Exercises — Corrosion in aerospace environments — stress corrosion cracking, hydrogen embrittlement

The three legs are: (1) a susceptible material, (2) sustained tensile stress, (3) a specific corrosive species. Remove any one leg and the "stool" falls — SCC stops. This is the single most useful design fact, because you only have to defeat the cheapest leg.

The figure below draws this as a literal 3-legged stool: the seat labelled "SCC occurs" only stays up while all three black legs are standing. The red leg on the right has been kicked out — the arrow shows the collapse. What to take from it: you don't have to defeat every cause of corrosion; snapping the one cheapest-to-remove leg (here, the corrosive-species leg) drops the whole stool and stops SCC.

(a) and (c). Both put the part in contact with $H^{+}$ ions (hydrogen ions from acid) that get reduced at the metal surface to atomic hydrogen. We write that surface-stuck atom as $H_{ads}$, where the subscript "ads" means adsorbed — a lone $H$ atom clinging to the metal surface, before it either pairs up into $H_2$ gas or diffuses into the lattice: $2H^{+}+2e^{-}⟶2H_{ads}$ Pickling is an acid dip (lots of $H^{+}$); Cd-plating is a cathodic bath where the bolt itself is the cathode, so $H^{+}$ discharges on it. Sand-blasting (b) is dry mechanical — no $H^{+}$. Epoxy paint (d) is a barrier, no electrochemistry.

$K_{op}<K_{ISCC}\ll K_{IC}$ You operate below the SCC threshold (so cracks don't creep), and that threshold sits far below the fast-fracture toughness. The whole danger of SCC is that $K_{ISCC}$ can be a small fraction of $K_{IC}$ — the part fails "early."

Step 1 — compute $K$. Why? $K$ is the driving force; we compare it to the threshold. Note $Y=1.12$ because this is a surface flaw (see the geometry definition above). First convert $a$: $1.5\text{mm}=0.0015\text{m}$. $K=1.12\times 140\times \sqrt{\pi \times 0.0015}=1.12\times 140\times 0.06865=10.76\text{MPa}\sqrt{\text{m}}$ Step 2 — compare. Why? SCC grows only if $K>K_{ISCC}$. $10.76<12\Rightarrow \text{no SCC propagation (yet).}$ The margin is thin — a slightly deeper flaw or higher stress flips it. This is a pass with warning.

Step 1 — track the units of current density. Why? This is where the "per area, per second" comes from, and getting it explicit stops all confusion. An ampere is a coulomb per second, so $i=40\frac{\text{A}}{{\text{m}}^2}=40\frac{\text{C}}{{\text{m}}^2\cdot \text{s}}.$ So over one second the charge delivered to each square metre is $Q/\text{area}=40{\text{C/m}}^2$. Step 2 — feed that charge into Faraday. Why? $m=MQ/(zF)$ converts charge to dissolved mass; feed it the charge-per-area and you get mass-per-area. Watch the units cancel: $\frac{m}{\text{area}\cdot \text{s}}=\frac{Mi}{zF}=\frac{(\text{g/mol})(\text{C}{\text{m}}^{-2}{\text{s}}^{-1})}{(\text{C/mol})}=\frac{\text{g}}{{\text{m}}^2\cdot \text{s}}✓$ $\frac{m}{\text{area}\cdot \text{s}}=\frac{56\times 40}{2\times 96485}=0.01161{\text{g/m}}^2\text{s}$ So about $0.0116$ grams of iron per square metre every second at the tip — a mass-rate per unit area, exactly what the units demanded.

Dissolved metal ions trapped inside the tight crack hydrolyse water, releasing $H^{+}$: $M^{n+}+H_{2}O⟶M(OH{)}^{(n-1)+}+H^{+}$ Because the crack is occluded (cut off from the bulk), the $H^{+}$ can't diffuse away and piles up — driving the local pH down to 2–3 even while the ocean outside is neutral. That acid pocket is exactly the hydrogen factory that feeds HE.

Step 1 — divide mass rate by density. Why? $\frac{\text{mass/area/time}}{\text{mass/volume}}=\frac{\text{volume/area}}{\text{time}}=\frac{\text{length}}{\text{time}}$ — the units literally give a velocity. $v=\frac{0.01161}{7.87\times 10^6}=1.475\times 10^{-9}\text{m/s}$ Step 2 — convert to mm/day. Multiply by $1000\text{mm/m}$ and by $86400\text{s/day}$: $v=1.475\times 10^{-9}\times 1000\times 86400=0.1274\text{mm/day}$ Per second it's invisible; over a week of static load it's a 0.9 mm crack — that's the treachery of SCC.

Step 1 — invert the K equation. Why? We want the $a$ that makes $K=K_{ISCC}$; rearrange $K=Y\sigma \sqrt{\pi a}$. $\sqrt{\pi a^{\ast }}=\frac{K_{ISCC}}{Y\sigma }\Rightarrow a^{\ast }=\frac{1}{\pi }{\left(\frac{K_{ISCC}}{Y\sigma }\right)}^2$ Step 2 — plug numbers. $a^{\ast }=\frac{1}{\pi }{\left(\frac{12}{1.12\times 140}\right)}^2=\frac{1}{\pi }(0.07653{)}^2=\frac{0.005857}{\pi }=0.001864\text{m}=1.86\text{mm}$ Step 3 — interpret. The flaw only needs to grow from $1.5\text{mm}$ to $1.86\text{mm}$ — a mere 0.36 mm — before SCC becomes self-sustaining. At the L3.1 rate ($0.127\text{mm/day}$) that's under 3 days if any dissolution starts. "Passes today" is not "safe."

Cathodic protection works by making the whole part a cathode, suppressing the anodic dissolution $M\to M^{n+}+n{e}^{-}$. Good so far. But every cathode must run some cathodic reaction, and if you drive the potential too negative, the dominant one becomes hydrogen discharge: $2H^{+}+2e^{-}⟶2H_{ads}$ Those adsorbed $H$ atoms ($H_{ads}$) diffuse into the lattice and embrittle it (HE). So over-protecting a high-strength steel trades a corrosion problem for a hydrogen-cracking problem. The fix is to hold the potential in a controlled window — protective, but not so negative that hydrogen evolves.

(a) Hydrogen embrittlement. Three clues: Cd-plating is an acid cathodic bath that co-deposits $H$; high-strength steel (high ${\sigma }_y$) has a small cohesion margin so a little $H$ drops ${\sigma }_{coh}$ below the working stress; no visible corrosion + fast failure rules out slow anodic-dissolution SCC, which needs an active corroding environment and shows attack. The hydrogen was baked in during plating, not grown in service. (b) Hydrogen-relief bake: heat to $190^{\circ }\text{C}$ for $\ge 8\text{h}$ within hours of plating and before loading. Why it works: atomic hydrogen is highly mobile; gentle heat gives it the thermal energy to diffuse back out of the lattice and escape as $H_2$ before the part is stressed. Restore cohesion, then load. This is a mandatory aerospace spec, not optional.

Map each fix to a leg (recall L1.1):

• (i) Drop stress 20% → attacks the tensile-stress leg. New $K=0.8\times 10.76=8.6\text{MPa}\sqrt{\text{m}}$, comfortably below 12. Direct and predictable, but you lose load capacity.
• (ii) Shot-peening → also attacks the stress leg, but locally. Why does compression help? $K$ responds only to the tensile stress that pulls the crack faces apart. Shot-peening hammers the surface layer, locking in a compressive residual stress ${\sigma }_{res}<0$ that adds to the applied stress at the flaw: the crack now feels ${\sigma }_{applied}+{\sigma }_{res}$. Because ${\sigma }_{res}$ is negative, this net opening stress is smaller, so $K=Y({\sigma }_{applied}+{\sigma }_{res})\sqrt{\pi a}$ drops. If the net stress at the surface even goes negative, the crack faces are squeezed shut and cannot open at all — $K$ effectively falls to zero there. Cheap, no redesign — usually the winner.
• (iii) Barrier coat / passivation → attacks the environment leg by keeping the corrosive species off the metal.

Best single answer: shot-peening, because it defeats a leg without sacrificing strength and is cheap to apply. The lesson: you never have to fix all three — knock out one.

At the tip, stress ruptures the passive film, exposing bare metal that anodically dissolves ($M\to M^{n+}+n{e}^{-}$) — that's the SCC slip-dissolution mechanism, and it sharpens the crack. The dissolved $M^{n+}$ then hydrolyses ($M^{n+}+H_{2}O\to M(OH{)}^{(n-1)+}+H^{+}$), acidifying the occluded tip to pH 2–3. That $H^{+}$ is reduced to atomic hydrogen ($2H^{+}+2e^{-}\to 2H_{ads}$), which diffuses ahead of the tip and lowers cohesion — that's HE. So dissolution manufactures the hydrogen: one tip, two coupled damage modes. This is why the parent note teaches SCC and HE together.

(a) Threshold check at $a_0=0.0005\text{m}$. Why? Compute $K$ at the starting flaw and compare to $K_{ISCC}$. $K_0=1.0\times 200\times \sqrt{\pi \times 0.0005}=200\times 0.03963=7.93\text{MPa}\sqrt{\text{m}}$ $7.93<8$ → just below threshold. It hasn't started yet, but barely. (b) Fast-fracture length. Why? Set $K=K_{IC}$ and invert the master equation to find the crack length that snaps the part. $a_c=\frac{1}{\pi }{\left(\frac{K_{IC}}{Y\sigma }\right)}^2=\frac{1}{\pi }{\left(\frac{60}{200}\right)}^2=\frac{0.09}{\pi }=0.02865\text{m}=28.6\text{mm}$ (c) Threshold length first. Why? The crack only advances by SCC once $K=K_{ISCC}$; invert to get that starting length, then count growth from there to $a_c$. $a_{th}=\frac{1}{\pi }{\left(\frac{K_{ISCC}}{Y\sigma }\right)}^2=\frac{1}{\pi }{\left(\frac{8}{200}\right)}^2=\frac{0.0016}{\pi }=0.000509\text{m}=0.509\text{mm}$ The crack must then travel from $a_{th}=0.509\text{mm}$ to $a_c=28.6\text{mm}$, a distance of $\Delta a=28.6-0.509=28.1\text{mm}.$ At the steady $0.13\text{mm/day}$: $t=\frac{\Delta a}{v}=\frac{28.1}{0.13}\approx 216\text{days}.$ Conclusion. Notice $a_0=0.5\text{mm}$ and $a_{th}=0.509\text{mm}$ are almost identical — the part sits a hair below threshold from day one, so the moment any corrosion nudges it over, the ~216-day (≈ 7-month) countdown begins. That window is invisible to a strength-only check, because $K$ never approached $K_{IC}=60$ until the very last moment. The lesson of the whole page: for corrosive sustained-load duty you design against $K_{ISCC}$ and inspect on a schedule shorter than 216 days, never against $K_{IC}$ alone.

A defensible spec:

1. Cap the operating stress so $K<K_{ISCC}$ with margin, verified from the largest inspectable flaw. → kills the tensile-stress leg of SCC.
2. Shot-peen the surface to impose compressive residual stress. → drives surface net stress down so cracks can't open (stress leg, locally).
3. Barrier coat / passivation to exclude chlorides. → kills the environment leg.
4. Mandatory hydrogen-relief bake ($190^{\circ }\text{C}$, $\ge 8\text{h}$) after any plating/pickling, before loading, and avoid over-negative cathodic potentials. → removes lattice hydrogen and prevents HE injection. Together: legs 1–3 defeat SCC; point 4 defeats HE. No single leg carries the whole load.

Tripod's three legs ::: susceptible material, sustained tensile stress, specific corrosive species Driving-force equation ::: $K=Y\sigma \sqrt{\pi a}$ Must-convert-first pitfall in $K$ ::: put crack length $a$ in metres, not mm When is $Y\approx 1$ valid ::: small crack, far from any edge, in a wide plate; use $\approx 1.12$ for a surface flaw Crack velocity from mass rate ::: divide mass-per-area-per-time by density $\rho$ Fix for hydrogen embrittlement after plating ::: hydrogen-relief bake, $190^{\circ }\text{C}$ for $\ge 8$ h before loading Why over-protection is dangerous on high-strength steel ::: too-negative potential discharges $H^{+}$ to atomic H, causing HE Design limit for corrosive sustained load ::: $K_{ISCC}$, not $K_{IC}$ Meaning of the subscript in $H_{ads}$ ::: adsorbed atomic hydrogen — a lone H atom stuck on the metal surface