Fatigue — S-N curves, Miner's rule: Question bank

Symbol & concept refresher (read before the traps)

Before hunting traps you must know exactly what every symbol means and be able to picture it. Nothing below is used before it is defined here.

Definition The symbols this page uses

$S$ = stress amplitude = half the peak-to-peak stress swing in one cycle, $S = σ_{a} = \frac{σ _{ma x} - σ _{min}}{2}$ . Picture the height of the wiggle in a stress-vs-time graph.
$N$ = cycles to failure = how many identical wiggles at amplitude $S$ the part survives before snapping.
$σ_{m}$ = mean stress = the centre-line the wiggle oscillates about, $σ_{m} = \frac{σ _{ma x} + σ _{min}}{2}$ .
$R$ = stress ratio $= σ_{min} / σ_{ma x}$ . It labels the shape of the cycle: $R = - 1$ = fully reversed (mean zero), $R = 0$ = zero-to-tension, $R \to 1$ = tiny swing about a big mean.
$σ_{f}^{'}$ = fatigue strength coefficient = the stress the Basquin line predicts at $N = 1$ (a curve-fit intercept, not $σ_{U T S}$ ).
$b$ = Basquin exponent = magnitude of the slope of the $lo g S$ – $lo g N$ line, a small number ( $\approx 0.05$ – $0.12$ ).
$S_{e}$ = endurance limit = the flat stress floor below which some steels last effectively forever.
$n_{i}$ = number of applied cycles at level $i$ ; $n_{i} / N_{i}$ = the damage fraction (fraction of life used at that level).
$D$ = cumulative damage $= \sum_{i} n_{i} / N_{i}$ , a dimensionless "battery drain" that reaches $1$ at failure.
$K_{t}$ = stress-concentration factor = how many times higher the local stress at a notch/hole is than the nominal stress.

Look at figure s01: one stress cycle drawn on a time axis, with $σ_{ma x}$ , $σ_{min}$ , the mean line $σ_{m}$ , and the amplitude arrow $S = σ_{a}$ all labelled. The amplitude is the vertical swing; the mean is where the swing is centred.

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

Look at figure s02 — the S–N curve on log–log axes. It shows three regions you must recognise: the low-cycle (plastic) knee on the left where Basquin's straight line bends and breaks down, the straight high-cycle Basquin line in the middle, and the endurance floor $S_{e}$ on the right (present for steel, absent for Al/Ti). Every trap about "which regime?" refers back to this picture.

Look at figure s03 — the Goodman / Gerber mean-stress diagram. Amplitude $S$ is on the vertical axis, mean $σ_{m}$ on the horizontal. A design point inside the line is safe; the same amplitude becomes unsafe as the mean pushes right toward $σ_{U T S}$ . This is how nonzero-mean data gets corrected before you ever enter the S–N curve.

Look at figure s04 — a notch raising local stress: the nominal field is uniform, but streamlines crowd at the hole so the peak local stress is $K_{t}$ times higher. That local peak, not the nominal, drives crack initiation — the reason surface finish and stress concentration dominate real fatigue.

True or false — justify

A static stress safely below yield guarantees the part never breaks.

False — fatigue cracks nucleate and grow under cyclic loading even in the fully elastic range, so a part safe under one static load can still snap after enough cycles.

The mean stress

σ_{m}

has no effect on fatigue life, only the amplitude matters.

False — amplitude dominates, but a positive (tensile) mean holds the crack slightly ajar so each swing "bites harder", shortening life; that is exactly what the Goodman line of figure s03 corrects for.

A true endurance limit means aluminium struts can be designed for infinite life.

False — a true endurance limit is mainly a plain-carbon / low-alloy steel feature; aluminium and common Ti alloys keep sloping down past

1 0^{7}

cycles (no floor in figure s02), so you design to a specified high-cycle strength instead.

On the S–N curve, a higher stress amplitude corresponds to more cycles to failure.

False — it is the opposite: higher amplitude

S

means fewer cycles

N

, because each harsh cycle opens the crack more; the curve slopes downward to the right.

Miner's rule says failure occurs exactly when

D = 1

every time.

False —

D = 1

is only the average linear-damage criterion; real tests scatter roughly

0.3

–

3

because loading order and overloads change the physics, which is why engineers keep safety factors.

Basquin's law

S = σ_{f}^{'} N^{- b}

is a fundamental physical law derived from atoms.

False — it is an empirical straight-line fit to

lo g S

lo g N

in the high-cycle regime (the middle segment of figure s02); the power law is just what a line in log–log coordinates looks like.

Thermal cycling on orbit cannot cause fatigue because the spacecraft is not vibrating.

False — each ~90-minute day/night thermal cycle strains joints as materials expand and contract, so it delivers thousands of stress cycles just like mechanical loading.

Basquin's straight line describes fatigue life all the way down to

N = 1

False — at low

N

the material yields plastically and the curve bends into the knee (left of figure s02); this low-cycle regime is governed by strain-based (Coffin–Manson) behaviour, not the high-cycle Basquin line.

Spot the error

"To combine two load levels, add the stresses:

S_{1} + S_{2}

, then read one

N

off the curve."

Wrong tool — you sum damage fractions

n_{i} / N_{i}

(dimensionless life-usages), not stresses; go to the S–N curve for each level separately to get each

N_{i}

first.

"The Basquin exponent

b

is a large number like 5, since fatigue is very sensitive to stress."

The slope magnitude

b

is small (

\approx 0.05

–

0.12

); it is its reciprocal

m = 1/ b

(~8–20) that is large, so a small drop in

S

multiplies

N

enormously.

"Since

D_{1} = 0.10

and

D_{2} = 0.40

give

D = 0.50

, we've used up half the stress."

We've used up half the life budget (damage

D

), not stress;

D

is a dimensionless fraction of the total number of tolerable cycles, unrelated to any stress magnitude.

σ_{f}^{'}

is the ultimate tensile strength

σ_{U T S}

No —

σ_{f}^{'}

is the fatigue strength coefficient, the stress the Basquin line predicts at

N = 1

; it is a curve-fit intercept, generally different from the static

σ_{U T S}

"A part at

D = 0.99

is basically fine, plenty of margin left."

It has spent 99% of its fatigue life with only 1% of the budget remaining; it is on the brink, and given the scatter in

D_{f ai l}

it may already have failed.

"A smooth polished bar and a bar with a bolt-hole have the same fatigue life at equal nominal stress."

Wrong — the hole raises the local stress by the factor

K_{t}

(figure s04), so cracks initiate far sooner; surface finish and stress concentration are first-order fatigue drivers, not details.

"Random launch vibration isn't fatigue, it's just one big load."

Random vibration delivers thousands of stress cycles of varying amplitude in minutes, precisely the mixed-amplitude spectrum Miner's rule was built to handle.

Why questions

Why is the S–N curve's

N

-axis plotted on a logarithmic scale?

Because

N

spans many orders of magnitude (

1 0^{3}

1 0^{8}

+); a log axis turns the power-law relationship into the readable straight line seen in figure s02 and keeps all regimes visible at once.

Why does amplitude

σ_{a}

drive damage more than the peak stress

σ_{ma x}

alone?

A crack grows by being repeatedly pulled open and released; it is the swing

S = σ_{a}

(figure s01) that cycles the crack tip, whereas a steady peak with no swing would not repeatedly work it.

Why can Miner's rule give

D_{f ai l} > 1

(part outlives the prediction)?

A large tensile overload can leave a compressive residual stress at the crack tip that clamps it shut and slows later crack growth, letting the part survive past

D = 1

Why do we still use safety factors if Miner's rule already predicts failure?

Because

D = 1

is an average with real-world scatter (~0.3–3) from order effects and material variability; the margin covers this uncertainty so the design stays conservative.

Why does a straight line on a

lo g S

–

lo g N

plot imply a power law?

A line

lo g S = lo g σ_{f}^{'} - b lo g N

becomes

S = σ_{f}^{'} N^{- b}

when exponentiated; multiplying inside a log corresponds to a power outside, so linear-in-logs is always a power law.

Why check fatigue life even when the static stress margin is positive?

Static margin only guarantees no immediate yield or rupture; fatigue is a time/cycle failure mode that a static check cannot see, so both must be satisfied independently.

Why does a Goodman correction lower the allowable amplitude as

σ_{m}

increases?

Because tensile mean stress holds the crack open, so less swing is tolerable; the Goodman line of figure s03 slides the safe amplitude down toward zero as the mean approaches

σ_{U T S}

Edge cases

If a load level has stress amplitude below the endurance limit

S_{e}

(for a steel), what is its damage contribution?

Effectively zero — those cycles sit on the flat floor of figure s02 where

N \to \infty

, so

n_{i} / N_{i} \to 0

and they consume negligible life budget.

What happens to predicted life

N

as the applied amplitude

S

approaches

σ_{f}^{'}

N

approaches

1

, since Basquin's line passes through

S = σ_{f}^{'}

N = 1

; but there the material is really in the plastic low-cycle knee, so Basquin is only nominal.

For an aluminium part, why can you never legitimately quote an "infinite-life" stress?

Aluminium's S–N curve keeps sloping downward past

1 0^{7}

cycles with no flat floor, so any stress will eventually cause failure; you quote a strength at a fixed large

N

(e.g.

5 \times 1 0^{8}

) instead.

How does the stress ratio

R

change fatigue life for the same amplitude

S

Raising

R

toward

1

raises the mean stress

σ_{m}

(crack held more open), shortening life;

R = - 1

(fully reversed, zero mean) is the benign baseline against which S–N data is usually quoted.

What about extreme ratios like

R < - 1

(compression-dominated) or

R > 1

R < - 1

means the cycle is mostly compressive with a small tensile blip, so crack driving is weak and life is long;

R > 1

occurs in fully compressive cycling where crack faces stay pressed shut and initiation from the bulk stress is strongly suppressed.

Does the environment (corrosion, vacuum, temperature) shift the S–N curve?

Yes — a corrosive medium removes any true endurance limit and lowers the whole curve (corrosion-fatigue), while high temperature adds creep; so laboratory S–N data must be de-rated for the on-orbit environment.

If two identical spectra are applied but in reversed order, does Miner's rule predict a different

D

No — the sum

\sum n_{i} / N_{i}

is order-independent, which is exactly Miner's blind spot; real life does differ because an early overload leaves residual stress that changes later crack growth.

What does

D = 0

physically represent, and can damage ever be negative?

D = 0

is a pristine part with zero cycles applied; damage in Miner's model cannot go negative — it only accumulates upward toward the failure budget, never healing.

A part sees only compressive cyclic stress with zero tension — is fatigue cracking likely?

Much less likely, because crack faces are pressed together rather than pulled open; tensile swings are what propagate cracks, so pure compression is comparatively benign.

As the crack from repeated cycles grows large, does the S–N framework still describe the final failure?

Not well — once a dominant crack exists, growth is governed by Paris' law and stress intensity, not the total-life S–N curve, which mainly covers crack initiation.