Young's, Bulk & Shear Moduli; Poisson's Ratio

Here is a definition of a liquid you have probably never heard: a liquid is a material whose shear modulus is zero. Push your hand slowly through water and it offers no lasting resistance to the change of shape — yet try to compress water and it fights back almost like a solid, which is why a depth charge is lethal and why hydraulic brakes work. One material, two completely different stiffness ratings depending on how you deform it. That is the insight this article builds: a material does not have one stiffness — it has one for every style of deformation. The previous article established the language — stress, strain, and the law $\sigma = E\varepsilon$ that links them within the elastic limit. Now we meet the three constants $E$ can stand for. Young's modulus $Y$ rates resistance to stretching and compressing along a line. The shear modulus $G$ rates resistance to shape change. The bulk modulus $B$ rates resistance to volume change — the only one of the three that fluids possess. A fourth number, Poisson's ratio, is not a modulus at all but a shape-memory fact: pull a wire longer and it gets thinner, and the ratio of thinning to lengthening is fixed for each material. Exam-wise this article is a numbers game: CBSE and ISC ask definitional derivations and single-formula numericals; JEE and NEET love cross-modulus comparisons, Poisson's-ratio volume tricks, and the "which modulus applies here?" classification question. One warning for CIE A-Level students: your syllabus demands Young's modulus only — read the rest as enrichment.

Key takeaways

Young's modulus $Y = \dfrac{F/A}{\Delta L/L} = \dfrac{FL}{A\,\Delta L}$ rates stretch/compression stiffness; solids only.
Shear modulus $G = \dfrac{F/A}{\theta}$ rates shape-change stiffness; zero for fluids — that is what makes them fluids.
Bulk modulus $B = -\dfrac{\Delta P}{\Delta V/V}$ rates volume stiffness; all matter has one. Compressibility $\kappa = 1/B$ .
Poisson's ratio $\sigma_p = -\dfrac{\text{lateral strain}}{\text{longitudinal strain}}$ is dimensionless, typically 0.2–0.4 for metals, 0.5 for incompressible materials.
Energy stored per unit volume of a strained wire: $u = \tfrac{1}{2}\,\text{stress}\times\text{strain} = \tfrac{1}{2}Y\varepsilon^2$ .
Key idea: one material carries a separate stiffness rating for each style of deformation — length ( $Y$ ), shape ( $G$ ), and volume ( $B$ ).

The core insight

Each style of elastic deformation defines its own modulus — Young's for length, shear for shape, bulk for volume — as the ratio of that style's stress to its strain; the trio (plus Poisson's ratio) is the complete elastic ID card of an isotropic material.

Where this lives today

The single most spectacular application of "fluids have $G = 0$ " is a map of the inside of the Earth. Earthquakes broadcast two body waves: P-waves (compressions, riding on $B$ ) and S-waves (shears, riding on $G$ ). Seismometers worldwide record P-waves arriving through every path — but S-waves vanish beyond about 104° from any quake. Something in the deep interior transmits squeezes but not shears: a liquid. That S-wave shadow zone is how we know the outer core is molten iron, a discovery made with exactly the two moduli in this article. The same pairing sets the speed of every mechanical wave — sound in air runs on $\sqrt{B/\rho}$ , seismic shear waves on $\sqrt{G/\rho}$ — so moduli are literally audible. And Poisson's ratio has gone designer: auxetic metamaterials with $\sigma_p < 0$ get fatter when stretched, and are being engineered into running-shoe midsoles, blast curtains, and stents that widen as arteries flex them.

The physical picture

Take a rubber eraser and play materials-testing lab. Pull its ends: it lengthens — Young's modulus territory. Press your palm flat on it and drag sideways: the rectangular side-face leans into a parallelogram, volume unchanged — shear. Now imagine it deep underwater, squeezed equally from every direction: it shrinks slightly in size without changing shape — bulk. Three experiments, three strains, three moduli. And while you pulled it longer in the first test, did you notice it getting thinner? That cross-talk between directions is Poisson's ratio.

Figure

Three deformations, three moduli — the complete elastic ID card

Why do solids resist all three deformations while fluids resist only the third? Back to the bond picture. A solid's atoms sit in a fixed lattice: stretching, tilting, or squeezing the lattice all stretch bonds, so all three cost energy and all three moduli are nonzero. A fluid's molecules have no fixed neighbours — slide one layer over another slowly and the molecules simply rearrange with no elastic penalty ( $G = 0$ ; resistance to fast sliding is viscosity, a different beast covered later in this unit). But pack the molecules closer together and their electron clouds protest immediately — hence a healthy bulk modulus even for water. This single sentence — fluids have no shear stiffness but real volume stiffness — is the bridge on which this chapter walks from solids to fluids.

The math: the three moduli

Each modulus is built the same way: pick a deformation style, take that style's stress, divide by that style's strain. Start with the one every board examines.

Young's modulus answers: how much tensile stress does this material develop per unit of longitudinal strain? For a wire of length $L$ and cross-section $A$ stretched by $\Delta L$ under force $F$ ,

Y = \frac{F/A}{\Delta L/L} = \frac{FL}{A\,\Delta L} \qquad (1)

where: $Y$ — Young's modulus, in Pa; the rest as defined in the previous article. Steel's $Y \approx 2\times10^{11}$ Pa means a stress of $2\times10^{11}$ Pa would double a steel wire's length ( $\varepsilon = 1$ ) — except of course steel fractures at a strain near $10^{-3}$ , so read $Y$ instead as "each 0.1% of strain costs $2\times10^8$ Pa of stress." Rearranged, equation (1) is the workhorse of the whole unit: $\Delta L = FL/AY$ .

The shear modulus (also called the modulus of rigidity, symbol $G$ or $\eta$ ) answers: how much tangential stress per unit of tilt? For a tangential force $F$ on a face of area $A$ producing shear angle $\theta$ ,

G = \frac{F/A}{\theta} \qquad (2)

where: $\theta$ — the shear strain in radians ( $\approx \Delta x/h$ for small tilts). For metals $G$ is roughly $Y/3$ : it is easier to tilt a lattice than to stretch it. For any liquid or gas, $G = 0$ exactly — which is the cleanest definition of "fluid" in all of physics, and the reason a lake cannot hold the shape of a sandcastle.

The bulk modulus answers: how much extra all-round pressure per unit of fractional volume shrink? If increasing the surrounding pressure by $\Delta P$ compresses volume $V$ by $\Delta V$ ,

B = -\frac{\Delta P}{\Delta V/V} \qquad (3)

where: the minus sign makes $B$ positive — increased pressure ( $\Delta P > 0$ ) decreases volume ( $\Delta V < 0$ ). Its reciprocal $\kappa = 1/B$ is the compressibility: big $B$ , hard to squeeze; big $\kappa$ , easy. Water's $B = 2.2\times10^9$ Pa is only a hundred times less than steel's — for most purposes "water is incompressible" is a superb lie, and the next few articles will tell it freely. Gases are the opposite extreme: air's bulk modulus at atmospheric pressure is around $10^5$ Pa (it equals the pressure itself for slow, isothermal squeezing) — squishy by a factor of ten thousand, which is why bubbles pulse and tyres cushion.

material	$Y$ (Pa)	$G$ (Pa)	$B$ (Pa)
steel	$2\times10^{11}$	$8\times10^{10}$	$1.6\times10^{11}$
copper	$1.2\times10^{11}$	$4.4\times10^{10}$	$1.4\times10^{11}$
aluminium	$7\times10^{10}$	$2.5\times10^{10}$	$7.5\times10^{10}$
water	—	$0$	$2.2\times10^{9}$
air (1 atm, slow)	—	$0$	$\sim10^{5}$

Poisson's ratio is the odd one out — dimensionless, not a stiffness. Stretch a wire and it thins: the lateral strain $\Delta d/d$ is negative while the longitudinal strain $\Delta L/L$ is positive, and their ratio is fixed per material:

\sigma_p = -\frac{\Delta d/d}{\Delta L/L} \qquad (4)

where: $\sigma_p$ — Poisson's ratio (no units). Metals cluster around $0.2$ – $0.4$ ; cork sits near $0$ (squeeze a cork lengthwise and it barely bulges — exactly why it enters a wine bottle so cooperatively); rubber approaches the theoretical ceiling $0.5$ , the value for a material whose volume never changes. A stretched wire's fractional volume change follows from the definition: $\dfrac{\Delta V}{V} = \varepsilon\,(1 - 2\sigma_p)$ — a JEE favourite derived in worked example 3, and instantly checkable: at $\sigma_p = 0.5$ the volume change vanishes.

Figure

Poisson's ratio — stretch it longer and it gets thinner, in a fixed proportion

Derivation AEnergy stored in a stretched wire — and the half that always surprises

We derive the elastic potential energy stored in a wire stretched within Hooke's law, aiming for both the total $U = \tfrac12 F\,\Delta L$ and the per-volume form $u = \tfrac12\,\sigma\varepsilon$ . The strategy is the same area-under-the-graph argument that gave $\tfrac12 kx^2$ for springs: the stretching force grows from zero as the wire extends, so the work is the average force times the distance, not the final force times the distance. The factor of ½ is the entire content — forget it and every energy answer doubles.

Step 1. At intermediate extension $x$ , the force needed is $F(x) = \dfrac{YA}{L}x$ (the wire-as-spring result, $k = YA/L$ ).

Step 2. Work done to reach final extension $\Delta L$ = area under the $F$ – $x$ line — a triangle:

U = \tfrac{1}{2}\,F_{\max}\,\Delta L = \tfrac{1}{2}\,\frac{YA}{L}\,(\Delta L)^2

Why the average: the first millimetre of stretch is cheap (force is small), the last is expensive; the linear ramp means the honest average force is exactly half the final force.

Step 3. Convert to energy per unit volume by dividing by the wire's volume $AL$ :

u = \frac{U}{AL} = \tfrac{1}{2}\,\frac{Y (\Delta L)^2}{L^2} = \tfrac{1}{2}\,Y\varepsilon^2 = \tfrac{1}{2}\,\sigma\,\varepsilon

Why per volume: $u = \tfrac12(\text{stress})(\text{strain})$ contains no geometry at all — it is the material-level statement, valid for every elastic body, and the form exams quote as "energy density = ½ × stress × strain". *Verified: units — Pa × (dimensionless) = J/m³ ✓. Limit — substituting $\sigma = Y\varepsilon$ gives the three equivalent forms $\tfrac12 Y\varepsilon^2 = \tfrac12\sigma\varepsilon = \sigma^2/2Y$ ✓. Numbers — the previous article's wire ( $\sigma = 9.8\times10^7$ Pa, $\varepsilon = 4.9\times10^{-4}$ ): $u = \tfrac12(9.8\times10^7)(4.9\times10^{-4}) = 2.4\times10^4\ \mathrm{J/m^3}$ ; times volume $2\times10^{-6}\ \mathrm{m^3}$ gives $U = 0.048$ J — equal to $\tfrac12 F\Delta L = \tfrac12(98)(9.8\times10^{-4})$ ✓.*

Limiting cases

case	result	meaning
fluid (liquid or gas)	$G = 0$ , $Y$ undefined, $B > 0$	fluids resist only volume change — the definition of fluidity
$\sigma_p = 0.5$	$\Delta V/V = 0$ under stretch	incompressible material (rubber, ideally); the theoretical maximum
$\sigma_p \approx 0$ (cork)	no lateral bulge	squeezing lengthwise doesn't fatten it — why corks seal bottles
$B \to \infty$	$\Delta V \to 0$ at any pressure	the "incompressible fluid" idealization used throughout fluid statics
gas, slow compression	$B \approx P$	squishiness grows with altitude: thin air is easier to compress fractionally
metals generally	$G \approx Y/3$ , $B \sim Y$	one Y-sized scale rules all three; ratios set by $\sigma_p$

Worked examples

1 · Board level — measuring Young's modulus (CBSE/ISC pattern, 3 marks)

A wire of length $2.0\ \mathrm{m}$ and diameter $1.0\ \mathrm{mm}$ is fixed at the ceiling. When a $4.0\ \mathrm{kg}$ mass is suspended, the wire extends by $0.50\ \mathrm{mm}$ . Find Young's modulus of the material. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Area from the diameter — halve first, then square: $r = 0.50\ \mathrm{mm} = 5.0\times10^{-4}\ \mathrm{m}$ , so $A = \pi r^2 = 3.14\times(5.0\times10^{-4})^2 = 7.85\times10^{-7}\ \mathrm{m^2}$ .

Step 2. Load: $F = mg = 4.0\times9.8 = 39.2\ \mathrm{N}$ .

Step 3. Apply equation (1): $Y = \dfrac{FL}{A\,\Delta L} = \dfrac{39.2 \times 2.0}{7.85\times10^{-7} \times 5.0\times10^{-4}} = \dfrac{78.4}{3.93\times10^{-10}}$ .

Step 4. $Y = \boxed{2.0\times10^{11}\ \mathrm{Pa}}$ — the textbook value for steel, so the material is likely steel. Quoting the identification earns the last half-mark on many ISC schemes.

2 · Competitive level — the ocean squeezes back (NEET/JEE-Main pattern, ~2 minutes)

The pressure at the bottom of an ocean trench is $1.1\times10^{8}\ \mathrm{Pa}$ . The bulk modulus of water is $2.2\times10^{9}\ \mathrm{Pa}$ . (a) Find the fractional compression of water taken to that depth. (b) Is the density at the bottom higher or lower, and by roughly what fraction?

Step 1. Fractional compression from equation (3): $\left|\dfrac{\Delta V}{V}\right| = \dfrac{\Delta P}{B} = \dfrac{1.1\times10^8}{2.2\times10^9} = \boxed{0.050 = 5.0\%}$ .

Step 2. Same mass in 5% less volume ⇒ density is higher by (to first order) the same fraction: $\dfrac{\Delta\rho}{\rho} \approx +5\%$ , about $1050\ \mathrm{kg/m^3}$ instead of $1000$ . Timing note: the entire problem is one division — the marks are for direction words ("compression", "density increases"). Distractor to dodge: using $B$ upside down and reporting 20× compression, which fails the smell test instantly (water is not a gas).

3 · Competitive level — Poisson's ratio and the volume of a stretched wire (JEE-Adv flavour, ~3 minutes)

A wire of Poisson's ratio $\sigma_p = 0.30$ is stretched to a longitudinal strain of $1.0\times10^{-3}$ . (a) Find the lateral strain. (b) Show that the fractional volume change is $\Delta V/V = \varepsilon(1 - 2\sigma_p)$ and evaluate it. (c) For what $\sigma_p$ would the volume not change at all?

Step 1. (a) From equation (4): lateral strain $= -\sigma_p\,\varepsilon = -0.30\times10^{-3} = \boxed{-3.0\times10^{-4}}$ (the wire thins).

Step 2. (b) Volume of a cylindrical wire $V = \dfrac{\pi d^2}{4}L$ . For small changes, add fractional changes with their powers: $\dfrac{\Delta V}{V} = 2\dfrac{\Delta d}{d} + \dfrac{\Delta L}{L}$ .

Step 3. Substitute $\Delta d/d = -\sigma_p \varepsilon$ and $\Delta L/L = \varepsilon$ : $\dfrac{\Delta V}{V} = \varepsilon(1 - 2\sigma_p) = 10^{-3}(1 - 0.60) = \boxed{4.0\times10^{-4}}$ — the wire's volume grows slightly.

Step 4. (c) Zero volume change needs $1 - 2\sigma_p = 0 \Rightarrow \sigma_p = \boxed{0.5}$ — the incompressible ceiling, approached by rubber. Timing note: step 2's "add fractional changes weighted by powers" is the error-propagation reflex from the measurement chapter working a second job — if you drilled products-and-powers there, this is a 90-second question. The answer's sign (volume up, since real $\sigma_p < 0.5$ ) is itself a one-mark MCQ.

Exam traps

Choosing the wrong modulus. Stretching a wire → $Y$ . Tilting/twisting/dragging a face sideways (or a cube glued to the floor pushed on top) → $G$ . Submerging, pressurizing, explosions underwater → $B$ . Exams disguise the classification: "a cube at the bottom of the sea" is $B$ ; "a cube bolted down with a horizontal force on its top face" is $G$ .

$Y$ for a liquid. Liquids have no Young's modulus — there is no restoring force along a line for a material that flows. Any option computing $Y_{\text{water}}$ is an auto-eliminate.

The minus sign in $B$ . Defined with $-\Delta P/(\Delta V/V)$ so $B > 0$ . If your bulk modulus comes out negative, you dropped the sign convention, not discovered exotic matter.

Modulus vs compressibility. $\kappa = 1/B$ . "Most compressible" = smallest $B$ . Ranking questions flip between the two mid-sentence — underline which one is asked.

Poisson's ratio is not in pascals. It is a strain ratio — dimensionless, between 0 and 0.5 for ordinary materials. An option quoting $\sigma_p = 0.3\ \mathrm{Pa}$ (or 3.0) is noise.

Energy: the missing half. $U = \tfrac12 F\Delta L$ , never $F\Delta L$ ; density $u = \tfrac12\sigma\varepsilon$ . The "work = force × distance" reflex overpays by exactly 2× because the force ramps from zero.

A-Level scope note. CIE 9702 demands the Young modulus, stress–strain graphs, and $E_p = \tfrac12 F x$ only — bulk, shear, and Poisson's ratio are not on the CIE paper (they ARE on JEE/NEET). Know your own syllabus before drilling.

$G \approx Y/3$ is metals folklore, not law. Fine for estimates and sanity checks, not a substitute for given data.

Origins · 1 minThomas Young and how this was discovered

Thomas Young

1773–1829 · London

Thomas Young was the last person plausibly described as knowing everything: physician, physicist, and philologist who read fluently in a dozen ancient languages. In physics he gave the double-slit experiment that established light's wave nature; in physiology, the three-colour theory of vision; in Egyptology, the first real cracks in the hieroglyphic code that Champollion finished. His 1807 Lectures on Natural Philosophy recast Hooke's proportionality per unit area and per unit length, creating the first true material constant — the modulus that bears his name. Characteristically, his original definition was so convoluted that contemporaries complained they could not understand it; the clean $\sigma/\varepsilon$ form is the polished version posterity kept. He also introduced the word "energy" in its modern physics sense.

History

1678

— Hooke publishes "ut tensio, sic vis": force ∝ extension, the object-level law.

1807

— Thomas Young defines the modulus, turning Hooke's proportionality into a property of the material itself.

1811

— Siméon Denis Poisson begins the elasticity analyses that lead to his ratio of lateral to longitudinal strain.

1822

— Cauchy's continuum mechanics gives stress and strain their general mathematical form, housing all the moduli in one framework.

1906

— Richard Oldham uses the absence of shear waves through Earth's core to argue it is fluid — the shear modulus as a planetary diagnostic.

Going further

Texts. NCERT Physics Class 11, Ch. 8 (Mechanical Properties of Solids) · HC Verma Vol. 1, Ch. 14 · Resnick–Halliday–Walker Ch. 12.

Provenance. All results standard. A verifier should recompute: example 1's $Y = 78.4/(3.93\times10^{-10}) = 2.0\times10^{11}$ Pa; example 2's $\Delta V/V = 0.050$ ; example 3's lateral strain $-3.0\times10^{-4}$ and $\Delta V/V = 4.0\times10^{-4}$ ; the derivation's $u = 2.4\times10^4\ \mathrm{J/m^3}$ and $U = 0.048$ J cross-check.

Prerequisites. Stress, Strain & Hooke's Law for Solids. Leads to. The Stretched Wire (archetype) · Fluids at Rest: Pressure & Pascal's Law.

Why do metals have three different moduli instead of one number?

Because stretching, tilting, and squeezing bend the atomic bonds in different ways. Stretching a lattice elongates bonds along one axis while the sides pull inward; shearing changes bond angles more than lengths; uniform squeezing shortens every bond equally. Each geometry has its own energy cost, hence its own stiffness. For an isotropic material the three are related (only two are independent — given $Y$ and $\sigma_p$ you can compute $G$ and $B$ ), which is why metals' $G \approx Y/3$ and $B \sim Y$ aren't coincidences.

If water's bulk modulus is so high, why do sonar and depth charges compress it at all?

High is not infinite. $B = 2.2\times10^9$ Pa means a pressure change of 2.2 MPa (about 22 atmospheres) compresses water by 0.1%. Sonar doesn't need much: a sound wave is a travelling ripple of exactly such tiny compressions, and $v = \sqrt{B/\rho} \approx 1480$ m/s is its speed. The "incompressible" idealization used in the coming fluid-statics articles just means: for pressure changes of a few atmospheres, a 0.01% density change is beneath our accounting.

Is Poisson's ratio ever negative? What would that even look like?

Yes — engineered "auxetic" materials (and a few natural ones, like some crystalline SiO₂ phases and cat skin, unexpectedly) have $\sigma_p < 0$ : stretch them and they get wider. The trick is internal geometry — re-entrant honeycomb cells that unfold sideways when pulled. Ordinary materials sit between 0 and 0.5 because their volume shouldn't grow without bound under stretch ( $\sigma_p > 0.5$ would mean volume decreases when pulled, which costs impossible energy in the other loading directions). Exam answers: metals 0.2–0.4, rubber →0.5, cork ≈0.

Why does cork have Poisson's ratio near zero, and why does that matter for wine bottles?

Cork's cells are collapsed, wrinkled air-filled boxes: compress the cork lengthwise and the wrinkles fold further without pushing the walls outward, so almost no lateral bulge. That is precisely what a bottle-stopper needs — you can push it into the neck without it jamming fatter as you push (a rubber stopper, $\sigma_p \approx 0.5$ , fights back by bulging). Same reason cork sheets seal machine joints under bolt pressure.

Which of Y, G, B decides how "hard" a material feels?

None of them directly — "hardness" (resistance to scratching/denting) is a surface, plasticity-dominated property, not an elastic one. What the moduli DO set: how much a structure flexes under load ( $Y$ for beams and cables, $G$ for twisting shafts), and how fast sound travels through it. Diamond's record $Y \approx 1.2\times10^{12}$ Pa is why diamond anvils transmit pressure so rigidly, but hardness tests measure a different, plastic quantity. On exams, stick to the deformation-style classification and never write "hardness" for stiffness.

Why is the bulk modulus of a gas roughly equal to its pressure?

Squeeze a gas slowly at constant temperature: Boyle's law $PV =$ const gives $P\,\Delta V + V\,\Delta P = 0$ , so $-\Delta P/(\Delta V/V) = P$ . The bulk modulus IS the pressure (for isothermal changes; fast, adiabatic squeezes give $\gamma P$ , about 40% stiffer for air — that version sets the speed of sound). Consequence worth storing: gases get fractionally squishier as pressure drops, and a gas at 1 atm is ~20,000× more compressible than water.

Do I need all four constants for JEE, or are some just trivia?

All four are live ammunition for JEE/NEET: $Y$ dominates numericals (wire problems), $B$ appears in ocean/pressure items and sound-speed crossovers, $G$ in classification MCQs ("which modulus?"), and Poisson's ratio in the $\Delta V/V = \varepsilon(1-2\sigma_p)$ trick and range questions (0 to 0.5). For CBSE boards, $Y$ derivations and definitions of the other two suffice. For CIE A-Level, Young only. Match your drilling to your paper.