Ceramics — alumina, zirconia, silicon carbide, silicon nitride; properties at high T: Worked examples

This page is a drill sheet. The parent topic gave you four formulas; here we push them into every corner — the safe case, the failure case, the degenerate case (zero jump, zero conductivity), the limiting case (huge stiffness), a word problem, and an exam twist. Before we compute anything, let us agree on what every symbol means, so a reader who has never seen them can follow line one.

Definition The five symbols we will keep reusing

$Δ T$ = temperature jump in kelvin (K). "How many degrees hotter one face suddenly got than the interior." A jump of 200 K means 200 degrees difference — same size in °C or K, since only the difference matters.
$α$ = linear thermal expansion coefficient, units $1/ K$ . "Fractional length gained per degree." If $α = 8 \times 1 0^{- 6} / K$ , then heating 1 metre by 1 K lengthens it by $8$ micrometres.
$E$ = Young's modulus (stiffness), units Pa. "How hard the material fights being stretched." Big $E$ = very stiff, so a tiny forced strain becomes a huge stress.
$ν$ = Poisson's ratio (dimensionless, ~0.2 for ceramics). It appears as $(1 - ν)$ because a hot surface is squeezed in two directions at once (biaxial), not one.
$σ_{f}$ = fracture strength in Pa. "The stress at which a crack runs and the part breaks."

The scenario matrix

Every question this topic can ask is one of these cells:

Cell	What it tests	Degenerate/limit knob	Example
A. Safe shock	$σ_{t h} < σ_{f}$ → survives	—	Ex 1
B. Failing shock	$σ_{t h} > σ_{f}$ → cracks	—	Ex 2
C. Zero jump	$Δ T = 0$	$σ_{t h} \to 0$	Ex 3
D. Rank materials	compare $R$ across the four ceramics	which wins & why	Ex 4
E. Free expansion	tolerance from $Δ L$	—	Ex 5
F. Coating heat block	Fourier $q$ , low- $k$ payoff	$k \to$ small	Ex 6
G. Limiting stiffness	$E \to$ huge / $ν \to 0.5$	asymptotic $R$	Ex 7
H. Word problem	pick the right ceramic for a duty	—	Ex 8
I. Exam twist	transformation toughening $K_{I c}$	crack-tip case	Ex 9

The examples below hit every cell.

Worked examples

Worked example (1) CELL A — Does silicon nitride survive a 500 K shock?

Data: $σ_{f} = 800$ MPa, $E = 310$ GPa, $α = 3.0 \times 1 0^{- 6} / K$ , $ν = 0.27$ , $Δ T = 500$ K.

Forecast: Si₃N₄ is famous for shock resistance — I bet the induced stress stays below 800 MPa. Guess before reading on.

Compute the induced stress. $σ_{t h} = \frac{E α Δ T}{1 - ν} = \frac{310 \times 1 0 ^{9} \cdot 3.0 \times 1 0 ^{- 6} \cdot 500}{1 - 0.27} .$ Why this step? The cold interior restrains the hot face; that forbidden strain $α Δ T$ becomes stress through Hooke's law, boosted by $1/ (1 - ν)$ for biaxial squeeze. Numerator $= 310 \times 1 0^{9} \cdot 1.5 \times 1 0^{- 3} = 4.65 \times 1 0^{8}$ . Divide by $0.73$ : $σ_{t h} \approx 6.37 \times 1 0^{8}$ Pa $= 637$ MPa.
Compare to fracture strength. $637 < 800$ MPa → survives. Why this step? A crack only runs once stress reaches $σ_{f}$ ; we are safely under it.

Verify: Independently, $R = \frac{σ _{f} ( 1 - ν )}{E α} = \frac{800 \times 1 0 ^{6} \cdot 0.73}{310 \times 1 0 ^{9} \cdot 3.0 \times 1 0 ^{- 6}} \approx 628$ K. Our jump 500 K $<$ 628 K, consistent with "survives." Units: Pa·(–)/[Pa·(1/K)] = K. ✓

Worked example (3) CELL C — The degenerate case: zero temperature jump.

Data: any ceramic, but $Δ T = 0$ K.

Forecast: No jump, no fight — stress should be exactly zero. This is the sanity-check cell every formula must pass.

Plug in. $σ_{t h} = \frac{E α \cdot 0}{1 - ν} = 0$ Pa. Why this step? Thermal stress comes only from a restrained temperature difference. Uniform temperature = every atom expands the same = no atom is forced against a neighbour = zero stress.
Interpretation. A part heated slowly and uniformly (interior and surface at the same $T$ ) develops no thermal stress, even at 1500 °C. Thermal shock is about the gradient, not the absolute temperature. Why: this is why a furnace ramp is gentle and a quench is lethal.

Verify: $σ_{t h} (0) = 0$ trivially; and $R$ for $Δ T = 0$ is undefined as a jump but the survivable margin is infinite — you can hold zero-gradient parts forever. ✓

Worked example (4) CELL D — Rank all four ceramics by shock resistance

$R$ . Data (approx):

Ceramic	$σ_{f}$ /MPa	$E$ /GPa	$α$ /( $1 0^{- 6}$ K $^{- 1}$ )	$ν$
Alumina	300	380	8.0	0.22
Zirconia (PSZ)	600	200	10.0	0.30
SiC	400	410	4.5	0.17
Si₃N₄	800	310	3.0	0.27

Forecast: $R = σ_{f} (1 - ν) / (E α)$ rewards high strength, low stiffness, low expansion. Si₃N₄ has tiny $α$ and high $σ_{f}$ → I predict it wins; alumina loses.

Compute each $R$ (units K).
- Alumina: $\frac{300 \cdot 0.78}{380 \cdot 8.0} \times \frac{1 0 ^{6}}{1 0 ^{9} \cdot 1 0 ^{- 6}} = \frac{234}{3040} \times 1 0^{3} \approx 77$ K.
- Zirconia: $\frac{600 \cdot 0.70}{200 \cdot 10.0} \times 1 0^{3} = \frac{420}{2000} \times 1 0^{3} = 210$ K.
- SiC: $\frac{400 \cdot 0.83}{410 \cdot 4.5} \times 1 0^{3} = \frac{332}{1845} \times 1 0^{3} \approx 180$ K.
- Si₃N₄: $\frac{800 \cdot 0.73}{310 \cdot 3.0} \times 1 0^{3} = \frac{584}{930} \times 1 0^{3} \approx 628$ K. Why this step? Direct application of the figure-of-merit; the $1 0^{6} / (1 0^{9} \cdot 1 0^{- 6}) = 1 0^{3}$ factor keeps units in K.
Rank: Si₃N₄ (628) $>$ Zirconia (210) $>$ SiC (180) $>$ Alumina (77). Why: Si₃N₄'s tiny $α = 3 \times 1 0^{- 6}$ is the dominant reason — expansion sits in the denominator. See figure.

Verify: All four numbers rechecked below; ordering Si₃N₄ > ZrO₂ > SiC > Al₂O₃ confirmed. This matches the parent's claim that Si₃N₄ has the "best thermal-shock resistance." ✓

Worked example (6) CELL F — Heat blocked by a zirconia thermal-barrier coating.

A $0.3 mm$ zirconia TBC (see Thermal Barrier Coatings on Superalloys): outer face $1200^{\circ} C$ , metal side $950^{\circ} C$ , $k = 2$ W/m·K.

Forecast: Thin layer but very low $k$ . I expect a heat flux of order a million W/m².

Jump across the coating. $Δ T = 1200 - 950 = 250$ K. Why this step? Fourier's law uses the temperature drop over the layer, and 250 °C = 250 K.
Flux. $q = k \frac{Δ T}{L} = 2 \cdot \frac{250}{3 \times 1 0 ^{- 4}}$ . Why: heat flows down the gradient; a thin $L$ with small $k$ still passes some heat, and we quantify it. $= 2 \cdot 8.33 \times 1 0^{5} = 1.67 \times 1 0^{6}$ W/m².
Payoff vs alumina. Same geometry with $k = 30$ gives $q = 30 \cdot 250/3 \times 1 0^{- 4} = 2.5 \times 1 0^{7}$ W/m² — 15× more heat. Why: the entire reason zirconia coats blades is its low $k$ ; alumina would dump too much heat into the metal.

Verify: $1.67 \times 1 0^{6}$ W/m² (matches parent Ex 3); ratio $30/2 = 15$ confirms the 15× claim. Units: (W/m·K)(K)/(m) = W/m². ✓

Worked example (7) CELL G — Limiting behaviour: what if the material were infinitely stiff, or

$ν \to 0.5$ ? Forecast: Both $E$ and $(1 - ν)$ live in $R$ . Very large $E$ should drive $R \to 0$ ; $ν \to 0.5$ (incompressible) should also shrink $R$ .

Stiffness limit. $R = \frac{σ _{f} ( 1 - ν )}{E α}$ ; as $E \to \infty$ , $R \to 0$ . Why this step? An infinitely stiff bond turns even a microscopic restrained strain into infinite stress — the part shatters at the tiniest gradient. This is why very stiff ceramics (SiC, alumina) are shock-prone despite being "strong."
Poisson limit. As $ν \to 0.5$ , $(1 - ν) \to 0.5$ , halving $R$ relative to $ν \to 0$ . Why: higher $ν$ means the biaxial squeeze is fiercer, raising $σ_{t h}$ and lowering the survivable jump.
Numeric feel: take Si₃N₄ but pretend $E$ doubled to $620$ GPa. $R = \frac{800 \times 1 0 ^{6} \cdot 0.73}{620 \times 1 0 ^{9} \cdot 3.0 \times 1 0 ^{- 6}} \approx 314$ K — exactly half of 628 K. Why: $R \propto 1/ E$ , so doubling $E$ halves $R$ . This isolates stiffness as the villain.

Verify: $314$ K $= 628/2$ within rounding confirms the $1/ E$ scaling. ✓

Worked example (8) CELL H — Word problem: choose a ceramic for a rocket nozzle throat.

Requirement: survives repeated $\approx 400$ K quench cycles, sits in oxygen-rich combustion, must not peel, keep tight shape.

Forecast: "Oxygen-rich" hints at an oxide (won't burn). "Survive 400 K quench" needs $R ≳ 400$ K. Let me see which oxide qualifies.

Screen by oxidation. Oxidising atmosphere → favour oxides (Al₂O₃, ZrO₂); non-oxides SiC/Si₃N₄ can work via their protective SiO₂ skin but risk active oxidation if O₂ is low — here O₂ is plentiful, so passive skin is fine too. Keep all four as candidates. Why this step? Chemistry gates the shortlist before mechanics does.
Screen by shock $R \geq 400$ K (from Ex 4): only Si₃N₄ (628 K) clears 400 K comfortably; zirconia (210) and SiC (180) and alumina (77) fail the repeated-quench bar. Why: the cycling stress must stay below $σ_{f}$ every cycle.
Decision: pick Si₃N₄. Its high $σ_{f}$ and tiny $α$ give margin; its SiO₂ passive layer handles the O₂-rich gas up to ~1600 °C. Why: it is the only candidate meeting both chemistry and shock — the classic reason Si₃N₄ is chosen for the harshest thermal-cycling parts.

Verify: $R_{Si_{3} N_{4}} = 628$ K $> 400$ K margin factor $1.57$ ; all others $< 400$ K, so the choice is forced. ✓

Worked example (9) CELL I — Exam twist: fracture toughness of transformation-toughened zirconia.

The matrix toughness is $K_{I c}^{matrix} = 3.2 MPa\cdotm^{1/2}$ ; transformation toughening at the crack tip adds $Δ K = 6.3 MPa\cdotm^{1/2}$ (see Phase Transformations and Thermal Stress and Fracture Mechanics).

Forecast: The tetragonal→monoclinic $+ 4%$ volume jump slams the crack shut. I expect total $K_{I c}$ near the $8$ – $12$ range the parent quotes.

Add the contributions. $K_{I c}^{tough} = K_{I c}^{matrix} + Δ K = 3.2 + 6.3 = 9.5 MPa\cdotm^{1/2}$ . Why this step? The extra work the crack must do to force the local transformation adds to the intrinsic resistance — a superposition of the crack-tip shielding.
Compare to alumina ( $K_{I c} \approx 4$ ): ratio $9.5/4 \approx 2.4$ → zirconia is ~2.4× tougher. Why: explains why zirconia — despite modest $σ_{f}$ — is the "toughest ceramic": the crack tries to open a door, the crystal slams a bigger one shut.

Verify: $3.2 + 6.3 = 9.5$ lands inside the quoted $8$ – $12$ MPa·m $^{1/2}$ band; ratio to alumina $\approx 2.4$ . ✓

Recall Quick self-test

Why does zero temperature jump give zero thermal stress? ::: Stress needs a restrained gradient; if all atoms expand equally there is nothing to fight, so $σ_{t h} = E α \cdot 0/ (1 - ν) = 0$ . In $R = σ_{f} (1 - ν) / (E α)$ , which single property makes Si₃N₄ win? ::: Its tiny $α \approx 3 \times 1 0^{- 6} / K$ sitting in the denominator. A stiffer ceramic (bigger $E$ ) is more or less shock-resistant? ::: Less — $R \propto 1/ E$ , so stiffness lowers the survivable jump.

Related deep material: Ceramic Matrix Composites (CMCs), Nickel Superalloys, Crystal Bonding — Ionic vs Covalent.