Part of a series on
Continuum mechanics
${\displaystyle J=-D{\frac {d\varphi }{dx))}$ Fick's laws of diffusion
Laws Conservations Mass Momentum Energy Inequalities Clausius–Duhem (entropy)
Solid mechanics Deformation Elasticity linear Plasticity Hooke's law Stress Strain Finite strain Infinitesimal strain Compatibility Bending Contact mechanics frictional Material failure theory Fracture mechanics
Fluid mechanics Fluids Statics · Dynamics Archimedes' principle · Bernoulli's principle Navier–Stokes equations Poiseuille equation · Pascal's law Viscosity (Newtonian · non-Newtonian) Buoyancy · Mixing · Pressure Liquids Adhesion Capillary action Chromatography Cohesion (chemistry) Surface tension Gases Atmosphere Boyle's law Charles's law Combined gas law Fick's law Gay-Lussac's law Graham's law Plasma
Rheology Viscoelasticity Rheometry Rheometer Smart fluids Electrorheological Magnetorheological Ferrofluids
Scientists Bernoulli Boyle Cauchy Charles Euler Fick Gay-Lussac Graham Hooke Newton Navier Noll Pascal Stokes Truesdell
v t e

In continuum mechanics, the infinitesimal strain theory is a mathematical approach to the description of the deformation of a solid body in which the displacements of the material particles are assumed to be much smaller (indeed, infinitesimally smaller) than any relevant dimension of the body; so that its geometry and the constitutive properties of the material (such as density and stiffness) at each point of space can be assumed to be unchanged by the deformation.

With this assumption, the equations of continuum mechanics are considerably simplified. This approach may also be called small deformation theory, small displacement theory, or small displacement-gradient theory. It is contrasted with the finite strain theory where the opposite assumption is made.

The infinitesimal strain theory is commonly adopted in civil and mechanical engineering for the stress analysis of structures built from relatively stiff elastic materials like concrete and steel, since a common goal in the design of such structures is to minimize their deformation under typical loads. However, this approximation demands caution in the case of thin flexible bodies, such as rods, plates, and shells which are susceptible to significant rotations, thus making the results unreliable.^[1]

Infinitesimal strain tensor

For infinitesimal deformations of a continuum body, in which the displacement gradient tensor (2nd order tensor) is small compared to unity, i.e. ${\displaystyle \|\nabla \mathbf {u} \|\ll 1}$, it is possible to perform a geometric linearization of any one of the finite strain tensors used in finite strain theory, e.g. the Lagrangian finite strain tensor ${\displaystyle \mathbf {E} }$, and the Eulerian finite strain tensor ${\displaystyle \mathbf {e} }$. In such a linearization, the non-linear or second-order terms of the finite strain tensor are neglected. Thus we have

$${\displaystyle \mathbf {E} ={\frac {1}{2))\left(\nabla _{\mathbf {X} }\mathbf {u} +(\nabla _{\mathbf {X} }\mathbf {u} )^{T}+(\nabla _{\mathbf {X} }\mathbf {u} )^{T}\nabla _{\mathbf {X} }\mathbf {u} \right)\approx {\frac {1}{2))\left(\nabla _{\mathbf {X} }\mathbf {u} +(\nabla _{\mathbf {X} }\mathbf {u} )^{T}\right)}$$

$${\displaystyle E_{KL}={\frac {1}{2))\left({\frac {\partial U_{K)){\partial X_{L))}+{\frac {\partial U_{L)){\partial X_{K))}+{\frac {\partial U_{M)){\partial X_{K))}{\frac {\partial U_{M)){\partial X_{L))}\right)\approx {\frac {1}{2))\left({\frac {\partial U_{K)){\partial X_{L))}+{\frac {\partial U_{L)){\partial X_{K))}\right)}$$

$${\displaystyle \mathbf {e} ={\frac {1}{2))\left(\nabla _{\mathbf {x} }\mathbf {u} +(\nabla _{\mathbf {x} }\mathbf {u} )^{T}-\nabla _{\mathbf {x} }\mathbf {u} (\nabla _{\mathbf {x} }\mathbf {u} )^{T}\right)\approx {\frac {1}{2))\left(\nabla _{\mathbf {x} }\mathbf {u} +(\nabla _{\mathbf {x} }\mathbf {u} )^{T}\right)}$$

$${\displaystyle e_{rs}={\frac {1}{2))\left({\frac {\partial u_{r)){\partial x_{s))}+{\frac {\partial u_{s)){\partial x_{r))}-{\frac {\partial u_{k)){\partial x_{r))}{\frac {\partial u_{k)){\partial x_{s))}\right)\approx {\frac {1}{2))\left({\frac {\partial u_{r)){\partial x_{s))}+{\frac {\partial u_{s)){\partial x_{r))}\right)}$$

This linearization implies that the Lagrangian description and the Eulerian description are approximately the same as there is little difference in the material and spatial coordinates of a given material point in the continuum. Therefore, the material displacement gradient tensor components and the spatial displacement gradient tensor components are approximately equal. Thus we have

$${\displaystyle \mathbf {E} \approx \mathbf {e} \approx {\boldsymbol {\varepsilon ))={\frac {1}{2))\left((\nabla \mathbf {u} )^{T}+\nabla \mathbf {u} \right)}$$

$${\displaystyle E_{KL}\approx e_{rs}\approx \varepsilon _{ij}={\frac {1}{2))\left(u_{i,j}+u_{j,i}\right)}$$

${\displaystyle \varepsilon _{ij))$

${\displaystyle {\boldsymbol {\varepsilon ))}$

$${\displaystyle {\begin{aligned}\varepsilon _{ij}&={\frac {1}{2))\left(u_{i,j}+u_{j,i}\right)\\&={\begin{bmatrix}\varepsilon _{11}&\varepsilon _{12}&\varepsilon _{13}\\\varepsilon _{21}&\varepsilon _{22}&\varepsilon _{23}\\\varepsilon _{31}&\varepsilon _{32}&\varepsilon _{33}\\\end{bmatrix))\\&={\begin{bmatrix}{\frac {\partial u_{1)){\partial x_{1))}&{\frac {1}{2))\left({\frac {\partial u_{1)){\partial x_{2))}+{\frac {\partial u_{2)){\partial x_{1))}\right)&{\frac {1}{2))\left({\frac {\partial u_{1)){\partial x_{3))}+{\frac {\partial u_{3)){\partial x_{1))}\right)\\{\frac {1}{2))\left({\frac {\partial u_{2)){\partial x_{1))}+{\frac {\partial u_{1)){\partial x_{2))}\right)&{\frac {\partial u_{2)){\partial x_{2))}&{\frac {1}{2))\left({\frac {\partial u_{2)){\partial x_{3))}+{\frac {\partial u_{3)){\partial x_{2))}\right)\\{\frac {1}{2))\left({\frac {\partial u_{3)){\partial x_{1))}+{\frac {\partial u_{1)){\partial x_{3))}\right)&{\frac {1}{2))\left({\frac {\partial u_{3)){\partial x_{2))}+{\frac {\partial u_{2)){\partial x_{3))}\right)&{\frac {\partial u_{3)){\partial x_{3))}\\\end{bmatrix))\end{aligned))}$$

$${\displaystyle {\begin{bmatrix}\varepsilon _{xx}&\varepsilon _{xy}&\varepsilon _{xz}\\\varepsilon _{yx}&\varepsilon _{yy}&\varepsilon _{yz}\\\varepsilon _{zx}&\varepsilon _{zy}&\varepsilon _{zz}\\\end{bmatrix))={\begin{bmatrix}{\frac {\partial u_{x)){\partial x))&{\frac {1}{2))\left({\frac {\partial u_{x)){\partial y))+{\frac {\partial u_{y)){\partial x))\right)&{\frac {1}{2))\left({\frac {\partial u_{x)){\partial z))+{\frac {\partial u_{z)){\partial x))\right)\\{\frac {1}{2))\left({\frac {\partial u_{y)){\partial x))+{\frac {\partial u_{x)){\partial y))\right)&{\frac {\partial u_{y)){\partial y))&{\frac {1}{2))\left({\frac {\partial u_{y)){\partial z))+{\frac {\partial u_{z)){\partial y))\right)\\{\frac {1}{2))\left({\frac {\partial u_{z)){\partial x))+{\frac {\partial u_{x)){\partial z))\right)&{\frac {1}{2))\left({\frac {\partial u_{z)){\partial y))+{\frac {\partial u_{y)){\partial z))\right)&{\frac {\partial u_{z)){\partial z))\\\end{bmatrix))}$$

Furthermore, since the deformation gradient can be expressed as ${\displaystyle {\boldsymbol {F))={\boldsymbol {\nabla ))\mathbf {u} +{\boldsymbol {I))}$ where ${\displaystyle {\boldsymbol {I))}$ is the second-order identity tensor, we have

$${\displaystyle {\boldsymbol {\varepsilon ))={\frac {1}{2))\left({\boldsymbol {F))^{T}+{\boldsymbol {F))\right)-{\boldsymbol {I))}$$

Also, from the general expression for the Lagrangian and Eulerian finite strain tensors we have

$${\displaystyle {\begin{aligned}\mathbf {E} _{(m)}&={\frac {1}{2m))(\mathbf {U} ^{2m}-{\boldsymbol {I)))={\frac {1}{2m))[({\boldsymbol {F))^{T}{\boldsymbol {F)))^{m}-{\boldsymbol {I))]\approx {\frac {1}{2m))[\((\boldsymbol {\nabla ))\mathbf {u} +({\boldsymbol {\nabla ))\mathbf {u} )^{T}+{\boldsymbol {I))\}^{m}-{\boldsymbol {I))]\approx {\boldsymbol {\varepsilon ))\\\mathbf {e} _{(m)}&={\frac {1}{2m))(\mathbf {V} ^{2m}-{\boldsymbol {I)))={\frac {1}{2m))[({\boldsymbol {F)){\boldsymbol {F))^{T})^{m}-{\boldsymbol {I))]\approx {\boldsymbol {\varepsilon ))\end{aligned))}$$

Geometric derivation

Consider a two-dimensional deformation of an infinitesimal rectangular material element with dimensions ${\displaystyle dx}$ by ${\displaystyle dy}$ (Figure 1), which after deformation, takes the form of a rhombus. From the geometry of Figure 1 we have

$${\displaystyle {\begin{aligned}{\overline {ab))&={\sqrt {\left(dx+{\frac {\partial u_{x)){\partial x))dx\right)^{2}+\left({\frac {\partial u_{y)){\partial x))dx\right)^{2))}\\&=dx{\sqrt {1+2{\frac {\partial u_{x)){\partial x))+\left({\frac {\partial u_{x)){\partial x))\right)^{2}+\left({\frac {\partial u_{y)){\partial x))\right)^{2))}\\\end{aligned))}$$

For very small displacement gradients, i.e., ${\displaystyle \|\nabla \mathbf {u} \|\ll 1}$, we have

$${\displaystyle {\overline {ab))\approx dx+{\frac {\partial u_{x)){\partial x))dx}$$

The normal strain in the ${\displaystyle x}$-direction of the rectangular element is defined by

$${\displaystyle \varepsilon _{x}={\frac ((\overline {ab))-{\overline {AB))}{\overline {AB))))$$

${\displaystyle {\overline {AB))=dx}$

$${\displaystyle \varepsilon _{x}={\frac {\partial u_{x)){\partial x))}$$

Similarly, the normal strain in the ${\displaystyle y}$-direction, and ${\displaystyle z}$-direction, becomes

$${\displaystyle \varepsilon _{y}={\frac {\partial u_{y)){\partial y))\quad ,\qquad \varepsilon _{z}={\frac {\partial u_{z)){\partial z))}$$

The engineering shear strain, or the change in angle between two originally orthogonal material lines, in this case line ${\displaystyle {\overline {AC))}$ and ${\displaystyle {\overline {AB))}$, is defined as

$${\displaystyle \gamma _{xy}=\alpha +\beta }$$

From the geometry of Figure 1 we have

$${\displaystyle \tan \alpha ={\frac ((\dfrac {\partial u_{y)){\partial x))dx}{dx+{\dfrac {\partial u_{x)){\partial x))dx))={\frac {\dfrac {\partial u_{y)){\partial x)){1+{\dfrac {\partial u_{x)){\partial x))))\quad ,\qquad \tan \beta ={\frac ((\dfrac {\partial u_{x)){\partial y))dy}{dy+{\dfrac {\partial u_{y)){\partial y))dy))={\frac {\dfrac {\partial u_{x)){\partial y)){1+{\dfrac {\partial u_{y)){\partial y))))}$$

For small rotations, i.e., ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are ${\displaystyle \ll 1}$ we have

$${\displaystyle \tan \alpha \approx \alpha \quad ,\qquad \tan \beta \approx \beta }$$

$${\displaystyle \alpha ={\frac {\partial u_{y)){\partial x))\quad ,\qquad \beta ={\frac {\partial u_{x)){\partial y))}$$

$${\displaystyle \gamma _{xy}=\alpha +\beta ={\frac {\partial u_{y)){\partial x))+{\frac {\partial u_{x)){\partial y))}$$

${\displaystyle x}$

${\displaystyle y}$

${\displaystyle u_{x))$

${\displaystyle u_{y))$

${\displaystyle \gamma _{xy}=\gamma _{yx))$

Similarly, for the ${\displaystyle y}$-${\displaystyle z}$ and ${\displaystyle x}$-${\displaystyle z}$ planes, we have

$${\displaystyle \gamma _{yz}=\gamma _{zy}={\frac {\partial u_{y)){\partial z))+{\frac {\partial u_{z)){\partial y))\quad ,\qquad \gamma _{zx}=\gamma _{xz}={\frac {\partial u_{z)){\partial x))+{\frac {\partial u_{x)){\partial z))}$$

It can be seen that the tensorial shear strain components of the infinitesimal strain tensor can then be expressed using the engineering strain definition, ${\displaystyle \gamma }$, as

$${\displaystyle {\begin{bmatrix}\varepsilon _{xx}&\varepsilon _{xy}&\varepsilon _{xz}\\\varepsilon _{yx}&\varepsilon _{yy}&\varepsilon _{yz}\\\varepsilon _{zx}&\varepsilon _{zy}&\varepsilon _{zz}\\\end{bmatrix))={\begin{bmatrix}\varepsilon _{xx}&\gamma _{xy}/2&\gamma _{xz}/2\\\gamma _{yx}/2&\varepsilon _{yy}&\gamma _{yz}/2\\\gamma _{zx}/2&\gamma _{zy}/2&\varepsilon _{zz}\\\end{bmatrix))}$$

Physical interpretation

From finite strain theory we have

$${\displaystyle d\mathbf {x} ^{2}-d\mathbf {X} ^{2}=d\mathbf {X} \cdot 2\mathbf {E} \cdot d\mathbf {X} \quad {\text{or))\quad (dx)^{2}-(dX)^{2}=2E_{KL}\,dX_{K}\,dX_{L))$$

For infinitesimal strains then we have

$${\displaystyle d\mathbf {x} ^{2}-d\mathbf {X} ^{2}=d\mathbf {X} \cdot 2\mathbf {\boldsymbol {\varepsilon )) \cdot d\mathbf {X} \quad {\text{or))\quad (dx)^{2}-(dX)^{2}=2\varepsilon _{KL}\,dX_{K}\,dX_{L))$$

Dividing by ${\displaystyle (dX)^{2))$ we have

$${\displaystyle {\frac {dx-dX}{dX)){\frac {dx+dX}{dX))=2\varepsilon _{ij}{\frac {dX_{i)){dX)){\frac {dX_{j)){dX))}$$

For small deformations we assume that ${\displaystyle dx\approx dX}$, thus the second term of the left hand side becomes: ${\displaystyle {\frac {dx+dX}{dX))\approx 2}$.

Then we have

$${\displaystyle {\frac {dx-dX}{dX))=\varepsilon _{ij}N_{i}N_{j}=\mathbf {N} \cdot {\boldsymbol {\varepsilon ))\cdot \mathbf {N} }$$

${\displaystyle N_{i}={\frac {dX_{i)){dX))}$

${\displaystyle d\mathbf {X} }$

${\displaystyle e_{(\mathbf {N} )))$

${\displaystyle \mathbf {N} }$

${\displaystyle \mathbf {N} }$

${\displaystyle X_{1))$

${\displaystyle \mathbf {N} =\mathbf {I} _{1))$

$${\displaystyle e_{(\mathbf {I} _{1})}=\mathbf {I} _{1}\cdot {\boldsymbol {\varepsilon ))\cdot \mathbf {I} _{1}=\varepsilon _{11}.}$$

Similarly, for ${\displaystyle \mathbf {N} =\mathbf {I} _{2))$ and ${\displaystyle \mathbf {N} =\mathbf {I} _{3))$ we can find the normal strains ${\displaystyle \varepsilon _{22))$ and ${\displaystyle \varepsilon _{33))$, respectively. Therefore, the diagonal elements of the infinitesimal strain tensor are the normal strains in the coordinate directions.

Strain transformation rules

If we choose an orthonormal coordinate system (${\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3))$) we can write the tensor in terms of components with respect to those base vectors as

$${\displaystyle {\boldsymbol {\varepsilon ))=\sum _{i=1}^{3}\sum _{j=1}^{3}\varepsilon _{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j))$$

$${\displaystyle {\underline {\underline {\boldsymbol {\varepsilon ))))={\begin{bmatrix}\varepsilon _{11}&\varepsilon _{12}&\varepsilon _{13}\\\varepsilon _{12}&\varepsilon _{22}&\varepsilon _{23}\\\varepsilon _{13}&\varepsilon _{23}&\varepsilon _{33}\end{bmatrix))}$$

${\displaystyle {\hat {\mathbf {e} ))_{1},{\hat {\mathbf {e} ))_{2},{\hat {\mathbf {e} ))_{3))$

$${\displaystyle {\boldsymbol {\varepsilon ))=\sum _{i=1}^{3}\sum _{j=1}^{3}{\hat {\varepsilon ))_{ij}{\hat {\mathbf {e} ))_{i}\otimes {\hat {\mathbf {e} ))_{j}\quad \implies \quad {\underline {\underline {\hat {\boldsymbol {\varepsilon ))))}={\begin{bmatrix}{\hat {\varepsilon ))_{11}&{\hat {\varepsilon ))_{12}&{\hat {\varepsilon ))_{13}\\{\hat {\varepsilon ))_{12}&{\hat {\varepsilon ))_{22}&{\hat {\varepsilon ))_{23}\\{\hat {\varepsilon ))_{13}&{\hat {\varepsilon ))_{23}&{\hat {\varepsilon ))_{33}\end{bmatrix))}$$

$${\displaystyle {\hat {\varepsilon ))_{ij}=\ell _{ip}~\ell _{jq}~\varepsilon _{pq))$$

${\displaystyle \ell _{ij}={\hat {\mathbf {e} ))_{i}\cdot {\mathbf {e} }_{j))$

$${\displaystyle {\underline {\underline {\hat {\boldsymbol {\varepsilon ))))}={\underline {\underline {\mathbf {L} ))}~{\underline {\underline {\boldsymbol {\varepsilon ))))~{\underline {\underline {\mathbf {L} ))}^{T))$$

$${\displaystyle {\begin{bmatrix}{\hat {\varepsilon ))_{11}&{\hat {\varepsilon ))_{12}&{\hat {\varepsilon ))_{13}\\{\hat {\varepsilon ))_{21}&{\hat {\varepsilon ))_{22}&{\hat {\varepsilon ))_{23}\\{\hat {\varepsilon ))_{31}&{\hat {\varepsilon ))_{32}&{\hat {\varepsilon ))_{33}\end{bmatrix))={\begin{bmatrix}\ell _{11}&\ell _{12}&\ell _{13}\\\ell _{21}&\ell _{22}&\ell _{23}\\\ell _{31}&\ell _{32}&\ell _{33}\end{bmatrix)){\begin{bmatrix}\varepsilon _{11}&\varepsilon _{12}&\varepsilon _{13}\\\varepsilon _{21}&\varepsilon _{22}&\varepsilon _{23}\\\varepsilon _{31}&\varepsilon _{32}&\varepsilon _{33}\end{bmatrix)){\begin{bmatrix}\ell _{11}&\ell _{12}&\ell _{13}\\\ell _{21}&\ell _{22}&\ell _{23}\\\ell _{31}&\ell _{32}&\ell _{33}\end{bmatrix))^{T))$$

Strain invariants

Certain operations on the strain tensor give the same result without regard to which orthonormal coordinate system is used to represent the components of strain. The results of these operations are called strain invariants. The most commonly used strain invariants are

$${\displaystyle {\begin{aligned}I_{1}&=\mathrm {tr} ({\boldsymbol {\varepsilon )))\\I_{2}&={\tfrac {1}{2))\{[\mathrm {tr} ({\boldsymbol {\varepsilon )))]^{2}-\mathrm {tr} ({\boldsymbol {\varepsilon ))^{2})\}\\I_{3}&=\det({\boldsymbol {\varepsilon )))\end{aligned))}$$

$${\displaystyle {\begin{aligned}I_{1}&=\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}\\I_{2}&=\varepsilon _{11}\varepsilon _{22}+\varepsilon _{22}\varepsilon _{33}+\varepsilon _{33}\varepsilon _{11}-\varepsilon _{12}^{2}-\varepsilon _{23}^{2}-\varepsilon _{31}^{2}\\I_{3}&=\varepsilon _{11}(\varepsilon _{22}\varepsilon _{33}-\varepsilon _{23}^{2})-\varepsilon _{12}(\varepsilon _{21}\varepsilon _{33}-\varepsilon _{23}\varepsilon _{31})+\varepsilon _{13}(\varepsilon _{21}\varepsilon _{32}-\varepsilon _{22}\varepsilon _{31})\end{aligned))}$$

Principal strains

It can be shown that it is possible to find a coordinate system (${\displaystyle \mathbf {n} _{1},\mathbf {n} _{2},\mathbf {n} _{3))$) in which the components of the strain tensor are

$${\displaystyle {\underline {\underline {\boldsymbol {\varepsilon ))))={\begin{bmatrix}\varepsilon _{1}&0&0\\0&\varepsilon _{2}&0\\0&0&\varepsilon _{3}\end{bmatrix))\quad \implies \quad {\boldsymbol {\varepsilon ))=\varepsilon _{1}\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\varepsilon _{2}\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\varepsilon _{3}\mathbf {n} _{3}\otimes \mathbf {n} _{3))$$

${\displaystyle \mathbf {n} _{1},\mathbf {n} _{2},\mathbf {n} _{3))$

${\displaystyle \mathbf {n} _{i))$

If we are given the components of the strain tensor in an arbitrary orthonormal coordinate system, we can find the principal strains using an eigenvalue decomposition determined by solving the system of equations

$${\displaystyle ({\underline {\underline {\boldsymbol {\varepsilon ))))-\varepsilon _{i}~{\underline {\underline {\mathbf {I} ))})~\mathbf {n} _{i}={\underline {\mathbf {0} ))}$$

${\displaystyle \mathbf {n} _{i))$

Volumetric strain

The volumetric strain, also called bulk strain, is the relative variation of the volume, as arising from dilation or compression; it is the first strain invariant or trace of the tensor:

$${\displaystyle \delta ={\frac {\Delta V}{V_{0))}=I_{1}=\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33))$$

${\displaystyle a\cdot (1+\varepsilon _{11})\times a\cdot (1+\varepsilon _{22})\times a\cdot (1+\varepsilon _{33})}$

$${\displaystyle {\frac {\Delta V}{V_{0))}={\frac {\left(1+\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}+\varepsilon _{11}\cdot \varepsilon _{22}+\varepsilon _{11}\cdot \varepsilon _{33}+\varepsilon _{22}\cdot \varepsilon _{33}+\varepsilon _{11}\cdot \varepsilon _{22}\cdot \varepsilon _{33}\right)\cdot a^{3}-a^{3)){a^{3))))$$

$${\displaystyle 1\gg \varepsilon _{ii}\gg \varepsilon _{ii}\cdot \varepsilon _{jj}\gg \varepsilon _{11}\cdot \varepsilon _{22}\cdot \varepsilon _{33))$$

In case of pure shear, we can see that there is no change of the volume.

Strain deviator tensor

The infinitesimal strain tensor ${\displaystyle \varepsilon _{ij))$, similarly to the Cauchy stress tensor, can be expressed as the sum of two other tensors:

1. a mean strain tensor or volumetric strain tensor or spherical strain tensor, ${\displaystyle \varepsilon _{M}\delta _{ij))$, related to dilation or volume change; and
2. a deviatoric component called the strain deviator tensor, ${\displaystyle \varepsilon '_{ij))$, related to distortion.

$${\displaystyle \varepsilon _{ij}=\varepsilon '_{ij}+\varepsilon _{M}\delta _{ij))$$

${\displaystyle \varepsilon _{M))$

$${\displaystyle \varepsilon _{M}={\frac {\varepsilon _{kk)){3))={\frac {\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33)){3))={\tfrac {1}{3))I_{1}^{e))$$

The deviatoric strain tensor can be obtained by subtracting the mean strain tensor from the infinitesimal strain tensor:

$${\displaystyle {\begin{aligned}\ \varepsilon '_{ij}&=\varepsilon _{ij}-{\frac {\varepsilon _{kk)){3))\delta _{ij}\\{\begin{bmatrix}\varepsilon '_{11}&\varepsilon '_{12}&\varepsilon '_{13}\\\varepsilon '_{21}&\varepsilon '_{22}&\varepsilon '_{23}\\\varepsilon '_{31}&\varepsilon '_{32}&\varepsilon '_{33}\\\end{bmatrix))&={\begin{bmatrix}\varepsilon _{11}&\varepsilon _{12}&\varepsilon _{13}\\\varepsilon _{21}&\varepsilon _{22}&\varepsilon _{23}\\\varepsilon _{31}&\varepsilon _{32}&\varepsilon _{33}\\\end{bmatrix))-{\begin{bmatrix}\varepsilon _{M}&0&0\\0&\varepsilon _{M}&0\\0&0&\varepsilon _{M}\\\end{bmatrix))\\&={\begin{bmatrix}\varepsilon _{11}-\varepsilon _{M}&\varepsilon _{12}&\varepsilon _{13}\\\varepsilon _{21}&\varepsilon _{22}-\varepsilon _{M}&\varepsilon _{23}\\\varepsilon _{31}&\varepsilon _{32}&\varepsilon _{33}-\varepsilon _{M}\\\end{bmatrix))\\\end{aligned))}$$

Octahedral strains

Let (${\displaystyle \mathbf {n} _{1},\mathbf {n} _{2},\mathbf {n} _{3))$) be the directions of the three principal strains. An octahedral plane is one whose normal makes equal angles with the three principal directions. The engineering shear strain on an octahedral plane is called the octahedral shear strain and is given by

$${\displaystyle \gamma _{\mathrm {oct} }={\tfrac {2}{3)){\sqrt {(\varepsilon _{1}-\varepsilon _{2})^{2}+(\varepsilon _{2}-\varepsilon _{3})^{2}+(\varepsilon _{3}-\varepsilon _{1})^{2))))$$

${\displaystyle \varepsilon _{1},\varepsilon _{2},\varepsilon _{3))$

The normal strain on an octahedral plane is given by

$${\displaystyle \varepsilon _{\mathrm {oct} }={\tfrac {1}{3))(\varepsilon _{1}+\varepsilon _{2}+\varepsilon _{3})}$$

Equivalent strain

A scalar quantity called the equivalent strain, or the von Mises equivalent strain, is often used to describe the state of strain in solids. Several definitions of equivalent strain can be found in the literature. A definition that is commonly used in the literature on plasticity is

$${\displaystyle \varepsilon _{\mathrm {eq} }={\sqrt ((\tfrac {2}{3)){\boldsymbol {\varepsilon ))^{\mathrm {dev} }:{\boldsymbol {\varepsilon ))^{\mathrm {dev} ))}={\sqrt ((\tfrac {2}{3))\varepsilon _{ij}^{\mathrm {dev} }\varepsilon _{ij}^{\mathrm {dev} ))}~;~~{\boldsymbol {\varepsilon ))^{\mathrm {dev} }={\boldsymbol {\varepsilon ))-{\tfrac {1}{3))\mathrm {tr} ({\boldsymbol {\varepsilon )))~{\boldsymbol {I))}$$

$${\displaystyle \sigma _{\mathrm {eq} }={\sqrt ((\tfrac {3}{2)){\boldsymbol {\sigma ))^{\mathrm {dev} }:{\boldsymbol {\sigma ))^{\mathrm {dev} ))))$$

Compatibility equations

For prescribed strain components ${\displaystyle \varepsilon _{ij))$ the strain tensor equation ${\displaystyle u_{i,j}+u_{j,i}=2\varepsilon _{ij))$ represents a system of six differential equations for the determination of three displacements components ${\displaystyle u_{i))$, giving an over-determined system. Thus, a solution does not generally exist for an arbitrary choice of strain components. Therefore, some restrictions, named compatibility equations, are imposed upon the strain components. With the addition of the three compatibility equations the number of independent equations are reduced to three, matching the number of unknown displacement components. These constraints on the strain tensor were discovered by Saint-Venant, and are called the "Saint Venant compatibility equations".

The compatibility functions serve to assure a single-valued continuous displacement function ${\displaystyle u_{i))$. If the elastic medium is visualised as a set of infinitesimal cubes in the unstrained state, after the medium is strained, an arbitrary strain tensor may not yield a situation in which the distorted cubes still fit together without overlapping.

In index notation, the compatibility equations are expressed as

$${\displaystyle \varepsilon _{ij,km}+\varepsilon _{km,ij}-\varepsilon _{ik,jm}-\varepsilon _{jm,ik}=0}$$

In engineering notation,

• ${\displaystyle {\frac {\partial ^{2}\epsilon _{x)){\partial y^{2))}+{\frac {\partial ^{2}\epsilon _{y)){\partial x^{2))}=2{\frac {\partial ^{2}\epsilon _{xy)){\partial x\partial y))}$
• ${\displaystyle {\frac {\partial ^{2}\epsilon _{y)){\partial z^{2))}+{\frac {\partial ^{2}\epsilon _{z)){\partial y^{2))}=2{\frac {\partial ^{2}\epsilon _{yz)){\partial y\partial z))}$
• ${\displaystyle {\frac {\partial ^{2}\epsilon _{x)){\partial z^{2))}+{\frac {\partial ^{2}\epsilon _{z)){\partial x^{2))}=2{\frac {\partial ^{2}\epsilon _{zx)){\partial z\partial x))}$
• ${\displaystyle {\frac {\partial ^{2}\epsilon _{x)){\partial y\partial z))={\frac {\partial }{\partial x))\left(-{\frac {\partial \epsilon _{yz)){\partial x))+{\frac {\partial \epsilon _{zx)){\partial y))+{\frac {\partial \epsilon _{xy)){\partial z))\right)}$
• ${\displaystyle {\frac {\partial ^{2}\epsilon _{y)){\partial z\partial x))={\frac {\partial }{\partial y))\left({\frac {\partial \epsilon _{yz)){\partial x))-{\frac {\partial \epsilon _{zx)){\partial y))+{\frac {\partial \epsilon _{xy)){\partial z))\right)}$
• ${\displaystyle {\frac {\partial ^{2}\epsilon _{z)){\partial x\partial y))={\frac {\partial }{\partial z))\left({\frac {\partial \epsilon _{yz)){\partial x))+{\frac {\partial \epsilon _{zx)){\partial y))-{\frac {\partial \epsilon _{xy)){\partial z))\right)}$

Special cases

Plane strain

In real engineering components, stress (and strain) are 3-D tensors but in prismatic structures such as a long metal billet, the length of the structure is much greater than the other two dimensions. The strains associated with length, i.e., the normal strain ${\displaystyle \varepsilon _{33))$ and the shear strains ${\displaystyle \varepsilon _{13))$ and ${\displaystyle \varepsilon _{23))$ (if the length is the 3-direction) are constrained by nearby material and are small compared to the cross-sectional strains. Plane strain is then an acceptable approximation. The strain tensor for plane strain is written as:

$${\displaystyle {\underline {\underline {\boldsymbol {\varepsilon ))))={\begin{bmatrix}\varepsilon _{11}&\varepsilon _{12}&0\\\varepsilon _{21}&\varepsilon _{22}&0\\0&0&0\end{bmatrix))}$$

$${\displaystyle {\underline {\underline {\boldsymbol {\sigma ))))={\begin{bmatrix}\sigma _{11}&\sigma _{12}&0\\\sigma _{21}&\sigma _{22}&0\\0&0&\sigma _{33}\end{bmatrix))}$$

${\displaystyle \sigma _{33))$

${\displaystyle \epsilon _{33}=0}$

Antiplane strain

Antiplane strain is another special state of strain that can occur in a body, for instance in a region close to a screw dislocation. The strain tensor for antiplane strain is given by

$${\displaystyle {\underline {\underline {\boldsymbol {\varepsilon ))))={\begin{bmatrix}0&0&\varepsilon _{13}\\0&0&\varepsilon _{23}\\\varepsilon _{13}&\varepsilon _{23}&0\end{bmatrix))}$$

Relation to infinitesimal rotation tensor

The infinitesimal strain tensor is defined as

$${\displaystyle {\boldsymbol {\varepsilon ))={\frac {1}{2))[{\boldsymbol {\nabla ))\mathbf {u} +({\boldsymbol {\nabla ))\mathbf {u} )^{T}]}$$

$${\displaystyle {\boldsymbol {\nabla ))\mathbf {u} ={\boldsymbol {\varepsilon ))+{\boldsymbol {W))}$$

$${\displaystyle {\boldsymbol {W)):={\frac {1}{2))[{\boldsymbol {\nabla ))\mathbf {u} -({\boldsymbol {\nabla ))\mathbf {u} )^{T}]}$$

${\displaystyle {\boldsymbol {W))}$

${\displaystyle {\boldsymbol {W))}$

${\displaystyle |W_{ij}|\ll 1}$

The axial vector

A skew symmetric second-order tensor has three independent scalar components. These three components are used to define an axial vector, ${\displaystyle \mathbf {w} }$, as follows

$${\displaystyle W_{ij}=-\epsilon _{ijk}~w_{k}~;~~w_{i}=-{\tfrac {1}{2))~\epsilon _{ijk}~W_{jk))$$

${\displaystyle \epsilon _{ijk))$

$${\displaystyle {\underline {\underline {\boldsymbol {W))))={\begin{bmatrix}0&-w_{3}&w_{2}\\w_{3}&0&-w_{1}\\-w_{2}&w_{1}&0\end{bmatrix))~;~~{\underline {\mathbf {w} ))={\begin{bmatrix}w_{1}\\w_{2}\\w_{3}\end{bmatrix))}$$

$${\displaystyle \mathbf {w} ={\tfrac {1}{2))~{\boldsymbol {\nabla ))\times \mathbf {u} }$$

$${\displaystyle w_{i}={\tfrac {1}{2))~\epsilon _{ijk}~u_{k,j))$$

${\displaystyle \lVert {\boldsymbol {W))\rVert \ll 1}$

${\displaystyle {\boldsymbol {\varepsilon ))={\boldsymbol {0))}$

${\displaystyle |\mathbf {w} |}$

${\displaystyle \mathbf {w} }$

Relation between the strain tensor and the rotation vector

Given a continuous, single-valued displacement field ${\displaystyle \mathbf {u} }$ and the corresponding infinitesimal strain tensor ${\displaystyle {\boldsymbol {\varepsilon ))}$, we have (see Tensor derivative (continuum mechanics))

$${\displaystyle {\boldsymbol {\nabla ))\times {\boldsymbol {\varepsilon ))=e_{ijk}~\varepsilon _{lj,i}~\mathbf {e} _{k}\otimes \mathbf {e} _{l}={\tfrac {1}{2))~e_{ijk}~[u_{l,ji}+u_{j,li}]~\mathbf {e} _{k}\otimes \mathbf {e} _{l))$$

${\displaystyle u_{l,ji}=u_{l,ij))$

$${\displaystyle e_{ijk}u_{l,ji}=(e_{12k}+e_{21k})u_{l,12}+(e_{13k}+e_{31k})u_{l,13}+(e_{23k}+e_{32k})u_{l,32}=0}$$

$${\displaystyle {\tfrac {1}{2))~e_{ijk}~u_{j,li}=\left({\tfrac {1}{2))~e_{ijk}~u_{j,i}\right)_{,l}=\left({\tfrac {1}{2))~e_{kij}~u_{j,i}\right)_{,l}=w_{k,l))$$

$${\displaystyle {\boldsymbol {\nabla ))\times {\boldsymbol {\varepsilon ))=w_{k,l}~\mathbf {e} _{k}\otimes \mathbf {e} _{l}={\boldsymbol {\nabla ))\mathbf {w} }$$

Relation between rotation tensor and rotation vector

From an important identity regarding the curl of a tensor we know that for a continuous, single-valued displacement field ${\displaystyle \mathbf {u} }$,

$${\displaystyle {\boldsymbol {\nabla ))\times ({\boldsymbol {\nabla ))\mathbf {u} )={\boldsymbol {0)).}$$

${\displaystyle {\boldsymbol {\nabla ))\mathbf {u} ={\boldsymbol {\varepsilon ))+{\boldsymbol {W))}$

$${\displaystyle {\boldsymbol {\nabla ))\times {\boldsymbol {W))=-{\boldsymbol {\nabla ))\times {\boldsymbol {\varepsilon ))=-{\boldsymbol {\nabla ))\mathbf {w} .}$$

Strain tensor in non-Cartesian coordinates

Strain tensor in cylindrical coordinates

In cylindrical polar coordinates (${\displaystyle r,\theta ,z}$), the displacement vector can be written as

$${\displaystyle \mathbf {u} =u_{r}~\mathbf {e} _{r}+u_{\theta }~\mathbf {e} _{\theta }+u_{z}~\mathbf {e} _{z))$$

$${\displaystyle {\begin{aligned}\varepsilon _{rr}&={\cfrac {\partial u_{r)){\partial r))\\\varepsilon _{\theta \theta }&={\cfrac {1}{r))\left({\cfrac {\partial u_{\theta )){\partial \theta ))+u_{r}\right)\\\varepsilon _{zz}&={\cfrac {\partial u_{z)){\partial z))\\\varepsilon _{r\theta }&={\cfrac {1}{2))\left({\cfrac {1}{r)){\cfrac {\partial u_{r)){\partial \theta ))+{\cfrac {\partial u_{\theta )){\partial r))-{\cfrac {u_{\theta )){r))\right)\\\varepsilon _{\theta z}&={\cfrac {1}{2))\left({\cfrac {\partial u_{\theta )){\partial z))+{\cfrac {1}{r)){\cfrac {\partial u_{z)){\partial \theta ))\right)\\\varepsilon _{zr}&={\cfrac {1}{2))\left({\cfrac {\partial u_{r)){\partial z))+{\cfrac {\partial u_{z)){\partial r))\right)\end{aligned))}$$

Strain tensor in spherical coordinates

In spherical coordinates (${\displaystyle r,\theta ,\phi }$), the displacement vector can be written as

$${\displaystyle \mathbf {u} =u_{r}~\mathbf {e} _{r}+u_{\theta }~\mathbf {e} _{\theta }+u_{\phi }~\mathbf {e} _{\phi ))$$