Hooke's Law, Elastic Constants and Thermal Stresses

Relations among elastic constants

Hooke's law for a linear elastic, isotropic material is governed by two independent elastic constants, so every other modulus can be expressed from any pair. The standard relations are G = \dfrac{E}{2(1+\nu)}, K = \dfrac{E}{3(1-2\nu)}, and equivalently E = 2G(1+\nu) = 3K(1-2\nu). Poisson's ratio must satisfy -1 < \nu \le 0.5 for a physically admissible isotropic material, and K becomes infinite as \nu \to 0.5, which describes an incompressible material. The usual exam method is to recover \nu from the given ratio or pair of moduli and then compute the third quantity. For example, if E = 140 GPa and G = 50 GPa, then \nu = \dfrac{E}{2G} - 1 = 0.40. The ratio E/K determines 1-2\nu directly, since E/K = 3(1-2\nu). A ratio E/K = 3/2 therefore gives \nu = 1/4, and only then is G/E = \dfrac{1}{2(1+\nu)} = \dfrac{2}{5} obtained. Note that G/E cannot be read off from E/K alone; the Poisson ratio must be found first. In a general stress state, the principal strain along direction 1 is \varepsilon_1 = [\sigma_1 - \nu(\sigma_2 + \sigma_3)]/E. A torsion test produces pure shear, whose principal stresses are equal and opposite, \sigma_1 = \tau and \sigma_2 = -\tau. The principal strain is then \varepsilon_1 = (1+\nu)\tau/E, and the Poisson term enters with a plus sign because the compressive second principal stress adds to the tensile strain. Under plane stress, \sigma_{zz} = 0, the normal strain through the thickness is \varepsilon_{zz} = -\nu(\sigma_{xx}+\sigma_{yy})/E. A measured thickness strain therefore gives the sum of the in-plane stresses, and an unknown in-plane component follows by subtracting the known one. When a cylinder is confined by a rigid, frictionless container that forbids lateral strain, the lateral stress is found by setting the lateral strain to zero, which gives \sigma_L = \nu\sigma_z/(1-\nu). Substituting back into the axial strain gives \varepsilon_{zz} = \dfrac{\sigma_z(1+\nu)(1-2\nu)}{(1-\nu)E}, so for an applied pressure p (with \sigma_z = -p) the axial strain is compressive with magnitude \dfrac{p(1+\nu)(1-2\nu)}{(1-\nu)E}. The factor (1+\nu)(1-2\nu)/(1-\nu) must not be confused with 1-2\nu alone. Under uniaxial stress \sigma, the principal strains are \sigma/E, -\nu\sigma/E and -\nu\sigma/E. The maximum engineering shear strain is the difference of the largest and smallest principal strains, \gamma_{max} = \varepsilon_1 - \varepsilon_3 = (1+\nu)\dfrac{\sigma}{E}. The tensor shear strain is half of this value, so the two definitions must not be mixed.