University Physics I · Static Equilibrium & Elasticity · 12.9
Shear & Bulk Deformation
Push along a face and a solid changes shape; squeeze it from all sides and it changes volume. Shear modulus and bulk modulus put numbers on both, and Poisson's ratio ties them back to Young modulus.
Build the model
Connect the measurement to the mechanism.
Young modulus answers one question: how far does a rod stretch when you pull on its ends? Two other loadings need answers of their own. Drag the force along a face instead of across it and the block leans into a parallelogram — its shape changes, its volume barely does — and the stiffness resisting that is the shear modulus, G = τ/γ.
Squeeze from every direction at once and nothing leans, but the volume falls, and the stiffness resisting that is the bulk modulus, B = −Δp/(ΔV/V₀). Each modulus is the same ratio, stress over strain; what differs is which stress you apply and which strain you measure. For an isotropic solid the three are not independent: E = 2G(1 + ν) = 3B(1 − 2ν), so any two of them fix the rest.
- Simple definition
- Shear modulus is shear stress divided by shear strain and measures a solid's resistance to changing shape; bulk modulus is pressure change divided by fractional volume decrease and measures resistance to changing volume.
- Example
- Push an eraser's top sideways and its square face tilts into a parallelogram: that is shear. Sink it in the sea and it holds its shape but loses volume: that is bulk compression.
Force spread over the face it drags along, not the face it pulls away from.
Pa; F∥ lies in the face, A is the area of that face
How far the top slides divided by the height it slides over — the tilt angle for small tilts.
Dimensionless; h is measured perpendicular to F∥, θ in radians
Stiffness against changing shape at nearly constant volume.
Pa, often GPa; G = 0 for any fluid at rest
Stiffness against changing volume when pressure rises on every face.
Pa; the minus sign makes B positive, since ΔV < 0 when Δp > 0
The same fact upside down: how much volume you lose per pascal of extra pressure.
Pa⁻¹; for water k ≈ 4.5 × 10⁻¹⁰ Pa⁻¹
Only two of E, G, B and ν are independent — fix any two and the others follow.
Isotropic solids only; ν is Poisson's ratio, always below 0.5 and near 0.3 for metals
Shear puts the force in the face
Tensile loading pulls perpendicular to the area it acts on; shear loading drags parallel to it. Clamp the bottom of a block, push its top face sideways with a force F∥, and the square cross-section leans into a parallelogram. The stress is still force per area, τ = F∥/A, but A is now the face the force slides along. The strain is the sideways displacement of the top face divided by the block's height, γ = Δx/h. That ratio is tan θ for the tilt angle θ, and at the strains a solid survives, tan θ ≈ θ, so shear strain is simply the tilt in radians. Both quantities keep the units of their tensile cousins: pascals for stress, nothing at all for strain.
Shear modulus is stiffness against changing shape
Divide one by the other and you have the shear modulus, G = τ/γ. Rearranged, Δx = F∥h/(AG): a taller block shears further, a wider one less, exactly as Young modulus predicts for stretching. In pure shear the volume hardly changes at all — the block leans, it does not swell — so G measures resistance to change of shape alone. That makes G the sharpest test of what counts as a solid: a fluid at rest has G = 0, because any tangential stress, however small, keeps it moving. It is also why seismic S-waves, which shear the rock they pass through, stop at the Earth's liquid outer core, while P-waves, which compress it, carry on.
Pressure loads every face at once
Now apply the same normal stress to all faces. The load is a pressure increase Δp, and the response is a fractional volume change ΔV/V₀ called volume strain — the only one of the three strains that is not a ratio of lengths. The bulk modulus is B = −Δp/(ΔV/V₀); the minus sign is there because raising the pressure shrinks the volume, and it keeps B positive. Its reciprocal, the compressibility k = 1/B, states the same fact from the other end. Water has B ≈ 2.2 GPa, so k ≈ 4.5 × 10⁻¹⁰ Pa⁻¹. At 1000 m depth the water pressure is about ρgh = 1025 × 9.81 × 1000 ≈ 1.0 × 10⁷ Pa, giving ΔV/V₀ ≈ −(1.0 × 10⁷)/(2.2 × 10⁹) ≈ −0.46 %. Water is nearly, but not quite, incompressible.
Read the modulus off the loading geometry
The three moduli are not three theories. They are one definition — stress over strain — applied to three loadings. Pull or push along a single axis and use E with ε = ΔL/L₀. Drag one face across another and use G with γ = Δx/h; twisting a shaft is that same shear wrapped around an axis, so G governs it too. Squeeze from all directions and use B with ΔV/V₀. Choosing wrongly costs a whole factor: for steel, E ≈ 200 GPa, G ≈ 79 GPa and B ≈ 160 GPa, so using E where B belongs understates the volume strain by about 20 %, and using G overstates it by a factor of two. A fluid at rest has only B, since G = 0 and there is no way to pull it along one axis.
Two constants, not three
For an isotropic material — one with no built-in grain direction — linear elasticity has only two independent constants. Stretching a bar lengthways also thins it sideways, and Poisson's ratio ν is the fractional sideways contraction divided by the fractional lengthways extension: a ratio of two strains, not of two distances. Once E and ν are known the rest follow: G = E/[2(1 + ν)] and B = E/[3(1 − 2ν)]. Take steel with E = 200 GPa and ν = 0.30. The relations give G = 200/2.6 = 77 GPa and B = 200/1.2 = 167 GPa, close to the measured 79 GPa and 160 GPa. They also explain the extremes: as ν → 0.5 the factor 3(1 − 2ν) → 0 and B runs away. That is rubber, which changes shape freely (G of order 1 MPa) while resisting compression almost like water (B of order 2 GPa).
Where the linear model runs out
Every relation here assumes small strains and a response that is proportional and recoverable — the same restriction that limits Hooke's law. Push past the elastic limit and no single modulus describes the material. Three further limits matter. Anisotropic materials such as wood, fibre composites and single crystals need more than two constants, because their stiffness depends on direction. A gas has no one bulk modulus: for an ideal gas held at constant temperature B = p itself, about 10⁵ Pa at sea level, some four orders of magnitude below water's, and it climbs as the gas is compressed. And even for a liquid the number depends on how fast you squeeze it, since the adiabatic bulk modulus is larger than the isothermal one.
Change one variable at a time
Make the relationship visible.
Push Poisson's ratio towards 0.5 and watch the pressure panel freeze while the shear panel leans further — B runs away as 3(1 − 2ν) → 0, but G only softens.
SHEAR MODULUS G77 GPa
BULK MODULUS B167 GPa
SHEAR STRAIN γ = τ/G2.34 × 10⁻³
VOLUME STRAIN |ΔV/V₀| = Δp/B1.08 × 10⁻³
Live interpretationSHEAR MODULUS G: 77 GPa. BULK MODULUS B: 167 GPa. SHEAR STRAIN γ = τ/G: 2.34 × 10⁻³. VOLUME STRAIN |ΔV/V₀| = Δp/B: 1.08 × 10⁻³
Catch the common trap
Explain before calculating.
A steel cube sits where the water pressure is 40 MPa above atmospheric. Steel has E = 200 GPa, G = 79 GPa, B = 160 GPa. What is the magnitude of ΔV/V₀?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA rubber block is bonded to a bench. Its top face measures 80 mm × 50 mm, the block is 0.12 m tall, and a horizontal force of 24 N slides that top face 2.4 mm sideways. Find the shear stress, the shear strain and the shear modulus.
- The area is the face the force slides along, not one it pulls away from: A = 0.080 × 0.050 = 4.0 × 10⁻³ m².
- τ = F∥/A = 24/(4.0 × 10⁻³) = 6.0 × 10³ Pa.
- γ = Δx/h = (2.4 × 10⁻³)/0.12 = 0.020 — a ratio of two lengths, so no units; the tilt is arctan 0.020 = 1.1°.
- G = τ/γ = (6.0 × 10³)/0.020 = 3.0 × 10⁵ Pa = 0.30 MPa, soft as rubber should be and five orders below a metal's.
Answerτ = 6.0 kPa, γ = 0.020 (a 1.1° tilt), G = 0.30 MPa
MediumA 0.500 m³ aluminium billet (B = 76 GPa, E = 70 GPa) is lowered to a depth where the water pressure is 4.5 MPa above the surface value. Find the volume strain and the change in volume.
- Pressure loads every face at once, so this is bulk loading and B is the modulus: ΔV/V₀ = −Δp/B.
- ΔV/V₀ = −(4.5 × 10⁶ Pa)/(76 × 10⁹ Pa) = −5.92 × 10⁻⁵ — dimensionless, and negative because the volume falls.
- ΔV = (ΔV/V₀)V₀ = (−5.92 × 10⁻⁵)(0.500 m³) = −2.96 × 10⁻⁵ m³, about −30 cm³.
- Dividing by E instead would give −6.43 × 10⁻⁵, overstating the loss by about 9 %: E belongs to one-axis loading, not to pressure on every face.
AnswerΔV/V₀ = −5.9 × 10⁻⁵; ΔV = −3.0 × 10⁻⁵ m³ (about −30 cm³)
HardCopper has E = 110 GPa and Poisson's ratio ν = 0.34. Find its shear and bulk moduli, then find the shear strain in a copper rivet of 6.0 mm diameter carrying a 4.2 kN shear force across its circular section.
- An isotropic solid has only two independent constants, so E and ν fix the rest: G = E/[2(1 + ν)] and B = E/[3(1 − 2ν)].
- G = 110/(2 × 1.34) = 110/2.68 = 41 GPa.
- B = 110/(3 × 0.32) = 110/0.96 = 115 GPa. G is much the smallest, so copper gives way to shape change most easily; B clears E only because ν > 1/3 here — steel, with ν = 0.30, has B = E/1.2 ≈ 167 GPa, below its E = 200 GPa.
- The rivet is sheared across its circular section, so A = πr² = π(3.0 × 10⁻³)² = 2.83 × 10⁻⁵ m².
- τ = F∥/A = 4200/(2.83 × 10⁻⁵) = 1.49 × 10⁸ Pa.
- γ = τ/G = (1.49 × 10⁸)/(41.0 × 10⁹) = 3.6 × 10⁻³ rad, a tilt of 0.21°.
AnswerG = 41 GPa, B = 115 GPa; τ = 1.49 × 10⁸ Pa and γ = 3.6 × 10⁻³ (0.21°)