Stress and strain

Pick up a fresh piece of chalk and pull its two ends apart. Nothing happens for a while — then, quite suddenly, it snaps. Now do the same to a rubber band: it stretches visibly, and when you let go it springs right back to its original length. Both objects were "resisting" your pull the whole time, but they resisted in completely different ways, and physics needs a language precise enough to tell the difference between a material that springs back and one that shatters. That language is built from exactly two ideas: stress, the internal force per unit area that a stretched, squeezed, or sheared body develops to fight back against whatever is deforming it, and strain, the fractional change in shape that results. Every bridge cable, every submarine hull, every fishing rod and every bone in your own skeleton is, at every instant, silently balancing an external load against an internal stress — and the strain it suffers in the process is what an engineer must keep small enough that the structure survives. This article opens the CBSE Class 11 chapter on the Mechanical Properties of Solids. Before you can make sense of Hooke's law, Young's modulus, or the dramatic stress-strain curve that separates a ductile metal from a brittle ceramic, you need rock-solid, unit-correct definitions of stress and strain themselves — and a feel for the small but very real elastic energy a stretched body stores while it resists. That is exactly what the next few pages build, using nothing beyond the algebra of forces, areas, and ratios you already own.

Key takeaways

Elasticity is the property of a solid that lets it recover its original shape and size once a deforming force is removed; a body that does not fully recover is plastic and keeps a permanent set.
Stress $\sigma = F/A$ is the internal restoring force per unit area that a deformed body develops; it has the same dimensions as pressure and SI unit the pascal (Pa).
Strain is the fractional deformation — longitudinal strain $\Delta L/L$ , volumetric strain $\Delta V/V$ , or shear strain $\theta$ — and is always a pure, dimensionless number.
Three flavours of stress and three matching flavours of strain: normal (longitudinal), volumetric (bulk), and shearing (tangential) — each caused by a different geometry of applied force.
A stretched or compressed elastic body stores elastic potential energy; per unit volume this works out to $u = \tfrac12\sigma\varepsilon = \tfrac12 Y\varepsilon^2$ , a formula you will derive in this very article from first principles.
The stress-strain curve of a real material has an elastic region, a yield point, a plastic region, and a fracture point — a map you must recognise even before you meet the numerical moduli that describe its slope.
Key idea: stress is the internal force per unit area a deformed solid develops to resist a load, strain is the resulting dimensionless fractional deformation, and the work done stretching an elastic body is stored as elastic potential energy of density u = ½σε.

The core insight

Stress $\sigma = F/A$ is the internal restoring force per unit area a solid develops when deformed, strain is the resulting dimensionless fractional change in length, volume, or shape, and the two together fix the elastic energy $u = \tfrac12\sigma\varepsilon$ stored per unit volume of the deformed body.

Where it fits in the syllabus

This topic opens the CBSE Class 11 Physics unit Mechanical Properties of Solids, immediately before Hooke's law and the individual elastic moduli (Young's, bulk, and the qualitative shear modulus) that occupy the rest of the chapter. Everything that follows — the linear $\sigma = Y\varepsilon$ relation, the numerical value of Young's modulus for steel or aluminium, the shape of the stress-strain curve — is built directly on the definitions of stress and strain given here, so this article is deliberately definitional and conceptual rather than a moduli-crunching exercise. Quantitative shear-modulus problems, the full elastic tensor, and dislocation or fatigue theory are outside the 2025-26 CBSE prescription and are not touched here; where a numerical modulus is needed to illustrate elastic energy, only the bulk modulus (in scope) is used.

The physical picture

A solid is not a rigid, unyielding block at the atomic level — it is a lattice of atoms held in place by interatomic bonds that behave, to a good approximation, like tiny springs. Pull, squeeze, or shear the solid and you stretch, compress, or tilt those bonds; they push back. Elasticity is the name for a solid's tendency to restore its original configuration once the external force is removed, because the interatomic bonds simply relax back to their equilibrium spacing. If you push the deformation past a certain point, some bonds break or atomic planes slip past one another permanently, and the body does not fully recover — it is then behaving plastically, and the leftover deformation is a permanent set. At mechanical equilibrium, the internal restoring force generated across any internal cross-section of the body must exactly balance the external deforming force. The magnitude of this internal force per unit area is called stress:

\sigma = \frac{F}{A}

where: $F$ is the applied deforming force and $A$ is the area of the cross-section over (or across) which it acts. Stress carries the dimensions $[\text{ML}^{-1}\text{T}^{-2}]$ , identical to pressure, and its SI unit is the pascal, $1\ \text{Pa} = 1\ \text{N/m}^2$ — in practice, stresses in structural materials are usually quoted in megapascals (MPa) or gigapascals (GPa) because a single pascal is a tiny stress. Notice immediately that stress is not the same thing as force: doubling the cross-sectional area of a rod while keeping the force fixed halves the stress, which is exactly why a thick cable can carry a load that would snap a thin wire of the same material. How the force acts relative to the surface decides which of three families the stress belongs to. A normal (longitudinal) stress arises when the force acts perpendicular to a cross-section, stretching or compressing a rod or wire along its length. A volumetric (bulk) stress arises when a force — typically a fluid pressure — acts uniformly over the entire surface of a body, squeezing it from every direction at once and changing only its volume, not its shape. A shearing (tangential) stress arises when the force acts parallel to a surface, causing one layer of the material to slide sideways relative to the layer beneath it, changing shape without necessarily changing volume. Strain measures the resulting deformation as a pure ratio, so it carries no units at all:

\text{longitudinal strain}\ \varepsilon = \frac{\Delta L}{L}, \qquad \text{volume strain}\ \varepsilon_V = \frac{\Delta V}{V}, \qquad \text{shear strain} = \tan\theta \approx \theta

where: $\Delta L$ is the change in length of a wire or rod of original length $L$ , $\Delta V$ is the change in volume of a body of original volume $V$ , and $\theta$ is the small angle (in radians) through which a face of a sheared body tilts relative to the opposite face. Longitudinal and volume strain are ratios of two lengths or two volumes and so are automatically dimensionless; shear strain is an angle, and for the small deformations met in this chapter $\tan\theta \approx \theta$ in radians, which is also dimensionless. This is why, once you meet Hooke's law in the next article, a modulus (stress divided by strain) always carries the same units as stress itself — strain contributes no units of its own. A useful companion definition, used later when comparing how a stretched wire changes shape sideways, is Poisson's ratio,

\nu = -\frac{\text{lateral strain}}{\text{longitudinal strain}}

which compares the (usually negative) sideways contraction of a stretched rod to its lengthwise extension; like strain itself, $\nu$ is a pure number with no units, and its value for common engineering metals lies roughly between $0.2$ and $0.4$ . Finally, it helps to have a mental map of what happens as the applied stress on a real solid is increased from zero. For small stresses the material is perfectly elastic — stress and strain grow together and the body fully recovers on unloading. Beyond a certain elastic limit the material enters a plastic region, where a permanent set remains even after the load is removed; increase the stress further and the material eventually reaches its fracture point and breaks. This full curve, and the numerical slope of its initial straight portion (Young's modulus), is the subject of the very next article in this chapter — here we only need the map, not the numbers.

Figure

Three identical rods drawn side by side showing normal stress stretching a rod along its length, volumetric stress squeezing a cube uniformly from all faces, and shearing stress sliding the top face of a block sideways relative to the bottom face, with force arrows and the resulting strain labelled under each

Derivation AElastic potential energy stored in a stretched wire

The goal is to find how much elastic potential energy is stored, per unit volume, inside a wire that has been stretched by an external force within its elastic limit — the quantity $u = \tfrac12\sigma\varepsilon = \tfrac12 Y\varepsilon^2$ quoted in the syllabus. The physical idea is simple: as you stretch the wire slowly from zero extension to its final extension $\Delta L$ , you do work against the wire's growing internal restoring force at every stage, and because the process is elastic (no permanent set, no heat lost to friction between atomic planes), every joule of that work is stored as recoverable potential energy in the stretched bonds — exactly like a spring wound up rather than a battery discharged. We build the answer using only the definition of work and the linear stress-strain relation $\sigma = Y\varepsilon$ (Hooke's law, quoted here as given), with no calculus beyond the area of a triangle.

Step 1. Set up the wire. Let the wire have natural length $L$ and cross-sectional area $A$ , so its volume is $A L$ . As an external force stretches it, let $x$ be the extension at some intermediate instant, running from $x = 0$ up to the final extension $x = \Delta L$ . At extension $x$ the strain is $x/L$ , and by Hooke's law the stress at that instant is $\sigma(x) = Y\,(x/L)$ , so the internal (and, at each slow quasi-static instant, the applied) force is $F(x) = \sigma(x)\,A = \dfrac{YA}{L}\,x$ . This is a force that grows linearly with extension — the wire behaves exactly like a spring of effective stiffness $k = YA/L$ .

Step 2. Compute the work done for a linearly growing force. Because $F(x)$ rises in a straight line from $F(0) = 0$ to $F(\Delta L) = \dfrac{YA}{L}\,\Delta L$ , the work done stretching the wire from $0$ to $\Delta L$ is the area of the force-versus-extension triangle, exactly as for an ordinary spring: $W = \tfrac12 \times \text{base} \times \text{height} = \tfrac12\,\Delta L \times F(\Delta L) = \tfrac12\,\Delta L \times \dfrac{YA}{L}\,\Delta L = \dfrac{YA\,(\Delta L)^2}{2L}$ .

Step 3. Convert total work to energy density. This work is entirely stored as elastic potential energy $U$ (no energy is dissipated for an ideal elastic deformation), so $U = \dfrac{YA(\Delta L)^2}{2L}$ . Dividing by the wire's volume $V = AL$ gives the energy stored per unit volume: $u = \dfrac{U}{AL} = \dfrac{YA(\Delta L)^2}{2L \cdot AL} = \dfrac{Y}{2}\left(\dfrac{\Delta L}{L}\right)^2 = \tfrac12\,Y\varepsilon^2$ , using $\varepsilon = \Delta L/L$ .

Step 4. Rewrite in terms of stress alone. Since the final stress is $\sigma = Y\varepsilon$ , we can substitute $Y = \sigma/\varepsilon$ into the boxed result to get the equally useful form $u = \tfrac12\left(\dfrac{\sigma}{\varepsilon}\right)\varepsilon^2 = \tfrac12\,\sigma\varepsilon$ . Both forms are the same physical quantity — the shaded triangular area under the straight-line stress-strain graph up to the working point — and either can be used depending on which of $\sigma$ , $\varepsilon$ , $Y$ you are given in a problem. Result: $u = \tfrac12\sigma\varepsilon = \tfrac12 Y\varepsilon^2$ , and the total stored energy in a wire of volume $V = AL$ is $U = u \times AL$ .

Worked examples

Board level — stress, strain, and stored elastic energy in a stretched steel wire

A steel wire of original length $L = 2.5\ \text{m}$ and cross-sectional area $A = 1.5\ \text{mm}^2$ is stretched by a force of $F = 60\ \text{N}$ , producing an elongation $\Delta L = 0.6\ \text{mm}$ . Find the longitudinal stress, the longitudinal strain, and the elastic potential energy stored in the wire.

Step 1. Convert every given quantity to SI units before touching a formula. $A = 1.5\ \text{mm}^2 = 1.5\times10^{-6}\ \text{m}^2$ , and $\Delta L = 0.6\ \text{mm} = 0.6\times10^{-3}\ \text{m}$ .

Step 2. Compute the longitudinal stress directly from its definition. $\sigma = \dfrac{F}{A} = \dfrac{60}{1.5\times10^{-6}} = 4.0\times10^{7}\ \text{Pa}$ , that is, $40\ \text{MPa}$ .

Step 3. Compute the longitudinal strain. $\varepsilon = \dfrac{\Delta L}{L} = \dfrac{0.6\times10^{-3}}{2.5} = 2.4\times10^{-4}$ , a dimensionless number — the wire has stretched by only $0.024\%$ of its length.

Step 4. Find the elastic potential energy stored, using the energy-density result derived above, $u = \tfrac12\sigma\varepsilon$ , and then multiplying by the wire's volume. $u = \tfrac12 \times (4.0\times10^{7}) \times (2.4\times10^{-4}) = 4800\ \text{J/m}^3$ . The wire's volume is $V = AL = (1.5\times10^{-6})(2.5) = 3.75\times10^{-6}\ \text{m}^3$ , so the total stored energy is $U = u \times V = 4800 \times 3.75\times10^{-6} = 1.8\times10^{-2}\ \text{J}$ .

Step 5. Sanity-check with the elementary $U = \tfrac12 F\,\Delta L$ formula for a force that has grown linearly to its final value $F$ : $U = \tfrac12 \times 60 \times 0.6\times10^{-3} = 1.8\times10^{-2}\ \text{J}$ — an exact match, confirming the derivation. Answer: $\sigma = 4.0\times10^{7}\ \text{Pa}$ , $\varepsilon = 2.4\times10^{-4}$ , and $U = 1.8\times10^{-2}\ \text{J}$ (about $18\ \text{mJ}$ ) of elastic energy is stored in the stretched wire.

Competitive level — volumetric stress, strain, bulk modulus, and stored energy for a submerged cube

A metal cube of side $10\ \text{cm}$ is lowered into the sea to a depth where the extra pressure is $2.0\times10^{6}\ \text{Pa}$ . This causes its volume to decrease by $90\ \text{mm}^3$ . Find the volumetric stress, the volumetric strain, the bulk modulus of the metal, and the elastic potential energy stored per unit volume of the cube.

Step 1. Find the original volume of the cube. $V = (0.10\ \text{m})^3 = 1.0\times10^{-3}\ \text{m}^3$ .

Step 2. Identify the volumetric stress. The extra water pressure acts uniformly and perpendicularly on every face of the cube, so the volumetric stress is simply that pressure: $\sigma_V = 2.0\times10^{6}\ \text{Pa}$ .

Step 3. Convert the volume change to SI and compute the volumetric strain. $\Delta V = 90\ \text{mm}^3 = 90\times10^{-9}\ \text{m}^3 = 9.0\times10^{-8}\ \text{m}^3$ , so $\varepsilon_V = \dfrac{\Delta V}{V} = \dfrac{9.0\times10^{-8}}{1.0\times10^{-3}} = 9.0\times10^{-5}$ .

Step 4. Compute the bulk modulus, defined (in scope for this chapter) as $B = -\dfrac{\Delta P}{\Delta V/V}$ ; using only magnitudes since we already know the volume shrinks under the extra pressure, $B = \dfrac{\sigma_V}{\varepsilon_V} = \dfrac{2.0\times10^{6}}{9.0\times10^{-5}} = 2.22\times10^{10}\ \text{Pa}$ , about $22.2\ \text{GPa}$ — a large number, consistent with how rigid solids are against volume change compared with how easily they stretch.

Step 5. Compute the elastic energy density stored, using $u = \tfrac12\sigma_V\varepsilon_V$ (the same energy relation as before, now applied to volumetric rather than longitudinal deformation): $u = \tfrac12 \times (2.0\times10^{6}) \times (9.0\times10^{-5}) = 90\ \text{J/m}^3$ , so the total energy stored in the whole cube is $U = u \times V = 90 \times 1.0\times10^{-3} = 9.0\times10^{-2}\ \text{J}$ . Answer: $\sigma_V = 2.0\times10^{6}\ \text{Pa}$ , $\varepsilon_V = 9.0\times10^{-5}$ , $B \approx 2.22\times10^{10}\ \text{Pa}$ , and about $90\ \text{mJ}$ of elastic energy is stored in the compressed cube. Timing note: aim for under three minutes. The chain is always the same regardless of which stress family the question uses — identify $\sigma$ and $\varepsilon$ from the given data, divide one by the other for a modulus if asked, and multiply $\tfrac12\sigma\varepsilon$ by the original volume for stored energy. The one place marks are lost under time pressure is unit slips: keep every length in metres and every volume in cubic metres before you touch a ratio, since mixing millimetres and metres inside the same fraction silently changes the answer by a power of ten or a million.

Common traps

Reporting strain with units, such as writing $0.00024\ \text{m/m}$ instead of the pure number $2.4\times10^{-4}$ , is one of the most common mark-losing habits — strain is always a dimensionless ratio, full stop. A closely related error is confusing stress with force outright: two wires under the identical applied force but different cross-sectional areas experience different stress, so "the force is the same" is never a valid reason to claim "the stress is the same." Students also frequently forget that shear strain is an angle, expressed in radians in the small-angle approximation $\tan\theta \approx \theta$ , and mistakenly try to compute it as a length ratio like longitudinal strain. Another frequent slip is placing the changed length, volume, or area in the denominator of a strain formula instead of the original one — strain always divides by the undeformed dimension, never the deformed one. Finally, when computing elastic energy, students sometimes use the final stress or strain value together with the wrong volume (for example, the deformed rather than the original volume), which introduces small but avoidable errors; for the strains typical of this chapter, always use the original, undeformed volume.

Common trapStrain is never a length, a volume, or a force — it is always a pure ratio, and using the deformed value instead of the original value in its denominator changes the answer

The single biggest misconception in this topic is treating strain as if it were "how much the thing changed" in absolute terms — a length in millimetres, a volume in cubic millimetres — rather than what it actually is: the fraction of the original dimension that the change represents. A $0.6\ \text{mm}$ stretch means very different things for a $2.5\ \text{m}$ wire (a tiny strain of $2.4\times10^{-4}$ ) and for a $6\ \text{mm}$ paperclip wire (a huge, likely plastic, strain of $0.1$ ). Always divide the change by the original length, volume, or reference dimension — never by the changed value, and never leave the ratio un-simplified as a fraction with units still attached. Getting this wrong silently propagates into every later calculation that uses strain, including Young's modulus, the bulk modulus, and the elastic energy density formulas met in this article.

You should be able to…

State the definitions of elasticity and plasticity, with an everyday example of each (a rubber band versus a bent, unrecoverable paperclip).
Write the formula for stress, name its SI unit, and compute it correctly from given force and cross-sectional area data.
Distinguish normal (longitudinal), volumetric, and shearing stress and strain, and identify which type applies to a described physical situation — a stretched cable, a submerged ball, a sheared block of jelly.
Explain, in one sentence, why strain has no units while stress does.
Derive, from the definition of work done against a linearly growing restoring force, the elastic energy density $u = \tfrac12\sigma\varepsilon = \tfrac12 Y\varepsilon^2$ , and use it to find the total elastic potential energy stored in a stretched or compressed body.
Sketch, in words, the shape of a stress-strain curve — elastic region, elastic limit, plastic region, fracture point — even before meeting the numerical moduli that describe its slope.

Ask the tutor: Not sure whether a given situation involves normal, volumetric, or shear stress — or how to picture the elastic energy stored in a stretched cable? Ask the AI tutor to walk through a few real-world examples, such as a suspension bridge cable, an inflated balloon, and a deck of cards being sheared, one at a time with you.

Why does stress have units but strain does not, when both come from the same stretched wire?

Stress is a force divided by an area, so it inherits units of newtons per square metre, which we call the pascal. Strain, on the other hand, is a change in length divided by an original length (or a change in volume divided by an original volume, or an angle) — a length divided by a length, or a volume divided by a volume. Whatever units the top and bottom carry, they are identical and cancel exactly, leaving a pure number. That is why a modulus like Young's modulus, which is stress divided by strain, ends up carrying the same units as stress alone.

If a wire returns to its original length after I remove the load, does that mean no energy was involved at all?

No — energy was very much involved, it just came back to you (mostly) rather than being lost. While you stretched the wire, you did real work against its internal restoring force, and that work was stored as elastic potential energy in the deformed atomic bonds, exactly like winding up a spring. When you release the load, that stored energy is returned as the wire snaps back, doing work to restore its own shape. Only a small fraction is lost as heat due to internal friction between atomic layers; for an ideal elastic solid within its elastic limit, essentially all of it is recoverable.

How is volumetric stress different from the normal stress on a stretched wire, since both are still just force over area?

The formula looks identical, force over area, but the geometry is completely different. Normal stress on a wire comes from a force applied at the two ends, along the length, so the wire elongates in one direction. Volumetric stress comes from a force such as fluid pressure pushing inward on every face of a body simultaneously and uniformly, so the body shrinks in every direction at once without changing its overall shape. That is why normal stress pairs with longitudinal strain (a length ratio) while volumetric stress pairs with volume strain (a volume ratio) — the type of strain produced always matches the geometry of the stress that caused it.

Is the stress-strain curve, with its elastic limit and yield point, something I need to calculate numbers from in this chapter?

Not in this particular article — here you only need to recognise the shape of the curve conceptually: a straight elastic region where the material obeys Hooke's law and fully recovers, an elastic limit beyond which permanent deformation begins, a plastic region, and finally a fracture point. The numerical slope of that initial straight line is Young's modulus, and working with it quantitatively is exactly what the next article in this chapter, on Hooke's law and the elastic moduli, builds on top of the definitions you have just learned here.