Heat treatment — annealing, normalising, quenching, tempering; precipitation hardening: Visual walkthrough

Intuition What we are actually building

The parent note dropped a small but powerful formula for precipitation-hardened alloys: $Δ τ = \frac{G b}{L}$ Here $Δ τ$ (read "delta-tau") is the extra shear stress the precipitates add — the strengthening increment on top of the bare crystal. It is exactly the critical push we will derive and label $τ_{crit}$ in Steps 4–6; the two names are the same quantity, and we will say so explicitly the moment we reach it. This formula explains the single most important fact of age hardening — that hardness rises, hits a peak, then falls again if you age too long. This page earns that formula from nothing. We will draw a moving line, hang some obstacles in its path, watch it bend, and count how hard it is to push. By the end the "hump" shape of the hardness curve will be obvious from a picture, not a claim.

Everything here rests on one prerequisite idea we build first: what a dislocation is and why moving it is what we call "deforming a metal" — see Dislocations and Plastic Deformation.

Step 1 — What "strength" secretly means: stopping one moving line

WHAT. Before any formula, let's agree what we are measuring. When you bend a paperclip, whole sheets of atoms slide over each other. But they do not slide all at once — that would take enormous force. Instead a tiny defect called a dislocation — a single half-plane of atoms that doesn't quite line up — glides across the crystal like a ruck moving under a carpet. Move the ruck all the way across and the carpet has shifted by one step, with almost no effort.

WHY this matters. "Making a metal stronger" therefore means exactly one thing: making that line harder to push. Every strengthening trick in the parent note — grain boundaries (Hall-Petch Strengthening), trapped carbon (quenching), precipitates (ageing) — is just a different way to block this line. So our whole derivation is about the force on one line.

PICTURE. The carpet-ruck. The blue line is the dislocation; pushing it across the crystal by a shear stress $τ$ (a sideways force per area) shifts the top of the crystal by one atomic step $b$ .

Figure — Heat treatment — annealing, normalising, quenching, tempering; precipitation hardening

Figure s01 — Look at the magenta half-plane: it is the whole dislocation. Follow the orange arrow (the push $τ$ ) and watch the violet arrow at the bottom show the crystal shifting by one atomic step $b$ . The point to absorb: only this one little line moves, not the whole crystal at once.

Step 2 — Drop obstacles in the path: the precipitates

WHAT. Now recall from the parent note what ageing does: it grows tiny hard particles — precipitates (GP zones, θ″ in Al–Cu) — scattered through the metal. On our slip plane these appear as a field of dots the moving line must get past.

WHY. A bare crystal offers almost no resistance (the line just glides). The precipitates are the obstacles that create the resistance we call hardness. If the line cannot pass them, the metal does not deform — it is strong. So the entire question of age-hardening reduces to: how much extra push does it take to force the line past a row of dots?

PICTURE. Same slip plane, now seeded with orange precipitate dots spaced a distance $L$ apart. The blue line is pinned where it touches them.

Figure s02 — Look at the violet double-arrow labelled $L$ : that spacing between two orange dots is the single quantity ageing controls. The magenta line lies across the dots; picture it having to squeeze past them. Fine dots close together = small $L$ ; that is what we are chasing.

Step 3 — The line bows: a string held at two nails

WHAT. Push the line with stress $τ$ but keep two neighbouring precipitates pinning it. The middle of the line is free, so it bows out into a curved arc between the two pinned points — exactly like a guitar string plucked between two frets, or a clothesline sagging between two poles.

WHY a curve and not a straight bend? A dislocation resists being lengthened. Stretching it costs energy, so it behaves like an elastic string under tension. Under a uniform sideways push a string of constant tension bows into a circular arc — the shape that balances the push against the tension everywhere. That circular arc is the geometric heart of the whole derivation.

PICTURE. Two pinned points a distance $L$ apart; the line bulges between them into a circular arc of radius $R$ . The push $τ b$ acts outward; the line tension $T$ pulls back along the line at each end.

Figure s03 — Look at the violet dashed line marked $R$ : it runs from the arc's centre out to the bowed magenta line and is the radius of curvature. Notice the two navy arrows labelled $T$ at the pinned ends pulling back along the line, and the orange arrow $τ b$ pushing the belly of the arc outward. The bigger the push, the tighter the arc, the smaller $R$ .

Step 4 — Balance the forces on the arc (this is the real derivation)

WHAT. We now write down the tug-of-war on a small piece of the bowed line and solve for how hard we must push. Take a tiny segment of arc that subtends a small angle $2 d θ$ at the centre of the circle.

WHY a force balance? The line sits still (in equilibrium) at each push level until the push wins. At the balance point, the outward force from the applied stress exactly equals the inward pull of the two ends of the tension trying to straighten the segment. Setting them equal tells us the push needed.

PICTURE. The zoomed-in arc segment. Outward: the pressure-like force $τ b$ acting on the segment's length. Inward: the two tension vectors $T$ at each end, each tilted by angle $d θ$ , whose sideways components add up.

Figure s04 — Look at the violet wedge angle $2 d θ$ at the centre: it opens up the little piece of magenta arc we are analysing. The orange arrow at the bottom is the outward push; the two navy arrows at the ends are the tension pulling inward. The whole derivation is just: these must balance.

Figure s05 — Look how the red arc, the straight chord, and the green vertical drop $sin (d θ)$ nearly overlap for the small wedge. That visual overlap is the approximation $sin (d θ) \approx d θ$ ; nothing more mysterious than "a tiny bit of a circle looks straight."

Now the algebra, term by term.

The outward force on a segment of arc length $ℓ = R (2 d θ)$ is the push per length $τ b$ times that length: $F_{out} = push/area τ step size b arc length ℓ R (2 d θ)$

$τ b$ is force per unit length on the line (stress $\times$ Burgers vector).
$R (2 d θ) = ℓ$ is just the length of that little piece of arc (radius $\times$ angle).

The inward pull: each end has tension $T$ ; the two ends are each tilted by $d θ$ from the segment's midline, so each contributes an inward component $T sin (d θ) \approx T d θ$ by the small-angle rule above. Two ends: $F_{in} = 2 T sin (d θ) \approx 2 T d θ$

Set them equal ( $F_{out} = F_{in}$ ): $τ b R (2 d θ) = 2 T d θ$

The $2 d θ$ cancels on both sides — why we can cancel it: it appears identically in both forces, so the answer doesn't depend on how big we chose the little segment. Left with the clean result: $τ b R = T ⟹ τ = \frac{T}{b R}$

Read it: the push $τ$ needed to hold a given curvature grows as the arc gets tighter (small $R$ ). To bend the line more sharply you must push harder. Hold that thought.

Step 5 — The tightest possible bow sets the strength

WHAT. As we keep raising $τ$ , the line bows more and more — $R$ shrinks. What is the smallest $R$ the line can reach while still stuck between two precipitates spaced $L$ apart?

WHY there is a limit. Look at the geometry: the arc must still touch both pinning points a distance $L$ apart. The most sharply curved circle that fits between two points a distance $L$ apart is the semicircle whose diameter is $L$ — i.e. radius $R = L /2$ . Push any harder and the line bulges past the semicircle, the two halves meet behind the obstacle, pinch off, and the line escapes to sweep onward. So the hardest moment is at $R = L /2$ : that is the critical stress the metal can resist.

PICTURE. Three snapshots of the same line at rising push: gentle bow (large $R$ ), semicircle ( $R = L /2$ , the critical moment, drawn in magenta), and break-away (line pinches off past the dots).

Figure s06 — Scan the three panels left to right. In the first the violet arc is gentle (large $R$ ). In the middle the magenta arc is a perfect semicircle — its radius is exactly $L /2$ , the tightest bow that still fits between the two orange dots. In the third the line has pinched off behind the dots and escaped. The middle panel is the moment of maximum resistance.

Substitute the critical radius $R = L /2$ into $τ = T / (b R)$ : $τ_{crit} = \frac{T}{b ( L /2 )} = \frac{2 T}{b L}$

$L /2$ is the semicircle radius — the tightest arc that fits between the two pins.
The result: the extra push needed to break past the precipitates is $\frac{2 T}{b L}$ .

Step 6 — Put a value to the tension: reach the parent's formula

WHAT. One symbol is still "unearned": the line tension $T$ . We replace it with things we can measure about the crystal.

WHY. A dislocation's stored energy per length is, to a good approximation, $T \approx \frac{1}{2} G b^{2}$ — where $G$ is the shear modulus (how stiff the metal is against shear, in Pa) and $b$ is our Burgers vector again. The $b^{2}$ appears because the elastic strain field around the line scales with $b$ , and its energy with $b$ squared. This is the standard line-tension estimate.

Common mistake What the "

$\approx$ " is quietly dropping — and how big it really is The honest, full line tension is $T = \frac{G b ^{2}}{4 π ( 1 - ν )} ln (\frac{R _{out}}{r _{0}}),$ where $ν$ (nu) is Poisson's ratio (how much a material bulges sideways when squeezed, $\approx 0.3$ for metals), $r_{0}$ is the atomic core radius where the elastic picture breaks down, and $R_{out}$ is the outer cut-off of the strain field. Note the new subscript: this $R_{out}$ is a completely different radius from the $R$ of Steps 3–5 (which was the radius of curvature of the bow). To avoid overloading one letter we call this elastic cut-off $R_{out}$ and keep plain $R$ strictly for the bowing radius.

How big is the dropped prefactor, quantitatively? With $ν = 0.3$ , the front factor is $\frac{1}{4 π ( 1 - ν )} = \frac{1}{4 π ( 0.7 )} \approx 0.114$ . The logarithm $ln (R_{out} / r_{0})$ runs over typical values $R_{out} / r_{0} \approx 1 0^{3}$ – $1 0^{4}$ , giving $ln \approx 6.9$ – $9.2$ . Multiplying: $0.114 \times (6.9 to 9.2) \approx 0.79$ to $1.05$ — i.e. the true prefactor sits near $0.8$ – $1.0$ , not exactly the $\frac{1}{2}$ we use. Across metals it varies only mildly: $ν$ ranges about $0.27$ – $0.35$ (changes the front factor by $\sim \pm 10$ ) and the log changes by well under a factor of $2$ even if the cut-off ratio shifts tenfold. Why it's safe here: every factor we drop is a slowly-varying constant of the material — it rescales $Δ τ$ by a fixed number of order one but leaves the one thing we care about, the $1/ L$ dependence, completely untouched. So our numeric answers below are good to a factor of order two in the constant, exact in their scaling.

PICTURE. The energy-in-the-strain-field idea: rings of distorted lattice around the line, denser (more energy) for a bigger step $b$ .

Figure s07 — Compare the two panels: the right one (bigger step $b$ ) has denser, more crowded strain rings than the left. More crowded rings = more stored elastic energy per length, i.e. larger tension $T$ . That is the visual meaning of $T \approx \frac{1}{2} G b^{2}$ : tension grows with the square of the step size.

Substitute $T = \frac{1}{2} G b^{2}$ into $Δ τ = τ_{crit} = 2 T / (b L)$ : $Δ τ = \frac{2 \cdot \frac{1}{2} G b ^{2}}{b L} = \frac{G b ^{2}}{b L} = \frac{G b}{L}$

$Δ τ = \frac{G b}{L}$

$G$ = shear modulus — the metal's intrinsic stiffness. Stiffer metal, harder to bow the line.
$b$ = Burgers vector — the atomic step size.
$L$ = precipitate spacing — the ONE thing ageing changes.

This is exactly the Orowan bowing stress the parent note asserted. We did not assume it — we watched a string bow between two nails and counted the force.

Step 7 — Why the spacing $L$ is tied to particle size (the geometry behind coarsening)

WHAT. The bowing formula rewards small $L$ , but ageing does not change $L$ directly — it grows the particles. We need the bridge: how does spacing $L$ depend on particle radius $r$ ? We derive it from one conserved quantity — the volume fraction $f$ of precipitate, which is roughly fixed once all the solute has come out of solution.

WHY volume fraction is the anchor. Ageing longer doesn't create more precipitate material (the alloy only has so much Cu); it just rearranges the same volume into fewer, bigger particles (Ostwald ripening — see Diffusion in Solids). So $f$ stays put while $r$ grows.

Derivation of $L$ vs $r$ . Put one particle of radius $r$ in each cubic cell of side $L$ (the spacing is, by definition, the cell size). The volume fraction is particle volume over cell volume: $f = \frac{\frac{4}{3} π r ^{3}}{L ^{3}} .$

$\frac{4}{3} π r^{3}$ = volume of one spherical precipitate of radius $r$ .
$L^{3}$ = volume of the little box that contains it (its share of the crystal).

Solve for $L$ : $L^{3} = \frac{4 π}{3} \frac{r ^{3}}{f} ⟹ L = (\frac{4 π}{3 f})^{1/3} r \propto \frac{r}{f ^{1/3}} .$ So at fixed $f$ , spacing grows in direct proportion to particle radius: bigger particles sit farther apart. (The exact constant depends on how you idealise the array; the key, robust result is $L \propto r$ at fixed $f$ .) Feeding this into $Δ τ_{bow} = G b / L$ : $Δ τ_{bow} = \frac{G b}{L} \propto \frac{G b f ^{1/3}}{r} \propto \frac{1}{r} .$ As particles coarsen ( $r$ grows), the bowing stress falls as $1/ r$ . That is the falling (over-aged) side of the story, now derived, not asserted.

Figure s08 — Both boxes hold the same orange volume of precipitate (same $f$ ). Left: many small particles, closely spaced (small $L$ ). Right: the same material gathered into a few big particles, widely spaced (large $L$ ). Watch the violet spacing arrow stretch as particles coarsen — that is $L \propto r$ , and it is why over-ageing softens.

Step 8 — The other way past an obstacle: cutting, and why it scales as $r$

WHAT. A dislocation has a second option besides bowing around a particle: if the particle is small and soft (coherent with the matrix), the line can slice straight through it, shearing the particle in two. We now estimate that cutting stress and show it rises with particle size — the opposite trend to bowing.

WHY it rises with $r$ (physical derivation). When the line cuts a particle it creates two fresh slices of new interface between particle and matrix. Creating interface costs energy: an energy per unit area $γ$ (the interfacial energy). The work the applied stress must supply equals the extra interface area created as the line sweeps through.

Derivation sketch.

A line of length equal to the spacing $L$ sweeps forward; each particle it meets has cross-section of order $π r^{2}$ , and cutting it makes new interface of area $\approx π r^{2} γ$ in energy.
The applied shear stress $τ$ does work $τ b$ per unit area swept. Balancing the work done against the interface energy created per particle encountered along the line gives $Δ τ_{cut} \approx \frac{γ r}{b L} .$
Now insert the spacing law from Step 7, $L \propto r / f^{1/3}$ : $Δ τ_{cut} \propto \frac{γ r}{b ( r / f ^{1/3} )} = \frac{γ f ^{1/3}}{b} (no r ?) .$ A sharper treatment tracks that the line only bends by the chord it cuts, which grows as $r$ , and keeps the full geometry; the standard result for coherent, shearable particles is $Δ τ_{cut} \propto r f^{1/2}$ The essential physics to remember: cutting gets harder as the particle grows (more material to shear, longer cut chord $\sim r$ ), whereas bowing gets easier ( $\propto 1/ r$ ). The two trends point in opposite directions.

WHY this guarantees a peak. Below some size the line prefers to cut (cheap when particles are tiny), and cutting resistance climbs as $r$ . Above that size it prefers to bow (cheap when particles are far apart), and bowing resistance falls as $1/ r$ . The metal always takes the easier route, so the strength it actually shows is the lower of the two curves at every size. A rising curve and a falling curve must cross at one particle size $r^{*}$ — and that crossing is the peak hardness. Age to $r^{*}$ and you sit on top; age past it and you slide down the bowing branch.

Figure s09 — This is the promised hump, drawn from the two derived branches. The orange dotted curve rising as $r$ is cutting; the magenta dashed curve falling as $1/ r$ is bowing. The thick violet curve is the lower of the two — the strength the metal actually shows — and the navy star at the crossing $r^{*}$ is peak hardness. Under-aged is the rising left side; over-aged is the falling right side.

The one-picture summary

Everything above, on one canvas: the pinned line bowing to a semicircle ( $R = L /2$ ), the force balance $τ b R = T$ , the substitution $T = \frac{1}{2} G b^{2}$ , the result $Δ τ = G b / L$ (with $Δ τ \equiv τ_{crit}$ ), the spacing law $L \propto r$ , and the cutting-vs-bowing crossing that bends it into the hardness hump.

Figure s10 — Read it left to right: the semicircle bow (radius $L /2$ ) feeds the force balance $τ b R = T$ ; substitute $R = L /2$ and $T = \frac{1}{2} G b^{2}$ to get $Δ τ = G b / L$ ; then the inset hump shows how $L \propto r$ plus cutting-vs-bowing produces a peak. One glance = the whole page.

Recall Feynman retelling — say it in plain words

Deforming a metal means dragging a little defect-line (a "ruck in the carpet") across the crystal. Ageing sprinkles tiny hard dots in its path, spaced a distance $L$ apart. The line can't pass them straight — it bows out between two dots like a clothesline between two poles. The line hates being stretched (it has tension $T$ ), so bowing costs force. Balance the push against the tension on a little piece of the arc and you get $τ b R = T$ : tighter bows need more push. The tightest bow that still fits between two dots is a semicircle, radius $L /2$ ; push past that and the line pinches off and escapes. That breaking push is the strengthening increment $Δ τ$ (same thing as $τ_{crit}$ ). Plug $R = L /2$ in, replace the tension with $\frac{1}{2} G b^{2}$ (a tidy stand-in for a messier constant that is really about $0.8$ to $1.0$ but doesn't change the $1/ L$ shape), and out pops $Δ τ = G b / L$ — closely spaced dots (small $L$ ) = strong. Now, ageing doesn't move $L$ directly; it grows the dots, and since the total dot volume is fixed, bigger dots sit farther apart ( $L \propto r$ ). So early on dots multiply and $L$ shrinks — strength climbs. But there's a second escape: when dots are tiny and soft the line just cuts through them, and that cost grows with dot size like $r$ . The metal always takes the easy route, so real strength follows the lower of the two curves — cutting on the left, bowing on the right — and they cross at one dot size $r^{*}$ . That crossing is the peak. Age past it and dots coarsen, $L$ grows, bowing gets easy, strength falls. Two opposite trends → a peak. That hump is peak-ageing, and now you can see exactly why it has to be there.

Recall Quick self-test

Why does the line bow into a circular arc and not some other shape? ::: Because a dislocation acts like a string of constant tension under a uniform sideways push; that balance is met everywhere only by a circular arc. What critical radius sets the breakaway stress, and why? ::: $R = L /2$ — the semicircle is the tightest arc that still fits between two pins spaced $L$ apart; push harder and the line pinches off. What is the relationship between $τ_{crit}$ and $Δ τ$ ? ::: They are the same quantity — the critical breakaway push equals the extra strength the precipitates add, since a bare crystal needs almost no push. In $Δ τ = G b / L$ , which symbol does ageing actually change? ::: $L$ , the precipitate spacing — indirectly, because ageing grows the particle radius $r$ and $L \propto r$ at fixed volume fraction. Why does over-ageing SOFTEN despite building more precipitate volume? ::: Coarsening (Ostwald ripening) merges particles into fewer, larger, more widely spaced ones — $L$ grows as $L \propto r$ , so $G b / L$ falls. What sets the exact peak hardness in the cutting-vs-bowing picture? ::: The crossover particle size $r^{*}$ where cutting strength ( $\propto r$ ) equals bowing strength ( $\propto 1/ r$ ); the metal follows the lower curve, whose maximum is that crossing.

Related: Heat treatment — annealing, normalising, quenching, tempering; precipitation hardening · Nickel Superalloys · TTT and CCT Diagrams · Iron-Carbon Phase Diagram