In the theory of finite deformations, extension and rotations of line elements are unrestricted as to size. For an infinitesimal fibre that deforms from an initial point given by the vector dX to the vector dx in the time t, the deformation gradient is defined by F_ij = ∂x_i(X, t)/∂X_j; the 3 × 3 matrix [F], with components F_ij, may be represented as a pure deformation, characterized by a symmetric matrix [U], followed by a rigid rotation [R]. This result is called the polar decomposition theorem and takes the form, in matrix notation, [F] = [R][U]. For an arbitrary deformation, there exist three mutually orthogonal principal stretch directions at each point of the material; call these directions in the reference configuration N^(I), N^(II), N^(III), and let the stretch ratios be λ_I, λ_II, λ_III. Fibres in these three principal strain directions undergo extensional strain but have no shearing between them. Those three fibres in the deformed configuration remain orthogonal but are rotated by the operation [R].

As noted earlier, an extensional strain may be defined by E = g(λ), where g(1) = 0 and g′(1) = 1, with examples for g(λ) given above. A finite strain tensor E_ij may then be defined based on any particular function g(λ) by E_ij = g(λ_I)N_i^(I)N_j^(I) + g(λ_II)N_i^(II)N_j^(II) + g(λ_III)N_i^(III)N_j^(III). Usually, it is rather difficult to actually solve for the λ’s and N’s associated with any general [F], so it is not easy to use this strain definition. However, for the special choice identified as g^M(λ) = (λ² − 1)/2 above, it may be shown thatwhich, like the finite strain generated by any other g(λ), reduces to ε_ij when linearized in [∂u/∂X].

Linear elastic isotropic solid

The simplest type of stress-strain relation is that of the linear elastic solid, considered in circumstances for which |∂u_i/∂X_j|<< 1 and for isotropic materials, whose mechanical response is independent of the direction of stressing. If a material point sustains a stress state σ₁₁ = σ, with all other σ_ij = 0, it is subjected to uniaxial tensile stress. This can be realized in a homogeneous bar loaded by an axial force. The resulting strain may be rewritten as ε₁₁ = σ/E, ε₂₂ = ε₃₃ = −νε₁₁ = −νσ/E, ε₁₂ = ε₂₃ = ε₃₁ = 0. Two new parameters have been introduced here, E and ν. E is called Young’s modulus, and it has dimensions of [force]/[length]² and is measured in units such as the pascal (1 Pa = 1 N/m²), dyne/cm², or pounds per square inch (psi); ν, which equals the ratio of lateral strain to axial strain, is dimensionless and is called the Poisson ratio.

If the isotropic solid is subjected only to shear stress τ—i.e., σ₁₂ = σ₂₁ = τ, with all other σ_ij = 0—then the response is shearing strain of the same type, ε₁₂ = τ/2G, ε₂₃ = ε₃₁ = ε₁₁ = ε₂₂ = ε₃₃ = 0. Notice that because 2ε₁₂ = γ₁₂, this is equivalent to γ₁₂ = τ/G. The constant G introduced is called the shear modulus. (Frequently, the symbol μ is used instead of G.) The shear modulus G is not independent of E and ν but is related to them by G = E/2(1 + ν), as follows from the tensor nature of stress and strain. The general stress-strain relations are thenwhere δ_ij is defined as 1 when its indices agree and 0 otherwise.

These relations can be inverted to read σ_ij = λδ_ij(ε₁₁ + ε₂₂ + ε₃₃) + 2με_ij, where μ has been used rather than G as the notation for the shear modulus, following convention, and where λ = 2νμ/(1 − 2ν). The elastic constants λ and μ are sometimes called the Lamé constants. Since ν is typically in the range 1/4 to 1/3 for hard polycrystalline solids, λ falls often in the range between μ and 2μ. (Navier’s particle model with central forces leads to λ = μ for an isotropic solid.)

Another elastic modulus often cited is the bulk modulus K, defined for a linear solid under pressure p(σ₁₁ = σ₂₂ = σ₃₃ = −p) such that the fractional decrease in volume is p/K. For example, consider a small cube of side length L in the reference state. If the length along, say, the 1 direction changes to (1 + ε₁₁)L, the fractional change of volume is (1 + ε₁₁)(1 + ε₂₂)(1 + ε₃₃) − 1 = ε₁₁ + ε₂₂ + ε₃₃, neglecting quadratic and cubic order terms in the ε_ij compared to the linear, as is appropriate when using linear elasticity. Thus, K = E/3(1 − 2ν) = λ + 2μ/3.

Thermal strains

Temperature change can also cause strain. In an isotropic material the thermally induced extensional strains are equal in all directions, and there are no shear strains. In the simplest cases, these thermal strains can be treated as being linear in the temperature change θ − θ₀ (where θ₀ is the temperature of the reference state), writing ε_ij^thermal = δ_ijα(θ − θ₀) for the strain produced by temperature change in the absence of stress. Here α is called the coefficient of thermal expansion. Thus, in cases of temperature change, ε_ij is replaced in the stress-strain relations above with ε_ij − ε_ij^thermal, with the thermal part given as a function of temperature. Typically, when temperature changes are modest, the small dependence of E and ν on temperature can be neglected.