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

Definition Symbols you will meet (all in plain words first)

$σ$ (sigma) — stress: force per unit area a material carries, in MPa (megapascals = millions of newtons per square metre). Think "how hard you push per unit of cross-section."
$σ_{y}$ — yield strength: the stress at which the metal starts to deform permanently (stops springing back).
$σ_{0}$ (sigma-zero) — friction stress: the baseline stress to move a dislocation inside a grain, before any boundary help. It is the intercept of the Hall–Petch line.
$ρ$ (rho) — density: mass packed into each cm³, in g/cm³. Steel ≈ 7.9 means one sugar-cube of steel weighs 7.9 g.
$\frac{σ}{ρ}$ — specific strength: strength per unit mass. The single number aerospace lives by.
$g$ — gravitational acceleration $\approx 9.81 m/s^{2}$ : converts a mass into the weight (downward force) it pulls with. Used only in the "reach" picture below.
$d$ — grain diameter: the width of one tiny crystal "tile" in the metal, usually in micrometres (µm, millionths of a metre).
$k$ — Hall–Petch coefficient: how strongly grain boundaries stiffen the metal; the slope of $σ_{y}$ against $d^{- 1/2}$ , in units of $MPa\cdotm^{1/2}$ .
$τ$ (tau) — shear stress: the sideways sliding stress that actually moves dislocations.
$G$ — shear modulus, $b$ — Burgers vector (the atomic step-size of a dislocation), $L$ — precipitate spacing (gap between pinning particles).
$T$ — line tension of a dislocation: the energy stored per unit length of the dislocation line, so it behaves like a stretched elastic string that resists being bent ( $T \approx G b^{2}$ ).
$M$ — Taylor factor ( $\approx 3$ ): the number that converts a shear stress on the slip plane into the tensile stress measured in a pull test.

Intuition The three pictures this whole page runs on — and how to read each one

Three figures carry the reasoning; every exercise leans on one, so learn to read them once, now. Each appears again right beside the level that needs it.

Figure 1 (below) — a bar chart of specific strength $σ_{y} / ρ$ . How to read it: each bar is one alloy family; taller bar = more strength carried per kilogram. Notice the Ti bar towers over the steel bar even though steel's raw strength is close — that height gap is the reason Ti flies. The colours are not a temperature scale; the caption warns that bar height does not track the temperature ceiling.
Figure 2 (before L2.2) — the Hall–Petch curve. How to read it: the horizontal axis is grain size (small grains on the left), the vertical axis is yield strength. The blue curve rises to the left — smaller grains, stronger metal — and the two marked dots are the exact grain sizes in problem L2.2. The red dashed branch at the far left is the breakdown: below ~30 nm the curve turns over and falls, warning you never to extrapolate the blue law that far.
Figure 3 (before L4.1) — a stacked bar of the three strengthening contributions. How to read it: the total bar height is $σ_{y}$ ; each coloured block is one additive term (baseline, grain-boundary, precipitate). The second stack shows what happens when precipitates coarsen — watch the orange block shrink while the others stay fixed.

Definition Units of specific strength — make it a "reach" number

We wrote $σ_{y} / ρ$ as $MPa\cdotcm^{3} / g$ because the inputs came in those units. That combination is awkward, so here is the standard conversion to build intuition. $1 MPa = 1 0^{6} Pa = 1 0^{6} N/m^{2}$ and $1 g/cm^{3} = 1000 kg/m^{3}$ , so $1 \frac{MPa\cdotcm ^{3}}{g} = \frac{1 0 ^{6} N/m ^{2}}{1000 kg/m ^{3}} = 1000 \frac{N\cdotm}{kg} = 1 \frac{kN\cdotm}{kg} .$ Since $1 N\cdotm = 1 J$ (a newton-metre of work is a joule of energy), this same number is also $1 kJ/kg$ . Read the "kJ/kg" form as energy-per-mass, not torque — the kN·m/kg form is the more natural "force × length per mass" reading for a hanging bar, and both are the identical number. So Ti's $\approx 199 MPa\cdotcm^{3} / g$ is $\approx 199 kN\cdotm/kg$ — read it as "the length of material (in km) that can hang under its own weight before snapping": using $g \approx 9.81 m/s^{2}$ (gravity, defined in the symbol list), $σ_{y} / (ρ g) \approx 199 000/9.81 \approx 20000 m = 20$ km of self-supporting titanium. That "reach" picture is why $σ / ρ$ is the master variable.

Level 1 — Recognition

L1.1

Rank these four families by service temperature ceiling, lowest first: Ni superalloy, Al 7075, stainless steel, Ti-6Al-4V.

Recall Solution

Read the ceilings from the parent table: Al ≈ 150 °C, Ti ≈ 400 °C, stainless ≈ 600 °C, Ni ≈ 1000 °C+. Order (cold → hot): Al 7075 → Ti-6Al-4V → stainless steel → Ni superalloy. This is exactly the mnemonic "All Tigers Snore Nightly."

L1.2

For each designation, name the main alloying element: (a) Al 2024, (b) Al 7075.

Recall Solution

The leading digit of an aluminium 4-digit code names the main alloy family. (a) 2xxx = Copper (Cu) → 2024 is Al–Cu–Mg. (b) 7xxx = Zinc (Zn) → 7075 is Al–Zn–Mg–Cu. Memory hook: "2 Cu, 7 Zn."

L1.3

In Ti-6Al-4V, which element stabilizes the α (HCP) phase and which stabilizes the β (BCC) phase?

Recall Solution

Aluminium (Al) is the α-stabilizer → strong, creep-resistant, HCP.
Vanadium (V) is the β-stabilizer → ductile, formable, BCC. "6-4" is literally 6% Al + 4% V. See Phase Diagrams (α–β Titanium).

Level 2 — Application

L2.1

Compute the specific strength $σ_{y} / ρ$ for Ti-6Al-4V using $σ_{y} = 880$ MPa, $ρ = 4.43$ g/cm³. Then do the same for a stainless steel with $σ_{y} = 700$ MPa, $ρ = 7.9$ g/cm³. Which wins per unit mass, and by what factor? Give the winner's value in standard kN·m/kg too.

Recall Solution

WHAT we compute: specific strength = strength divided by density. WHY: aerospace pays fuel to carry mass, so the fair contest is strength per unit mass, not raw strength. $(\frac{σ _{y}}{ρ})_{Ti} = \frac{880}{4.43} \approx 198.6 MPa\cdotcm^{3} / g$ $(\frac{σ _{y}}{ρ})_{SS} = \frac{700}{7.9} \approx 88.6 MPa\cdotcm^{3} / g$ Ti wins. Ratio $= 198.6/88.6 \approx 2.24$ — titanium delivers ~2.2× more strength per kilogram even though the steel is nearly as strong in raw terms. In standard units (using $1 MPa\cdotcm^{3} / g = 1 kN\cdotm/kg$ from the conversion box), Ti $\approx 198.6 kN \cdot m/kg$ . This is the whole reason Ti exists in airframes. Read this off Figure 1: find the Ti bar and the steel bar and note the Ti bar is more than twice as tall — that height ratio is exactly the 2.24 you just computed.

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

L2.2

A metal has grain diameter $d_{1} = 100 μ m$ and yield strength $σ_{y, 1} = 200$ MPa. Its friction stress is $σ_{0} = 50$ MPa. Using Hall–Petch $σ_{y} = σ_{0} + k d^{- 1/2}$ , find $k$ , then predict $σ_{y}$ when the grains are refined to $d_{2} = 25 μ m$ .

Recall Solution

WHAT Hall–Petch says: yield strength = a baseline $σ_{0}$ (the friction stress, defined in the symbol list — resistance inside a grain) plus a boundary bonus $k d^{- 1/2}$ , where $k$ is the Hall–Petch coefficient. Smaller grains → more boundaries → bigger bonus. See Hall–Petch Strengthening and the blue curve in Figure 2. WHY $d^{- 1/2}$ : dislocations pile up at a boundary; the stress at the pile-up tip grows like $d$ , and solving for the applied stress needed to punch into the next grain flips it to $d^{- 1/2}$ . HOW to read it on Figure 2: locate the two green/orange dots — they sit on the blue curve at $d = 100$ µm and $d = 25$ µm. Moving left (smaller grain) climbs the curve; the vertical jump between the dots is the strength gain you're about to compute.

Work in metres: $d_{1} = 100 \times 1 0^{- 6} m$ , so $d_{1}^{- 1/2} = (1 0^{- 4})^{- 1/2} = 100 m^{- 1/2}$ . $200 = 50 + k (100) \Rightarrow k = \frac{150}{100} = 1.5 MPa\cdotm^{1/2}$ Now $d_{2} = 25 \times 1 0^{- 6} m$ , $d_{2}^{- 1/2} = (25 \times 1 0^{- 6})^{- 1/2} = 200 m^{- 1/2}$ . $σ_{y, 2} = 50 + 1.5 (200) = 50 + 300 = 350 MPa$ Quartering the grain size (÷4) doubled the boundary bonus (150 → 300 MPa), because $d^{- 1/2}$ of $d /4$ is $2 \times$ larger. Total strength rose from 200 to 350 MPa.

Edge case you must respect (the red dashed branch in Figure 2): the $d^{- 1/2}$ law only holds while grains are big enough to host a dislocation pile-up (roughly $d ≳ 20$ – $50$ nm). Below that — the nanocrystalline regime — there is no room for a pile-up; deformation switches to grain-boundary sliding, and the curve inverts: making grains even smaller now makes the metal weaker. This is the famous inverse Hall–Petch breakdown. Never extrapolate the straight blue line into the nanometre regime.

L2.3

Two alloys have the same shear modulus $G$ and Burgers vector $b$ . Alloy A has precipitate spacing $L_{A} = 200$ nm, alloy B has $L_{B} = 50$ nm. Using the Orowan law $τ_{O r o w an} = \frac{G b}{L}$ , which resists dislocation bowing better, and by what factor?

Recall Solution

WHAT Orowan says: a dislocation forced to bow around particles it cannot cut needs shear stress $τ = G b / L$ . See Precipitation Hardening. WHY the $1/ L$ scaling — the geometry of bowing (this is the real intuition): picture the dislocation as a stretched elastic string pinned at two particles a distance $L$ apart. Push on it and it bows out into an arc. A line under tension resists bending with a line tension $T$ (defined in the symbol list: energy stored per unit length, $T \approx G b^{2}$ ). To bend a segment of length $L$ into a half-circle, the sideways force per unit length the applied stress must supply is $τ b$ , and balancing bowing force against line tension gives $τ b \cdot L \sim 2 T = 2 G b^{2}$ , i.e. $τ \sim G b / L$ . The shorter the gap $L$ , the tighter the arc the string is forced into, and a tighter curve costs more stress — exactly like it takes more force to bend a wire around a small peg than a big one. That is why closer particles (smaller $L$ ) are harder to bow around → stronger. With $G$ and $b$ identical, $τ \propto 1/ L$ : $\frac{τ _{B}}{τ _{A}} = \frac{1/ L _{B}}{1/ L _{A}} = \frac{L _{A}}{L _{B}} = \frac{200}{50} = 4$ Alloy B is 4× harder to bow through. Halving spacing doubles strength; quartering it quadruples it — this is why fine, closely spaced precipitates are the goal of peak-ageing.

Edge case — the cut/bypass transition (why smaller is not always better): the $G b / L$ bowing law assumes the dislocation goes around the particles. But if the particles are very small and coherent, the dislocation finds it easier to slice straight through them; that "cutting" strength instead grows with particle radius $r$ (roughly $\propto r$ ), so it rises as particles grow. Real strength is the lower of the two paths the dislocation can take. At small $r$ /small $L$ cutting wins (strength climbs with size); at large $L$ bowing wins (strength falls with size). Peak strength sits exactly at the crossover — this is the microscopic reason a peak-aged condition exists and why both under-ageing and over-ageing lose strength.

Level 3 — Analysis

L3.1

A design team must choose between 2024-T3 ( $σ_{y} \approx 350$ MPa, high fracture toughness, excellent fatigue) and 7075-T6 ( $σ_{y} \approx 500$ MPa, lower toughness, worse stress-corrosion) for (a) the lower fuselage skin (tension-dominated, must tolerate cracks) and (b) the upper wing spar (compression-dominated, strength-limited). Assign each and justify.

Recall Solution

WHAT drives the choice: not raw strength, but how the part fails.

(a) Lower fuselage — 2024-T3. The lower skin is loaded in tension during flight; tension makes cracks grow. This part must be damage-tolerant — survive with a small crack until inspection catches it. 2024's superior fracture toughness and fatigue resistance win here even though it is weaker. See Fatigue and Fracture Toughness.
(b) Upper wing spar — 7075-T6. Under compression, cracks tend to close, so toughness matters less; the limit is how much stress the material can carry before yielding. 7075's higher $σ_{y}$ (≈500 vs 350 MPa, a 43% strength gain) is decisive. Principle: match the alloy's strong property to the part's failure mode — strength for compression-limited parts, toughness for tension/fatigue-critical parts.

L3.2

At room temperature, refining grains raises strength (Hall–Petch). Yet the hottest turbine blades are made as single crystals with no grain boundaries. Explain, in terms of what boundaries do at high vs low temperature, why the rule reverses.

Recall Solution

Low T: dislocation glide is the only deformation path, and grain boundaries block it — pile-ups can't cross easily, so more boundaries = stronger (Hall–Petch, $σ_{y} = σ_{0} + k d^{- 1/2}$ ). High T (>0.7× melting point): a new deformation path opens — creep: atoms diffuse and boundaries slide against each other and act as fast diffusion highways and crack-nucleation sites. Now boundaries are the weakest link. See Creep and High-Temperature Deformation and Single-Crystal Turbine Blades. Reversal: removing all boundaries (single crystal) removes the sliding/diffusion paths, so it kills creep. The strengthening philosophy inverts with temperature — the winning microstructure at 20 °C is the losing one at 1000 °C.

L3.3

Titanium and aluminium both age-harden, yet Ti parts cost far more to make. Give two physical/chemical reasons rooted in titanium's reactivity, and state the design consequence.

Recall Solution

Reason 1 — hot reactivity with air: at processing temperatures Ti eagerly absorbs oxygen and nitrogen, forming a brittle surface layer, so it must be melted and heat-treated under vacuum or inert gas — expensive equipment and slow cycles. Reason 2 — poor machinability: Ti's low thermal conductivity concentrates cutting heat at the tool tip and it work-hardens/galls, wearing tools fast → slow feeds, more scrap. Design consequence: because Ti processing is costly, engineers use cheaper aluminium wherever the temperature ceiling (~150 °C) allows, reserving Ti only for parts that are too hot or too weight-critical for Al but that cannot afford steel's mass. In other words, titanium's reactivity is why the material-selection ladder still starts at aluminium rather than jumping straight to the stronger metal — cost, driven by chemistry, is a first-class design constraint alongside strength and temperature.

Level 4 — Synthesis

L4.1

An alloy is strengthened by both grain refinement and precipitates, and the two contributions add: $σ_{y} = σ_{0} + k d^{- 1/2} + M τ_{O r o w an}, τ_{O r o w an} = \frac{G b}{L} .$ Given $σ_{0} = 60$ MPa, $k = 0.6$ MPa·m $^{1/2}$ , $d = 36 μ m$ , $G = 26, 000$ MPa, $b = 0.29$ nm, $L = 100$ nm, and Taylor factor $M = 3$ (converts shear stress to tensile stress), compute $σ_{y}$ . Show each additive piece.

Recall Solution

WHAT the equation says: total yield = intrinsic baseline + grain-boundary bonus + precipitate bonus. Two independent obstacle systems, so their stress contributions add. WHY $M$ : the Orowan law gives a shear stress $τ$ ; multiplying by the Taylor factor $M \approx 3$ (defined in the symbol list) converts it to the tensile yield contribution measured in a pull test. The three pieces are exactly the three stacked blocks in the first stack of Figure 3 — read the total bar height as $σ_{y}$ and each colour as one term.

Piece 1 — baseline: $σ_{0} = 60$ MPa.

Piece 2 — Hall–Petch. $d = 36 \times 1 0^{- 6} m$ , so $d^{- 1/2} = (36 \times 1 0^{- 6})^{- 1/2} = \frac{1}{6 \times 1 0 ^{- 3}} \approx 166.7 m^{- 1/2}$ . $k d^{- 1/2} = 0.6 \times 166.7 = 100 MPa .$

Piece 3 — Orowan. $b = 0.29 \times 1 0^{- 9} m$ , $L = 100 \times 1 0^{- 9} m$ : $τ_{O r o w an} = \frac{G b}{L} = \frac{26000 \times 0.29 \times 1 0 ^{- 9}}{100 \times 1 0 ^{- 9}} = \frac{26000 \times 0.29}{100} = 75.4 MPa .$ $M τ_{O r o w an} = 3 \times 75.4 = 226.2 MPa .$

Total: $σ_{y} = 60 + 100 + 226.2 = 386.2 MPa .$ The precipitates carry the biggest share here (226 MPa, the tallest colour block) — showing why age-hardening is the dominant lever in high-strength 7075-type alloys.

L4.2

For the alloy in L4.1, over-ageing coarsens precipitates so the spacing grows from $L = 100$ nm to $L = 250$ nm while everything else stays fixed. Compute the new $σ_{y}$ and the strength lost. Which mechanism's contribution changed?

Recall Solution

Only the Orowan term depends on $L$ (as $1/ L$ ). Grain size and baseline are untouched. $τ_{O r o w an}^{'} = \frac{26000 \times 0.29 \times 1 0 ^{- 9}}{250 \times 1 0 ^{- 9}} = \frac{7540}{250} = 30.16 MPa, M τ^{'} = 3 \times 30.16 = 90.48 MPa .$ $σ_{y}^{'} = 60 + 100 + 90.48 = 250.48 MPa .$ Strength lost $= 386.2 - 250.5 = 135.7$ MPa — a 35% drop, entirely from the precipitation (Orowan) term as particles spread apart. On Figure 3 this is the second stack: the orange (precipitate) block has shrunk while the baseline and grain-boundary blocks are unchanged. This is precisely why alloys are held at peak-aged (finest useful spacing) and why over-ageing is avoided. See Precipitation Hardening.

Level 5 — Mastery

L5.1

A bracket sits near an engine at 350 °C, carries a fixed tensile load $F = 40$ kN, and every gram costs fuel. You may pick Al 7075 ( $σ_{y} = 500$ MPa, $ρ = 2.8$ , ceiling 150 °C) or Ti-6Al-4V ( $σ_{y} = 880$ MPa, $ρ = 4.43$ , ceiling 400 °C). Using a safety factor of 1.5 (design to $σ_{d es i g n} = σ_{y} /1.5$ ): (a) eliminate any material that cannot survive the temperature; (b) for the survivor, size the minimum cross-section $A$ ; (c) compute the mass per metre of bar and confirm the choice is sensible.

Recall Solution

(a) Temperature screen. The bracket runs at 350 °C. Al 7075's ceiling is 150 °C → it would soften and creep; eliminate Al. Ti-6Al-4V's ceiling is 400 °C > 350 °C → survives. (Note: the first filter in any selection is "does it survive the environment?", before any strength arithmetic — a lighter material that melts is worth nothing.)

(b) Size the section (Ti). Design stress: $σ_{d es i g n} = \frac{σ _{y}}{1.5} = \frac{880}{1.5} = 586.7 MPa .$ Required area from $σ = F / A$ , with $F = 40000$ N and $586.7$ MPa $= 586.7 \times 1 0^{6}$ Pa: $A = \frac{F}{σ _{d es i g n}} = \frac{40000}{586.7 \times 1 0 ^{6}} = 6.82 \times 1 0^{- 5} m^{2} = 68.2 mm^{2} .$

(c) Mass per metre. Volume of a 1 m bar $= A \times L = 6.82 \times 1 0^{- 5} m^{2} \times 1 m = 6.82 \times 1 0^{- 5} m^{3} = 68.2 cm^{3}$ . With $ρ = 4.43$ g/cm³: $m = 68.2 \times 4.43 \approx 302 g/m .$ Sensible? Yes. Ti is the only candidate that survives 350 °C and its high specific strength keeps the bar to ~0.30 kg/m. Aluminium's lower density can't help if it can't take the heat — the temperature filter is decisive, and among survivors specific strength (the tallest bar for Ti in Figure 1) picks the winner.

L5.2

Same bracket, but now redesigned to run cooler, at 120 °C. Re-run part (a): does the answer flip? If so, resize with Al 7075 and compare mass per metre to the titanium result. What general design lesson does the flip teach?

Recall Solution

(a) Temperature screen at 120 °C. Now both survive (Al ceiling 150 °C > 120 °C; Ti ceiling 400 °C). The decision moves from "who survives?" to "who's lighter for the load?"

Size Al 7075. $σ_{d es i g n} = 500/1.5 = 333.3$ MPa. $A = \frac{40000}{333.3 \times 1 0 ^{6}} = 1.20 \times 1 0^{- 4} m^{2} = 120 mm^{2} .$ Al needs a bigger section (120 vs 68.2 mm²) because it is weaker. Mass per metre: volume $= 1.20 \times 1 0^{- 4} m^{3} = 120 cm^{3}$ ; with $ρ = 2.8$ : $m = 120 \times 2.8 = 336 g/m .$ Compare: Al 336 g/m vs Ti 302 g/m — Ti is still slightly lighter (~10% less mass) because its specific strength ( $880/4.43 \approx 199$ ) beats Al's ( $500/2.8 \approx 179$ ). But Al is far cheaper to make (no vacuum processing, easy machining). Lesson: once temperature stops being the deciding filter, the choice becomes a specific-strength vs cost trade. Ti wins on mass but Al often wins overall because the 10% weight penalty is cheaper than titanium's processing cost. Selection is layered: survive the environment first, then optimise strength-per-mass against cost.

Recall Full mark-scheme summary (one line each)

L1.1 order low→high T ::: Al → Ti → Stainless → Ni L1.2 main elements ::: 2024 = Cu, 7075 = Zn L1.3 stabilizers ::: Al stabilizes α (HCP), V stabilizes β (BCC) L2.1 specific strengths ::: Ti ≈ 198.6, SS ≈ 88.6, ratio ≈ 2.24 (Ti ≈ 198.6 kN·m/kg) L2.2 Hall–Petch ::: k = 1.5 MPa·m^(1/2), σ_y2 = 350 MPa L2.3 Orowan ratio ::: Alloy B is 4× stronger L3.1 assignment ::: lower fuselage → 2024-T3, upper spar → 7075-T6 L3.2 reversal ::: boundaries block slip cold but slide/diffuse (creep) hot → single crystal L3.3 Ti cost ::: vacuum/inert processing + poor machinability → use Al where T allows L4.1 total yield ::: 60 + 100 + 226.2 = 386.2 MPa L4.2 over-aged ::: 250.5 MPa, loss ≈ 135.7 MPa (~35%) L5.1 Ti bracket ::: A = 68.2 mm², m ≈ 302 g/m L5.2 Al bracket ::: A = 120 mm², m = 336 g/m; Ti still ~10% lighter

See also: Dislocations and Slip · Corrosion and Passivation · Phase Diagrams (α–β Titanium)