The objective in this section is to express the three primary measures of strain \( \epsilon \), \( \gamma \), and \( \epsilon_v \) (see definitions in the previous section) by means of the Green strain tensor \( \boldsymbol{E} \), which will be defined below. Before ending up with the Green strain tensor, other commonly used tensors will also be introduced, such as the deformation gradient tensor \( \bf{F} \), Green{'s} deformation tensor \( \bf{C} \), the displacement gradient tensor \( \bf{H} \).
The length \( s_0 \) of the material line \( P_0Q_0 \) in direction \( \boldsymbol{n} \), will now be used to parameterize the length \( s \) in the current configuration \( K \), i.e., \( s(s_0) \) is a function of \( s_0 \). The coordinates of the endpoints \( Q_0 \) and \( Q \) are \( X_i + s_0 n_i \) and \( x_i\left( X + s_0 \boldsymbol{n}, t \right) \), respectively. Here the directional unit vector is expressed by its components \( \bf{n} = n_i \bf{e}_i \). The length \( s \) calculated with \( s_0 \) as curve parameter by the arc length formula: $$ \begin{equation} s = \int\limits_0^{s_0 } {\sqrt {\frac{{\partial x_i }}{{\partial s_0 }}\frac{{\partial x_i }}{{\partial s_0 }}} \;d s_0 } \tag{3.5} \end{equation} $$ which yields $$ \begin{equation} d s^2 = \left( {\frac{{\partial x_i }}{{\partial s_0 }}\frac{{\partial x_i }}{{\partial s_0 }}} \right)d s_0^2 \tag{3.6} \end{equation} $$ Let \( d s_0 \) be the length of \( d \mathbf{r}_0 \) in direction of \( \boldsymbol{n} \) in \( K_0 \) (Fig. 18). Then it follows that: $$ \begin{equation} d \boldsymbol{r}_0 = \boldsymbol{n}\;d s_0 = d\!X_k \boldsymbol{e}_k \tag{3.7} \end{equation} $$ which yields: $$ \begin{equation} \tag{3.8} \left| {d \mathbf{r}_{\bf{0}} } \right| = d s_0 , \quad d\!X_k = n_k d s_0 \Leftrightarrow n_k = \frac{{d\!X_k }}{{d s_0 }} \end{equation} $$ The expression for the components of the directional unit vector \( n_k \) in (3.8), will be useful, in the following derivation. Similarly, let \( d \mathbf{r} \) be a line element from \( P \) in \( K \), which due to our parameterization may be expressed: $$ \begin{equation} d \mathbf{r} = \frac{{\partial {\bf{r}}}}{{\partial s_0 }}\;d s_0 \quad \Leftrightarrow \quad dx_i = \frac{{\partial x_i }}{{\partial s_0 }}d s_0 \tag{3.9} \end{equation} $$ From (3.6), we see that \( d s \) is the length of \( d \mathbf{r} \). As the current position \( \boldsymbol{r} \) is a function of the reference position \( \boldsymbol{r}_0 \), the relation between \( d\bf{r} \) in \( K \) and its corresponding representation \( d\bf{r}_0 \) in \( K_0 \) may be presented: $$ \begin{equation} d \mathbf{r} = \frac{{\partial {\bf{r}}}}{{\partial {\bf{r}}_0 }} \cdot \;d \mathbf{r}_0 = {\bf{F}} \cdot \;d \mathbf{r}_0 \quad \Leftrightarrow \quad dx_i = \frac{{\partial x_i }}{{\partial X_k }}d\!X_k = F_{ik} d\!X_k \tag{3.10} \end{equation} $$ where the deformation gradient \( \bf{F} \) has been introduced: $$ \begin{equation} {\bf{F}} = \frac{{\partial {\bf{r}}}}{{\partial {\bf{r}}_0 }}\quad \Leftrightarrow \quad F_{ik} = \frac{{\partial x_i }}{{\partial X_k }} \tag{3.11} \end{equation} $$ Loosely speaking, one may say that the deformation gradient holds information about the difference in the current locations of neighboring particles in \( K_0 \), which is unity when they are displaced equally i.e., for no deformation.
Now, consider an expression for the directional derivative \( \partial x_i/\partial s_0 \) of the components of \( \boldsymbol{r} \) in direction of \( \boldsymbol{n} \): $$ \begin{equation} \left. \frac{\partial x_i \left( X + s_0 \boldsymbol{n},t \right)}{{\partial s_0 }} \right|_{s_0 = 0} = \left. {\frac{{\partial x_i \left( {X,t} \right)}}{{\partial X_k }}\frac{{d\left( {X_k + s_0 n_k ,t} \right)}}{{d s_0 }}} \right|_{s_0 = 0} \tag{3.12} \end{equation} $$ which may be simplified by substitution of (3.8): $$ \begin{equation} \frac{{\partial x_i }}{{\partial s_0 }} = F_{ik} \frac{{d\!X_k }}{{d s_0 }} = F_{ik} n_k \tag{3.13} \end{equation} $$ The result in (3.13) may be substituted into (3.6) to yield: $$ \begin{equation} \left( {\frac{{d s}}{{d s_0 }}} \right)^2 = \frac{{\partial x_i }}{{\partial s_0 }}\frac{{\partial x_i }}{{\partial s_0 }} = F_{ij} n_j F_{ik} n_k = {\boldsymbol{n}} \cdot \left( {{\boldsymbol{F}}^T {\boldsymbol{F}}} \right) \cdot {\boldsymbol{n}} = {\boldsymbol{n}} \cdot {\boldsymbol{C}} \cdot {\boldsymbol{n}} \tag{3.14} \end{equation} $$ where the Green deformation tensor is introduced as a second order symmetric tensor: $$ \begin{equation} \tag{3.15} \boldsymbol{C} = \boldsymbol{F}^T \boldsymbol{F}\quad \Leftrightarrow \quad C_{ij} = F_{ki} F_{kj} \end{equation} $$
This tensor is also commonly called the Left Green deformation tensor as the transpose \( \boldsymbol{F}^T \) occurs on the left hand side of equation (3.15), as opposed to its equivalent Right Green deformation tensor: $$ \begin{equation} \tag{3.16} \boldsymbol{B} = \boldsymbol{F} \boldsymbol{F}^T\quad \Leftrightarrow \quad B_{ij} = F_{ik} F_{jk} \end{equation} $$ which plays an important role the modelling of isotropic materials (see e.g., 4.3.5 Example 12: The Mooney-Rivlin material).
Longitudinal strain \( \epsilon \)
From (3.3) and (3.14) an expression for the longitudinal strain \( \epsilon \) in the direction \( \boldsymbol{n} \) may be presented: $$ \begin{equation} \epsilon = \sqrt{\boldsymbol{n \cdot C \cdot n}} - 1 \tag{3.17} \end{equation} $$ However, in (3.15) \( \boldsymbol{C} \) is implicitly related to the deformation gradient \( \boldsymbol{F} \), which in some applications is not a preferable way of representing deformation, as it is based on the current location \( \mathbf{r} \) and not the displacement \( \mathbf{u}(\mathbf{r}_0,t) \). From Eq. (3.2) and Figure 17, we see that $$ \begin{equation} \tag{3.18} \mathbf{r} = \mathbf{r}_0 + \mathbf{u}(\mathbf{r}_0,t) \qquad \Leftrightarrow \qquad x_i(X,t) = X_i + u_i(X,t) \end{equation} $$ Thus, to obtain strain tensors based on displacement rather than the current location, the displacement gradient \( \boldsymbol{H} \) is commonly introduced: $$ \begin{equation} H_{ik} = \frac{{\partial u_i }}{{\partial X_k }}\quad \Leftrightarrow \quad {\bf{H}} = \frac{{\partial {\bf{u}}}}{{\partial {\bf{r}}_0 }} \tag{3.19} \end{equation} $$ A relation between the displacement gradient \( \boldsymbol{H} \) and the deformation gradient \( \boldsymbol{F} \) by combining \rthreeeqs{eq:54}{eq:60}{eq:67}: $$ \begin{equation} F_{ik} = \delta _{ik} + H_{ik} \quad \Leftrightarrow \quad {\bf{F}} = {\bf{1}} + {\bf{H}} \tag{3.20} \end{equation} $$ as: $$ \begin{equation} \frac{{\partial x_i }}{{\partial X_k }} = \delta _{ik} + \frac{{\partial u_i }}{{\partial X_k }} \tag{3.21} \end{equation} $$ Thus, from (3.15) and (3.20) an alternative representation of the Green deformation tensor is found based on displacements: $$ \begin{equation} {\boldsymbol{C}} = {\boldsymbol{F}}^{T} {\boldsymbol{F}} = \left( {{\boldsymbol{1}} + {\boldsymbol{H}}^T } \right)\left( {{\boldsymbol{1}} + {\boldsymbol{H}}} \right) = {\boldsymbol{1}} + {\boldsymbol{H}} + {\boldsymbol{H}}^T + {\boldsymbol{H}}^T {\boldsymbol{H}} \tag{3.22} \end{equation} $$ The expression for \( \boldsymbol{C} \) in (3.22) may be simplified further by introducing the definition of the Green strain tensor: $$ \begin{equation} {\boldsymbol{E}} = \frac{1}{2}\left( {{\boldsymbol{H}} + {\boldsymbol{H}}^T + {\bf{H}}^T {\boldsymbol{H}}} \right) \Leftrightarrow E_{kl} = \frac{1}{2} \, \left ( \partd{u_k}{X_l} + \partd{u_l}{X_k} + \partd{u_i}{X_k} \partd{u_i}{X_l} \right ) \tag{3.23} \end{equation} $$
such that by substitution of (3.23) into (3.22) the expression for the Green deformation tensor reduces to: $$ \begin{equation} \boldsymbol{C} = \boldsymbol{1} + 2\boldsymbol{E} \tag{3.24} \end{equation} $$ An expression for the longitudinal strain \( \epsilon \) based on the Green strain tensor is then obtained by substitution of (3.24) into (3.17): $$ \begin{equation} \epsilon = \sqrt{1 + 2 \, \boldsymbol{n \cdot E \cdot n}} - 1 \tag{3.25} \end{equation} $$
Shear strain \( \gamma \)
An expression for the shear strain \( \gamma \) may be found by considering two differential material elements, which were perpendicular in \( K_0 \), in the current configuration \( K \): $$ \begin{align} d \mathbf{r} & = \boldsymbol{F} \cdot d \mathbf{r}_0 = (\boldsymbol{1 + H} ) \cdot \boldsymbol{n} \, d s_0 \tag{3.26}\\ d \bar{\boldsymbol{r}} & = \boldsymbol{F} \cdot d \bar{\boldsymbol{r}}_0 = (\boldsymbol{1 + H} ) \cdot \bar{\boldsymbol{n}} \, d \bar{s}_0 \tag{3.27} \end{align} $$
The scalar product between the two vectors in (3.26) and (3.27) takes the form: $$ \begin{align} d \mathbf{r} \cdot d \bar{\boldsymbol{r}} = d \bar{s}_0 \bar{\boldsymbol{n}} \cdot (\boldsymbol{1} + \boldsymbol{H}^T )(\boldsymbol{1} + \boldsymbol{H}) \cdot \boldsymbol{n} d s_0 = \bar{\boldsymbol{n}} \cdot \boldsymbol{C} \cdot \boldsymbol{n} \, d \bar{s}_0 d s_0 \tag{3.28} \end{align} $$ From (3.14) the length of the two differential vectors are found to be: $$ \begin{equation} | d \mathbf{r} | = d s = \sqrt{\boldsymbol{n \cdot C \cdot n}}\, d s_0, \quad | d \bar{\boldsymbol{r}} | = d \bar{s} = \sqrt{\boldsymbol{\bar{n} \cdot C \cdot \bar{n}}}\, d \bar{s}_0 \tag{3.29} \end{equation} $$ The definition of a scalar product along with (3.28) and (3.29) may then be used to find an expression for the shear strain: $$ \begin{equation} \sin \gamma = \frac{\boldsymbol{\bar{n} \cdot C \cdot n}}{\sqrt{\boldsymbol{n \cdot C \cdot n}} \, \sqrt{\boldsymbol{\bar{n} \cdot C \cdot \bar{n}}}} = \frac{\boldsymbol{\bar{n} \cdot C \cdot n}}{(1+\epsilon) \, (1+\bar{\epsilon})} \tag{3.30} \end{equation} $$ As \( \bar{\boldsymbol{n}} \) and \( \boldsymbol{n} \) are orthogonal, an expression for the shear strain based on the Green strain tensor is obtained from (3.24) and (3.30): $$ \begin{equation} \sin \gamma = \frac{\boldsymbol{\bar{n} \cdot C \cdot n}}{(1+\epsilon) \, (1+\bar{\epsilon})} =\frac{2 \, \boldsymbol{\bar{n} \cdot E \cdot n}}{(1+\epsilon) \, (1+\bar{\epsilon})} \tag{3.31} \end{equation} $$
Volumetric strain \( \epsilon_v \)
The volume of a differential volume element \( d V_0 \) in \( K_0 \) is: $$ \begin{equation} d V_0 = d\!X_1\, d\!X_2 \, d\!X_3 \tag{3.32} \end{equation} $$ the corresponding deformed volume \( d V \) in \( K \) by the volume spanned by the the three vectors \( d \mathbf{r}_i \): $$ \begin{equation} d \mathbf{r}_1 = \partd{\boldsymbol{r}}{X_1} \, d\!X_1 = \partd{x_i}{X_1} \, \boldsymbol{e}_i \, d\!X_1, \quad d \mathbf{r}_2 = \partd{x_j}{X_2} \, \boldsymbol{e}_j \, d\!X_2, \quad % d \mathbf{r}_3 = \partd{x_k}{X_3} \, \boldsymbol{e}_k \, d\!X_3 \tag{3.33} \end{equation} $$ which may be expressed by the box product (see (7.11)) of the vectors in (3.33): $$ \begin{align} d V &= [d \mathbf{r}_1 \, d \mathbf{r}_2 \, d \mathbf{r}_3] = (d \mathbf{r}_1 \times d \mathbf{r}_2) \cdot d \mathbf{r}_3 = e_{ijk}\, \partd{x_j}{X_1} \, d\!X_1 \partd{x_j}{X_2} \, d\!X_2\, \partd{x_j}{X_3} \, d\!X_3 \tag{3.34} \end{align} $$ In the latter expression of (3.34), the determinant (see (7.12)) of the deformation gradient is identified, and thus (3.34) may be represented: $$ \begin{equation} d V = \det \boldsymbol{F} \, d V_0 = \det \left (\boldsymbol{1 + H} \right ) \, d V_0 = \sqrt{\det \boldsymbol{C}} \, d V_0 \tag{3.35} \end{equation} $$ the latter equalities follow from (3.20) and (3.22) and determinants of products. Various expressions for the volumetric strain \( \epsilon_v \) may then be found from (3.32) and (3.35): $$ \begin{equation} \epsilon_v = \frac{d V - d V_0}{d V_0} = \det {\boldsymbol{F}} - 1 = \det (\boldsymbol{1} + \boldsymbol{H}) - 1 = \sqrt{\det\boldsymbol{C}} -1 = \sqrt{\det\boldsymbol{1} + 2\boldsymbol{E}} - 1 \tag{3.36} \end{equation} $$
In this section, all measures of strain, \( \epsilon,\; \gamma, \; \epsilon_v \), have been expressed by the Green strain tensor \( \boldsymbol{E} \) and the Green deformation tensor \( \boldsymbol{C} \), which in turn are related to the deformation gradient \( \boldsymbol{F} \) or the displacement gradient \( \boldsymbol{H} \). In the following chapters it will be shown that material models relevant for biomechanical applications are mostly defined by relations between a stress tensor (\( \boldsymbol{T} \) or other appropriate stresses) and the Green strain tensor \( \boldsymbol{E} \) or the Green deformation tensor \( \boldsymbol{C} \).