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

This is the central result behind Carbon–Carbon composites. The parent note stated $\sigma =E\alpha \Delta T$; here we build it from absolute zero, one picture per step.

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Step 1 — What "expansion" even means (a heated bar grows)

WHAT. Take a straight bar whose length before we touch it is $L_0$ (the little zero means "the starting value, before heating"). Warm it up. It gets a bit longer; call this new, after-heating length simply $L$ (no subscript = "final length, once it is hot").

WHY start here. Before we can talk about stress from being blocked, we must know how much the bar wants to move when nothing blocks it. That "wanting" is the raw ingredient of everything below.

PICTURE. Look at the figure: the cool bar (teal) of length $L_0$ sits above the warm bar (orange), now length $L$, which has stretched a touch to the right. The extra bit at the end is the growth $\Delta L=L-L_0$.

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Step 2 — Turning growth into a fair number: strain

WHAT. A $1\text{ mm}$ growth is huge for a matchstick but nothing for a bridge. To compare fairly we divide the growth by the original length. That fraction is called strain:

$\epsilon =\underset{\text{growth per unit of original length}}{\underbrace{\frac{\Delta L}{L_0}}}$

WHY this tool and not raw $\Delta L$. We want a number that describes the material, not the size of the sample. Dividing out $L_0$ does exactly that — $\epsilon$ is "fractional stretch", pure number, no units. A rubber band and a steel wire can now be compared on one scale.

PICTURE. The bar is drawn as $10$ equal boxes. Heating pushes each box to grow by the same fraction; strain is that single fraction, shown magnified on one box.

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Step 3 — How much does heat stretch it? The coefficient $\alpha$

WHAT. Experiment shows: double the temperature rise, double the stretch. So strain is proportional to $\Delta T$. The proportionality constant is the material's coefficient of thermal expansion, written $\alpha$:

${\epsilon }_{th}=\underset{\begin{array}{c}\text{stretch per}\\ \text{degree, per unit length}\end{array}}{\underbrace{\alpha }}\times \underset{\text{degrees of heating}}{\underbrace{\Delta T}}$

WHY a single number $\alpha$ works. The graph of "free stretch" against "temperature rise" is a straight line through the origin over the range we care about. A straight line through the origin is fully described by one number — its slope. That slope is $\alpha$. We use a slope, not a curve, because the physics is linear here.

PICTURE. Two straight lines from the origin: steep orange (steel, big $\alpha$) and shallow plum (carbon, tiny $\alpha$). Same $\Delta T$ on the x-axis gives very different free strain on the y-axis.

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Step 4 — The trap: what if the bar is NOT allowed to grow?

WHAT. On a real nose cone the hot skin is welded to cooler structure around it. It wants to grow by ${\epsilon }_{th}=\alpha \Delta T$, but the surroundings hold its ends fixed. Final length must stay $L_0$. So the net strain is zero.

WHY this is the whole game. Re-entry heating is not the danger by itself — it's heating plus constraint. A free bar that grows and shrinks feels no stress at all. Trouble only appears when growth is forbidden. So we ask: what has to happen inside the bar so it ends up its original length even though heat tried to stretch it?

PICTURE. Top: free bar grows past the wall (dotted outline = where it wanted to reach). Bottom: rigid walls (grey) shove it back to $L_0$. The red arrows are the walls pushing inward.

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Step 5 — Bookkeeping the two strains (they must cancel)

WHAT. Split the total strain into the heat part and the mechanical (force) part:

$\underset{=0\text{(ends fixed)}}{\underbrace{{\epsilon }_{total}}}=\underset{+\alpha \Delta T\text{(heat pushes out)}}{\underbrace{{\epsilon }_{th}}}+\underset{\text{(walls push in)}}{\underbrace{{\epsilon }_{mech}}}$

Set ${\epsilon }_{total}=0$ and solve for the mechanical strain:

${\epsilon }_{mech}=-{\epsilon }_{th}=-\alpha \Delta T$

WHY the minus sign. The heat strain is outward (positive, a stretch). To land back at zero, the mechanical strain must be exactly inward (negative, a squeeze). The minus sign literally says "compression" — the bar is being crushed along its length.

PICTURE. A number line of strain: green tick at $+\alpha \Delta T$ (thermal), red tick at $-\alpha \Delta T$ (mechanical), and they meet at $0$ (total). Equal and opposite, like a tug-of-war ending in a tie.

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Step 6 — Turning squeeze into stress: Hooke's law and $E$

WHAT. A squeeze inside a solid means the atoms are pushed closer than they'd like, and they push back. That internal push-per-area is stress, $\sigma$. For small strains, stress is proportional to strain — Hooke's law:

$\sigma =\underset{\text{stiffness}}{\underbrace{E}}\times \underset{\text{how much squeezed}}{\underbrace{{\epsilon }_{mech}}}$

WHY this tool. We need to convert a geometric fact (how much it's squeezed) into a mechanical danger (force per area that could crack it). Hooke's law is the exact bridge, and it is linear for the same reason Step 3 was: small deformations behave like stiff springs.

PICTURE. Two springs pushed in by the same amount. The stiff spring (orange, big $E$) pushes back hard; the soft spring (plum, small $E$) barely pushes. Same squeeze, very different force.

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Step 7 — Assemble the result

WHAT. Substitute Step 5's ${\epsilon }_{mech}=-\alpha \Delta T$ into Step 6's Hooke's law:

$\boxed{\sigma =E{\epsilon }_{mech}=E(-\alpha \Delta T)=-E\alpha \Delta T}$

Read each symbol where it stands:

$\sigma =\underset{\begin{array}{c}\text{stiffness}\\ \text{(Pa)}\end{array}}{\underbrace{E}}\cdot \underset{\begin{array}{c}\text{wants-to-grow}\\ ({\text{K}}^{-1})\end{array}}{\underbrace{\alpha }}\cdot \underset{\begin{array}{c}\text{heating}\\ (\text{K})\end{array}}{\underbrace{\Delta T}}$

The minus sign just labels it compressive; engineers quote the magnitude $E\alpha \Delta T$.

WHY this is the answer we wanted. It tells us before building anything how badly a constrained hot part is stressed, from three numbers you can look up. To survive, we want $\sigma$ small — and the only material knobs are $E$ and $\alpha$.

PICTURE. A "stress machine": three dials ($E$, $\alpha$, $\Delta T$) feeding one output gauge $\sigma$. Turning any dial up turns the output up.

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Step 8 — Carbon vs steel: the payoff (worked numbers)

WHAT. Same brutal heating $\Delta T=1200\text{K}$, both fully constrained.

${\sigma }_{\text{carbon}}=\underset{E}{\underbrace{30\times 10^9}}\cdot \underset{\alpha }{\underbrace{2\times 10^{-6}}}\cdot \underset{\Delta T}{\underbrace{1200}}=7.2\times 10^7\text{Pa}=72\text{MPa}$

${\sigma }_{\text{steel}}=\underset{E}{\underbrace{200\times 10^9}}\cdot \underset{\alpha }{\underbrace{12\times 10^{-6}}}\cdot \underset{\Delta T}{\underbrace{1200}}=2.88\times 10^9\text{Pa}=2880\text{MPa}$

WHY it matters. Steel is hit by 40× more stress — because it is both stiffer and far more eager to expand. That number is past steel's breaking point; carbon's $72\text{MPa}$ is easily survived. This single ratio is why the nose cone is carbon.

PICTURE. Two bars from the same $\Delta T$: carbon bar whole (teal, small stress arrow), steel bar cracked (orange, huge stress arrow). Bar heights show the 40× stress gap.

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Step 9 — Edge & degenerate cases (don't get ambushed)

WHAT & WHY. A formula you trust only in the "nice" case is a trap. Walk every boundary:

PICTURE. A sign map: horizontal axis $\Delta T$ (negative→positive), vertical axis $\sigma$, a straight line through the origin. Left half = tension (plum), right half = compression (orange), the origin = the safe zero. Carbon's shallow line vs steel's steep line both pass through the same origin.

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The one-picture summary

Everything above collapses into one flow: heat → wants to grow ($\alpha \Delta T$) → walls forbid it → equal squeeze → Hooke turns squeeze into stress ($\times E$) → $\sigma =E\alpha \Delta T$; make $\alpha$ tiny and the stress vanishes.

Recall Feynman retelling (cover and re-explain)

Heat a metal bar and it grows a tiny bit — that fractional growth is $\alpha$ times the temperature rise. Now imagine the bar is trapped between two walls so it can't grow. It still wants to, so the walls crush it back to size. That crushing is a real internal push — a stress — and its size is stiffness ($E$) times the squeeze ($\alpha \Delta T$), giving $\sigma =E\alpha \Delta T$. Steel is both stiff and eager to grow, so trapping it while it's hot builds up an enormous crushing stress that cracks it. Carbon barely wants to grow ($\alpha$ almost zero), so even when trapped and white-hot it feels almost no stress — which is exactly why the spaceship's nose is carbon and not metal. And two twists: if it cools while trapped, the push reverses into a pull that can tear it; and if it's free to grow, there's no stress at all no matter how hot — you need heat and a clamp for danger.

Recall One-line self-tests

Formula for constrained thermal stress? ::: $\sigma =E\alpha \Delta T$ Why the minus sign in ${\epsilon }_{mech}=-\alpha \Delta T$? ::: the wall's squeeze must cancel the outward thermal stretch → compression. Free (unclamped) hot bar — what stress? ::: zero; no constraint, no mechanical strain. Which material property should be small to survive re-entry? ::: $\alpha$, the expansion coefficient. Carbon-vs-steel stress ratio at the same $\Delta T$? ::: about $40\times$ (steel worse).

Related: Thermal stress and α (coefficient of expansion) · Graphite structure and sublimation · Silicon Carbide and oxidation-resistant ceramics · Space Shuttle Columbia disaster — materials case study