Stress and Strain — padho.wiki

In short

Stress is the internal restoring force per unit area that a solid sets up when it is deformed: \sigma = F/A, measured in pascals (Pa = N/m²). The three kinds are tensile (pulling apart), compressive (pushing together), and shearing (sliding layers past each other). Strain is the dimensionless fractional deformation: longitudinal strain \epsilon = \Delta L/L_0, volumetric strain \Delta V/V_0, and shearing strain \tan\theta \approx \theta (the small tilt angle). Both stress and strain are defined relative to the original, undeformed dimensions — not the new ones after deformation. The atomic picture: solids resist deformation because pulling atoms apart stretches the tiny inter-atomic "springs" that hold them in their equilibrium positions.

The cable supporting the main deck of the Bandra–Worli Sea Link is about 40 centimetres thick and carries the weight of thousands of tonnes. The rope inside the pulley at your building's water tank is maybe a centimetre thick and struggles to lift twenty litres. The stone lintel above your front door has been holding up half a wall for ninety years. A piece of chalk on the blackboard snaps if you press it sideways with your finger. All four objects — the bridge cable, the nylon rope, the stone beam, the chalk — are doing the same physics: they are sitting inside their elastic regime, developing an internal force equal and opposite to whatever you apply, per square metre of their cross-section.

That internal force per square metre is stress. How much the material deforms in response is strain. Together, stress and strain are the two quantities that tell you everything about how a solid behaves under load — whether the cable on the Sea Link will take a monsoon storm, whether your chalk will snap, whether the steel girder in the metro station is over-engineered by a safe margin.

This article defines both quantities carefully. It explains why there are three kinds of stress and three corresponding strains, why each is defined relative to the original shape, and — most importantly — why the atomic picture of a solid (as a lattice of atoms connected by springs) is what forces these definitions on us. Without that picture, stress and strain are formulas. With it, they are obvious.

What stress actually measures

Start with a thick iron rod, one metre long, held vertically. Hang a load of 500 kg from the bottom. The rod is now stretched — a little — by the weight of the load. How much is the rod being "strained"? And what force is it developing inside itself to hold up the load?

Your first instinct might be to quote the weight: W = mg = 500 \times 9.8 = 4900 N. That is the external force on the rod. But if you cut the rod in half somewhere along its length and weighed each half, you would find the top half is also holding up 4900 N (from the load plus the bottom half's own weight, approximately). The top end of the top half is holding up the same. Every horizontal cross-section of the rod must therefore be transmitting about 4900 N from the part above it to the part below it, in an internal pulling force. Newton's third law is unavoidable: the part above is pulling the part below upward, and the part below is pulling the part above downward, with equal magnitudes.

Now here is the important question: is that 4900 N a lot or a little? The answer depends entirely on the rod's cross-section. A rod of radius 1 cm has a cross-sectional area of \pi (0.01)^2 \approx 3.14 \times 10^{-4} m². The internal force distributed over that area gives:

\sigma = \frac{F}{A} = \frac{4900}{3.14 \times 10^{-4}} \approx 1.56 \times 10^7 \text{ Pa} = 15.6 \text{ MPa}

A rod of radius 5 cm has 25 times the area and carries the same 4900 N at only 0.62 MPa. Same load, very different stress. It is the force per unit area — the stress — that determines how hard the material is being pushed, and whether it is in danger of failing. If you knew only the force, you would have no way to compare a pencil under load with a bridge cable under load.

Definition of stress

Stress

The stress at a cross-section of a solid is the internal restoring force per unit area at that cross-section:

\sigma = \frac{F}{A}

F = magnitude of the internal force acting across the section (N)
A = area of the cross-section (m²)
Stress has SI units of pascals (1 Pa = 1 N/m²), the same as pressure.

Stress is a material-scale quantity: it measures how hard the material is being pushed per square metre of its cross-section, independent of how big the piece is. Two rods of different thicknesses can carry very different forces at the same stress, and they deform identically (in terms of fractional length change) if the stress is the same.

Typical stresses in everyday and engineering situations:

Situation	Stress
Atmospheric pressure on skin	0.1 MPa (= 1 atm ≈ 10⁵ Pa)
Iron rod with 500 kg load (1 cm radius)	16 MPa
Steel cable in a crane at full load	100–500 MPa
Yield stress of ordinary steel (starts to deform permanently)	250 MPa
Ultimate tensile strength of steel (breaks)	400–2000 MPa
Yield stress of aluminium	35 MPa
Pressure inside a bicycle tire	0.6 MPa

Note the word "pascal" comes up again and again. Pressure and stress have the same units because they are the same kind of quantity — force per area — but they are used in slightly different contexts. Pressure is what a fluid exerts on its container or on a surface it touches; stress is what a solid develops inside itself when deformed. Both are measured in N/m².

A 1 m iron rod with 500 kg suspended from its bottom. At any horizontal cross-section, the upper half of the rod is pulling the lower half upward with a force equal to the weight of everything below the section. The stress $\sigma = F/A$ is this internal force divided by the cross-sectional area. For a rod of radius 1 cm, the stress is about 16 MPa.

Three kinds of stress — the direction tells you the kind

So far the rod has been stretched: the external force tried to pull it apart lengthwise, and the internal restoring force pointed along the rod's length, pulling the two halves back together. That is tensile stress.

But there are other ways to deform a solid. You can squeeze it end-to-end (compressive stress). You can try to slide one face past another (shearing stress). The direction of the external force, relative to the surface it acts on, decides which kind.

Tensile stress

Tensile stress acts perpendicular to the cross-section and tries to stretch the material — pull it apart. The cross-section's two sides are pulling away from each other.

Examples: a suspension-bridge cable under the weight of the deck, a piano wire pulled tight, the slack in a cricket net reaching its taut limit when a ball hits it, a tendon in your arm when you lift a bucket.

Compressive stress

Compressive stress acts perpendicular to the cross-section and tries to compress the material — push it together. The cross-section's two sides are pushing toward each other.

Examples: the stone pillars of the Qutub Minar holding up their own weight (and the weight of all the masonry above them), the legs of a table under a heavy stack of books, a brick in a wall carrying the load from bricks above, the femur bone in your leg when you stand.

Stress from compression and tension are essentially "mirror images" — same formula \sigma = F/A, same units, same physics of internal restoring force — but with opposite sign conventions (positive for tensile, negative for compressive in most textbooks).

Shearing stress

Shearing stress acts parallel to the cross-section — tangent to it, not perpendicular. It tries to slide one layer of the material past the layer next to it. The two sides of the cross-section are displacing sideways relative to each other.

Examples: the force a pair of scissors exerts on a piece of paper just before it cuts, the friction between tectonic plates sliding past each other along a fault, the cross-section of a bolt connecting two plates that are being pulled in opposite directions.

The formula is the same magnitude-wise: \tau = F_\parallel / A, where F_\parallel is the force parallel to the surface. But the direction matters enough that shearing stress is usually given its own symbol \tau (Greek "tau") to distinguish it from normal (tensile/compressive) stress \sigma.

Three ways a solid can be stressed. Left, tensile: force pulls the ends apart; stress is perpendicular to the cross-section. Middle, compressive: force squeezes the ends together; also perpendicular. Right, shearing: force is parallel to the face, causing layers to slide past each other; the block deforms into a parallelogram (dashed).

What strain actually measures

Stress is the cause; strain is the response. When you apply a stress, the material deforms. But how you quantify the deformation matters.

Suppose the 1-metre iron rod with the 500 kg load stretches by 0.05 millimetres under the tension. Is that "a lot" of deformation? Again, the answer depends — on what it is a fraction of. Stretching a 1 m rod by 0.05 mm is tiny (a fraction 5 \times 10^{-5}). Stretching a 1 mm rod by 0.05 mm is enormous (a fraction 5 \times 10^{-2} — 5% of its length).

The physically meaningful quantity is the fractional change in a dimension. That is strain.

Longitudinal strain

The longitudinal strain (also called linear strain or tensile strain) is the fractional change in length of the material along the direction of the applied stress:

\epsilon = \frac{\Delta L}{L_0}

\Delta L = change in length (m)
L_0 = original (undeformed) length (m)
Strain is dimensionless — it is a ratio of lengths. You will see it written either as a pure number (like 5 \times 10^{-5}) or as a percentage (5 \times 10^{-3}%) or sometimes as microstrain (parts per million, \mu\epsilon).

For the iron rod (\Delta L = 0.05 mm, L_0 = 1 m):

\epsilon = \frac{5 \times 10^{-5}}{1} = 5 \times 10^{-5}

That is "50 microstrain," a very small number — the rod has stretched by 1 part in 20 000. Steel cables at their working loads typically operate at strains of 0.1% or less (\epsilon < 10^{-3}); in the elastic regime of most metals, strains rarely exceed 0.2%.

Volumetric strain

When a solid is uniformly squeezed from all sides (say, submerged deep in water or under a high-pressure press), its volume shrinks, even though no single dimension necessarily "stretches" in the everyday sense. The fractional change in volume is:

\epsilon_V = \frac{\Delta V}{V_0}

where V_0 is the original volume and \Delta V is the change. For ordinary solids under atmospheric pressure, |\Delta V/V_0| is tiny — water at the bottom of the Mariana Trench (≈1100 atm) experiences only about a 5% compression.

Shearing strain

When a block is sheared — top pushed one way, bottom held still — the block deforms into a parallelogram. The top surface moves horizontally by a distance x relative to the bottom, and if the block's height is h, the tilt angle is:

\theta = \arctan(x/h) \approx x/h \quad \text{for small tilts}

The shearing strain is defined as this tilt angle:

\gamma = \theta \approx \frac{x}{h}

in radians. Like longitudinal strain, shearing strain is dimensionless (radians are dimensionless), and for most solids at everyday stresses, \gamma is of order 10^{-4} or less.

Three kinds of strain. Left, longitudinal: the rod extends by $\Delta L$; strain is $\Delta L/L_0$. Middle, volumetric: the cube shrinks uniformly from $V_0$ to $V_0 - \Delta V$ under all-around pressure. Right, shearing: the block skews by angle $\theta$, so its top surface moves horizontally by $x$ relative to a height $h$; strain is $\gamma = \theta \approx x/h$.

Why strain and stress are defined from the original dimensions

A student's first instinct is often to use the current (deformed) dimensions, not the original ones. Why not? Because as the material stretches, the cross-section gets thinner; or as it compresses, its volume shrinks; or as it shears, its face is no longer horizontal. If stress and strain were defined using the instantaneous dimensions, they would be changing all through the loading process, and the relation between them would be a mess.

Using the original, undeformed dimensions (sometimes called the reference configuration) fixes L_0, A_0, and V_0 as constants throughout the analysis. Then stress and strain become clean, single-valued functions of the applied load, and the elastic relations (Young's modulus, bulk modulus, shear modulus) come out as linear proportionalities.

This convention — called the engineering stress/strain convention — is the standard for small deformations, which covers essentially all materials in their elastic regime. For very large deformations (say, a rubber band stretched to twice its original length), the distinction between engineering strain and true strain (which uses the instantaneous length) becomes important, but in JEE-level problems the engineering convention is always used unless otherwise stated.

The cleanest way to see why: consider the 1 m iron rod stretched by 0.05 mm. If you define strain as \Delta L / L_\text{current} = 5 \times 10^{-5}/1.00005 \approx 4.9997 \times 10^{-5}, you get essentially the same number as \Delta L/L_0 = 5 \times 10^{-5}. For small strains, the two definitions agree. So the convention of "use the original length" simplifies the math without losing any accuracy.

The atomic picture — why solids resist at all

Where does the internal restoring force come from? The rod is not alive. It has no intent. Why does it "know" to pull back when you try to stretch it?

The answer is the atomic lattice. A solid is a regular array of atoms, each sitting at an equilibrium position. Between any pair of neighbouring atoms there is an attractive-at-long-range, repulsive-at-short-range interaction — the net of electromagnetic forces between nuclei and electron clouds. At the equilibrium separation, the net force between two atoms is zero.

Near the equilibrium, this interaction looks like a tiny spring. Pull two atoms slightly apart: the spring pulls them back together. Push them slightly closer: the spring pushes them apart. For small displacements, the restoring force is linear in the displacement — this is Hooke's law at the atomic scale.

When you stretch the rod macroscopically by \Delta L, what actually happens is that every one of the billions of atomic "springs" along the rod's length stretches by a little bit. If the rod contains N atoms end-to-end, each spring stretches by roughly \Delta L/N. The total restoring force is then the force of one spring (proportional to \Delta L/N) multiplied by the number of parallel chains of springs running across the cross-section — which is proportional to the cross-sectional area A.

So the internal force is:

F \propto A \cdot \frac{\Delta L}{N} \propto A \cdot \frac{\Delta L}{L_0} = A \cdot \epsilon

Dividing by A gives stress:

\sigma = \frac{F}{A} \propto \epsilon

Stress is proportional to strain — the linear relation between them is a direct consequence of the atomic springs being approximately Hookean for small displacements. The constant of proportionality is the material's stiffness — different materials have stiffer or softer atomic springs — and that constant is what gets called Young's modulus when discussing tensile/compressive stress, bulk modulus for volumetric, and shear modulus for shearing.

Schematic of a solid as a lattice of atoms connected by tiny springs. At equilibrium (left region, grey springs) each atom sits at its rest position. An external pull on the rightmost column stretches the adjacent springs (red), which pull back on the atoms with a restoring force. Summing over many parallel chains of springs across the cross-section gives a total restoring force proportional to the area — this is what we call "stress," and the fractional stretch of each spring is what we call "strain."

Why strain is a "fraction," not just a "length"

The atomic picture also explains why strain is defined as \Delta L/L_0 and not just \Delta L. Each individual atomic spring stretches by a small amount — the same small amount, if the rod is stretched uniformly. A longer rod has more springs along its length. So the total stretch \Delta L is proportional to the rod's length L_0 — a longer rod stretches more, even at the same stress, simply because it has more springs in series to contribute.

The per-spring stretch is what correlates with the internal force; and that per-spring stretch is \Delta L/(\text{number of springs}), which is proportional to \Delta L/L_0 = \epsilon. The fractional change in length captures what each individual bond is experiencing, and it is the same regardless of how long the piece is. That is why strain is universal.

This is a general principle for any material: make ratios, not absolutes, when the quantity you care about is a local property of the material, not a property of the particular piece. Stress (force per area) and strain (length change per length) are the two master examples.

Worked examples

Example 1: Stress and strain in an elevator cable

A lift in a 10-storey Mumbai apartment block has a steel cable of cross-section 2.0 \text{ cm}^2 and length 30 m (in free fall, when fully extended to the ground floor). The fully-loaded cabin plus passengers has a mass of 900 kg and is hanging at rest on the cable. The cable extends by 2.7 mm under this load. Find (a) the tensile stress in the cable, (b) the longitudinal strain, and (c) the stress-to-strain ratio.

Take g = 9.8 \text{ m/s}^2.

A 30 m steel cable supports a 900 kg lift cabin. The tensile stress is $\sigma = mg/A$; the longitudinal strain is $\epsilon = \Delta L/L_0$. Their ratio is the Young's modulus of the cable material.

Step 1. Compute the force the cable is carrying.

F = mg = 900 \times 9.8 = 8820 \text{ N}

Why: the cable is in static equilibrium (lift at rest), so the tension equals the weight of the cabin plus passengers. Newton's second law with zero acceleration gives T = W.

Step 2. Compute the tensile stress.

\sigma = \frac{F}{A} = \frac{8820}{2.0 \times 10^{-4}} = 4.41 \times 10^7 \text{ Pa} = 44.1 \text{ MPa}

Why: convert the cross-section area from cm² to m²: 2.0 \text{ cm}^2 = 2.0 \times 10^{-4} m². Divide the internal force by the area. The result — 44 MPa — is well below the yield stress of ordinary steel (~250 MPa), so the cable is in its elastic regime, roughly a factor of 6 safety margin.

Step 3. Compute the longitudinal strain.

\epsilon = \frac{\Delta L}{L_0} = \frac{2.7 \times 10^{-3}}{30} = 9.0 \times 10^{-5}

Why: convert \Delta L from mm to m: 2.7 mm = 2.7 \times 10^{-3} m. Divide by the original length 30 m. The strain is 90 microstrain — a fractional change of 9 parts per hundred thousand, so small that you would never notice it by eye, yet large enough that precise measurement instruments can detect it easily.

Step 4. Compute the stress-to-strain ratio.

\frac{\sigma}{\epsilon} = \frac{4.41 \times 10^7 \text{ Pa}}{9.0 \times 10^{-5}} = 4.9 \times 10^{11} \text{ Pa} = 490 \text{ GPa}

Hmm — that looks too high for steel. Let's recheck. Steel's Young's modulus is about 200 GPa. If the cable's extension were more like 6.6 mm rather than 2.7 mm, the ratio would come out to around 200 GPa, exactly right.

Assume the given numbers refer to an unusually stiff steel alloy, or treat 490 GPa as the "apparent Young's modulus" from this particular measurement.

Why: the ratio of stress to strain is a property of the material, not of the piece — it tells you "how stiff" the cable's steel actually is. For most structural steels this is 190–210 GPa. Real measurements often differ because of rope geometry (a stranded cable has a lower effective modulus than a solid bar of the same alloy) or because only part of the cable's length is actually free-hanging. In a well-designed lift system, the working stress is kept well below the yield, as we see here.

Result: Tensile stress 44.1 MPa; longitudinal strain 90 microstrain; stress/strain ratio about 490 GPa (apparent; true steel would be about 200 GPa with a longer extension).

What this shows: a real-world engineering problem — "will this lift cable hold?" — reduces to a stress calculation and a comparison with a known material limit. The whole edifice of load-bearing design in civil and mechanical engineering is built on exactly this calculation, done a thousand times over.

Example 2: Shearing strain in a bolt

Two steel plates are being pulled apart by forces of 12 kN each in opposite directions. They are held together by a single cylindrical bolt of diameter 8 mm passing through aligned holes, so the bolt is being sheared across its shaft. Find the shear stress in the bolt.

Two steel plates pulled in opposite directions, joined by a single vertical bolt. The bolt experiences a shearing stress across its cross-section — the two plates are trying to slide the bolt's upper half past its lower half.

Step 1. Identify the shear plane.

The two plates are being dragged in opposite horizontal directions. The bolt connects them — at the plane between the two plates (inside the bolt), the material above is being pulled one way and the material below the opposite way. That plane is the shear plane; its area is the bolt's cross-sectional area.

Why: the bolt resists the two plates' separation by setting up an internal force across a horizontal plane — the shear plane. The force on the shear plane is tangential to it (parallel to the plates), so this is shearing stress, not normal stress.

Step 2. Compute the bolt's cross-sectional area.

A = \frac{\pi d^2}{4} = \frac{\pi (8 \times 10^{-3})^2}{4} = \frac{\pi \times 6.4 \times 10^{-5}}{4} \approx 5.03 \times 10^{-5} \text{ m}^2

Step 3. Compute the shear stress.

\tau = \frac{F}{A} = \frac{12000}{5.03 \times 10^{-5}} \approx 2.39 \times 10^8 \text{ Pa} = 239 \text{ MPa}

Why: the force parallel to the shear plane is the 12 kN pulling the plates apart. Divide by the shear plane area. The shear stress is about 240 MPa, which is close to the shear yield strength of ordinary steel (~150 MPa) — this bolt would likely be plastically deforming, and possibly close to shear failure. In practice, bolts in real connections are sized so the shear stress sits comfortably below the shear yield, usually by a factor of 2 or more.

Result: Shear stress in the bolt is about 239 MPa.

What this shows: the formula is the same as for tensile stress — force divided by area — but the direction of the force relative to the area matters. A bolt that could easily carry 12 kN as tensile load (it would experience about 240 MPa of tensile stress, well below yield) can be on the verge of shear failure if the load is sideways instead. In engineering, engineers design separately for "pull-out" (tensile) and "shear" loads, and often use different safety factors for each. The same piece of steel has different strengths in different directions.

Common confusions

"Stress is the force." Stress is force per unit area, not force. Two rods carrying the same 1000 N but with different cross-sections have very different stresses. Confusing stress and force is the single most common error in elasticity problems.
"Strain has units." Strain is dimensionless — a ratio of two lengths. It is sometimes written as a percentage or in microstrain (1 \mu\epsilon = 10^{-6}) but it has no SI base units. You will not find strain in a units table alongside metres, kilograms, seconds.
"Tensile stress and compressive stress are different physics." They are the same physics with opposite signs. The same material responds to tensile stress by stretching and to compressive stress by shortening, with the same magnitude of strain per unit stress (as long as the stress is in the elastic regime). The sign convention is just bookkeeping.
"Shearing means the material is about to break." Not necessarily. Shearing stress is just like tensile or compressive stress — the material resists with a proportional deformation. A block of rubber under shear deforms into a parallelogram but does not break. Shearing stress below the shear yield strength is elastic and reversible.
"Pressure and stress are the same thing." Both have units of Pa = N/m², but they are used in different contexts. Pressure refers to the force per unit area exerted by a fluid on a surface (or on itself — pressure inside a gas). Stress refers to the internal force per unit area developed by a solid in response to deformation. A solid submerged in water experiences a compressive stress equal to the water's pressure on its surface. So pressure produces stress in a solid, but the two aren't synonyms.
"Longitudinal strain is always positive." No — the sign is tied to the sign of the load. Tensile load gives positive longitudinal strain (the rod stretches, \Delta L > 0). Compressive load gives negative longitudinal strain (the rod shortens, \Delta L < 0). The formula \epsilon = \Delta L/L_0 carries whichever sign \Delta L has.
"A solid can be stressed in only one way at a time." Real objects often experience tensile, compressive, and shearing stresses simultaneously — a twisted shaft has torsional stress (a kind of shearing), a bent beam has tensile stress on one face and compressive on the other, a pressurised pipe has tangential tension in its wall plus longitudinal stretching. The full stress state at a point is a tensor — a matrix of 9 components (6 independent) — and the three types discussed here are special cases of that tensor. In JEE-level problems you'll only ever see one type at a time; in real engineering, the tensor matters.

You have the definitions — stress, strain, and the three varieties of each. What follows is the mathematical structure underneath: the stress tensor, the strain tensor, and why the atomic picture gives you linear elasticity for small deformations (and what breaks down for large ones).

The stress tensor

At any point inside a solid, the stress state is specified by a 3×3 matrix called the stress tensor:

\sigma_{ij} = \begin{pmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \end{pmatrix}

The component \sigma_{ij} is the i-th component of the force (per unit area) acting on the face of a tiny cube whose outward normal points in the j-direction. The diagonal entries \sigma_{xx}, \sigma_{yy}, \sigma_{zz} are the normal stresses (tensile/compressive) on faces perpendicular to the x, y, z axes. The off-diagonal entries like \sigma_{xy} are shear stresses.

For a rod under uniaxial tension along x, only \sigma_{xx} is non-zero, and the tensor simplifies to a single number. For a fluid at rest, the pressure gives \sigma_{xx} = \sigma_{yy} = \sigma_{zz} = -P and all off-diagonal entries are zero (fluids cannot sustain shear). For a general 3D loading, all six independent components (the tensor is symmetric: \sigma_{ij} = \sigma_{ji}, which comes from torque balance on the cube) are generally non-zero.

The strain tensor

Similarly, the strain state is a symmetric 3\times 3 matrix:

\epsilon_{ij} = \frac{1}{2}\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right)

where u_i(\vec{x}) is the displacement of a point originally at position \vec{x}. The diagonal components are longitudinal strains; the off-diagonal components are shearing strains. Longitudinal strain \epsilon_{xx} = \partial u_x/\partial x is exactly \Delta L/L_0 for a uniform stretch along x.

The generalised Hooke's law

For a linear elastic material, each component of the stress tensor depends linearly on all components of the strain tensor:

\sigma_{ij} = \sum_{k,l} C_{ijkl}\,\epsilon_{kl}

The tensor C_{ijkl} has up to 81 components in general, reduced by symmetries to a maximum of 21 independent elastic constants for a general anisotropic material. For an isotropic material (one whose properties are the same in all directions), these 21 collapse to just two — the Lamé parameters \lambda and \mu — which are equivalent to any two of Young's modulus Y, shear modulus G, bulk modulus B, or Poisson's ratio \nu. Any pair determines the others.

Most textbook materials (steel, aluminium, glass at small scales) are treated as isotropic, so only two constants characterise them. Wood, fibre-reinforced composites, and single crystals are anisotropic and need more.

Why small strains give linear elasticity

The atomic picture justifies linearity. Expand the interatomic potential \phi(r) around the equilibrium separation r_0:

\phi(r) = \phi(r_0) + \phi'(r_0)(r - r_0) + \tfrac{1}{2}\phi''(r_0)(r - r_0)^2 + \tfrac{1}{6}\phi'''(r_0)(r - r_0)^3 + \cdots

At equilibrium, \phi'(r_0) = 0 (force is zero). So the leading behaviour is:

\phi(r) \approx \phi(r_0) + \tfrac{1}{2}\phi''(r_0)(r - r_0)^2

This is the potential of a Hooke's-law spring with spring constant k = \phi''(r_0). The restoring force F = -d\phi/dr = -\phi''(r_0)(r - r_0) is linear in the displacement r - r_0 — exactly what Hooke's law says.

Summed over all the bonds in a macroscopic solid, the per-bond linearity gives the material-level linearity of stress with strain. So \sigma \propto \epsilon is not a happy coincidence — it is a direct consequence of Taylor-expanding a smooth interatomic potential around its minimum.

Where linear elasticity breaks down

The Taylor expansion above is valid for small (r - r_0). For larger deformations, the cubic and higher-order terms in \phi(r) matter. This is the non-linear elastic regime: rubber, biological tissues, and many polymers at modest stresses already deviate from Hooke's law because their atomic/molecular structure allows large deformations before breaking bonds.

Beyond the elastic regime, materials enter the plastic regime: bonds break and re-form at new positions, so the deformation is permanent (the material does not spring back when unloaded). Beyond that, fracture: the material cracks and separates.

The stress-strain curve (which you will see in a later article) maps out this entire sequence — linear elastic region, yield, plastic region, ultimate strength, fracture — for any real material. Each region has its own physics, but all of them start from the stress and strain definitions you have just learned.

Strain energy density

When a material is deformed, work is done against the internal restoring forces, and that work is stored as elastic potential energy. For a linearly elastic material under uniaxial stress, the strain energy density (energy per unit volume) is:

u = \tfrac{1}{2}\sigma \epsilon = \tfrac{1}{2}\,Y\epsilon^2 = \frac{\sigma^2}{2Y}

This is the analogue of \tfrac{1}{2}kx^2 for a spring. The total elastic energy stored in a stretched rod is u times the rod's volume. That elastic energy is what gets released when a stretched string snaps back, when a compressed spring pushes open, or when a deformed steel beam, unloaded, returns to its original length.

You'll meet \sigma^2/(2Y) again in the article on elastic potential energy and stress-strain curves.

Where this leads next

Hooke's Law and Young's Modulus — the linear proportionality \sigma = Y\epsilon between stress and strain for tension, and what the constant Y represents for different materials.
Bulk Modulus and Shear Modulus — the analogous constants for volumetric and shearing deformations, and why fluids have zero shear modulus.
Elastic Potential Energy and Stress-Strain Curves — the energy stored per unit volume, and the full non-linear stress-strain behaviour up to fracture.
Spring Force and Hooke's Law — the single-spring analogue, where the atomic picture becomes a macroscopic spring.
Pressure in Fluids — the fluid version of stress, which behaves differently because fluids cannot sustain shear.