Carbon-carbon composites (RCC for nose cone - leading edges): Worked examples

Before any symbol appears, here is the toolbox we lean on (each earned in the parent note or built here):

$E$ — Young's modulus: how stiff a material is. Think "how hard you must push to stretch it 1 unit". Units of pressure, $Pa$ .
$α$ — coefficient of thermal expansion: how much a material grows per degree of heating (see Thermal stress and α (coefficient of expansion)). Tiny for carbon.
$Δ T$ — the temperature change, "final minus start". Positive = heating, negative = cooling.
$σ = E α Δ T$ — the thermal stress built up when a part is heated but not allowed to expand.

The scenario matrix

Every problem in this topic falls into one of these cells. The examples below are tagged with the cell they cover.

Cell	Case class	What's special about it	Example
A	Thermal stress, positive $Δ T$ (heating)	compressive stress, the re-entry case	Ex 1
B	Thermal stress, negative $Δ T$ (cooling)	sign flips → tensile stress	Ex 2
C	Zero / degenerate input ( $α = 0$ or $Δ T = 0$ )	limiting behaviour, stress vanishes	Ex 3
D	Comparison / ratio (carbon vs metal)	the "40×" selling point	Ex 4
E	Char-yield chain (pyrolysis + densification cycles)	diminishing-returns series	Ex 5
F	CVI stoichiometry (mass balance of a gas→solid)	moles, off-gas $H_{2}$	Ex 6
G	Real-world word problem (Columbia breach energy)	translate a scenario into oxidation chemistry	Ex 7
H	Exam twist / limiting case (what $Δ T$ makes carbon reach steel's stress?)	solve the formula backwards	Ex 8

The figure below lays these eight cells out as a colour-coded grid — read it as your map for the page: the two purple cells (A, E) and the coral cells (B, F) come in matched pairs (heating↔cooling, char-yield↔mass-balance), and the mint/butter cells are the degenerate, comparison, word-problem and exam-twist corners. Glance back at it whenever you finish an example to see which corner you have covered.

Figure — Carbon-carbon composites (RCC for nose cone - leading edges)

Cell A — Heating: compressive thermal stress

Worked example Example 1 (Cell A)

An RCC leading edge ( $E = 30 GPa$ , $α = 2 \times 1 0^{- 6} K^{- 1}$ ) is clamped in a rig so it cannot expand. Re-entry raises it by $Δ T = + 1400 K$ . Find the thermal stress and say whether it pushes or pulls.

Forecast: Guess first — is the stress squeezing (compressive) or stretching (tensile)? And will the number be closer to $50 MPa$ or $500 MPa$ ?

Step 1. Identify the free thermal strain the part wants: $ε_{t h} = α Δ T = 2 \times 1 0^{- 6} \times 1400 = 2.8 \times 1 0^{- 3}$ . Why this step? The material always tries to grow by $α Δ T$ first; the stress only exists because we forbid that growth.

Step 2. Because it's clamped, an equal-and-opposite mechanical strain appears: $ε_{m ec h} = - 2.8 \times 1 0^{- 3}$ (negative = compressed). Why this step? The rig physically shoves the material back to its old length — that shove is a compression.

Step 3. Apply Hooke's law $σ = E ε_{m ec h}$ : $σ = 30 \times 1 0^{9} \times (- 2.8 \times 1 0^{- 3}) = - 8.4 \times 1 0^{7} Pa = - 84 MPa .$ Why this step? $E$ converts a strain (a fraction) into a stress (a pressure). The minus sign is the physics telling us it's compression.

Verify: Units: $Pa \times (dimensionless) = Pa$ ✓. Magnitude $84 MPa$ is well below RCC's ~ $100$ – $200 MPa$ strength, so it survives — exactly why we chose carbon. Sign negative = compressive as forecast.

Cell B — Cooling: the sign flips to tensile

Worked example Example 2 (Cell B)

The same part, now cooling down after re-entry: $Δ T = - 1400 K$ , still clamped. Find the stress.

Forecast: If heating gave compression, what should cooling give? Guess the sign before reading on.

Step 1. Free thermal strain: $ε_{t h} = α Δ T = 2 \times 1 0^{- 6} \times (- 1400) = - 2.8 \times 1 0^{- 3}$ . Why this step? A negative $Δ T$ means the part wants to shrink; the strain it wants is negative.

Step 2. Clamped, so mechanical strain is $- ε_{t h} = + 2.8 \times 1 0^{- 3}$ (positive = stretched). Why this step? The rig won't let it shrink, so it holds the part stretched relative to where it wants to be — tension.

Step 3. $σ = E ε_{m ec h} = 30 \times 1 0^{9} \times (+ 2.8 \times 1 0^{- 3}) = + 8.4 \times 1 0^{7} Pa = + 84 MPa$ (tensile). Why this step? Same Hooke's-law conversion as Ex 1 — $E$ turns the mechanical strain into a stress — but now the mechanical strain is positive, so the sign comes out positive, telling us the part is being pulled apart (tension) rather than squeezed.

Verify: Same magnitude, opposite sign to Ex 1 ✓ — because $σ = E α Δ T$ is linear in $Δ T$ , flipping the sign of $Δ T$ flips the sign of $σ$ . Physically: cooling cracks are tensile cracks, which is why cooldown is watched as carefully as heat-up.

Cell C — Degenerate inputs: the stress vanishes

Worked example Example 3 (Cell C)

Two "what if" limits: (a) a hypothetical material with $α = 0$ heated by $Δ T = 1400 K$ ; (b) any material with $Δ T = 0$ . Find $σ$ in each.

Forecast: Should either of these have any stress at all? Guess yes/no for each.

Step 1 (case a). $σ = E \times 0 \times 1400 = 0 Pa$ . Why this step? If a material doesn't try to expand ( $α = 0$ ), clamping it changes nothing — no fight, no stress. This is the ideal re-entry material, and carbon's tiny $α$ is the closest real thing.

Step 2 (case b). $σ = E \times α \times 0 = 0 Pa$ . Why this step? No temperature change = no thermal driving force at all. A part sitting at constant temperature carries no thermal stress no matter how stiff.

Verify: Both give exactly $0$ ✓. The formula $σ = E α Δ T$ is a product — kill any factor and the whole thing dies. This is the sanity check that any thermal-stress answer must pass: no $Δ T$ or no expansion ⇒ no stress.

Cell D — Comparison: why carbon and not steel

Worked example Example 4 (Cell D)

Heat carbon ( $E = 30 GPa$ , $α = 2 \times 1 0^{- 6} K^{- 1}$ ) and steel ( $E = 200 GPa$ , $α = 12 \times 1 0^{- 6} K^{- 1}$ ) by the same $Δ T$ , both clamped. Find the ratio $σ_{s t ee l} / σ_{c a r b o n}$ — without plugging in $Δ T$ .

Forecast: Guess the ratio — is steel 10×, 40× or 100× worse?

Step 1. Write both stresses: $σ_{s t ee l} = E_{s} α_{s} Δ T$ , $σ_{c a r b o n} = E_{c} α_{c} Δ T$ . Why this step? Keeping them symbolic lets $Δ T$ cancel — the ratio is a material property, independent of temperature.

Step 2. Divide: $\frac{σ _{s t ee l}}{σ _{c a r b o n}} = \frac{E _{s} α _{s}}{E _{c} α _{c}} = \frac{200 \times 12}{30 \times 2} = \frac{2400}{60} = 40.$ Why this step? The product $E α$ is the "thermal-stress index" of a material; the ratio of indices is the whole story.

Verify: Re-derive both stresses at $Δ T = 1200 K$ on this page to check the ratio. Carbon: $σ_{c} = 30 \times 1 0^{9} \times 2 \times 1 0^{- 6} \times 1200 = 7.2 \times 1 0^{7} Pa = 72 MPa$ . Steel: $σ_{s} = 200 \times 1 0^{9} \times 12 \times 1 0^{- 6} \times 1200 = 2.88 \times 1 0^{9} Pa = 2880 MPa$ . Then $2880/72 = 40$ ✓, matching the symbolic answer. Steel builds 40× more stress for the same heating — it cracks, carbon shrugs.

Cell E — Char-yield chain (densification cycles)

Worked example Example 5 (Cell E)

A pyrolysed char is 30 % porous (i.e. 30 % of the target volume is empty). Each re-impregnation cycle fills roughly half of whatever porosity remains (this models the taper in the parent note). How much porosity is left after 4 cycles, and how many cycles to get below 5 % porosity?

Forecast: Guess — does it take 3, 5, or 10 cycles to drop below 5 %?

Step 1. Start porosity $p_{0} = 0.30$ . "Fill half of what remains" means after each cycle porosity is multiplied by $\frac{1}{2}$ : $p_{n} = 0.30 \times (\frac{1}{2})^{n} .$ Why this step? Each cycle only reaches into the remaining pores, and pores shrink each time — that's exactly a geometric (halving) decay, matching the "diminishing returns" the parent describes.

Step 2. After 4 cycles: $p_{4} = 0.30 \times (1/2)^{4} = 0.30 \times \frac{1}{16} = 0.01875 = 1.875%$ . Why this step? Plug $n = 4$ to answer the first question directly.

Step 3. Solve $0.30 \times (1/2)^{n} < 0.05$ for the smallest whole $n$ . Divide: $(1/2)^{n} < 0.1667$ , so $n > lo g_{2} (0.30/0.05) = lo g_{2} 6 \approx 2.585$ . Smallest whole number is $n = 3$ . Why this step? We invert the formula to find when a target is reached — a "solve backwards" move you'll reuse in Ex 8.

Verify: Check $n = 3$ : $0.30 \times (1/2)^{3} = 0.30 \times 0.125 = 0.0375 = 3.75% < 5%$ ✓, while $n = 2$ gives $7.5% > 5%$ ✓. So 3 cycles clear the bar, and 4 cycles leave $1.875%$ — real RCC uses ~4–6 cycles, consistent with this taper.

Cell F — CVI gas→solid mass balance

Worked example Example 6 (Cell F)

In the densification step, methane cracks: $CH_{4} (g) \to C (s) + 2 H_{2} (g)$ (see Chemical Vapour Infiltration / Deposition (CVI/CVD)). To deposit 60 g of carbon, how much $CH_{4}$ is consumed and how much $H_{2}$ vented? (Molar masses: $C = 12$ , $CH_{4} = 16$ , $H_{2} = 2 g mol^{- 1}$ .)

Forecast: Guess whether the $H_{2}$ off-gas mass is bigger or smaller than the carbon deposited.

Step 1. Moles of carbon: $n_{C} = \frac{60}{12} = 5 mol$ . Why this step? Stoichiometry works in moles, not grams — convert first.

Step 2. From the 1:1:2 ratio: $n_{CH_{4}} = 5 mol$ , $n_{H_{2}} = 2 \times 5 = 10 mol$ . Why this step? The equation's coefficients are a mole recipe; scale them by 5.

Step 3. Back to grams: $CH_{4} = 5 \times 16 = 80 g$ ; $H_{2} = 10 \times 2 = 20 g$ . Why this step? The answer is asked in grams.

Verify: Conservation of mass: in $= 80 g$ ; out $= 60 (C) + 20 (H_{2}) = 80 g$ ✓. The vented $H_{2}$ (20 g) is heavier than intuition suggests — and it's flammable, which is why CVI furnaces manage the off-gas carefully.

Cell G — Real-world word problem (the Columbia breach)

Worked example Example 7 (Cell G)

A breach in the SiC coating exposes bare carbon (see Space Shuttle Columbia disaster — materials case study). Suppose $30 g$ of that carbon burns completely in the oxygen-rich plasma: $C (s) + O_{2} (g) \to CO_{2} (g)$ , $Δ H = - 394 kJ mol^{- 1}$ . How much heat is released, and how much $O_{2}$ is consumed? ( $C = 12$ , $O_{2} = 32 g mol^{- 1}$ .)

Forecast: Will the energy released be nearer $500 kJ$ or $1000 kJ$ ? (For scale, $1000 kJ$ boils ~3 L of water.)

Step 1. Moles of carbon burnt: $n_{C} = \frac{30}{12} = 2.5 mol$ . Why this step? $Δ H$ is per mole, so we need moles.

Step 2. Heat released: $Q = n_{C} \times ∣Δ H ∣ = 2.5 \times 394 = 985 kJ$ . Why this step? Multiply the per-mole energy by the number of moles. The minus sign in $Δ H$ just means "given out"; the amount released is the magnitude.

Step 3. Oxygen consumed: 1:1 with carbon, so $n_{O_{2}} = 2.5 mol = 2.5 \times 32 = 80 g$ . Why this step? The stoichiometry ties $O_{2}$ directly to the carbon burnt.

Verify: $985 kJ$ from just 30 g of carbon is a huge self-feeding energy release — this is the runaway that once the coating fails, the burning carbon heats its neighbours and the damage cascades. Units: $mol \times kJ mol^{- 1} = kJ$ ✓.

Cell H — Exam twist: solve the formula backwards

Worked example Example 8 (Cell H)

"How much would you have to heat a carbon part before its thermal stress equals the $84 MPa$ that clamped steel reaches at just $Δ T = 35 K$ ?" Use carbon $E = 30 GPa$ , $α = 2 \times 1 0^{- 6} K^{- 1}$ .

Forecast: Guess — a few hundred K, or thousands of K?

Step 1. First confirm steel's target: $σ = 200 \times 1 0^{9} \times 12 \times 1 0^{- 6} \times 35 = 8.4 \times 1 0^{7} Pa = 84 MPa$ . Why this step? We need a concrete stress level to aim carbon at.

Step 2. Now invert $σ = E α Δ T$ for carbon, solving for $Δ T$ : $Δ T = \frac{σ}{E α} = \frac{8.4 \times 1 0 ^{7}}{30 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 6}} = \frac{8.4 \times 1 0 ^{7}}{6 \times 1 0 ^{4}} = 1400 K .$ Why this step? Rearranging the formula turns "find the stress" into "find the temperature that causes a stress" — the exam's favourite reversal.

Verify: Plug $Δ T = 1400 K$ back into Ex 1's forward calculation → $84 MPa$ ✓. So carbon needs a 1400 K rise to reach a stress steel hits at only 35 K — a 40× temperature headroom, the mirror image of the ratio in Ex 4.

Active recall

Recall Cover the answers

Heating a clamped part gives which sign of stress? ::: Compressive (negative).
Cooling a clamped part gives which sign? ::: Tensile (positive).
What makes thermal stress vanish entirely? ::: $α = 0$ or $Δ T = 0$ — any zero factor kills the product.
Ratio of steel to carbon thermal stress? ::: $\frac{E _{s} α _{s}}{E _{c} α _{c}} = 40$ .
To deposit 60 g C by CVI, how much $CH_{4}$ and $H_{2}$ ? ::: 80 g $CH_{4}$ in, 20 g $H_{2}$ out.
Heat from burning 30 g carbon ( $Δ H = - 394$ kJ/mol)? ::: $985 kJ$ .

Return to the parent: Carbon–carbon composites.