Chapter 8 · Mechanical Properties of Solids

8.1 Stress, Strain and Hooke's Law

Every solid gives a little when it is loaded. Stress $\sigma=F/A$ measures how hard a solid is being pulled, pushed or sheared, strain measures how much it deforms, and up to a limit the two are proportional — Hooke's law. The stress–strain curve tells the rest of the story, all the way to fracture.

A lift cable carrying a full car stretches by a few millimetres; the steel cables of a suspension bridge lengthen by centimetres when rush-hour traffic rolls on, and shorten again at night. A bungee cord stretches to several times its length and still springs back, but a glass rod snaps after barely changing length at all, and a copper wire pulled hard enough simply flows, thins and never returns. In Chapters 4–6 we treated bodies as perfectly rigid. That was a useful idealisation — but no real solid is rigid, and engineers, surgeons and materials scientists need to know exactly how much a solid gives, and when it will give way.

This section builds the language for that. Stress measures how hard a solid is being loaded — force per unit area, so that a thick cable and a thin wire can be compared fairly. Strain measures the resulting deformation as a fraction of the original size. For small deformations stress is proportional to strain (Hooke's law), and the constant of proportionality — a modulus of elasticity — is a property of the material alone. Beyond that region the stress–strain curve shows the elastic limit, permanent stretching, the greatest stress a material can bear, and fracture; its shape separates ductile metals from brittle glass and from rubbery elastomers.

Elastic and plastic behaviour

A solid is a huge array of atoms or molecules, each sitting where the forces from its neighbours balance. A useful picture is a three-dimensional lattice of balls joined by tiny springs. Pull on the solid and the atoms are dragged slightly apart; the interatomic forces then pull back. Push and they are squeezed together; the interatomic forces push back. These internal forces are the solid's restoring force (Section 5.2), and when the load is removed they return the atoms — and the whole body — to the original shape.

📘 Definition 8.1 — Deforming force, elasticity and plasticity

A deforming force is an external force that changes the size or shape of a body (while the body stays in equilibrium, so it is always balanced by other external forces).

Elasticity is the property of a body by which it regains its original size and shape when the deforming force is removed. A body that does so completely is perfectly elastic.

Plasticity is the property of a body by which it shows no tendency to regain its original size and shape, so that the deformation is permanent. A body that behaves like this is perfectly plastic.

No body is perfectly elastic or perfectly plastic — these are limiting cases. A quartz fibre comes very close to perfect elasticity; putty, wet clay and chewing gum come close to perfect plasticity. Most solids are elastic for small deformations and become plastic when deformed too far: bend a steel paper clip slightly and it springs back, bend it a lot and it stays bent. Where that change happens, and what follows, is the subject of the last part of this section.

Stress and strain

Hang the same $100\ \text{N}$ load from a thin wire and from a thick rod of the same steel. The thin wire is far closer to breaking — the load is shared among fewer atomic "springs". So the useful measure of loading is not the force itself but the force per unit area of the surface over which it acts. When a body is deformed and in equilibrium, an internal restoring force develops across every cross-section, equal in magnitude to the applied force.

📘 Definition 8.2 — Stress

The stress on a surface of a body is the restoring force per unit area that develops across it; in equilibrium its magnitude equals the applied force $F$ divided by the area $A$ over which it acts:

$$\sigma=\frac{F}{A}.$$

SI unit: $\text{N/m}^2$, called the pascal ($1\ \text{Pa}=1\ \text{N/m}^2$); dimensions $[\text{M L}^{-1}\text{T}^{-2}]$. There are three kinds, according to the direction of the force relative to the surface:

  • Longitudinal stress — force perpendicular (normal) to the cross-section: tensile stress when the body is stretched, compressive stress when it is squashed.
  • Shearing (tangential) stress — force parallel to the surface, $F_\parallel/A$, tending to slide one face of the body relative to the opposite face.
  • Hydraulic stress — a body immersed in a fluid is pushed inwards by equal normal forces on every face; the hydraulic stress equals the pressure $p$ of the fluid.

Stress has the same unit as pressure, but it is a more general idea: a wire under tension carries a tensile stress across its cross-section that has nothing to do with any fluid. You will study pressure in fluids properly in Chapter 9. Because loads on wires are often given in kilograms and areas in mm², remember $1\ \text{mm}^2=10^{-6}\ \text{m}^2$, and for a wire of diameter $d$, $A=\pi d^2/4$.

📘 Definition 8.3 — Strain

Strain is the fractional change in size or shape produced by a stress. It is a ratio of two like quantities, so it has no unit and no dimensions.

  • Longitudinal strain: a length $L$ changes by $\Delta L$; $\ \varepsilon=\dfrac{\Delta L}{L}$ (tensile if $\Delta L\gt0$, compressive if $\Delta L\lt0$).
  • Shearing strain: two opposite faces a distance $L$ apart are displaced by $\Delta x$ relative to each other; $\ \text{shearing strain}=\dfrac{\Delta x}{L}=\tan\theta\approx\theta$, where $\theta$ (in radians, small) is the angle through which the sides perpendicular to those faces turn.
  • Volume strain: a volume $V$ changes by $\Delta V$; $\ \text{volume strain}=\dfrac{\Delta V}{V}$ (negative under a hydraulic stress, since the volume decreases).

Tensile and compressive stress produce longitudinal strain, shearing stress produces shearing strain (a change of shape at constant volume), and hydraulic stress produces volume strain (a change of size with no change of shape). For a typical metal in everyday use, strains are tiny — of order $10^{-4}$ to $10^{-3}$ — which is why we usually treat metal structures as rigid.

✏️ Example 1 — Stress in a lift cable and in a column
(a) A lift car of total mass $800\ \text{kg}$ hangs from a steel cable of diameter $1.6\ \text{cm}$. Find the tensile stress in the cable when the car is at rest, and when it accelerates upward at $1.2\ \text{m/s}^2$. (b) A marble column of square cross-section $0.50\ \text{m}\times0.50\ \text{m}$ supports a load of $3.0\times10^5\ \text{N}$. Find the compressive stress in it (ignore the column's own weight).
Area $A=\dfrac{\pi d^2}{4}=\dfrac{\pi(0.016)^2}{4}=2.01\times10^{-4}\ \text{m}^2$.
(a) at rest The tension equals the weight: $T=800\times9.8=7840\ \text{N}$, so $\sigma=\dfrac{7840}{2.01\times10^{-4}}=3.9\times10^7\ \text{Pa}$.
(a) accelerating Newton's second law for the car: $T-mg=ma$, so $T=800(9.8+1.2)=8800\ \text{N}$ and $\sigma=\dfrac{8800}{2.01\times10^{-4}}=4.4\times10^7\ \text{Pa}$ — about $12\%$ more.
(b) $A=0.50\times0.50=0.25\ \text{m}^2$; $\ \sigma=\dfrac{3.0\times10^5}{0.25}=1.2\times10^6\ \text{Pa}$ (compressive).
Check Units: $\text{N}/\text{m}^2=\text{Pa}$ ✓. The thin cable carries a stress about $30$ times larger than the massive column, although its load is nearly $40$ times smaller — stress, not force, decides how close a material is to failing.
✏️ Example 2 — Three kinds of strain
(a) A wire $2.0\ \text{m}$ long stretches by $1.2\ \text{mm}$. (b) A rubber block $4.0\ \text{cm}$ thick is fixed at its base, and a tangential force on its top face displaces the top by $0.30\ \text{cm}$ relative to the base. (c) A sealed $1.000\ \text{L}$ sample of sea water, lowered into the deep ocean, shrinks to $0.9977\ \text{L}$. Find the strain in each case and say what kind it is.
(a) Longitudinal (tensile) strain $\dfrac{\Delta L}{L}=\dfrac{1.2\times10^{-3}}{2.0}=6.0\times10^{-4}$.
(b) Shearing strain $\dfrac{\Delta x}{L}=\dfrac{0.30}{4.0}=0.075$. The angle of shear is $\theta=\tan^{-1}0.075=0.0749\ \text{rad}=4.3^\circ$ — so the approximation $\theta\approx\tan\theta$ is excellent even at this fairly large strain.
(c) Volume strain $\dfrac{\Delta V}{V}=\dfrac{0.9977-1.000}{1.000}=-2.3\times10^{-3}$, produced by the hydraulic stress of the deep water.
Note All three are pure numbers: the units cancel. Keep the same unit top and bottom (mm with mm, or m with m) and the strain comes out right.
Interactive 1 — Four Ways to Deform a Block

A $10\ \text{cm}$ cube of soft rubber (each face $0.010\ \text{m}^2$) is loaded with a force $F$ on each face shown. Choose the kind of stress and watch the kind of strain it produces; the dashed outline is the original cube. The deformation is drawn exaggerated by the factor shown — real strains are small, and the hydraulic one is tiny. (How strain follows stress — Hooke's law — comes next; the widget uses typical moduli for rubber.)

1000 N
stress $F/A$—
kind of strain—
strain—
change in the cube—

Hooke's law

In 1676 Robert Hooke summed up his experiments on springs and wires in a Latin anagram meaning "as the extension, so the force". You used his law for a spring in Section 5.2 as $F=-kx$. For a solid material the same idea is best stated with stress and strain, because then the constant no longer depends on how long or thick the specimen is.

📐 Law 8.1 — Hooke's law

For small deformations — up to a limiting stress called the proportional limit, which depends on the material — the stress in a body is directly proportional to the strain it produces:

$$\text{stress}\propto\text{strain},\qquad \text{stress}=(\text{modulus of elasticity})\times\text{strain}.$$

The law holds separately for each kind of deformation (longitudinal, shearing, volume), each with its own constant. It is an experimental law, obeyed by most solids for small strains; it fails for large deformations, and some materials (rubber, many biological tissues) hardly obey it at all.

📘 Definition 8.4 — Modulus of elasticity

The modulus of elasticity of a material for a given kind of deformation is the ratio

$$\text{modulus of elasticity}=\frac{\text{stress}}{\text{strain}}$$

in the region where Hooke's law holds. Since strain has no unit, a modulus has the unit of stress, $\text{Pa}$ ($\text{N/m}^2$), and dimensions $[\text{M L}^{-1}\text{T}^{-2}]$. It depends only on the material (and its temperature), not on the size or shape of the specimen. The three moduli — Young's modulus $Y$ for stretching or compressing, the shear modulus $G$ for shearing and the bulk modulus $B$ for hydraulic compression — are studied in Section 8.2.

A large modulus means a large stress is needed for a small strain: the material is stiff. For stretching, steel has a modulus of about $2.0\times10^{11}\ \text{Pa}$, while soft rubber has one of only a few times $10^6\ \text{Pa}$. In physics, the body that develops the larger restoring stress for the same strain is called the more elastic one — so steel is far more elastic than rubber, even though rubber stretches much further before it is damaged. Everyday language ("elastic band") uses the word differently; in exams, use the physics meaning.

A wire is a spring. For a wire of length $L$ and cross-section $A$, Hooke's law with modulus $Y$ reads $\dfrac{F}{A}=Y\dfrac{\Delta L}{L}$, i.e.

$$F=\left(\frac{YA}{L}\right)\Delta L .$$

This is $F=k\,\Delta L$ with an effective spring constant $k=YA/L$: a thicker wire is stiffer, a longer one is softer. That is how the material property $Y$ and the object property $k$ of Section 5.2 are connected.

✏️ Example 3 — Hooke's law for a steel wire
A steel wire $3.0\ \text{m}$ long with cross-section $1.0\ \text{mm}^2$ hangs vertically and carries a $10\ \text{kg}$ load. The modulus of elasticity of steel for stretching is $2.0\times10^{11}\ \text{Pa}$. Find (a) the stress, (b) the strain, (c) the extension, (d) the effective spring constant of the wire. (e) A rubber cord with the same dimensions and a modulus of $5.0\times10^6\ \text{Pa}$ carries a stress of the same size. What is its strain?
(a) $F=10\times9.8=98\ \text{N}$, $A=1.0\times10^{-6}\ \text{m}^2$: $\ \sigma=\dfrac{98}{1.0\times10^{-6}}=9.8\times10^7\ \text{Pa}$.
(b) $\varepsilon=\dfrac{\sigma}{Y}=\dfrac{9.8\times10^7}{2.0\times10^{11}}=4.9\times10^{-4}$.
(c) $\Delta L=\varepsilon L=4.9\times10^{-4}\times3.0=1.5\times10^{-3}\ \text{m}=1.5\ \text{mm}$.
(d) $k=\dfrac{YA}{L}=\dfrac{2.0\times10^{11}\times1.0\times10^{-6}}{3.0}=6.7\times10^4\ \text{N/m}$. Check: $\Delta L=\dfrac{F}{k}=\dfrac{98}{6.67\times10^4}=1.47\times10^{-3}\ \text{m}$ ✓.
(e) $\varepsilon=\dfrac{9.8\times10^7}{5.0\times10^6}=19.6$ — a strain of nearly $2000\%$! Long before that the rubber would have left the Hooke's-law region (and snapped). A stress that steel shrugs off is far beyond what rubber can carry: steel is the more elastic material.

The stress–strain curve

What happens beyond the proportional limit is found by a tensile test. A specimen — a wire or a machined rod — is clamped at both ends and stretched slowly while the force and the extension are recorded. The stress is computed as $F/A$ using the original cross-section, the strain as $\Delta L/L$ with the original length, and stress is plotted against strain. For a ductile metal such as mild steel or copper the curve looks like the one in Interactive 2:

📘 Definition 8.5 — Landmarks on the stress–strain curve
  • Proportional limit (A): the greatest stress up to which stress is proportional to strain.
  • Elastic limit or yield point (B): the greatest stress at which the body still returns to its original size and shape when the load is removed; the stress there is the yield strength $\sigma_y$.
  • Permanent set: the strain that remains at zero stress after a body loaded beyond its elastic limit is unloaded.
  • Ultimate tensile strength (D): the greatest stress the material can sustain, $\sigma_u$.
  • Fracture point (E): the point at which the specimen breaks; the stress there is the breaking stress (fracture strength).

For many materials A and B are so close that they are treated as one point. The shape of the curve after B depends very much on the material, and three broad families emerge.

📘 Definition 8.6 — Ductile, brittle and elastomeric materials
  • Ductile materials have a large plastic region: the fracture point E lies far beyond the elastic limit B. They can be drawn into wires and hammered into sheets, and they give warning (they stretch, bend and neck) before they break. Examples: copper, aluminium, gold, mild steel.
  • Brittle materials break soon after — or even before reaching — the elastic limit, with almost no plastic region and no warning. Examples: glass, cast iron, ceramics, chalk.
  • Elastomers can be stretched to very large strains (rubber to several times its length) and still return to their original shape, but have no straight Hooke's-law region of any size. Examples: rubber, the elastic tissue of the aorta and other large arteries.

Whether a wire survives a load depends on the stress, so a simple rule follows for the largest load it can carry.

📐 Result 8.2 — Breaking load of a wire

A wire (or rod) of cross-sectional area $A$ under tension breaks when the stress reaches the material's ultimate tensile strength $\sigma_u$. The greatest load it can carry, the breaking load, is

$$F_\text{max}=\sigma_u A=\sigma_u\,\frac{\pi d^2}{4}.$$

It is proportional to the cross-sectional area (to $d^2$) and independent of the length, provided the wire's own weight is negligible compared with the load.

Engineers never load a structure close to its breaking load. They choose a working stress equal to the ultimate tensile strength (or the yield strength) divided by a factor of safety — commonly between about $2$ and $10$ depending on how dangerous failure would be, how well the loads are known, and whether the load varies. Lift cables use large factors because the cable also carries jerks and wears over the years.

✏️ Example 4 — Reading a stress–strain curve
A tensile test on a metal wire shows a straight line up to a stress of $3.0\times10^8\ \text{Pa}$ at a strain of $0.0015$; the elastic limit is $3.2\times10^8\ \text{Pa}$; the greatest stress reached is $4.5\times10^8\ \text{Pa}$, at a strain of $0.20$; the wire breaks at a strain of $0.30$. (a) Find the modulus of elasticity. (b) Is the metal ductile or brittle? (c) Find the breaking load of a wire of this metal $1.0\ \text{mm}$ in diameter. (d) The wire is loaded to a stress of $2.0\times10^8\ \text{Pa}$ and unloaded. Is it permanently stretched?
(a) Slope of the straight part: $\dfrac{3.0\times10^8}{0.0015}=2.0\times10^{11}\ \text{Pa}$ (a typical value for steel).
(b) The proportional region ends at a strain of $0.0015$ but the wire breaks only at $0.30$ — two hundred times further. A huge plastic region: ductile.
(c) $A=\dfrac{\pi(1.0\times10^{-3})^2}{4}=7.85\times10^{-7}\ \text{m}^2$; $\ F_\text{max}=\sigma_uA=4.5\times10^8\times7.85\times10^{-7}=3.5\times10^2\ \text{N}$ — the weight of about $36\ \text{kg}$.
(d) $2.0\times10^8\ \text{Pa}$ is below the proportional limit, so the wire is in the Hooke's-law region and returns to its original length: no permanent set.
Check This wire is the "ductile metal" setting of Interactive 2: slide into the plastic region and read the permanent set, then compare with glass and rubber.
Interactive 2 — Stress–Strain Curve Explorer

Stretch a specimen in a tensile test. The slider sets the strain; the red dot rides the stress–strain curve and the sketch on the right shows the specimen (stretch exaggerated). The dashed line shows what happens if you remove the load now: the specimen relaxes along a line parallel to the Hooke's-law slope, and any strain left at zero stress is the permanent set. Switch materials, and zoom into the first few tenths of a per cent to see the straight Hooke's-law part of the metal's curve.

strain—
stress—
region—
unload now → permanent set—
🧭 Problem-solving strategy — stress, strain and Hooke's law
  1. Find the force on the surface. For a hanging load at rest it is the weight; if the system accelerates, use Newton's second law first ($T=m(g+a)$ for upward acceleration).
  2. Find the area the force acts on — for longitudinal stress the cross-section, $\pi d^2/4$ for a wire. Convert everything to SI: mm → m, mm² → $10^{-6}\ \text{m}^2$, cm² → $10^{-4}\ \text{m}^2$.
  3. Stress $=F/A$; strain $=$ (change)/(original), same units top and bottom.
  4. If the modulus is known and the stress is below the proportional limit, strain $=$ stress/modulus. Compare the stress with the yield strength and ultimate strength to decide whether the body returns, stays deformed or breaks.
  5. Check orders of magnitude: metal strains are usually $10^{-4}$–$10^{-3}$; a strain above $0.01$ for a metal means plastic flow (or an arithmetic slip).
✏️ Example 5 — Designing a crane cable
A crane is to lift loads of up to $5.0\ \text{t}$ ($5.0\times10^3\ \text{kg}$) with a single steel cable whose ultimate tensile strength is $1.5\times10^9\ \text{Pa}$. The design code demands a factor of safety of $5$. (a) What working stress may the cable carry? (b) What is the least diameter of the cable? (c) The site manager suggests a longer cable "to spread the load". Will a longer cable of the same diameter carry a bigger load?
(a) Working stress $=\dfrac{\sigma_u}{5}=\dfrac{1.5\times10^9}{5}=3.0\times10^8\ \text{Pa}$.
(b) Load $W=5.0\times10^3\times9.8=4.9\times10^4\ \text{N}$. Area needed: $A=\dfrac{W}{3.0\times10^8}=1.63\times10^{-4}\ \text{m}^2$. Then $d=\sqrt{\dfrac{4A}{\pi}}=\sqrt{\dfrac{4\times1.63\times10^{-4}}{\pi}}=1.44\times10^{-2}\ \text{m}$, so the cable must be at least $1.4\ \text{cm}$ in diameter — in practice the next standard size up, say $1.6\ \text{cm}$.
(c) No. By Result 8.2 the breaking load $\sigma_uA$ depends on the area, not the length. A longer cable stretches more (extension $\propto L$) but carries the same stress for the same load — and its own extra weight slightly reduces the safe load.
Check With $d=1.44\ \text{cm}$ the stress at full load is $\dfrac{4.9\times10^4}{1.63\times10^{-4}}=3.0\times10^8\ \text{Pa}$, one fifth of $\sigma_u$ ✓.

Applications of elastic behaviour. Engineers use these ideas everywhere a structure carries a load. Crane ropes are sized, as in Example 5, so that the working stress stays well below the elastic limit; a rope is made of many thin wires twisted together rather than one solid rod, which makes it flexible and strong at the same time. Beams in bridges and buildings bend under a load: the upper part is compressed and the lower part stretched. For a beam supported at its ends and loaded in the middle, the sag grows rapidly with the length and falls with the stiffness $Y$ of the material, and depends far more on the beam's depth (its size in the direction of the load) than on its breadth — doubling the depth cuts the sag about eightfold, doubling the breadth only halves it. A very deep, thin beam, however, may buckle sideways, so a compromise is used: the I-shaped section of steel girders puts most of the metal at the top and bottom, where the stresses are greatest, with a thin web between. Mountains cannot be arbitrarily tall: the weight of rock above presses on the rock at the base, and if that stress exceeds the elastic limit of rock, the base flows and the mountain sinks.

✏️ Example 6 — How high can a mountain be?
Rock has a density of about $2.7\times10^3\ \text{kg/m}^3$ and an elastic limit of about $3.0\times10^8\ \text{Pa}$. (a) Treating a mountain as a column of rock of uniform cross-section, estimate the greatest height it can have on the Earth. (b) Compare with Mount Everest ($8.85\ \text{km}$). (c) Estimate the greatest height on Mars, where $g=3.7\ \text{m/s}^2$.
Idea A column of height $h$ and cross-section $A$ has weight $\rho Ahg$, so the stress at its base is $\dfrac{\rho Ahg}{A}=h\rho g$ — independent of $A$. The base just holds when $h\rho g$ equals the elastic limit.
(a) $h_\text{max}=\dfrac{3.0\times10^8}{2.7\times10^3\times9.8}=1.1\times10^4\ \text{m}$ — about $10\ \text{km}$.
(b) Everest, at $8.85\ \text{km}$, is close to this limit: the stress at its base is about $2.7\times10^3\times9.8\times8850=2.3\times10^8\ \text{Pa}$. No mountain on the Earth is much higher than about $10\ \text{km}$.
(c) $h\propto\dfrac1g$: $\dfrac{3.0\times10^8}{2.7\times10^3\times3.7}=3.0\times10^4\ \text{m}=30\ \text{km}$. Mars's Olympus Mons does rise more than $20\ \text{km}$ above its surroundings.
Check Units: $\dfrac{\text{Pa}}{\text{kg m}^{-3}\times\text{m s}^{-2}}=\dfrac{\text{kg m}^{-1}\text{s}^{-2}}{\text{kg m}^{-2}\text{s}^{-2}}=\text{m}$ ✓. This is an order-of-magnitude estimate: real mountains are wider at the base than at the top, and rock fails mainly by shear.
⚠️ Common mistakes

1. Wrong area. For tensile and compressive stress use the cross-section perpendicular to the force, not the curved surface of the wire. For shearing stress use the area of the face the force is parallel to.

2. Diameter used as radius. $A=\pi r^2=\pi d^2/4$. Using $d$ in place of $r$ makes the area four times too large and the stress four times too small.

3. Unit slips. $1\ \text{mm}^2=10^{-6}\ \text{m}^2$ (not $10^{-3}$), $1\ \text{cm}^2=10^{-4}\ \text{m}^2$. A load "of $10\ \text{kg}$" is a force of $98\ \text{N}$.

4. Giving strain a unit. Strain is a pure number; "a strain of $2\ \text{mm}$" is meaningless.

5. "Rubber is more elastic than steel." In physics the more elastic material is the one with the larger modulus — steel. Rubber has a larger elastic range of strain, not a larger elasticity.

6. Proportional limit = elastic limit. They are close for most metals but not the same: between them the body still recovers, although stress is no longer proportional to strain.

7. Thinking the breaking load depends on length. It depends only on the cross-section and the material (Result 8.2) — unless the wire's own weight matters.

Exercises

Take $g=9.8\ \text{m/s}^2$. Where a modulus of elasticity for stretching is needed, take steel $2.0\times10^{11}\ \text{Pa}$ unless stated otherwise.

I
Stress and strain.
  • 1 A $2.0\ \text{kg}$ mass hangs from a wire of cross-section $0.50\ \text{mm}^2$. Find the tensile stress in the wire.
  • 2 A copper wire of diameter $0.80\ \text{mm}$ carries a tension of $50\ \text{N}$. Find the stress.
  • 3 A wire $1.50\ \text{m}$ long is stretched by $0.60\ \text{mm}$. Find the longitudinal strain.
  • 4 A cube of gelatin of edge $5.0\ \text{cm}$ stands on a table. A tangential force of $0.50\ \text{N}$ on its top face displaces the top by $0.40\ \text{cm}$ relative to the base. Find the shearing stress, the shearing strain and the angle of shear in degrees.
  • 5 A rubber ball of volume $120\ \text{cm}^3$ shrinks by $0.30\ \text{cm}^3$ when taken to the bottom of a deep pool. Find the volume strain.
  • 6 Write the dimensional formulae of stress and of strain. Which other physical quantity has the same dimensions as stress?
  • 7 A concrete pillar of radius $0.30\ \text{m}$ supports a load of $4.5\times10^5\ \text{N}$. Find the compressive stress.
  • 8 Name the kind of stress in (a) a rope in a tug of war, (b) the legs of a loaded table, (c) a bolt holding two plates that are pulled in opposite directions along their surfaces, (d) the hull of a submarine deep under water.
1 $\sigma=\dfrac{2.0\times9.8}{0.50\times10^{-6}}=3.9\times10^7\ \text{Pa}$.
2 $A=\pi(0.40\times10^{-3})^2=5.03\times10^{-7}\ \text{m}^2$; $\sigma=\dfrac{50}{5.03\times10^{-7}}=9.9\times10^7\ \text{Pa}$.
3 $\dfrac{0.60\times10^{-3}}{1.50}=4.0\times10^{-4}$.
4 Stress $=\dfrac{0.50}{(0.050)^2}=2.0\times10^2\ \text{Pa}$; strain $=\dfrac{0.40}{5.0}=0.080$; $\theta=\tan^{-1}0.080=0.080\ \text{rad}=4.6^\circ$.
5 $\dfrac{\Delta V}{V}=\dfrac{-0.30}{120}=-2.5\times10^{-3}$.
6 Stress: $[\text{M L}^{-1}\text{T}^{-2}]$; strain: dimensionless, $[\text{M}^0\text{L}^0\text{T}^0]$. Pressure (and the moduli of elasticity, and energy per unit volume) have the dimensions of stress.
7 $A=\pi(0.30)^2=0.283\ \text{m}^2$; $\sigma=\dfrac{4.5\times10^5}{0.283}=1.6\times10^6\ \text{Pa}$.
8 (a) tensile; (b) compressive; (c) shearing; (d) hydraulic (compressive from every side).
II
Hooke's law and the modulus of elasticity.
  • 1 A wire obeying Hooke's law stretches by $0.40\ \text{mm}$ under a load of $20\ \text{N}$. Find its extension under $35\ \text{N}$, and the load that stretches it by $1.0\ \text{mm}$.
  • 2 A steel wire $2.5\ \text{m}$ long with cross-section $2.0\ \text{mm}^2$ carries a load of $400\ \text{N}$. Find the stress, the strain and the extension.
  • 3 What load would stretch the wire of part 2 by exactly $1.0\ \text{mm}$?
  • 4 Find the effective spring constant $k=YA/L$ of the wire of part 2.
  • 5 A rubber cord $0.50\ \text{m}$ long with cross-section $4.0\ \text{mm}^2$ obeys Hooke's law for small strains with a modulus of $5.0\times10^6\ \text{Pa}$. It carries a load of $2.0\ \text{N}$. Find the stress, the strain and the extension.
  • 6 A student says, "Rubber is more elastic than steel because it stretches more." Correct the statement.
  • 7 Wires A and B are of the same material. B is twice as long as A and has twice the diameter. They carry equal loads. Find the ratio of (a) their stresses $\sigma_B/\sigma_A$, (b) their extensions $\Delta L_B/\Delta L_A$.
  • 8 A wire is cut in half and one half carries the same load as the whole wire did. What happens to the stress, the strain and the extension?
1 Extension $\propto$ load: $0.40\times\dfrac{35}{20}=0.70\ \text{mm}$; load $=20\times\dfrac{1.0}{0.40}=50\ \text{N}$.
2 $\sigma=\dfrac{400}{2.0\times10^{-6}}=2.0\times10^8\ \text{Pa}$; $\varepsilon=\dfrac{2.0\times10^8}{2.0\times10^{11}}=1.0\times10^{-3}$; $\Delta L=1.0\times10^{-3}\times2.5=2.5\times10^{-3}\ \text{m}=2.5\ \text{mm}$.
3 $F=\dfrac{YA\,\Delta L}{L}=\dfrac{2.0\times10^{11}\times2.0\times10^{-6}\times1.0\times10^{-3}}{2.5}=1.6\times10^2\ \text{N}$ (or $400\times\tfrac{1.0}{2.5}$).
4 $k=\dfrac{2.0\times10^{11}\times2.0\times10^{-6}}{2.5}=1.6\times10^5\ \text{N/m}$. Check: $\dfrac{400}{1.6\times10^5}=2.5\ \text{mm}$ ✓.
5 $\sigma=\dfrac{2.0}{4.0\times10^{-6}}=5.0\times10^5\ \text{Pa}$; $\varepsilon=\dfrac{5.0\times10^5}{5.0\times10^6}=0.10$; $\Delta L=0.10\times0.50=0.050\ \text{m}=5.0\ \text{cm}$.
6 Elasticity is measured by the modulus — the stress needed per unit strain. Steel's modulus ($\sim10^{11}\ \text{Pa}$) is about $10^4$–$10^5$ times rubber's, so steel is more elastic. Rubber has a larger range of strain over which it recovers, which is a different property.
7 Area $\propto d^2$: $A_B=4A_A$, so (a) $\sigma_B/\sigma_A=\tfrac14$. (b) $\Delta L=\dfrac{FL}{AY}\propto\dfrac{L}{A}$: $\dfrac{\Delta L_B}{\Delta L_A}=\dfrac{2}{4}=\tfrac12$.
8 Same force and area: stress unchanged, so strain unchanged (same material); the extension $\varepsilon L$ halves because the length halves.
III
The stress–strain curve.
  • 1 Distinguish between the proportional limit and the elastic limit. On a stress–strain curve, which comes first as the load increases?
  • 2 Material P breaks at a strain of $0.40$; material Q breaks at a strain of $0.002$, just after its elastic limit. Classify each and give an example of each.
  • 3 A steel wire $2.0\ \text{m}$ long is stretched beyond its elastic limit to a strain of $0.050$, where the stress is $3.6\times10^8\ \text{Pa}$. It is then unloaded. Find the permanent set and the permanent increase in length.
  • 4 The ultimate tensile strength of copper is $2.2\times10^8\ \text{Pa}$. Find the breaking load of a copper wire $1.0\ \text{mm}$ in diameter, and the largest mass it can hold at rest.
  • 5 A wire has a breaking load of $400\ \text{N}$. What is the breaking load of a wire of the same material with twice the diameter and three times the length?
  • 6 After the highest point of the curve for a ductile metal, the stress falls although the wire is still being stretched. Explain why.
  • 7 Describe how the stress–strain curve of rubber differs from that of steel, and give one place where elastomeric behaviour is useful.
1 Proportional limit: the greatest stress for which stress $\propto$ strain. Elastic limit: the greatest stress after which the body still fully recovers. The proportional limit comes first (or the two coincide); between them the body recovers although the curve is no longer straight.
2 P: ductile (large plastic region) — copper, aluminium, mild steel. Q: brittle (breaks with almost no plastic flow) — glass, cast iron, ceramics.
3 On unloading the wire recovers along a line of slope $2.0\times10^{11}\ \text{Pa}$: elastic recovery $=\dfrac{3.6\times10^8}{2.0\times10^{11}}=0.0018$. Permanent set $=0.050-0.0018=0.048$; permanent increase $=0.0482\times2.0=0.096\ \text{m}=9.6\ \text{cm}$.
4 $A=7.85\times10^{-7}\ \text{m}^2$; $F_\text{max}=2.2\times10^8\times7.85\times10^{-7}=1.7\times10^2\ \text{N}$ ($173$); mass $=\dfrac{173}{9.8}=18\ \text{kg}$.
5 Breaking load $\propto d^2$, independent of length: $400\times4=1.6\times10^3\ \text{N}$.
6 Beyond the ultimate strength the wire necks: its cross-section shrinks sharply at one place, so a smaller force is enough to keep stretching it. The plotted stress is force divided by the original area, so it falls; the true stress in the neck is still rising until fracture.
7 Rubber's curve has no long straight region, reaches strains of several hundred per cent, and the material returns to its original length even from large strains; steel's curve is straight up to a strain of about $0.001$, then plastic. Uses: the elastic walls of the aorta smooth the pulsing blood flow; tyres, shock absorbers, bungee cords, rubber seals.
IV
Applications and harder problems.
  • 1 (Modelling) A climbing rope of diameter $10\ \text{mm}$ holds a $75\ \text{kg}$ climber at rest. Treat the rope as a solid cylinder obeying Hooke's law with a modulus of $1.2\times10^8\ \text{Pa}$. Find the stress, the strain and the stretch of $20\ \text{m}$ of rope.
  • 2 A lift of mass $1200\ \text{kg}$ is held by a steel cable of cross-section $3.0\ \text{cm}^2$ with ultimate tensile strength $1.2\times10^9\ \text{Pa}$. Find the stress in the cable while the lift accelerates upward at $1.5\ \text{m/s}^2$, and the factor of safety at that moment.
  • 3 A long steel wire hangs vertically from its top end and carries only its own weight. Show that the stress at the top is $\rho gL$ ($\rho$ = density, $L$ = length), and find the longest wire that will not break. Take $\rho=7.9\times10^3\ \text{kg/m}^3$ and $\sigma_u=4.5\times10^8\ \text{Pa}$.
  • 4 A steel wire and a brass wire of the same length and diameter are joined end to end and hang a load. The modulus for stretching is $2.0\times10^{11}\ \text{Pa}$ for steel and $1.0\times10^{11}\ \text{Pa}$ for brass. Compare their stresses and their strains.
  • 5 (Modelling) The shaft of a human thigh bone has about $6.0\ \text{cm}^2$ of solid bone in its cross-section, and bone fails in compression at about $1.7\times10^8\ \text{Pa}$. Find the compressive force that would break it, and express it as a multiple of the weight of a $70\ \text{kg}$ person.
  • 6 A strain gauge on a steel tie-rod $0.40\ \text{m}$ long shows that it has shortened by $0.020\ \text{mm}$. Find the strain, the stress, and the compressive force if the rod's cross-section is $5.0\ \text{cm}^2$.
  • 7 The sag of a beam supported at its ends and loaded in the middle is inversely proportional to $bd^3$ ($b$ = breadth, $d$ = depth in the direction of the load). A wooden beam $10\ \text{cm}$ broad and $20\ \text{cm}$ deep is laid on its side ($20\ \text{cm}$ broad, $10\ \text{cm}$ deep). By what factor does its sag change? Why are steel girders made I-shaped?
  • 8 Using the data of Example 6, estimate the greatest height of a mountain on the Moon ($g=1.62\ \text{m/s}^2$).
1 $A=\dfrac{\pi(0.010)^2}{4}=7.85\times10^{-5}\ \text{m}^2$; $\sigma=\dfrac{735}{7.85\times10^{-5}}=9.4\times10^6\ \text{Pa}$; $\varepsilon=\dfrac{9.36\times10^6}{1.2\times10^8}=0.078$; stretch $=0.078\times20=1.6\ \text{m}$ — climbing ropes are deliberately stretchy, to soften a fall.
2 $T=1200(9.8+1.5)=1.356\times10^4\ \text{N}$; $\sigma=\dfrac{1.356\times10^4}{3.0\times10^{-4}}=4.5\times10^7\ \text{Pa}$; factor of safety $=\dfrac{1.2\times10^9}{4.52\times10^7}=27$.
3 The top cross-section supports the whole wire, of weight $\rho ALg$, so $\sigma_\text{top}=\dfrac{\rho ALg}{A}=\rho gL$ — independent of $A$. Breaks when $\rho gL=\sigma_u$: $L_\text{max}=\dfrac{4.5\times10^8}{7.9\times10^3\times9.8}=5.8\times10^3\ \text{m}$ ($5.8\ \text{km}$).
4 Same tension and same area, so equal stresses. Strain $=$ stress/modulus, so brass has twice the strain of steel (and twice the extension).
5 $F=1.7\times10^8\times6.0\times10^{-4}=1.0\times10^5\ \text{N}$; weight $=686\ \text{N}$, so about $1.5\times10^2$ body weights ($149$). Bones are rarely broken by steady compression — falls load them by bending and twisting.
6 $\varepsilon=\dfrac{0.020\times10^{-3}}{0.40}=5.0\times10^{-5}$; $\sigma=2.0\times10^{11}\times5.0\times10^{-5}=1.0\times10^7\ \text{Pa}$; $F=1.0\times10^7\times5.0\times10^{-4}=5.0\times10^3\ \text{N}$.
7 Sag $\propto\dfrac1{bd^3}$: $\dfrac{\text{new}}{\text{old}}=\dfrac{10\times20^3}{20\times10^3}=4$ — laid on its side the beam sags four times as much. An I-section puts most of the metal at the top and bottom, far from the middle, where the compression and tension of bending are greatest: it is deep (little sag) yet light, and its flanges stop it buckling sideways.
8 $h\propto\dfrac1g$: $h_\text{max}=\dfrac{3.0\times10^8}{2.7\times10^3\times1.62}=6.9\times10^4\ \text{m}$, about $70\ \text{km}$.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Define stress and strain. State Hooke's law and give the SI unit of the modulus of elasticity.

Definitions Stress: restoring force per unit area, $F/A$. Strain: fractional change in dimension (length, shape or volume), e.g. $\Delta L/L$; it has no unit. [1]
Law Within the proportional limit, stress is directly proportional to strain; stress/strain $=$ modulus of elasticity, unit $\text{Pa}$ ($\text{N/m}^2$). [1]
Exam tip Always add "within the proportional (elastic) limit" — examiners deduct the half mark for leaving it out.

E2. NEET Two wires A and B of the same material carry equal loads. A has length $L$ and radius $2r$; B has length $2L$ and radius $r$. The ratio of their extensions $\Delta L_A:\Delta L_B$ is (a) $1:2$ (b) $1:4$ (c) $1:8$ (d) $8:1$

Idea $\Delta L=\dfrac{FL}{AY}\propto\dfrac{L}{r^2}$ for the same $F$ and material.
Ratio $\dfrac{\Delta L_A}{\Delta L_B}=\dfrac{L/(2r)^2}{2L/r^2}=\dfrac{1/4}{2}=\dfrac18$.
Answer (c).
Exam tip (b) forgets that B is also twice as long; (d) inverts the ratio. Write the proportionality first, then substitute.

E3. JEE Main (Numerical value) A uniform steel wire hangs vertically from a fixed support. Steel has density $8.0\times10^3\ \text{kg/m}^3$ and breaking stress $8.0\times10^8\ \text{Pa}$. Taking $g=10\ \text{m/s}^2$, find the greatest length of wire, in km, that can hang without breaking under its own weight.

Set-up The top cross-section carries the whole weight $\rho ALg$, so the greatest stress is $\rho gL$ (the area cancels).
Solve $\rho gL_\text{max}=\sigma_b$: $\ L_\text{max}=\dfrac{8.0\times10^8}{8.0\times10^3\times10}=1.0\times10^4\ \text{m}$.
Answer $10$ (km).
Exam tip If a question gives the diameter, it is a distractor here — the breaking length of a hanging wire does not depend on its thickness.

E4. JEE Main NEET A metal wire is stretched beyond its elastic limit and the load is then removed. Which statement is correct? (a) The wire returns to its original length. (b) The unloading curve retraces the loading curve. (c) The wire contracts along a line roughly parallel to the initial straight part of the curve and is left permanently longer. (d) The wire's modulus of elasticity becomes zero.

Idea Beyond the elastic limit some strain is plastic (atomic planes have slipped). On unloading only the elastic part is recovered, with the same stiffness as at the start.
Answer (c) — the leftover strain is the permanent set.
Exam tip (a) and (b) describe loading within the elastic limit; the unloading slope shows the modulus is unchanged, so (d) is false.

E5. JEE Advanced (One or more correct) A uniform rod of length $L$, density $\rho$ and cross-sectional area $A$ hangs vertically from its top end and carries no other load. Which statements are correct? (A) The tensile stress at height $y$ above the lower end is $\rho gy$. (B) The greatest stress is $\rho gL$, at the top. (C) The greatest length that can hang without breaking is independent of $A$. (D) Doubling $A$ doubles the greatest length that can hang without breaking.

(A) The cross-section at height $y$ supports the part below it, of weight $\rho Ayg$: stress $=\rho gy$ ✓.
(B) Largest at $y=L$: $\rho gL$ ✓.
(C), (D) Breaking when $\rho gL=\sigma_b$ gives $L_\text{max}=\sigma_b/(\rho g)$, free of $A$: (C) ✓, (D) ✗.
Answer (A), (B), (C).
Exam tip The stress grows linearly from zero at the free end, so the average stress is $\tfrac12\rho gL$ — useful when you later find the extension of such a rod.

E6. Board · Case study A hospital lift car with its load has a mass of $1500\ \text{kg}$. It hangs from steel cables with a total cross-section of $4.0\ \text{cm}^2$; the steel has an ultimate tensile strength of $1.6\times10^9\ \text{Pa}$. (i) Find the stress in the cables when the lift is at rest, and when it accelerates upward at $2.0\ \text{m/s}^2$. [2] (ii) Find the factor of safety during the upward acceleration. [1] (iii) Lift cables are made of many thin strands twisted together rather than one solid rod of the same total area. Give one reason. [1]

(i) At rest: $\sigma=\dfrac{1500\times9.8}{4.0\times10^{-4}}=3.7\times10^7\ \text{Pa}$. Accelerating: $T=1500(9.8+2.0)=1.77\times10^4\ \text{N}$, $\sigma=4.4\times10^7\ \text{Pa}$. [2]
(ii) $\dfrac{1.6\times10^9}{4.425\times10^7}=36$. [1]
(iii) The breaking load depends only on the total area, but many thin strands make the cable flexible enough to bend round the pulley without large bending stresses, and a crack in one strand cannot run through the whole cable — the failure is gradual and visible at inspection. [1]
Exam tip In "accelerating lift" questions, find the tension from $T=m(g+a)$ before dividing by the area.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. Is spider silk really "stronger than steel"? Which materials would make the strongest cable for its weight?

Why it matters. Materials scientists in many countries are trying to spin artificial spider silk for surgical sutures, lightweight body armour and climbing gear, and aerospace engineers dream of cables light and strong enough for very tall structures. For anything that must carry its own weight — a hanging cable, a tall tower, an aircraft wing — what matters is not the ultimate tensile strength alone but strength divided by density.

What to do. Use the hanging-wire result of Exercise IV.3: a cable hanging from its top breaks under its own weight when its length reaches the breaking length $L_b=\dfrac{\sigma_u}{\rho g}$, independent of its thickness.

  1. Find published values of $\sigma_u$ and $\rho$ for mild steel, high-strength steel wire, aluminium alloy, nylon, an aramid fibre (such as Kevlar), carbon fibre and spider dragline silk. Record the source of each.
  2. Compute $L_b$ for each and rank the materials. Compare the ranking by $\sigma_u$ alone with the ranking by $L_b$.
  3. Is "stronger than steel" true for silk by $\sigma_u$, by $L_b$, or both?
  4. Find the breaking strains of the same materials. Which could absorb the energy of a falling climber or a flying insect without snapping, and why does that need both strength and a large strain?
  5. A cable from the ground to geostationary height ($3.6\times10^4\ \text{km}$) has been proposed. Using your $L_b$ values, explain why no material listed could hang that far in uniform gravity, and why tapering the cable helps.

Tools: Spreadsheet; materials-property databases and review articles on spider silk and fibres.

Hand in: A table of $\sigma_u$, $\rho$, breaking strain and $L_b$ with sources, a bar chart of $L_b$, and a 150-word answer to the research question.

Going further. Real gravity weakens with height and the cable would rotate with the Earth (Chapter 7). Find out how these change the requirement — the needed "breaking length" is usually quoted in thousands of kilometres.

Instructor note: with typical values ($\sigma_u$, $\rho$): mild steel $4.5\times10^8\ \text{Pa}$, $7.9\times10^3\ \text{kg/m}^3$ → $L_b\approx5.8\ \text{km}$; dragline silk $\approx1.1\times10^9\ \text{Pa}$, $1.3\times10^3$ → $\approx86\ \text{km}$; aramid $\approx3.6\times10^9\ \text{Pa}$, $1.44\times10^3$ → $\approx2.6\times10^2\ \text{km}$. Silk's $\sigma_u$ is comparable to high-strength steel, but by $L_b$ it wins easily; its breaking strain ($\sim0.3$) is far larger than steel's, so it absorbs much more energy before breaking. Accept a range of values with sources.

🧪 Self-check quiz

0/5 correct

Q1. The SI unit of stress is

Q2. A wire $1.5\ \text{m}$ long stretches by $0.30\ \text{mm}$. The strain is

Q3. On a stress–strain curve, the proportional limit is

Q4. A material that is drawn easily into thin wires and stretches a great deal before it breaks is

Q5. The same load hangs from two wires of the same material, one with twice the diameter of the other. The stress in the thicker wire is

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