finite strain theory

In continuum mechanics, the finite strain theory—also called large strain theory, or large deformation theory—deals with deformations in which strains and/or rotations are large enough to invalidate assumptions inherent in infinitesimal strain theory. In this case, the undeformed and deformed configurations of the continuum are significantly different, requiring a clear distinction between them. This is commonly the case with elastomers, plastically-deforming materials and other fluids and biological soft tissue.

Displacement

The displacement of a body has two components: a rigid-body displacement and a deformation.
* A rigid-body displacement consists of a simultaneous translation (physics) and rotation of the body without changing its shape or size.
* Deformation implies the change in shape and/or size of the body from an initial or undeformed configuration $\kappa_0(\mathcal B)$ to a current or deformed configuration $\kappa_t(\mathcal B)$ (Figure 1).

A change in the configuration of a continuum body can be described by a displacement field. A ''displacement field'' is a vector field of all displacement vectors for all particles in the body, which relates the deformed configuration with the undeformed configuration. The distance between any two particles changes if and only if deformation has occurred. If displacement occurs without deformation, then it is a rigid-body displacement.

Material coordinates (Lagrangian description)

The displacement of particles indexed by variable may be expressed as follows. The vector joining the positions of a particle in the undeformed configuration $P_i$ and deformed configuration $p_i$ is called the displacement vector. Using $\mathbf$ in place of $P_i$ and $\mathbf$ in place of $p_i\,\!$, both of which are vectors from the origin of the coordinate system to each respective point, we have the Lagrangian description of the displacement vector:
$\mathbf u(\mathbf X,t) = u_i \mathbf e_i$
where $\mathbf e_i$ are the orthonormal unit vectors that define the basis of the spatial (lab-frame) coordinate system.

Expressed in terms of the material coordinates, i.e. $\mathbf u$ as a function of $\mathbf X$, the displacement field is:
$\mathbf u(\mathbf X, t) = \mathbf b(t)+\mathbf x(\mathbf X,t) - \mathbf X \qquad \text\qquad u_i = \alpha_ b_J + x_i - \alpha_ X_J$
where $\mathbf b(t)$ is the displacement vector representing rigid-body translation.

The partial derivative of the displacement vector with respect to the material coordinates yields the ''material displacement gradient tensor'' $\nabla_ \mathbf u\,\!$. Thus we have,
$\nabla_\mathbf u = \nabla_\mathbf x - \mathbf R = \mathbf F - \mathbf R \qquad \text \qquad \frac = \frac - \alpha_ = F_ - \alpha_$
where $\mathbf F$ is the ''deformation gradient tensor''.

Spatial coordinates (Eulerian description)

In the Eulerian description, the vector extending from a particle $P$ in the undeformed configuration to its location in the deformed configuration is called the displacement vector:
$\mathbf U(\mathbf x,t) = U_J\mathbf E_J$
where $\mathbf E_i$ are the unit vectors that define the basis of the material (body-frame) coordinate system.

Expressed in terms of spatial coordinates, i.e. $\mathbf U$ as a function of $\mathbf x$, the displacement field is:
$\mathbf U(\mathbf x, t) = \mathbf b(t) + \mathbf x - \mathbf X(\mathbf x,t) \qquad \text\qquad U_J = b_J + \alpha_ x_i - X_J$

The partial derivative of the displacement vector with respect to the spatial coordinates yields the ''spatial displacement gradient tensor'' $\nabla_ \mathbf U\,\!$. Thus we have,
$\nabla_\mathbf U = \mathbf R^ - \nabla_\mathbf X = \mathbf R^ -\mathbf F^ \qquad \text \qquad \frac = \alpha_ - \frac = \alpha_ - F^_ \,.$

Relationship between the material and spatial coordinate systems

$\alpha_$ are the direction cosines between the material and spatial coordinate systems with unit vectors $\mathbf E_J$ and $\mathbf e_i\,\!$, respectively. Thus
$\mathbf E_J \cdot \mathbf e_i = \alpha_ = \alpha_$

The relationship between $u_i$ and $U_J$ is then given by
$u_i=\alpha_U_J \qquad \text \qquad U_J=\alpha_ u_i$

Knowing that
$\mathbf e_i = \alpha_ \mathbf E_J$
then
$\mathbf u(\mathbf X, t) = u_i\mathbf e_i = u_i(\alpha_\mathbf E_J) = U_J \mathbf E_J = \mathbf U(\mathbf x, t)$

Combining the coordinate systems of deformed and undeformed configurations

It is common to superimpose the coordinate systems for the deformed and undeformed configurations, which results in $\mathbf b = 0\,\!$, and the direction cosines become Kronecker deltas, i.e.,
$\mathbf E_J \cdot \mathbf e_i = \delta_ = \delta_$

Thus in material (undeformed) coordinates, the displacement may be expressed as:
$\mathbf u(\mathbf X, t) = \mathbf x(\mathbf X,t) - \mathbf X \qquad \text\qquad u_i = x_i - \delta_ X_J$

And in spatial (deformed) coordinates, the displacement may be expressed as:
$\mathbf U(\mathbf x, t) = \mathbf x - \mathbf X(\mathbf x,t) \qquad \text\qquad U_J = \delta_ x_i - X_J$

Deformation gradient tensor

The deformation gradient tensor $\mathbf F(\mathbf X,t) = F_ \mathbf e_j \otimes \mathbf I_K$ is related to both the reference and current configuration, as seen by the unit vectors $\mathbf e_j$ and $\mathbf I_K\,\!$, therefore it is a '' two-point tensor''.

Due to the assumption of continuity of $\chi(\mathbf X,t)\,\!$, $\mathbf F$ has the inverse $\mathbf H = \mathbf F^\,\!$, where $\mathbf H$ is the ''spatial deformation gradient tensor''. Then, by the implicit function theorem, the Jacobian determinant $J(\mathbf X,t)$ must be nonsingular, i.e. $J(\mathbf X,t) = \det \mathbf F(\mathbf X,t) \neq 0$

The ''material deformation gradient tensor'' $\mathbf F(\mathbf X,t) = F_ \mathbf e_j\otimes\mathbf I_K$ is a second-order tensor that represents the gradient of the mapping function or functional relation $\chi(\mathbf X,t)\,\!$, which describes the motion of a continuum. The material deformation gradient tensor characterizes the local deformation at a material point with position vector $\mathbf X\,\!$, i.e., deformation at neighbouring points, by transforming (linear transformation) a material line element emanating from that point from the reference configuration to the current or deformed configuration, assuming continuity in the mapping function $\chi(\mathbf X,t)\,\!$, i.e. differentiable function of $\mathbf$ and time $t\,\!$, which implies that cracks and voids do not open or close during the deformation. Thus we have,
$\begin d\mathbf &= \frac \,d\mathbf \qquad &\text& \qquad dx_j =\frac\,dX_K \\ &= \nabla \chi(\mathbf X,t) \,d\mathbf \qquad &\text& \qquad dx_j =F_\,dX_K \,. \\ & = \mathbf F(\mathbf X,t) \,d\mathbf \end$

Relative displacement vector

Consider a particle or material point $P$ with position vector $\mathbf X = X_I \mathbf I_I$ in the undeformed configuration (Figure 2). After a displacement of the body, the new position of the particle indicated by $p$ in the new configuration is given by the vector position $\mathbf = x_i \mathbf e_i\,\!$. The coordinate systems for the undeformed and deformed configuration can be superimposed for convenience.

Consider now a material point $Q$ neighboring $P\,\!$, with position vector $\mathbf+ \Delta \mathbf = (X_I+\Delta X_I) \mathbf I_I\,\!$. In the deformed configuration this particle has a new position $q$ given by the position vector $\mathbf+ \Delta \mathbf\,\!$. Assuming that the line segments $\Delta X$ and $\Delta \mathbf x$ joining the particles $P$ and $Q$ in both the undeformed and deformed configuration, respectively, to be very small, then we can express them as $d\mathbf X$ and $d\mathbf x\,\!$. Thus from Figure 2 we have
$\begin \mathbf+ d\mathbf&= \mathbf+d\mathbf+\mathbf(\mathbf+d\mathbf) \\ d\mathbf &= \mathbf-\mathbf+d\mathbf+ \mathbf(\mathbf+d\mathbf) \\ &= d\mathbf+\mathbf(\mathbf+d\mathbf)- \mathbf(\mathbf) \\ &= d\mathbf+d\mathbf \\ \end$

where $\mathbf$ is the relative displacement vector, which represents the relative displacement of $Q$ with respect to $P$ in the deformed configuration.

Taylor approximation

For an infinitesimal element $d\mathbf X\,\!$, and assuming continuity on the displacement field, it is possible to use a Taylor series expansion around point $P\,\!$, neglecting higher-order terms, to approximate the components of the relative displacement vector for the neighboring particle $Q$ as
$\begin \mathbf(\mathbf+d\mathbf)&=\mathbf(\mathbf)+d\mathbf \quad & \text & \quad u_i^* = u_i+du_i \\ &\approx \mathbf(\mathbf)+\nabla_\mathbf u\cdot d\mathbf X \quad & \text & \quad u_i^* \approx u_i + \fracdX_J \,. \end$
Thus, the previous equation $d\mathbf x = d\mathbf + d\mathbf$ can be written as
$\begin d\mathbf x&=d\mathbf X+d\mathbf u \\ &=d\mathbf X+\nabla_\mathbf u\cdot d\mathbf X\\ &=\left(\mathbf I + \nabla_\mathbf u\right)d\mathbf X\\ &=\mathbf F d\mathbf X \end$

Time-derivative of the deformation gradient

Calculations that involve the time-dependent deformation of a body often require a time derivative of the deformation gradient to be calculated. A geometrically consistent definition of such a derivative requires an excursion into differential geometryA. Yavari, J.E. Marsden, and M. Ortiz
On spatial and material covariant balance laws in elasticity
Journal of Mathematical Physics, 47, 2006, 042903; pp. 1–53. but we avoid those issues in this article.

The time derivative of $\mathbf$ is
\dot = \frac = \frac \left frac\right= \frac\left frac\right= \frac\left mathbf(\mathbf, t)\right
where $\mathbf$ is the (material) velocity. The derivative on the right hand side represents a material velocity gradient. It is common to convert that into a spatial gradient by applying the chain rule for derivatives, i.e.,

\dot =
\frac\left mathbf(\mathbf, t)\right=
\frac\left mathbf(\mathbf(\mathbf, t),t)\right=
\left.\frac\left mathbf(\mathbf,t)\right_ \cdot
\frac
= \boldsymbol\cdot\mathbf

where $\boldsymbol$ is the spatial velocity gradient and where $\mathbf(\mathbf,t) = \mathbf(\mathbf,t)$ is the spatial (Eulerian) velocity at $\mathbf = \mathbf(\mathbf, t)$. If the spatial velocity gradient is constant in time, the above equation can be solved exactly to give
$\mathbf = e^$
assuming $\mathbf = \mathbf$ at $t = 0$. There are several methods of computing the exponential above.

Related quantities often used in continuum mechanics are the rate of deformation tensor and the spin tensor defined, respectively, as:
$\boldsymbol = \tfrac \left(\boldsymbol + \boldsymbol^T\right) \,,~~ \boldsymbol = \tfrac \left(\boldsymbol - \boldsymbol^T\right) \,.$
The rate of deformation tensor gives the rate of stretching of line elements while the spin tensor indicates the rate of rotation or vorticity of the motion.

The material time derivative of the inverse of the deformation gradient (keeping the reference configuration fixed) is often required in analyses that involve finite strains. This derivative is
$\frac \left(\mathbf^\right) = - \mathbf^ \cdot \dot \cdot \mathbf^ \,.$
The above relation can be verified by taking the material time derivative of $\mathbf^ \cdot d\mathbf = d\mathbf$ and noting that $\dot = 0$.

Transformation of a surface and volume element

To transform quantities that are defined with respect to areas in a deformed configuration to those relative to areas in a reference configuration, and vice versa, we use Nanson's relation, expressed as $da~\mathbf = J~dA ~\mathbf^ \cdot \mathbf$ where $da$ is an area of a region in the deformed configuration, $dA$ is the same area in the reference configuration, and $\mathbf$ is the outward normal to the area element in the current configuration while $\mathbf$ is the outward normal in the reference configuration, $\mathbf$ is the deformation gradient, and $J = \det\mathbf\,\!$.

The corresponding formula for the transformation of the volume element is $dv = J~dV$

Polar decomposition of the deformation gradient tensor

The deformation gradient $\mathbf\,\!$, like any invertible second-order tensor, can be decomposed, using the polar decomposition theorem, into a product of two second-order tensors (Truesdell and Noll, 1965): an orthogonal tensor and a positive definite symmetric tensor, i.e., $\mathbf = \mathbf \mathbf = \mathbf \mathbf$ where the tensor $\mathbf$ is a proper orthogonal tensor, i.e., $\mathbf R^ = \mathbf R^T$ and $\det \mathbf R = +1\,\!$, representing a rotation; the tensor $\mathbf$ is the ''right stretch tensor''; and $\mathbf$ the ''left stretch tensor''. The terms ''right'' and ''left'' means that they are to the right and left of the rotation tensor $\mathbf\,\!$, respectively. $\mathbf$ and $\mathbf$ are both positive definite, i.e. $\mathbf x \cdot \mathbf U \cdot \mathbf x > 0$ and $\mathbf x\cdot\mathbf V \cdot \mathbf x > 0$ for all non-zero $\mathbf x \in \R^3$, and symmetric tensors, i.e. $\mathbf U = \mathbf U^T$ and $\mathbf V = \mathbf V^T\,\!$, of second order.

This decomposition implies that the deformation of a line element $d\mathbf X$ in the undeformed configuration onto $d\mathbf x$ in the deformed configuration, i.e., $d\mathbf x = \mathbf F \,d\mathbf X\,\!$, may be obtained either by first stretching the element by $\mathbf U\,\!$, i.e. $d\mathbf x' = \mathbf U \,d\mathbf X\,\!$, followed by a rotation $\mathbf R\,\!$, i.e., $d\mathbf x' = \mathbf R \,d\mathbf x\,\!$; or equivalently, by applying a rigid rotation $\mathbf R$ first, i.e., $d\mathbf x' = \mathbf R \, d\mathbf X\,\!$, followed later by a stretching $\mathbf V\,\!$, i.e., $d\mathbf x' = \mathbf V \, d\mathbf x$ (See Figure 3).

Due to the orthogonality of $\mathbf R$
$\mathbf V = \mathbf R \cdot \mathbf U \cdot \mathbf R^T$
so that $\mathbf U$ and $\mathbf V$ have the same eigenvalues or ''principal stretches'', but different eigenvectors or ''principal directions'' $\mathbf_i$ and $\mathbf_i\,\!$, respectively. The principal directions are related by
$\mathbf_i = \mathbf \mathbf_i.$

This polar decomposition, which is unique as $\mathbf F$ is invertible with a positive determinant, is a corrolary of the singular-value decomposition.

Deformation tensors

Several rotation-independent deformation tensors are used in mechanics. In solid mechanics, the most popular of these are the right and left Cauchy–Green deformation tensors.

Since a pure rotation should not induce any strains in a deformable body, it is often convenient to use rotation-independent measures of deformation in continuum mechanics. As a rotation followed by its inverse rotation leads to no change ($\mathbf\mathbf^T=\mathbf^T\mathbf=\mathbf\,\!$) we can exclude the rotation by multiplying $\mathbf$ by its transpose.

The right Cauchy–Green deformation tensor

In 1839, George Green introduced a deformation tensor known as the ''right Cauchy–Green deformation tensor'' or ''Green's deformation tensor'', defined as:The IUPAC recommends that this tensor be called the Cauchy strain tensor.

$\mathbf C=\mathbf F^T\mathbf F=\mathbf U^2 \qquad \text \qquad C_=F_~F_ = \frac \frac .$

Physically, the Cauchy–Green tensor gives us the square of local change in distances due to deformation, i.e. $d\mathbf x^2=d\mathbf X\cdot\mathbf C \cdot d\mathbf X$

Invariants of $\mathbf$ are often used in the expressions for strain energy density functions. The most commonly used invariants are

\begin
I_1^C & := \text(\mathbf) = C_ = \lambda_1^2 + \lambda_2^2 + \lambda_3^2 \\
I_2^C & := \tfrac\left \text~\mathbf)^2 - \text(\mathbf^2) \right = \tfrac\left C_)^2 - C_C_\right= \lambda_1^2\lambda_2^2 + \lambda_2^2\lambda_3^2 + \lambda_3^2\lambda_1^2 \\
I_3^C & := \det(\mathbf) = J^2 = \lambda_1^2\lambda_2^2\lambda_3^2.
\end

where $J:=\det\mathbf$ is the determinant of the deformation gradient $\mathbf$ and $\lambda_i$ are stretch ratios for the unit fibers that are initially oriented along the eigenvector directions of the right (reference) stretch tensor (these are not generally aligned with the three axis of the coordinate systems).

The Finger deformation tensor

The IUPAC recommends that the inverse of the right Cauchy–Green deformation tensor (called the Cauchy tensor in that document), i. e., $\mathbf C^$, be called the Finger tensor. However, that nomenclature is not universally accepted in applied mechanics.

$\mathbf=\mathbf C^=\mathbf F^\mathbf F^ \qquad \text \qquad f_=\frac \frac$

The left Cauchy–Green or Finger deformation tensor

Reversing the order of multiplication in the formula for the right Green–Cauchy deformation tensor leads to the ''left Cauchy–Green deformation tensor'' which is defined as:
$\mathbf B = \mathbf F\mathbf F^T = \mathbf V^2 \qquad \text \qquad B_ = \frac \frac$

The left Cauchy–Green deformation tensor is often called the ''Finger deformation tensor'', named after Josef Finger (1894).The IUPAC recommends that this tensor be called the Green strain tensor.

Invariants of $\mathbf$ are also used in the expressions for strain energy density functions. The conventional invariants are defined as

\begin
I_1 & := \text(\mathbf) = B_ = \lambda_1^2 + \lambda_2^2 + \lambda_3^2\\
I_2 & := \tfrac\left \text~\mathbf)^2 - \text(\mathbf^2)\right = \tfrac\left(B_^2 - B_B_\right) = \lambda_1^2\lambda_2^2 + \lambda_2^2\lambda_3^2 + \lambda_3^2\lambda_1^2 \\
I_3 & := \det\mathbf = J^2 = \lambda_1^2\lambda_2^2\lambda_3^2
\end

where $J:=\det\mathbf$ is the determinant of the deformation gradient.

For compressible materials, a slightly different set of invariants is used:
$(\bar_1 := J^ I_1 ~;~~ \bar_2 := J^ I_2 ~;~~ J\neq 1) ~.$

The Cauchy deformation tensor

Earlier in 1828, Augustin-Louis Cauchy introduced a deformation tensor defined as the inverse of the left Cauchy–Green deformation tensor, $\mathbf B^\,\!$. This tensor has also been called the Piola tensor and the Finger tensor in the rheology and fluid dynamics literature.

$\mathbf=\mathbf B^=\mathbf F^\mathbf F^ \qquad \text \qquad c_=\frac \frac$

Spectral representation

If there are three distinct principal stretches $\lambda_i \,\!$, the spectral decompositions of $\mathbf$ and $\mathbf$ is given by

$\mathbf = \sum_^3 \lambda_i^2 \mathbf_i \otimes \mathbf_i \qquad \text \qquad \mathbf = \sum_^3 \lambda_i^2 \mathbf_i \otimes \mathbf_i$

Furthermore,

$\mathbf U = \sum_^3 \lambda_i \mathbf N_i \otimes \mathbf N_i ~;~~ \mathbf V = \sum_^3 \lambda_i \mathbf n_i \otimes \mathbf n_i$
$\mathbf R = \sum_^3 \mathbf n_i \otimes \mathbf N_i ~;~~ \mathbf F = \sum_^3 \lambda_i \mathbf n_i \otimes \mathbf N_i$

Observe that
$\mathbf = \mathbf~\mathbf~\mathbf^T = \sum_^3 \lambda_i~\mathbf~(\mathbf_i\otimes\mathbf_i)~\mathbf^T = \sum_^3 \lambda_i~(\mathbf~\mathbf_i)\otimes(\mathbf~\mathbf_i)$
Therefore, the uniqueness of the spectral decomposition also implies that $\mathbf_i = \mathbf~\mathbf_i \,\!$. The left stretch ($\mathbf\,\!$) is also called the ''spatial stretch tensor'' while the right stretch ($\mathbf\,\!$) is called the ''material stretch tensor''.

The effect of $\mathbf$ acting on $\mathbf_i$ is to stretch the vector by $\lambda_i$ and to rotate it to the new orientation $\mathbf_i\,\!$, i.e.,
$\mathbf~\mathbf_i = \lambda_i~(\mathbf~\mathbf_i) = \lambda_i~\mathbf_i$
In a similar vein,
$\mathbf^~\mathbf_i = \cfrac~\mathbf_i ~;~~ \mathbf^T~\mathbf_i = \lambda_i~\mathbf_i ~;~~ \mathbf^~\mathbf_i = \cfrac~\mathbf_i ~.$

Examples

; Uniaxial extension of an incompressible material
: This is the case where a specimen is stretched in 1-direction with a stretch ratio of $\mathbf\,\!$. If the volume remains constant, the contraction in the other two directions is such that $\mathbf$ or $\mathbf\,\!$. Then: $\mathbf=\begin \alpha & 0 & 0 \\ 0 & \alpha^ & 0 \\ 0 & 0 & \alpha^ \end$ $\mathbf = \mathbf = \begin \alpha^2 & 0 & 0 \\ 0 & \alpha^ & 0 \\ 0 & 0 & \alpha^ \end$
; Simple shear
:$\mathbf=\begin 1 & \gamma & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end$ $\mathbf = \begin 1+\gamma^2 & \gamma & 0 \\ \gamma & 1 & 0 \\ 0 & 0 & 1 \end$ $\mathbf = \begin 1 & \gamma & 0 \\ \gamma & 1+\gamma^2 & 0 \\ 0 & 0 & 1 \end$
; Rigid body rotation
:$\mathbf = \begin \cos \theta & \sin \theta & 0 \\ - \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end$ $\mathbf = \mathbf = \begin 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end = \mathbf$

Derivatives of stretch

Derivatives of the stretch with respect to the right Cauchy–Green deformation tensor are used to derive the stress-strain relations of many solids, particularly hyperelastic materials. These derivatives are
$\cfrac = \cfrac~\mathbf_i\otimes\mathbf_i = \cfrac~\mathbf^T~(\mathbf_i\otimes\mathbf_i)~\mathbf ~;~~ i=1,2,3$
and follow from the observations that
$\mathbf:(\mathbf_i\otimes\mathbf_i) = \lambda_i^2 ~;~~~~\cfrac = \mathsf^ ~;~~~~ \mathsf^:(\mathbf_i\otimes\mathbf_i)=\mathbf_i\otimes\mathbf_i.$

Physical interpretation of deformation tensors

Let $\mathbf = X^i~\boldsymbol_i$ be a Cartesian coordinate system defined on the undeformed body and let $\mathbf = x^i~\boldsymbol_i$ be another system defined on the deformed body. Let a curve $\mathbf(s)$ in the undeformed body be parametrized using s \in ,1/math>. Its image in the deformed body is $\mathbf(\mathbf(s))$.

The undeformed length of the curve is given by
$l_X = \int_0^1 \left, \cfrac \~ds = \int_0^1 \sqrt~ds = \int_0^1 \sqrt~ds$
After deformation, the length becomes
$\begin l_x & = \int_0^1 \left, \cfrac \~ds = \int_0^1 \sqrt~ds = \int_0^1 \sqrt~ds \\ & = \int_0^1 \sqrt~ds \end$
Note that the right Cauchy–Green deformation tensor is defined as
$\boldsymbol := \boldsymbol^T\cdot\boldsymbol = \left(\cfrac\right)^T\cdot \cfrac$
Hence,
$l_x = \int_0^1 \sqrt~ds$
which indicates that changes in length are characterized by $\boldsymbol$.

Finite strain tensors

The concept of ''strain'' is used to evaluate how much a given displacement differs locally from a rigid body displacement. One of such strains for large deformations is the ''Lagrangian finite strain tensor'', also called the ''Green-Lagrangian strain tensor'' or ''Green – St-Venant strain tensor'', defined as

$\mathbf E=\frac(\mathbf C - \mathbf I)\qquad \text \qquad E_=\frac\left( \frac\frac-\delta_\right)$

or as a function of the displacement gradient tensor
\mathbf E =\frac\left (\nabla_\mathbf u)^T + \nabla_\mathbf u + (\nabla_\mathbf u)^T \cdot\nabla_\mathbf u\right/math>
or
$E_=\frac\left(\frac+\frac+\frac\frac\right)$

The Green-Lagrangian strain tensor is a measure of how much $\mathbf C$ differs from $\mathbf I\,\!$.

The ''Eulerian-Almansi finite strain tensor'', referenced to the deformed configuration, i.e. Eulerian description, is defined as

$\mathbf e=\frac(\mathbf I - \mathbf c)=\frac(\mathbf I - \mathbf B ^) \qquad \text \qquad e_ = \frac \left(\delta_ - \frac \frac\right)$

or as a function of the displacement gradients we have
$e_ = \frac \left(\frac + \frac - \frac \frac\right)$

d\mathbf=\mathbf F^ \, d\mathbf=\mathbf \,d\mathbf \qquad \text \qquad dX_M=\frac\, dx_n
where $\frac$ are the components of the ''spatial deformation gradient tensor'', $\mathbf\,\!$. Thus we have

$\begin d\mathbf^2 &= d\mathbf X \cdot d\mathbf X \\ &= \mathbf F^ \cdot d\mathbf x \cdot \mathbf F^ \cdot d\mathbf x \\ &= d\mathbf x \cdot \mathbf F^\mathbf F^ \cdot d\mathbf x \\ &= d\mathbf x\cdot\mathbf c\cdot d\mathbf x \end \qquad \text \qquad \begin (dX)^2&=dX_M\,dX_M \\ &= \frac\frac\,dx_r\,dx_s \\ &= c_\,dx_r\,dx_s \\ \end$
where the second order tensor $c_$ is called ''Cauchy's deformation tensor'', $\mathbf c=\mathbf F^\mathbf F^\,\!$. Then we have,

$\begin d\mathbf^2 - d\mathbf^2 &= d\mathbf x\cdot d\mathbf x-d\mathbf x\cdot\mathbf c\cdot d\mathbf x \\ &=d\mathbf x\cdot (\mathbf I - \mathbf c)\cdot d\mathbf x \\ &= d\mathbf x \cdot 2\mathbf e \cdot d\mathbf x \\ \end$
or
$\begin (dx)^2 - (dX)^2 &= dx_j\,dx_j-\frac\frac\,dx_r\,dx_s \\ &= \left(\delta_ - \frac\frac \right)\,dx_r\,dx_s \\ &=2e_\,dx_r\,dx_s \end$

where $e_\,\!$, are the components of a second-order tensor called the ''Eulerian-Almansi finite strain tensor'',
$\mathbf e=\frac(\mathbf I - \mathbf c) \qquad \text \qquad e_=\frac\left(\delta_ - \frac\frac \right)$

Both the Lagrangian and Eulerian finite strain tensors can be conveniently expressed in terms of the ''displacement gradient tensor''. For the Lagrangian strain tensor, first we differentiate the displacement vector $\mathbf u(\mathbf X, t)$ with respect to the material coordinates $X_M$ to obtain the ''material displacement gradient tensor'', $\nabla_\mathbf u$

$\begin \mathbf u(\mathbf X,t) &= \mathbf x(\mathbf X,t) - \mathbf X \\ \nabla_\mathbf u &= \mathbf F - \mathbf I \\ \mathbf F &= \nabla_\mathbf u + \mathbf I \\ \end \qquad \text \qquad \begin u_i& = x_i-\delta_X_J \\ \delta_U_J &= x_i-\delta_X_J \\ x_i&=\delta_\left(U_J+X_J\right) \\ \frac&=\delta_\left(\frac+\delta_\right) \\ \end$

Replacing this equation into the expression for the Lagrangian finite strain tensor we have
$\begin \mathbf E &= \frac\left(\mathbf F^T\mathbf F-\mathbf I\right) \\ &=\frac\left$