Corrosion in aerospace environments — stress corrosion cracking, hydrogen embrittlement: Worked examples

The scenario matrix

Before touching numbers, let us list every kind of case the three tools can produce. Each row is a "cell"; every worked example below is tagged with the cell it fills.

#	Cell (scenario class)	Governing tool	What makes it tricky
A	$K > K_{I S C C}$ — crack grows	$K = Y σ π a$	positive margin, part unsafe
B	$K < K_{I S C C}$ — crack dormant	same	negative margin, part safe for now
C	$K = K_{I S C C}$ — exactly at threshold	same	boundary / degenerate case
D	$a \to 0$ — no crack limit	same	$K \to 0$ , what does "safe" even mean?
E	Solve for the critical flaw size $a_{c}$	invert the formula	rearranging, not just plugging
F	Faraday crack velocity (per-second → per-day)	$m = M Q / z F$	unit chain, density division
G	Life prediction: how long until fast fracture?	combine F + A	time = distance / velocity
H	Hydrogen bake (degenerate: no mechanical fix works)	diffusion / spec rule	over-protection injects $H$
I	Exam twist: cathodic protection causes failure	potential vs $H$ evolution	"more protection" backfires

The tool set:

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

The figure above is the map for cells A–D: the same curve $K = Y σ π a$ , with the horizontal threshold line $K_{I S C C}$ cutting it. Above the line the crack runs; below it sleeps; the crossing point is the critical flaw. Keep this picture in mind — every example either sits on this curve or feeds numbers into it.

Worked examples

Example 1 — Cell A: crack grows ( $K > K_{I S C C}$ )

Worked example 7075-T6 wing fitting, visible flaw

$σ = 150 MPa$ , flaw $a = 2 mm$ , $Y = 1.12$ , and for 7075-T6 in salt-laden air $K_{I S C C} = 12 MPa m$ . Forecast: yield is ~500 MPa and we sit at 150 MPa — one-third of yield. Guess: safe or not?

Step 1 — convert the crack to metres. Why? The formula's $π a$ must come out in $m$ to match $K$ in $MPa m$ . $a = 2 mm = 0.002 m$ .

Step 2 — compute the driving force $K$ . Why? $K$ is the single number that says how hard the crack is being pulled open — see the rising curve in the figure. $K = 1.12 \times 150 \times π \times 0.002 = 1.12 \times 150 \times 0.07927 = 13.3 MPa m$

Step 3 — compare to threshold. Why? This is the horizontal line in the figure; crossing it means propagation. $13.3 > 12 \Rightarrow SCC propagates. Unsafe.$

Verify: intuition-check — being at 30 % of yield gives zero protection against SCC, exactly the parent note's warning that failure happens "far below yield." Units: $MPa \times m = MPa m$ ✓. Numeric answer $13.3$ confirmed in VERIFY.

Example 2 — Cell B: crack dormant ( $K < K_{I S C C}$ )

Example 3 — Cell C: exactly at threshold ( $K = K_{I S C C}$ )

Worked example What stress puts us exactly on the line?

Keep $a = 2 mm$ , $Y = 1.12$ , $K_{I S C C} = 12 MPa m$ . Find the stress $σ^{*}$ at which $K$ equals the threshold. Forecast: it must lie between the 100 MPa (safe) and 150 MPa (unsafe) of Examples 1–2. Guess a value.

Step 1 — set $K = K_{I S C C}$ and solve for $σ$ . Why? The threshold is a boundary, not a plug-in; we invert the formula. $σ^{*} = \frac{K _{I S C C}}{Y π a} = \frac{12}{1.12 \times 0.07927} = 135.2 MPa$

Step 2 — interpret the degenerate boundary. Why care? At exactly $K = K_{I S C C}$ the crack is on a knife-edge — theory says "just barely stationary," but real life has scatter, so engineers apply a safety factor and never design to the boundary.

Verify: $135.2$ lies between $100$ and $150$ ✓, and $150 > 135.2$ (Ex 1 unsafe) while $100 < 135.2$ (Ex 2 safe) — fully consistent. Plug back: $1.12 \times 135.2 \times 0.07927 = 12.0$ ✓ (VERIFY).

Example 4 — Cell D: the no-crack limit ( $a \to 0$ )

Worked example A pristine surface — does the formula behave?

Same alloy, $σ = 150 MPa$ , but the part is freshly polished: what is $K$ as $a \to 0$ ? Forecast: with no crack, is the part infinitely safe, or is there a hidden catch?

Step 1 — take the limit. Why? We must know the formula does something sane at the degenerate input. $lim_{a \to 0} K = Y σ π a a \to 0 0$ $K \to 0$ : no crack means no crack-driving force. Good — see the curve starting at the origin in the figure.

Step 2 — the catch. Why not declare victory? SCC and pitting can initiate their own flaw. A pit from pitting corrosion grows until $a$ is large enough to enter Cell A. So $a = 0$ today ≠ $a = 0$ next month.

Step 3 — a tiny but finite flaw. With $a = 0.01 mm = 1 0^{- 5} m$ : $K = 1.12 \times 150 \times π \times 1 0^{- 5} = 1.12 \times 150 \times 0.005605 = 0.94 MPa m$ Far below $12$ — a truly tiny flaw is harmless until it grows.

Verify: monotone in $a$ ✓ ( $0.94$ at $a = 1 0^{- 5}$ < $13.3$ at $a = 2 \times 1 0^{- 3}$ ). $K \to 0$ as $a \to 0$ ✓ (VERIFY). Answer $0.94$ confirmed.

Example 5 — Cell E: critical flaw size (invert for $a_{c}$ )

Worked example The largest flaw you may tolerate

$σ = 150 MPa$ , $Y = 1.12$ , $K_{I S C C} = 12 MPa m$ . What crack length $a_{c}$ first reaches threshold? (This sets the inspection limit — flaws bigger than $a_{c}$ must be rejected.) Forecast: Example 1 showed $a = 2 mm$ already exceeds threshold, so $a_{c}$ must be smaller than 2 mm. Guess.

Step 1 — set $K = K_{I S C C}$ and solve for $a$ . Why? We want the crack length where the curve crosses the line — the crossing point in the figure. $K_{I S C C} = Y σ π a_{c} \Rightarrow a_{c} = \frac{1}{π} (\frac{K _{I S C C}}{Y σ})^{2}$

Step 2 — plug in. Why square? $K$ depends on $a$ , so inverting brings a square. $a_{c} = \frac{1}{π} (\frac{12}{1.12 \times 150})^{2} = \frac{1}{π} (0.07143)^{2} = \frac{0.005102}{π} = 1.62 \times 1 0^{- 3} m = 1.62 mm$

Step 3 — read it off the inversion figure. Why a picture? Inverting a square root is the step readers fumble; the figure below plots $a_{c}$ against stress so you see that the tolerable flaw shrinks fast as stress rises.

Verify: $a_{c} = 1.62 mm < 2 mm$ ✓ (consistent with Ex 1 being over threshold). Plug back: $1.12 \times 150 \times π \times 0.00162 = 12.0$ ✓ (VERIFY).

Example 6 — Cell F: Faraday crack velocity

Worked example How fast does the bare tip dissolve?

A crack tip on iron passes current density $i = 50 A/m^{2}$ . Iron: $M = 56 g/mol$ , $z = 2$ , $ρ = 7.87 g/cm^{3} = 7.87 \times 1 0^{6} g/m^{3}$ . See Faraday's law. Forecast: a whole ampere per few cm² sounds fast. Is the crack cm/day or micron/day?

Step 1 — mass dissolved per area per second. Why? Faraday says charge → mass; per second, $Q$ per area is just $i$ . $\frac{m}{A t} = \frac{M i}{z F} = \frac{56 \times 50}{2 \times 96485} = 0.01451 g m^{- 2} s^{- 1}$

Step 2 — convert mass-rate to a length-rate. Why divide by $ρ$ ? $\frac{g/m ^{2} s}{g/m ^{3}} = m/s$ — mass per area removed, divided by mass per volume, leaves the depth removed per second. $v = \frac{0.01451}{7.87 \times 1 0 ^{6}} = 1.844 \times 1 0^{- 9} m/s$

Step 3 — put it in human units. Why? $1.8 \times 1 0^{- 9} m/s$ is impossible to feel; converting to mm/day lets us compare against a flight-hour maintenance interval, which is how the answer is actually used. Multiply by $86400 s/day$ : $v = 1.594 \times 1 0^{- 4} m/day = 0.159 mm/day$ .

Verify: units chain closes to m/s ✓. $1.844 \times 1 0^{- 9}$ m/s and $0.159$ mm/day confirmed in VERIFY. "Slow per second, catastrophic per flight-budget," as the parent note warns.

Example 7 — Cell G: life prediction (combine F + A)

Worked example Days until fast fracture

Take the iron crack of Example 6 advancing at $v = 1.844 \times 1 0^{- 9} m/s$ by dissolution. It starts at $a_{0} = 1 mm$ and, for the given stress/geometry, fast fracture occurs once $a$ reaches $a_{f} = 5 mm$ . How long? Forecast: it must grow 4 mm at ~0.16 mm/day. Rough guess before computing.

Step 1 — crack distance to cover. Why? Life = distance ÷ speed; the "distance" is how far the tip must travel. $Δ a = a_{f} - a_{0} = 5 - 1 = 4 mm = 0.004 m$

Step 2 — divide by velocity. Why constant $v$ ? We approximate the dissolution rate as steady (tip current density roughly constant), the simplest life model. $t = \frac{Δ a}{v} = \frac{0.004}{1.844 \times 1 0 ^{- 9}} = 2.169 \times 1 0^{6} s$

Step 3 — human units. Why? Seconds are meaningless to a maintenance planner; days match the inspection calendar. $t = 2.169 \times 1 0^{6} /86400 = 25.1 days$ .

Verify: cross-check with Ex 6's per-day rate: $4 mm /0.159 mm/day = 25.1$ days ✓ (two routes agree). Answer $25.1$ days confirmed in VERIFY. Interpretation: a flaw invisible in a walk-around inspection can end the part in under a month.

Example 8 — Cell H: hydrogen bake, the only fix (degenerate: mechanics can't help)

Worked example Cd-plated high-strength steel bolts

Bolts of $σ_{y} = 1800 MPa$ are cadmium-electroplated; hydrogen is co-deposited. The spec demands a hydrogen-relief bake: hold at $19 0^{\circ} C$ for $t \geq 8 h$ . Suppose diffusion out of the bolt follows a characteristic time $τ$ and the retained fraction is $f = e^{- t / τ}$ with $τ = 3.0 h$ . What fraction of hydrogen remains after the mandated 8 h? Forecast: 8 h is under three time-constants. Guess: does most $H$ leave, or barely any?

Step 1 — why baking and not a mechanical fix. Why this is a degenerate cell? No stress reduction or coating removes $H$ already dissolved in the lattice — the only lever is to let it diffuse out before loading. Note also the cathodic-protection link: over-protection would re-inject $H$ .

Step 2 — apply the decay. Why an exponential? Diffusive loss with a single dominant path gives first-order decay $f = e^{- t / τ}$ . $f = e^{- 8/3.0} = e^{- 2.667} = 0.0697$

Step 3 — interpret. Why 8 h is enough: about $93%$ of the hydrogen has escaped ( $1 - 0.0697 = 0.930$ ), dropping $C_{H}$ far enough that the reduced cohesion $σ_{co h} (C_{H})$ stays above the 1800 MPa working stress.

Verify: $f = 0.0697$ and removed fraction $0.930$ confirmed in VERIFY. Sanity: less bake time (say 3 h $= 1 τ$ ) leaves $e^{- 1} = 0.368$ , clearly worse — monotone with time ✓.

Example 9 — Cell I: exam twist, cathodic protection causes the failure

Worked example Over-protection backfires

An engineer applies cathodic protection to a high-strength steel part to stop anodic dissolution, and drives the potential very negative. Days later it cracks. The cathodic reaction $2 H^{+} + 2 e^{-} \to 2 H_{a d s}$ deposits atomic hydrogen at a rate proportional to the (magnitude of the) cathodic current. The engineer doubles the driving current from $i_{1} = 20 A/m^{2}$ to $i_{2} = 40 A/m^{2}$ . By what factor does the hydrogen-generation rate change? Forecast: "more protection = safer" is the intuitive answer. Guess whether the hydrogen supply goes up or down.

Step 1 — name the coupled reaction. Why? Suppressing the anode forces the electrons somewhere; the only reaction available in acidified crack-tip water is hydrogen evolution. So more protection current means more hydrogen at exactly the crack tip.

Step 2 — relate current to hydrogen using Faraday. Why the same Faraday law? Atoms of $H$ produced $\propto$ charge passed $\propto$ current $i$ (identical logic to the mass in Example 6, just counting $H$ instead of dissolved metal). Therefore the hydrogen-rate ratio is simply the current ratio: $\frac{n ˙ _{H, 2}}{n ˙ _{H, 1}} = \frac{i _{2}}{i _{1}} = \frac{40}{20} = 2$

Step 3 — the punchline. Why this is a failure, not a fix? Doubling the "protection" current doubles the hydrogen injected into an 1800 MPa lattice — trading a slow dissolution problem for a faster embrittlement one. The correct practice is controlled potential, not maximal current.

Verify: factor $= 2$ confirmed in VERIFY. Consistency with Example 8: more $H$ in ⇒ need longer bake or lower stress — the two hydrogen cells tell one coherent story.

Recall Self-test (cover the answers)

Which matrix cell asks you to invert the $K$ formula? ::: Cell E — solve $a_{c} = \frac{1}{π} (K_{I S C C} / Y σ)^{2}$ . Why divide the Faraday mass-rate by density? ::: (g/m²·s)/(g/m³) = m/s — it converts metal removed into crack-tip depth per time. In Cell I, doubling cathodic current does what to hydrogen supply? ::: Doubles it — more "protection" can embrittle high-strength steel. At $a \to 0$ what does $K$ do, and why is the part still not permanently safe? ::: $K \to 0$ , but pitting can initiate a new flaw that grows into Cell A.