Visual walkthrough — Corrosion in aerospace environments — stress corrosion cracking, hydrogen embrittlement

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Step 1 — A scratch is a stress magnifier

WHAT. Imagine pulling on a flat metal bar. The pulling force is spread evenly across the whole width, so every little patch of metal feels the same gentle "tug". We call that tug the applied stress, written $\sigma$ (the Greek letter sigma). Stress just means force divided by the area it is spread over — like pressure.

Now cut a tiny notch into the edge. The lines of force can no longer go straight through — they have to bend around the notch, and they crowd together at the sharp bottom of it. Where lines crowd, the local pull is much bigger than $\sigma$.

WHY. This is the seed of everything. A crack does not need a huge average stress; it only needs the local stress at its tip to be huge. The notch is a lens that focuses the far-away gentle pull into a sharp spike.

PICTURE. Look at how the flow lines (think of water flowing past a rock) squeeze together at the notch root — the red region is where the metal is being pulled hardest.

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Step 2 — How big is the spike? The $1/\sqrt{r}$ law

WHAT. Let $r$ be the distance you stand away from the very tip of the crack, measured straight ahead of it. Elasticity theory (the maths of how solids stretch) says the local stress ${\sigma }_{\text{local}}$ grows as you walk toward the tip like this:

$\underset{\text{stress you feel at distance }r}{\underbrace{{\sigma }_{\text{local}}}}\propto \sigma \frac{\sqrt{a}}{\sqrt{r}}$

Reading it term by term: $\sigma$ is the faraway tug (Step 1); $a$ is the crack length; $r$ is your distance ahead of the tip; the symbol $\propto$ means "grows in proportion to". As $r$ shrinks toward $0$ (you approach the tip), $\frac{1}{\sqrt{r}}$ blows up — the stress heads to infinity.

WHY the square root, and why does it matter? We ask: how fast does the crowding of force-lines fade as we back away from the tip? The answer that the geometry of a sharp slit forces on us is "like one-over-root-r" — not $1/r$, not $1/r^2$. This particular fade rate is the whole reason cracks are special, and it is where the $\sqrt{\cdot }$ in every crack formula is born.

PICTURE. The curve rockets up near $r=0$. A longer crack (bigger $a$) lifts the whole curve — that is the $\sqrt{a}$ on top.

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Step 3 — Bottling the whole spike into one number: $K$

WHAT. Every term that does not change as you move toward the tip — the applied stress $\sigma$, the crack length through $\sqrt{a}$, and a shape number $Y$ — gets bundled into one quantity called the stress-intensity factor $K$:

$K=\underset{\begin{array}{c}\text{geometry}\\ \text{factor}\approx 1\end{array}}{\underbrace{Y}}\underset{\begin{array}{c}\text{applied}\\ \text{tug}\end{array}}{\underbrace{\sigma }}\underset{\begin{array}{c}\text{crack-size}\\ \text{term}\end{array}}{\underbrace{\sqrt{\pi a}}}$

• $Y$ — a dimensionless "shape correction". For a small edge crack $Y\approx 1.12$; for an ideal internal crack $Y=1$. It just tweaks for geometry.
• $\sigma$ — the same average stress from Step 1.
• $\sqrt{\pi a}$ — the crack-length term inherited from the $\sqrt{a}$ of Step 2 (the $\pi$ pops out when you do the integral of the $1/\sqrt{r}$ field cleanly).

WHY combine them? Because the $1/\sqrt{r}$ shape of the stress field near any crack tip is always the same — only its overall height changes from case to case. $K$ is precisely that height. Two totally different parts with the same $K$ have identical stress fields at the tip, so they crack the same way. $K$ is the "one number that matters".

PICTURE. Same $1/\sqrt{r}$ curve as Step 2, but now shaded to show that its area/scale is what $K$ measures — a taller field means a bigger $K$.

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Step 4 — Two thresholds, not one: $K_{IC}$ vs $K_{ISCC}$

WHAT. There is a value of $K$ so large the crack runs at the speed of sound and the part snaps in one shot. That value is the fracture toughness $K_{IC}$ — a fixed material property. If $K\ge K_{IC}$: instant fast fracture.

But add a corrosive environment, and a second, much lower threshold appears: $K_{ISCC}$ ("K, I, stress-corrosion cracking"). Above $K_{ISCC}$ but below $K_{IC}$, the crack does not snap — it creeps forward slowly, helped by chemistry.

$K_{ISCC}\ll K_{IC}$

The symbol $\ll$ means "much less than". Typically $K_{ISCC}$ is only a fraction of $K_{IC}$.

WHY two thresholds? Fast fracture is purely mechanical — it needs enough stored elastic energy. Slow SCC growth only needs the tip to be reactive enough for corrosion chemistry to eat a little metal there. Chemistry is patient; it accepts a much smaller driving force. That is why $K_{ISCC}$ sits far below $K_{IC}$ — and why "$K<K_{IC}$, so I'm safe" is a deadly mistake.

PICTURE. A crack-growth-rate axis: nothing happens below $K_{ISCC}$, a slow plateau between the two thresholds, then a cliff at $K_{IC}$.

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Step 5 — Zoom into the tip: the occluded acid cell

WHAT. Slide a microscope inside the crack. The narrow mouth chokes off fresh solution, so the tip becomes an occluded (sealed-off) pocket. Metal keeps dissolving there:

$\text{M}⟶{\text{M}}^{n+}+n{e}^{-}$

• $\text{M}$ — a metal atom in the wall.
• ${\text{M}}^{n+}$ — that atom, now a positive ion, floating free in the trapped water. $n$ is how many electrons it left behind.
• $n{e}^{-}$ — the freed electrons, which travel away to a cathode.

Those trapped ions then react with water (hydrolysis):

${\text{M}}^{n+}+{\text{H}}_2\text{O}⟶{\text{M(OH)}}^{(n-1)+}+{\text{H}}^{+}$

The right-hand ${\text{H}}^{+}$ is a loose acid proton. It piles up because it cannot escape the narrow crack — so the pH inside crashes to 2–3 while the bulk sea-spray outside stays near neutral.

WHY does this matter? The tip has quietly built its own acid micro-reactor. Acid does two nasty jobs at once: it strips the protective oxide film (feeding Step 6, dissolution) and it manufactures hydrogen atoms (feeding Step 7, embrittlement).

PICTURE. The crack drawn as a bottleneck: bulk neutral water at the mouth, a red acidic pocket at the tip, arrows showing ${\text{M}}^{n+}$ out and ${\text{H}}^{+}$ accumulating.

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Step 6 — Mechanism A: slip-dissolution eats the tip (Faraday)

WHAT. Under the stress spike of Step 2, the metal at the tip slips, cracking the brittle oxide film and baring fresh metal. Bare metal in acid dissolves fast until the film re-heals; then the cycle repeats. How much metal vanishes per burst? Faraday's law, which we build from atoms:

1. To dissolve $N$ atoms, each giving up $z$ electrons of charge $e$, the charge that must flow is $Q=Nze$.
2. Those $N$ atoms are $N/N_A$ moles ($N_A$ = Avogadro's number), so their mass is $m=\frac{N}{N_A}M$, where $M$ is the molar mass.
3. Solve step 1 for $N=\frac{Q}{ze}$ and use $F=N_{A}e$ (the Faraday constant). Substitute:

$\boxed{m=\frac{MQ}{zF}}$

• $M$ — grams per mole of the metal.
• $Q$ — total charge that flowed before the film re-sealed.
• $z$ — electrons per atom dissolved.
• $F=96485\text{C/mol}$ — the charge of one mole of electrons.

Since $Q$ = current $\times$ time, more current at the tip = more metal removed = faster crack.

WHY this tool? We wanted to connect chemistry (how much dissolves) to electricity (the corrosion current we can actually measure). Faraday's law is the exact bridge between charge and mass — no other law does that. See Electrochemistry — galvanic cells & Faraday's law.

PICTURE. The tip in four frames: intact film → slip step ruptures it → bare metal dissolves (current spike) → film re-heals, crack a notch deeper.

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Step 7 — Mechanism B: hydrogen weakens the bonds ahead

WHAT. In the same acid pocket, some protons grab electrons and become atomic hydrogen stuck to the surface:

$2{\text{H}}^{+}+2e^{-}⟶2{\text{H}}_{\text{ads}}$

${\text{H}}_{\text{ads}}$ means hydrogen adsorbed (clinging) on the metal. A single H atom is tiny — small enough to squeeze between the metal atoms and diffuse ahead of the crack tip into the still-solid lattice. There it collects where the stress is highest and pries the atomic bonds apart, lowering the cohesive strength ${\sigma }_{coh}$ — the pull the bonds can survive before letting go:

${\sigma }_{\text{local}}\ge {\sigma }_{coh}(C_H),{\sigma }_{coh}\downarrow \text{ as }{C}_H\uparrow$

• ${\sigma }_{\text{local}}$ — the focused stress from Step 2.
• ${\sigma }_{coh}(C_H)$ — the bond strength, now a function of the hydrogen concentration $C_H$.
• The arrows: more dissolved hydrogen ($C_H\uparrow$) means weaker bonds (${\sigma }_{coh}\downarrow$).

The crack jumps forward the instant the local stress overtakes the lowered cohesion.

WHY does this overlap with Step 6? The very same acidified tip that dissolves metal (Mechanism A) is the factory that breeds this hydrogen (Mechanism B). In high-strength alloys the two run together — that is why SCC and HE are one story, not two.

PICTURE. Two bond-strength bars: a tall clean-metal bond vs a short hydrogen-charged bond, with H atoms drawn wedged into the lattice ahead of the tip.

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Step 8 — Putting it together: the worked threshold check

WHAT. Now use the whole chain on a real part (parent Example 1). A 7075-T6 wing fitting: $\sigma =150\text{MPa}$, flaw $a=2\text{mm}=0.002\text{m}$, $Y=1.12$, and for this alloy in salt air $K_{ISCC}=12\text{MPa}\sqrt{\text{m}}$.

Compute the driving force with the Step 3 formula:

$K=\underset{Y}{\underbrace{1.12}}\times \underset{\sigma }{\underbrace{150}}\times \underset{\sqrt{\pi a}}{\underbrace{\sqrt{\pi \times 0.002}}}=1.12\times 150\times 0.0793=13.3\text{MPa}\sqrt{\text{m}}$

Compare to the threshold: $13.3>12$, so $K>K_{ISCC}$ → the crack will grow. And yet $150\text{MPa}$ is far below the ~$500\text{MPa}$ yield strength — the part passes every static test and still fails. That is the parent note's whole warning, now earned from atoms up.

WHY. Steps 1–4 told us $K$ is the driving force and $K_{ISCC}$ the door; Steps 5–7 told us the chemistry that lets slow growth happen once the door is open. Step 8 simply reads the verdict.

PICTURE. A single number line: mark $K_{ISCC}=12$, place the computed $K=13.3$ just past it in the "grows" zone, with the safe zone shaded on the left.

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The one-picture summary

The whole derivation on one canvas: a scratch focuses stress ($1/\sqrt{r}$) → that field's height is $K=Y\sigma \sqrt{\pi a}$ → cross $K_{ISCC}$ and the sealed acid tip both dissolves metal (Faraday) and breeds hydrogen (embrittlement) → the crack creeps until $a$ grows enough that $K$ finally reaches $K_{IC}$ and the part snaps.

Recall Feynman retelling — say it back in plain words

Start with a bar you're gently pulling. A scratch bends the force-lines so they pile up at its tip — the closer you look to the tip, the harder the metal is pulled, blowing up like one-over-root-distance. We can't track an infinite peak, so we bottle the whole spike into a single height number, $K=Y\sigma \sqrt{\pi a}$: bigger stress or a longer crack both make $K$ bigger. There are two danger lines. The high one, $K_{IC}$, is where the part snaps instantly. The low one, $K_{ISCC}$, is where slow chemistry-driven creep begins — and it sits far below the snap line, which is the sneaky part. Why can chemistry work at such a low push? Because the tip is a sealed pocket: metal ions can't escape, they turn the trapped water acidic, and acid does two things — it strips the protective film so bare metal dissolves (Faraday's law tells us how much per unit of current), and it makes tiny hydrogen atoms that wriggle into the metal ahead of the tip and weaken the bonds there. Both push the crack forward a hair at a time. It keeps creeping, $a$ keeps growing, so $K$ keeps rising, until one day $K$ hits $K_{IC}$ and — snap. For our 7075 wing fitting the numbers gave $K=13.3$ against a threshold of $12$: over the line, so it grows, even though the stress was nowhere near the yield strength. To stop it, kick any leg of the tripod: less stress, no flaw, or no corrosive environment.

Recall

What single number captures the whole tip stress field, and what is its formula? ::: The stress-intensity factor $K=Y\sigma \sqrt{\pi a}$. Why is $K_{ISCC}$ much smaller than $K_{IC}$? ::: Fast fracture needs enough stored elastic energy; slow SCC only needs the tip reactive enough for corrosion chemistry, which is patient and accepts a far smaller driving force. Where does the $\sqrt{a}$ in every crack formula come from? ::: The near-tip stress fades like $1/\sqrt{r}$; integrating that singular field makes the driving force scale with $\sqrt{a}$. Why does a crack tip go acidic even in neutral seawater? ::: The narrow (occluded) tip traps metal ions, which hydrolyse to release ${\text{H}}^{+}$ that cannot flush out, dropping the local pH to 2–3. How can over-doing cathodic protection cause failure? ::: Very negative potential accelerates $2{\text{H}}^{+}+2e^{-}\to \text{H}$, injecting hydrogen and embrittling high-strength steel.

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Parent: main topic note · See also: Passivation and oxide films, High-strength aluminium alloys (7075) and tempering.