ARCHETYPE: Thermal Stress & Bimetallic Strips

Heat a steel rod on a bench and it quietly grows a few millimetres longer. Now clamp that rod between two immovable walls and heat it the same amount. It cannot grow — so where does the expansion go? Nowhere, and that is the whole story: the growth the rod was denied turns into a colossal internal push, a thermal stress large enough to buckle railway lines and crack engine blocks. A thirty-degree warm-up on a clamped steel rod builds up seventy-two million pascals — roughly a third of what it takes to permanently deform the steel. This is one of the most reliably examined ideas in the thermal syllabus, and it hides a counter-intuition: the stress does not depend on how long the rod is. A one-centimetre stub and a hundred-metre span, same steel, same clamps, same temperature rise, feel exactly the same stress; only the force on the clamps knows about the rod's thickness. Miss that and you lose marks in CBSE, ICSE, JEE and NEET alike. So this article treats the setup itself as the concept — an ARCHETYPE. One picture — a rod that wants to expand by $\alpha\,\Delta T$ but is prevented — generates the master result, and every variant examiners pose (the gap-and-wall problem, the bending bimetallic strip that runs your thermostat, and the composite rod — the same strain balance, drilled to destruction in Irodov) is that one picture with a twist. Learn the machine once, and a whole family of problems becomes fill-in-the-blanks.

Key takeaways

A free rod heated by $\Delta T$ expands by $\Delta L = L_0\alpha\,\Delta T$ . Clamp it so it cannot expand, and the prevented strain $\varepsilon = \alpha\,\Delta T$ becomes stress instead.
Master result: the thermal stress is $\sigma = Y\alpha\,\Delta T$ — and it is independent of the rod's length. Only the force $F = YA\alpha\,\Delta T$ scales, and it scales with area, never length.
Heating a clamped rod puts it in compression (it shoves the walls apart); cooling a clamped rod puts it in tension (it pulls them in).
A bimetallic strip — two metals with $\alpha_1 < \alpha_2$ bonded together — bends toward the lower- $\alpha$ metal on heating, with radius of curvature $R \approx t/[(\alpha_2-\alpha_1)\,\Delta T]$ .
Variants are all one idea: fully clamped rod, rod with a gap $g$ (stress $\sigma = Y(\alpha\,\Delta T - g/L)$ , zero until the gap closes), composite rod between rigid walls, and the bending strip.
Key idea: a body that is prevented from expanding freely converts its frustrated strain $\alpha\,\Delta T$ into stress $\sigma = Y\alpha\,\Delta T$ — the geometry of the constraint IS the physics.

The core insight

When thermal expansion is prevented, the strain the material was denied, $\varepsilon = \alpha\,\Delta T$ , reappears as an elastic stress $\sigma = Y\varepsilon = Y\alpha\,\Delta T$ — a quantity set only by the material and the temperature change, never by the length — and the same principle, applied to two bonded metals that want to expand by different amounts, makes a bimetallic strip curl.

Where this lives today

The bending strip and the clamped rod are everywhere once you know their shape. Every mechanical thermostat — in an iron, an oven, a kettle's cut-off — is a bimetallic strip curling to make or break a contact, John Harrison's eighteenth-century trick miniaturised; coiled up it becomes a dial thermometer's needle, and a snap-acting version is the heart of a thermal circuit breaker. On the railways, the expansion gap behind the classic clack-clack is the clamped-rod problem solved by not clamping — leaving the $5.76\ \mathrm{mm}$ per twelve metres the steel demands; modern continuous welded rail instead pre-stresses the steel to a neutral temperature so summer compression and winter tension both stay survivable. Long bridges ride on toothed expansion joints for the same reason, and power lines are strung with deliberate summer sag so winter contraction cannot snap them. Harrison's gridiron pendulum and compensated chronometer were the first triumphs of the idea; your thermostat is its quiet descendant.

The setup

Geometry first, because in an archetype the geometry is the physics. Take a straight rod of length $L_0$ , cross-sectional area $A$ , of a material with linear-expansion coefficient $\alpha$ and Young's modulus $Y$ . Left free, a temperature rise $\Delta T$ lengthens it by $\Delta L = L_0\alpha\,\Delta T$ — ordinary thermal expansion, nothing stressful. The drama begins the instant you forbid that expansion by holding both ends against rigid walls. Here is the mental move that unlocks everything. Let the rod expand freely first — it grows by $L_0\alpha\,\Delta T$ and pokes past the wall. Now squeeze it back to $L_0$ so it fits between the walls. Squeezing a rod by $\Delta L$ costs a compressive stress $\sigma = Y\varepsilon$ with strain $\varepsilon = \Delta L / L_0$ — and $\Delta L / L_0 = \alpha\,\Delta T$ exactly. So the walls push on the rod (and the rod back) with a stress $\sigma = Y\alpha\,\Delta T$ . Temperature never mechanically "pushed" anything: the rod simply wants a length the walls refuse to grant, and the refusal is the stress.

Figure

Heating a clamped rod — the expansion it is denied turns into compressive stress

Two facts about that argument deserve to be pinned down, because they are where marks live.

The stress carries no memory of length. The strain $\alpha\,\Delta T$ is a fractional squeeze — a percentage, not a distance. A long rod is denied a large absolute expansion and a short rod a small one, but the fraction denied is $\alpha\,\Delta T$ in both, so the stress is identical. This is the most tested feature of the topic.

Heating clamps means compression; cooling clamps means tension. Heating makes the rod want to grow, so the walls squeeze it into compression (it shoves them apart). Cool a clamped rod and it wants to shrink; the walls hold it stretched, so it is in tension and pulls them inward. Winter, not summer, is when restrained cables and rails snap. Now bond two metals into a strip and the same "frustrated expansion" logic produces motion instead of a static push. Give them coefficients $\alpha_1 < \alpha_2$ , glued so they must stay the same length everywhere. On heating, the high- $\alpha_2$ metal wants to be longer than the low- $\alpha_1$ metal; stuck together, the only way both can be satisfied is to curl, putting the eager (high- $\alpha$ ) metal on the outside of the bend where the arc is longer and the reluctant (low- $\alpha$ ) metal on the inside. The strip bends toward the lower- $\alpha$ metal; reverse the temperature and it curls the other way. That curl, wired to a contact, is a thermostat.

Figure

A bimetallic strip curls on heating — the high-α metal takes the longer outer arc, so the strip bends toward the low-α metal

The master results

Everything on this page flows from four short statements; number them and you can point to whichever one a question is really asking. The prevented rod is squeezed by exactly the fraction it wanted to grow:

\varepsilon = \frac{\Delta L}{L_0} = \alpha\,\Delta T \qquad (1)

where: $\varepsilon$ — the strain the clamp forces on the rod (dimensionless); $\alpha$ — coefficient of linear expansion (K⁻¹); $\Delta T$ — temperature change (K, identical to °C for a difference). Read it as a fractional squeeze: the clamp never asks how long the rod is, only what fraction of its length it must give back. Feed that strain through Hooke's law $\sigma = Y\varepsilon$ and the thermal stress appears:

\sigma = Y\varepsilon = Y\alpha\,\Delta T \qquad (2)

where: $\sigma$ — thermal stress (Pa, i.e. N m⁻²); $Y$ — Young's modulus of the material (Pa). Read the headline: the material ( $Y$ , $\alpha$ ) and the temperature change set the stress alone — the length has vanished. Two rods of wildly different lengths, same steel, same $\Delta T$ , carry the same stress, and "it depends on $L$ " is the distractor examiners plant almost every time. Multiply the stress by the area to get the force the rod exerts on each clamp:

F = \sigma A = Y A \alpha\,\Delta T \qquad (3)

where: $A$ — cross-sectional area (m²); $F$ — force on each support (N), compressive (outward) when heated. Read it against (2): length is still absent, but now the area enters. A thicker rod threatens its clamps with a bigger force; a longer rod of the same thickness does not. For the bonded strip, the two metals' length disagreement, $(\alpha_2 - \alpha_1)\Delta T$ , is resolved by curling to a radius:

R \approx \frac{t}{(\alpha_2 - \alpha_1)\,\Delta T} \qquad (4)

where: $R$ — radius of curvature of the curled strip; $t$ — the strip's thickness; $\alpha_2 > \alpha_1$ the two coefficients. Read it: a bigger mismatch $\Delta\alpha = \alpha_2-\alpha_1$ or bigger $\Delta T$ curls the strip tighter (smaller $R$ ); a thicker strip resists curling. The concave side is always the lower- $\alpha$ metal. This is a first-order estimate — the exact bilayer formula carries factors of order one — but it is the relation boards and JEE expect.

The master solution pattern

Every problem on this page is the same four questions asked in a costume. Answer them in order and the algebra writes itself.

Step 1 — What does the body want to do? Compute the free thermal strain it would undergo, $\alpha\,\Delta T$ (or, for two bodies, the difference in their strains). This is the "desire."

Step 2 — How much of that is prevented? A fully clamped rod is denied all of it (prevented strain $= \alpha\,\Delta T$ ). A rod with a gap $g$ over length $L$ is allowed $g/L$ and denied the rest (prevented strain $= \alpha\,\Delta T - g/L$ ). A free rod is denied nothing (no stress).

Step 3 — Turn prevented strain into stress. Multiply the prevented strain by $Y$ : that is the stress. Multiply by $A$ if a force is wanted.

Step 4 — Sanity-check direction and size. Heating clamps → compression; cooling clamps → tension. Confirm the stress is a believable fraction of the material's yield (for steel, tens to a couple of hundred MPa; anything near $10^{10}$ Pa is an arithmetic slip). The variants below are just different answers to Step 2.

variant	key relation	what changed at Step 2
fully clamped rod	$\sigma = Y\alpha\,\Delta T$ , $\;F = YA\alpha\,\Delta T$	all expansion prevented; no length dependence
rod with a gap $g$	$\sigma = Y\!\left(\alpha\,\Delta T - \dfrac{g}{L}\right)$	a slice $g/L$ of strain is allowed; zero stress until the gap closes
composite rod between rigid walls	one shared force $F$ ; total length fixed	each segment's thermal minus elastic strain must sum to zero
bimetallic strip / thermostat	$R \approx \dfrac{t}{(\alpha_2-\alpha_1)\,\Delta T}$	the two metals' strains disagree by $\Delta\alpha\,\Delta T$ ; the strip curls

Derivation AThe master result — thermal stress in a fully clamped rod

We derive equation (2), $\sigma = Y\alpha\,\Delta T$ , by the "expand-then-squeeze" route that every marking scheme rewards, because it separates the two pieces of physics — thermal expansion and Hooke's-law elasticity — instead of blurring them into one formula. The strategy: let the rod expand freely, then apply the mechanical stress needed to compress it back to its clamped length, and recognise that this stress is what the walls supply. Watch how the length $L_0$ enters in the expansion step and cancels in the strain step.

Step 1. Free expansion. Unclamp the rod and raise its temperature by $\Delta T$ . Its length grows by $\Delta L = L_0\,\alpha\,\Delta T$ . Nothing is stressed yet — a free body under uniform heating carries no internal force. Why: this isolates the "desire" — the rod's natural, unstressed length at the new temperature is $L_0 + \Delta L$ , which the clamps are about to deny it.

Step 2. Mechanically squeeze it back. To fit between the immovable walls the rod must be shortened from $L_0+\Delta L$ back to $L_0$ — a compression of $\Delta L$ . The strain of this compression is

\varepsilon = \frac{\Delta L}{L_0} = \frac{L_0\,\alpha\,\Delta T}{L_0} = \alpha\,\Delta T.

Why: here is the cancellation. The absolute squeeze $\Delta L$ is proportional to $L_0$ , so dividing by $L_0$ erases the length. The fractional squeeze is $\alpha\,\Delta T$ for a rod of any length.

Step 3. Convert strain to stress with Hooke's law. Young's modulus is defined by $\sigma = Y\varepsilon$ , so

\sigma = Y\,\varepsilon = Y\alpha\,\Delta T.

The walls provide exactly this compressive stress, and by Newton's third law the rod pushes back on them with it.

Step 4. Force, if asked. Multiply the uniform stress by the cross-section: $F = \sigma A = Y A \alpha\,\Delta T$ . Only now does a geometric dimension — the area — appear; the length never returns. Verified: steel with $Y = 2.0\times10^{11}\ \mathrm{Pa}$ , $\alpha = 1.2\times10^{-5}\ \mathrm{K^{-1}}$ , $\Delta T = 30\ \mathrm{K}$ gives $\varepsilon = 3.6\times10^{-4}$ , $\sigma = 7.2\times10^{7}\ \mathrm{Pa}$ , and $F = 7200\ \mathrm{N}$ for $A = 1\ \mathrm{cm^2}$ , all recomputed in verify_hs2-thermal-stress-bimetallic.py.

Go deeper · Derivation B · 1 minThe gap variant — a rod that can breathe a little before it fights

Derivation BThe gap variant — a rod that can breathe a little before it fights

We derive the residual stress in a rod between two rigid walls with a small clearance gap $g$ , obtaining $\sigma = Y(\alpha\,\Delta T - g/L)$ , because this is the most common way JEE and NEET dress up the clamped-rod result and the point where students either forget the gap or, worse, subtract it as a length rather than a strain. The strategy: let the rod expand freely until it closes the gap, then treat only the remaining expansion as prevented — the gap simply shifts where the clamped-rod story begins.

Step 1. Spend the free expansion on the gap first. Heating by $\Delta T$ , the rod would freely lengthen by $L\alpha\,\Delta T$ . The first $g$ of that growth is allowed — it merely carries the rod across the gap to touch the far wall. No stress accumulates during this phase. Why: below the temperature that closes the gap the rod is effectively free, so $\sigma = 0$ — the gap buys a stress-free head start.

Step 2. Identify the prevented part. Once the rod touches, any further expansion is denied. The prevented elongation is the total desired growth minus the gap: $L\alpha\,\Delta T - g$ . Divide by the length to get the prevented strain:

\varepsilon_{\text{prevented}} = \frac{L\alpha\,\Delta T - g}{L} = \alpha\,\Delta T - \frac{g}{L}.

Why: the gap must be turned into a strain by dividing by $L$ before it can meet $\alpha\,\Delta T$ . Subtracting $g$ straight from a strain — a units mismatch — is the classic wrong turn.

Step 3. Convert to stress. Apply Hooke's law to the prevented strain only:

\sigma = Y\left(\alpha\,\Delta T - \frac{g}{L}\right),\qquad \text{valid once } L\alpha\,\Delta T > g.

If the free expansion never reaches $g$ , the rod never touches and $\sigma = 0$ . The gap closes exactly at $\Delta T_{\text{close}} = g/(L\alpha)$ . Verified: steel rod $L = 1\ \mathrm{m}$ , $g = 0.5\ \mathrm{mm}$ , $\Delta T = 60\ \mathrm{K}$ gives $\alpha\,\Delta T = 7.2\times10^{-4}$ , $g/L = 5\times10^{-4}$ , so $\sigma = 2.0\times10^{11}(2.2\times10^{-4}) = 4.4\times10^{7}\ \mathrm{Pa}$ ; the gap closes at $\Delta T_{\text{close}} = 41.7\ \mathrm{K}$ — recomputed in verify_hs2-thermal-stress-bimetallic.py.

Go deeper · Derivation C · 1 minWhy a bimetallic strip curls — the curvature of the bend

Derivation CWhy a bimetallic strip curls — the curvature of the bend

We derive equation (4), $R \approx t/[(\alpha_2-\alpha_1)\Delta T]$ , for a strip of two bonded metals using nothing beyond arc length — no calculus — because a board answer must explain the curl, not merely quote a formula, and the geometry makes the "bends toward the lower- $\alpha$ metal" rule self-evident. The strategy: model the curled strip as two concentric arcs, one per metal centre-line, force them to share the same subtended angle, and let their length disagreement fix the radius.

Step 1. Two arcs, one angle. When the heated strip curls to radius $R$ , the two metals' centre-lines become concentric arcs separated by the strip thickness $t$ : the inner (low- $\alpha_1$ ) metal at radius $R - t/2$ , the outer (high- $\alpha_2$ ) metal at radius $R + t/2$ . Being bonded, both arcs subtend the same angle $\theta$ at the centre.

Step 2. Write each arc length from its own expansion. Starting from a common length $L_0$ , each metal reaches the length its own coefficient demands:

\left(R+\tfrac{t}{2}\right)\theta = L_0(1+\alpha_2\Delta T),\qquad \left(R-\tfrac{t}{2}\right)\theta = L_0(1+\alpha_1\Delta T).

Why: the outer metal is longer precisely because it has the larger $\alpha$ — which is exactly why the high- $\alpha$ metal must sit on the outside, and the strip bends toward the low- $\alpha$ side.

Step 3. Subtract to kill the common terms. The two equations differ only in the $t/2$ offset and the $\alpha\Delta T$ term:

t\,\theta = L_0(\alpha_2-\alpha_1)\Delta T.

Step 4. Eliminate the angle. The mean arc is essentially the original length, $R\theta \approx L_0$ (the expansions are tiny), so $\theta \approx L_0/R$ . Substitute:

t\cdot\frac{L_0}{R} = L_0(\alpha_2-\alpha_1)\Delta T \;\;\Rightarrow\;\; R \approx \frac{t}{(\alpha_2-\alpha_1)\,\Delta T}.

Verified: $t = 0.5\ \mathrm{mm}$ , $\Delta\alpha = 1.0\times10^{-5}\ \mathrm{K^{-1}}$ , $\Delta T = 100\ \mathrm{K}$ gives $R = 0.5\ \mathrm{m}$ ; a real steel–brass pair ( $\Delta\alpha = 0.7\times10^{-5}$ ) gives $R \approx 0.71\ \mathrm{m}$ — both recomputed in verify_hs2-thermal-stress-bimetallic.py.

Limiting cases

case	result	meaning
free rod (no clamp)	$\sigma = 0$	unprevented expansion stores no stress — it just gets longer
double the length, same $\Delta T$	$\sigma$ unchanged, $F$ unchanged	stress and force ignore length; only $A$ changes $F$
double the area, same $\Delta T$	$\sigma$ unchanged, $F$ doubles	thicker rod, same stress but twice the push on the clamps
cooling a clamped rod ( $\Delta T < 0$ )	$\sigma < 0$ → tension	rod wants to shrink; walls hold it stretched, it pulls inward
gap wide enough ( $g > L\alpha\,\Delta T$ )	$\sigma = 0$	rod never reaches the far wall; behaves as if free
equal coefficients ( $\alpha_1 = \alpha_2$ )	$R \to \infty$	no mismatch, no curl: a "bimetal" of identical metals stays straight

Common trapA longer rod builds up more thermal stress

It does not — and this is the mark-losing instinct the whole topic is built to catch. Because a long rod expands by a larger absolute amount, students assume it must be more stressed. But stress responds to strain, and strain is the fractional squeeze $\alpha\,\Delta T$ , which is the same for a one-centimetre stub and a hundred-metre span of the same steel at the same $\Delta T$ . The formula $\sigma = Y\alpha\,\Delta T$ contains no length, full stop. What does scale is the force on the clamps, $F = \sigma A$ — and even that scales with cross-sectional area, not length. So the correct sentence is: "thermal stress is independent of length; only the force depends on the rod's thickness." Say it that way and the examiner's distractor — an option that scales the stress with $L$ — has nothing to catch.

Worked examples

1 · Board level — the clamped steel rod (CBSE/ICSE/JEE-Main pattern, 3 marks)

A steel rod of cross-sectional area $1.0\ \mathrm{cm^2}$ is clamped rigidly between two immovable walls and then heated through $\Delta T = 30\ \mathrm{K}$ . Find (a) the prevented strain, (b) the thermal stress, (c) the force on the walls. (d) State how each answer changes if the rod were ten times longer. Take $Y = 2.0\times10^{11}\ \mathrm{Pa}$ , $\alpha = 1.2\times10^{-5}\ \mathrm{K^{-1}}$ .

Step 1. Prevented strain — the fraction the rod is denied: $\varepsilon = \alpha\,\Delta T = (1.2\times10^{-5})(30) = \boxed{3.6\times10^{-4}}$ (that is $0.036\%$ ).

Step 2. Thermal stress by Hooke's law: $\sigma = Y\varepsilon = (2.0\times10^{11})(3.6\times10^{-4}) = \boxed{7.2\times10^{7}\ \mathrm{Pa}} = 72\ \mathrm{MPa}$ .

Step 3. Force on each wall: convert the area, $A = 1.0\ \mathrm{cm^2} = 1.0\times10^{-4}\ \mathrm{m^2}$ , then $F = \sigma A = (7.2\times10^{7})(1.0\times10^{-4}) = \boxed{7200\ \mathrm{N}}$ — compressive, since heating pushes the rod outward.

Step 4. Length dependence: none of (a)–(c) changes. Strain, stress and force all lack any $L$ , so a ten-times-longer rod of the same steel and area feels the identical $72\ \mathrm{MPa}$ and pushes with the identical $7200\ \mathrm{N}$ . For perspective, $72\ \mathrm{MPa}$ is about $29\%$ of mild steel's $\approx 250\ \mathrm{MPa}$ yield strength — a serious stress from a modest warm-up.

2 · Board level — the railway expansion gap (CBSE/ICSE/NEET pattern)

A steel railway rail is $12\ \mathrm{m}$ long and can face a summer temperature rise of $\Delta T = 40\ \mathrm{K}$ above the temperature at which it was laid. (a) If the rail is free to expand, how much does it lengthen — i.e. what gap must be left between rails? (b) If instead the rail were rigidly prevented from expanding, what stress would build up? Take $\alpha = 1.2\times10^{-5}\ \mathrm{K^{-1}}$ , $Y = 2.0\times10^{11}\ \mathrm{Pa}$ .

Step 1. Free expansion sets the gap. $\Delta L = L\alpha\,\Delta T = (12)(1.2\times10^{-5})(40) = \boxed{5.76\times10^{-3}\ \mathrm{m}} = 5.76\ \mathrm{mm}$ . Leaving at least this clearance lets the rail grow harmlessly.

Step 2. The stress if the gap were absent. With the rail fully clamped, $\sigma = Y\alpha\,\Delta T = (2.0\times10^{11})(1.2\times10^{-5})(40) = \boxed{9.6\times10^{7}\ \mathrm{Pa}} = 96\ \mathrm{MPa}$ .

Step 3. Read the moral. $96\ \mathrm{MPa}$ is nearly $40\%$ of the $\approx 250\ \mathrm{MPa}$ yield strength — enough compression to bow the track into a "sun kink." Notice the length again: the $12\ \mathrm{m}$ entered the gap (Step 1) but vanished from the stress (Step 2). Engineers size the gap to the length but quote the stress without one.

3 · Competitive level — the rod with a clearance gap (JEE-Main/Advanced pattern, ~2.5 min)

A steel rod of length $1.0\ \mathrm{m}$ and area $2.0\ \mathrm{cm^2}$ is mounted between two rigid walls with a clearance gap of $g = 0.5\ \mathrm{mm}$ to the far wall. It is heated through $\Delta T = 60\ \mathrm{K}$ . Find (a) the temperature rise at which the rod first touches the far wall, (b) the residual stress at $\Delta T = 60\ \mathrm{K}$ , and (c) the force on the walls. Take $\alpha = 1.2\times10^{-5}\ \mathrm{K^{-1}}$ , $Y = 2.0\times10^{11}\ \mathrm{Pa}$ .

Step 1. When does it touch? The gap closes when the free expansion equals $g$ : $L\alpha\,\Delta T_{\text{close}} = g \Rightarrow \Delta T_{\text{close}} = g/(L\alpha) = (0.5\times10^{-3})/[(1.0)(1.2\times10^{-5})] = \boxed{41.7\ \mathrm{K}}$ . Below this, $\sigma = 0$ .

Step 2. Prevented strain at $\Delta T = 60\ \mathrm{K}$ . Desired strain $\alpha\,\Delta T = (1.2\times10^{-5})(60) = 7.2\times10^{-4}$ ; allowed strain $g/L = (0.5\times10^{-3})/(1.0) = 5.0\times10^{-4}$ ; prevented $= 7.2\times10^{-4} - 5.0\times10^{-4} = 2.2\times10^{-4}$ .

Step 3. Residual stress: $\sigma = Y(\alpha\,\Delta T - g/L) = (2.0\times10^{11})(2.2\times10^{-4}) = \boxed{4.4\times10^{7}\ \mathrm{Pa}} = 44\ \mathrm{MPa}$ — well below the $72\ \mathrm{MPa}$ the same $60\ \mathrm{K}$ would give with no gap, because the gap donated $5.0\times10^{-4}$ of strain for free.

Step 4. Force on the walls: $A = 2.0\ \mathrm{cm^2} = 2.0\times10^{-4}\ \mathrm{m^2}$ , so $F = \sigma A = (4.4\times10^{7})(2.0\times10^{-4}) = \boxed{8800\ \mathrm{N}}$ . Timing note: about 2.5 minutes. The only real decision is Step 2 — subtract the gap <strong>as a strain</strong> ( $g/L$ ), never as a length. Compute $\alpha\,\Delta T$ and $g/L$ side by side so the units are visibly the same before you subtract; that single habit kills the commonest error on this problem.

4 · Competitive level — the bimetallic thermostat strip (JEE-Advanced pattern, ~3 min)

A bimetallic strip of total thickness $t = 0.5\ \mathrm{mm}$ and free length $\ell = 5.0\ \mathrm{cm}$ is straight at room temperature. Its two metals differ in expansion coefficient by $\Delta\alpha = 1.0\times10^{-5}\ \mathrm{K^{-1}}$ . It is heated through $\Delta T = 100\ \mathrm{K}$ . Find (a) the radius of curvature, (b) which way the free tip moves, and (c) the tip deflection.

Step 1. Radius of curvature from equation (4): $R = \dfrac{t}{\Delta\alpha\,\Delta T} = \dfrac{0.5\times10^{-3}}{(1.0\times10^{-5})(100)} = \dfrac{0.5\times10^{-3}}{1.0\times10^{-3}} = \boxed{0.5\ \mathrm{m}}$ .

Step 2. Direction. The higher- $\alpha$ metal wants the longer arc, so it lies on the outside of the bend; the free tip swings toward the lower- $\alpha$ metal. Wire a contact on that side and heating trips it — a thermostat.

Step 3. Tip deflection. For a gently curved strip the free-end deflection is the sagitta $\delta \approx \dfrac{\ell^2}{2R} = \dfrac{(0.05)^2}{2(0.5)} = \dfrac{2.5\times10^{-3}}{1.0} = \boxed{2.5\times10^{-3}\ \mathrm{m}} = 2.5\ \mathrm{mm}$ — a small strip moves a visible, switch-closing distance.

Step 4. Reality check with real metals. A genuine steel–brass strip has $\Delta\alpha = (1.9 - 1.2)\times10^{-5} = 0.7\times10^{-5}\ \mathrm{K^{-1}}$ , giving $R = (0.5\times10^{-3})/[(0.7\times10^{-5})(100)] \approx 0.71\ \mathrm{m}$ — a gentler curl, because brass and steel disagree a little less than $1.0\times10^{-5}$ . Timing note: about 3 minutes. Do part (a) in one line — do not derive the arc formula under exam pressure, just quote $R = t/(\Delta\alpha\,\Delta T)$ . The tempting slip is to say the strip bends toward the metal that expands more; it is the opposite, and a one-word wrong answer forfeits the direction mark.

Exam traps

the trap	why it is wrong	the fix
"a longer rod has more thermal stress"	$\sigma = Y\alpha\,\Delta T$ has no $L$ in it; only $F = \sigma A$ scales, and with area	quote " $\sigma$ is length-independent; $F$ scales with $A$ "
"the strip bends toward the metal that expands more"	the high- $\alpha$ metal takes the longer outer arc	it bends toward the lower- $\alpha$ metal (inner arc)
"heating a clamped rod stretches it (tension)"	prevented growth means the rod is compressed, pushing the walls out	heating → compression; only cooling a clamped rod gives tension
" $Y = 200\ \mathrm{GPa}$ " left as GPa	mixing $\mathrm{GPa}$ or $\mathrm{N\,mm^{-2}}$ with SI throws a factor of $10^{9}$	put $Y$ in pascals: $2.0\times10^{11}\ \mathrm{Pa}$
"the gap multiplies the stress by some factor"	the gap subtracts a strain $g/L$ : $\sigma = Y(\alpha\,\Delta T - g/L)$	below $\Delta T = g/(L\alpha)$ the stress is exactly zero
mixing up $\alpha$ , $Y$ or the $\Delta T$ scale	boards quote slightly different constants	use the paper's values; $\Delta T$ in K equals $\Delta T$ in °C

Origins · 1 minGabriel Lamé and how this was discovered

Gabriel Lamé

1795–1870 · École Polytechnique, Paris

Gabriel Lamé was the mathematician who turned "a body that cannot expand freely" into exact equations. Trained at the École Polytechnique and seasoned by a decade building roads and bridges in Russia, he returned to Paris to create much of the modern mathematical theory of elasticity, introducing the two elastic constants (the Lamé parameters) that still bear his name. In his 1852 Leçons sur la théorie mathématique de l'élasticité des corps solides he analysed the stresses set up inside elastic solids by constrained deformation — including the thermal stresses that appear when heating or cooling is prevented from becoming free expansion. That is exactly the $\sigma = Y\alpha\,\Delta T$ story of this article, generalised to three dimensions.

History

1726

— John Harrison builds his gridiron pendulum: brass and steel rods of unequal expansion cancel each other, holding the pendulum's effective length steady as temperature swings.

1759

— Harrison completes the H4 marine chronometer, whose temperature compensation keeps it accurate at sea and helps crack the longitude problem.

1830

— Andrew Ure patents a bimetallic-strip heat regulator, the direct ancestor of every modern thermostat.

1837

— Jean-Marie Duhamel formulates the equations of thermoelastic stress, tying a temperature change to the stress it induces in an elastic solid.

1852

— Gabriel Lamé publishes his lectures on the mathematical theory of elasticity, systematising the stresses produced by constrained thermal expansion.

1950

s — continuous welded rail replaces the old gapped joints, pre-stressing the steel to a neutral temperature so it neither buckles in summer nor snaps in winter.

Going further

Texts. NCERT Physics Class 11, Ch. 11 (Thermal Properties of Matter) · HC Verma, Concepts of Physics Vol. 2 for worked thermal-stress problems · Irodov, Problems in General Physics, for competitive gap-and-composite variants · Cambridge AS & A-Level Physics 9702 (Deformation of solids + Thermal physics). For the exact bimetallic curvature, S. P. Timoshenko's Theory of Elasticity.

Provenance. All results are standard; a verifier should recompute: prevented strain $\alpha\,\Delta T = 3.6\times10^{-4}$ and stress $\sigma = 7.2\times10^{7}\ \mathrm{Pa}$ with force $F = 7200\ \mathrm{N}$ (steel, $\Delta T = 30$ , $A = 1\ \mathrm{cm^2}$ ); the rail's free expansion $\Delta L = 5.76\ \mathrm{mm}$ and clamped stress $\sigma = 9.6\times10^{7}\ \mathrm{Pa}$ ( $L = 12\ \mathrm{m}$ , $\Delta T = 40$ ); the gap-rod's closing temperature $\Delta T = 41.7\ \mathrm{K}$ , residual stress $\sigma = 4.4\times10^{7}\ \mathrm{Pa}$ and force $F = 8800\ \mathrm{N}$ ( $L = 1\ \mathrm{m}$ , $g = 0.5\ \mathrm{mm}$ , $\Delta T = 60$ , $A = 2\ \mathrm{cm^2}$ ); the bimetal radius $R = 0.5\ \mathrm{m}$ and tip deflection $\delta = 2.5\ \mathrm{mm}$ ( $t = 0.5\ \mathrm{mm}$ , $\Delta\alpha = 1.0\times10^{-5}$ , $\Delta T = 100$ , $\ell = 5\ \mathrm{cm}$ ), plus the steel–brass check $\Delta\alpha = 0.7\times10^{-5}$ , $R \approx 0.71\ \mathrm{m}$ .

Prerequisites. Thermal Expansion of Solids & Liquids ( $\Delta L = L_0\alpha\,\Delta T$ ) · Young's, Bulk & Shear Moduli; Poisson's Ratio ( $\sigma = Y\varepsilon$ ). Leads to. This archetype is a capstone — it feeds engineering applications (thermostats, expansion joints, thermal fatigue) rather than a further school topic. Go deeper. The three-dimensional generalisation — how a temperature field sets up a full stress tensor in an elastic solid — is thermoelasticity, built on Lamé's elastic constants; it is where the university theory of elasticity picks up this thread.

Why doesn't the length of the rod affect the thermal stress?

Because stress is driven by strain, and strain is a fraction, not a distance. A long rod is denied a big absolute expansion and a short rod a small one — but each is denied the same fraction of its own length, $\alpha\,\Delta T$ . Since $\sigma = Y\alpha\,\Delta T$ , the length cancels completely. What does grow with size is the force on the clamps, $F = \sigma A$ — and even that responds to cross-sectional area, not length. Remember it as: material and temperature set the stress; area only sets the force.

Is a heated clamped rod in tension or compression?

Compression. Heating makes the rod want to get longer; the walls refuse, so they squeeze it and it pushes back outward — compressive stress. The reverse is just as often tested: cool a clamped rod and it wants to shrink, but the walls hold it stretched, so it is in tension and pulls the walls inward. That is why cracks and snapped cables tend to appear on the coldest nights, not the hottest days — winter contraction, not summer expansion, puts a restrained structure into dangerous tension.

Which way does a bimetallic strip bend, and how do I remember it?

It bends toward the metal with the lower expansion coefficient. On heating, the high- $\alpha$ metal wants to be much longer, the low- $\alpha$ metal only a little. The only bonded shape that lets the eager metal be longer is a curl with the high- $\alpha$ metal on the outside, where the arc is longer — which automatically puts the reluctant low- $\alpha$ metal on the inside. Memory hook: "the keen one goes long on the outside; the strip leans to the lazy metal."

Both metals get longer when heated — so why does the strip bend at all?

Exactly — both lengthen, but by different amounts. If they were free they would become two rods of slightly different length. Bonded, they must share one shape, and the only way two bonded layers can have different natural lengths is to curve, the outer layer tracing a longer arc than the inner. The bend is the geometric compromise between two layers that each "want" a different length. Equal coefficients mean equal wants, no compromise, and the strip stays straight.

What is the difference between the fully clamped case and the gap case?

Only where the story starts. A fully clamped rod is denied all of its expansion from the very first degree, so stress builds immediately: $\sigma = Y\alpha\,\Delta T$ . A rod with a gap $g$ is allowed to expand freely until it has crossed the gap and touched the far wall; only the expansion beyond that is prevented. So its stress is zero until the closing temperature $\Delta T = g/(L\alpha)$ , and after that it follows $\sigma = Y(\alpha\,\Delta T - g/L)$ . The gap subtracts a strain $g/L$ ; the fully clamped case is just the special case $g = 0$ .

Do I need calculus for the bimetallic-strip formula?

No. The result $R \approx t/[(\alpha_2-\alpha_1)\Delta T]$ comes out of pure geometry — two concentric arcs sharing an angle, plus arc length $=$ radius $\times$ angle. The derivation on this page uses only algebra and the expansion law, which is why it is fair game at board and JEE level. The exact engineering formula (Timoshenko's) carries factors of order one depending on the metals' thicknesses and stiffnesses — that is where calculus and beam theory come in, enrichment for later, not something an exam here demands.

Which values of α and Y should I use in the exam?

Use exactly what the question states, and never mix conventions mid-solution. This article uses the standard board set: $\alpha_{\text{steel}} = 1.2\times10^{-5}\ \mathrm{K^{-1}}$ , $\alpha_{\text{brass}} = 1.9\times10^{-5}\ \mathrm{K^{-1}}$ , $Y_{\text{steel}} = 2.0\times10^{11}\ \mathrm{Pa}$ . Some papers quote steel's $Y$ as $200\ \mathrm{GPa}$ — convert $\mathrm{GPa}$ to pascals ( $\times10^{9}$ ) before substituting. And remember a temperature difference is the same number in kelvin and Celsius, so $\Delta T = 40\ \mathrm{K} = 40\ \mathrm{^\circ C}$ ; only absolute temperatures need the $273$ shift, and they never appear in $\sigma = Y\alpha\,\Delta T$ .