
Physics I course sequence · Interactive lesson
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.
UPI-12.09 · Lesson 9 of 10 in this unit. Position is not a completion score.Build the model
Change of shape, change of volume
The key idea
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.
The complete physical model and assumptions
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.
1Shear 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.
2Shear 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.
3Pressure 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.
4Read 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.
5Two 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).
6Where 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.
Key equations
Read the units, assumptions and limits before substituting.
Shear stress
τ = F∥/A
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 faceShear strain
γ = Δx/h = tan θ ≈ θ
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 radiansShear modulus
G = τ/γ = F∥h/(A Δx)
Stiffness against changing shape at nearly constant volume.
Pa, often GPa; G = 0 for any fluid at restBulk modulus
B = −Δp/(ΔV/V₀)
Stiffness against changing volume when pressure rises on every face.
Pa; the minus sign makes B positive, since ΔV < 0 when Δp > 0Compressibility
k = 1/B, so ΔV/V₀ = −k Δp
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⁻¹Linking the three moduli
E = 2G(1 + ν) = 3B(1 − 2ν)
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 metalsInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. 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.
Live results
- SHEAR 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
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Applied stress τ and Δp: 180 MPa; Young modulus E: 200 GPa; Poisson's ratio ν: 0.30
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Uniform pressure is often read as "the forces cancel, so nothing happens". They do cancel as a net force — nothing accelerates or tilts — but every face is still pushed inward, and ΔV/V₀ = −Δp/B. A second slip treats shear strain as a distance: γ = Δx/h is a ratio, the tilt angle in radians, so the metres cancel as they do in tensile strain.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- 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.
Answer. τ = 6.0 kPa, γ = 0.020 (a 1.1° tilt), G = 0.30 MPa
MediumExample 2
A 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- 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.
Answer. ΔV/V₀ = −5.9 × 10⁻⁵; ΔV = −3.0 × 10⁻⁵ m³ (about −30 cm³)
HardExample 3
Copper 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- 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°.
Answer. G = 41 GPa, B = 115 GPa; τ = 1.49 × 10⁸ Pa and γ = 3.6 × 10⁻³ (0.21°)
Quick check
Select an answer and explain it before checking.
Review & continue
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.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.