Chapter 11 · Properties of Matter · Full book

Elasticity: When Solids Push Back

Core intuition: Chapter Roadmap

A rigid body is a convenient fiction. Every real solid deforms when you push, pull, twist, or squeeze it—and then pushes back. This chapter is about that pushing back: how much a solid yields, how hard it resists, where the energy goes, and when the resistance finally fails. The single organising idea is that the response is set not by force and stretch directly, but by stress and strain—quantities scrubbed of the body's size and shape.

Stress and strain — the size-independent measures of internal force (\(\sigma = F/\mathcal{A}\)) and fractional deformation (\(\varepsilon = \Delta L/L\)) that let one number describe a material rather than a particular specimen.
Young's modulus and the stress–strain curve — the stiffness \(Y = \sigma/\varepsilon\) of axial loading, and the full journey of a material from elastic, through yield and plastic flow, to fracture.
Shear and bulk moduli — the response to tangential stress (\(G\)) and to all-around pressure (\(B\)): the three moduli that between them describe any small deformation.
Elastic potential energy — the work stored in a stretched solid, the energy density \(\tfrac12\sigma\varepsilon\), and the difference between energy stored and work done.
Poisson's ratio and the inter-modulus relations — the lateral contraction that accompanies stretching, the volume change it implies, and the equations that tie \(Y\), \(G\), \(B\), and \(\nu\) together.
Putting elasticity to work — thermal stress in clamped members, hoop stress in spinning rings, composite and tapered bars, and the elastic limits that govern real structures.
Beyond Hooke — where linear elasticity ends: nonlinear, anisotropic, and biological materials whose behaviour the simple modulus picture only begins to capture.

Master stress and strain, and the rest of the chapter is a single idea applied six ways: define a deformation, write down the modulus that resists it, and track where the energy is stored.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Equal Wires: Two wires are made of the same steel, have the same length, and are subjected to the same hanging load. One wire has twice the cross-sectional area of the other. The thicker wire stretches half as much. Why should area matter? The load is identical, the material is identical, the length is identical. What does the cross-section have to do with how much the wire stretches?
The Longer Wire: Take the same wire and double its length, keeping everything else the same—same material, same cross-section, same load. The wire now stretches twice as much. The restoring force has not changed. The material has not changed. Yet the wire is twice as compliant. Why does the length of the wire determine how much it stretches?
The Stress–Strain Curve: Pull a steel wire with increasing force. For a long time, it stretches proportionally—double the force, double the extension. Then, suddenly, it begins to stretch much faster for small increases in force. Then it necks and eventually breaks. A rubber band, pulled in the same way, behaves completely differently: it stretches easily at first, then becomes increasingly stiff as you pull harder. Both are solids. Why do they deform so differently?
The Speed of Sound in Steel: Sound travels through steel at roughly \(5000\,\mathrm{m\,s^{-1}}\)— fifteen times faster than in air. Steel is far denser than air. Intuitively, a heavier medium should be harder to disturb and should carry waves more slowly. Yet the denser solid is dramatically faster. What determines the speed of a wave through a material, and why does steel win despite its weight?

Each question targets a trap that catches students every year.
By the end of this chapter, every one of them will be transparent.

Why Solids Need Their Own Framework

Every chapter of this book has treated objects as either perfectly rigid or as point masses with no internal structure. A block on an incline, a planet in orbit, a rigid body spinning about a fixed axis: in each case we assumed the object did not deform under the forces applied to it.

This assumption is an idealization. Every real solid deforms when forces act on it. A bridge sags under traffic. A bone flexes under the weight of the body. A steel cable stretches when it supports a load. An earthquake compresses and dilates the rock of the Earth's crust.

The question is not whether solids deform—they always do. The question is how much, and what happens when the deformation becomes too large.

Why Force Alone Is Not Enough

The first perplexing question already reveals the central issue. Two wires, same material, same length, same load—yet one stretches twice as much as the other because it has half the cross-sectional area. If we described the situation simply as “a force \(F\) is applied,” we would have no way to distinguish the two wires.

The resolution is to replace force with a quantity that accounts for how the force is distributed over the cross-section: the stress, defined as force per unit area. The thicker wire has twice the area, so the same force produces half the stress. Half the stress produces half the deformation. Area matters because stress—not force—is the physical agent that drives deformation inside the material.

Why Extension Alone Is Not Enough

The second perplexing question reveals an equally important point. A wire twice as long stretches twice as much under the same load. But every small element of the wire stretches by the same amount— the total extension is simply the sum of contributions from twice as many elements. If we described the response simply as “extension \(\Delta L\),” we would conflate a wire that is intrinsically compliant with one that is compliant merely because it is long.

The resolution is to replace extension with a quantity that measures fractional deformation: the strain, defined as extension per unit original length \(\Delta L / L\). The strain is the same in both wires because each unit length contributes equally to the total extension. Length matters because strain—not raw extension—captures the intrinsic response of the material.

The Molecular Picture

Why do solids resist deformation at all? The answer lies in the interatomic force law.

Atoms in a solid sit near the minima of their mutual potential energy curves. When the solid is at its natural (unstressed) configuration, each atom is at the equilibrium separation from its neighbours, where the net interatomic force is zero. When the solid is stretched, atoms are pulled slightly apart. The interatomic force becomes attractive, pulling them back. When compressed, atoms are pushed together, and the force becomes repulsive. The solid resists both stretching and compression by the same mechanism: atoms displaced from equilibrium experience restoring forces directed back toward that equilibrium.

At small displacements, the restoring force is approximately linear in the displacement—exactly as in simple harmonic motion. This is why Hooke's Law (\(F \propto x\)) holds for small deformations. The spring constant of the interatomic bond is an intrinsic property of the chemical bond: strong bonds (diamond, steel) give high stiffness; weak bonds (rubber, biological tissues) give low stiffness.

Core intuition: The Molecular Spring

A solid is a three-dimensional lattice of atoms connected by interatomic bonds that behave, at small displacements, like springs. Macroscopic elasticity is collective Hooke's Law applied to \(\sim 10^{23}\) atomic springs in parallel and in series.

Springs in parallel (more area): stiffer—larger force for same extension. This explains why thicker wires stretch less.
Springs in series (more length): more compliant—same force, more total extension. This explains why longer wires stretch more.

Both observations follow from one picture.

The Elastic Limit: When the Springs Break

Hooke's Law holds only as long as the deformation is small enough that atoms remain near their equilibrium positions and the restoring force remains approximately linear. Beyond a critical deformation, the interatomic bonds begin to slip or break, the response becomes nonlinear, and the material does not return to its original shape when the load is removed. This is the elastic limit.

Below the elastic limit: deformation is proportional to load, and the material fully recovers when the load is removed. Above the elastic limit: permanent (plastic) deformation occurs. The material is permanently changed.

This boundary—and the rich physics of the stress–strain curve between zero load and fracture—is the subject of a later section.

Connections to Earlier Chapters

This chapter extends and completes several threads left open earlier.

Newton's Laws (Ch. 3). Stress is simply internal force per unit area—Newton's third law applied to a cross-section inside the solid. The stress at any cross-section equals the force one part of the solid exerts on the adjacent part divided by the area of the interface.

Work–Energy (Ch. 4). When a wire is stretched, work is done against the interatomic restoring forces. This work is stored as elastic potential energy in the material. The energy density stored in a deformed solid is a central result of this chapter, and it connects directly to the \(U = \frac{1}{2}kx^2\) of a spring.

Rotational Mechanics (Ch. 7). The shear modulus, which we will meet later, governs the twisting of shafts and cylinders—the torsion problems that appear in engineering applications of rotational mechanics.

Waves and Sound. The speed of a longitudinal wave in a solid is \(v = \sqrt{Y/\rho}\), where \(Y\) is the Young's modulus defined earlier and \(\rho\) is the density. In the Waves volume we used this formula without deriving where \(Y\) comes from. This chapter provides that foundation, and with it the answer to the fourth perplexing question: steel is fast not despite its density, but because its stiffness (\(Y \sim 2\times10^{11}\,\mathrm{Pa}\)) outweighs its density by an enormous factor compared with air.

Fluid Mechanics (Ch. 11). The bulk modulus \(B\)—resistance to uniform compression—applies to both solids and fluids. In fluids we wrote \(v_{\text{sound}} = \sqrt{B/\rho}\). In this chapter we place the bulk modulus in its proper family alongside Young's modulus and the shear modulus, and we find the relations connecting them.

What This Chapter Covers

We begin with the two fundamental variables: stress and strain. We then treat the three elastic moduli in turn: Young's modulus for stretching and compression, the shear modulus for twisting and sliding, and the bulk modulus for uniform compression. Energy stored in elastic deformation follows, then Poisson's ratio and the relations connecting all the moduli. Structural applications—composite wires, thermal stress, rotating rings—occupy a later section. The chapter closes with pitfalls, summary, an EXTRA section on nonlinear and biological elasticity, an IKS reflection on Indian metallurgy, and the exercise bank.

✓ Checkpoint · Before We Begin

Two wires \(A\) and \(B\) are made of the same material. Wire \(A\) has length \(L\) and cross-sectional area \(\mathcal{A}\). Wire \(B\) has length \(2L\) and cross-sectional area \(2\mathcal{A}\). The same load is hung from each. Without any formula, predict which wire stretches more, and by what factor.
Wire \(B\) has twice the length (each unit length contributes the same extension, so total extension doubles) but also twice the area (so stress is halved, halving the extension per unit length). The two effects exactly cancel: both wires stretch by the same absolute amount.
A steel wire is loaded beyond its elastic limit and the load is then removed. What do you expect to observe, and why, in terms of the molecular picture?
The wire does not return to its original length. Beyond the elastic limit, atomic planes in the crystal have slipped past each other (plastic deformation), creating a new equilibrium configuration. The interatomic bonds have been permanently rearranged, and the restoring mechanism that would return the wire to its original length no longer operates.
Sound travels at \(\approx 5000\,\mathrm{m\,s^{-1}}\) in steel and at \(\approx 340\,\mathrm{m\,s^{-1}}\) in air. Steel is roughly 8000 times denser than air. What does this imply about the ratio of the elastic stiffness of steel to that of air? (A qualitative estimate is sufficient.)
Since \(v = \sqrt{\text{stiffness}/\rho}\), we have \(\text{stiffness} \propto v^2 \rho\). The ratio of stiffnesses is \((v_\text{steel}/v_\text{air})^2 \times (\rho_\text{steel}/\rho_\text{air}) \approx (5000/340)^2 \times 8000 \approx 216 \times 8000 \approx 1.7\times10^6\). Steel's elastic modulus exceeds air's by a factor of nearly two million—far more than compensating for its greater density.

Stress and Strain: The Right Variables

◈ Stress, strain & Young’s modulus

live bench

Pick a material and hang a load: the wire stretches by \(\Delta L\), and the readouts show stress \(\sigma=F/A\), strain \(\varepsilon=\Delta L/L\), and their ratio \(Y\) — a number that describes the material, not the specimen.

◈ loading the interactive bench…

↗ open in the Simulations page

Fig 12.1 The three independent modes of small deformation. Axial loading is resisted by Young’s modulus \(Y\), a shear by the shear modulus \(G\), and all-around pressure by the bulk modulus \(B\). Between them they describe any elastic strain.

The previous section argued, from first principles, that the natural variables for describing elastic deformation are not force and extension but force-per-unit-area and fractional-extension. This section makes those definitions precise, classifies the different geometries of deformation, and establishes the vocabulary that every subsequent section will use.

Why Not Force and Extension?

Consider two experiments. In the first, a steel wire of cross-sectional area \(\mathcal{A} = 1\,\mathrm{mm^2}\) and length \(L = 1\,\mathrm{m}\) is loaded with \(F = 100\,\mathrm{N}\) and extends by \(\Delta L = 0.5\,\mathrm{mm}\).

In the second experiment, we replace the wire with one of identical material but area \(\mathcal{A}' = 2\,\mathrm{mm^2}\) and length \(L' = 2\,\mathrm{m}\), and apply the same \(F = 100\,\mathrm{N}\).

The extension is now \(\Delta L' = 1\,\mathrm{mm}\)—twice as large. Yet the material is identical. If we report the result as (force \(100\,\mathrm{N}\), extension \(\Delta L'\)), we would conclude the material is twice as compliant. This is wrong. The material's intrinsic response has not changed; only the geometry has.

To extract a material property that is independent of geometry, we must normalise:

normalise force by area \(\to\) stress
normalise extension by original length \(\to\) strain

When expressed in terms of stress and strain, identical materials give identical responses, regardless of geometry. The ratio of stress to strain is a material property—an elastic modulus.

Normal Stress

When a force acts perpendicular to the cross-section of a body, the resulting stress is called a normal stress, denoted \(\sigma\) (sigma):

\[ \sigma = \frac{F_\perp}{\mathcal{A}} \]

where \(F_\perp\) is the component of force perpendicular to the area \(\mathcal{A}\). The SI unit of stress is the pascal: \(1\,\mathrm{Pa} = 1\,\mathrm{N\,m^{-2}}\). Typical elastic stresses in engineering materials range from \(10^6\,\mathrm{Pa}\) (soft rubber near failure) to \(10^9\,\mathrm{Pa}\) (high-strength steel near failure).

Normal stress is tensile when the force tends to pull the cross-section apart (the wire being stretched) and compressive when the force tends to push the faces together (a column bearing a load). By convention, tensile stress is positive and compressive stress is negative, though some engineering texts reverse this convention.

Internal stress. It is essential to understand that stress is an internal quantity. The stress \(\sigma\) at a cross-section is the force that the material on one side of the section exerts on the material on the other side, divided by the area of the section. For a wire in static equilibrium under an external load \(F\), the internal normal stress at every cross-section equals \(F/\mathcal{A}\) by Newton's third law applied to any imaginary cut through the wire. This is the connection to Newton's Laws noted earlier.

Shear Stress

When a force acts parallel to a cross-section (tangent to the surface), it tends to slide one layer of the material over the adjacent layer. The resulting stress is called a shear stress, denoted \(\tau\) (tau):

\[ \tau = \frac{F_\parallel}{\mathcal{A}} \]

where \(F_\parallel\) is the force component parallel to the area \(\mathcal{A}\). The units are identical to normal stress (Pa).

Shear stress appears whenever a body resists sliding, twisting, or bending. A bolt holding two plates together is in shear. A shaft transmitting torque is in shear. A book resting on a table, pushed from the side so that each page slides slightly relative to the next, is in shear.

Longitudinal Strain

When a body of original length \(L\) is stretched by \(\Delta L\), the longitudinal strain (or tensile strain) is:

\[ \varepsilon = \frac{\Delta L}{L} \]

Strain is dimensionless—it is a pure ratio. A strain of \(10^{-3}\) means each metre of the wire has stretched by \(1\,\mathrm{mm}\). This is a typical elastic strain for steel under working loads. Rubber, by contrast, can sustain strains of several hundred percent before failing.

The strain \(\varepsilon\) as defined above is the engineering strain. It is adequate for small deformations and is the standard for all JEE and NEET calculations. For large deformations (rubber, biological tissues) a more careful definition—the logarithmic or true strain—is needed, but this is beyond the scope of this chapter.

Compressive strain. If the body is compressed by \(\Delta L\) (shortening), the engineering strain is \(-\Delta L / L\) by the sign convention that tensile strain is positive. In most JEE problems, sign is handled by context and the magnitude is used directly.

Shear Strain

When a shear stress acts on a body, it deforms by sliding or twisting. The appropriate measure of deformation is the shear strain, defined as the angular deformation \(\phi\) (in radians) of a line that was originally perpendicular to the shearing force:

\[ \gamma = \tan\phi \approx \phi \qquad (\phi \ll 1) \]

Consider a rectangular block whose bottom face is fixed to a surface. A force \(F\) is applied tangentially to the top face of area \(\mathcal{A}\). The top face moves sideways by \(\Delta x\) while the bottom face stays fixed. If the block has height \(h\), then \(\tan\phi = \Delta x / h \approx \phi\) for small deformations. The shear strain \(\gamma\) measures how much the block has been “leaned over,” normalised by its height—analogous to how longitudinal strain normalises extension by original length.

Volumetric Strain

When a body is subjected to uniform pressure \(\Delta P\) on all surfaces, every dimension changes proportionally. The volumetric strain is:

\[ \frac{\Delta V}{V} \]

where \(V\) is the original volume and \(\Delta V\) is the change in volume. For a compression, \(\Delta V \lt 0\) and the volumetric strain is negative. This deformation mode is the subject of a later section.

Core intuition: Three Deformation Modes, Three Moduli

Each type of deformation has its own pair of variables and its own elastic modulus:

Mode	Stress	Strain	Modulus
Stretching / compression	\(\sigma = F_\perp/\mathcal{A}\)	\(\varepsilon = \Delta L/L\)	Young's \(Y\)
Shearing / twisting	\(\tau = F_\parallel/\mathcal{A}\)	\(\gamma \approx \phi\)	Shear \(G\)
Uniform compression	\(-\Delta P\)	\(\Delta V/V\)	Bulk \(B\)

Each modulus is defined as the ratio \(\text{stress}/\text{strain}\) for its mode. Units: Pa in every case.

The Range of Elastic Moduli

The elastic moduli span an enormous range across materials. Diamond, with the stiffest interatomic bonds known, has \(Y \approx 10^{12}\,\mathrm{Pa}\). Steel: \(Y \approx 2\times10^{11}\,\mathrm{Pa}\). Bone: \(Y \approx 10^{10}\,\mathrm{Pa}\). Rubber: \(Y \approx 10^6\)–\(10^7\,\mathrm{Pa}\). Soft tissue: \(Y \sim 10^3\)–\(10^5\,\mathrm{Pa}\).

This six-order-of-magnitude range reflects the enormous diversity of interatomic bond strengths. It is what makes materials science a subject of its own.

Hooke's Law: The Proportionality of Stress and Strain

For any of the three deformation modes, and for sufficiently small deformations, the stress is proportional to the strain. This generalised statement is Hooke's Law:

\[ \text{stress} = \text{modulus} \times \text{strain} \]

Specifically: \[\begin{aligned} \sigma &= Y\,\varepsilon \\ \tau &= G\,\gamma \\ -\Delta P &= B\,\frac{\Delta V}{V} \end{aligned}\]

Each of these is an empirical law that holds within the elastic regime. Each breaks down near and beyond the elastic limit, where the stress–strain relationship becomes nonlinear and irreversible.

The macroscopic Hooke's Law \(F = kx\) for a spring (from an earlier chapter) is a consequence of these material-level relations: combining \(\sigma = Y\varepsilon\) with \(\sigma = F/\mathcal{A}\) and \(\varepsilon = \Delta L / L\) gives \(F = (Y\mathcal{A}/L)\,\Delta L\), so the spring constant of a wire is:

\[ k = \frac{Y\mathcal{A}}{L} \]

This one equation encodes both perplexing questions from the chapter opener: \(k\) increases with area \(\mathcal{A}\) (thicker wire stretches less) and decreases with length \(L\) (longer wire stretches more). Both effects are built into the geometry of the modulus relation.

Example 1 · Stress and strain in a loaded wire

A steel wire of length \(L = 2.0\,\mathrm{m}\) and diameter \(d = 1.0\,\mathrm{mm}\) is loaded with a mass of \(m = 5.0\,\mathrm{kg}\). Take \(Y_\text{steel} = 2.0\times10^{11}\,\mathrm{Pa}\) and \(g = 9.8\,\mathrm{m\,s^{-2}}\). (a) Find the stress in the wire. (b) Find the strain. (c) Find the extension \(\Delta L\). (d) Find the effective spring constant \(k\) of the wire.

Solution. (a) Stress. Cross-sectional area: \[ \mathcal{A} = \pi\left(\frac{d}{2}\right)^2 = \pi\left(\frac{1.0\times10^{-3}}{2}\right)^2 = \frac{\pi}{4}\times10^{-6} \approx 7.854\times10^{-7}\,\mathrm{m^2} \] The load is \(F = mg = 5.0 \times 9.8 = 49\,\mathrm{N}\). \[ \sigma = \frac{F}{\mathcal{A}} = \frac{49}{7.854\times10^{-7}} \approx 6.24\times10^7\,\mathrm{Pa} \approx 62\,\mathrm{MPa} \] (b) Strain. \[ \varepsilon = \frac{\sigma}{Y} = \frac{6.24\times10^7}{2.0\times10^{11}} = 3.12\times10^{-4} \] This means each metre of the wire has stretched by \(0.312\,\mathrm{mm}\). The strain is small (much less than 1), confirming we are well within the elastic regime. (c) Extension. \[ \Delta L = \varepsilon \cdot L = 3.12\times10^{-4} \times 2.0 = 6.24\times10^{-4}\,\mathrm{m} \approx 0.62\,\mathrm{mm} \] (d) Effective spring constant. \[ k = \frac{Y\mathcal{A}}{L} = \frac{2.0\times10^{11} \times 7.854\times10^{-7}}{2.0} \approx 7.85\times10^4\,\mathrm{N\,m^{-1}} \] Verification: \(\Delta L = F/k = 49 / 7.85\times10^4 \approx 6.24\times10^{-4}\,\mathrm{m}\). ✓

✓ Checkpoint · Stress and Strain

A copper wire (\(Y = 1.2\times10^{11}\,\mathrm{Pa}\)) of length \(3.0\,\mathrm{m}\) and cross-sectional area \(2.0\times10^{-6}\,\mathrm{m^2}\) is stretched by a force of \(120\,\mathrm{N}\). Find (a) the stress, (b) the strain, and (c) the extension.
(a) \(\sigma = F/\mathcal{A} = 120/(2.0\times10^{-6}) = 6.0\times10^7\,\mathrm{Pa}\). (b) \(\varepsilon = \sigma/Y = 6.0\times10^7/1.2\times10^{11} = 5.0\times10^{-4}\). (c) \(\Delta L = \varepsilon L = 5.0\times10^{-4}\times3.0 = 1.5\times10^{-3}\,\mathrm{m} = 1.5\,\mathrm{mm}\).
Wire \(P\) has Young's modulus \(Y\), length \(L\), and area \(\mathcal{A}\). Wire \(Q\) has Young's modulus \(2Y\), length \(2L\), and area \(\mathcal{A}/2\). Both are subjected to the same tensile force \(F\). Find the ratio of their extensions \(\Delta L_P / \Delta L_Q\).
\(\Delta L = FL/(Y\mathcal{A})\). \(\Delta L_P = FL/(Y\mathcal{A})\). \(\Delta L_Q = F(2L)/((2Y)(\mathcal{A}/2)) = F(2L)/(Y\mathcal{A}) = 2FL/(Y\mathcal{A})\). Ratio: \(\Delta L_P/\Delta L_Q = 1/2\). Despite being made of a softer material, wire \(Q\) stretches more because its geometry (longer, thinner) overcomes the stiffer modulus.
A block of rubber (\(G = 5\times10^5\,\mathrm{Pa}\)) has a square top face of side \(0.10\,\mathrm{m}\) and height \(h = 0.05\,\mathrm{m}\). A tangential force of \(F = 20\,\mathrm{N}\) is applied to the top face while the bottom is fixed. Find the shear strain \(\gamma\) and the lateral displacement \(\Delta x\) of the top face.
\(\tau = F/\mathcal{A} = 20/(0.10)^2 = 2000\,\mathrm{Pa}\). \(\gamma = \tau/G = 2000/(5\times10^5) = 4.0\times10^{-3}\,\mathrm{rad}\). \(\Delta x = \gamma h = 4.0\times10^{-3} \times 0.05 = 2.0\times10^{-4}\,\mathrm{m} = 0.20\,\mathrm{mm}\).

Young's Modulus: Stretching, Compression, and the Stress–Strain Curve

◈ The stress–strain curve, walked

live bench

Drag the strain past the elastic limit, the yield point and ultimate strength into necking and fracture. Unload and watch the permanent set — the whole life of a ductile material in one trace.

◈ loading the interactive bench…

↗ open in the Simulations page

Fig 12.2 The stress–strain curve of a ductile material. \(O\!-\!A\) is the linear Hookean region (slope \(Y\)); past the elastic limit \(A\) the material yields at \(B\), work-hardens to its ultimate strength \(C\), then necks and fractures at \(D\). Unloading from the plastic region leaves a permanent set.

Young's modulus is the simplest and most widely used elastic modulus. It describes the response of a material to uniaxial stress—a force directed along one axis, either stretching or compressing the material. But the real physics of a material under tension is richer than a single number. The stress–strain curve—the graph of stress against strain as a wire is loaded to failure—reveals the full story of how a solid deforms, yields, and ultimately breaks.

Definition of Young's Modulus

Within the linear elastic regime—where stress and strain are proportional—Young's modulus \(Y\) (also written \(E\) in engineering) is defined as:

\[ \boxed{Y = \frac{\sigma}{\varepsilon} = \frac{F/\mathcal{A}}{\Delta L / L} = \frac{FL}{\mathcal{A}\,\Delta L}} \]

The SI unit is the pascal (Pa). Since strain is dimensionless, \(Y\) has the same units as stress.

The definition contains an important qualifier: “within the linear elastic regime.” \(Y\) is not the ratio of stress to strain at any point on the curve; it is the slope of the stress–strain curve in the region where that curve is a straight line through the origin. Outside this region, \(Y\) is not defined.

The Stress–Strain Curve: A Complete Picture

Loading a ductile material (such as mild steel or copper) to failure, while recording stress and strain, produces the following landmark regions:

O to A: Elastic (Hookean) region. Stress is strictly proportional to strain. The curve is a straight line through the origin with slope \(Y\). If the load is removed at any point in this region, the wire returns exactly to its original length—no permanent deformation. Point \(A\) is the proportionality limit: the last point at which Hooke's Law holds exactly.

A to B: Elastic but nonlinear. The curve begins to deviate from a straight line. Stress and strain are no longer proportional, but if the load is removed, the material still returns to its original dimensions. Point \(B\) is the elastic limit: the last point at which full recovery is possible. For many materials, the difference between \(A\) and \(B\) is small enough that the two are treated as coincident in introductory calculations.

B to C: Yield region (plastic onset). Beyond the elastic limit, permanent (plastic) deformation begins. At the yield point \(C\), the material may elongate noticeably with little or no increase in stress—in some materials the curve actually dips slightly. This dramatic change is caused by the motion of dislocations—defects in the crystal lattice—which allows atomic planes to slide over each other. The deformation is no longer recoverable.

C to D: Strain hardening. After yielding, the material becomes harder to deform further—the curve rises again. This is work hardening: the dislocation density increases so much that dislocations impede each other's motion. The material is stronger in this region than it was at the yield point, but it is permanently deformed.

D: Ultimate tensile strength (UTS). Point \(D\) is the maximum on the stress–strain curve—the ultimate tensile strength. It is the maximum stress the material can sustain. Beyond \(D\), a localised constriction (necking) forms in the wire: the cross-sectional area decreases rapidly in a small region, so the true local stress rises sharply even as the applied force decreases.

D to E: Fracture. The necked region elongates rapidly and the wire breaks at point \(E\). The fracture stress (stress at \(E\)) is lower than the UTS because the cross-section at the neck has decreased. (If stress were computed using the actual, reduced area rather than the original area, the true fracture stress would exceed the UTS.)

Birth of the formula: Reading the Stress–Strain Curve

Three material properties are directly readable from the curve: Young's modulus \(Y\): the slope of the straight portion \(OA\). A steeper initial slope means a stiffer material. Elastic limit: the strain at point \(B\). Materials with a large elastic-limit strain (rubber: hundreds of percent) can be deformed greatly without permanent damage. Materials with a small elastic-limit strain (glass: \(\sim 0.1\%\)) fail with very little warning. Toughness: the area under the entire curve from \(O\) to \(E\). This is the energy per unit volume required to fracture the material— the most important quantity for applications where impact resistance matters (armour, crash structures, bone). A material can be strong (high UTS) but brittle (small area under curve) or weak but tough (large area). Do not confuse:

Strength: maximum stress sustainable.
Stiffness: slope of elastic region (\(Y\)).
Hardness: resistance to localised surface deformation.
Toughness: total energy absorbed before fracture.

These are independent properties. A material can be stiff but brittle (glass), tough but soft (rubber), or strong and tough (high-carbon steel).

Ductile Versus Brittle Materials

Ductile materials (mild steel, copper, aluminium, gold) show a long plastic region between the elastic limit and fracture. They deform substantially before breaking, giving visible warning of impending failure. This is why structural steel is preferred over cast iron in buildings: a steel beam deforms visibly before it fails, whereas a cast-iron beam breaks without warning.

Brittle materials (glass, ceramics, cast iron, high-carbon steel) have virtually no plastic region. The stress–strain curve is almost a straight line up to the fracture point. Brittle materials fail suddenly and without deformation. They are often strong in compression but weak in tension—a key reason why concrete (brittle) is reinforced with steel (ductile) to handle tensile loads.

Elastomers (rubber, elastin) show a completely different curve: low initial stiffness that increases sharply at large strains (the curve bends upward, away from linearity, rather than downward). This arises from an entropic mechanism rather than interatomic bond stretching—the third perplexing question, fully addressed later.

Elastic Moduli of Common Materials

Material	\(Y\) (GPa)	Character
Diamond	\(\sim 1000\)	Stiff, brittle
Steel	\(\sim 200\)	Stiff, ductile
Copper	\(\sim 120\)	Moderately stiff, ductile
Bone (compact)	\(\sim 17\)	Moderately stiff, somewhat tough
Wood (along grain)	\(\sim 10\)	Anisotropic, tough
Rubber	\(0.001\)–\(0.01\)	Very soft, very tough

Wave Speed in a Solid: Resolving the Fourth Perplexing Question

In the Waves volume, the speed of a longitudinal wave in a solid rod was stated to be \(v = \sqrt{Y/\rho}\) without derivation. We can now make this formula fully transparent.

A longitudinal wave in a rod is a travelling pattern of compressions and rarefactions. Consider a thin slice of the rod of thickness \(\Delta x\) and cross-sectional area \(\mathcal{A}\). When the wave passes, the slice is compressed by stress \(\sigma\), producing strain \(\varepsilon = \sigma/Y\).

Newton's second law applied to the slice gives, after standard manipulation, the wave equation with speed:

\[ v_\text{rod} = \sqrt{\frac{Y}{\rho}} \]

For steel: \(Y = 2\times10^{11}\,\mathrm{Pa}\), \(\rho = 7.8\times10^3\,\mathrm{kg\,m^{-3}}\), so \[ v = \sqrt{\frac{2\times10^{11}}{7.8\times10^3}} = \sqrt{2.56\times10^7} \approx 5060\,\mathrm{m\,s^{-1}} \] For air: \(B \approx 1.4\times10^5\,\mathrm{Pa}\), \(\rho_\text{air} \approx 1.2\,\mathrm{kg\,m^{-3}}\), so \(v_\text{air} = \sqrt{1.4\times10^5/1.2} \approx 342\,\mathrm{m\,s^{-1}}\).

The ratio is \(5060/342 \approx 15\). Steel is 15 times faster than air not because it is light but because its stiffness (\(Y_\text{steel}/B_\text{air} \approx 1.4\times10^6\)) overwhelmingly outweighs its density advantage for air (\(\rho_\text{steel}/\rho_\text{air} \approx 6500\)). Stiffness wins.

Example 2 · Two-wire composite: which breaks first?

A steel wire (\(Y_s = 2.0\times10^{11}\,\mathrm{Pa}\), breaking stress \(\sigma_\text{break,s} = 4.0\times10^8\,\mathrm{Pa}\), area \(\mathcal{A}_s = 1.0\times10^{-6}\,\mathrm{m^2}\), length \(L_s = 1.0\,\mathrm{m}\)) and a copper wire (\(Y_c = 1.2\times10^{11}\,\mathrm{Pa}\), breaking stress \(\sigma_\text{break,c} = 2.5\times10^8\,\mathrm{Pa}\), area \(\mathcal{A}_c = 2.0\times10^{-6}\,\mathrm{m^2}\), length \(L_c = 1.0\,\mathrm{m}\)) are connected in series (end-to-end) and a gradually increasing load \(F\) is hung from the lower end. Which wire breaks first? At what load does it break, and what is the extension of the composite system at that moment?

Solution. Identifying the configuration. In series, both wires carry the same tensile force \(F\) at all times (by Newton's third law applied to the junction). The stress in each wire is therefore: \[ \sigma_s = \frac{F}{\mathcal{A}_s} = \frac{F}{1.0\times10^{-6}} \qquad \sigma_c = \frac{F}{\mathcal{A}_c} = \frac{F}{2.0\times10^{-6}} \] Since \(\mathcal{A}_s \lt \mathcal{A}_c\), the stress in the steel wire is always twice that in the copper wire for the same \(F\). Which reaches breaking stress first? Steel breaks when \(\sigma_s = 4.0\times10^8\,\mathrm{Pa}\): \[ F_s = \sigma_\text{break,s}\cdot\mathcal{A}_s = 4.0\times10^8 \times 1.0\times10^{-6} = 400\,\mathrm{N} \] Copper breaks when \(\sigma_c = 2.5\times10^8\,\mathrm{Pa}\): \[ F_c = \sigma_\text{break,c}\cdot\mathcal{A}_c = 2.5\times10^8 \times 2.0\times10^{-6} = 500\,\mathrm{N} \] The steel wire breaks first at \(F = 400\,\mathrm{N}\). Exam trap: Students often reason “steel is stronger, so copper breaks first.” The breaking strength of a wire is not the breaking stress alone—it is \(\sigma_\text{break} \times \mathcal{A}\). The copper wire has a larger area that more than compensates for its lower breaking stress. Extension at the moment of breaking. Just before the steel wire snaps, \(F = 400\,\mathrm{N}\). Extension of steel wire: \[ \Delta L_s = \frac{FL_s}{Y_s \mathcal{A}_s} = \frac{400 \times 1.0}{2.0\times10^{11} \times 1.0\times10^{-6}} = \frac{400}{2.0\times10^5} = 2.0\times10^{-3}\,\mathrm{m} = 2.0\,\mathrm{mm} \] Extension of copper wire: \[ \Delta L_c = \frac{FL_c}{Y_c \mathcal{A}_c} = \frac{400 \times 1.0}{1.2\times10^{11} \times 2.0\times10^{-6}} = \frac{400}{2.4\times10^5} \approx 1.67\times10^{-3}\,\mathrm{m} \approx 1.67\,\mathrm{mm} \] Total extension of composite system: \[ \Delta L_\text{total} = 2.0 + 1.67 = 3.67\,\mathrm{mm} \]

Misconception alert

“A stiffer material is a stronger material.” False — stiffness and strength are independent properties. Young's modulus \(Y\) measures stiffness: how much stress is needed to produce a given elastic strain, i.e. the initial slope of the stress–strain curve. Strength (the yield or breaking stress) is a different number entirely—the height at which the curve ends. A steep initial slope says nothing about how far the curve extends before it stops. Glass illustrates the gap: with \(Y \approx 70\,\mathrm{GPa}\) it is as stiff as aluminium, yet it is brittle and fails at a low stress with almost no plastic give. Mild steel is far stiffer (\(Y \approx 200\,\mathrm{GPa}\)) but also far tougher than cast iron, which is stiffer still yet shatters. Rubber sits at the opposite corner: a tiny \(Y\) (very compliant) but able to survive enormous strains before tearing. A diving board must be stiff enough not to sag under a standing diver and strong enough not to snap when she jumps—two separate requirements that a single number cannot capture. (See the resilience and toughness comparison in Exercise 27.)

✓ Checkpoint · Young's Modulus and the Stress–Strain Curve

A stress–strain curve for an unknown material shows a linear region up to a strain of \(0.002\) at a stress of \(6\times10^8\,\mathrm{Pa}\), followed immediately by fracture with no visible plastic region. (a) Find Young's modulus for the material. (b) Classify the material as ductile or brittle, and identify which of the following it is most likely to be: mild steel, glass, copper, rubber.
(a) \(Y = \sigma/\varepsilon = 6\times10^8/0.002 = 3\times10^{11}\,\mathrm{Pa} = 300\,\mathrm{GPa}\)—close to diamond or tungsten carbide. (b) Brittle (no plastic region, fractures at proportionality limit). Of the options given, glass is brittle; but \(Y = 300\,\mathrm{GPa}\) is too high for glass (\(\sim 70\,\mathrm{GPa}\)); this is more consistent with a very hard ceramic.
A steel rod (\(Y = 2\times10^{11}\,\mathrm{Pa}\), \(\mathcal{A} = 4.0\times10^{-4}\,\mathrm{m^2}\), \(L = 0.50\,\mathrm{m}\)) is compressed by a force of \(8\times10^5\,\mathrm{N}\) applied axially. Find (a) the compressive stress, (b) the compressive strain, and (c) the shortening \(|\Delta L|\).
(a) \(\sigma = F/\mathcal{A} = 8\times10^5/4\times10^{-4} = 2\times10^9\,\mathrm{Pa}\). (b) \(\varepsilon = \sigma/Y = 2\times10^9/2\times10^{11} = 0.01\). (c) \(|\Delta L| = \varepsilon L = 0.01\times0.50 = 5.0\times10^{-3}\,\mathrm{m} = 5.0\,\mathrm{mm}\). Note: the stress \(2\,\mathrm{GPa}\) exceeds the yield strength of mild steel (\(\sim 250\,\mathrm{MPa}\)); in practice this rod would have yielded. This is a mathematical exercise, not a realistic scenario.
Two identical wires \(P\) and \(Q\) (same material, same geometry) are connected in parallel (side by side, sharing the same load symmetrically) and support a load \(W\). Compare the stress, strain, and extension of each wire with the case where a single wire supports the same load \(W\).
In parallel, each wire carries \(F = W/2\). Stress in each wire: \(\sigma_\text{parallel} = (W/2)/\mathcal{A} = \sigma_\text{single}/2\). Strain: \(\varepsilon_\text{parallel} = \sigma_\text{parallel}/Y = \varepsilon_\text{single}/2\). Extension: \(\Delta L_\text{parallel} = \varepsilon_\text{parallel} L = \Delta L_\text{single}/2\). The parallel pair is half as stressed and stretches half as much as a single wire—equivalent to a single wire of double the area. This is the mechanical analogue of resistors in parallel.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Elongation of a loaded wire

A vertical steel wire of length \(L = 2.0\,\mathrm{m}\) and cross-sectional area \(\mathcal{A} = 1.0\times10^{-6}\,\mathrm{m^2}\) carries a load \(F = 100\,\mathrm{N}\). Young's modulus of steel is \(Y = 2.0\times10^{11}\,\mathrm{Pa}\). Find the extension.

Solution. Given: \(L = 2.0\,\mathrm{m}\), \(\mathcal{A} = 1.0\times10^{-6}\,\mathrm{m^2}\), \(F = 100\,\mathrm{N}\), \(Y = 2.0\times10^{11}\,\mathrm{Pa}\).
Find: the extension \(\Delta L\).
Setup: The wire is in uniaxial tension, so \(\Delta L = FL/(\mathcal{A}Y)\) from the definition \(Y = (\,F/\mathcal{A})/(\Delta L/L)\).
Solve: \[\begin{aligned} \Delta L &= \frac{FL}{\mathcal{A}Y} = \frac{(100)(2.0)}{(1.0\times10^{-6})(2.0\times10^{11})} = \frac{200}{2.0\times10^{5}} = 1.0\times10^{-3}\,\mathrm{m}. \end{aligned}\] Answer: \(\boxed{\Delta L = 1.0\,\mathrm{mm}}\)
Check: Dimensionally, \([FL/\mathcal{A}Y] = (\mathrm{N\cdot m})/(\mathrm{m^2\cdot Pa}) = (\mathrm{N\cdot m})/\mathrm{N} = \mathrm{m}\). The stress is \(\sigma = F/\mathcal{A} = 1.0\times10^{8}\,\mathrm{Pa} = 100\,\mathrm{MPa}\), well below steel's yield (\(\sim\)250 MPa), so Hooke's law applies and a sub-millimetre stretch on a 2 m wire is physically reasonable. ✓

Example 2 · Maximum safe load from ultimate strength

A steel cable of diameter \(d = 2.0\,\mathrm{mm}\) has ultimate tensile strength \(\sigma_\text{ult} = 4.0\times10^{8}\,\mathrm{Pa}\). What is the largest load it can carry before fracture, and the corresponding suspended mass?

Solution. Given: \(d = 2.0\,\mathrm{mm} \Rightarrow r = 1.0\times10^{-3}\,\mathrm{m}\), \(\sigma_\text{ult} = 4.0\times10^{8}\,\mathrm{Pa}\).
Find: the breaking load \(F_{\max}\) and mass \(m\).
Setup: Fracture occurs when the stress reaches \(\sigma_\text{ult}\), so \(F_{\max} = \sigma_\text{ult}\,\mathcal{A}\) with \(\mathcal{A} = \pi r^2\).
Solve: \[\begin{aligned} \mathcal{A} &= \pi (1.0\times10^{-3})^2 = 3.14\times10^{-6}\,\mathrm{m^2},\\ F_{\max} &= \sigma_\text{ult}\,\mathcal{A} = (4.0\times10^{8})(3.14\times10^{-6}) = 1.26\times10^{3}\,\mathrm{N},\\ m &= \frac{F_{\max}}{g} = \frac{1256}{10} \approx 126\,\mathrm{kg}. \end{aligned}\] Answer: \(\boxed{F_{\max} \approx 1.26\,\mathrm{kN},\quad m \approx 126\,\mathrm{kg}}\)
Check: Strength scales with area, hence with \(d^2\): doubling the diameter to \(4\,\mathrm{mm}\) would quadruple \(F_{\max}\) to \(\sim\)5 kN—consistent with \(\mathcal{A}\propto d^2\). The elastic strain just before fracture, \(\sigma_\text{ult}/Y \approx 4\times10^8/2\times10^{11} = 0.2\%\), is the right order for a metal near its limit. ✓

Example 3 · Compression of a solid sphere at depth

A solid copper sphere is lowered to an ocean depth where the pressure exceeds the surface value by \(\Delta P = 4.0\times10^{7}\,\mathrm{Pa}\). Copper has bulk modulus \(B = 1.4\times10^{11}\,\mathrm{Pa}\). Find the fractional change in volume.

Solution. Given: \(\Delta P = 4.0\times10^{7}\,\mathrm{Pa}\), \(B = 1.4\times10^{11}\,\mathrm{Pa}\).
Find: \(\Delta V/V\).
Setup: Uniform pressure produces volumetric strain \(\Delta V/V = -\Delta P/B\) (negative: the sphere shrinks).
Solve: \[\begin{aligned} \frac{\Delta V}{V} = -\frac{\Delta P}{B} = -\frac{4.0\times10^{7}}{1.4\times10^{11}} = -2.9\times10^{-4}. \end{aligned}\] Answer: \(\boxed{\Delta V/V = -2.9\times10^{-4}\ \ (\text{a }0.029\%\text{ contraction})}\)
Check: Via the compressibility \(k = 1/B = 7.1\times10^{-12}\,\mathrm{Pa^{-1}}\): \(|\Delta V/V| = k\,\Delta P = (7.1\times10^{-12})(4.0\times10^{7}) = 2.9\times10^{-4}\), matching. The tiny result is expected—solids are nearly incompressible. ✓

Example 4 · Shear of a fixed metal block

A solid aluminium cube of side \(s = 0.10\,\mathrm{m}\) is cemented to the floor. A tangential force \(F_\parallel = 2.5\times10^{5}\,\mathrm{N}\) is applied to its top face. Aluminium has shear modulus \(G = 2.5\times10^{10}\,\mathrm{Pa}\). Find the shear angle and the horizontal displacement of the top face.

Solution. Given: \(s = 0.10\,\mathrm{m}\), \(F_\parallel = 2.5\times10^{5}\,\mathrm{N}\), \(G = 2.5\times10^{10}\,\mathrm{Pa}\); top-face area \(\mathcal{A} = s^2 = 1.0\times10^{-2}\,\mathrm{m^2}\).
Find: shear angle \(\phi\) and displacement \(\Delta x\).
Setup: Shear stress \(\tau = F_\parallel/\mathcal{A}\), and \(\phi = \tau/G\); the displacement is \(\Delta x = \phi\,s\).
Solve: \[\begin{aligned} \tau &= \frac{F_\parallel}{\mathcal{A}} = \frac{2.5\times10^{5}}{1.0\times10^{-2}} = 2.5\times10^{7}\,\mathrm{Pa},\\ \phi &= \frac{\tau}{G} = \frac{2.5\times10^{7}}{2.5\times10^{10}} = 1.0\times10^{-3}\,\mathrm{rad},\\ \Delta x &= \phi\, s = (1.0\times10^{-3})(0.10) = 1.0\times10^{-4}\,\mathrm{m}. \end{aligned}\] Answer: \(\boxed{\phi = 1.0\times10^{-3}\,\mathrm{rad},\quad \Delta x = 0.10\,\mathrm{mm}}\)
Check: Combine into one relation: \(\Delta x = F_\parallel s/(G s^2) = F_\parallel/(Gs) = 2.5\times10^{5}/[(2.5\times10^{10})(0.10)] = 1.0\times10^{-4}\,\mathrm{m}\), matching. That a quarter-meganewton produces only a tenth-millimetre shift is exactly the point: metals resist shear stiffly. ✓

Example 5 · Pressure to compress a hydraulic fluid

Hydraulic oil has bulk modulus \(B = 1.7\times10^{9}\,\mathrm{Pa}\). What gauge pressure is needed to reduce its volume by \(0.50\%\)?

Solution. Given: \(B = 1.7\times10^{9}\,\mathrm{Pa}\), \(\Delta V/V = -5.0\times10^{-3}\).
Find: required pressure increase \(\Delta P\).
Setup: Invert \(\Delta V/V = -\Delta P/B\) to get \(\Delta P = -B\,(\Delta V/V)\).
Solve: \[\begin{aligned} \Delta P = -B\,\frac{\Delta V}{V} = -(1.7\times10^{9})(-5.0\times10^{-3}) = 8.5\times10^{6}\,\mathrm{Pa}. \end{aligned}\] Answer: \(\boxed{\Delta P = 8.5\,\mathrm{MPa}\ \ (\approx 84\,\mathrm{atm})}\)
Check: Re-substitute: at \(\Delta P = 8.5\times10^{6}\,\mathrm{Pa}\), the predicted strain is \(-\Delta P/B = -8.5\times10^{6}/1.7\times10^{9} = -5.0\times10^{-3} = -0.50\%\), recovering the requirement. A few tens of atmospheres for half-a-percent compression is sensible for a liquid. ✓