Chapter 8 · Mechanical Properties of Solids

8.2 Elastic Moduli and Poisson's Ratio

Each kind of deformation has its own modulus: Young's modulus $Y$ for stretching and compressing, the shear modulus $G$ for a change of shape, and the bulk modulus $B$ for a change of volume. Poisson's ratio links a stretch to the sideways contraction that comes with it, and a stretched wire stores elastic potential energy $\tfrac12F\,\Delta L$.

Hang the same $5\ \text{kg}$ load from three wires of identical length and thickness — steel, copper and aluminium — and they stretch by different amounts: the aluminium about three times as much as the steel. Under a highway bridge the deck rests on thick rubber pads that let it slide a centimetre or two as it expands on a hot afternoon, by changing shape rather than size. At the bottom of the deepest ocean trench the water is squeezed by more than a thousand atmospheres, yet its volume shrinks by only about five per cent. And pull a rubber band: it gets longer and visibly thinner at the same time.

Section 8.1 showed that, within the proportional limit, stress $=$ modulus $\times$ strain. This section gives each kind of deformation its own modulus. Young's modulus describes stretching and compressing, and you will see how it is measured in the laboratory from a load–extension graph. The shear modulus describes changes of shape, the bulk modulus describes changes of volume, and Poisson's ratio tells how much a stretched body narrows. Together they let you predict the deformation of a wire, a beam, a bearing pad or a deep-sea vessel from a short table of material properties.

Young's modulus

When a wire or rod is stretched or compressed along its length, the stress is longitudinal and so is the strain. Their ratio is the material's stiffness for this kind of deformation.

📘 Definition 8.7 — Young's modulus

The Young's modulus $Y$ of a material is the ratio of tensile (or compressive) stress to the longitudinal strain it produces, within the proportional limit:

$$Y=\frac{\sigma}{\varepsilon}=\frac{F/A}{\Delta L/L}=\frac{FL}{A\,\Delta L}.$$

Here a force $F$ acts along a wire or rod of original length $L$ and cross-sectional area $A$, changing its length by $\Delta L$. SI unit: $\text{Pa}$ ($\text{N/m}^2$); dimensions $[\text{M L}^{-1}\text{T}^{-2}]$. For metals the same value applies to stretching and to compressing.

Rearranging the definition gives the result you will use most often. It is simply Hooke's law for a wire, written with the material constant $Y$.

📐 Result 8.3 — Extension of a wire or rod

A wire or rod of length $L$, cross-section $A$ (diameter $d$) and Young's modulus $Y$, carrying a longitudinal force $F$ within its proportional limit, changes in length by

$$\Delta L=\frac{FL}{AY}=\frac{4FL}{\pi d^2Y}.$$

So the extension is proportional to the load and to the length, and inversely proportional to the square of the diameter and to Young's modulus. (The wire's own weight is neglected.)

Typical values of the elastic moduli are below (they vary somewhat with composition, heat treatment and temperature). Notice how close together the metals are, how much softer rubber is, and that water — like every fluid — has a bulk modulus only: a fluid cannot be stretched or sheared into a new static shape.

Material$Y$ ($10^9\ \text{Pa}$)$G$ ($10^9\ \text{Pa}$)$B$ ($10^9\ \text{Pa}$)$\nu$
steel200781600.29
copper120441400.36
aluminium7026700.33
glass6526400.23
soft rubber0.0050.00172.0≈ 0.50
water——2.2—
✏️ Example 1 — Finding Young's modulus from a measurement
A wire $2.0\ \text{m}$ long and $0.50\ \text{mm}$ in diameter stretches by $2.0\ \text{mm}$ when a $4.0\ \text{kg}$ mass is hung from it. Find Young's modulus of the wire and use the table to suggest what it is made of.
Data $F=4.0\times9.8=39.2\ \text{N}$; $\ A=\dfrac{\pi(0.50\times10^{-3})^2}{4}=1.96\times10^{-7}\ \text{m}^2$; $\ L=2.0\ \text{m}$; $\ \Delta L=2.0\times10^{-3}\ \text{m}$.
Modulus $Y=\dfrac{FL}{A\,\Delta L}=\dfrac{39.2\times2.0}{1.96\times10^{-7}\times2.0\times10^{-3}}=2.0\times10^{11}\ \text{Pa}$.
Material This matches steel.
Check Stress $=\dfrac{39.2}{1.96\times10^{-7}}=2.0\times10^8\ \text{Pa}$ and strain $=\dfrac{2.0\times10^{-3}}{2.0}=1.0\times10^{-3}$; their ratio is $2.0\times10^{11}\ \text{Pa}$ ✓. This is the default wire in Interactive 1.
✏️ Example 2 — Steel and copper wires compared
A steel wire and a copper wire are each $1.5\ \text{m}$ long and $1.0\ \text{mm}$ in diameter, and each carries a $10\ \text{kg}$ load. (a) Find the extension of each. (b) What diameter must a copper wire of the same length have to stretch no more than the steel one under this load?
Area $A=\dfrac{\pi(1.0\times10^{-3})^2}{4}=7.85\times10^{-7}\ \text{m}^2$; $\ F=98\ \text{N}$.
(a) steel $\Delta L=\dfrac{98\times1.5}{7.85\times10^{-7}\times2.0\times10^{11}}=9.4\times10^{-4}\ \text{m}=0.94\ \text{mm}$.
(a) copper $\Delta L=\dfrac{98\times1.5}{7.85\times10^{-7}\times1.2\times10^{11}}=1.6\times10^{-3}\ \text{m}=1.6\ \text{mm}$ — larger in the ratio $\dfrac{200}{120}=1.67$.
(b) Equal extensions need $d^2Y$ equal: $d_\text{Cu}=d_\text{steel}\sqrt{\dfrac{Y_\text{steel}}{Y_\text{Cu}}}=1.0\sqrt{1.67}=1.3\ \text{mm}$.
✏️ Example 3 — Two wires joined end to end
A steel wire $1.2\ \text{m}$ long is joined end to end to a copper wire $0.80\ \text{m}$ long; both have cross-section $1.0\ \text{mm}^2$. The combination hangs from a ceiling and carries a $100\ \text{N}$ load at its lower end. Neglecting the weights of the wires, find the stress in each wire, the extension of each and the total extension.
Idea Each wire, and the joint between them, is in equilibrium, so the tension is $100\ \text{N}$ all the way up. Equal areas: stress $=\dfrac{100}{1.0\times10^{-6}}=1.0\times10^8\ \text{Pa}$ in both.
Steel $\Delta L_1=\dfrac{100\times1.2}{1.0\times10^{-6}\times2.0\times10^{11}}=6.0\times10^{-4}\ \text{m}$.
Copper $\Delta L_2=\dfrac{100\times0.80}{1.0\times10^{-6}\times1.2\times10^{11}}=6.7\times10^{-4}\ \text{m}$.
Total $\Delta L=6.0\times10^{-4}+6.7\times10^{-4}=1.3\times10^{-3}\ \text{m}=1.3\ \text{mm}$.
Check Each wire is a spring of constant $k=YA/L$: $k_1=1.67\times10^5$, $k_2=1.5\times10^5\ \text{N/m}$. In series the extensions add: $\dfrac{100}{1.67\times10^5}+\dfrac{100}{1.5\times10^5}=1.27\times10^{-3}\ \text{m}$ ✓.

Measuring $Y$ in the laboratory. The extensions here are fractions of a millimetre, so a careful method is needed. In the classic Searle's method, two long, thin wires of the same material hang side by side from one rigid support. The reference wire carries a fixed load to keep it straight; the experimental wire carries a hanger to which slotted masses are added. A spirit level bridges the two wires, and a micrometer screw (least count $0.01\ \text{mm}$) is turned until the level is horizontal again after each new load. The reading change is the extension. Because both wires hang from the same support and share the same room, any sag of the support or thermal expansion affects both equally and cancels out. The wire is long and thin so that the extension is large enough to measure; readings are taken while loading and again while unloading, and averaged, to check that the wire stayed within its elastic limit.

By Result 8.3, load $F$ is proportional to extension $\Delta L$, so a graph of $F$ against $\Delta L$ is a straight line through the origin with slope $\dfrac{AY}{L}$. Hence

$$Y=\frac{L}{A}\times(\text{slope of the }F\text{–}\Delta L\text{ graph}).$$

Using the slope of the best-fit line rather than a single reading averages out the small random errors of the individual readings.

Interactive 1 — Young's Modulus Lab (Searle-style)

Press Play to add $0.50\ \text{kg}$ slotted masses one at a time to the experimental wire (E). Each extension is read to the micrometer's least count of $0.01\ \text{mm}$ and plotted on the load–extension graph, and the best-fit straight line gives $Y=\dfrac{L}{A}\times\text{slope}$. Change the wire's metal, length or diameter and run the experiment again. The stretch of E is drawn exaggerated; the reference wire R never moves.

2.0 m
load on E—
micrometer reading—
slope of best-fit line—
$Y$ from the graph—

Shear modulus and bulk modulus

A shearing stress changes the shape of a body without changing its volume: a cube becomes a slanted box. The stiffness against this kind of deformation is a separate property of the material.

📘 Definition 8.8 — Shear modulus (modulus of rigidity)

The shear modulus $G$ of a material is the ratio of shearing stress to shearing strain, within the proportional limit:

$$G=\frac{F/A}{\Delta x/L}=\frac{FL}{A\,\Delta x}=\frac{F}{A\theta},$$

where a tangential force $F$ acts on a face of area $A$, displacing it by $\Delta x$ relative to the opposite face a distance $L$ away, and $\theta\approx\Delta x/L$ is the angle of shear in radians. Unit: $\text{Pa}$. It is also called the modulus of rigidity.

The table shows that for most solids $G$ is roughly one third to two fifths of $Y$: it is easier to shear a solid than to stretch it. Only solids have a shear modulus. A liquid at rest cannot resist a shearing stress at all — however gently you push the top layer sideways, it keeps flowing — so a fluid's shear modulus is zero.

✏️ Example 4 — Shearing a copper block and a bridge bearing
(a) A copper cube of side $10\ \text{cm}$ has its base fixed. A tangential force of $2.0\times10^5\ \text{N}$ acts on its top face. Find the shearing stress, the shearing strain and the displacement of the top face. (b) A bridge deck rests on a rubber bearing pad $0.30\ \text{m}\times0.40\ \text{m}$ in area and $5.0\ \text{cm}$ thick, with shear modulus $1.0\times10^6\ \text{Pa}$. On a hot day the deck expands and drags the top of the pad $1.0\ \text{cm}$ sideways relative to its base. Find the shearing strain, the shearing stress and the horizontal force on the pad.
(a) stress $A=(0.10)^2=0.010\ \text{m}^2$; shearing stress $=\dfrac{2.0\times10^5}{0.010}=2.0\times10^7\ \text{Pa}$.
(a) strain $\theta=\dfrac{\text{stress}}{G}=\dfrac{2.0\times10^7}{4.4\times10^{10}}=4.5\times10^{-4}\ \text{rad}$; $\ \Delta x=\theta L=4.5\times10^{-4}\times0.10=4.5\times10^{-5}\ \text{m}$ — less than a twentieth of a millimetre.
(b) strain $\theta=\dfrac{\Delta x}{L}=\dfrac{1.0}{5.0}=0.20\ \text{rad}$ (about $11^\circ$).
(b) force Stress $=G\theta=1.0\times10^6\times0.20=2.0\times10^5\ \text{Pa}$; $\ F=\text{stress}\times A=2.0\times10^5\times0.12=2.4\times10^4\ \text{N}$.
Why rubber? A low shear modulus lets the pad follow the deck's expansion with a modest force on the piers, while its large bulk modulus means it hardly compresses under the deck's weight.

Bulk modulus. When a body is immersed in a fluid, the fluid presses equally on every part of its surface. The body keeps its shape but shrinks in volume. If the pressure rises by $p$ (the hydraulic stress) and the volume $V$ changes by $\Delta V$ (negative), the volume strain is $\Delta V/V$.

📘 Definition 8.9 — Bulk modulus and compressibility

The bulk modulus $B$ of a material is the ratio of hydraulic stress to volume strain, within the proportional limit:

$$B=-\frac{p}{\Delta V/V}=-\frac{pV}{\Delta V}.$$

The minus sign makes $B$ positive, because an increase in pressure produces a decrease in volume. Unit: $\text{Pa}$. The reciprocal of the bulk modulus is the compressibility, $\ k=\dfrac1B$, the fractional decrease in volume per unit increase in pressure (unit $\text{Pa}^{-1}$).

Solids and liquids have large bulk moduli: their atoms are already packed close together, and squeezing them closer is resisted by strong repulsive forces. Water's bulk modulus, $2.2\times10^9\ \text{Pa}$, means that even $1000$ atmospheres shrinks it by only about $5\%$ — which is why liquids are treated as incompressible in hydraulic machines (Chapter 9). Gases are thousands of times more compressible; their bulk modulus is of the order of the gas pressure itself (about $10^5\ \text{Pa}$ for air at room conditions), as you will see in Chapters 11 and 12.

✏️ Example 5 — The deepest ocean
At the bottom of the deepest ocean trench the pressure is about $1.1\times10^8\ \text{Pa}$ above atmospheric. (a) By what percentage is a sample of water there compressed, and by what percentage is its density increased? ($B_\text{water}=2.2\times10^9\ \text{Pa}$.) (b) A solid steel sphere of radius $10\ \text{cm}$ ($B=1.6\times10^{11}\ \text{Pa}$) is lowered to that depth. Find the decrease in its volume.
(a) $\dfrac{\Delta V}{V}=-\dfrac{p}{B}=-\dfrac{1.1\times10^8}{2.2\times10^9}=-0.050$: the water is compressed by $5.0\%$.
(a) density The mass is unchanged, so $\rho\propto\dfrac1V$: $\dfrac{\rho'}{\rho}=\dfrac{1}{1-0.050}=1.053$ — about $5.3\%$ denser (for small strains, the density rises by the same fraction as the volume falls).
(b) $V=\tfrac43\pi(0.10)^3=4.19\times10^{-3}\ \text{m}^3$; $\ \dfrac{\Delta V}{V}=-\dfrac{1.1\times10^8}{1.6\times10^{11}}=-6.9\times10^{-4}$; $\ \Delta V=-2.9\times10^{-6}\ \text{m}^3$, a decrease of $2.9\ \text{cm}^3$.
Check Steel's bulk modulus is about $70$ times water's, so its volume strain is about $70$ times smaller: $\dfrac{0.050}{6.9\times10^{-4}}=73$ ✓.

Poisson's ratio

Stretch a rubber band and watch it narrow; a steel wire does the same, though far too little to see. A longitudinal strain is always accompanied by a lateral strain $\Delta d/d$ — a fractional change in width or diameter — of the opposite sign.

📘 Definition 8.10 — Poisson's ratio

For a body stretched (or compressed) along its length within the proportional limit, Poisson's ratio $\nu$ is the ratio of the lateral strain to the longitudinal strain, with a minus sign so that $\nu$ is positive:

$$\nu=-\frac{\Delta d/d}{\Delta L/L}.$$

It is a ratio of two strains, so it has no unit and no dimensions. It depends only on the material: about $0.2$–$0.4$ for most metals and glasses, very close to $0.5$ for rubber.

Does a stretched wire change its volume? For a wire of diameter $d$ and length $L$, $V=\tfrac{\pi}{4}d^2L$. For small changes, the fractional change of a product is the sum of the fractional changes of its factors (because $(1+a)(1+b)\approx1+a+b$ when $a$ and $b$ are small), so

$$\frac{\Delta V}{V}=2\,\frac{\Delta d}{d}+\frac{\Delta L}{L}=-2\nu\varepsilon+\varepsilon .$$
📐 Result 8.4 — Volume change of a stretched wire

A wire or rod with Poisson's ratio $\nu$, given a small longitudinal strain $\varepsilon=\Delta L/L$, has lateral strain $-\nu\varepsilon$ and changes in volume by

$$\frac{\Delta V}{V}=\varepsilon\,(1-2\nu).$$

For ordinary materials $\nu\le0.5$, so stretching never decreases the volume. When $\nu=0.5$ (rubber, very nearly) the volume stays constant to first order: the body simply changes shape.

✏️ Example 6 — How much does a stretched wire narrow?
A steel wire $3.0\ \text{m}$ long and $2.0\ \text{mm}$ in diameter is stretched by $1.5\ \text{mm}$. Steel's Poisson's ratio is $0.29$. Find (a) the longitudinal strain, (b) the change in diameter, (c) the fractional change in volume. (d) What tension produced the stretch?
(a) $\varepsilon=\dfrac{1.5\times10^{-3}}{3.0}=5.0\times10^{-4}$.
(b) Lateral strain $=-\nu\varepsilon=-0.29\times5.0\times10^{-4}=-1.45\times10^{-4}$; $\ \Delta d=-1.45\times10^{-4}\times2.0\ \text{mm}=-2.9\times10^{-4}\ \text{mm}$ — the diameter shrinks by $0.29\ \mu\text{m}$, about one three-hundredth of the thickness of a human hair.
(c) $\dfrac{\Delta V}{V}=\varepsilon(1-2\nu)=5.0\times10^{-4}\times0.42=2.1\times10^{-4}$: a slight increase.
(d) $F=\dfrac{YA\,\Delta L}{L}=\dfrac{2.0\times10^{11}\times\pi(1.0\times10^{-3})^2\times1.5\times10^{-3}}{3.0}=3.1\times10^2\ \text{N}$.
Interactive 2 — Stretch, Shear, Squeeze: the Three Moduli

Apply the same kind of stress to blocks of different materials. Stretching uses Young's modulus and also narrows the block by Poisson's ratio; shearing uses the shear modulus and keeps the volume fixed; squeezing in a fluid uses the bulk modulus and keeps the shape. The dashed outline is the undeformed block and the deformation is drawn exaggerated by the factor shown. The stress slider runs up to a safe value for each material.

50 MPa
modulus used—
main strain—
side effect—
drawn exaggerated—

Elastic potential energy

Stretching a wire takes work, and within the elastic limit that work is not lost: it is stored in the strained bonds between the atoms and given back when the load is removed, as a catapult's rubber shows. When a wire is stretched slowly from its natural length to an extension $\Delta L$ within the proportional limit, the force needed grows in proportion to the extension, from $0$ to $F$ (Result 8.3). The work done is the area under the straight force–extension line — a triangle, exactly as for the spring of Section 5.2.

📐 Result 8.5 — Elastic potential energy of a stretched wire

A wire of length $L$ and cross-section $A$, stretched by $\Delta L$ within its proportional limit by a final force $F$, stores elastic potential energy

$$U=\tfrac12F\,\Delta L=\tfrac12\,\frac{YA}{L}(\Delta L)^2 .$$

Dividing by the wire's volume $AL$ gives the energy per unit volume, which depends only on the stress $\sigma$ and the strain $\varepsilon$:

$$u=\tfrac12\,\sigma\varepsilon=\tfrac12\,Y\varepsilon^2=\frac{\sigma^2}{2Y}.$$

The factor $\tfrac12$ is there because the force was smaller than $F$ during most of the stretch. For the steel wire of Example 1 ($F=39.2\ \text{N}$, $\Delta L=2.0\ \text{mm}$): $U=\tfrac12(39.2)(2.0\times10^{-3})=3.9\times10^{-2}\ \text{J}$. Per unit volume, $\sigma=2.0\times10^8\ \text{Pa}$ and $\varepsilon=1.0\times10^{-3}$ give $u=1.0\times10^5\ \text{J/m}^3$; times the volume $AL=3.93\times10^{-7}\ \text{m}^3$ this is again $3.9\times10^{-2}\ \text{J}$ ✓. Because $u=\dfrac{\sigma^2}{2Y}$, at the same stress a less stiff material stores more energy — which is why springs and bows are designed to strain a lot.

🧭 Problem-solving strategy — choosing the modulus
  1. Identify the deformation: change of length along the force → $Y$; sliding of one face relative to the opposite face → $G$; change of volume under pressure from all sides → $B$; sideways narrowing of a stretched body → $\nu$.
  2. Write the definition of that modulus as stress/strain and solve for the unknown. For wires use $\Delta L=\dfrac{FL}{AY}$.
  3. For two wires in series (end to end) the tension is the same and the extensions add; side by side (sharing a load through a rigid bar kept level) the extensions are equal and the loads add.
  4. For ratio questions write the proportionality first — e.g. $\Delta L\propto\dfrac{FL}{d^2Y}$ — then substitute the ratios.
  5. Check the size: metal strains of $10^{-4}$–$10^{-3}$, volume strains of liquids of order $10^{-3}$–$10^{-2}$ even at enormous pressures.
⚠️ Common mistakes

1. Wrong length in the shear formula. In $G=\dfrac{FL}{A\,\Delta x}$, $L$ is the distance between the two sheared faces (the thickness of the pad), and $A$ is the area of the face the force is parallel to.

2. Dropping the minus sign in $B$. $\Delta V$ is negative when $p$ is positive; $B$ itself is always positive.

3. Using Young's modulus for a liquid. Liquids and gases have only a bulk modulus.

4. Poisson's ratio with the strains upside down. $\nu$ = lateral strain ÷ longitudinal strain (without the sign), never the other way round; it is less than $0.5$ for ordinary materials.

5. Squaring the wrong thing. In $\Delta L=\dfrac{4FL}{\pi d^2Y}$ the extension depends on $d^2$, not on $d$ — doubling the diameter quarters the extension.

6. Assuming a larger modulus means a stronger material. The modulus measures stiffness (stress per unit strain); strength is the stress at which the material fails (Section 8.1). Glass is about as stiff as aluminium but far weaker in tension.

Exercises

Take $g=9.8\ \text{m/s}^2$ and use the table of moduli in this section unless other values are given.

I
Young's modulus.
  • 1 A wire $2.0\ \text{m}$ long with cross-section $1.0\ \text{mm}^2$ stretches by $2.0\ \text{mm}$ under a load of $200\ \text{N}$. Find its Young's modulus.
  • 2 A copper wire $3.0\ \text{m}$ long and $1.0\ \text{mm}$ in diameter carries a load of $50\ \text{N}$. Find its extension.
  • 3 An aluminium rod $1.0\ \text{m}$ long and $2.0\ \text{cm}$ in diameter is compressed by a force of $1.0\times10^4\ \text{N}$ along its length. By how much does it shorten?
  • 4 What load stretches a steel wire $0.80\ \text{mm}$ in diameter by $0.10\%$ of its length?
  • 5 Write the dimensional formula of Young's modulus. What would be the Young's modulus of a perfectly rigid body?
  • 6 Wires A and B are of the same metal. A is $1.0\ \text{m}$ long and $0.50\ \text{mm}$ in diameter; B is $2.0\ \text{m}$ long and $1.0\ \text{mm}$ in diameter. Find the ratio of the loads $F_B/F_A$ that give them equal extensions.
  • 7 In a Searle's experiment on a wire $2.0\ \text{m}$ long and $0.50\ \text{mm}$ in diameter, the load–extension graph has a slope of $1.0\times10^4\ \text{N/m}$. Find Young's modulus.
  • 8 In Searle's method, why is a reference wire hung from the same support, and why is the experimental wire long and thin?
1 $Y=\dfrac{FL}{A\,\Delta L}=\dfrac{200\times2.0}{1.0\times10^{-6}\times2.0\times10^{-3}}=2.0\times10^{11}\ \text{Pa}$.
2 $A=7.85\times10^{-7}\ \text{m}^2$; $\Delta L=\dfrac{50\times3.0}{7.85\times10^{-7}\times1.2\times10^{11}}=1.6\times10^{-3}\ \text{m}=1.6\ \text{mm}$.
3 $A=\pi(0.010)^2=3.14\times10^{-4}\ \text{m}^2$; $\Delta L=\dfrac{1.0\times10^4\times1.0}{3.14\times10^{-4}\times7.0\times10^{10}}=4.5\times10^{-4}\ \text{m}=0.45\ \text{mm}$.
4 Strain $1.0\times10^{-3}$, stress $=Y\varepsilon=2.0\times10^8\ \text{Pa}$; $A=5.03\times10^{-7}\ \text{m}^2$; $F=1.0\times10^2\ \text{N}$ ($101$).
5 $[\text{M L}^{-1}\text{T}^{-2}]$. A perfectly rigid body has zero strain under any stress, so its Young's modulus would be infinite.
6 Equal $\Delta L$ needs $F\propto\dfrac{d^2}{L}$: $\dfrac{F_B}{F_A}=\dfrac{(1.0)^2/2.0}{(0.50)^2/1.0}=\dfrac{0.50}{0.25}=2$.
7 $A=1.96\times10^{-7}\ \text{m}^2$; $Y=\dfrac{L}{A}\times\text{slope}=\dfrac{2.0\times1.0\times10^4}{1.96\times10^{-7}}=1.0\times10^{11}\ \text{Pa}$.
8 Any sag of the support or thermal expansion of the wires affects both wires equally, so it cancels in the comparison. Extension $\propto L/d^2$, so a long, thin wire gives an extension large enough to measure accurately with a micrometer.
II
Shear modulus and bulk modulus.
  • 1 A tangential force of $4.0\times10^4\ \text{N}$ acts on the top face, of area $25\ \text{cm}^2$, of a copper block $4.0\ \text{cm}$ high whose base is fixed. Find the shearing stress, the shearing strain and the displacement of the top face.
  • 2 A rubber bearing pad of area $0.20\ \text{m}\times0.30\ \text{m}$ and thickness $4.0\ \text{cm}$ has shear modulus $8.0\times10^5\ \text{Pa}$. A horizontal force of $2.4\times10^3\ \text{N}$ acts on its top face. Find the shearing strain and the sideways displacement of the top.
  • 3 The top face of a $5.0\ \text{cm}$ cube moves $0.020\ \text{mm}$ relative to the base under a tangential force of $1.0\times10^4\ \text{N}$. Find the shear modulus of the material.
  • 4 By how much does the volume of $1.0\ \text{L}$ of water decrease when the pressure on it is increased by $1.0\times10^7\ \text{Pa}$?
  • 5 What pressure is needed to reduce the volume of a copper block by $0.10\%$?
  • 6 Find the compressibility of water.
  • 7 Why is the bulk modulus of a gas so much smaller than that of a solid or a liquid? Why does a liquid have no shear modulus?
  • 8 A glass sphere's volume decreases by $0.010\%$ when it is placed in a pressure chamber. Find the increase in pressure.
1 Stress $=\dfrac{4.0\times10^4}{2.5\times10^{-3}}=1.6\times10^7\ \text{Pa}$; $\theta=\dfrac{1.6\times10^7}{4.4\times10^{10}}=3.6\times10^{-4}\ \text{rad}$; $\Delta x=\theta L=3.64\times10^{-4}\times0.040=1.5\times10^{-5}\ \text{m}$.
2 Stress $=\dfrac{2.4\times10^3}{0.060}=4.0\times10^4\ \text{Pa}$; $\theta=\dfrac{4.0\times10^4}{8.0\times10^5}=0.050\ \text{rad}$; $\Delta x=0.050\times4.0\ \text{cm}=0.20\ \text{cm}=2.0\ \text{mm}$.
3 Stress $=\dfrac{1.0\times10^4}{2.5\times10^{-3}}=4.0\times10^6\ \text{Pa}$; strain $=\dfrac{0.020}{50}=4.0\times10^{-4}$; $G=1.0\times10^{10}\ \text{Pa}$.
4 $\Delta V=-\dfrac{pV}{B}=-\dfrac{1.0\times10^7\times1.0\times10^{-3}}{2.2\times10^9}=-4.5\times10^{-6}\ \text{m}^3$: a decrease of $4.5\ \text{cm}^3$.
5 $p=B\times\dfrac{|\Delta V|}{V}=1.4\times10^{11}\times1.0\times10^{-3}=1.4\times10^8\ \text{Pa}$ (about $1400$ atmospheres).
6 $k=\dfrac1B=\dfrac{1}{2.2\times10^9}=4.5\times10^{-10}\ \text{Pa}^{-1}$.
7 Gas molecules are far apart with weak forces between them, so a small pressure rise produces a large fractional volume change. A liquid's molecules slide past one another freely, so a liquid cannot hold a static shear strain: under any shearing stress it flows.
8 $p=B\times\dfrac{|\Delta V|}{V}=4.0\times10^{10}\times1.0\times10^{-4}=4.0\times10^6\ \text{Pa}$.
III
Poisson's ratio.
  • 1 A wire $2.0\ \text{m}$ long and $1.0\ \text{mm}$ in diameter is stretched by $1.0\ \text{mm}$, and its diameter decreases by $1.5\times10^{-4}\ \text{mm}$. Find Poisson's ratio.
  • 2 A rubber strip ($\nu=0.50$) is given a small longitudinal strain of $0.020$. Find its lateral strain and its fractional change in volume.
  • 3 A steel rod is given a longitudinal strain of $1.0\times10^{-3}$. Find its fractional change in volume.
  • 4 Use Result 8.4 to explain why Poisson's ratio of an ordinary material cannot exceed $0.5$.
  • 5 A copper wire $1.5\ \text{mm}$ in diameter carries a stress of $6.0\times10^7\ \text{Pa}$. Find the longitudinal strain and the decrease in diameter.
  • 6 State the unit and dimensions of Poisson's ratio, and explain why it is defined with a minus sign.
1 Longitudinal strain $=\dfrac{1.0}{2000}=5.0\times10^{-4}$; lateral strain $=-\dfrac{1.5\times10^{-4}}{1.0}=-1.5\times10^{-4}$; $\nu=\dfrac{1.5\times10^{-4}}{5.0\times10^{-4}}=0.30$.
2 Lateral strain $=-0.50\times0.020=-0.010$; $\dfrac{\Delta V}{V}=0.020(1-1.0)=0$ — the volume is unchanged (to first order).
3 $\dfrac{\Delta V}{V}=1.0\times10^{-3}(1-0.58)=4.2\times10^{-4}$.
4 If $\nu\gt0.5$, then $1-2\nu\lt0$ and stretching ($\varepsilon\gt0$) would decrease the volume — pulling would compress the material, which ordinary materials never do. So $\nu\le0.5$, with $0.5$ meaning no volume change.
5 $\varepsilon=\dfrac{6.0\times10^7}{1.2\times10^{11}}=5.0\times10^{-4}$; lateral strain $=-0.36\times5.0\times10^{-4}=-1.8\times10^{-4}$; decrease in diameter $=1.8\times10^{-4}\times1.5\ \text{mm}=2.7\times10^{-4}\ \text{mm}$.
6 No unit, dimensionless (a ratio of two strains). The lateral strain is negative when the longitudinal strain is positive; the minus sign makes $\nu$ a positive number.
IV
Applications and harder problems.
  • 1 (Modelling) A mine lift hangs from a steel cable $50\ \text{m}$ long with cross-section $2.0\ \text{cm}^2$. By how much does the cable stretch when a $2.0\ \text{t}$ load is suspended from it at rest?
  • 2 A steel wire and a copper wire, each $2.0\ \text{m}$ long with cross-section $1.0\ \text{mm}^2$, hang side by side from a ceiling and are joined at the bottom by a light rigid bar that stays horizontal. A $160\ \text{N}$ load hangs from the bar at the point that keeps it level. Find the tension in each wire and their common extension.
  • 3 (Modelling) The shaft of a thigh bone is $0.45\ \text{m}$ long with $6.0\ \text{cm}^2$ of solid bone in its cross-section; bone's Young's modulus is $1.8\times10^{10}\ \text{Pa}$. By how much does the bone shorten when a $70\ \text{kg}$ person stands on one leg?
  • 4 (Modelling) A deep-sea vessel is a hollow steel sphere of outer radius $1.0\ \text{m}$. At a depth of $4.0\ \text{km}$ the water pressure is $\rho gh$ above atmospheric (take $\rho=1.0\times10^3\ \text{kg/m}^3$). Treating the sphere's outer volume as if it were solid steel, estimate the decrease in that volume.
  • 5 Show that when a liquid is compressed by a small pressure $p$, its density increases by the fraction $\dfrac{\Delta\rho}{\rho}\approx\dfrac{p}{B}$. Find the percentage increase in the density of water at a depth of $4.0\ \text{km}$.
  • 6 A metal wire stretches by $1.2\ \text{mm}$ under a certain load. A wire of the same metal with twice the length and three times the diameter carries twice the load. Find its extension.
  • 7 A copper wire $1.5\ \text{m}$ long and $1.0\ \text{mm}$ in diameter carries a $10\ \text{kg}$ load. Find its extension, the elastic potential energy stored in it, and the energy stored per unit volume.
1 $F=2.0\times10^3\times9.8=1.96\times10^4\ \text{N}$; $\Delta L=\dfrac{1.96\times10^4\times50}{2.0\times10^{-4}\times2.0\times10^{11}}=2.5\times10^{-2}\ \text{m}=2.5\ \text{cm}$.
2 Equal extensions and equal $L$, $A$: $\dfrac{T_1}{Y_1}=\dfrac{T_2}{Y_2}$, so the load divides in the ratio $200:120$. $T_\text{steel}=160\times\dfrac{200}{320}=100\ \text{N}$, $T_\text{Cu}=60\ \text{N}$. $\Delta L=\dfrac{100\times2.0}{1.0\times10^{-6}\times2.0\times10^{11}}=1.0\times10^{-3}\ \text{m}=1.0\ \text{mm}$ (copper: $\dfrac{60\times2.0}{1.2\times10^5}=1.0\ \text{mm}$ ✓).
3 $\Delta L=\dfrac{686\times0.45}{6.0\times10^{-4}\times1.8\times10^{10}}=2.9\times10^{-5}\ \text{m}$ — about $0.03\ \text{mm}$.
4 $p=1.0\times10^3\times9.8\times4000=3.92\times10^7\ \text{Pa}$; $\dfrac{\Delta V}{V}=-\dfrac{3.92\times10^7}{1.6\times10^{11}}=-2.45\times10^{-4}$; $V=4.19\ \text{m}^3$, so $\Delta V=-1.0\times10^{-3}\ \text{m}^3$ (about one litre). A real hollow hull deforms more than a solid ball would.
5 $\rho V$ is constant, so $\dfrac{\Delta\rho}{\rho}\approx-\dfrac{\Delta V}{V}=\dfrac{p}{B}$. At $4.0\ \text{km}$: $\dfrac{3.92\times10^7}{2.2\times10^9}=0.018$, i.e. $1.8\%$ denser.
6 $\Delta L\propto\dfrac{FL}{d^2}$: factor $\dfrac{2\times2}{3^2}=\dfrac49$; extension $=1.2\times\dfrac49=0.53\ \text{mm}$.
7 As in Example 2, $\Delta L=\dfrac{98\times1.5}{7.85\times10^{-7}\times1.2\times10^{11}}=1.56\times10^{-3}\ \text{m}$. $U=\tfrac12F\,\Delta L=\tfrac12(98)(1.56\times10^{-3})=7.6\times10^{-2}\ \text{J}$. $\sigma=\dfrac{98}{7.85\times10^{-7}}=1.25\times10^8\ \text{Pa}$, so $u=\dfrac{\sigma^2}{2Y}=\dfrac{(1.25\times10^8)^2}{2.4\times10^{11}}=6.5\times10^4\ \text{J/m}^3$ (check, unrounded: $u\times AL=6.49\times10^4\times1.178\times10^{-6}=7.6\times10^{-2}\ \text{J}$ ✓).
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Define Young's modulus, shear modulus and bulk modulus. Which of them can be defined for a liquid, and why?

Y Ratio of longitudinal (tensile or compressive) stress to longitudinal strain, $Y=\dfrac{FL}{A\,\Delta L}$, within the proportional limit. [1]
G, B $G$ = shearing stress / shearing strain $=\dfrac{F}{A\theta}$; $B$ = hydraulic stress / volume strain $=-\dfrac{p}{\Delta V/V}$. [1]
Liquids Only the bulk modulus. A liquid has no fixed shape: it cannot sustain a static shearing stress (it flows) and cannot be stretched into a longer fixed length, so $Y$ and $G$ are not defined; it does resist compression, so $B$ is. [1]
Exam tip Every definition needs "within the elastic (proportional) limit" and the sign convention for $B$.

E2. JEE Main (Numerical value) A steel wire ($Y=2.0\times10^{11}\ \text{Pa}$) and a copper wire ($Y=1.2\times10^{11}\ \text{Pa}$) of equal length and equal cross-section are joined end to end and hung from a ceiling. When a load is attached to the lower end, the total extension is $1.6\ \text{mm}$. Find the extension of the steel wire in mm.

Set-up Series: same tension, same $L$ and $A$, so $\Delta L\propto\dfrac1Y$: $\dfrac{\Delta L_s}{\Delta L_c}=\dfrac{1.2}{2.0}=0.60$.
Solve $\Delta L_s+\Delta L_c=1.6$ with $\Delta L_c=\dfrac{\Delta L_s}{0.60}$: $\ \Delta L_s\left(1+\dfrac{1}{0.60}\right)=1.6$, $\ \Delta L_s=0.60\ \text{mm}$ (and $\Delta L_c=1.0\ \text{mm}$).
Answer $0.60$.
Exam tip The stiffer wire stretches less. Splitting $1.6\ \text{mm}$ in the ratio $2.0:1.2$ instead of $1.2:2.0$ gives the trap answer $1.0$.

E3. JEE Advanced (One or more correct) A uniform rod of length $L$, cross-section $A$, density $\rho$ and Young's modulus $Y$ hangs vertically from its top end under its own weight $W$. Which statements are correct? (A) Its extension is $\dfrac{\rho gL^2}{2Y}$. (B) Its extension equals that produced in the same rod, if weightless, by a load $\tfrac12W$ at its lower end. (C) Its extension does not depend on $A$. (D) Doubling the length doubles the extension.

Idea The stress at height $y$ above the lower end is $\rho gy$ (Section 8.1, Exam corner E5), rising linearly from $0$ to $\rho gL$. A thin slice of length $dy$ stretches by $\dfrac{\rho gy}{Y}dy$; adding up (the area under a straight line) gives $\Delta L=\dfrac{\rho g}{Y}\cdot\dfrac{L^2}{2}$ — the same as using the average stress $\tfrac12\rho gL$ over the whole length.
(A), (B) $\Delta L=\dfrac{\rho gL^2}{2Y}$ ✓. With $W=\rho ALg$, this is $\dfrac{(W/2)L}{AY}$ — the extension due to a load $\tfrac12W$ at the end ✓.
(C), (D) $A$ does not appear in (A) ✓. $\Delta L\propto L^2$, so doubling $L$ quadruples it ✗.
Answer (A), (B), (C).
Exam tip Limiting check: a rod with $\rho\to0$ does not stretch ✓; dimensions of $\rho gL^2/Y$: $\dfrac{[\text{M L}^{-3}][\text{L T}^{-2}][\text{L}^2]}{[\text{M L}^{-1}\text{T}^{-2}]}=[\text{L}]$ ✓.

E4. NEET The pressure needed to reduce the volume of water by $0.50\%$ is ($B=2.2\times10^9\ \text{Pa}$) (a) $1.1\times10^9\ \text{Pa}$ (b) $4.4\times10^{11}\ \text{Pa}$ (c) $1.1\times10^7\ \text{Pa}$ (d) $2.2\times10^7\ \text{Pa}$

Idea $p=B\,\dfrac{|\Delta V|}{V}=2.2\times10^9\times0.0050=1.1\times10^7\ \text{Pa}$.
Answer (c).
Exam tip (a) uses $0.50$ instead of $0.0050$ (per cent not converted); (b) divides by the strain instead of multiplying; (d) is the pressure for $1\%$.

E5. JEE Main NEET A wire made of a material with Poisson's ratio $0.50$ is stretched slightly within its elastic limit. Its volume (a) increases (b) decreases (c) stays the same, to first order in the strain (d) first decreases and then increases

Idea Result 8.4: $\dfrac{\Delta V}{V}=\varepsilon(1-2\nu)=\varepsilon(1-1)=0$.
Answer (c).
Exam tip For $\nu\lt0.5$ (metals) the volume increases slightly on stretching; (b) would need $\nu\gt0.5$.

E6. Board · Case study In a Searle's experiment a wire $2.50\ \text{m}$ long and $0.40\ \text{mm}$ in diameter gives these extensions for loads of $1.0$, $2.0$, $3.0$ and $4.0\ \text{kg}$: $0.97$, $1.95$, $2.92$ and $3.90\ \text{mm}$. (i) Use the first and last readings to find the slope of the load–extension graph and hence Young's modulus. [2] (ii) What is the purpose of the reference wire? [1] (iii) Why are readings also taken while the masses are removed? [1]

(i) Slope $=\dfrac{(4.0-1.0)\times9.8}{(3.90-0.97)\times10^{-3}}=\dfrac{29.4}{2.93\times10^{-3}}=1.00\times10^4\ \text{N/m}$. $A=\dfrac{\pi(0.40\times10^{-3})^2}{4}=1.26\times10^{-7}\ \text{m}^2$; $\ Y=\dfrac{L}{A}\times\text{slope}=\dfrac{2.50\times1.00\times10^4}{1.26\times10^{-7}}=2.0\times10^{11}\ \text{Pa}$ (steel). [2]
(ii) It hangs from the same support, so any sag of the support or change in room temperature affects both wires equally; measuring E relative to R cancels these effects. [1]
(iii) If the unloading readings agree with the loading readings, the wire has stayed within its elastic limit (no permanent set), and averaging the two sets reduces random errors. [1]
Exam tip The equal steps of about $0.98\ \text{mm}$ per kilogram already show the graph is a straight line — Hooke's law holds.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. Why can a very stiff metal hip implant make the bone around it grow weaker — and how do biomedical engineers choose implant materials to prevent this?

Why it matters. Millions of hip and knee replacements are fitted worldwide every year. Bone is a living material that strengthens where it is loaded and thins where it is not. When a metal stem is fixed inside the thigh bone, the two share the load, and the much stiffer metal takes most of it. The bone around the stem is "shielded" from stress, can weaken over the years, and the implant may loosen. Implant designers in orthopaedic research groups compare materials using exactly the Young's-modulus ideas of this section.

What to do. Model the bone and the implant stem as two rods of equal length side by side, fixed together so that they shorten by the same amount — like the two wires of Exercise IV.2 joined by a rigid bar.

  1. Show that, for equal strains, the load divides between bone and stem in the ratio $Y_\text{bone}A_\text{bone}:Y_\text{stem}A_\text{stem}$.
  2. Take a ring of bone of cross-section $4.0\ \text{cm}^2$ with $Y_\text{bone}=18\ \text{GPa}$ and a stem of cross-section $1.5\ \text{cm}^2$. Find the fraction of the load carried by the bone for stems of cobalt–chromium alloy ($210\ \text{GPa}$), titanium alloy ($110\ \text{GPa}$) and a stiff polymer such as PEEK ($4\ \text{GPa}$).
  3. For a $2.0\ \text{kN}$ load while walking, find the stress in the bone in each case and compare with the stress if there were no implant at all.
  4. Which material would you choose to reduce stress shielding? What other properties (strength, fatigue, corrosion, how the body reacts to it) must the engineers also consider?
  5. Find out how porous or "lattice" metal stems made by 3-D printing lower the effective Young's modulus of an implant.

Tools: Calculator or spreadsheet; review articles on stress shielding and implant materials.

Hand in: Your derivation, a table of load fractions and bone stresses for the three materials, and a one-page recommendation to a surgical team.

Going further. The real loads include bending and twisting, so the shear modulus matters too. Find out how finite-element computer models refine this simple parallel-rod picture.

Instructor note: $Y_\text{bone}A_\text{bone}=18\times4.0=72$ (GPa cm²). Co–Cr: $210\times1.5=315$, bone fraction $72/387=0.19$; Ti: $165$, fraction $72/237=0.30$; PEEK: $6$, fraction $72/78=0.92$. With $2.0\ \text{kN}$: bone stress without implant $5.0\ \text{MPa}$; with Co–Cr $0.93\ \text{MPa}$, Ti $1.5\ \text{MPa}$, PEEK $4.6\ \text{MPa}$. Lower-modulus stems shield less but may be too weak or too flexible at the joint — a genuine design trade-off.

🧪 Self-check quiz

0/5 correct

Q1. A wire stretches by $1.0\ \text{mm}$ under a certain load. A wire of the same material, twice as long and twice the diameter, carries the same load. Its extension is

Q2. Which elastic modulus can be defined for water?

Q3. A rubber pad $5.0\ \text{cm}$ thick has its top face pushed $1.0\ \text{mm}$ sideways relative to its base. The shearing strain is

Q4. The pressure on a sample of water ($B=2.2\times10^9\ \text{Pa}$) is increased by $2.2\times10^7\ \text{Pa}$. Its volume decreases by

Q5. When a steel wire ($\nu=0.29$) is stretched slightly within its elastic limit,

← 8.1 Stress, Strain and Hooke's LawChapter 9 — Mechanical Properties of Fluids →