Elastic constants
An isotropic elastic material, one that behaves the same in every direction, is fully described by just two independent constants. The four commonly used are Young's modulus E (axial stress over axial strain), the shear modulus G (shear stress over shear strain), the bulk modulus K (pressure over volumetric strain) and Poisson's ratio ν (lateral strain over axial strain, taken as a positive number). Because only two are independent, the others follow from relationships between them.
Poisson's ratio describes how a bar gets thinner when it is stretched. For structural steel it is about 0.3, for concrete roughly 0.15 to 0.2, and for rubber close to 0.5, which means rubber hardly changes volume. A value of 0.5 is the theoretical upper limit for a stable isotropic material; above it the bulk modulus would be negative, meaning the volume would shrink under tension. The theoretical lower limit is −1; a few engineered foams and lattices have negative Poisson's ratios, but ordinary construction materials lie between about 0.1 and 0.5.
The relations E = 2G(1 + ν) and E = 3K(1 − 2ν) are used to get G for torsion and shear deflection when only E is known, and to understand confined behaviour, such as a bearing pad squeezed between steel plates or soil under a wide footing. Volumetric strain of a bar under axial stress is ε(1 − 2ν).
Real materials are not perfectly isotropic. Timber is far stiffer along the grain than across it; rolled steel plates and fibre composites show directional properties, and concrete's modulus changes with age, stress level and moisture. Codes therefore give design values for E and, where needed, G, rather than expecting designers to measure them.
E = σ / ε; G = τ / γ; ν = −ε_lateral / ε_axialE = 2 G (1 + ν)E = 3 K (1 − 2ν)Volumetric strain (uniaxial): ε_v = ε (1 − 2ν)
- The relations E = 2G(1 + ν) and E = 3K(1 − 2ν) hold only for isotropic, linear elastic materials; they do not apply to timber or fibre composites.
- Concrete is neither isotropic nor linear after cracking; code values of E for concrete are approximate short-term values.
Problem. A steel bar has E = 200 GPa and ν = 0.3. Find G and K. If a 25 mm bar carries an axial strain of 0.0008, find the change in diameter.
- G = E / [2(1 + ν)] = 200 / 2.6 = 76.9 GPa.
- K = E / [3(1 − 2ν)] = 200 / 1.2 = 166.7 GPa.
- Lateral strain = 0.3 × 0.0008 = 0.00024, so the diameter shrinks by 25 × 0.00024 = 0.006 mm.
Answer. G ≈ 77 GPa, K ≈ 167 GPa, diameter decreases by 0.006 mm
Quick quiz
1How many independent elastic constants does an isotropic material have?
Two.
2What is the upper limit of Poisson's ratio?
0.5.
3Give the relation between E, G and ν.
E = 2G(1 + ν).
- IS 456 (India): Plain and reinforced concrete: general rules for materials, durability, design and construction.
- IS 800 (India): General construction in steel: limit state design of steel members and connections.
Clauses and documents to read
- IS 456:2000 cl. 6.2.3.1 (modulus of elasticity of concrete)
- IS 800:2007 cl. 2.2.4.1 (physical properties of structural steel)
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- J. M. Gere, B. J. Goodno, Mechanics of Materials, Cengage, latest edition.
- F. P. Beer, E. R. Johnston, J. T. DeWolf, Mechanics of Materials, McGraw Hill, latest edition.
- E. P. Popov, Engineering Mechanics of Solids, Prentice Hall, 2nd edition.
Free and official sources
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