Metals & alloys — Al alloys (2024, 7075), Ti alloys (Ti-6Al-4V), Ni superalloys (Inconel, Hastelloy), stainless steels: Question bank

Symbol legend (read this first)

Definition Every symbol used on this page, in plain words

Before any trap uses a symbol, here is what each one means and its picture.

$σ$ (sigma) — stress: force spread over area, in megapascals (MPa). Picture pulling on a bar; $σ$ is how hard you pull per unit of cross-section.
$σ_{y}$ — yield stress: the stress at which the metal stops springing back and starts to deform permanently. The "breaking-point of elasticity."
$ρ$ (rho) — density: mass packed into a volume, in g/cm³. Picture how heavy a fixed-size cube of the metal feels.
$σ / ρ$ — specific strength: strength divided by density. With $σ$ in MPa and $ρ$ in g/cm³, the unit is $MPa \cdot cm^{3} / g$ (equivalently $kN \cdot m/kg$ ). Bigger = more strength per kilogram carried.
$σ_{0}$ — the base strength of a single grain's interior (the Hall–Petch constant when grains are huge).
$k$ — the Hall–Petch slope: how strongly grain size affects strength (a material constant).
$d$ — the grain diameter: the width of one crystal grain.
$G$ — the shear modulus: how stiff the metal is against sliding/shearing.
$b$ — the Burgers vector: the size of one atomic step a dislocation carries (roughly one atom spacing).
$L$ — the precipitate spacing: the gap between neighbouring hardening particles.
$τ$ (tau) — shear stress: the sideways stress that actually drives a dislocation to glide.
$r$ — the precipitate radius: how big one hardening particle is.

Quick reference — the numbers behind the traps

Recall Embedded specific-strength table (so you needn't chase the parent)

All $σ_{y}$ are typical peak-aged / standard-temper room-temperature values; $σ_{y} / ρ$ has units $MPa \cdot cm^{3} / g$ (≡ $kN \cdot m/kg$ ). "Max T" is safe service temperature.

Family	$ρ$ (g/cm³)	$σ_{y}$ (MPa)	$σ_{y} / ρ$ (MPa·cm³/g)	Max service T
Al 2024/7075	~2.8	350–500	~150	~150 °C
Ti-6Al-4V	~4.43	~880	~200	~400 °C
Ni superalloy	~8.2	~1000 (hot)	~120	~1000+ °C
Stainless steel	~7.9	250–1000	~50–125	~600 °C

Read: titanium's ~200 MPa·cm³/g beats aluminium's ~150 and the superalloy's ~120 — but only where the temperature stays within its ~400 °C limit.

Visual context (skim before the traps)

Figure 1 — dislocations trip over obstacles. This is the master idea behind Dislocations and Slip and every strengthening trick: a dislocation (the moving "wrinkle") is easy to slide through a clean lattice but snags on foreign atoms, boundaries and particles.

Figure — Metals & alloys — Al alloys (2024, 7075), Ti alloys (Ti-6Al-4V), Ni superalloys (Inconel, Hastelloy), stainless steels

Figure 2 — Hall–Petch: why the exponent is $- \frac{1}{2}$ . Dislocations pile up against a grain boundary. A bigger grain ( $d$ ) holds more of them, and the stress at the pile-up tip grows like $d$ . To break into the next grain you need a fixed critical tip stress, so solving for the applied stress flips the $d$ into $d^{- 1/2}$ . See Hall–Petch Strengthening.

Figure 3 — cutting vs bowing = the peak-aged sweet spot. Small close particles get cut (strength rises with $r$ ); large far-apart particles get bowed around (strength $= G b / L$ falls as spacing grows). The two lines cross at an intermediate size — that peak is the T6 condition of Precipitation Hardening.

Figure 4 — γ′ anomalous strength. In Ni₃(Al,Ti), dislocations lock up by cross-slip onto planes where they can't move (Kear–Wilsdorf locks); more heat makes more locks, so strength climbs with temperature over a wide range before finally falling. This is why superalloys keep their grip in the hot section.

Figure 5 — how a blade beats melting. A ceramic thermal-barrier coating insulates the surface while cool air bled through internal passages leaks out of tiny holes to form an insulating film over the skin. The metal runs far below the gas temperature — the trick behind Creep and High-Temperature Deformation survival and Single-Crystal Turbine Blades.

True or false — justify

Steel is the strongest metal, so it makes the best aircraft, false or true?

False. Aircraft select on specific strength

σ / ρ

(MPa·cm³/g), not raw

σ

. Steel's density (

\approx 7.9

) is nearly double titanium's (

\approx 4.5

), so per kilogram titanium usually wins even when steel is stronger in absolute MPa. See Precipitation Hardening and the table above.

Making grains smaller always makes a metal stronger, true or false?

False — only at low temperature. Hall–Petch Strengthening (

σ_{y} = σ_{0} + k d^{- 1/2}

) holds when boundaries block dislocations (Figure 2). But near turbine heat, grain boundaries slide and diffuse (Creep and High-Temperature Deformation), becoming the weak link — which is why the hottest blades are single crystals with no boundaries.

7075 aluminium is stronger than 2024, so it belongs on every aircraft part, true or false?

False. 7075 has higher yield strength but worse fracture toughness and stress-corrosion resistance. Damage-tolerant, fatigue-critical structure (lower fuselage) uses 2024; compression-loaded upper wing uses 7075. Right tool, right load — see Fatigue and Fracture Toughness.

γ′ precipitate in nickel superalloys gets weaker as temperature rises, like most strengthening, true or false?

False — the opposite. Ni₃(Al,Ti) γ′ shows anomalous yield (Figure 4): cross-slip forms immobile Kear–Wilsdorf locks, and more heat makes more locks, so strength rises with temperature over a wide range, keeping dislocations pinned exactly when you need it hot.

Since both aluminium and titanium age-harden, titanium is roughly as cheap to process, true or false?

False. Titanium is reactive — it burns and greedily absorbs O and N when hot — so it demands vacuum or inert-gas processing and is hard to machine. That expense is precisely why aluminium is preferred wherever the temperature allows.

Adding more chromium always improves a stainless steel's overall performance, true or false?

False. Cr ≥ 10.5% gives the self-healing Cr₂O₃ passive film (Corrosion and Passivation), but excess Cr can promote brittle intermetallic phases and alter the crystal structure (ferrite vs austenite). Composition is a balance, not "more is better."

The "6-4" in Ti-6Al-4V refers to a temper condition, true or false?

False. It is literally the composition: 6% aluminium + 4% vanadium by weight. Al stabilizes the α (HCP) phase, V stabilizes β (BCC) — see Phase Diagrams (α–β Titanium).

Spot the error

"Al 2024 is a 7xxx alloy because it contains zinc." — find the mistake.

The first digit encodes the main alloying element: ==2xxx = copper==, 7xxx = zinc(+Mg). 2024 is Al–Cu–Mg, so it is a 2xxx copper alloy, not a zinc alloy.

"A dislocation always strengthens a metal by cutting through precipitates." — find the mistake.

A dislocation can either cut through (force

\propto r

) or bow around a precipitate (Orowan,

τ = G b / L

) — see Figure 3. Cutting is not "strengthening" — it's how soft, under-aged material yields. Peak strength sits at the crossover between cutting and bowing. See Precipitation Hardening.

"To maximise Al strength, age it as long as possible so the precipitates grow large." — find the mistake.

Over-ageing coarsens precipitates, increasing spacing

L

, which lowers the Orowan stress

G b / L

→ material gets weaker. Maximum strength is the peak-aged (T6) condition at intermediate particle size (the peak in Figure 3), not the longest ageing time.

"Hall–Petch predicts strength

\propto d^{- 1}

because smaller grains help linearly." — find the mistake.

The exponent is

- 1/2

, not

- 1

. As Figure 2 shows, the pile-up tip stress grows like

d

; setting it to a fixed critical value and solving for the applied stress gives the ==

d^{- 1/2}

== dependence, not

d^{- 1}

"Single-crystal turbine blades are weak because Hall–Petch says no boundaries means no strength." — find the mistake.

Hall–Petch's boundary-strengthening applies at low temperature. At turbine heat, boundaries are the creep weak link; removing them (single crystal) increases high-temperature life. The strengthening rules flip with temperature — see Single-Crystal Turbine Blades.

"Stainless steel is stainless because chromium is a noble, unreactive metal." — find the mistake.

Chromium is actually quite reactive — that's the point. It reacts fast with oxygen to form a thin, adherent, self-healing Cr₂O₃ film that seals the surface. Passivation comes from reactivity, not nobility.

"Nickel superalloys are strong hot only because of γ′ precipitates." — find the mistake.

γ′ is central but not alone. Strengthening also comes from solid-solution elements (Cr, Mo, Co) distorting the FCC γ matrix and from carbides pinning grain boundaries. It's a combined system.

Why questions

Why does aerospace use

σ / ρ

instead of just

σ

Because every kilogram lifted to altitude burns fuel for the whole flight, so the real question is "strong for its weight?" The gravitational

g

in the true weight-specific strength

σ / (ρ g)

cancels across all materials, so engineers compare with

σ / ρ

Why can turbine gas be hotter than the metal's melting point without the blade melting?

As Figure 5 shows, a ceramic thermal-barrier coating plus a cool-air film over the surface keep the metal far below the gas temperature, while alloys retain useful strength above

0.7 \times

their melting temperature. Superalloys resist creep, not melting, in that regime — see Creep and High-Temperature Deformation.

Why is 2024 chosen for the fatigue-critical lower fuselage rather than the stronger 7075?

The lower fuselage is damage-tolerant, fatigue-loaded, so what matters is resistance to crack growth (fracture toughness), where 2024 beats 7075. Raw yield strength is not the governing property there.

Why does titanium resist corrosion so well?

It forms a dense, adherent, self-healing TiO₂ passive film that instantly re-seals if scratched, blocking further attack — much like Cr₂O₃ on stainless steel.

Why does adding vanadium make Ti-6Al-4V more formable?

V is a β-stabilizer, promoting the BCC β phase, which is more ductile and easier to work than the HCP α phase (stabilized by Al). Balancing the two gives a tunable α+β microstructure.

Why do dislocations govern strength at all, rather than breaking every bond at once?

A dislocation lets a crystal slip one row of bonds at a time (like moving a rug by walking a wrinkle across it, Figure 1), so real yield stress is far below the ideal. Strength then just means "how hard is it to move that wrinkle?" — see Dislocations and Slip.

Edge cases

If two aluminium tempers have identical composition, can they have very different strengths?

Yes. Strength depends on microstructure, not just composition: T3 (solution + cold-worked) versus T6 (peak-aged) of the same alloy differ markedly because precipitate size/spacing and dislocation density differ.

At what point does grain refinement stop helping and start hurting?

When service temperature is high enough that grain-boundary creep and sliding dominate. There the extra boundaries add diffusion paths and sliding sites, so coarse-grained or single-crystal material outlasts fine-grained.

What happens to the Orowan strength if you keep shrinking precipitate spacing

L

toward zero?

τ_{Orowan} = G b / L

grows without bound as

L \to 0

, so bowing becomes impossibly hard — dislocations then switch to cutting through the (necessarily tiny) particles instead, and the real limit is set by that crossover (Figure 3), not by Orowan alone.

Is a superalloy the right choice for a cold, weight-critical wing spar?

No. At low temperature its specific strength (~120 MPa·cm³/g) is beaten by titanium (~200) and even aluminium (~150), and its density (~8.2) is punishing. Superalloys only justify their weight where the temperature is extreme.

Does stainless steel's ordering "less dense than Ni superalloy" mean the temperature ladder is ordered by density?

No. The service-temperature ladder (Al → Ti → Stainless → Ni) is ordered by heat tolerance, not density. Stainless (~7.9) is actually slightly less dense than Ni superalloys (~8.2), so density and the temperature order genuinely disagree.

Can a metal with lower absolute strength be the correct structural choice over a stronger one?

Yes — routinely. Selection weighs specific strength, toughness, corrosion, temperature and cost together, so 2024 over 7075, or titanium over steel, are cases where the "weaker in MPa" metal wins the real design problem.

Recall One-line survival summary

Every trap on this page reduces to one habit: never rank metals by a single number. Ask "which property, at which temperature, under which load?" — strength ranking, the role of grain boundaries, and even the cost story all flip depending on the answer.