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

Before we start, here is the toolbox — every symbol you will meet, defined once in plain words.

Definition Symbol dictionary (read this first)

$σ_{t h}$ — thermal stress: the internal push/pull (in pascals, Pa) that builds up inside a part when one part wants to expand more than another. $1 MPa = 1 0^{6} Pa$ , $1 GPa = 1 0^{9} Pa$ .
$σ_{f}$ — fracture strength: the stress at which the material cracks (we compare it against tensile stress — see the sign convention below).
$E$ — Young's modulus (stiffness): how hard it is to stretch the material. Big $E$ = very stiff.
$α$ — linear thermal expansion coefficient: fractional length change per degree, units 1/K.
$ν$ (nu) — Poisson's ratio: a dimensionless number (~0.2–0.3) describing sideways squeeze when you stretch; it appears as $1 - ν$ when a hot surface is held flat (biaxial constraint).
$Δ T$ — a temperature difference (kelvin, K). A change of 1 K equals a change of 1 °C.
$k$ — thermal conductivity (W/m·K): how easily heat flows through the material. Small $k$ = good insulator.
$q$ — heat flux (W/m²): heat power crossing one square metre.
$L$ , $L_{0}$ — a thickness or an original length (metres).
$R = Δ T_{ma x}$ — the thermal-shock figure of merit: the biggest sudden temperature jump the part survives.
$K_{I c}$ — fracture toughness (MPa·m $^{1/2}$ ): resistance to a crack running. Bigger = tougher.

Definition Sign convention for stress (fix this in your mind now)

We call stress positive when it is tensile (pulling atoms apart, opening cracks) and negative when it is compressive (pushing atoms together, closing cracks). This matters because ceramics fail in tension, not compression — a tensile stress is dangerous, a compressive one is usually harmless. In L2–L3 we quote only magnitudes of $σ_{t h}$ , but the physical sign is stated with each case below. In L5 you will have to declare the sign yourself, so learn to read it: a surface heated faster than its surroundings ends up in compression (it is pushed by the cold, restrained material around it); a surface cooled faster ends up in tension (it wants to shrink but is held stretched). Because ceramics are weak in tension, the deadly thermal shock is a sudden cooling (quench), not a sudden heating.

Look at the map below step by step: the orange arrows on the left are the hot face trying to grow; the teal arrows are the cold interior pushing back; the purple zig-zag is the crack that appears only once the built-up stress beats $σ_{f}$ . Read the caption's sign note — this exact picture (a heated surface held flat) is what puts the $1/ (1 - ν)$ into every stress formula on this page.

Figure — Ceramics — alumina, zirconia, silicon carbide, silicon nitride; properties at high T

Level 1 — Recognition

L1.1 Name the property

Problem. For each aerospace job, name the single property that makes the chosen ceramic right: (a) a thermal barrier coating on a turbine blade needs a low ___ ; (b) a precision bearing that must not crack on sudden heating needs a high ___ ; (c) an oxide chosen because it cannot burn further in hot air is already ___ .

Recall Solution

(a) low ==thermal conductivity $k$ == — a thin insulating skin only works if heat struggles to cross it ( $q = k Δ T / L$ , small $k$ → small $q$ ). Zirconia's $k \approx 2$ W/m·K wins. (b) high ==thermal-shock resistance $R$ == — set by $R = σ_{f} (1 - ν) / (E α)$ ; silicon nitride's tiny $α$ makes $R$ large. (c) already oxidised — oxides like $A l_{2} O_{3}$ and $Z r O_{2}$ have already combined with oxygen, so hot oxygen-rich air cannot attack them further.

L1.2 True or false

Problem. True or false: "A ceramic with a very high melting point automatically resists thermal shock."

Recall Solution

False. Thermal shock depends on $R = σ_{f} (1 - ν) / (E α)$ — strength, stiffness, and expansion — not on melting point. Graphite melts (sublimes) extremely high and shrugs off shock; dense alumina also has a high melting point but cracks easily under shock. Melting (bonds giving way) and cracking (stress running a crack) are two different failures with two different formulas.

Level 2 — Application

L2.1 Thermal stress in alumina

Problem. An alumina part sees a temperature jump $Δ T = 150$ K. Data: $E = 380$ GPa, $α = 8 \times 1 0^{- 6}$ /K, $ν = 0.22$ . Find the induced thermal stress $σ_{t h}$ .

Recall Solution

What/why: the hot surface wants to grow by strain $α Δ T$ , but the cold body holds it back, converting that would-be strain into stress via $σ_{t h} = E α Δ T / (1 - ν)$ . Trace this on figure s01: the orange "wants to expand" arrows are the strain $α Δ T$ ; the teal "pushes back" arrows are the constraint that turns it into stress. Build the numerator step by step: $380 \times 1 0^{9} \times 8 \times 1 0^{- 6} = 3.04 \times 1 0^{6}$ ; times $150 = 4.56 \times 1 0^{8}$ Pa. Then divide by $1 - 0.22 = 0.78$ : $σ_{t h} = \frac{( 380 \times 1 0 ^{9} ) ( 8 \times 1 0 ^{- 6} ) ( 150 )}{1 - 0.22} = \frac{4.56 \times 1 0 ^{8}}{0.78} = 5.85 \times 1 0^{8} Pa \approx 585 MPa .$ Sign: this is the heated face, so it is held in compression (magnitude 585 MPa). Ceramics are strong in compression, so a pure heating shock is the less dangerous case; the same 585 MPa on a cooled (quenched) face would be tensile and far more likely to crack.

L2.2 Growth of a silicon nitride rotor

Problem. A silicon nitride shaft is $L_{0} = 80$ mm long. Heated from 20 °C to 1020 °C, how much longer does it get? Use $α = 3.0 \times 1 0^{- 6}$ /K.

Recall Solution

What/why: with a (nearly) constant $α$ , expansion is simply $Δ L = α L_{0} Δ T$ — integrating a constant stretch-rate over the temperature rise. $Δ T = 1020 - 20 = 1000$ K. $L_{0} = 0.080$ m. $Δ L = (3.0 \times 1 0^{- 6}) (0.080) (1000) = 2.4 \times 1 0^{- 4} m = 0.24 mm .$ A quarter of a millimetre over 8 cm at white heat — this tiny growth is why $S i_{3} N_{4}$ holds tolerances in hot rotating parts.

L2.3 Heat blocked by a TBC

Problem. A zirconia thermal barrier coating is $L = 0.25$ mm thick. Its outer face is at 1250 °C, the metal side at 1000 °C, and $k = 2$ W/m·K. Find the heat flux $q$ .

Recall Solution

What/why: steady heat crossing a slab is $q = k Δ T / L$ — thicker or less conductive means less heat leaks through. $Δ T = 1250 - 1000 = 250$ K, $L = 2.5 \times 1 0^{- 4}$ m. $q = 2 \cdot \frac{250}{2.5 \times 1 0 ^{- 4}} = 2 \cdot 1 0^{6} = 2.0 \times 1 0^{6} W/m^{2} .$ When this simple model breaks down: $q = k Δ T / L$ assumes steady state (temperatures have settled) and a constant $k$ . During the first milliseconds of a shock the heat is transient (you would need the heat-diffusion equation, not this algebra), and real ceramic $k$ falls as temperature rises — so at true engine temperatures the effective $k$ , and thus $q$ , is somewhat lower than a room-temperature $k$ predicts.

Level 3 — Analysis

L3.1 Does alumina survive?

Problem. Using the L2.1 alumina data plus $σ_{f} = 300$ MPa, decide whether the $Δ T = 150$ K jump cracks the part, and find the maximum safe jump $R$ .

Recall Solution

From L2.1, $σ_{t h} \approx 585$ MPa (in magnitude). Compare to $σ_{f} = 300$ MPa: since $585 > 300$ , the stress exceeds strength, so it cracks — provided the stress is tensile, i.e. this is the dangerous quench (sudden-cooling) case. On the heating side the same magnitude is compressive and far safer; the design limit $R$ is written for the worst (tensile) case. The safe limit is where the tensile $σ_{t h} = σ_{f}$ , i.e. $R = \frac{σ _{f} ( 1 - ν )}{E α}$ : $R = \frac{( 300 \times 1 0 ^{6} ) ( 0.78 )}{( 380 \times 1 0 ^{9} ) ( 8 \times 1 0 ^{- 6} )} = \frac{2.34 \times 1 0 ^{8}}{3.04 \times 1 0 ^{6}} \approx 77 K .$ Alumina tolerates only ~77 K sudden jumps — never use it for thermal-shock duty.

L3.2 Which ceramic wins the shock contest?

Problem. Rank alumina and silicon nitride by thermal-shock resistance $R$ . Take $σ_{f} = 300$ MPa for both to isolate the effect of expansion and stiffness. Alumina: $E = 380$ GPa, $α = 8 \times 1 0^{- 6}$ /K, $ν = 0.22$ . Silicon nitride: $E = 300$ GPa, $α = 3.0 \times 1 0^{- 6}$ /K, $ν = 0.27$ .

Recall Solution

What/why: $R$ combines strength, stiffness, and expansion; comparing at equal $σ_{f}$ shows how $E$ and $α$ alone shift the answer. Alumina: $R_{A l} = \frac{300 \times 1 0 ^{6} \cdot 0.78}{380 \times 1 0 ^{9} \cdot 8 \times 1 0 ^{- 6}} = 77 K$ (as above). Silicon nitride: $R_{S i_{3} N_{4}} = \frac{300 \times 1 0 ^{6} \cdot ( 1 - 0.27 )}{300 \times 1 0 ^{9} \cdot 3.0 \times 1 0 ^{- 6}} = \frac{300 \times 1 0 ^{6} \cdot 0.73}{9.0 \times 1 0 ^{5}} = \frac{2.19 \times 1 0 ^{8}}{9.0 \times 1 0 ^{5}} \approx 243 K$ . Silicon nitride survives ~3× the temperature jump. The driver is its small $α$ (and lower $E$ ) in the denominator — less expansion means less trapped stress.

Now read figure s02 as the proof of that last sentence. Both bars are computed from the same $σ_{f} = 300$ MPa (the dashed note on the chart), so their labelled heights — $77$ K for alumina, $243$ K for silicon nitride, printed on each bar against the K-scale on the vertical axis — differ only because of the $E α$ denominator. The purple arrow marks the exact ratio you just computed: the taller bar is $243/77 \approx 3.15$ times the shorter one. That vertical gap is the shock advantage of a low- $α$ ceramic, drawn to scale.

Level 4 — Synthesis

L4.1 Design a TBC thickness

Problem. A cooled superalloy blade can tolerate a metal-side temperature of at most 1000 °C. Hot gas heats the coating's outer face to 1300 °C. The cooling system can pull away at most $q = 1.5 \times 1 0^{6}$ W/m² of heat. Using zirconia ( $k = 2$ W/m·K), find the minimum coating thickness $L$ that keeps the metal at or below 1000 °C.

Recall Solution

What/why: the coating must drop $Δ T = 1300 - 1000 = 300$ K while passing no more heat than the cooling can remove. Rearrange $q = k Δ T / L$ for $L$ : $L = \frac{k Δ T}{q} = \frac{2 \cdot 300}{1.5 \times 1 0 ^{6}} = \frac{600}{1.5 \times 1 0 ^{6}} = 4.0 \times 1 0^{- 4} m = 0.40 mm .$ Check the logic: thicker than 0.40 mm → the metal stays cooler (bigger $Δ T$ drop) but adds weight and stress. Exactly 0.40 mm is the minimum that meets the 1000 °C limit at the available cooling. Why zirconia and not alumina? With alumina ( $k = 30$ ) the same job would need $L = 30 \cdot 300/1.5 \times 1 0^{6} = 6.0 \times 1 0^{- 3}$ m $= 6$ mm — an absurd, heavy, easily-spalled slab. Low $k$ is the whole reason.

L4.2 Transformation toughening boost

Problem. Partially stabilised zirconia's matrix has $K_{I c} (matrix) = 4.0$ MPa·m $^{1/2}$ . Transformation toughening at the crack tip adds $Δ K_{t r an s f or ma t i o n} = 6.0$ MPa·m $^{1/2}$ . (a) Find the total $K_{I c}$ . (b) By what factor is it tougher than the un-toughened matrix?

Recall Solution

What/why: the parent note's rule $K_{I c} (toughened) = K_{I c} (matrix) + Δ K_{t r an s f or ma t i o n}$ — the crack must do extra work because the 4% volume expansion at its tip clamps it shut. (a) $K_{I c} = 4.0 + 6.0 = 10.0 MPa\cdotm^{1/2}$ . (b) Factor $= 10.0/4.0 = 2.5 \times$ tougher. This is why zirconia ( $\approx 8$ – $12$ ) is several times tougher than alumina ( $\approx 4$ ): the crack tries to open the door, the crystal slams a bigger door in its face.

Level 5 — Mastery

L5.1 Derive the two-material stress-match condition

Problem. A zirconia coating (subscript c) sits on a superalloy blade (subscript m). On cooling by $Δ T$ , each would shrink by its own $α Δ T$ . Because they are bonded, the mismatch in shrinkage strain, $Δ ε = (α_{m} - α_{c}) Δ T$ , is forced into the coating as stress. (a) Write the mismatch stress in the coating, $σ_{c} = \frac{E _{c} Δ ε}{1 - ν _{c}}$ , and justify why the biaxial factor $1/ (1 - ν_{c})$ appears. (b) With $E_{c} = 200$ GPa, $ν_{c} = 0.25$ , $α_{c} = 10 \times 1 0^{- 6}$ /K, $α_{m} = 14 \times 1 0^{- 6}$ /K, and cooling $Δ T = 1000$ K, compute $σ_{c}$ and state whether it is tension or compression. (c) Explain why matching $α_{c}$ to $α_{m}$ is the design goal.

Recall Solution

(a) The forced strain is the difference in free shrinkages, $Δ ε = (α_{m} - α_{c}) Δ T$ ; converted to biaxial stress by Hooke's law with the $1/ (1 - ν_{c})$ constraint factor: $σ_{c} = \frac{E _{c} ( α _{m} - α _{c} ) Δ T}{1 - ν _{c}} .$ Why the $1/ (1 - ν_{c})$ appears (re-derived here): a coating bonded to a blade is a thin sheet clamped in its own plane in two perpendicular directions at once, not stretched along a single wire. Squeeze it in direction $x$ and Poisson's ratio $ν_{c}$ makes it want to bulge in direction $y$ ; but it is clamped in $y$ too, so that bulge is also resisted — and vice-versa. Each direction stiffens the other. Working the two-direction Hooke's law $ε_{x} = \frac{1}{E _{c}} (σ_{x} - ν_{c} σ_{y})$ with the symmetric equal-biaxial case $σ_{x} = σ_{y} = σ_{c}$ and forced strain $ε_{x} = Δ ε$ gives $Δ ε = \frac{σ _{c}}{E _{c}} (1 - ν_{c})$ , i.e. $σ_{c} = E_{c} Δ ε / (1 - ν_{c})$ . This is the exact same reason the single-material thermal stress carried $1/ (1 - ν)$ — a heated surface and a bonded coating are both in-plane biaxially constrained sheets, so they share the factor. (b) $α_{m} - α_{c} = (14 - 10) \times 1 0^{- 6} = 4.0 \times 1 0^{- 6}$ /K. Numerator: $E_{c} \cdot Δ ε = 200 \times 1 0^{9} \times 4.0 \times 1 0^{- 6} \times 1000 = 8.0 \times 1 0^{8}$ Pa. Divide by $1 - 0.25 = 0.75$ : $σ_{c} = \frac{8.0 \times 1 0 ^{8}}{0.75} = 1.067 \times 1 0^{9} Pa = 1067 MPa$ . Sign (tension or compression?): the metal shrinks more than the coating ( $α_{m} > α_{c}$ ), so on cooling the metal contracts around the coating and squeezes it inward — the coating ends in compression (magnitude 1067 MPa). By our convention that is a negative stress. This is actually favourable for a brittle ceramic, since ceramics are strong in compression; a mismatch of the opposite sign ( $α_{c} > α_{m}$ ) would put the coating in dangerous tension and crack it. Either way a 1067 MPa mismatch is large enough to risk spalling at the interface over many cycles. (c) If $α_{c} = α_{m}$ , then $Δ ε = 0$ and $σ_{c} = 0$ — no mismatch stress at all, so the coating never peels. Real TBCs choose zirconia partly because its $α$ is close to the superalloy's, keeping $σ_{c}$ tolerable across thousands of heat/cool cycles.

Intuition When does the

$1/ (1 - ν)$ factor stop being valid? The factor is exact only for a thin, flat, equal-biaxially constrained sheet (heated surface layer or a thin coating firmly bonded to a much thicker, stiffer substrate). Its limits:

As $ν \to 0.5$ (incompressible, rubber-like — not ceramics, which sit near 0.2–0.3) the factor $1/ (1 - ν) \to 2$ and blows up beyond; the thin-plate assumption also fails, so the formula must be replaced by the full 3-D elasticity solution.
For a thick coating (thickness comparable to the part), the substrate no longer fully clamps it, the stress varies through the thickness, and this single-number result over-estimates the surface stress.
It assumes fully bonded, purely elastic behaviour; once creep, cracking, or interface sliding begins at high $T$ , the stored stress relaxes and the simple $E Δ ε / (1 - ν)$ over-predicts.

L5.2 Prove silicon nitride's shock edge is anharmonicity

Problem. The parent note derived $α = \frac{3 g k _{B}}{4 c ^{2} x _{0}}$ , where $c$ is bond stiffness and $g$ is the anharmonicity (asymmetry) of the bond well. Two ceramics have the same $x_{0}$ and $k_{B}$ . Ceramic A has $(c, g) = (c_{0}, g_{0})$ ; ceramic B is stiffer and more symmetric: $(c, g) = (2 c_{0}, 0.5 g_{0})$ . (a) Find the ratio $α_{B} / α_{A}$ . (b) If both start at equal $σ_{f}$ , $E$ , $ν$ , by what factor is B's shock resistance $R$ larger? Interpret.

Recall Solution

What/why: only the combination $g / c^{2}$ carries the difference between the two ceramics, because $k_{B}$ and $x_{0}$ are identical and cancel. So we compare $α \propto g / c^{2}$ directly. (a) Form the ratio, letting the shared $3 k_{B} / (4 x_{0})$ cancel top and bottom: $\frac{α _{B}}{α _{A}} = \frac{g _{B} / c _{B}^{2}}{g _{A} / c _{A}^{2}} = \frac{( 0.5 g _{0} ) / ( 2 c _{0} ) ^{2}}{g _{0} / c _{0}^{2}} = \frac{0.5 g _{0} / ( 4 c _{0}^{2} )}{g _{0} / c _{0}^{2}} = \frac{0.5}{4} = 0.125 .$ So ceramic B expands only 1/8 as much as A. (b) With $σ_{f}, E, ν$ equal, $R = \frac{σ _{f} ( 1 - ν )}{E α} \propto \frac{1}{α}$ , so $\frac{R _{B}}{R _{A}} = \frac{α _{A}}{α _{B}} = \frac{1}{0.125} = 8 \times .$ Interpretation: deeper, stiffer, more symmetric bond wells (large $c$ , small $g$ ) give tiny $α$ — and because $R \propto 1/ α$ , that tiny expansion translates into a huge jump in thermal-shock survival. This is exactly why the low- $α$ covalent ceramics like silicon nitride shrug off shocks that shatter oxides.

See also: Thermal Stress and Fracture Mechanics, Thermal Barrier Coatings on Superalloys, Nickel Superalloys, Ceramic Matrix Composites (CMCs), Phase Transformations, Crystal Bonding — Ionic vs Covalent.