Stress, Strain & Hooke's Law for Solids

Every certified batch of structural steel, aircraft aluminium, or surgical-implant titanium ships with a tensile test report — a stress–strain curve drawn by a machine that stretches a standard sample to destruction, exactly as this article describes. Design codes then command engineers to keep working stresses a safety factor below yield: typically a third of yield stress for buildings, stricter for lifts and pressure vessels. The 2010s-to-2020s materials headline is at the nano end: graphene's Young's modulus of about 1 TPa — five times steel's — was measured by pressing an atomic-force-microscope tip into a suspended one-atom-thick sheet and reading its stress–strain response, elasticity's oldest experiment re-run on the thinnest possible sample. And metallurgists deliberately use the plastic region this article warns you away from: car crumple zones are engineered to deform plastically, spending crash energy as the metal yields — the stress–strain curve's area put to work saving lives.

Why does a solid resist stretching at all? Zoom in. A crystalline solid is a lattice of atoms held at preferred spacings by interatomic forces. Pull the lattice and every bond stretches slightly; each bond pulls back like a microscopic spring. Push, and the bonds compress and push back. A solid is Avogadro's number of tiny springs in a 3-D mesh — which is exactly why the macroscopic behaviour ends up looking like Hooke's spring law, and why it stops looking like it when you pull hard enough for atoms to start sliding past each other and re-bonding elsewhere (that is plastic deformation — the paper clip that stays bent).

Now the two definitions that carry the whole chapter, and the physical reason each is built the way it is.

Stress spreads the force over the area that carries it. Hang a 10 kg mass from a steel wire of cross-section $1\ \mathrm{mm^2}$ and every square millimetre of that cross-section carries the full 98 N. Hang it from a wire of $2\ \mathrm{mm^2}$ and each unit of area carries half as much — the material is working half as hard even though the force is identical. Think of lanes on a highway: the same traffic across more lanes means less congestion per lane. What decides whether the wire deforms — or breaks — is the load per lane, not the total load. Strictly, stress is defined through the internal restoring force the deformed material develops across the section; in equilibrium that internal force equals the applied one, which is why we compute it as applied force over area.

Strain compares the deformation to the original size. A 1 mm extension is an enormous deformation for a 10 cm sample and a negligible one for a 100 m cable. Dividing $\Delta L$ by $L$ produces a fair, size-free measure: both the lab wire and the bridge cable at $\varepsilon = 10^{-3}$ are in the same mechanical state, every bond stretched by the same fraction. That is why strain — a pure, unitless number — is the honest x-axis of material behaviour. And the three ways to load a solid, which you must be able to tell apart on sight: pull or push along the axis (tensile/compressive — length changes), slide the top face relative to the bottom like pushing a book sideways (shear — shape changes, angles tilt), or squeeze uniformly from all directions as fluid pressure does (hydraulic/volume — volume shrinks). Each style gets its own stress, its own strain, and — next article — its own modulus.

The normal (tensile or compressive) stress answers: how hard is each unit of cross-section being pulled? For a force $F$ applied perpendicular to a section of area $A$,

where: $\sigma$ — normal stress, in pascals ($1\ \mathrm{Pa} = 1\ \mathrm{N/m^2}$); $F$ — the magnitude of the deforming force normal to the section; $A$ — the cross-sectional area carrying it. The pascal is tiny — real steel stresses run in megapascals ($10^6$ Pa) and moduli in gigapascals ($10^9$ Pa). Same unit as pressure, different job: pressure in a fluid pushes on external surfaces from all sides; stress lives inside a solid and has a direction and a style (tensile, compressive, shear). The matching deformation measure is the longitudinal strain. A rod of natural length $L$ that extends by $\Delta L$ has

where: $\varepsilon$ — longitudinal strain, dimensionless (often quoted as a percentage: $\varepsilon = 0.002$ is "0.2% strain"); $\Delta L$ — the change in length; $L$ — the original length. Typical exam strains are $10^{-4}$ to $10^{-3}$: solid metals under working loads deform by less than a tenth of a percent — invisible to the eye, which is why wires "feel" rigid. For shear, the force is applied parallel to a face of area $A$: shearing stress $= F_{\parallel}/A$, and shearing strain is the small tilt angle $\theta \approx \Delta x/h$ (displacement of the sheared face over the height between faces). For hydraulic loading, the stress is just the pressure $P$, and the volume strain is $\Delta V/V$. Keep the table:

With the per-unit variables built, Hooke's discovery takes its modern, material-level form. For small deformations,

where: $E$ — the modulus of elasticity for that loading style (Young's modulus $Y$ for stretching — the star of the next article; shear and bulk moduli for the other two styles). Because $\varepsilon$ is dimensionless, $E$ carries the units of stress, Pa. Read the equation as a stiffness rating: at the same strain, steel ($Y \approx 2\times10^{11}$ Pa) develops about $10^5$ times the stress of rubber ($\sim 10^6$ Pa) — steel fights back harder per unit of deformation. Hooke's law is not a law of nature like Newton's laws; it is an approximation valid for small strains, and the stress–strain curve below shows exactly where it expires.

This one result is worth memorizing with its shape, not just its symbols: $k = YA/L$. It converts every spring-chapter skill you own into elasticity-chapter currency — wires in series and parallel, energy stored, oscillations on a wire — all of it returns in the Stretched Wire archetype.

Put a metal sample in a testing machine, pull slowly until it breaks, and plot stress against strain. The resulting curve is the material's mechanical biography, and its labelled landmarks are permanent members of the JEE/NEET question bank.

Walk the curve left to right. From the origin to the proportional limit (A), stress and strain climb together in a straight line — this is Hooke's law's entire kingdom, and the slope of this line is the modulus. Between A and the elastic limit or yield point (B), the curve bends: stress and strain are no longer proportional, but the material still snaps back completely when unloaded — elastic, just not linear. Beyond B the character changes: atoms slide into new positions, and unloading leaves a permanent set — the sample returns along a different path and stays deformed. That is the plastic region. The stress peaks at the ultimate tensile strength (D) — the biggest load per area the material can ever carry — after which the sample begins to thin ("necking") and the curve droops to fracture (E). Two vocabulary words classify materials by the width of the plastic region. Ductile materials (steel, copper, aluminium) have a long B-to-E run: they yield, stretch, warn you, and only then break — which is why they can be drawn into wires and why bridges are made of them. Brittle materials (glass, cast iron, chalk) break almost as soon as they leave the straight line: no warning, no plastic buffer. And elastomers (rubber, the aorta's wall tissue) are the curve's rebels: they stretch to strains of 5 or more and return, but their curve is never straight — elastic without obeying Hooke's law, a favourite trick-question fact.

1 · Board level — stress, strain and extension of a loaded wire (CBSE/ISC pattern, 3 marks)

A structural steel wire of length $2.0\ \mathrm{m}$ and cross-sectional area $1.0\ \mathrm{mm^2}$ hangs vertically. A mass of $10\ \mathrm{kg}$ is suspended from its end. Find the stress, the strain, and the elongation. Young's modulus of steel $= 2.0\times10^{11}\ \mathrm{Pa}$; $g = 9.8\ \mathrm{m/s^2}$; neglect the wire's own weight.

Step 1. Convert the area first — the classic mark-loser: $A = 1.0\ \mathrm{mm^2} = 1.0\times10^{-6}\ \mathrm{m^2}$.

Step 2. Stress: the load is $F = mg = 10 \times 9.8 = 98\ \mathrm{N}$, so $\sigma = F/A = 98/10^{-6} = \boxed{9.8\times10^{7}\ \mathrm{Pa}}$ — 98 MPa, comfortably below steel's yield (~250 MPa): the wire is safe.

Step 3. Strain from Hooke's law: $\varepsilon = \sigma/Y = \dfrac{9.8\times10^7}{2.0\times10^{11}} = \boxed{4.9\times10^{-4}}$ — about 0.05%, dimensionless, no units to write.

Step 4. Elongation: $\Delta L = \varepsilon L = 4.9\times10^{-4} \times 2.0 = 9.8\times10^{-4}\ \mathrm{m} = \boxed{0.98\ \mathrm{mm}}$. A two-metre steel wire under a 10 kg load stretches less than a millimetre — that is what a modulus of $10^{11}$ Pa feels like.

2 · Competitive level — scaling with geometry (JEE-Main pattern, ~90 seconds)

Wire 1 stretches by $1.0\ \mathrm{mm}$ under a load $W$. Wire 2 is of the same material but twice the length and twice the diameter. What is its extension under the same load $W$?

Step 1. Build the extension formula once from $\sigma = Y\varepsilon$: $\Delta L = \dfrac{FL}{AY}$. Under fixed $F$ and fixed material, $\Delta L \propto \dfrac{L}{A} \propto \dfrac{L}{d^2}$.

Step 2. Apply the ratios: $L \to 2L$ multiplies the extension by 2; $d \to 2d$ multiplies the area by 4, dividing the extension by 4.

Step 3. Net factor $= 2/4 = 1/2$, so $\Delta L_2 = \boxed{0.5\ \mathrm{mm}}$. Timing note: never recompute stresses in a scaling question — write $\Delta L \propto L/d^2$ and multiply factors; this is a 30-second problem dressed as a 3-minute one. Watch the diameter–radius trap: doubling the diameter doubles the radius too, so area goes up 4×, not 2× or 16×.

3 · Competitive level — reading the curve (NEET pattern, ~60 seconds)

The stress–strain graphs of materials P and Q, drawn on the same axes, are both straight lines up to their proportional limits. P's line is steeper; Q's line extends much farther before fracture, while P fractures just past its yield. (a) Which material has the larger Young's modulus? (b) Which is more ductile? (c) Which is stronger, if Q's fracture occurs at a higher stress value than P's?

Step 1. (a) The slope of the straight region is the modulus, $\sigma/\varepsilon$. Steeper line → larger modulus: $\boxed{P}$.

Step 2. (b) Ductility is the length of the run between yield and fracture. P breaks just past yield (brittle); Q keeps deforming plastically: $\boxed{Q}$ is more ductile.

Step 3. (c) "Strength" in curve language means the maximum stress reached. Given Q fractures at the higher stress, $\boxed{Q}$ is stronger — despite being less stiff. Stiff, strong, and ductile are three independent virtues; conflating them is the question's whole trap. Timing note: for any curve MCQ, decode the three words first — stiffness = slope, strength = height of the peak, ductility = horizontal run after yield. Then read the graph. Sixty seconds includes checking axes aren't swapped (a strain–stress plot inverts every conclusion).

mm² → m² is a factor of $10^{-6}$, not $10^{-3}$. The single most common elasticity error. $1\ \mathrm{mm^2} = (10^{-3}\ \mathrm{m})^2 = 10^{-6}\ \mathrm{m^2}$. Get this wrong and your stress is off by a thousand — usually flagged by a modulus-sized stress or a metre-sized extension. Sanity band: exam stresses land between $10^6$ and $10^9$ Pa.

Diameter given, radius needed. Questions quote diameter ("a wire of diameter 2 mm"); $A = \pi d^2/4 = \pi r^2$. Halve before squaring. In ratio problems, remember $A \propto d^2$, so doubling diameter quarters the extension.

Strain with units. Writing $\varepsilon = 4.9\times10^{-4}\ \mathrm{m}$ costs the answer mark on boards. Strain is a ratio of lengths — a pure number (or a percentage).

$L$ is the original length, $\Delta L$ the change. In "a 2 m wire stretches to 2.001 m", the strain is $0.001/2$, not $0.001/2.001$ and not $2.001/2$. Slow down on the subtraction.

Elastic limit ≠ proportional limit ≠ yield of the same number. See the pitfall box: proportionality dies first, elasticity a little later. Assertion–reason and statement-I/II formats exploit exactly this gap.

Stress is not the applied force, and not pressure. Two wires with the same load carry different stresses if their areas differ; and unlike fluid pressure, stress has a style (tensile/shear) and lives inside the solid. If a question mixes "pressure applied to a wire's end" with "stress", translate carefully: numerically equal here, conceptually distinct.

Which $g$? CBSE/ISC: $9.8\ \mathrm{m/s^2}$ unless stated. JEE-Main/NEET: almost always "take $g = 10$" — obey the paper, never mix values mid-solution. A-Level: $9.81$.

Elastic ≠ obeys Hooke's law. Rubber recovers from 500% strain (elastic) with a thoroughly curved graph (non-Hookean). The converse trap: within the straight region every material is both.

Texts. NCERT Physics Class 11, Ch. 8 (Mechanical Properties of Solids) · HC Verma, Concepts of Physics Vol. 1, Ch. 14 · Resnick–Halliday–Walker, Ch. 12 (Equilibrium and Elasticity).

Provenance. All results are standard. A verifier should recompute: $k = YA/L = 10^5$ N/m for the 2 m/1 mm² steel wire; example 1's $\sigma = 9.8\times10^7$ Pa, $\varepsilon = 4.9\times10^{-4}$, $\Delta L = 0.98$ mm; example 2's factor $2/4 = 1/2 \Rightarrow 0.5$ mm.

Prerequisites. Spring Force & Hooke's Law · Equilibrium of Concurrent Forces. Leads to. Young's, Bulk & Shear Moduli · The Stretched Wire (archetype).