--- title: "Generalised Hooke's law for isotropic materials" subject: "Mechanics of Solids" chapter: 4 key: "semester_5/mechanics_of_solids:generalised-hooke-law-isotropic" source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/generalised-hooke-law-isotropic site: BCT Engineering --- # Generalised Hooke's law for isotropic materials Mechanics of Solids · Chapter 4: General Hooke’s Law - Asked in: year not given, 6 marks - Video: [08.4 Generalized Hooke's Law](https://www.youtube.com/watch?v=UDMNrUt-0cE) — Introductory Engineering Mechanics - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/generalised-hooke-law-isotropic - Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-4 ## Question State the generalised Hooke's law. Write the stress-strain relations for a linear isotropic material, define the elastic constants $E$, $G$, $\nu$ and $K$ and show the relations between them. How many independent constants are needed for anisotropic, orthotropic and isotropic materials? ## Short answer Generalized Hooke’s law: σ_ij = C_ijkl ε_kl (21 independent constants for anisotropy). Isotropic form: ε_x = [σ_x-ν(σ_y+σ_z)]/E, γ_xy = τ_xy/G. Elastic constants: E (Young’s), ν (Poisson’s), G (shear), K (bulk). Relations: G = E/[2(1+ν)], K = E/[3(1-2ν)]. Two independent constants. Anisotropic: 21, orthotropic: 9, isotropic: 2. ## Answer #### Generalised Hooke's law For a linear elastic body each stress component is a linear function of all strain components: $\sigma_{ij} = C_{ijkl}\,\varepsilon_{kl}$, where $C_{ijkl}$ is the elasticity (stiffness) tensor. Because $\sigma_{ij}$ and $\varepsilon_{ij}$ are symmetric and a strain energy function exists, the 81 constants reduce to **21** for the most general anisotropic body. #### Isotropic material $$ \begin{aligned} \varepsilon_x &= \frac{1}{E}\left[\sigma_x - \nu(\sigma_y+\sigma_z)\right] &\qquad \gamma_{xy} &= \frac{\tau_{xy}}{G} \\ \varepsilon_y &= \frac{1}{E}\left[\sigma_y - \nu(\sigma_z+\sigma_x)\right] & \gamma_{yz} &= \frac{\tau_{yz}}{G} \\ \varepsilon_z &= \frac{1}{E}\left[\sigma_z - \nu(\sigma_x+\sigma_y)\right] & \gamma_{zx} &= \frac{\tau_{zx}}{G} \end{aligned} $$ Inverse (Lamé form): $\sigma_{ij} = \lambda\,\varepsilon_{kk}\delta_{ij} + 2G\varepsilon_{ij}$, with $\lambda = \dfrac{\nu E}{(1+\nu)(1-2\nu)}$. #### Elastic constants | Constant | Definition | |---|---| | Young's modulus $E$ | $\sigma/\varepsilon$ in uniaxial tension | | Poisson's ratio $\nu$ | $-\varepsilon_{lateral}/\varepsilon_{axial}$ | | Shear modulus $G$ | $\tau/\gamma$ in pure shear | | Bulk modulus $K$ | $p/\varepsilon_v$ under hydrostatic pressure | #### Relations Adding the three normal strain equations gives the volumetric strain $\varepsilon_v = \dfrac{1-2\nu}{E}(\sigma_x+\sigma_y+\sigma_z)$. For a hydrostatic pressure $p$ (taking the magnitude, with compression positive) the sum of the three stresses is $3p$ in magnitude, so $$ K = \frac{p}{\varepsilon_v} = \frac{E}{3(1-2\nu)} $$ Pure shear: the principal stresses are $\pm\tau$, so $\varepsilon_1 = \dfrac{\tau}{E}(1+\nu)$. Since $\gamma = 2\varepsilon_1$ (the principal axes are at $45^\circ$ to the shear planes) and $\gamma = \tau/G$, $$ G = \frac{E}{2(1+\nu)} $$ Hence $E = 9KG/(3K+G)$ and $\nu = (3K-2G)/(2(3K+G))$. Only **two** of the four constants are independent. #### Number of independent constants | Material | Independent constants | |---|---| | General anisotropic | 21 | | Orthotropic (three perpendicular symmetry planes, e.g. timber) | 9 | | Transversely isotropic | 5 | | Isotropic | 2 | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the generalised Hooke's law state? A. Each stress component is a linear function of all strain components, via the stiffness tensor C_ijkl B. Stress equals strain squared C. Each stress depends only on its own strain D. Strain depends on stress rate Answer: A. sigma_ij = C_ijkl eps_kl, with 81 constants reducing to 21 for general anisotropy. (Key point: Generalised Hooke's law) 2. Which expression gives K? A. E/(2(1 + nu)) B. E/(3(1 - 2nu)) C. E/(1 - 2nu) D. 3E(1 - 2nu) Answer: B. From eps_v = (1 - 2nu) 3p / E, K = p/eps_v = E/(3(1 - 2nu)). (Key point: Bulk modulus) 3. How is G = E/(2(1+nu)) obtained? A. From hydrostatic pressure B. From uniaxial tension with eps_lateral C. From pure shear: principal stresses +/- tau, eps_1 = tau(1 + nu)/E, and gamma = 2 eps_1 = tau/G D. From the Lame constant only Answer: C. Pure shear is equivalent to principal stresses +/- tau, and gamma = 2 eps_1 at 45 degrees equals tau/G. (Key point: Shear modulus relation) 4. Which list of independent constants is correct (anisotropic, orthotropic, isotropic)? A. 9, 21, 2 B. 21, 5, 2 C. 21, 2, 9 D. 21, 9, 2 Answer: D. General anisotropic has 21, orthotropic 9, isotropic 2. 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