Elastic constants and generalised three-dimensional Hooke's law

Poisson's ratio, the four elastic constants and their relations, the generalised 3-D Hooke's law and volumetric strain for isotropic materials.

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Why it matters

Real aircraft parts are rarely loaded in one direction only: a pressurised fuselage skin is stretched both along and around the body, and a lug is squeezed and pulled at once. To find strains, dimension changes and volume changes in such parts you need the full set of elastic constants and the three-dimensional form of Hooke's law. The same relations are used to back out material properties from strain-gauge tests.

Key ideas

Poisson effect. A bar stretched along its axis gets thinner sideways. For a uniaxial stress, the lateral strain is a fixed fraction of the axial strain. Poisson's ratio ν = −(lateral strain)/(axial strain) is positive for ordinary materials: about 0.3 for steel, 0.33 for aluminium alloys, close to 0.5 for rubber. Thermodynamics limits an isotropic material to −1 < ν ≤ 0.5.

The four elastic constants of an isotropic, linear elastic material:

  • Young's modulus E, the ratio of normal stress to normal strain in uniaxial loading.
  • Shear modulus (modulus of rigidity) G, the ratio of shear stress to shear strain.
  • Bulk modulus K, the ratio of hydrostatic pressure to the magnitude of volumetric strain.
  • Poisson's ratio ν.

Only two are independent. If you know any two, the other two follow from the relations below. A general anisotropic solid needs up to 21 independent constants, an orthotropic one (such as a unidirectional composite ply) needs 9, and an isotropic one needs 2.

Generalised Hooke's law (isotropic). By superposition, each normal stress produces its own axial strain σ/E plus lateral strains −νσ/E in the other two directions. Shear stresses produce only shear strains in their own plane, with no coupling to the normal strains. This holds only for isotropic materials, small strains and stresses below the proportional limit.

Volumetric strain. For small strains, the change in volume per unit volume is ε_v = ΔV/V = εx + εy + εz. Substituting Hooke's law gives ε_v = (1 − 2ν)(σx + σy + σz)/E. For ν = 0.5 the volumetric strain is zero, so the material is incompressible and K becomes infinite. For a bar in uniaxial tension, ε_v = ε(1 − 2ν).

Hydrostatic stress. When σx = σy = σz = −p, ε_v = −3p(1 − 2ν)/E = −p/K. This is where K = E/[3(1 − 2ν)] comes from.

Formulas

ν = − ε_lateral / ε_axial ν = Poisson's ratio (–). Uniaxial stress only.

εx = [σx − ν·(σy + σz)] / E εy = [σy − ν·(σz + σx)] / E εz = [σz − ν·(σx + σy)] / E ε = normal strain (–), σ = normal stress (Pa), tension positive; E in Pa.

γxy = τxy / G, γyz = τyz / G, γzx = τzx / G γ = engineering shear strain (rad), τ = shear stress (Pa), G = shear modulus (Pa).

ε_v = εx + εy + εz = (1 − 2ν)·(σx + σy + σz) / E Volumetric strain (–), small strains.

E = 2·G·(1 + ν) E = 3·K·(1 − 2ν) E = 9·K·G / (3·K + G) ν = (3·K − 2·G) / (6·K + 2·G) K = bulk modulus (Pa). Isotropic, linear elastic material.

ΔV = ε_v · V Change in volume (m³) of a body of original volume V (m³) under uniform strain.

Worked examples

Example 1 (standard). A bar of diameter 25 mm and gauge length 250 mm carries an axial tensile load of 60 kN. The gauge length increases by 0.15 mm and the diameter decreases by 0.0045 mm. Find E, ν, G and K.

  1. Area: A = π·d²/4 = π × 25² / 4 = 490.87 mm².
  2. Stress: σ = P/A = 60 000 / 490.87 = 122.23 MPa.
  3. Axial strain: ε = δ/L = 0.15 / 250 = 6.0 × 10⁻⁴.
  4. Young's modulus: E = σ/ε = 122.23 / 6.0 × 10⁻⁴ = 203 718 MPa ≈ 203.7 GPa.
  5. Lateral strain: −0.0045 / 25 = −1.8 × 10⁻⁴, so ν = −ε_lat/ε = 1.8 × 10⁻⁴ / 6.0 × 10⁻⁴ = 0.30.
  6. G = E / [2(1 + ν)] = 203.7 / 2.6 = 78.35 GPa.
  7. K = E / [3(1 − 2ν)] = 203.7 / 1.2 = 169.8 GPa.

Answer: E ≈ 203.7 GPa, ν = 0.30, G ≈ 78.4 GPa, K ≈ 169.8 GPa.

Example 2 (GATE level). A steel cube of side 100 mm is subjected to σx = +120 MPa, σy = −80 MPa and σz = −50 MPa. E = 200 GPa, ν = 0.3. Find the three normal strains and the change in volume.

  1. εx = [σx − ν(σy + σz)]/E = [120 − 0.3(−130)] / 200 000 = 159 / 200 000 = 7.95 × 10⁻⁴.
  2. εy = [σy − ν(σz + σx)]/E = [−80 − 0.3(70)] / 200 000 = −101 / 200 000 = −5.05 × 10⁻⁴.
  3. εz = [σz − ν(σx + σy)]/E = [−50 − 0.3(40)] / 200 000 = −62 / 200 000 = −3.10 × 10⁻⁴.
  4. ε_v = εx + εy + εz = (7.95 − 5.05 − 3.10) × 10⁻⁴ = −2.0 × 10⁻⁵.
  5. Check: (1 − 2ν)(σx + σy + σz)/E = 0.4 × (−10) / 200 000 = −2.0 × 10⁻⁵. Agrees.
  6. ΔV = ε_v·V = −2.0 × 10⁻⁵ × 100³ mm³ = −20 mm³.

Answer: εx = 7.95 × 10⁻⁴, εy = −5.05 × 10⁻⁴, εz = −3.10 × 10⁻⁴; volume decreases by 20 mm³.

Common mistakes

  • Dropping the minus signs on compressive stresses inside the bracket of the generalised Hooke's law.
  • Assuming a direction with zero stress has zero strain. In uniaxial tension the lateral strains are −νσ/E, not zero.
  • Writing E = 2G(1 − ν) or K = E/[3(1 + 2ν)]. Check with ν = 0.3: G should be about 0.38E and K about 0.83E for steel.
  • Using the isotropic relations for composites or other anisotropic materials.
  • Taking ν > 0.5 for an isotropic material; it would give a negative bulk modulus.

For GATE AE

Expect numericals that ask for strains or dimension changes under biaxial or triaxial stress, the stress needed to prevent strain in one direction (for example a block constrained between rigid walls), volume change of a bar or cube, and conversion between E, G, K and ν. Conceptual questions test the number of independent constants and the limits on ν. Practise writing the three strain equations quickly and checking the volumetric strain by both routes.

Quick check

  1. How many independent elastic constants does an isotropic material have?
  2. For E = 200 GPa and ν = 0.25, what is G?
  3. What is the volumetric strain of a material with ν = 0.5?
  4. A bar in uniaxial tension has ε = 0.001 and ν = 0.3. What is its volumetric strain?

Answers: 1. Two. 2. 80 GPa. 3. Zero (incompressible). 4. 4 × 10⁻⁴.

Interview questions

All Mechanics of Solids interview questions

Try answering each one aloud before you open it.

  1. 1.What are elastic constants, and why are they important in the mechanics of solids?Concept

    Elastic constants are material properties that define the relationship between stress and strain in a material under elastic deformation. They are important because they help predict how materials will behave under different loading conditions. The primary elastic constants include Young's modulus, shear modulus, bulk modulus, and Poisson's ratio. These constants are crucial for designing structures and components to ensure they can withstand applied loads without permanent deformation.

  2. 2.Explain the generalized three-dimensional Hooke's law.Concept

    For an isotropic linear elastic material, each normal stress causes a direct strain σ/E in its own direction and lateral strains −νσ/E in the other two, so by superposition εx = [σx − ν(σy + σz)]/E, and similarly for εy and εz. Shear stresses cause only their own shear strains, γxy = τxy/G, with no coupling to normal strains. It assumes small strains, stresses below the proportional limit and isotropy; anisotropic materials need a full stiffness matrix instead.

  3. 3.How is Young's modulus related to the stiffness of a material?Concept

    Young's modulus, also known as the modulus of elasticity, measures a material's stiffness. It is defined as the ratio of tensile stress to tensile strain in the linear elastic region of the material's stress-strain curve. A higher Young's modulus indicates a stiffer material, meaning it deforms less under the same applied stress. This property is crucial for selecting materials in applications where minimal deformation is desired.

  4. 4.Why is Poisson's ratio significant in material science?Concept

    Poisson's ratio ν is the negative ratio of lateral strain to axial strain under uniaxial stress, so it tells you how much a part narrows when stretched. It couples the three directions in the generalised Hooke's law and fixes the volume change, ε_v = ε(1 − 2ν) for a bar. Together with E it gives G = E/[2(1 + ν)] and K = E/[3(1 − 2ν)]. For isotropic materials −1 < ν ≤ 0.5; metals are about 0.3 and rubber is close to 0.5.

  5. 5.What happens to a material if its Poisson's ratio is close to 0.5?Application

    If a material's Poisson's ratio is close to 0.5, it behaves nearly incompressibly, meaning it undergoes very little volume change when subjected to stress. This is typical of rubber-like materials. In such cases, when the material is stretched, it will exhibit significant lateral expansion. This property is important in applications where maintaining volume is crucial, such as in seals and gaskets.

  6. 6.Why is the shear modulus important in the analysis of materials?Application

    The shear modulus, also known as the modulus of rigidity, measures a material's response to shear stress. It is important because it helps predict how a material will deform under torsional or shear loads. This property is crucial in applications involving shafts, beams, and other components subjected to twisting or shear forces. A higher shear modulus indicates a material that is more resistant to shear deformation.

  7. 7.How do you calculate the bulk modulus of a material, and what does it signify?Application

    The bulk modulus is calculated as the ratio of volumetric stress to the corresponding change in volume (volumetric strain) of a material. It signifies a material's resistance to uniform compression. A higher bulk modulus indicates that the material is less compressible. This property is important in applications where materials are subjected to high-pressure environments, such as in deep-sea or aerospace applications.

  8. 8.A steel rod with a Young's modulus of 210 GPa is subjected to a tensile stress of 100 MPa. Calculate the tensile strain experienced by the rod.Numerical

    To calculate the tensile strain, use the formula: strain = stress / Young's modulus. Here, stress = 100 MPa = 100 × 10^6 Pa, and Young's modulus = 210 GPa = 210 × 10^9 Pa. Thus, strain = (100 × 10^6) / (210 × 10^9) = 0.000476. The tensile strain experienced by the rod is 0.000476 (dimensionless).

  9. 9.A material has a shear modulus of 80 GPa and a Poisson's ratio of 0.3. Calculate its Young's modulus.Numerical

    Young's modulus (E) can be calculated using the relationship: E = 2G(1 + ν), where G is the shear modulus and ν is Poisson's ratio. Here, G = 80 GPa = 80 × 10^9 Pa, and ν = 0.3. Thus, E = 2 × 80 × 10^9 × (1 + 0.3) = 208 × 10^9 Pa = 208 GPa. The Young's modulus of the material is 208 GPa.

  10. 10.Explain how the stiffness matrix is used in the generalized Hooke's law.Concept

    Writing the six stress components and six strain components as vectors, Hooke's law becomes {σ} = [C]{ε}, where [C] is a symmetric 6 × 6 stiffness matrix (its inverse is the compliance matrix). Symmetry leaves at most 21 independent constants for a fully anisotropic solid, 9 for an orthotropic one such as a composite ply, and only 2 (for example E and ν) for an isotropic material. The matrix form is what finite-element codes and laminate theory use.

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