---
title: "Continuum Mechanics and Elasticity"
book: "University Physics — Year 3"
subject: physics
language: en
chapter: 3
exercises: 12
source: https://one-course.com/books/physics/5/en/chapter/3-continuum-mechanics-and-elasticity
---

# Chapter 3 — Continuum Mechanics and Elasticity

Press your ear to a long steel rail while a distant worker strikes it: you hear two clangs — one through the steel, one through the air, a second or more apart. Rock, steel, bone and rubber all carry forces and waves the way a fluid carries pressure, but with something no fluid has: they resist *changes of shape*, not only changes of volume. The Year 2 volume built the mechanics of fluids on two ideas, the material particle and the pressure field; this chapter extends them to solids. The displacement of each particle becomes a field, its local distortion a *strain*, the internal forces a *stress*, and between them stands the solid’s identity card, [Hooke’s law](#thm-b3-continuum-elasticity-hooke) with its two elastic constants. The payoff runs from the everyday — why cables stretch, why bottles burst, how a diving board bends — to the planetary: earthquakes send through the Earth exactly the two kinds of elastic wave this chapter predicts, and their arrival times were the first sound ever taken of our planet’s interior.

## 3.1 Strain: describing deformation

**Definition 3.1 (Displacement field and strain tensor).**

When a solid deforms, the particle initially at $\vect r$ moves to $\vect r + \vect u(\vect r)$: $\vect u$ is the *displacement field*. For small deformations, the local distortion is measured by the *strain tensor*, the symmetric array

$$
\varepsilon_{ij} = \frac12\Big(
 \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}
\Big) , \qquad i, j \in \{x, y, z\} .
$$

A diagonal component $\varepsilon_{xx}$ is the relative elongation of the material along $x$ (dimensionless, e.g. $10^{-3}$ for a steel cable in service); an off-diagonal component $\varepsilon_{xy}$ is half the closing of the angle between the $x$ and $y$ material directions — a *shear*. The trace is the relative change of volume, the *dilatation*:

$$
\frac{\delta V}{V} = \varepsilon_{xx} + \varepsilon_{yy} +
\varepsilon_{zz} = \operatorname{div}\vect u .
$$

**Example 3.2 (Three elementary deformations).**

Uniform dilation $\vect u = \alpha\vect r$: $\varepsilon_{ij} =
\alpha\,\delta_{ij}$, volume change $3\alpha$, no shear. Simple shear $\vect u = (\gamma y, 0, 0)$: $\varepsilon_{xy} = \gamma/2$, all diagonal components zero — shape changes, volume does not. Rigid rotation $\vect u = (-\omega y, \omega x, 0)$: $\varepsilon_{ij} = 0$ identically — the antisymmetric part of $\partial u_i/\partial x_j$, discarded by the symmetrisation, is exactly the part that rotates without deforming. Strain measures deformation only.

![The strain tensor sees only true deformation: stretching (diagonal components), shearing (off-diagonal components) — and a rigid rotation not at all.](https://one-course.com/images/onecourse/chapters/physics-5/b3-continuum-elasticity/fig-905fb482f598.svg)

*The [strain tensor](#def-b3-continuum-elasticity-strain) sees only true deformation: stretching (diagonal components), shearing (off-diagonal components) — and a rigid rotation not at all.*

## 3.2 Stress: describing internal forces

**Definition 3.3 (Traction and the stress tensor).**

Cut the solid, in thought, along a small surface $\dd S$ of normal $\vect n$: the material on the $+\vect n$ side pulls on the other side with a force $\dd\vect F = \boldsymbol\sigma(\vect n)\,\dd S$ — the *traction*. This force depends linearly on $\vect n$ (Cauchy), so it is encoded by the *stress tensor* $\sigma_{ij}$:

$$
\dd F_i = \sum_j \sigma_{ij}\,n_j\,\dd S ,
$$

in pascals. $\sigma_{xx}$ is a pull ($> 0$: tension) or push ($< 0$: compression) across a face normal to $x$; $\sigma_{xy}$ is a force along $x$ carried by a face normal to $y$ — a shear stress. A fluid at rest is the special case $\sigma_{ij} = -p\,\delta_{ij}$: pressure pushes equally on every face and shears on none, which is why fluids flow — they cannot carry static shear.

**Proposition 3.4 (Equilibrium and symmetry).**

In a solid at equilibrium under a body force $\vect f$ per unit volume (e.g. $\rho\vect g$),

$$
\sum_j\frac{\partial\sigma_{ij}}{\partial x_j} + f_i = 0
$$

at every interior point; and the [stress tensor](#def-b3-continuum-elasticity-stress) is symmetric, $\sigma_{ij} = \sigma_{ji}$.

**Partial proof.** Balance the forces on a small cube of side $a$: the [tractions](#def-b3-continuum-elasticity-stress) on the two faces normal to $x$ differ by $a\,\partial_x\sigma_{ix}$ per unit area, and likewise for the other pairs; the net surface force per unit volume is $\sum_j\partial_j\sigma_{ij}$, which must cancel $f_i$ — the same bookkeeping that gave $-\vect\nabla p + \rho\vect g = \vect 0$ in the fluid statics of the Year 1 volume. Symmetry: the torque of the shear stresses about the cube’s centre is $(\sigma_{xy} -
\sigma_{yx})a^3$ at leading order, while its moment of inertia scales as $a^5$; an asymmetric stress would spin small cubes infinitely fast. The full continuum argument is admitted. ∎

![The three components of the traction on one face of a material cube: one normal (tension or compression), two tangential (shear). The nine components over the three faces form the stress tensor.](https://one-course.com/images/onecourse/chapters/physics-5/b3-continuum-elasticity/fig-70cdcbbd65b5.svg)

*The three components of the [traction](#def-b3-continuum-elasticity-stress) on one face of a material cube: one normal (tension or compression), two tangential (shear). The nine components over the three faces form the [stress tensor](#def-b3-continuum-elasticity-stress).*

## 3.3 Hooke’s law and the elastic constants

**Theorem 3.5 (Linear isotropic elasticity).**

For small strains, an isotropic solid responds linearly: stress is proportional to strain,

$$
\sigma_{ij} = \lambda\,(\varepsilon_{xx} + \varepsilon_{yy} +
\varepsilon_{zz})\,\delta_{ij} + 2\mu\,\varepsilon_{ij} ,
$$

with two material constants, the *Lamé coefficients* $\lambda$ and $\mu$ ($\mu$ is also written $G$, the *shear modulus*). In the simple [traction](#def-b3-continuum-elasticity-stress) test — a bar pulled along $x$, free on its sides — the same law takes the engineer’s form

$$
\varepsilon_{xx} = \frac{\sigma_{xx}}{E} , \qquad
\varepsilon_{yy} = \varepsilon_{zz} = -\nu\,\varepsilon_{xx} ,
$$

defining *[Young’s modulus](#thm-b3-continuum-elasticity-hooke)* $E$ (stiffness against stretching) and *[Poisson’s ratio](#thm-b3-continuum-elasticity-hooke)* $\nu$ (lateral contraction), with

$$
\mu = \frac{E}{2(1 + \nu)} , \qquad
\lambda = \frac{E\nu}{(1 + \nu)(1 - 2\nu)} .
$$

This is Hooke’s “as the extension, so the force” (1678), promoted to a tensor. It holds up to a material-dependent *elastic limit*.

**Proof.** *Admitted at this level.* ∎

**Remark 3.6 (Orders of magnitude).**

$E \approx 200\,\mathrm{GPa}$ for steel, $70\,\mathrm{GPa}$ for aluminium and window glass, $50\,\mathrm{GPa}$ for granite, $15\,\mathrm{GPa}$ for bone and concrete, $10\,\mathrm{GPa}$ for wood along the grain, and only a few MPa for rubber — five orders of magnitude, the span between a bridge and an elastic band. $\nu$ lies between $0$ (cork, nearly) and $1/2$ (rubber): $\nu = 1/2$ means volume-preserving deformation, and $\nu \approx 0.3$ is typical of metals. A strain of $10^{-3}$ in steel already means $\sigma = 200\,\mathrm{MPa}$, close to the yield stress of ordinary grades: everyday elasticity lives below one part in a thousand.

**Example 3.7 (An elevator cable).**

A $60\,\mathrm{m}$ steel cable of cross-section $2.0\,\mathrm{cm}^{2}$ carries a $1000\,\mathrm{kg}$ car: $\sigma = mg/S = 49\,\mathrm{MPa}$, $\varepsilon =
\sigma/E = 2.5 \times 10^{-4}$, stretch $\varepsilon L = 15\,\mathrm{mm}$ — and the cable behaves as a spring of stiffness $k = ES/L =
6.7 \times 10^{5}\,\mathrm{N}/\mathrm{m}$: the formula $k = ES/L$ is how a continuum hands back the springs of the Year 1 volume.

**Proposition 3.8 (Elastic energy).**

A strained solid stores, per unit volume, the energy density

$$
w = \frac12\sum_{i,j}\sigma_{ij}\,\varepsilon_{ij}
\qquad\text{(simple traction: } w = \tfrac12 E\varepsilon^2 =
\sigma^2/2E\text{)} ,
$$

the three-dimensional $\tfrac12 kx^2$.

**Partial proof.** In simple [traction](#def-b3-continuum-elasticity-stress), bringing the stress from $0$ to $\sigma$ does the work per unit volume $\int_0^\varepsilon\sigma'\,\dd\varepsilon' =
\int_0^\varepsilon E\varepsilon'\,\dd\varepsilon' = \tfrac12
E\varepsilon^2$ — the area under the stress–strain line, exactly as for a spring. The general quadratic form follows by superposing the components; admitted. ∎

![Left: the traction test defines E (relative elongation) and (lateral thinning, exaggerated). Right: the stress–strain curve of a metal — linear elasticity up to the elastic limit, then irreversible plastic flow, then fracture. This chapter lives on the straight part.](https://one-course.com/images/onecourse/chapters/physics-5/b3-continuum-elasticity/fig-853382b4e5bf.svg)

*Left: the [traction](#def-b3-continuum-elasticity-stress) test defines $E$ (relative elongation) and $\nu$ (lateral thinning, exaggerated). Right: the stress–strain curve of a metal — linear elasticity up to the elastic limit, then irreversible plastic flow, then fracture. This chapter lives on the straight part.*

**Method 3.9 (Solving small-elasticity problems).**

(1) Identify the geometry and the loading; guess which stress components are nonzero (a pulled bar: only $\sigma_{xx}$; a twisted wire: shear; a pressurised shell: tangential tensions). (2) Write force balance on a well-chosen piece — a half-cylinder, a slice, a cube. (3) Convert stress to strain by [Hooke’s law](#thm-b3-continuum-elasticity-hooke), strain to displacement by integrating. (4) Energies via $w = \sigma^2/2E$ (or $\sigma^2/2G$ in shear). (5) Always check the strain stays small and the stress below the elastic limit — otherwise the answer describes a solid that no longer exists.

## 3.4 Elastic waves

**Proposition 3.10 (The two sounds of a solid).**

In an unbounded elastic solid of density $\rho$, small disturbances propagate as two independent kinds of wave: *longitudinal* (P) waves, in which matter oscillates along the propagation direction by compression and dilation, at speed

$$
c_{\text{P}} = \sqrt{\frac{\lambda + 2\mu}{\rho}} ,
$$

and *transverse* (S) waves, in which matter shears sideways, at

$$
c_{\text{S}} = \sqrt{\frac{\mu}{\rho}} < c_{\text{P}} .
$$

A fluid has $\mu = 0$: no [S waves](#prop-b3-continuum-elasticity-waves) — shear cannot be transmitted — and the P speed reduces to the sound speed of the Year 2 volume. For $\nu = 1/4$ (typical rock), $\lambda = \mu$ and $c_{\text{P}} =
\sqrt3\,c_{\text{S}}$.

**Partial proof.** Take a plane disturbance $\vect u = \vect u(x, t)$. Newton’s law per unit volume is $\rho\,\partial_t^2u_i = \sum_j\partial_j\sigma_{ij}$ (the equilibrium equation with inertia restored). For the longitudinal component, [Hooke’s law](#thm-b3-continuum-elasticity-hooke) gives $\sigma_{xx} = (\lambda +
2\mu)\,\partial_xu_x$ (the lateral strains vanish in a plane wave, the neighbouring matter forbidding lateral release), so $\rho\,\partial_t^2u_x = (\lambda + 2\mu)\,\partial_x^2u_x$: d’Alembert at speed $c_{\text{P}}$. For the transverse component, $\sigma_{yx} =
2\mu\varepsilon_{yx} = \mu\,\partial_xu_y$, whence $\rho\,\partial_t^2u_y = \mu\,\partial_x^2u_y$: speed $c_{\text{S}}$. That a general disturbance splits into these two families is admitted. ∎

![The two elastic waves. P: planes of matter bunch and spread along the travel direction (compression wave — exists in solids, liquids and gases). S: matter shears sideways (exists only where shear is resisted: solids).](https://one-course.com/images/onecourse/chapters/physics-5/b3-continuum-elasticity/fig-f81f0dfb0824.svg)

*The two [elastic waves](#prop-b3-continuum-elasticity-waves). P: planes of matter bunch and spread along the travel direction (compression wave — exists in solids, liquids and gases). S: matter shears sideways (exists only where shear is resisted: solids).*

**Example 3.11 (The rail and the earthquake).**

Steel: $c_{\text{P}} \approx 5.9\,\mathrm{km}/\mathrm{s}$ — a strike on a rail $2\,\mathrm{km}$ away arrives through the steel in $0.34\,\mathrm{s}$ and through the air ($340\,\mathrm{m}/\mathrm{s}$) in $5.9\,\mathrm{s}$: two clangs. Granite ($E = 75\,\mathrm{GPa}$, $\nu = 1/4$, $\rho =
2.7 \times 10^{3}\,\mathrm{kg}/\mathrm{m}^{3}$): $c_{\text{P}} \approx 5.8\,\mathrm{km}/\mathrm{s}$, $c_{\text{S}} \approx 3.3\,\mathrm{km}/\mathrm{s}$. An earthquake therefore announces itself twice: a sharp P jolt, then, seconds later, the slower, stronger S shaking — and the delay measures the distance ([Problem 3.1](#pb-b3-continuum-elasticity-1)). Earthquake early-warning systems live inside that delay: the [P wave](#prop-b3-continuum-elasticity-waves), and the radio message it triggers, outrun the [S wave](#prop-b3-continuum-elasticity-waves) that does the damage.

**Remark 3.12 (Strings, rods and sound recovered).**

The waves of the Year 2 volume are all limits of this chapter. A thin rod, free to thin sideways, carries compression waves at $\sqrt{E/\rho}$ (slower than $c_{\text{P}}$: the lateral release softens the response); a stretched string carries transverse waves at $\sqrt{T/\rho_\ell}$, with tension standing in for stiffness; a fluid keeps only $\lambda$ — its bulk modulus — and the single sound speed $\sqrt{\lambda/\rho}$.

![A loaded diving board: stress and strain distributed through a continuum, bent into the cubic deflection profile of the cantilever — and a spring about to give its elastic energy back.](https://one-course.com/images/onecourse/chapters/physics-5/b3-continuum-elasticity/img-5763894027e1.jpg)

*A loaded diving board: stress and strain distributed through a continuum, bent into the cubic deflection profile of the cantilever — and a spring about to give its [elastic energy](#prop-b3-continuum-elasticity-energy) back.*

## 3.5 Exercises

**Exercise 3.1 ★.**

(a) Check the units: what is a pascal in terms of kg, m, s, and what is the unit of strain? (b) A steel cable works at $\sigma =
200\,\mathrm{MPa}$: compute its strain. (c) A rubber band ($E \approx
2\,\mathrm{MPa}$) is stretched to twice its length: what “strain” is that, and why does this chapter’s framework not really apply? (d) Rank by stored [elastic energy](#prop-b3-continuum-elasticity-energy) density at their working stress: steel at $200\,\mathrm{MPa}$, rubber at $1\,\mathrm{MPa}$.

**Solution of Exercise 3.1.**

(a) $\mathrm{Pa} = \mathrm{N}/\mathrm{m}^{2} = \mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-2}$; strain is a pure number. (b) $\varepsilon = \sigma/E = 10^{-3}$. (c) Doubling the length is $\varepsilon = 1$: a thousand times beyond “small”, and rubber’s response there is strongly nonlinear (and of a different, entropic origin) — [Hooke’s law](#thm-b3-continuum-elasticity-hooke) only opens the curve. (d) $w =
\sigma^2/2E$: steel $(2 \times 10^{8})^2/2(2 \times 10^{11}) =
1 \times 10^{5}\,\mathrm{J}/\mathrm{m}^{3}$; rubber $(10^{6})^2/2(2 \times 10^{6}) =
2.5 \times 10^{5}\,\mathrm{J}/\mathrm{m}^{3}$ — the soft material stores *more* per unit volume at working stress, which is why slingshots are rubber, not steel.

**Exercise 3.2 ★.**

Compute the [strain tensor](#def-b3-continuum-elasticity-strain) of each [displacement field](#def-b3-continuum-elasticity-strain) and describe the deformation: (a) $\vect u = \alpha(x, y, z)$; (b) $\vect u =
(\gamma y, 0, 0)$; (c) $\vect u = (-\omega y, \omega x, 0)$; (d) $\vect u = (\varepsilon x, -\nu\varepsilon y, -\nu\varepsilon z)$ — which experiment realises it?

**Solution of Exercise 3.2.**

(a) $\varepsilon_{ij} = \alpha\delta_{ij}$: isotropic dilation, $\delta V/V = 3\alpha$. (b) $\varepsilon_{xy} = \varepsilon_{yx} =
\gamma/2$, rest zero: pure shear, volume unchanged. (c) $\varepsilon_{ij} = 0$: rigid rotation, no deformation. (d) $\operatorname{diag}(\varepsilon, -\nu\varepsilon, -\nu\varepsilon)$: the [traction](#def-b3-continuum-elasticity-stress) test — stretch along $x$, Poisson contraction across.

**Exercise 3.3 ★.**

The base of a granite column of height $h$ carries the stress $\sigma
= \rho gh$. (a) Derive this from the equilibrium equation with gravity. (b) Granite crushes at about $200\,\mathrm{MPa}$: what is the tallest column? (c) Everest is $8.8\,\mathrm{km}$ high: comment. (d) Why can a mountain be a little taller than a column of its own rock (think about the shape)?

**Solution of Exercise 3.3.**

(a) With only $\sigma_{zz}(z)$ and weight: $\partial_z\sigma_{zz} =
\rho g$ (taking compression positive downward), $\sigma_{zz} = \rho
gh$ at the base. (b) $h_{\max} = \sigma_{\text{c}}/\rho g =
2 \times 10^{8}/(2700 \times 9.81) \approx 7.6\,\mathrm{km}$. (c) Everest is at the crushing limit of its own base — Earth’s mountains are as tall as rock strength allows, and no taller. (d) A cone of height $h$ loads its base with only $\rho gh/3$ (a third of the column: the mass grows with the section), so a mountain-shaped pile can stand about three times taller than a column.

**Exercise 3.4 ★.**

A steel rod ($E = 200\,\mathrm{GPa}$, $\nu = 0.30$) is stretched by $\varepsilon = 10^{-3}$. (a) Lateral strain? (b) Relative volume change? (c) For which $\nu$ would the volume not change at all, and which common material is close to it? (d) Why is $\nu > 1/2$ impossible (consider hydrostatic compression)?

**Solution of Exercise 3.4.**

(a) $-\nu\varepsilon = -3 \times 10^{-4}$. (b) $\delta V/V = (1 -
2\nu)\varepsilon = 4 \times 10^{-4}$. (c) $\nu = 1/2$: incompressible deformation — rubber. (d) For $\nu > 1/2$ the bulk modulus $K =
E/3(1 - 2\nu)$ would be negative: squeezing from all sides would *grow* the volume, and the material would release energy by collapsing — no stable solid can do it.

**Exercise 3.5 ★★.**

Hoop stress. A thin-walled cylinder (radius $R$, wall thickness $t \ll
R$) holds a pressure $p$. (a) Balancing forces on a half-cylinder of unit length, show the wall carries the tangential stress $\sigma_\theta = pR/t$. (b) Show the longitudinal stress (balance on a cross-section) is $pR/2t$: a cylinder is stressed twice as hard around as along — which way do sausages split? (c) A diving cylinder: $p =
200\,\mathrm{bar}$, $R = 9\,\mathrm{cm}$, $t = 5\,\mathrm{mm}$: compute $\sigma_\theta$ and compare with a steel yield stress of $700\,\mathrm{MPa}$. (d) Why do high-pressure tanks have hemispherical ends?

**Solution of Exercise 3.5.**

(a) On a half-cylinder of unit length, the pressure pushes with $p
\times 2R$ (projected area) and the two cut walls pull back with $2\sigma_\theta t$: $\sigma_\theta = pR/t$. (b) On a cross-section, $p\pi R^2 = \sigma_z\,2\pi Rt$: $\sigma_z = pR/2t$ — half. A sausage splits *lengthwise*: the bigger hoop stress tears the skin along the axis. (c) $\sigma_\theta = 2 \times 10^{7} \times
0.09/0.005 = 360\,\mathrm{MPa}$: a factor $2$ below yield — which is why cylinders are proof-tested and inspected. (d) A sphere carries $pR/2t$ in every direction: hemispherical ends halve the stress and avoid the corners where flat ends would concentrate it.

**Exercise 3.6 ★★.**

A rubber block ($E = 3.0\,\mathrm{MPa}$, $\nu \approx 0.5$), base $10
\times 10\,\mathrm{cm}$, height $2\,\mathrm{cm}$, is glued between two plates; the top plate is pushed sideways with $150\,\mathrm{N}$. (a) Compute $G$. (b) Shear stress and shear angle $\gamma = \sigma_{xy}/G$; sideways displacement of the top plate. (c) Why is $\nu \approx 1/2$ for rubber (what is hard and what is easy for a tangle of polymer chains)? (d) Such rubber blocks carry entire buildings in earthquake zones: which property of this mount protects the building, stiffness in compression or softness in shear?

**Solution of Exercise 3.6.**

(a) $G = E/2(1 + \nu) = 1.0\,\mathrm{MPa}$. (b) $\sigma_{xy} = F/S =
150/10^{-2} = 15\,\mathrm{kPa}$; $\gamma = \sigma_{xy}/G = 0.015$; displacement $\gamma h = 0.3\,\mathrm{mm}$. (c) A rubber network changes *shape* by mere reorientation of its chains (easy), but changing *volume* means packing the chains closer, as hard as compressing a liquid: $G \ll K$, hence $\nu \to 1/2$. (d) Softness in shear: the mount lets the ground shake horizontally underneath while transmitting little force, yet remains stiff enough in compression to hold the building’s weight.

**Exercise 3.7 ★★.**

A climbing rope, length $L = 20\,\mathrm{m}$, cross-section $S =
80\,\mathrm{mm}^{2}$, effective modulus $E = 1.2\,\mathrm{GPa}$. (a) Its stiffness $k = ES/L$. (b) A $80\,\mathrm{kg}$ climber falls freely $4\,\mathrm{m}$ before the rope engages: equating energies, find the maximum rope stretch (solve the quadratic; neglect the fall continued during braking at your first pass). (c) Deduce the peak force and the peak deceleration in $g$. (d) Why must a rope that has held a hard fall be retired (where did the energy go, and what does the stress–strain curve say)?

**Solution of Exercise 3.7.**

(a) $k = ES/L = 1.2 \times 10^{9} \times 8 \times 10^{-5}/20 = 4.8\,\mathrm{kN}/\mathrm{m}$. (b) $mg(h + x) = \tfrac12 kx^2$: $2400x^2 - 785x - 3140 = 0$, $x =
1.3\,\mathrm{m}$. (c) $F = kx \approx 6.3\,\mathrm{kN}$; deceleration $F/m
\approx 79\,\mathrm{m}/\mathrm{s}^{2} \approx 8g$ — the reason ropes are made deliberately stretchy. (d) Part of the absorbed energy went into breaking fibres and plastic rearrangement: the rope now sits on a degraded stress–strain curve, stiffer and weaker, and the next fall would be harder in both senses.

**Exercise 3.8 ★★.**

Torsion. A wire of radius $a$, length $L$, shear modulus $G$ is twisted by an angle $\theta$. (a) Show a tube of radius $r$ inside it suffers the shear angle $\gamma(r) = r\theta/L$. (b) Its shear stress is $G\gamma$: integrate $r \times$ stress over the section to get the restoring torque $\Gamma = (\pi Ga^4/2L)\,\theta$. (c) The $a^4$: halve the radius, and by what factor does the torsional stiffness drop? (d) This extreme softness is why Cavendish (1798) hung his balance from a fine wire: for $a = 25\,\text{µ}\mathrm{m}$, $L =
1\,\mathrm{m}$, $G = 40\,\mathrm{GPa}$, compute $C = \pi Ga^4/2L$ and the period with a rod of moment of inertia $I = 5 \times 10^{-4}\,\mathrm{kg}\,\mathrm{m}^{2}$.

**Solution of Exercise 3.8.**

(a) The top of a tube of radius $r$ turns by the arc $r\theta$ over the length $L$: $\gamma = r\theta/L$. (b) $\Gamma = \int_0^a r\,(G r\theta/L)\,
2\pi r\,\dd r = (\pi Ga^4/2L)\,\theta$. (c) By $2^4 = 16$. (d) $C =
\pi \times 4 \times 10^{10} \times (2.5 \times 10^{-5})^4/2 =
2.5 \times 10^{-8}\,\mathrm{N}\,\mathrm{m}/\mathrm{rad}$; $T = 2\pi\sqrt{I/C} = 2\pi\sqrt{5 \times 10^{-4}/
2.5 \times 10^{-8}} \approx 900\,\mathrm{s}$ — a quarter of an hour per swing: sensitivity enough to feel the gravity of lead spheres.

**Exercise 3.9 ★★.**

(a) From the plane-wave derivation of [Proposition 3.10](#prop-b3-continuum-elasticity-waves), compute $c_{\text{P}}$ and $c_{\text{S}}$ for steel ($E = 200\,\mathrm{GPa}$, $\nu = 0.29$, $\rho =
7.85 \times 10^{3}\,\mathrm{kg}/\mathrm{m}^{3}$). (b) Compare $c_{\text{P}}$ with the thin-rod speed $\sqrt{E/\rho}$ and explain the difference in one sentence. (c) For water ($\mu = 0$, bulk modulus $2.2\,\mathrm{GPa}$): the S speed and the P speed. (d) At what angle does a [P wave](#prop-b3-continuum-elasticity-waves)’s matter motion differ from an [S wave](#prop-b3-continuum-elasticity-waves)’s, and how does a seismometer with three components tell them apart?

**Solution of Exercise 3.9.**

(a) $\mu = E/2(1+\nu) = 78\,\mathrm{GPa}$, $\lambda = E\nu/(1+\nu)(1-2\nu)
= 107\,\mathrm{GPa}$: $c_{\text{P}} = \sqrt{262\times10^9/7850} =
5.8\,\mathrm{km}/\mathrm{s}$, $c_{\text{S}} = 3.1\,\mathrm{km}/\mathrm{s}$. (b) The thin rod gives $\sqrt{E/\rho} = 5.0\,\mathrm{km}/\mathrm{s}$: in the rod the sides bulge freely (Poisson release), softening the response; in the bulk the surrounding matter forbids it. (c) $c_{\text{S}} = 0$; $c_{\text{P}} = \sqrt{2.2 \times 10^{9}/1000} = 1.5\,\mathrm{km}/\mathrm{s}$ — the sound speed of water. (d) P moves the ground along the ray (near-vertical for a deep source), S across it: the three components of a seismometer separate the polarisations, and the P/S delay then dates the distance.

**Exercise 3.10 ★★★.**

Bending. A beam is bent to a radius of curvature $R$; its inner fibres shorten, its outer fibres stretch, and a *neutral surface* in between keeps its length. (a) Show the fibre at distance $y$ from the neutral surface has strain $\varepsilon = y/R$, hence stress $Ey/R$. (b) Summing moments over the cross-section, show the bending moment is $M = EI/R$ with $I = \int y^2\,\dd S$ (the *second moment*); compute $I$ for a rectangle of width $b$ and height $h$. (c) The $h^3$: why does a plank bent flat-wise sag visibly while the same plank on edge feels rigid? Compute the ratio for $b/h = 5$. (d) For a cantilever of length $L$ loaded by $F$ at its tip, the tip sag is $\delta = FL^3/3EI$ (admitted): estimate $\delta$ for a diving board ($L = 1.8\,\mathrm{m}$, $b = 50\,\mathrm{cm}$, $h = 3.5\,\mathrm{cm}$, wood $E
= 12\,\mathrm{GPa}$) under a $75\,\mathrm{kg}$ diver, and comment.

**Solution of Exercise 3.10.**

(a) An arc at radius $R + y$ has length $(R + y)\alpha$ against $R\alpha$ at the neutral surface: $\varepsilon = y/R$, $\sigma =
Ey/R$. (b) $M = \int y\sigma\,\dd S = (E/R)\int y^2\,\dd S = EI/R$; for the rectangle $I = bh^3/12$. (c) Flat: $I = bh^3/12$; on edge: $hb^3/12$ — ratio $(b/h)^2 = 25$ for $b/h = 5$: same wood, twenty-five times stiffer, purely by geometry. (d) $I = 0.50 \times
0.035^3/12 = 1.8 \times 10^{-6}\,\mathrm{m}^{4}$, $EI = 2.1 \times 10^{4}\,\mathrm{N}\,\mathrm{m}^{2}$; $\delta =
FL^3/3EI = 736 \times 1.8^3/(3 \times 2.1 \times 10^{4}) \approx
7\,\mathrm{cm}$: exactly the pleasant give of a diving board.

**Exercise 3.11 ★★★.**

Buckling. A slender column (length $L$, flexural rigidity $EI$), pinned at both ends, carries an axial load $P$. Suppose it bows sideways by $y(x)$. (a) Show the load then exerts the bending moment $M(x) = -Py(x)$ about the displaced axis, so $EI\,y'' = -Py$. (b) With $y(0) = y(L) = 0$, show a nonzero bow first becomes possible at Euler’s critical load $P_{\text{c}} = \pi^2EI/L^2$. (c) Compute $P_{\text{c}}$ for a metre rule ($b = 3\,\mathrm{cm}$, $h = 1.5\,\mathrm{mm}$, $E = 12\,\mathrm{GPa}$) and compare with your hand’s push. (d) Explain from $I$ why bones, bamboo and bicycle frames are *tubes*.

**Solution of Exercise 3.11.**

(a) In the bowed configuration the axial load $P$ acts with lever arm $y(x)$ about the section at $x$: $M = -Py$, and $M = EIy''$ gives $EIy'' = -Py$. (b) $y = A\sin(x\sqrt{P/EI})$ with $y(L) = 0$: nonzero $A$ first at $\sqrt{P/EI}\,L = \pi$, i.e. $P_{\text{c}} =
\pi^2EI/L^2$; below it the straight column is the only solution, above it bowing costs no force. (c) $I = 0.03 \times 0.0015^3/12 =
8.4 \times 10^{-12}\,\mathrm{m}^{4}$: $P_{\text{c}} = \pi^2 \times 12\times10^9 \times
8.4 \times 10^{-12}/1^2 \approx 1\,\mathrm{N}$ — the weight of an apple: a metre rule buckles under a finger. (d) $I$ grows as material moves away from the axis: a tube puts all its section at large $y$, maximising $I$ (hence $P_{\text{c}}$) for a given weight of material — the design of bones, bamboo and bicycle frames.

**Exercise 3.12 ★★★.**

The speed of sound in a solid, from atoms. Model a solid as chains of atoms, spacing $a$, connected by “springs” of stiffness $\kappa$; atomic mass $m$. (a) Show that stretching the chain maps onto continuum [traction](#def-b3-continuum-elasticity-stress) with $E = \kappa/a$ (count springs per unit area, stretch per spring). (b) Show $\rho = m/a^3$ and conclude $c =
\sqrt{E/\rho} = a\sqrt{\kappa/m}$. (c) Estimate $\kappa$ from the depth of an interatomic bond ($\sim3\,\mathrm{eV}$ over $\sim
0.1\,\mathrm{nm}$): $\kappa \sim 50\,\mathrm{N}/\mathrm{m}$; with $a =
2.5 \times 10^{-10}\,\mathrm{m}$ and $m = 1 \times 10^{-25}\,\mathrm{kg}$, estimate $c$. (d) Compare with measured metal sound speeds, and conclude what everyday elasticity is made of.

**Solution of Exercise 3.12.**

(a) One chain per area $a^2$; stretching by strain $\varepsilon$ stretches each spring by $\delta = \varepsilon a$, force $\kappa
\varepsilon a$ per chain, stress $\kappa\varepsilon/a$: $E =
\kappa/a$. (b) $\rho = m/a^3$, so $E/\rho = \kappa a^2/m$ and $c =
a\sqrt{\kappa/m}$. (c) $c = 2.5 \times 10^{-10}\sqrt{50/10^{-25}} \approx
5.6\,\mathrm{km}/\mathrm{s}$. (d) Right on the measured few km/s of metals: the stiffness of everything solid — and the speed of every earthquake wave — is the stiffness of the chemical bond.

## 3.6 Problem: Listening to the Earth

**Problem 3.1.**

Weekend problem — how earthquakes revealed the liquid core

A single earthquake rings the whole planet, and the two [elastic waves](#prop-b3-continuum-elasticity-waves) of this chapter, crossing it, x-ray it. This problem follows the physics from [Hooke’s law](#thm-b3-continuum-elasticity-hooke) in granite to Oldham’s 1906 discovery that the Earth has a core — with the straight-ray estimate of its size. Take for the crust and mantle rock $E = 75\,\mathrm{GPa}$, $\nu = 1/4$, $\rho = 2.7 \times 10^{3}\,\mathrm{kg}/\mathrm{m}^{3}$; Earth radius $R_{\text{E}} =
6371\,\mathrm{km}$.

**Part I — Rock as an elastic solid.**

1. Compute the Lamé coefficients $\mu$ and $\lambda$ of this rock, and observe that $\nu = 1/4$ makes them equal.
2. Compute $c_{\text{P}}$ and $c_{\text{S}}$ , and verify $c_{\text{P}} = \sqrt3\,c_{\text{S}}$ .
3. An earthquake slips two rock faces by metres in seconds: estimate the strain released if a $2\,\mathrm{m}$ slip relaxes rock over a $50\,\mathrm{km}$ scale, and the stress that had built up.
4. From the energy density $w = \tfrac12 E\varepsilon^2$ , estimate the [elastic energy](#prop-b3-continuum-elasticity-energy) per cubic metre, then the energy in a $50 \times 50 \times 20\,\mathrm{km}^{3}$ volume; compare with a megaton ( $4.2 \times 10^{15}\,\mathrm{J}$ ).
5. Why does the [S wave](#prop-b3-continuum-elasticity-waves) usually shake buildings harder than the [P wave](#prop-b3-continuum-elasticity-waves) (think about the direction of ground motion for a wave arriving from below)?
6. In which of the two waves does the ground change volume?

**Part II — The seismometer’s two arrivals.**

7. A station records the P arrival, then the S arrival $\Delta t$ later. For a source at distance $d$ (short enough for straight rays), show $$d = \frac{\Delta t}{\dfrac{1}{c_{\text{S}}} -  \dfrac{1}{c_{\text{P}}}} .$$
8. Evaluate the coefficient: how many kilometres per second of S–P delay?
9. A station measures $\Delta t = 28\,\mathrm{s}$ : how far is the earthquake?
10. One station gives a distance, not a place: how many stations determine the epicentre, and how, geometrically?
11. Japan’s early-warning sirens sound seconds before the strong shaking: for a quake $80\,\mathrm{km}$ below a city, how much warning does the P–S delay itself provide?
12. Modern warnings add the speed of light: the [P wave](#prop-b3-continuum-elasticity-waves) detected near the source is radioed ahead of both waves. For a city $200\,\mathrm{km}$ from the epicentre, how long after the rupture does the [S wave](#prop-b3-continuum-elasticity-waves) arrive, and what warning can a radio message sent at the P’s first arrival $20\,\mathrm{km}$ from the source give?

**Part III — Waves that cross the planet.**

13. At the pressures of the deep mantle the moduli grow: P speeds reach $13.7\,\mathrm{km}/\mathrm{s}$ at the mantle’s base. Taking a rough average $c_{\text{P}} \approx 10\,\mathrm{km}/\mathrm{s}$ , how long does a [P wave](#prop-b3-continuum-elasticity-waves) need to cross the Earth diametrically?
14. Stations register quakes from the far side of the globe: what does the mere existence of these arrivals say about the deep Earth (is it elastic? molten through?)?
15. Define the epicentral angle $\Delta$ (angle at the Earth’s centre between source and station). For straight rays, show that the chord length is $2R_{\text{E}}\sin(\Delta/2)$ .
16. Evaluate the chord and its straight-ray travel time for $\Delta = 60^\circ$ at the average $c_{\text{P}} \approx  10\,\mathrm{km}/\mathrm{s}$ .
17. Around 1900, Oldham noticed that [S waves](#prop-b3-continuum-elasticity-waves) are recorded up to $\Delta \approx 103^\circ$ and then *disappear* : no direct S beyond. What property of a region deep inside the Earth kills [S waves](#prop-b3-continuum-elasticity-waves) , and what state of matter has that property?
18. [P waves](#prop-b3-continuum-elasticity-waves) beyond $103^\circ$ are not absent but weakened, delayed and displaced (a “shadow zone” up to $\approx 142^\circ$ ): why does a liquid region delay and refract [P waves](#prop-b3-continuum-elasticity-waves) but not stop them?
19. Conclude in one sentence what sits at the centre of the Earth.

**Part IV — Weighing the core with a ruler.**

20. A straight S ray leaving the source grazes the core if its closest approach to the centre equals the core radius $R_{\text{c}}$ . Show that this grazing ray reaches the surface at the epicentral angle $\Delta^* =  2\arccos(R_{\text{c}}/R_{\text{E}})$ .
21. From $\Delta^* = 103^\circ$ , compute $R_{\text{c}}$ .
22. The modern value is $3480\,\mathrm{km}$ : compute your error, and explain its *sign* : real rays curve back toward the surface as speed grows with depth — argue whether straight rays over- or underestimate the core.
23. In 1936 Inge Lehmann found weak P arrivals *inside* the shadow zone: what did she conclude sits inside the liquid core?
24. The inner core’s radius is $1220\,\mathrm{km}$ and it is solid: propose the wave observation that could confirm solidity (which wave exists there that the outer core forbids?).
25. Summarise the named result: two elastic wave speeds in rock ( $5.8$ and $3.3\,\mathrm{km}/\mathrm{s}$ ), one missing wave beyond $103^\circ$ , and a ruler give a liquid core of radius $\approx 4000\,\mathrm{km}$ — within $15\%$ of the modern $3480\,\mathrm{km}$ , measured through $6000\,\mathrm{km}$ of solid rock.

**Solution of Problem 3.1.**

**1.** $\mu = E/2(1+\nu) = 30\,\mathrm{GPa}$; $\lambda =
E\nu/(1+\nu)(1-2\nu) = 30\,\mathrm{GPa}$: for $\nu = 1/4$, $\lambda =
\mu$. **2.** $c_{\text{P}} = \sqrt{3\mu/\rho} = \sqrt{9 \times 10^{10}/2700}
= 5.8\,\mathrm{km}/\mathrm{s}$; $c_{\text{S}} = \sqrt{\mu/\rho} =
3.3\,\mathrm{km}/\mathrm{s}$; ratio $\sqrt3$. **3.** $\varepsilon \sim 2/5 \times 10^{4} = 4 \times 10^{-5}$; $\sigma =
E\varepsilon \approx 3\,\mathrm{MPa}$ — the measured “stress drop” of real earthquakes is indeed a few MPa. **4.** $w = \tfrac12 E\varepsilon^2 \approx 60\,\mathrm{J}/\mathrm{m}^{3}$; over $5 \times 10^{13}{}\,\mathrm{m}^{3}$: $\sim3 \times 10^{15}\,\mathrm{J}$, about three-quarters of a megaton — a large earthquake. **5.** From below, P moves the ground vertically — buildings are built to carry vertical loads; S moves it *horizontally*, the direction in which buildings are weakest. **6.** Only the [P wave](#prop-b3-continuum-elasticity-waves): it is a compression wave; the [S wave](#prop-b3-continuum-elasticity-waves) is pure shear, at constant volume. **7.** $t_{\text{P}} = d/c_{\text{P}}$, $t_{\text{S}} =
d/c_{\text{S}}$: $\Delta t = d(1/c_{\text{S}} - 1/c_{\text{P}})$, inverted as stated. **8.** $1/(1/3.33 - 1/5.77)\,\mathrm{km}/\mathrm{s} \approx
7.9\,\mathrm{km}$ per second of delay — the field seismologist’s “eight kilometres per second”. **9.** $d \approx 7.9 \times 28 \approx 220\,\mathrm{km}$. **10.** Three (in general): each station knows a circle of possible epicentres; two circles cross in two points, the third decides. **11.** $t_{\text{P}} = 80/5.8 = 14\,\mathrm{s}$, $t_{\text{S}} =
80/3.3 = 24\,\mathrm{s}$: about $10\,\mathrm{s}$ of warning between the jolt and the destructive shaking. **12.** S arrives at $200/3.3 = 60\,\mathrm{s}$. The P reaches $20\,\mathrm{km}$ at $3.5\,\mathrm{s}$; a radio message is practically instantaneous, so the city can have almost a minute of warning — current systems achieve tens of seconds. **13.** $2R_{\text{E}}/c \approx 12742/10 \approx 1270\,\mathrm{s}$, some twenty minutes. **14.** That [elastic waves](#prop-b3-continuum-elasticity-waves) cross it at all: the deep Earth is not molten through — it transmits, and ([S waves](#prop-b3-continuum-elasticity-waves) crossing the mantle) even shears, like a solid. **15.** Two radii and the angle $\Delta$ between them: the chord is $2R_{\text{E}}\sin(\Delta/2)$. **16.** $2 \times 6371 \times \sin30^\circ = 6371\,\mathrm{km}$; $t \approx 640\,\mathrm{s} \approx 11\,\mathrm{min}$. **17.** Beyond $103^\circ$ every ray must pass through a deep central region; if that region is *fluid* ($\mu = 0$), it carries no shear wave: the [S waves](#prop-b3-continuum-elasticity-waves) die there. Fluids are the state of matter that cannot resist shear. **18.** A fluid still carries compression: [P waves](#prop-b3-continuum-elasticity-waves) cross the core, but slower — so they refract sharply at its boundary, and the bent rays leave an annular shadow instead of a clean cut. **19.** A liquid core sits at the centre of the Earth. **20.** The chord’s closest approach to the centre is $R_{\text{E}}\cos(\Delta/2)$; grazing means $R_{\text{E}}\cos(\Delta^*/2) = R_{\text{c}}$, i.e. $\Delta^* =
2\arccos(R_{\text{c}}/R_{\text{E}})$. **21.** $R_{\text{c}} = 6371\cos(51.5^\circ) \approx
3970\,\mathrm{km}$. **22.** $+14\%$ too large. Speed grows with depth, so real rays curve back toward the surface: a ray whose deepest point just touches the core comes up at a *smaller* angle than the straight chord through the same depth — so from the observed $103^\circ$, the straight-ray inversion places the tangent point too shallow, i.e. overestimates the core. **23.** Weak P energy inside the shadow means something inside the liquid core bends rays back out: a distinct *inner core* (Lehmann, 1936). **24.** Shear waves exist only in solids: a [P wave](#prop-b3-continuum-elasticity-waves) converting to a shear wave inside the inner core and back (the phase called PKJKP) would prove it solid — its detection, long sought, is the accepted evidence. **25.** Two speeds in rock ($5.8$ and $3.3\,\mathrm{km}/\mathrm{s}$), one missing wave beyond $103^\circ$, and the chord geometry yield a liquid core of $\approx4000\,\mathrm{km}$ — within $15\%$ of the modern $3480\,\mathrm{km}$: [Hooke’s law](#thm-b3-continuum-elasticity-hooke), read at planetary scale.