Hooke's law

Consider a simple helical spring that has one end attached to some fixed object, while the free end is being pulled by a force whose magnitude is F_s. Suppose that the spring has reached a state of equilibrium, where its length is not changing anymore. Let x be the amount by which the free end of the spring was displaced from its "relaxed" position (when it is not being stretched). Hooke's law states that

$${\displaystyle F_{s}=kx}$$

or, equivalently,

$${\displaystyle x={\frac {F_{s}}{k}}}$$

where k is a positive real number, characteristic of the spring. Moreover, the same formula holds when the spring is compressed, with F_s and x both negative in that case. According to this formula, the graph of the applied force F_s as a function of the displacement x will be a straight line passing through the origin, whose slope is k. Hooke's law for a spring is sometimes, but rarely, stated under the convention that F_s is the restoring force exerted by the spring on whatever is pulling its free end. In that case, the equation becomes

$${\displaystyle F_{s}=-kx}$$

since the direction of the restoring force is opposite to that of the displacement.

Hooke's spring law usually applies to any elastic object, of arbitrary complexity, as long as both the deformation and the stress can be expressed by a single number that can be both positive and negative.

For example, when a block of rubber attached to two parallel plates is deformed by shearing, rather than stretching or compression, the shearing force F_s and the sideways displacement of the plates x obey Hooke's law (for small enough deformations).

Hooke's law also applies when a straight steel bar or concrete beam (like the one used in buildings), supported at both ends, is bent by a weight F placed at some intermediate point. The displacement x in this case is the deviation of the beam, measured in the transversal direction, relative to its unloaded shape.

The law also applies when a stretched steel wire is twisted by pulling on a lever attached to one end. In this case the stress F_s can be taken as the force applied to the lever, and x as the distance traveled by it along its circular path. Or, equivalently, one can let F_s be the torque applied by the lever to the end of the wire, and x be the angle by which that end turns. In either case F_s is proportional to x (although the constant k is different in each case.)

In the case of a helical spring that is stretched or compressed along its axis, the applied (or restoring) force and the resulting elongation or compression have the same direction (which is the direction of said axis). Therefore, if F_s and x are defined as vectors, Hooke's equation still holds and says that the force vector is the elongation vector multiplied by a fixed scalar.

Some elastic bodies will deform in one direction when subjected to a force with a different direction. One example is a horizontal wood beam with non-square rectangular cross section that is bent by a transverse load that is neither vertical nor horizontal. In such cases, the magnitude of the displacement x will be proportional to the magnitude of the force F_s, as long as the direction of the latter remains the same (and its value is not too large); so the scalar version of Hooke's law F_s = −kx will hold. However, the force and displacement vectors will not be scalar multiples of each other, since they have different directions. Moreover, the ratio k between their magnitudes will depend on the direction of the vector F_s.

Yet, in such cases there is often a fixed linear relation between the force and deformation vectors, as long as they are small enough. Namely, there is a function κ from vectors to vectors, such that F = κ(X), and κ(αX₁ + βX₂) = ακ(X₁) + βκ(X₂) for any real numbers α, β and any displacement vectors X₁, X₂. Such a function is called a (second-order) tensor.

With respect to an arbitrary Cartesian coordinate system, the force and displacement vectors can be represented by 3 × 1 matrices of real numbers. Then the tensor κ connecting them can be represented by a 3 × 3 matrix κ of real coefficients, that, when multiplied by the displacement vector, gives the force vector:

$${\displaystyle \mathbf {F} \,=\,{\begin{bmatrix}F_{1}\\F_{2}\\F_{3}\end{bmatrix}}\,=\,{\begin{bmatrix}\kappa _{11}&\kappa _{12}&\kappa _{13}\\\kappa _{21}&\kappa _{22}&\kappa _{23}\\\kappa _{31}&\kappa _{32}&\kappa _{33}\end{bmatrix}}{\begin{bmatrix}X_{1}\\X_{2}\\X_{3}\end{bmatrix}}\,=\,{\boldsymbol {\kappa }}\mathbf {X} }$$

That is,

$${\displaystyle F_{i}=\kappa _{i1}X_{1}+\kappa _{i2}X_{2}+\kappa _{i3}X_{3}}$$

for i = 1, 2, 3. Therefore, Hooke's law F = κX can be said to hold also when X and F are vectors with variable directions, except that the stiffness of the object is a tensor κ, rather than a single real number k.

(a) Schematic of a polymer nanospring. The coil radius, R, pitch, P, length of the spring, L, and the number of turns, N, are 2.5 μm, 2.0 μm, 13 μm, and 4, respectively. Electron micrographs of the nanospring, before loading (b-e), stretched (f), compressed (g), bent (h), and recovered (i). All scale bars are 2 μm. The spring followed a linear response against applied force, demonstrating the validity of Hooke's law at the nanoscale.[4]

The stresses and strains of the material inside a continuous elastic material (such as a block of rubber, the wall of a boiler, or a steel bar) are connected by a linear relationship that is mathematically similar to Hooke's spring law, and is often referred to by that name.

However, the strain state in a solid medium around some point cannot be described by a single vector. The same parcel of material, no matter how small, can be compressed, stretched, and sheared at the same time, along different directions. Likewise, the stresses in that parcel can be at once pushing, pulling, and shearing.

In order to capture this complexity, the relevant state of the medium around a point must be represented by two-second-order tensors, the strain tensor ε (in lieu of the displacement X) and the stress tensor σ (replacing the restoring force F). The analogue of Hooke's spring law for continuous media is then

$${\displaystyle {\boldsymbol {\sigma }}=\mathbf {c} {\boldsymbol {\varepsilon }},}$$

where c is a fourth-order tensor (that is, a linear map between second-order tensors) usually called the stiffness tensor or elasticity tensor. One may also write it as

$${\displaystyle {\boldsymbol {\varepsilon }}=\mathbf {s} {\boldsymbol {\sigma }},}$$

where the tensor s, called the compliance tensor, represents the inverse of said linear map.

In a Cartesian coordinate system, the stress and strain tensors can be represented by 3 × 3 matrices

$${\displaystyle {\boldsymbol {\varepsilon }}\,=\,{\begin{bmatrix}\varepsilon _{11}&\varepsilon _{12}&\varepsilon _{13}\\\varepsilon _{21}&\varepsilon _{22}&\varepsilon _{23}\\\varepsilon _{31}&\varepsilon _{32}&\varepsilon _{33}\end{bmatrix}}\,;\qquad {\boldsymbol {\sigma }}\,=\,{\begin{bmatrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}&\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}\end{bmatrix}}}$$

Being a linear mapping between the nine numbers σ_ij and the nine numbers ε_kl, the stiffness tensor c is represented by a matrix of 3 × 3 × 3 × 3 = 81 real numbers c_ijkl. Hooke's law then says that

$${\displaystyle \sigma _{ij}=\sum _{k=1}^{3}\sum _{l=1}^{3}c_{ijkl}\varepsilon _{kl}}$$

where i,j = 1,2,3.

All three tensors generally vary from point to point inside the medium, and may vary with time as well. The strain tensor ε merely specifies the displacement of the medium particles in the neighborhood of the point, while the stress tensor σ specifies the forces that neighboring parcels of the medium are exerting on each other. Therefore, they are independent of the composition and physical state of the material. The stiffness tensor c, on the other hand, is a property of the material, and often depends on physical state variables such as temperature, pressure, and microstructure.

Due to the inherent symmetries of σ, ε, and c, only 21 elastic coefficients of the latter are independent.[5] This number can be further reduced by the symmetry of the material: 9 for an orthorhombic crystal, 5 for an hexagonal structure, and 3 for a cubic symmetry.[6] For isotropic media (which have the same physical properties in any direction), c can be reduced to only two independent numbers, the bulk modulus K and the shear modulus G, that quantify the material's resistance to changes in volume and to shearing deformations, respectively.

A rod of any elastic material may be viewed as a linear spring. The rod has length L and cross-sectional area A. Its tensile stress σ is linearly proportional to its fractional extension or strain ε by the modulus of elasticity E:

$${\displaystyle \sigma =E\varepsilon .}$$

The modulus of elasticity may often be considered constant. In turn,

$${\displaystyle \varepsilon ={\frac {\Delta L}{L}}}$$

(that is, the fractional change in length), and since

$${\displaystyle \sigma ={\frac {F}{A}}\,,}$$

it follows that:

$${\displaystyle \varepsilon ={\frac {\sigma }{E}}={\frac {F}{AE}}\,.}$$

The change in length may be expressed as

$${\displaystyle \Delta L=\varepsilon L={\frac {FL}{AE}}\,.}$$

The potential energy U_el(x) stored in a spring is given by

$${\displaystyle U_{\mathrm {el} }(x)={\tfrac {1}{2}}kx^{2}}$$

which comes from adding up the energy it takes to incrementally compress the spring. That is, the integral of force over displacement. Since the external force has the same general direction as the displacement, the potential energy of a spring is always non-negative.

This potential U_el can be visualized as a parabola on the Ux-plane such that U_el(x) = 1/2kx². As the spring is stretched in the positive x-direction, the potential energy increases parabolically (the same thing happens as the spring is compressed). Since the change in potential energy changes at a constant rate:

$${\displaystyle {\frac {d^{2}U_{\mathrm {el} }}{dx^{2}}}=k\,.}$$

Note that the change in the change in U is constant even when the displacement and acceleration are zero.

Relaxed force constants (the inverse of generalized compliance constants) are uniquely defined for molecular systems, in contradistinction to the usual "rigid" force constants, and thus their use allows meaningful correlations to be made between force fields calculated for reactants, transition states, and products of a chemical reaction. Just as the potential energy can be written as a quadratic form in the internal coordinates, so it can also be written in terms of generalized forces. The resulting coefficients are termed compliance constants. A direct method exists for calculating the compliance constant for any internal coordinate of a molecule, without the need to do the normal mode analysis.[7] The suitability of relaxed force constants (inverse compliance constants) as covalent bond strength descriptors was demonstrated as early as 1980. Recently, the suitability as non-covalent bond strength descriptors was demonstrated too.[8]

A mass suspended by a spring is the classical example of a harmonic oscillator

A mass m attached to the end of a spring is a classic example of a harmonic oscillator. By pulling slightly on the mass and then releasing it, the system will be set in sinusoidal oscillating motion about the equilibrium position. To the extent that the spring obeys Hooke's law, and that one can neglect friction and the mass of the spring, the amplitude of the oscillation will remain constant; and its frequency f will be independent of its amplitude, determined only by the mass and the stiffness of the spring:

$${\displaystyle f={\frac {1}{2\pi }}{\sqrt {\frac {k}{m}}}}$$

This phenomenon made possible the construction of accurate mechanical clocks and watches that could be carried on ships and people's pockets.

If the mass m were attached to a spring with force constant k and rotating in free space, the spring tension (F_t) would supply the required centripetal force (F_c):

$${\displaystyle F_{\mathrm {t} }=kx\,;\qquad F_{\mathrm {c} }=m\omega ^{2}r}$$

Since F_t = F_c and x = r, then:

$${\displaystyle k=m\omega ^{2}}$$

Given that ω = 2πf, this leads to the same frequency equation as above:

$${\displaystyle f={\frac {1}{2\pi }}{\sqrt {\frac {k}{m}}}}$$

Isotropic materials are characterized by properties which are independent of direction in space. Physical equations involving isotropic materials must therefore be independent of the coordinate system chosen to represent them. The strain tensor is a symmetric tensor. Since the trace of any tensor is independent of any coordinate system, the most complete coordinate-free decomposition of a symmetric tensor is to represent it as the sum of a constant tensor and a traceless symmetric tensor.[9] Thus in index notation:

$${\displaystyle \varepsilon _{ij}=\left({\tfrac {1}{3}}\varepsilon _{kk}\delta _{ij}\right)+\left(\varepsilon _{ij}-{\tfrac {1}{3}}\varepsilon _{kk}\delta _{ij}\right)}$$

where δ_ij is the Kronecker delta. In direct tensor notation:

$${\displaystyle {\boldsymbol {\varepsilon }}=\operatorname {vol} ({\boldsymbol {\varepsilon }})+\operatorname {dev} ({\boldsymbol {\varepsilon }})\,;\qquad \operatorname {vol} ({\boldsymbol {\varepsilon }})={\tfrac {1}{3}}\operatorname {tr} ({\boldsymbol {\varepsilon }})~\mathbf {I} \,;\qquad \operatorname {dev} ({\boldsymbol {\varepsilon }})={\boldsymbol {\varepsilon }}-\operatorname {vol} ({\boldsymbol {\varepsilon }})}$$

where I is the second-order identity tensor.

The first term on the right is the constant tensor, also known as the volumetric strain tensor, and the second term is the traceless symmetric tensor, also known as the deviatoric strain tensor or shear tensor.

The most general form of Hooke's law for isotropic materials may now be written as a linear combination of these two tensors:

$${\displaystyle \sigma _{ij}=3K\left({\tfrac {1}{3}}\varepsilon _{kk}\delta _{ij}\right)+2G\left(\varepsilon _{ij}-{\tfrac {1}{3}}\varepsilon _{kk}\delta _{ij}\right)\,;\qquad {\boldsymbol {\sigma }}=3K\operatorname {vol} ({\boldsymbol {\varepsilon }})+2G\operatorname {dev} ({\boldsymbol {\varepsilon }})}$$

where K is the bulk modulus and G is the shear modulus.

Using the relationships between the elastic moduli, these equations may also be expressed in various other ways. A common form of Hooke's law for isotropic materials, expressed in direct tensor notation, is [10] ${\displaystyle {\boldsymbol {\sigma }}=\lambda \operatorname {tr} ({\boldsymbol {\varepsilon }})\mathbf {I} +2\mu {\boldsymbol {\varepsilon }}={\mathsf {c}}:{\boldsymbol {\varepsilon }}\,;\qquad {\mathsf {c}}=\lambda \mathbf {I} \otimes \mathbf {I} +2\mu {\mathsf {I}}}$ where λ = K − 2/3G = c₁₁₁₁ − 2c₁₂₁₂ and μ = G = c₁₂₁₂ are the Lamé constants, I is the second-rank identity tensor, and I is the symmetric part of the fourth-rank identity tensor. In index notation:

$${\displaystyle \sigma _{ij}=\lambda \varepsilon _{kk}~\delta _{ij}+2\mu \varepsilon _{ij}=c_{ijkl}\varepsilon _{kl}\,;\qquad c_{ijkl}=\lambda \delta _{ij}\delta _{kl}+\mu \left(\delta _{ik}\delta _{jl}+\delta _{il}\delta _{jk}\right)}$$

The inverse relationship is[11]

$${\displaystyle {\boldsymbol {\varepsilon }}={\frac {1}{2\mu }}{\boldsymbol {\sigma }}-{\frac {\lambda }{2\mu (3\lambda +2\mu )}}\operatorname {tr} ({\boldsymbol {\sigma }})\mathbf {I} ={\frac {1}{2G}}{\boldsymbol {\sigma }}+\left({\frac {1}{9K}}-{\frac {1}{6G}}\right)\operatorname {tr} ({\boldsymbol {\sigma }})\mathbf {I} }$$

Therefore, the compliance tensor in the relation ε = s : σ is

$${\displaystyle {\mathsf {s}}=-{\frac {\lambda }{2\mu (3\lambda +2\mu )}}\mathbf {I} \otimes \mathbf {I} +{\frac {1}{2\mu }}{\mathsf {I}}=\left({\frac {1}{9K}}-{\frac {1}{6G}}\right)\mathbf {I} \otimes \mathbf {I} +{\frac {1}{2G}}{\mathsf {I}}}$$

In terms of Young's modulus and Poisson's ratio, Hooke's law for isotropic materials can then be expressed as

$${\displaystyle \varepsilon _{ij}={\frac {1}{E}}{\big (}\sigma _{ij}-\nu (\sigma _{kk}\delta _{ij}-\sigma _{ij}){\big )}\,;\qquad {\boldsymbol {\varepsilon }}={\frac {1}{E}}{\big (}{\boldsymbol {\sigma }}-\nu (\operatorname {tr} ({\boldsymbol {\sigma }})\mathbf {I} -{\boldsymbol {\sigma }}){\big )}={\frac {1+\nu }{E}}{\boldsymbol {\sigma }}-{\frac {\nu }{E}}\operatorname {tr} ({\boldsymbol {\sigma }})\mathbf {I} }$$

This is the form in which the strain is expressed in terms of the stress tensor in engineering. The expression in expanded form is

$${\displaystyle {\begin{aligned}\varepsilon _{11}&={\frac {1}{E}}{\big (}\sigma _{11}-\nu (\sigma _{22}+\sigma _{33}){\big )}\\\varepsilon _{22}&={\frac {1}{E}}{\big (}\sigma _{22}-\nu (\sigma _{11}+\sigma _{33}){\big )}\\\varepsilon _{33}&={\frac {1}{E}}{\big (}\sigma _{33}-\nu (\sigma _{11}+\sigma _{22}){\big )}\\\varepsilon _{12}&={\frac {1}{2G}}\sigma _{12}\,;\qquad \varepsilon _{13}={\frac {1}{2G}}\sigma _{13}\,;\qquad \varepsilon _{23}={\frac {1}{2G}}\sigma _{23}\end{aligned}}}$$

where E is Young's modulus and ν is Poisson's ratio. (See 3-D elasticity).

Derivation of Hooke's law in three dimensions

The three-dimensional form of Hooke's law can be derived using Poisson's ratio and the one-dimensional form of Hooke's law as follows. Consider the strain and stress relation as a superposition of two effects: stretching in direction of the load (1) and shrinking (caused by the load) in perpendicular directions (2 and 3),

$${\displaystyle {\begin{aligned}\varepsilon _{1}'&={\frac {1}{E}}\sigma _{1}\,,\\\varepsilon _{2}'&=-{\frac {\nu }{E}}\sigma _{1}\,,\\\varepsilon _{3}'&=-{\frac {\nu }{E}}\sigma _{1}\,,\end{aligned}}}$$

where ν is Poisson's ratio and E is Young's modulus.

We get similar equations to the loads in directions 2 and 3,

$${\displaystyle {\begin{aligned}\varepsilon _{1}''&=-{\frac {\nu }{E}}\sigma _{2}\,,\\\varepsilon _{2}''&={\frac {1}{E}}\sigma _{2}\,,\\\varepsilon _{3}''&=-{\frac {\nu }{E}}\sigma _{2}\,,\end{aligned}}}$$

and

$${\displaystyle {\begin{aligned}\varepsilon _{1}'''&=-{\frac {\nu }{E}}\sigma _{3}\,,\\\varepsilon _{2}'''&=-{\frac {\nu }{E}}\sigma _{3}\,,\\\varepsilon _{3}'''&={\frac {1}{E}}\sigma _{3}\,.\end{aligned}}}$$

Summing the three cases together (ε_i = ε_i′ + ε_i″ + ε_i‴) we get

$${\displaystyle {\begin{aligned}\varepsilon _{1}&={\frac {1}{E}}{\big (}\sigma _{1}-\nu (\sigma _{2}+\sigma _{3}){\big )}\,,\\\varepsilon _{2}&={\frac {1}{E}}{\big (}\sigma _{2}-\nu (\sigma _{1}+\sigma _{3}){\big )}\,,\\\varepsilon _{3}&={\frac {1}{E}}{\big (}\sigma _{3}-\nu (\sigma _{1}+\sigma _{2}){\big )}\,,\end{aligned}}}$$

or by adding and subtracting one νσ

$${\displaystyle {\begin{aligned}\varepsilon _{1}&={\frac {1}{E}}{\big (}(1+\nu )\sigma _{1}-\nu (\sigma _{1}+\sigma _{2}+\sigma _{3}){\big )}\,,\\\varepsilon _{2}&={\frac {1}{E}}{\big (}(1+\nu )\sigma _{2}-\nu (\sigma _{1}+\sigma _{2}+\sigma _{3}){\big )}\,,\\\varepsilon _{3}&={\frac {1}{E}}{\big (}(1+\nu )\sigma _{3}-\nu (\sigma _{1}+\sigma _{2}+\sigma _{3}){\big )}\,,\end{aligned}}}$$

and further we get by solving σ₁

$${\displaystyle \sigma _{1}={\frac {E}{1+\nu }}\varepsilon _{1}+{\frac {\nu }{1+\nu }}(\sigma _{1}+\sigma _{2}+\sigma _{3})\,.}$$

Calculating the sum

$${\displaystyle {\begin{aligned}\varepsilon _{1}+\varepsilon _{2}+\varepsilon _{3}&={\frac {1}{E}}{\big (}(1+\nu )(\sigma _{1}+\sigma _{2}+\sigma _{3})-3\nu (\sigma _{1}+\sigma _{2}+\sigma _{3}){\big )}={\frac {1-2\nu }{E}}(\sigma _{1}+\sigma _{2}+\sigma _{3})\\\sigma _{1}+\sigma _{2}+\sigma _{3}&={\frac {E}{1-2\nu }}(\varepsilon _{1}+\varepsilon _{2}+\varepsilon _{3})\end{aligned}}}$$

and substituting it to the equation solved for σ₁ gives

$${\displaystyle {\begin{aligned}\sigma _{1}&={\frac {E}{1+\nu }}\varepsilon _{1}+{\frac {E\nu }{(1+\nu )(1-2\nu )}}(\varepsilon _{1}+\varepsilon _{2}+\varepsilon _{3})\\&=2\mu \varepsilon _{1}+\lambda (\varepsilon _{1}+\varepsilon _{2}+\varepsilon _{3})\,,\end{aligned}}}$$

where μ and λ are the Lamé parameters.

Similar treatment of directions 2 and 3 gives the Hooke's law in three dimensions.

In matrix form, Hooke's law for isotropic materials can be written as

$${\displaystyle {\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\end{bmatrix}}\,=\,{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\\gamma _{23}\\\gamma _{13}\\\gamma _{12}\end{bmatrix}}\,=\,{\frac {1}{E}}{\begin{bmatrix}1&-\nu &-\nu &0&0&0\\-\nu &1&-\nu &0&0&0\\-\nu &-\nu &1&0&0&0\\0&0&0&2+2\nu &0&0\\0&0&0&0&2+2\nu &0\\0&0&0&0&0&2+2\nu \end{bmatrix}}{\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\end{bmatrix}}}$$

where γ_ij = 2ε_ij is the engineering shear strain. The inverse relation may be written as

$${\displaystyle {\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\end{bmatrix}}\,=\,{\frac {E}{(1+\nu )(1-2\nu )}}{\begin{bmatrix}1-\nu &\nu &\nu &0&0&0\\\nu &1-\nu &\nu &0&0&0\\\nu &\nu &1-\nu &0&0&0\\0&0&0&{\frac {1-2\nu }{2}}&0&0\\0&0&0&0&{\frac {1-2\nu }{2}}&0\\0&0&0&0&0&{\frac {1-2\nu }{2}}\end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\end{bmatrix}}}$$

which can be simplified thanks to the Lamé constants:

$${\displaystyle {\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\end{bmatrix}}\,=\,{\begin{bmatrix}2\mu +\lambda &\lambda &\lambda &0&0&0\\\lambda &2\mu +\lambda &\lambda &0&0&0\\\lambda &\lambda &2\mu +\lambda &0&0&0\\0&0&0&\mu &0&0\\0&0&0&0&\mu &0\\0&0&0&0&0&\mu \end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\end{bmatrix}}}$$

In vector notation this becomes

$${\displaystyle {\begin{bmatrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{12}&\sigma _{22}&\sigma _{23}\\\sigma _{13}&\sigma _{23}&\sigma _{33}\end{bmatrix}}\,=\,2\mu {\begin{bmatrix}\varepsilon _{11}&\varepsilon _{12}&\varepsilon _{13}\\\varepsilon _{12}&\varepsilon _{22}&\varepsilon _{23}\\\varepsilon _{13}&\varepsilon _{23}&\varepsilon _{33}\end{bmatrix}}+\lambda \mathbf {I} \left(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}\right)}$$

where I is the identity tensor.

Under plane stress conditions, σ₃₁ = σ₁₃ = σ₃₂ = σ₂₃ = σ₃₃ = 0. In that case Hooke's law takes the form

$${\displaystyle {\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{12}\end{bmatrix}}\,=\,{\frac {E}{1-\nu ^{2}}}{\begin{bmatrix}1&\nu &0\\\nu &1&0\\0&0&{\frac {1-\nu }{2}}\end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\2\varepsilon _{12}\end{bmatrix}}}$$

In vector notation this becomes

$${\displaystyle {\begin{bmatrix}\sigma _{11}&\sigma _{12}\\\sigma _{12}&\sigma _{22}\end{bmatrix}}\,=\,{\frac {E}{1-\nu ^{2}}}\left((1-\nu ){\begin{bmatrix}\varepsilon _{11}&\varepsilon _{12}\\\varepsilon _{12}&\varepsilon _{22}\end{bmatrix}}+\nu \mathbf {I} \left(\varepsilon _{11}+\varepsilon _{22}\right)\right)}$$

The inverse relation is usually written in the reduced form

$${\displaystyle {\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\2\varepsilon _{12}\end{bmatrix}}\,=\,{\frac {1}{E}}{\begin{bmatrix}1&-\nu &0\\-\nu &1&0\\0&0&2+2\nu \end{bmatrix}}{\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{12}\end{bmatrix}}}$$

Under plane strain conditions, ε₃₁ = ε₁₃ = ε₃₂ = ε₂₃ = ε₃₃ = 0. In this case Hooke's law takes the form

$${\displaystyle {\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{12}\end{bmatrix}}\,=\,{\frac {E}{(1+\nu )(1-2\nu )}}{\begin{bmatrix}1-\nu &\nu &0\\\nu &1-\nu &0\\0&0&{\frac {1-2\nu }{2}}\end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\2\varepsilon _{12}\end{bmatrix}}}$$

The symmetry of the Cauchy stress tensor (σ_ij = σ_ji) and the generalized Hooke's laws (σ_ij = c_ijklε_kl) implies that c_ijkl = c_jikl. Similarly, the symmetry of the infinitesimal strain tensor implies that c_ijkl = c_ijlk. These symmetries are called the minor symmetries of the stiffness tensor c. This reduces the number of elastic constants from 81 to 36.

If in addition, since the displacement gradient and the Cauchy stress are work conjugate, the stress–strain relation can be derived from a strain energy density functional (U), then

$${\displaystyle \sigma _{ij}={\frac {\partial U}{\partial \varepsilon _{ij}}}\quad \implies \quad c_{ijkl}={\frac {\partial ^{2}U}{\partial \varepsilon _{ij}\partial \varepsilon _{kl}}}\,.}$$

The arbitrariness of the order of differentiation implies that c_ijkl = c_klij. These are called the major symmetries of the stiffness tensor. This reduces the number of elastic constants from 36 to 21. The major and minor symmetries indicate that the stiffness tensor has only 21 independent components.

It is often useful to express the anisotropic form of Hooke's law in matrix notation, also called Voigt notation. To do this we take advantage of the symmetry of the stress and strain tensors and express them as six-dimensional vectors in an orthonormal coordinate system (e₁,e₂,e₃) as

$${\displaystyle [{\boldsymbol {\sigma }}]\,=\,{\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\end{bmatrix}}\,\equiv \,{\begin{bmatrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{bmatrix}}\,;\qquad [{\boldsymbol {\varepsilon }}]\,=\,{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\end{bmatrix}}\,\equiv \,{\begin{bmatrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _{6}\end{bmatrix}}}$$

Then the stiffness tensor (c) can be expressed as

$${\displaystyle [{\mathsf {c}}]\,=\,{\begin{bmatrix}c_{1111}&c_{1122}&c_{1133}&c_{1123}&c_{1131}&c_{1112}\\c_{2211}&c_{2222}&c_{2233}&c_{2223}&c_{2231}&c_{2212}\\c_{3311}&c_{3322}&c_{3333}&c_{3323}&c_{3331}&c_{3312}\\c_{2311}&c_{2322}&c_{2333}&c_{2323}&c_{2331}&c_{2312}\\c_{3111}&c_{3122}&c_{3133}&c_{3123}&c_{3131}&c_{3112}\\c_{1211}&c_{1222}&c_{1233}&c_{1223}&c_{1231}&c_{1212}\end{bmatrix}}\,\equiv \,{\begin{bmatrix}C_{11}&C_{12}&C_{13}&C_{14}&C_{15}&C_{16}\\C_{12}&C_{22}&C_{23}&C_{24}&C_{25}&C_{26}\\C_{13}&C_{23}&C_{33}&C_{34}&C_{35}&C_{36}\\C_{14}&C_{24}&C_{34}&C_{44}&C_{45}&C_{46}\\C_{15}&C_{25}&C_{35}&C_{45}&C_{55}&C_{56}\\C_{16}&C_{26}&C_{36}&C_{46}&C_{56}&C_{66}\end{bmatrix}}}$$

and Hooke's law is written as

$${\displaystyle [{\boldsymbol {\sigma }}]=[{\mathsf {C}}][{\boldsymbol {\varepsilon }}]\qquad {\text{or}}\qquad \sigma _{i}=C_{ij}\varepsilon _{j}\,.}$$

Similarly the compliance tensor (s) can be written as

$${\displaystyle [{\mathsf {s}}]\,=\,{\begin{bmatrix}s_{1111}&s_{1122}&s_{1133}&2s_{1123}&2s_{1131}&2s_{1112}\\s_{2211}&s_{2222}&s_{2233}&2s_{2223}&2s_{2231}&2s_{2212}\\s_{3311}&s_{3322}&s_{3333}&2s_{3323}&2s_{3331}&2s_{3312}\\2s_{2311}&2s_{2322}&2s_{2333}&4s_{2323}&4s_{2331}&4s_{2312}\\2s_{3111}&2s_{3122}&2s_{3133}&4s_{3123}&4s_{3131}&4s_{3112}\\2s_{1211}&2s_{1222}&2s_{1233}&4s_{1223}&4s_{1231}&4s_{1212}\end{bmatrix}}\,\equiv \,{\begin{bmatrix}S_{11}&S_{12}&S_{13}&S_{14}&S_{15}&S_{16}\\S_{12}&S_{22}&S_{23}&S_{24}&S_{25}&S_{26}\\S_{13}&S_{23}&S_{33}&S_{34}&S_{35}&S_{36}\\S_{14}&S_{24}&S_{34}&S_{44}&S_{45}&S_{46}\\S_{15}&S_{25}&S_{35}&S_{45}&S_{55}&S_{56}\\S_{16}&S_{26}&S_{36}&S_{46}&S_{56}&S_{66}\end{bmatrix}}}$$

If a linear elastic material is rotated from a reference configuration to another, then the material is symmetric with respect to the rotation if the components of the stiffness tensor in the rotated configuration are related to the components in the reference configuration by the relation[12]

$${\displaystyle c_{pqrs}=l_{pi}l_{qj}l_{rk}l_{sl}c_{ijkl}}$$

where l_ab are the components of an orthogonal rotation matrix [L]. The same relation also holds for inversions.

In matrix notation, if the transformed basis (rotated or inverted) is related to the reference basis by

$${\displaystyle [\mathbf {e} _{i}']=[L][\mathbf {e} _{i}]}$$

then

$${\displaystyle C_{ij}\varepsilon _{i}\varepsilon _{j}=C_{ij}'\varepsilon '_{i}\varepsilon '_{j}\,.}$$

In addition, if the material is symmetric with respect to the transformation [L] then

$${\displaystyle C_{ij}=C'_{ij}\quad \implies \quad C_{ij}(\varepsilon _{i}\varepsilon _{j}-\varepsilon '_{i}\varepsilon '_{j})=0\,.}$$

Orthotropic materials have three orthogonal planes of symmetry. If the basis vectors (e₁,e₂,e₃) are normals to the planes of symmetry then the coordinate transformation relations imply that

$${\displaystyle {\begin{bmatrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{bmatrix}}\,=\,{\begin{bmatrix}C_{11}&C_{12}&C_{13}&0&0&0\\C_{12}&C_{22}&C_{23}&0&0&0\\C_{13}&C_{23}&C_{33}&0&0&0\\0&0&0&C_{44}&0&0\\0&0&0&0&C_{55}&0\\0&0&0&0&0&C_{66}\end{bmatrix}}{\begin{bmatrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _{6}\end{bmatrix}}}$$

The inverse of this relation is commonly written as[13]

$${\displaystyle {\begin{bmatrix}\varepsilon _{xx}\\\varepsilon _{yy}\\\varepsilon _{zz}\\2\varepsilon _{yz}\\2\varepsilon _{zx}\\2\varepsilon _{xy}\end{bmatrix}}\,=\,{\begin{bmatrix}{\frac {1}{E_{x}}}&-{\frac {\nu _{yx}}{E_{y}}}&-{\frac {\nu _{zx}}{E_{z}}}&0&0&0\\-{\frac {\nu _{xy}}{E_{x}}}&{\frac {1}{E_{y}}}&-{\frac {\nu _{zy}}{E_{z}}}&0&0&0\\-{\frac {\nu _{xz}}{E_{x}}}&-{\frac {\nu _{yz}}{E_{y}}}&{\frac {1}{E_{z}}}&0&0&0\\0&0&0&{\frac {1}{G_{yz}}}&0&0\\0&0&0&0&{\frac {1}{G_{zx}}}&0\\0&0&0&0&0&{\frac {1}{G_{xy}}}\\\end{bmatrix}}{\begin{bmatrix}\sigma _{xx}\\\sigma _{yy}\\\sigma _{zz}\\\sigma _{yz}\\\sigma _{zx}\\\sigma _{xy}\end{bmatrix}}}$$

where

• E_i is the Young's modulus along axis i
• G_ij is the shear modulus in direction j on the plane whose normal is in direction i
• ν_ij is the Poisson's ratio that corresponds to a contraction in direction j when an extension is applied in direction i.

Under plane stress conditions, σ_zz = σ_zx = σ_yz = 0, Hooke's law for an orthotropic material takes the form

$${\displaystyle {\begin{bmatrix}\varepsilon _{xx}\\\varepsilon _{yy}\\2\varepsilon _{xy}\end{bmatrix}}\,=\,{\begin{bmatrix}{\frac {1}{E_{x}}}&-{\frac {\nu _{yx}}{E_{y}}}&0\\-{\frac {\nu _{xy}}{E_{x}}}&{\frac {1}{E_{y}}}&0\\0&0&{\frac {1}{G_{xy}}}\end{bmatrix}}{\begin{bmatrix}\sigma _{xx}\\\sigma _{yy}\\\sigma _{xy}\end{bmatrix}}\,.}$$

The inverse relation is

$${\displaystyle {\begin{bmatrix}\sigma _{xx}\\\sigma _{yy}\\\sigma _{xy}\end{bmatrix}}\,=\,{\frac {1}{1-\nu _{xy}\nu _{yx}}}{\begin{bmatrix}E_{x}&\nu _{yx}E_{x}&0\\\nu _{xy}E_{y}&E_{y}&0\\0&0&G_{xy}(1-\nu _{xy}\nu _{yx})\end{bmatrix}}{\begin{bmatrix}\varepsilon _{xx}\\\varepsilon _{yy}\\2\varepsilon _{xy}\end{bmatrix}}\,.}$$

The transposed form of the above stiffness matrix is also often used.

A transversely isotropic material is symmetric with respect to a rotation about an axis of symmetry. For such a material, if e₃ is the axis of symmetry, Hooke's law can be expressed as

$${\displaystyle {\begin{bmatrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{bmatrix}}\,=\,{\begin{bmatrix}C_{11}&C_{12}&C_{13}&0&0&0\\C_{12}&C_{11}&C_{13}&0&0&0\\C_{13}&C_{13}&C_{33}&0&0&0\\0&0&0&C_{44}&0&0\\0&0&0&0&C_{44}&0\\0&0&0&0&0&{\frac {C_{11}-C_{12}}{2}}\end{bmatrix}}{\begin{bmatrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _{6}\end{bmatrix}}}$$

More frequently, the x ≡ e₁ axis is taken to be the axis of symmetry and the inverse Hooke's law is written as [14]

$${\displaystyle {\begin{bmatrix}\varepsilon _{xx}\\\varepsilon _{yy}\\\varepsilon _{zz}\\2\varepsilon _{yz}\\2\varepsilon _{zx}\\2\varepsilon _{xy}\end{bmatrix}}\,=\,{\begin{bmatrix}{\frac {1}{E_{x}}}&-{\frac {\nu _{yx}}{E_{y}}}&-{\frac {\nu _{zx}}{E_{z}}}&0&0&0\\-{\frac {\nu _{xy}}{E_{x}}}&{\frac {1}{E_{y}}}&-{\frac {\nu _{zy}}{E_{z}}}&0&0&0\\-{\frac {\nu _{xz}}{E_{x}}}&-{\frac {\nu _{yz}}{E_{y}}}&{\frac {1}{E_{z}}}&0&0&0\\0&0&0&{\frac {1}{G_{yz}}}&0&0\\0&0&0&0&{\frac {1}{G_{xz}}}&0\\0&0&0&0&0&{\frac {1}{G_{xy}}}\\\end{bmatrix}}{\begin{bmatrix}\sigma _{xx}\\\sigma _{yy}\\\sigma _{zz}\\\sigma _{yz}\\\sigma _{zx}\\\sigma _{xy}\end{bmatrix}}}$$

To grasp the degree of anisotropy of any class, a universal elastic anisotropy index (AU)[15] was formulated. It replaces the Zener ratio, which is suited for cubic crystals.