Visual walkthrough — Fatigue — S-N curves, Miner's rule

Step 1 — What "one cycle of stress" actually looks like

WHAT. Before any formula, look at a single bar of metal being pulled (stretched) and pushed (squeezed) over and over. The quantity we track is stress — how hard the material is being pulled per unit of cross-section area. Think of it as "tightness inside the metal". We call it $\sigma$ (the Greek letter sigma, our symbol for stress).

WHY. Fatigue is about repeated loading, so the very first thing we must draw is what one repeat is. If we cannot name the top, the bottom, and the middle of one wiggle, we cannot say what "damage per wiggle" means later.

PICTURE. In the figure, stress $\sigma$ rises and falls like a wave as time goes on. Read off three heights:

• ${\sigma }_{max}$ — the highest point of the wave (metal pulled hardest).
• ${\sigma }_{min}$ — the lowest point (metal pushed, or least-pulled).
• The midline and the half-height of the swing get their own names next.

${\sigma }_m=\underset{\text{average of top and bottom}}{\underbrace{\frac{{\sigma }_{max}+{\sigma }_{min}}{2}}},{\sigma }_a=\underset{\text{half of the full up-down swing}}{\underbrace{\frac{{\sigma }_{max}-{\sigma }_{min}}{2}}}$

• ${\sigma }_m$ (mean) — the height of the midline. It says how far the whole wave is shifted up. Geometrically it is the average of the two dashed lines.
• ${\sigma }_a$ (amplitude) — half the vertical distance between top and bottom. It measures how big the swing is, regardless of where the midline sits.

We now follow ${\sigma }_a$ (the swing) all the way to the finish — it is the main character. The mean ${\sigma }_m$ returns for its own special step (Step 6), where we see exactly how a shifted midline changes the life.

Step 2 — One wiggle grows one crack: why amplitude is king

WHAT. Zoom into the surface of the metal. There is always a tiny flaw — a scratch, a grain boundary, a machining mark. When the bar is pulled, that flaw is pulled open a hair; when pushed, it closes again. Each open–close is one cycle, and each cycle leaves the crack a whisker longer.

WHY. This is the physical reason a bar can survive one big pull but die from a thousand small ones — and it tells us which number controls the damage. It is the swing ${\sigma }_a$ that opens and closes the crack, so ${\sigma }_a$ is the driver. The mean ${\sigma }_m$ only holds the crack slightly ajar so each swing bites a little deeper.

PICTURE. Watch the crack tip in the figure ratchet outward, one notch per cycle. The deeper physics of how fast a crack lengthens per cycle is Paris' Law; here we only need "more swing ⇒ more growth per cycle".

Step 3 — Collect the data: raw stress vs life

WHAT. Take many identical bars. Give each a different, fixed amplitude and wiggle it until it breaks, recording how many cycles $N$ it lasted. Plot one dot per bar: swing size up, life across.

WHY. We want a law connecting swing size to life. You cannot guess a law — you must first see the shape of the data. This step is pure experiment: no formula yet.

PICTURE. On ordinary (linear) axes the dots crash down steeply on the left and then hug the floor — a curve so bent it is useless to fit. Notice the dots stretch from tens of cycles to hundreds of millions: the $x$-values span many powers of ten.

• Vertical axis: $S={\sigma }_a$, the swing size we chose.
• Horizontal axis: $N$, the life we measured.
• The steep bend on the left is the whole problem we fix next.

Step 4 — Why we take logarithms (and what a log even is)

WHAT. A logarithm answers the question "ten to the what gives this number?" — writing ${\log }_{10}1000=3$ because $10^3=1000$. So ${\log }_{10}$ turns a number's size in powers of ten into a plain small number: $10,10^2,10^3$ become $1,2,3$, evenly spaced.

WHY THIS TOOL. Our $N$ values span from $\sim 10$ to $\sim 10^8$ — eight powers of ten. Linear axes cram the small numbers into a smear at the origin. Taking ${\log }_{10}N$ stretches the smear out evenly so we can actually see structure. We take ${\log }_{10}S$ too, for symmetry. This is the only tool that turns "spread across many orders of magnitude" into "spread evenly".

PICTURE. Re-plot the same dots with ${\log }_{10}N$ across and ${\log }_{10}S$ up. The bent curve snaps into a straight line (in the high-cycle region). A straight line is a gift: straight lines have a simple equation.

Step 5 — From the straight line to Basquin's law

WHAT. A straight line is $y=c-bx$. Put $y={\log }_{10}S$ and $x={\log }_{10}N$:

$\underset{\text{height}}{\underbrace{{\log }_{10}S}}=\underset{\begin{array}{c}c:\text{intercept}\\ \text{(value at }{\log }_{10}N=0)\end{array}}{\underbrace{{\log }_{10}{\sigma }_f^{\prime }}}-\underset{\begin{array}{c}\text{slope}\\ \text{magnitude}\end{array}}{\underbrace{b}}\underset{\text{across}}{\underbrace{{\log }_{10}N}}$

WHY. We found a line in Step 4; now we name its two numbers so we can compute with it. Every line needs exactly two: where it starts (intercept) and how steep it tilts (slope).

PICTURE. In the figure, the intercept is the line's height at ${\log }_{10}N=0$, i.e. at $N=1$ (one single cycle). The slope $-b$ is "down $b$ for every 1 across". Both are marked.

• ${\sigma }_f^{\prime }$ — fatigue strength coefficient: the stress the line predicts at $N=1$. It sets the line's height.
• $b$ — Basquin exponent: how fast life drops as swing rises (small, $\sim 0.05$–$0.12$). The minus sign means life falls as stress rises.

Now undo the logarithm by raising $10$ to both sides — the exact inverse move of Step 4, using the same base 10 we chose there:

$\boxed{S={\sigma }_f^{\prime }{N}^{-b}}⟺N={\left(\frac{S}{{\sigma }_f^{\prime }}\right)}^{-1/b}$

• Left form: give a life $N$, get the swing $S$ it can take.
• Right form: give a swing $S$, get the life $N$. (We just solved for $N$ by raising both sides to the power $-1/b$.)

Step 6 — What the mean stress ${\sigma }_m$ does to the curve

WHAT. Everything so far assumed the wave swings symmetrically about zero, i.e. ${\sigma }_m=0$. But a bolt pre-loaded in tension, or a pressurised tank, sits with the midline shoved up: ${\sigma }_m>0$. That shift matters, because a crack that is already held ajar by a positive mean opens further on each swing and grows faster. So the same swing $S$ gives shorter life when ${\sigma }_m$ is positive.

WHY THIS TOOL. Basquin's law only knows about the swing $S$; it is blind to the midline. To fold in the mean we convert a real $({\sigma }_a,{\sigma }_m)$ pair into an equivalent zero-mean amplitude ${\sigma }_{ar}$ — the swing that, with no mean, would do the same damage. Two standard conversions do this, using the material's ultimate strength ${\sigma }_{UTS}$ (the stress that snaps it in one pull, from Stress and Strain):

${\sigma }_{ar}=\underset{\text{Goodman (straight line)}}{\underbrace{\frac{{\sigma }_a}{1-\frac{{\sigma }_m}{{\sigma }_{UTS}}}}},{\sigma }_{ar}=\underset{\text{Gerber (parabola)}}{\underbrace{\frac{{\sigma }_a}{1-{\left(\frac{{\sigma }_m}{{\sigma }_{UTS}}\right)}^2}}}$

• ${\sigma }_a$ — the real swing you applied.
• ${\sigma }_m$ — the real midline height; if ${\sigma }_m=0$ both formulas collapse to ${\sigma }_{ar}={\sigma }_a$ and nothing changes.
• ${\sigma }_{UTS}$ — one-pull breaking stress; as ${\sigma }_m\to {\sigma }_{UTS}$ the denominator $\to 0$ and ${\sigma }_{ar}\to \infty$ (a mean near the breaking stress means the part is already almost gone — zero life).
• Feed ${\sigma }_{ar}$ into Basquin in place of $S$ to read the mean-corrected life.

PICTURE. The figure plots allowable amplitude ${\sigma }_a$ against mean ${\sigma }_m$. At ${\sigma }_m=0$ you get the full endurance amplitude; the Goodman line slopes straight down to zero at ${\sigma }_{UTS}$, the Gerber parabola bulges above it. Any $({\sigma }_m,{\sigma }_a)$ point under the curve survives; on or above it fails.

Step 7 — Edge case: does the line ever stop falling?

WHAT. Basquin's straight line, if trusted forever, says any swing eventually breaks the bar (life just gets astronomically large). Real materials split into two families here.

WHY. A derivation must cover all cases. The line is only a fit to the data we have; at very low stress the real curve may bend flat — or may not. Which happens depends on the material, and getting this wrong is a classic, dangerous mistake.

PICTURE. Two curves on one plot:

• Steel (plain-carbon / low-alloy): the curve flattens at a stress called the endurance limit $S_e$. Swing below $S_e$ ⇒ effectively infinite life. The Basquin line is replaced by a horizontal floor.
• Aluminium & most titanium (e.g. Ti-6Al-4V): no flat floor — the line keeps sloping down past $10^7$ cycles. For these you must quote a strength at a stated life (say $N=5\times 10^8$), never "infinite life".

Step 8 — Real life is a mix: the damage-fraction idea

WHAT. Real launch and orbit loading is not one steady swing — it is $n_1$ cycles at swing $S_1$, then $n_2$ at $S_2$, and so on. The S–N curve only knows what each swing does alone ($N_1$, $N_2$, …). We need to combine them.

WHY. Random vibration and thermal cycling deliver messy spectra of many amplitudes. Without a combining rule the S–N curve is useless for a real spacecraft.

PICTURE. Think of a fuel gauge for "fatigue life", starting full. One cycle at level $i$ drains the fraction $1/N_i$ of the tank (because $N_i$ such cycles would empty it exactly). So $n_i$ cycles drain

$D_i=\underset{\begin{array}{c}\text{cycles done}\\ ÷\text{cycles it would take alone}\end{array}}{\underbrace{\frac{n_i}{N_i}}}$

• $n_i$ — cycles you actually applied at level $i$.
• $N_i$ — cycles-to-failure at level $i$, read off the S–N curve (Step 5, mean-corrected via Step 6 if needed).
• $D_i$ — the dimensionless slice of the tank that level $i$ used up.

Step 9 — Summing the slices: the Palmgren–Miner rule

WHAT. Assume the slices are independent and just add. Total damage is the sum of all fractions; the tank empties (part fails) when the total reaches 1 (100 %).

WHY. "Linear, independent accumulation" is the simplest honest guess — it needs no extra data beyond the S–N curve. It is an average model, not a law of nature, which is exactly why we later multiply in safety factors.

PICTURE. Stack the fraction-bars end to end toward a line marked "1". Cross the line ⇒ failure.

Worked check (numbers that match the parent)

The one-picture summary

Everything on one canvas: a wiggling stress wave (Steps 1–2) feeds the log–log S–N line (Steps 3–5), the mean-stress correction bends the allowable swing (Step 6) and the steel-vs-Ti fork sets the low-stress tail (Step 7); reading lives off that line fills a fatigue tank whose slices sum to 1 (Steps 8–9).

Active Recall