 1 0 ^{5}} = 0.20$ . Why? Same rule, independent contribution.
Sum: $D = 0.15 + 0.20 = 0.35 < 1$ . Why? Failure only at $D \geq 1$ ; we sum fractions, never stresses.

Answer: survives, having used $35%$ of its fatigue life.

Verify: the two fractions are dimensionless (cycles ÷ cycles), and $0.35 < 1$ . ✓

Worked example Example 6 — Cell F: mixed spectrum that fails; find where

A joint runs repeated identical missions. Per mission it sees $n_{1} = 2 \times 1 0^{4}$ cycles at $N_{1} = 1 \times 1 0^{5}$ and $n_{2} = 1 \times 1 0^{4}$ cycles at $N_{2} = 4 \times 1 0^{4}$ . On which mission does $D$ first reach 1?

Forecast: Each mission adds a fixed chunk of damage. How many missions until the budget empties?

Damage per mission: $D_{mission} = \frac{2 \times 1 0 ^{4}}{1 \times 1 0 ^{5}} + \frac{1 \times 1 0 ^{4}}{4 \times 1 0 ^{4}} = 0.20 + 0.25 = 0.45$ . Why this step? Sum both blocks' fractions to get the per-mission life used.
Missions to $D = 1$ : $\frac{1}{0.45} = 2.22$ missions. Why? Total budget 1 divided by budget spent each mission.
Interpret the fraction: after 2 full missions $D = 0.90$ ; failure occurs during the 3rd, once cumulative $D$ crosses 1. Why? You cannot fly a fractional mission — failure lands mid-way through mission 3.

Answer: fails during the 3rd mission (predicted $D = 1$ at $2.22$ missions).

Verify: $2 \times 0.45 = 0.90 < 1$ (survives 2), $3 \times 0.45 = 1.35 \geq 1$ (fails by 3). ✓

Worked example Example 8 — Cell H: real-world word problem (launch + orbit)

A bracket must survive one launch and a 5-year orbit life. Two damage sources, from Launch Loads & Environments and Thermal Cycling on Orbit:

Launch vibration (see Random Vibration & PSD): $n_{1} = 8 \times 1 0^{3}$ cycles at a level with $N_{1} = 5 \times 1 0^{4}$ .
Thermal cycling on orbit: one cycle every 90 min. Each thermal cycle has $N_{2} = 1 \times 1 0^{5}$ . How much life does the whole mission use, and does it survive?

Forecast: 5 years of 90-minute cycles is a lot of thermal cycles. Will the slow thermal drip or the violent launch dominate the damage?

Count thermal cycles: cycles per year $\approx \frac{365 \times 24 \times 60}{90} = \frac{525600}{90} = 5840$ ; over 5 years, $n_{2} = 5 \times 5840 = 29200$ . Why this step? Miner needs an applied cycle count; convert "every 90 min for 5 years" into a number.
Launch damage: $D_{1} = \frac{8 \times 1 0 ^{3}}{5 \times 1 0 ^{4}} = 0.16$ . Why? Fraction of life spent by the launch block.
Thermal damage: $D_{2} = \frac{29200}{1 \times 1 0 ^{5}} = 0.292$ . Why? Same rule for the orbital block.
Total & margin: $D = 0.16 + 0.292 = 0.452 < 1$ . Why? Survives; margin against $D = 1$ is what Safety Factors & Margins of Safety would formalise (here we have $\approx 2.2 \times$ life margin).

Answer: uses $\approx 45.2%$ of fatigue life ⇒ survives the mission. Thermal cycling ( $0.292$ ) narrowly out-damages the launch ( $0.16$ ) here.

Verify: $525600/90 = 5840$ exactly; $5 \times 5840 = 29200$ ; $0.16 + 0.292 = 0.452 < 1$ . ✓

Worked example Example 9 — Cell I: mean stress, Goodman correction, and the "sum the stresses" trap

A titanium fitting has $σ_{U T S} = 950$ MPa, $σ_{f}^{'} = 1800$ MPa, $b = 0.10$ . It sees a cycle with $σ_{ma x} = 260$ MPa and $σ_{min} = - 60$ MPa. (i) Find amplitude $σ_{a}$ and mean $σ_{m}$ . (ii) Because $σ_{m} \neq = 0$ , apply the Goodman correction to get the equivalent fully-reversed amplitude $S_{eq}$ . (iii) Then, given two blocks $n_{1} = 1 \times 1 0^{4}$ at $N_{1} = 4 \times 1 0^{4}$ and $n_{2} = 3 \times 1 0^{4}$ at $N_{2} = 9 \times 1 0^{4}$ , find total damage. Trap: a classmate says "just add $S_{1} + S_{2}$ and compare to strength."

Forecast: Is the offset cycle more or less damaging than a fully-reversed one of the same amplitude? And is adding two stresses in MPa even a legal move?

Amplitude and mean: $σ_{a} = \frac{260 - ( - 60 )}{2} = \frac{320}{2} = 160$ MPa; $σ_{m} = \frac{260 + ( - 60 )}{2} = \frac{200}{2} = 100$ MPa. Why this step? Amplitude = half the peak-to-trough swing (it opens the crack); mean = the mid-level (holds it ajar). Since $σ_{m} = 100 \neq = 0$ , this is not fully reversed, so Basquin cannot be used raw.
Goodman correction: $S_{eq} = \frac{σ _{a}}{1 - σ _{m} / σ _{U T S}} = \frac{160}{1 - 100/950} = \frac{160}{1 - 0.10526} = \frac{160}{0.89474} = 178.8$ MPa. Why? The tensile mean holds the crack open, so the offset cycle bites like a larger fully-reversed one. Goodman inflates $160 \to 178.8$ MPa; that is the amplitude you'd carry into the S–N curve — giving a shorter life than the naive 160 MPa would suggest. (Gerber, using $(σ_{m} / σ_{U T S})^{2} = 0.01108$ , gives a gentler $S_{eq} = 160/0.98892 = 161.8$ MPa.)
The trap — why "add the stresses" is illegal: the classmate would compute $S_{1} + S_{2}$ (adding two amplitudes in MPa) and compare to strength. This is meaningless: MPa values are not life-usages, cannot be summed across separate loading blocks, and give a number with no failure criterion. Miner's rule sums dimensionless fractions $n_{i} / N_{i}$ , never stresses. Why? Elastic superposition adds stresses acting at the same instant on the same point — but these are different blocks over time, so only their damage fractions add.
Correct damage sum: $D = \frac{n _{1}}{N _{1}} + \frac{n _{2}}{N _{2}} = \frac{1 \times 1 0 ^{4}}{4 \times 1 0 ^{4}} + \frac{3 \times 1 0 ^{4}}{9 \times 1 0 ^{4}} = 0.25 + 0.333 \dots = 0.583$ . Why? Convert each block to a life fraction first, then add — the only legal accumulation.

Answer: $σ_{a} = 160$ MPa, $σ_{m} = 100$ MPa; Goodman-equivalent $S_{eq} \approx 178.8$ MPa (the mean makes it more damaging); total damage $D = 7/12 \approx 0.583 < 1$ ⇒ survives. The "add the stresses" method is invalid.

Verify: $(260 - (- 60)) /2 = 160$ ; $(260 + (- 60)) /2 = 100$ ; Goodman $160/ (1 - 100/950) = 178.82$ MPa $> 160$ (correctly inflated); $0.25 + 1/3 = 7/12 \approx 0.5833 < 1$ . ✓

Recall

Recall When is

$N$ infinite versus finite? $N$ is finite whenever you sit on Basquin's sloping line (any Ti/Al, any steel above $S_{e}$ ). $N$ is infinite only for (a) $S = 0$ as a limit, or (b) a steel stressed at or below its endurance limit, i.e. $S \leq S_{e}$ (equality included, but with zero margin).

Recall Miner's rule: what do you sum, and what is the failure number?

Sum the dimensionless damage fractions $n_{i} / N_{i}$ (cycles applied ÷ cycles-to-failure). Failure when the sum reaches $D = 1$ . Never sum stresses in MPa.

Recall What does the Basquin line here assume about mean stress, and how do you fix a non-zero mean?

Fully reversed loading, $R = - 1$ , so $σ_{m} = 0$ . If $σ_{m} \neq = 0$ , first convert to an equivalent fully-reversed amplitude with Goodman $S_{eq} = σ_{a} / (1 - σ_{m} / σ_{U T S})$ (or the gentler Gerber parabola), then enter the S–N curve — the true life is shorter.

Convert this trap

Adding

S_{1} + S_{2}

in MPa is illegal; convert each block to

n_{i} / N_{i}

first, then sum.

At or below

S_{e}

for steel,

N =

Infinite (design-safe); the boundary

S = S_{e}

is included but has zero margin.

S = 0

gives

N =

Infinite in the limit; never plug

0

into

0^{- 1/ b}

(undefined).

Non-zero mean fix?

Goodman

S_{eq} = σ_{a} / (1 - σ_{m} / σ_{U T S})

then enter the S–N curve; true life is shorter.