Physical interpretation of strain components \(\boldsymbol{\varepsilon}_{\boldsymbol{i j}}\)

For the line element parallel to the \(x_{1}\)-axis in the reference configuration, the components of the unit vector are \((n_{1}, n_{2}, n_{3})=(1,0,0)\), and from eq. ([eqA.18]) its strain is given by \(\varepsilon_{L}=\varepsilon_{11}\). For the line element parallel to the \(x_{2}\)-axis, the components of the unit vector are \((n_{1}, n_{2}, n_{3})=(0,1,0)\), and its strain is given by \(\varepsilon_{L}=\varepsilon_{22}\). For the line element parallel to the \(x_{3}\)-axis its strain is \(\varepsilon_{L}=\varepsilon_{33}\). The physical interpretation of component \(\varepsilon_{12}\) is determined from the passing of line elements \(d x_{1} \hat{i}_{1}\) and \(d x_{2} \hat{i}_{2}\) in region \(B_{0}\) to directions \((\hat{N})_{1}\) and \((\hat{N})_{2}\), respectively, in region \(B\). Unit vector \(\hat{N}=N_{1} \hat{i}_{1}+N_{2} \hat{i}_{2}+N_{3} \hat{i}_{3}\) where the components \(N_{i}\) are given by \[\begin{aligned} \label{eqA.26} N_{i}=\frac{d y_{i}}{d S}=\frac{d y_{i}}{d s} \frac{d s}{d S}=\frac{d y_{i}}{d s}\left(\frac{1}{\sqrt{1+2 \varepsilon_{L}}}\right)=\left(\frac{\partial y_{i}}{\partial x_{1}} n_{1}+\frac{\partial y_{i}}{\partial x_{2}} n_{2}+\frac{\partial y_{i}}{\partial x_{3}} n_{3}\right)\left(\frac{1}{\sqrt{1+2 \varepsilon_{L}}}\right), i=1,2,3.\end{aligned}\] Take \((n_{1}, n_{2}, n_{3})=(1,0,0)\) in eq. ([eqA.26]), then from eq. ([eqA.18]) \(\varepsilon_{L}=\varepsilon_{11}\). In the transition from region \(B_{0}\) to region \(B\) the unit vector \(\hat{i}_{1} \rightarrow(\hat{N})_{1}\), where the unit vector \((\hat{N})_{1}\) is given by \[\begin{aligned} \label{eqA.27} (\hat{N})_{1}=\left[\frac{\partial y_{1}}{\partial x_{1}} \hat{i}_{1}+\frac{\partial y_{2}}{\partial x_{1}} \hat{i}_{2}+\frac{\partial y_{3}}{\partial x_{1}} \hat{i}_{3}\right]\left(\frac{1}{\sqrt{1+2 \varepsilon_{11}}}\right).\end{aligned}\] Take \((n_{1}, n_{2}, n_{3})=(0,1,0)\) in eq. ([eqA.26]), then \(\varepsilon_{L}=\varepsilon_{22}\). In the transition from region \(B_{0}\) to region \(B\) the unit vector \(\hat{i}_{2} \rightarrow(\hat{N})_{2}\). The result for \((\hat{N})_{2}\) is \[\begin{aligned} \label{eqA.28} (\hat{N})_{2}=\left[\frac{\partial y_{1}}{\partial x_{2}} \hat{i}_{1}+\frac{\partial y_{2}}{\partial x_{2}} \hat{i}_{2}+\frac{\partial y_{3}}{\partial x_{2}} \hat{i}_{3}\right]\left(\frac{1}{\sqrt{1+2 \varepsilon_{22}}}\right).\end{aligned}\] The scalar product of the two unit vectors \((\hat{N})_{1}\) and \((\hat{N})_{2}\) is equal to the cosine of the angle between them. Let this angle be denoted by \(\pi / 2-\theta_{12}\) such that, \[\begin{aligned} \label{eqA.29} (\hat{N})_{1} \bullet (\hat{N})_{2}=\cos \left(\frac{\pi}{2}-\theta_{12}\right)=\sin \theta_{12}.\end{aligned}\] Angle \(\theta_{12}\) represents the reduction in the right angle. Substitute eq. ([eqA.27]) for \((\hat{N})_{1}\) and eq. ([eqA.28]) for \((\hat{N})_{2}\) in the left-hand side of eq. ([eqA.29]) to get \[\begin{aligned} \label{eqA.30} \sin \theta_{12}=\left[\frac{\partial y_{1}}{\partial x_{1}} \frac{\partial y_{1}}{\partial x_{2}}+\frac{\partial y_{2}}{\partial x_{1}} \frac{\partial y_{2}}{\partial x_{2}}+\frac{\partial y_{3}}{\partial x_{1}} \frac{\partial y_{3}}{\partial x_{2}}\right] \frac{1}{\sqrt{1+2 \varepsilon_{11}} \sqrt{1+2 \varepsilon_{22}}}.\end{aligned}\] Next substitute \(y_{i}=u_{i}+x_{i}\), \(i=1,2,3\), in the terms in the square brackets of the previous equation, and compare the result to eq. ([eqA.23]) to find \[\begin{aligned} \label{eqA.31} \sin \theta_{12}=\frac{2 \varepsilon_{12}}{\sqrt{1+2 \varepsilon_{11}} \sqrt{1+2 \varepsilon_{22}}}.\end{aligned}\] Thus, the right angle between line elements \(d x_{1} \hat{i}_{1}\) and \(d x_{2} \hat{i}_{2}\) in region \(B_{0}\) is changed in the transition to region \(B\) in direct proportion to the strain component \(\varepsilon_{12}\). If \(\varepsilon_{12}=0\), then \(\theta_{12}=0\), and the right angle is preserved in the deformed body. Similarly, the strain component \(\varepsilon_{13}\) is proportional to the change in the right angle between line elements \(dx_{1}\) and \(dx_{3}\) in the transition to the deformed body.

Engineering strain

Engineering strain is defined by \[\begin{aligned} \label{eqA.32} \varepsilon_{E}=\frac{d S-d s}{d s}=\frac{d S}{d s}-1.\end{aligned}\] Substitute for \(d S / d s\) from eq. ([eqA.32]) into eq. ([eqA.12]) to get \(\varepsilon_{L}=\varepsilon_{E}+\varepsilon_{E}^{2} / 2\). Since the product of the direction cosines commute, eq. ([eqA.18]) is rewritten as \[\begin{aligned} \label{eqA.33} \varepsilon_{L}=\varepsilon_{E}+\frac{1}{2} \varepsilon_{E}^{2}=\varepsilon_{11} n_{1}^{2}+\varepsilon_{22} n_{2}^{2}+\varepsilon_{33} n_{3}^{2}+\gamma_{12} n_{1} n_{2}+\gamma_{13} n_{1} n_{3}+\gamma_{23} n_{2} n_{3},\end{aligned}\] where the engineering shear strains are defined by \[\begin{aligned} \label{eqA.34} \gamma_{12}=\varepsilon_{12}+\varepsilon_{21} \quad \gamma_{13}=\varepsilon_{13}+\varepsilon_{31} \quad \gamma_{23}=\varepsilon_{23}+\varepsilon_{32}.\end{aligned}\]

Linear strain-displacement relations

For many materials the strains are very small in the elastic range. A linear theory of deformation is characterized by the magnitude of the nine displacement gradients \(\left|\frac{\partial u_{i}}{\partial x_{j}}\right| \sim 10^{-3}\). Hence, we neglect the quadratic terms in the displacement gradients with respect to the linear terms in the strain-displacement eqs. ([eqA.20]) to ([eqA.25]). The resulting linear strain-displacement relations are \[\begin{gathered} \varepsilon_{11}=\frac{\partial u_{1}}{\partial x_{1}} \quad \varepsilon_{22}=\frac{\partial u_{2}}{\partial x_{2}} \quad \varepsilon_{33}=\frac{\partial u_{3}}{\partial x_{3}},\label{eqA.35} \\ \gamma_{12}=\frac{\partial u_{1}}{\partial x_{2}}+\frac{\partial u_{2}}{\partial x_{1}} \quad \gamma_{13}=\frac{\partial u_{1}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{1}} \quad \gamma_{23}=\frac{\partial u_{2}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{2}}\mbox{, and}\label{eqA.36}\\ \varepsilon_{E}=\varepsilon_{11} n_{1}^{2}+\varepsilon_{22} n_{2}^{2}+\varepsilon_{33} n_{3}^{2}+\gamma_{12} n_{1} n_{2}+\gamma_{13} n_{1} n_{3}+\gamma_{23} n_{2} n_{3}.\label{eqA.37} \end{gathered}\]

Transformation of the strains between two Cartesian coordinate systems

In the reference configuration \(B_{0}\) consider the two orthogonal Cartesian coordinate systems \((x_{1}, x_{2}, x_{3})\) and \((x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime})\), which have the same origin. In the \((x_{1}, x_{2}, x_{3})\) system, or simply the \(x_{i}\)-system, the corresponding unit vectors are \((\hat{i}_{1}, \hat{i}_{2}, \hat{i}_{3})\). In the \((x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime})\) system, or simply the \(x_{i}^{\prime}\)-system, the corresponding unit vectors are \((\hat{i}_{1}^{\,\prime}, \hat{i}_{2}^{\,\prime}, \hat{i}_{3}^{\,\prime})\). The position vector \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{r}}}\) of a point \(P_{0}\) in the three-dimensional space is the same if written in the \(x_{i}\)-system or in the \(x_{i}^{\prime}\)-system. That is, \[\begin{aligned} \label{eqA.38} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{r}}}=x_{1} \hat{i}_{1}+x_{2} \hat{i}_{2}+x_{3} \hat{i}_{3}=x_{1}^{\prime} \hat{i}_{1}^{\,\prime}+x_{2}^{\prime} \hat{i}_{2}^{\,\prime}+x_{3}^{\prime} \hat{i}_{3}^{\,\prime}.\end{aligned}\] The linear form \(x_{1} \hat{i}_{1}+x_{2} \hat{i}_{2}+x_{3} \hat{i}_{3}\) is said to remain invariant under the transformation of variables. Take the scalar product, or dot product, of eq. ([eqA.38]) with unit vector \(\hat{i}_{1}^{\,\prime}\) to get \[\begin{aligned} \label{eqA.39} x_{1} \hat{i}_{1} \bullet \hat{i}_{1}^{\,\prime}+x_{2} \hat{i}_{2} \bullet \hat{i}_{1}^{\,\prime}+x_{3} \hat{i}_{3} \bullet \hat{i}_{1}^{\,\prime}=x_{1}^{\prime}.\end{aligned}\] Define the nine direction cosines \(\lambda_{i j}\), \(i, j=1,2,3\), by \[\begin{aligned} \label{eqA.40} \hat{i}_{i}^{\,\prime} \bullet \hat{i}_{j} \equiv \lambda_{i j}=\cos (x_{i}^{\prime}, x_{j}).\end{aligned}\] For example, the notation \(\lambda_{12}=\cos (x_{1}^{\prime}, x_{2})\) represents the cosine of the angle between the \(x_{1}^{\prime}\)-axis and the \(x_{2}\)-axis, and the notation \(\lambda_{21}=\cos (x_{2}^{\prime}, x_{1})\) represents the cosine of the angle between the \(x_{2}^{\prime}\)-axis and the \(x_{1}\)-axis. From the definition of \(\lambda_{i j}\) in eq. ([eqA.39]) becomes \(x_{1}^{\prime}=\lambda_{11} x_{1}+\lambda_{12} x_{2}+\lambda_{13} x_{3}\). If we take the dot product of eq. ([eqA.38]) with \(\hat{i}_{2}^{\,\prime}\) and use the definition ([eqA.40]), then we find \(x_{2}^{\prime}=\lambda_{21} x_{1}+\lambda_{22} x_{2}+\lambda_{23} x_{3}\). The relation between the \(x_{i}^{\prime}\)-coordinates and the \(x_{i}\)-coordinates at point \(P_{0}\) is given by the linear transformation \[\begin{aligned} \label{eqA.41} \left[\begin{array}{@{}l@{}} x_{1}^{\prime} \\[3pt] x_{2}^{\prime} \\[3pt] x_{3}^{\prime}\end{array}\right]=\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{12} & \lambda_{13} \\\lambda_{21} & \lambda_{22} & \lambda_{23} \\\lambda_{31} & \lambda_{32} & \lambda_{33}\end{array}\right]\left[\begin{array}{@{}l@{}}x_{1} \\x_{2} \\x_{3}\end{array}\right].\end{aligned}\] The compact form of matrix eq. ([eqA.41]) is \[\begin{gathered} \left\{x^{\prime}\right\}=[\lambda]\{x\}\mbox{, where}\label{eqA.42}\\ \left\{x^{\prime}\right\}=\left[\begin{array}{@{}l@{}}x_{1}^{\prime} \\[3pt] x_{2}^{\prime} \\[3pt] x_{3}^{\prime}\end{array}\right] \quad[\lambda]=\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{12} & \lambda_{13} \\\lambda_{21} & \lambda_{22} & \lambda_{23} \\\lambda_{31} & \lambda_{32} & \lambda_{33}\end{array}\right] \quad\{x\}=\left[\begin{array}{@{}l@{}}x_{1} \\x_{2} \\x_{3}\end{array}\right].\label{eqA.43}\end{gathered}\] The unit vectors in the \(x_{i}^{\prime}\)-system are related to those in the \(x_{i}\)-system by the same relation given in eq. ([eqA.41]): \[\begin{aligned} \label{eqA.44} \left[\begin{array}{@{}l@{}} \hat{i}_{1}^{\,\prime} \\[3pt] \hat{i}_{2} \\[3pt] \hat{i}_{3}^{\,\prime}\end{array}\right]=\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{12} & \lambda_{13} \\\lambda_{21} & \lambda_{22} & \lambda_{23} \\\lambda_{31} & \lambda_{32} & \lambda_{33}\end{array}\right]\left[\begin{array}{@{}l@{}}\hat{l}_{1} \\\hat{l}_{2} \\\hat{l}_{3}\end{array}\right].\end{aligned}\] Consider the dot product \(\hat{i}_{1}^{\,\prime} \bullet \hat{i}_{1}^{\,\prime}=1\), and from eq. ([eqA.44]) write it in terms of the unit vectors in the \(x_{i}\)-system; i.e., \[\begin{aligned} \label{eqA.45} 1=(\lambda_{11} \hat{i}_{1}+\lambda_{12} \hat{i}_{2}+\lambda_{13} \hat{i}_{3}) \bullet (\lambda_{11} \hat{i}_{1}+\lambda_{12} \hat{i}_{2}+\lambda_{13} \hat{i}_{3}).\end{aligned}\] Since unit vectors \((\hat{i}_{1}, \hat{i}_{2}, \hat{i}_{3})\) are mutually perpendicular we find the relation \[\begin{aligned} \label{eqA.46} 1=\lambda_{11}^{2}+\lambda_{12}^{2}+\lambda_{13}^{2}.\end{aligned}\] Consider the dot product \(\hat{i}_{1}^{\,\prime} \bullet \hat{i}_{2}^{\,\prime}=0\) and write \(\hat{i}_{1}^{\,\prime}\) and \(\hat{i}_{2}^{\,\prime}\) from eq. ([eqA.44]) to get \[\begin{aligned} \label{eqA.47} 0=(\lambda_{11} \hat{i}_{1}+\lambda_{12} \hat{i}_{2}+\lambda_{13} \hat{i}_{3}) \bullet (\lambda_{21} \hat{i}_{1}+\lambda_{22} \hat{i}_{2}+\lambda_{23} \hat{i}_{3}).\end{aligned}\] Again \((\hat{i}_1, \hat{i}_2, \hat{i}_3)\) are mutually perpendicular, so we find the relation \[\begin{aligned} \label{eqA.48} 0=\lambda_{11} \lambda_{21}+\lambda_{12} \lambda_{22}+\lambda_{13} \lambda_{23}.\end{aligned}\] We can proceed by performing the scalar products \(\hat{i}_{1}^{\,\prime} \bullet \hat{i}_{3}^{\,\prime}=0\), \(\hat{i}_{2}^{\,\prime} \bullet \hat{i}_{2}^{\,\prime}=1\), \(\hat{i}_{2}^{\,\prime} \bullet \hat{i}_{3}^{\,\prime}=0\), and \(\hat{i}_{3}^{\,\prime} \bullet \hat{i}_{3}^{\,\prime}=1\). Collectively we find the following relations between the direction cosines: \[\begin{aligned} \label{eqA.49} \begin{gathered} 1=\lambda_{11}^{2}+\lambda_{12}^{2}+\lambda_{13}^{2} \\ 0=\lambda_{11} \lambda_{21}+\lambda_{12} \lambda_{22}+\lambda_{13} \lambda_{23} \\ 0=\lambda_{11} \lambda_{31}+\lambda_{12} \lambda_{32}+\lambda_{13} \lambda_{33} \end{gathered},\quad \begin{gathered} 1=\lambda_{21}^{2}+\lambda_{22}^{2}+\lambda_{23}^{2} \\ 0=\lambda_{21} \lambda_{31}+\lambda_{22} \lambda_{32}+\lambda_{23} \lambda_{33} \end{gathered}\mbox{, and }1=\lambda_{31}^{2}+\lambda_{32}^{2}+\lambda_{33}^{2}.\end{aligned}\] There are six relations in eq. ([eqA.49]) between the nine direction cosines. Hence, only three of the direction cosines are independent. We show some interesting properties of the direction cosine matrix \([\lambda]\) beginning with the matrix product \([\lambda][\lambda]^{T}\). The result is \[\begin{aligned} \label{eqA.50} [\lambda][\lambda]^{T}=\left[\begin{array}{ccc}\lambda_{11}^{2}+\lambda_{12}^{2}+\lambda_{13}^{2} & \lambda_{11} \lambda_{21}+\lambda_{12} \lambda_{22}+\lambda_{13} \lambda_{23} & \lambda_{11} \lambda_{31}+\lambda_{12} \lambda_{32}+\lambda_{13} \lambda_{33} \\[4pt] \lambda_{11} \lambda_{21}+\lambda_{12} \lambda_{22}+\lambda_{13} \lambda_{23} & \lambda_{21}^{2}+\lambda_{22}^{2}+\lambda_{23}^{2} & \lambda_{21} \lambda_{31}+\lambda_{22} \lambda_{32}+\lambda_{23} \lambda_{33} \\[4pt] \lambda_{11} \lambda_{31}+\lambda_{12} \lambda_{32}+\lambda_{13} \lambda_{33} & \lambda_{21} \lambda_{31}+\lambda_{22} \lambda_{32}+\lambda_{23} \lambda_{33} & \lambda_{31}^{2}+\lambda_{32}^{2}+\lambda_{33}^{2}\end{array}\right].\end{aligned}\] Compare the elements of the matrix in eq. ([eqA.50]) to the relations in eq. ([eqA.49]) to find \[\begin{aligned} \label{eqA.51} [\lambda][\lambda]^{T}=\left[\begin{array}{@{}lll@{}}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]=[I].\end{aligned}\] Equation ([eqA.51]) shows that the matrix \([\lambda]\) is an orthogonal matrix. That is, the inverse \([\lambda]^{-1}\) is equal to its transpose \([\lambda]^{T}\). Also \(\operatorname{\it det}\left([\lambda][\lambda]^{T}\right)=\operatorname{\it det}[\lambda] \operatorname{\it det}[\lambda]^{T}\), but \(\operatorname{\it det}[\lambda]^{T}=\operatorname{\it det}[\lambda]\). Hence. \[\begin{aligned} \label{eqA.52} \operatorname{\it det}\big([\lambda][\lambda]^{T}\big)=\big(\operatorname{\it det}[\lambda]\big)^{2}=\big(\operatorname{\it det}[I]\big)^{2}=1.\end{aligned}\] The determinate of an orthogonal matrix is either 1 or \(-\)1. The inverse of eq. ([eqA.42]) is \(\{x\}=[\lambda]^{T}\{x^{\prime}\}\) which written in expanded form is \[\begin{aligned} \label{eqA.53} \left[\begin{array}{@{}l@{}}x_{1} \\x_{2} \\x_{3}\end{array}\right]=\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{21} & \lambda_{31} \\\lambda_{12} & \lambda_{22} & \lambda_{32} \\\lambda_{13} & \lambda_{23} & \lambda_{33}\end{array}\right]\left[\begin{array}{@{}l@{}}x_{1}^{\prime} \\[3pt] x_{2}^{\prime} \\[3pt] x_{3}^{\prime}\end{array}\right].\end{aligned}\] From eq. ([eqA.38]) the differential of the position vector in eq. ([eqA.38]) is \[\begin{aligned} \label{eqA.54} d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{r}}}=d x_{1} \hat{i}_{1}+d x_{2} \hat{i}_{2}+d x_{3} \hat{i}_{3}=d x_{1}^{\prime} \hat{i}_{1}^{\,\prime}+d x_{2}^{\prime} \hat{i}_{2}^{\,\prime}+d x_{3}^{\prime} \hat{i}_{3}^{\,\prime}.\end{aligned}\] The square of the length of \(d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{r}}}\) is \(d s^{2}=d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{r}}} \bullet d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{r}}}\). The length is the same in the \(x_{i}\)-system and the \(x_{i}^{\prime}\)-system, so \[\begin{aligned} \label{eqA.55} d s^{2}=(d x_{1})^{2}+(d x_{2})^{2}+(d x_{3})^{2}=(d x_{1}^{\prime})^{2}+(d x_{2}^{\prime})^{2}+(d x_{3}^{\prime})^{2}.\end{aligned}\] Hence, the unit vector \(\hat{n}\) in both the \(x_{i}\)-system and the \(x_{i}^{\prime}\)-system has the invariant forms \[\begin{aligned} \label{eqA.56} \hat{n}=\frac{d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{r}}}}{d s}=\frac{d x_{1}}{d s} \hat{i}_{1}+\frac{d x_{2}}{d s} \hat{i}_{2}+\frac{d x_{3}}{d s} \hat{i}_{3}=\frac{d x_{1}^{\prime}}{d s} \hat{i}_{1}^{\,\prime}+\frac{d x_{2}^{\prime}}{d s} \hat{i}_{2}^{\,\prime}+\frac{d x_{3}^{\prime}}{d s} \hat{i}_{3}^{\,\prime}.\end{aligned}\] The derivatives of the coordinates in eq. ([eqA.56]) are obtained from eq. ([eqA.41]). These derivatives are \[\begin{aligned} \label{eqA.57} \left[\begin{array}{@{}l@{}}\frac{d x_{1}^{\prime}}{d s} \\[3pt]\frac{d x_{2}^{\prime}}{d s} \\[3pt]\frac{d x_{3}^{\prime}}{d s}\end{array}\right]=\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{12} & \lambda_{13} \\\lambda_{21} & \lambda_{22} & \lambda_{23} \\\lambda_{31} & \lambda_{32} & \lambda_{33}\end{array}\right]\left[\begin{array}{@{}l@{}}\frac{d x_{1}}{d s} \\[3pt]\frac{d x_{2}}{d s} \\[3pt]\frac{d x_{3}}{d s}\end{array}\right].\end{aligned}\] Let \(n_{i}=\frac{d x_{i}}{d s}\) and \(n_{i}^{\prime}=\frac{d x_{1}^{\prime}}{d s}\), \(i=1,2,3\). The matrix form of eq. ([eqA.57]) is \[\begin{aligned} \label{eqA.58} \{n^{\prime}\}=[\lambda]\{n\},\end{aligned}\] where we write the unit vector in matrix notation as \(\{n\}=[n_{1}\ n_{2}\ n_{3}]^{T}\) and \(\{n^{\prime}\}=[n_{1}^{\prime}\ n_{2}^{\prime}\ n_{3}^{\prime}]^{T}\). The inverse of eq. ([eqA.58]) is \[\begin{aligned} \label{eqA.59} \{n\}=[\lambda]^{T}\{n^{\prime}\}.\end{aligned}\]

The strain components \(\varepsilon_{i j}\) are functions in the variables \((x_{1}, x_{2}, x_{3})\), and the strains \(\varepsilon_{i j}^{\prime}\) are functions in the variables \((x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime})\). These strain components are the elements in the matrices \[\begin{aligned} \label{eqA.60} [\varepsilon]=\left[\begin{array}{@{}lll@{}}\varepsilon_{11} & \varepsilon_{12} & \varepsilon_{13} \\\varepsilon_{21} & \varepsilon_{22} & \varepsilon_{23} \\\varepsilon_{31} & \varepsilon_{32} & \varepsilon_{33}\end{array}\right]\mbox{, and } \left[\varepsilon^{\prime}\right]=\left[\begin{array}{@{}lll@{}} \varepsilon_{11}^{\prime} & \varepsilon_{12}^{\prime} & \varepsilon_{13}^{\prime} \\[5pt]\varepsilon_{21}^{\prime} & \varepsilon_{22}^{\prime} & \varepsilon_{23}^{\prime} \\[5pt]\varepsilon_{31}^{\prime} & \varepsilon_{32}^{\prime} & \varepsilon_{33}^{\prime}\end{array}\right].\end{aligned}\] Equations ([eqA.23]) to ([eqA.25]) show that the strain matrix \([\varepsilon]\) is symmetric. The strain of line element \(\overset{\smash{\lower 3pt\hbox{$\frown$}}}{P_{0} Q_{0}}\) must the same, or invariant, if computed in the \(x_{i}\)-system or in the \(x_{i}^{\prime}\)-system. The expression for the strain ([eqA.19]) in matrix notation is \[\begin{aligned} \label{eqA.61} \varepsilon_{L}=\{n\}^{T}[\varepsilon]\{n\}.\end{aligned}\] Substitute eq. ([eqA.59]) for the unit vector \(\{n\}\) into eq. ([eqA.61]) to get \[\begin{aligned} &\varepsilon_{L}=\big([\lambda]^{T}\{n^{\prime}\}\big)^{T}[\varepsilon][\lambda]^{T}\{n^{\prime}\}\mbox{, or}\nonumber\\ &\varepsilon_{L}=\{n^{\prime}\}^{T}[\lambda][\varepsilon][\lambda]^{T}\{n^{\prime}\}.\label{eqA.62}\end{aligned}\] For eq. ([eqA.62]) to be an invariant form of eq. ([eqA.61]) we conclude \[\begin{aligned} \label{eqA.63} [\varepsilon^{\prime}]=[\lambda][\varepsilon][\lambda]^{T}.\end{aligned}\] Hence, \[\begin{aligned} \label{eqA.64} \varepsilon_{L}=\left\{n^{\prime}\right\}^{T}[\varepsilon^{\prime}]\{n^{\prime}\}.\end{aligned}\] Compare the forms of eq. ([eqA.61]) and eq. ([eqA.64]) to note their similarity. Also, the inverse transformation of eq. ([eqA.63]) is \[\begin{aligned} \label{eqA.65} [\varepsilon]=[\lambda]^{T}[\varepsilon^{\prime}][\lambda].\end{aligned}\] The transpose of eq. ([eqA.63]) is \([\lambda][\varepsilon]^{T}[\lambda]^{T}\), but \([\varepsilon]^{T}=[\varepsilon]\), so matrix \([\varepsilon^{\prime}]\) is also symmetric. From eq. ([eqA.34]) the engineering shear strain \(\gamma_{12}=\varepsilon_{12}+\varepsilon_{21}\). But \(\varepsilon_{12}=\varepsilon_{21}\), which means \(\varepsilon_{12}=\varepsilon_{21}=\gamma_{12} / 2\). In terms of engineering shear strains ([eqA.34]), the strain transformation ([eqA.63]) is \[\begin{aligned} \label{eqA.66} \left[\begin{array}{ccc}\varepsilon_{11}^{\prime} & \gamma_{12}^{\prime} / 2 & \gamma_{13}^{\prime} / 2 \\[4pt] \gamma_{12}^{\prime} / 2 & \varepsilon_{22}^{\prime} & \gamma_{23}^{\prime} / 2 \\[4pt] \gamma_{13}^{\prime} / 2 & \gamma_{23}^{\prime} / 2 & \varepsilon_{33}^{\prime}\end{array}\right]=[\lambda]\left[\begin{array}{ccc} \varepsilon_{11} & \gamma_{12} / 2 & \gamma_{13} / 2 \\ \gamma_{12} / 2 & \varepsilon_{22} & \gamma_{23} / 2 \\ \gamma_{13} / 2 & \gamma_{23} / 2 & \varepsilon_{33} \end{array}\right][\lambda]^{T}.\end{aligned}\]

Equilibrium differential equations

Consider the forces acting on a rectangular parallelepiped at point \(P\). The free body diagram is shown in figure [figA.7]. The vector sum of forces is \[\begin{aligned} &\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1} \Delta x_{2} \Delta x_{3}\big|_{x_{1} +\Delta x_{1}}-\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1} \Delta x_{2} \Delta x_{3}\big|_{x_{1}}+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2} \Delta x_{1} \Delta x_{3}\big|_{x_{2}+\Delta x_{2}}-\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2} \Delta x_{1} \Delta x_{3}\big|_{x_{2}}+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3} \Delta x_{1} \Delta x_{2}\big|_{x_{3}+\Delta x_{3}}\\ &\quad-\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3} \Delta x_{1} \Delta x_{2}\big|_{x_{3}}+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{B}}} \Delta x_{1} \Delta x_{2} \Delta x_{3}.\end{aligned}\] For small increments in \(\Delta x_{i}\), \(i=1,2,3\), use the Taylor series representation of surface forces results in the equilibrium equation to get eq. ([eqA.78]) below. \[\begin{aligned} \label{eqA.78} \frac{\partial}{\partial x_{1}}\left(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1} \Delta x_{2} \Delta x_{3}\right) \Delta x_{1}+\frac{\partial}{\partial x_{2}}\left(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2} \Delta x_{1} \Delta x_{3}\right) \Delta x_{2}+\frac{\partial}{\partial x_{3}}\left(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3} \Delta x_{1} \Delta x_{2}\right) \Delta x_{3}+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{B}}} \Delta x_{1} \Delta x_{2} \Delta x_{3}+O\big((\Delta x_{i})^{4}\big)=0.\end{aligned}\] Arrange the terms in eq. ([eqA.78]) to the form \[\begin{aligned} \label{eqA.79} \left(\frac{\partial \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1}}{\partial x_{1}}+\frac{\partial \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2}}{\partial x_{2}}+\frac{\partial \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3}}{\partial x_{3}}+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{B}}}\right) \Delta x_{1} \Delta x_{2} \Delta x_{3}+O\big((\Delta x_{i})^{4}\big)=0\end{aligned}\] Divide eq. ([eqA.79]) by the volume followed by the limit as \(\Delta x_{1} \Delta x_{2} \Delta x_{3} \rightarrow 0\) to get the vector differential equation of force equilibrium at point \(P\) as \[\begin{aligned} \label{eqA.80} \frac{\partial \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1}}{\partial x_{1}}+\frac{\partial \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2}}{\partial x_{2}}+\frac{\partial \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3}}{\partial x_{3}}+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{B}}}=0.\end{aligned}\] Substitute eq. ([eqA.70]) for the traction vectors in eq. ([eqA.80]) to write the equilibrium differential equations in the \(x_{i}\)-coordinate directions. In the order of \(x_{1}\), \(x_{2}\), \(x_{3}\) coordinate directions these equations are \[\begin{aligned} \label{eqA.81} \begin{aligned} &\frac{\partial \sigma_{11}}{\partial x_{1}}+\frac{\partial \sigma_{21}}{\partial x_{2}}+\frac{\partial \sigma_{31}}{\partial x_{3}}+B_{1}=0\\ &\frac{\partial \sigma_{12}}{\partial x_{1}}+\frac{\partial \sigma_{22}}{\partial x_{2}}+\frac{\partial \sigma_{32}}{\partial x_{3}}+B_{2}=0\\ &\frac{\partial \sigma_{13}}{\partial x_{1}}+\frac{\partial \sigma_{23}}{\partial x_{2}}+\frac{\partial \sigma_{33}}{\partial x_{3}}+B_{3}=0\end{aligned}.\end{aligned}\]Now consider moment equilibrium about the coordinate axes of the rectangular parallelepiped at point \(P\). For moment equilibrium about the \(x_{1}\)-axis refer to the free body diagram in figure [figA.8]. The moment arm from point \(P\) to the line of action of the normal force \((\sigma_{22} \Delta x_{1} \Delta x_{3})_{x_{2}+\Delta x_{2}}\) acting on the positive \(x_{2}\)-face is denoted by \(\varepsilon \Delta x_{3}\), where \(\varepsilon\) is a small numerical value. Parameter \(\varepsilon\) is not known, but this will not matter in the end result. The moment arm from point \(P\) to the line of action of the shear force \((\sigma_{23} \Delta x_{1} \Delta x_{3})_{x_{2}+\Delta x_{2}}\) acting on the positive \(x_{2}\) face is \(\Delta x_{2} / 2\). Including all the forces shown in figure [figA.8], the sum of moments about the \(x_{1}\)-axis through point \(P\), counterclockwise positive, is \[\begin{aligned} \label{eqA.82} &-\varepsilon \Delta x_{3}(\sigma_{22} \Delta x_{1} \Delta x_{3})_{x_{2}+\Delta x_{2}}+\varepsilon \Delta x_{3}(\sigma_{22} \Delta x_{1} \Delta x_{3})_{x_{2}}+\frac{\Delta x_{2}}{2}(\sigma_{23} \Delta x_{1} \Delta x_{3})_{x_{2}+\Delta x_{2}}+\frac{\Delta x_{2}}{2}(\sigma_{23} \Delta x_{1} \Delta x_{3})_{x_{2}}\nonumber\\ &\quad+\varepsilon \Delta x_{2}(\sigma_{33} \Delta x_{1} \Delta x_{2})_{x_{3}+\Delta x_{3}}-\varepsilon \Delta_{2}(\sigma_{33} \Delta x_{1} \Delta x_{2})_{x_{3}}-\frac{\Delta x_{3}}{2}(\sigma_{32} \Delta x_{1} \Delta x_{2})_{x_{3}+\Delta x_{3}}-\frac{\Delta x_{3}}{2}(\sigma_{32} \Delta x_{1} \Delta x_{2})_{x_{3}}=0.\end{aligned}\]

Use the Taylor series to expand the forces acting on the positive coordinate faces with respect to the forces acting on the negative coordinate faces to get \[\begin{aligned} \label{eqA.83} &-\varepsilon \Delta x_{3}\left[\frac{\partial \sigma_{22}}{\partial x_{2}} \Delta x_{2}+O(\Delta x_{2}^{2})\right] \Delta x_{1} \Delta x_{3}+\frac{\Delta x_{2}}{2}\left[2 \sigma_{23}+\frac{\partial \sigma_{23}}{\partial x_{2}} \Delta x_{2}+O(\Delta x_{2}^{2})\right] \Delta x_{1} \Delta x_{3}\nonumber\\ &\quad+\varepsilon \Delta x_{2}\left[\frac{\partial \sigma_{33}}{\partial x_{3}} \Delta x_{3}+O(\Delta x_{3}^{2})\right] \Delta x_{1} \Delta x_{2}-\frac{\Delta x_{3}}{2}\left[2 \sigma_{32}+\frac{\partial \sigma_{32}}{\partial x_{3}} \Delta x_{3}+O(\Delta x_{3}^{2})\right] \Delta x_{1} \Delta x_{2}=0.\end{aligned}\] Expand eq. ([eqA.83]) in powers of \(\Delta x_{i}\) to write it as \[\begin{aligned} \label{eqA.84} \left(\sigma_{23}-\sigma_{32}\right) \Delta x_{1} \Delta x_{2} \Delta x_{3}+\varepsilon\left[-\frac{\partial \sigma_{22}}{\partial x_{2}} \Delta x_{1} \Delta x_{2} \Delta x_{3}^{2}+\frac{\partial \sigma_{33}}{\partial x_{3}} \Delta x_{1} \Delta x_{2}^{2} \Delta x_{3}\right]+\text { H.O.T. }=0,\end{aligned}\] where H.O.T. means higher order terms, that is, terms of quartic powers and higher in the increments in the coordinates. Notice the terms multiplied by \(\varepsilon\) are quartic powers of the increments in the coordinates. Division of eq. ([eqA.84]) by \(\Delta x_{1} \Delta x_{2} \Delta x_{3}\), followed by the limit of \(\Delta x_{i} \rightarrow 0\) leads the condition of moment equilibrium about the \(x_{1}\)-axis that \(\sigma_{23}-\sigma_{32}=0\). Moment equilibrium about the \(x_{2}\)-axis leads to \(\sigma_{31}-\sigma_{13}=0\), and moment equilibrium about the \(x_{3}\)-axis leads to \(\sigma_{12}-\sigma_{21}=0\). The equations of moment equilibrium at point \(P\) are \[\begin{aligned} \label{eqA.85} \sigma_{12}=\sigma_{21} \quad \sigma_{13}=\sigma_{31} \quad \sigma_{23}=\sigma_{32}.\end{aligned}\] Hence, the stress matrix ([eqA.71]) is symmetric.

Transformation of stresses between two Cartesian coordinate systems

At point \(P\) coordinates \((x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime})\) are linearly related to coordinates \((x_{1}, x_{2}, x_{3})\) by eq. ([eqA.41]). The stress components \(\sigma_{i j}\) are functions in the variables \((x_{1}, x_{2}, x_{3})\), and the stresses \(\sigma_{i j}^{\prime}\) are functions in the variables\((x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime})\). The stress vectors acting on the \(x_{i}\)-faces are denoted by \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{i}\), and those acting on the \(x_{i}^{\prime}\)-faces are denoted by \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{i}^{\prime}\). These stress vectors are written in their respective coordinate systems by \[\begin{aligned} \label{eqA.86} \left[\begin{array}{@{}c@{}}\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1} \\[4pt] \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2} \\[4pt] \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3}\end{array}\right]=\left[\begin{array}{@{}lll@{}}\sigma_{11} & \sigma_{12} & \sigma_{13} \\\sigma_{21} & \sigma_{22} & \sigma_{23} \\\sigma_{31} & \sigma_{32} & \sigma_{33}\end{array}\right]\left[\begin{array}{@{}l@{}}\hat{i}_{1} \\[4pt] \hat{i}_{2} \\[4pt] \hat{i}_{3}\end{array}\right]\mbox{, and } \left[\begin{array}{@{}c@{}}\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1}^{\prime} \\[4pt] \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2}^{\prime} \\[4pt] \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3}^{\prime}\end{array}\right]= \left[\begin{array}{@{}lll@{}}\sigma_{11}^{\prime} & \sigma_{12}^{\prime} & \sigma_{13}^{\prime} \\[4pt]\sigma_{21}^{\prime} & \sigma_{22}^{\prime} & \sigma_{23}^{\prime} \\[4pt]\sigma_{31}^{\prime} & \sigma_{32}^{\prime} & \sigma_{33}^{\prime}\end{array}\right]\left[\begin{array}{@{}l@{}}\hat{i}_{1}^{\,\prime} \\[4pt] \hat{i}_{2}^{\,\prime} \\[4pt] \hat{i}_{3}^{\,\prime}\end{array}\right]\!.\end{aligned}\] In eq. ([eqA.86]) the stress matrices are \[\begin{aligned} \label{eqA.87} [\sigma]=\left[\begin{array}{@{}lll@{}}\sigma_{11} & \sigma_{12} & \sigma_{13} \\\sigma_{21} & \sigma_{22} & \sigma_{23} \\\sigma_{31} & \sigma_{32} & \sigma_{33}\end{array}\right]\mbox{, and }[\sigma^{\prime}]=\left[\begin{array}{@{}lll@{}}\sigma_{11}^{\prime} & \sigma_{12}^{\prime} & \sigma_{13}^{\prime} \\[4pt]\sigma_{21}^{\prime} & \sigma_{22}^{\prime} & \sigma_{23}^{\prime} \\[4pt]\sigma_{31}^{\prime} & \sigma_{32}^{\prime} & \sigma_{33}^{\prime}\end{array}\right].\end{aligned}\] Note that the stress matrix \([\sigma]\) is symmetric by eq. ([eqA.85]). The stress transformation equations between the Cartesian coordinate systems \((x_{1}, x_{2}, x_{3})\) and \((x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime})\) are determined by selecting the unit normal in eq. ([eqA.77]) to be either \(\hat{i}_1^{\,\prime}\), \(\hat{i}_2^{\,\prime}\), or \(\hat{i}_{3}^{\,\prime}\). First let \(\hat{n}=\hat{i}_{1}^{\,\prime}\) such that \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}^{(n)}=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1}^{\prime}\) in eq. ([eqA.77]). From eq. ([eqA.44]) we have \(\hat{n}=\lambda_{11} \hat{i}_{1}+\lambda_{12} \hat{i}_{2}+\lambda_{13} \hat{i}_{3}\). Hence, eq. ([eqA.77]) becomes \[\begin{aligned} \label{eqA.88} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1}^{\prime}=\lambda_{11} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1}+\lambda_{12} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2}+\lambda_{13} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3}.\end{aligned}\] Second, let \(\hat{n}=\hat{i}_{2}^{\,\prime}\) such that \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}^{(n)}=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2}^{\prime}\). From eq. ([eqA.44]) we have \(\hat{n}=\lambda_{21} \hat{i}_{1}+\lambda_{22} \hat{i}_{2}+\lambda_{23} \hat{i}_{3}\). Hence, eq. ([eqA.77]) becomes \[\begin{aligned} \label{eqA.89} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2}^{\prime}=\lambda_{21} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1}+\lambda_{22} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2}+\lambda_{23} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3}.\end{aligned}\] Third, let \(\hat{n}=\hat{i}_{3}^{\,\prime}\) that \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}^{(n)}=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3}^{\prime}\) and \(\hat{n}=\lambda_{31} \hat{i}_{1}+\lambda_{32} \hat{i}_{2}+\lambda_{33} \hat{i}_{3}\). Hence, \[\begin{aligned} \label{eqA.90} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3}^{\prime}=\lambda_{31} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1}+\lambda_{32} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2}+\lambda_{33} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3}.\end{aligned}\] The three selections for the unit normal in eq. ([eqA.77]) relate the tractions acting on the \(x_{i}^{\prime}\) coordinate faces to the tractions acting on the \(x_{i}\)-faces by \[\begin{aligned} \label{eqA.91} \left[\begin{array}{@{}c@{}}\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1}^{\prime} \\[4pt] \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2}^{\prime} \\[4pt] \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3}^{\prime}\end{array}\right]=\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{12} & \lambda_{13} \\\lambda_{21} & \lambda_{22} & \lambda_{23} \\\lambda_{31} & \lambda_{32} & \lambda_{33}\end{array}\right]\left[\begin{array}{@{}c@{}}\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{1} \\[4pt] \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{2} \\[4pt] \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{T}}}_{3}\end{array}\right].\end{aligned}\] Substitute the expressions for the stress vectors from eq. ([eqA.86]) into eq. ([eqA.91]) to get \[\begin{aligned} \label{eqA.92} \left[\begin{array}{@{}lll@{}} \sigma_{11}^{\prime} & \sigma_{12}^{\prime} & \sigma_{13} \\[5pt] \sigma_{21}^{\prime} & \sigma_{22}^{\prime} & \sigma_{23} \\[5pt] \sigma_{31}^{\prime} & \sigma_{32}^{\prime} & \sigma_{33}\end{array}\right]\left[\begin{array}{@{}l@{}}\hat{i}_{1}^{\,\prime} \\[4pt] \hat{i}_{2}^{\,\prime} \\[4pt] \hat{i}_{3}^{\,\prime}\end{array}\right]=\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{12} & \lambda_{13} \\\lambda_{21} & \lambda_{22} & \lambda_{23} \\\lambda_{31} & \lambda_{32} & \lambda_{33}\end{array}\right]\left[\begin{array}{@{}lll@{}}\sigma_{11} & \sigma_{12} & \sigma_{13} \\\sigma_{21} & \sigma_{22} & \sigma_{23} \\\sigma_{31} & \sigma_{32} & \sigma_{33}\end{array}\right]\left[\begin{array}{@{}l@{}}\hat{i}_{1} \\[4pt] \hat{i}_{2} \\[4pt] \hat{i}_{3}\end{array}\right].\end{aligned}\] The inverse eq. ([eqA.44]) is \[\begin{aligned} \label{eqA.93} \left[\begin{array}{@{}l@{}}\hat{i}_{1} \\[4pt] \hat{i}_{2} \\[4pt] \hat{i}_{3}\end{array}\right]=\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{21} & \lambda_{31} \\\lambda_{12} & \lambda_{22} & \lambda_{32} \\\lambda_{13} & \lambda_{23} & \lambda_{33}\end{array}\right]\left[\begin{array}{@{}l@{}}\hat{i}_{1}^{\,\prime} \\[4pt] \hat{i}_{2}^{\,\prime} \\[4pt] \hat{i}_{3}^{\,\prime}\end{array}\right].\end{aligned}\] Substitute eq. ([eqA.93]) into the right-hand side of eq. ([eqA.92]) and rearrange the result to find \[\begin{aligned} \label{eqA.94} \left(\left[\begin{array}{ccc} \sigma_{11}^{\prime} & \sigma_{12}^{\prime} & \sigma_{13}^{\prime} \\[5pt]\sigma_{21}^{\prime} & \sigma_{22}^{\prime} & \sigma_{23}^{\prime} \\[5pt]\sigma_{31}^{\prime} & \sigma_{32}^{\prime} & \sigma_{33}^{\prime}\end{array}\right]-\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{12} & \lambda_{13} \\\lambda_{21} & \lambda_{22} & \lambda_{23} \\\lambda_{31} & \lambda_{32} & \lambda_{33}\end{array}\right]\left[\begin{array}{@{}lll@{}}\sigma_{11} & \sigma_{12} & \sigma_{13} \\\sigma_{21} & \sigma_{22} & \sigma_{23} \\\sigma_{31} & \sigma_{32} & \sigma_{33}\end{array}\right]\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{21} & \lambda_{31} \\\lambda_{12} & \lambda_{22} & \lambda_{32} \\\lambda_{13} & \lambda_{23} & \lambda_{33}\end{array}\right]\right)\left[\begin{array}{@{}l@{}}\hat{i}_{1}^{\,\prime} \\[4pt] \hat{i}_{2}^{\,\prime} \\[4pt] \hat{i}_{3}^{\,\prime}\end{array}\right]=\left[\begin{array}{@{}l@{}}0 \\0 \\0\end{array}\right].\end{aligned}\] To satisfy eq. ([eqA.94]) we find that the stress components \(\sigma_{i j}^{\prime}\) in the \(x_{i}^{\prime}\)-system are related to the stress components \(\sigma_{i j}\) in the \(x_{i}\)-system by \[\begin{aligned} \label{eqA.95} \left[\begin{array}{@{}lll@{}}\sigma_{11}^{\prime} & \sigma_{12}^{\prime} & \sigma_{13} \\[3pt] \sigma_{21}^{\prime} & \sigma_{22}^{\prime} & \sigma_{23}^{\prime} \\[3pt] \sigma_{31}^{\prime} & \sigma_{32}^{\prime} & \sigma_{33}^{\prime}\end{array}\right]=\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{12} & \lambda_{13} \\\lambda_{21} & \lambda_{22} & \lambda_{23} \\\lambda_{31} & \lambda_{32} & \lambda_{33}\end{array}\right]\left[\begin{array}{@{}lll@{}}\sigma_{11} & \sigma_{12} & \sigma_{13} \\\sigma_{21} & \sigma_{22} & \sigma_{23} \\\sigma_{31} & \sigma_{32} & \sigma_{33}\end{array}\right]\left[\begin{array}{@{}lll@{}}\lambda_{11} & \lambda_{21} & \lambda_{31} \\\lambda_{12} & \lambda_{22} & \lambda_{32} \\\lambda_{13} & \lambda_{23} & \lambda_{33}\end{array}\right].\end{aligned}\] Equation ([eqA.95]) in compact form is \[\begin{aligned} \label{eqA.96} [\sigma^{\prime}]=[\lambda][\sigma][\lambda]^{T}.\end{aligned}\] Pre-multiply eq. ([eqA.96]) by \([\lambda]^{T}\), post-multiply it by \([\lambda]\), and note that \([\lambda]^{T}[\lambda]=[\lambda][\lambda]^{T}=[I]\) to find the inverse transformation \[\begin{aligned} \label{eqA.97} [\sigma]=[\lambda]^{T}[\sigma^{\prime}][\lambda].\end{aligned}\] The transpose of eq. ([eqA.96]) is \([\lambda][\sigma]^{T}[\lambda]^{T}\), but \([\sigma]^{T}=[\sigma]\), so the stress matrix \([\sigma^{\prime}]\) is also symmetric. Comparing the strain transformation eq. ([eqA.63]) to the stress transformation eq. ([eqA.96]), it is clear that the transformation of strains \(\varepsilon_{ij}\) is the same form as the transformation of the stresses \(\sigma_{ij}\).

Cartesian tensors

A tensor is a system of numbers or functions, whose components obey a certain law of transformation when the independent variables undergo a linear transformation. If the independent variables are the rectangular cartesian systems \(x_{i}\) and \(x_{i}^{\prime}\) transforming by the linear relations given by \(\{x^{\prime}\}=[\lambda]\{x\}\) at point \(P\), then the systems obeying certain laws of transformation are called Cartesian tensors.

Definition.A system of order two may be defined to have nine components \(\varepsilon_{i j}\) in \(x_{i}\) and nine components \(\varepsilon_{ij}^{\prime}\) in \(x_{i}^{\prime}\). If \[\begin{aligned} \label{eqA.98} [\varepsilon_{i j}^{\prime}]=[\lambda][\varepsilon_{i j}][\lambda]^{T}\end{aligned}\] then the functions \(\varepsilon_{ij}\) and \(\varepsilon_{ij}^{\prime}\) are the components in their respective variables of a second order Cartesian tensor. Similarly, functions \(\sigma_{ij}\) and \(\sigma_{ij}^{\prime}\) are the components in their respective variables of a second order Cartesian tensor.

First law of thermodynamics

The theoretical basis for an elastic material law is the first law of thermodynamics applied to an arbitrary infinitesimal rectangular parallelepiped isolated from the body. We assume the deformation process is adiabatic. That is, no heat is lost or gained in the body during the deformation. An alternate definition of elastic behavior is that the work expended in the transition from the reference state to the final deformed state is independent of the manner in which the process proceeds. The first law of thermodynamics states that work done on the rectangular parallelepiped is equal to the change in internal energy of the material contained in the parallelepiped.¹ In elasticity the internal energy is called the strain energy. The incremental work of the tractions and body force acting on the parallelepiped are formulated in terms of incremental displacements from the equilibrium state. These incremental displacement functions are denoted by \(\delta u_{i}(x_{1}, x_{2}, x_{3})\), \(i=1,2,3\), and to be kinematically admissible they are continuous and single-valued in the independent variables. In addition, functions \(\delta u_{i}\) are assumed to be infinitesimal in magnitude. The total displacement at point \(P\) is \(\tilde{u}_{i}=u_{i}+\delta u_{i}\), where \(u_{i}\) are the displacements components in the equilibrium state.

R158.14pt

Material symmetry

Consider a monoclinic material for which the \(x_{1} x_{2}\) plane at a point \(P\) is a plane of elastic symmetry. This means that the elastic constants at point \(P\) have the same values for a pair of Cartesian coordinate systems that are mirror images of one another in the elastic plane. The elastic constants \(S_{ij}\) are invariant under the reflection coordinate transformation \(x_{1}^{\prime}=x_{1}\), \(x_{2}^{\prime}=x_{2}\), and \(x_{3}^{\prime}=-x_{3}\). The direction cosines matrix ([eqA.41]) for this reflection transformation is \[\begin{aligned} \label{eqA.120} [\lambda]=\left[\begin{array}{ccc}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & -1\end{array}\right].\end{aligned}\] Consider the material law for \(\sigma_{1}\) in the \(x_{i}\)-system from eq. ([eqA.119]). It is \[\begin{aligned} \label{eqA.121} \sigma_{1}=-\beta_{1}(T-T_{0})+S_{11} \varepsilon_{1}+S_{12} \varepsilon_{2}+S_{13} \varepsilon_{3}+S_{14} \varepsilon_{4}+S_{15} \varepsilon_{5}+S_{16} \varepsilon_{6}.\end{aligned}\] The material law for \(\sigma_{1}^{\prime}\) in the \(x_{i}^{\prime}\)-system is written as \[\begin{aligned} \label{eqA.122} \sigma_{1}^{\prime}=-\beta_{1}(T-T_{0})+S_{11} \varepsilon_{1}^{\prime}+S_{12} \varepsilon_{2}^{\prime}+S_{13} \varepsilon_{3}^{\prime}+S_{14} \varepsilon_{4}^{\prime}+S_{15} \varepsilon_{5}^{\prime}+S_{16} \varepsilon_{6}^{\prime}.\end{aligned}\] From strain and stress transformations in eq. ([eqA.66]) and eq. ([eqA.95]), respectively, we find \[\begin{aligned} \label{eqA.123} \begin{gathered} \sigma_{i}^{\prime}=\sigma_{i} \quad \varepsilon_{i}^{\prime}=\varepsilon_{i} \quad i=1,2,3,6 \\ \sigma_{4}^{\prime}=-\sigma_{4} \quad \sigma_{5}^{\prime}=-\sigma_{5} \quad \varepsilon_{4}^{\prime}=-\varepsilon_{4} \quad \varepsilon_{5}^{\prime}=-\varepsilon_{5} \end{gathered}.\end{aligned}\] Substitute the stress and strain transformation relations ([eqA.123]) into eq. ([eqA.122]) to find \[\begin{aligned} \label{eqA.124} \sigma_{1}=-\beta_{1}(T-T_{0})+S_{11} \varepsilon_{1}+S_{12} \varepsilon_{2}+S_{13} \varepsilon_{3}-S_{14} \varepsilon_{4}-S_{15} \varepsilon_{5}+S_{16} \sigma_{6}.\end{aligned}\] Comparison of eq. ([eqA.124]) to eq. ([eqA.121]) requires that \(S_{14}=0\) and \(S_{15}=0\) if the material law for \(\sigma_{1}\) is to be the same with respect to the plane of elastic symmetry. Constructing the material law for \(\sigma_{2}^{\prime}\) and following a similar procedure used for the \(\sigma_{1}^{\prime}\) material law leads to \(S_{24}=0\) and \(S_{25}=0\). Constructing for the material law for \(\sigma_{3}^{\prime}\) leads to \(S_{34}=0\) and \(S_{35}=0\), constructing the material law for \(\sigma_{4}^{\prime}\) leads to \(\beta_{4}=0\) and \(S_{46}=0\), and constructing the material law for \(\sigma_{5}^{\prime}\) leads to \(\beta_{5}=0\) and \(S_{56}=0\). The material law for the \(x_{1} x_{2}\) plane of elastic symmetry is \[\begin{aligned} \label{eqA.125} \left[\begin{array}{@{}l@{}}\sigma_{1} \\\sigma_{2} \\\sigma_{3} \\\sigma_{4} \\\sigma_{5} \\\sigma_{6}\end{array}\right]=-\left[\begin{array}{@{}l@{}}\beta_{1} \\\beta_{2} \\\beta_{3} \\0 \\0 \\\beta_{6}\end{array}\right](T-T_{0})+\left[\begin{array}{@{}cccccc@{}}S_{11} & S_{12} & S_{13} & 0 & 0 & S_{16} \\S_{21} & S_{22} & S_{23} & 0 & 0 & S_{26} \\S_{31} & S_{32} & S_{33} & 0 & 0 & S_{36} \\0 & 0 & 0 & S_{44} & S_{45} & 0 \\0 & 0 & 0 & S_{54} & S_{55} & 0 \\S_{61} & S_{62} & S_{63} & 0 & 0 & S_{66}\end{array}\right]\left[\begin{array}{@{}l@{}}\varepsilon_{1} \\\varepsilon_{2} \\\varepsilon_{3} \\\varepsilon_{4} \\\varepsilon_{5} \\\varepsilon_{6}\end{array}\right].\end{aligned}\] There are thirteen independent elastic constants \(S_{i j}\), and four independent thermal moduli. Equation ([eqA.125]) is the material law for a monoclinic material.

Certain elastic constants in eq. ([eqA.125]) vanish if in addition to the \(x_{1} x_{2}\) plane of elastic symmetry the \(x_{2} x_{3}\) plane is a plane of elastic symmetry. The reflection coordinate transformation is \(x_{1}^{\prime}=-x_{1}\), \(x_{2}^{\prime}=x_{2}\), and \(x_{3}^{\prime}=x_{3}\). The direction cosines matrix ([eqA.41]) for this reflection transformation is \[\begin{aligned} \label{eqA.126} [\lambda]=\left[\begin{array}{ccc}-1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right].\end{aligned}\] Consider the material law for \(\sigma_{1}\) in the \(x_{i}\)-system from eq. ([eqA.125]). It is \[\begin{aligned} \label{eqA.127} \sigma_{1}=-\beta_{1}(T-T_{0})+S_{11} \varepsilon_{1}+S_{12} \varepsilon_{2}+S_{13} \varepsilon_{3}+S_{16} \varepsilon_{6}.\end{aligned}\] The material law for \(\sigma_{1}^{\prime}\) in the \(x_{i}^{\prime}\)-system is written as \[\begin{aligned} \label{eqA.128} \sigma_{1}^{\prime}=-\beta_{1}(T-T_{0})+S_{11} \varepsilon_{1}^{\prime}+S_{12} \varepsilon_{2}^{\prime}+S_{13} \varepsilon_{3}^{\prime}+S_{16} \varepsilon_{6}^{\prime}.\end{aligned}\] From strain and stress transformations in eq. ([eqA.66]) and eq. ([eqA.95]), respectively, we find \[\begin{aligned} \label{eqA.129} \begin{gathered} \sigma_{i}^{\prime}=\sigma_{1} \quad \varepsilon_{i}^{\prime}=\varepsilon_{i} \quad i=1,2,3,4 \\ \sigma_{5}^{\prime}=-\sigma_{5} \quad \sigma_{6}^{\prime}=-\sigma_{6} \quad \varepsilon_{5}^{\prime}=-\varepsilon_{5} \quad \varepsilon_{6}^{\prime}=-\varepsilon_{6} \end{gathered}.\end{aligned}\] Substitute the transformations for the stress and strain from eq. ([eqA.129]) into eq. ([eqA.128]) to get \[\begin{aligned} \label{eqA.130} \sigma_{1}=-\beta_{1}(T-T_{0})+S_{11} \varepsilon_{1}+S_{12} \varepsilon_{2}+S_{13} \varepsilon_{3}-S_{16} \varepsilon_{6}.\end{aligned}\]

Comparison of eq. ([eqA.130]) and eq. ([eqA.127]) leads to \(S_{16}=0\). Also, following the same procedure for the equation starting with \(\sigma_{2}^{\prime}\) leads to \(S_{26}=0\), and starting with the equation for \(\sigma_{3}^{\prime}\) leads to \(S_{36}=0\). The material law for \(\sigma_{4}^{\prime}=S_{44} \varepsilon_{4}^{\prime}+S_{45} \varepsilon_{5}^{\prime}\) transforms by eq. ([eqA.129]) to \(\sigma_{4}=S_{44} \varepsilon_{4}-S_{45} \varepsilon_{5}\). Hence \(S_{45}=0\). Finally, the material law for \(\sigma_{6}^{\prime}=-\beta_{6}(T-T_{0})+S_{66} \varepsilon_{6}^{\prime}\) transforms to \(-\sigma_{6}=-\beta_{6}(T-T_{0})- {S}_{66} \varepsilon_{6}\). Hence \(\beta_{6}=0\). The material law for two orthogonal elastic planes of symmetry is \[\begin{aligned} \label{eqA.131} \left[\begin{array}{@{}l@{}}\sigma_{1} \\\sigma_{2} \\\sigma_{3} \\\sigma_{4} \\\sigma_{5} \\\sigma_{6}\end{array}\right]=-\left[\begin{array}{@{}l@{}}\beta_{1} \\\beta_{2} \\\beta_{3} \\0 \\0 \\0\end{array}\right](T-T_{0})+\left[\begin{array}{@{}cccccc@{}}S_{11} & S_{12} & S_{13} & 0 & 0 & 0 \\S_{21} & S_{22} & S_{23} & 0 & 0 & 0 \\S_{31} & S_{32} & S_{33} & 0 & 0 & 0 \\0 & 0 & 0 & S_{44} & 0 & 0 \\0 & 0 & 0 & 0 & S_{55} & 0 \\0 & 0 & 0 & 0 & 0 & S_{66}\end{array}\right]\left[\begin{array}{@{}l@{}}\varepsilon_{1} \\\varepsilon_{2} \\\varepsilon_{3} \\\varepsilon_{4} \\\varepsilon_{5} \\\varepsilon_{6}\end{array}\right].\end{aligned}\] If we additonally impose that the \(x_{3} x_{1}\) plane is a plane of elastic symmetry, this reflection transformation does not change the results given in eq. ([eqA.131]). An orthotropic material has three mutually orthogonal planes of elastic symmetry, nine independent elastic constants \(S_{i j}\), and three independent thermal moduli. Equation ([eqA.131]) is the material law for an orthotropic material (e.g., wood).

The material properties are independent of direction for an isotropic material. Starting with the orthotropic material law ([eqA.131]) consider the following sequence of rotations from the \((x_{1}, x_{2}, x_{3})\) coordinates to the \((x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime})\) coordinates.

1. \(90^{\circ}\) rotation about the \(x_{1}\)-axis. The direction cosine matrix \([\lambda]=\left[\begin{array}{@{}ccc@{}}1 & 0 & 0 \\0 & 0 & 1 \\0 & -1 & 0\end{array}\right]\).

2. \(90^{\circ}\) rotation about the \(x_{3}\)-axis. The direction cosine matrix \([\lambda]=\left[\begin{array}{@{}ccc@{}}0 & 1 & 0 \\-1 & 0 & 0 \\0 & 0 & 1\end{array}\right]\).

3. \(45^{\circ}\) rotation about the \(x_{3}\)-axis. The direction cosine matrix \([\lambda]=\left[\begin{array}{@{}ccc@{}}1 / \sqrt{2} & 1 / \sqrt{2} & 0 \\-1 / \sqrt{2} & 1 / \sqrt{2} & 0 \\0 & 0 & 1\end{array}\right]\).

For the material law to be invariant from the first rotation we find \[\begin{aligned} \label{eqA.132} S_{12}=S_{13} \quad \beta_{3}=\beta_{2} \quad S_{33}=S_{22} \quad S_{55}=S_{66}.\end{aligned}\] For the material law to be invariant from the second rotation we find \[\begin{aligned} \label{eqA.133} \beta_{2}=\beta_{1} \quad S_{12}=S_{23} \quad S_{11}=S_{22} \quad S_{44}=S_{55}.\end{aligned}\] For the third rotation, the material law for \(\sigma_{1}^{\prime}\) in the \((x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime})\) coordinates is \[\begin{aligned} \label{eqA.134} \sigma_{1}^{\prime}=-\beta_{1}(T-T_{0})+S_{11} \varepsilon_{1}^{\prime}+S_{12} \varepsilon_{2}^{\prime}+S_{12} \varepsilon_{3}^{\prime},\end{aligned}\] since \(S_{13}=S_{12}\). The stress and the strain transformation relations are \[\begin{aligned} \label{eqA.135} \sigma_{1}^{\prime}=(\sigma_{1}+\sigma_{2}) / 2+\sigma_{6} \quad \varepsilon_{1}^{\prime}=\left(\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{6}\right) / 2 \quad \varepsilon_{2}^{\prime}=\left(\varepsilon_{1}+\varepsilon_{2}-\varepsilon_{6}\right) / 2 \quad \varepsilon_{3}^{\prime}=\varepsilon_{3}\end{aligned}\] Substitute the transformations in eq. ([eqA.135]) into eq. ([eqA.134]) to get \[\begin{aligned} \label{eqA.136} (\sigma_{1}+\sigma_{2}) / 2+\sigma_{6}=-\beta_{1}(T-T_{0})+\frac{1}{2}(S_{11}+S_{12}) \varepsilon_{1}+\frac{1}{2}(S_{11}+S_{12}) \varepsilon_{2}+S_{12} \varepsilon_{3}+\frac{1}{2}(S_{11}-S_{12}) \varepsilon_{6}.\end{aligned}\] In the \((x_{1}, x_{2}, x_{3})\) coordinates the formulation of the quantity \((\sigma_{1}+\sigma_{2}) / 2+\sigma_{6}\) is \[\begin{aligned} \label{eqA.137} (\sigma_{1}+\sigma_{2}) / 2+\sigma_{6}=-\beta_{1}(T-T_{0})+\frac{1}{2}(S_{11}+S_{12}) \varepsilon_{1}+\frac{1}{2}(S_{12}+S_{11}) \varepsilon_{2}+S_{12} \varepsilon_{3}+S_{44} \varepsilon_{6}.\end{aligned}\] For eqs. ([eqA.136]) and ([eqA.137]) to be identical we find \[\begin{aligned} \label{eqA.138} \frac{1}{2}(S_{11}-S_{12})=S_{44}.\end{aligned}\] For an isotropic material there are two independent elastic constants and one thermal modulus. Let \(S_{12}=\lambda\) and \(S_{44}=G\), where \(\lambda\) and \(G\) are called Lame’s elastic constants. From eq. ([eqA.138]) \(S_{11}=\lambda+2 G\). The isotropic material law is \[\begin{aligned} \label{eqA.139} \left[\begin{array}{@{}l@{}}\sigma_{1} \\\sigma_{2} \\\sigma_{3} \\\sigma_{4} \\\sigma_{5} \\\sigma_{6}\end{array}\right]=-\left[\begin{array}{@{}l@{}}\beta \\\beta \\\beta \\0 \\0 \\0\end{array}\right](T-T_{0})+\left[\begin{array}{@{}cccccc@{}}\lambda+2 G & \lambda & \lambda & 0 & 0 & 0 \\\lambda & \lambda+2 G & \lambda & 0 & 0 & 0 \\\lambda & \lambda & \lambda+2 G & 0 & 0 & 0 \\0 & 0 & 0 & G & 0 & 0 \\0 & 0 & 0 & 0 & G & 0 \\0 & 0 & 0 & 0 & 0 & G\end{array}\right]\left[\begin{array}{@{}l@{}}\varepsilon_{1} \\\varepsilon_{2} \\\varepsilon_{3} \\\varepsilon_{5} \\\varepsilon_{6}\end{array}\right].\end{aligned}\] Boresi (1965) gives the strain energy density for an isotropic material as \[\begin{aligned} \label{eqA.140} U=\lambda\left(\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3}\right)^{2} / 2+G\left(\varepsilon_{1}^{2}+\varepsilon_{2}^{2}+\varepsilon_{3}^{2}+\frac{\varepsilon_{4}^{2}}{2}+\frac{\varepsilon_{5}^{2}}{2}+\frac{\varepsilon_{6}^{2}}{2}\right)-\beta\left(\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3}\right)(T-T_{0}).\end{aligned}\] The isotropic material law ([eqA.139]) is retrieved from eq. ([eqA.140]) using the relation in eq. ([eqA.114]). The strain-stress form of eq. ([eqA.139]) is \[\begin{aligned} \label{eqA.141} \left[\begin{array}{@{}l@{}}\varepsilon_{1} \\\varepsilon_{2} \\\varepsilon_{3} \\\varepsilon_{5} \\\varepsilon_{6}\end{array}\right]=\left[\begin{array}{@{}cccccc@{}}C_{11} & C_{12} & C_{12} & 0 & 0 & 0 \\C_{12} & C_{11} & C_{12} & 0 & 0 & 0 \\C_{12} & C_{12} & C_{11} & 0 & 0 & 0 \\0 & 0 & 0 & G^{-1} & 0 & 0 \\0 & 0 & 0 & 0 & G^{-1} & 0 \\0 & 0 & 0 & 0 & 0 & G^{-1}\end{array}\right]\left[\begin{array}{@{}l@{}}\sigma_{1} \\\sigma_{2} \\\sigma_{3} \\\sigma_{4} \\\sigma_{5} \\\sigma_{6}\end{array}\right]+\left[\begin{array}{@{}c@{}}C_{11}+C_{12}+C_{12} \\C_{12}+C_{11}+C_{12} \\C_{12}+C_{12}+C_{11} \\0 \\0 \\0\end{array}\right] \beta(T-T_{0}),\end{aligned}\] where \[\begin{aligned} \label{eqA.142} C_{11}=\frac{\lambda+G}{3 \lambda G+2 G^{2}} \quad C_{12}=\frac{-\lambda}{2(3 \lambda G+2 G^{2})}.\end{aligned}\] Consider a uniaxial tension test conducted on a circular cylindrical bar made of an isotropic, homogenous material at the reference state temperature. The test apparatus is configured such that the applied axial normal force divided by the cross-sectional area of the bar is equal to the normal stress \(\sigma_{1}\), and the remaining stresses \(\sigma_{2}=\sigma_{3}=\sigma_{4}=\sigma_{5}=\sigma_{6}=0\). The normal strains \((\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3})\) are monitored and plotted with respect to the applied normal stress \(\sigma_{1}\). In the linear elastic range of the test data the following relationships are established: \(\varepsilon_{1}=\sigma_{1} / E\), \(\varepsilon_{2}=\varepsilon_{3}=-\nu \varepsilon_{1}=-\nu(\sigma_{1} / E)\), where \(E\) denotes Young’s modulus, or the modulus of elasticity, and \(\nu\) denotes Poisson’s ratio. The tension test results correspond to the first column of the 6X6 compliance matrix ([eqA.141]). Hence, \(C_{11}=1 / E\) and \(C_{12}=-v / E\). Substitute the values for \(C_{11}\) and \(C_{12}\) from the tensile test into eq. ([eqA.142]) to get \[\begin{aligned} \label{eqA.143} \frac{\lambda+G}{3 \lambda G+2 G^{2}}=\frac{1}{E} \quad \frac{-\lambda}{2\left(3 \lambda G+2 G^{2}\right)}=-\frac{\nu}{E}.\end{aligned}\] Solve eq. ([eqA.143]) to get Young’s modulus and Poisson’s ratio in terms of \(\lambda\) and \(G\): \[\begin{aligned} \label{eqA.144} E=\frac{3 \lambda G+2 G^{2}}{\lambda+G} \quad \nu=\frac{\lambda}{2(\lambda+G)}.\end{aligned}\] It can be shown from the two expressions in eq. ([eqA.144]) that the Lame’s elastic constants in terms of \(E\) and \(v\) are \[\begin{aligned} \label{eqA.145} G=\frac{E}{2(1+\nu)} \quad \lambda=\frac{\nu E}{(1-2 \nu)(1+\nu)}.\end{aligned}\] Also note that \((C_{11}+C_{12}+C_{12}) \beta=[(1-2 \nu) / E] \beta=\alpha\), where \(\alpha\) is the linear coefficient of thermal expansion. In shear tests of an isotropic, homogeneous material, Lame’s elastic constant \(G\) is called the shear modulus of the material. In terms of engineering constants the material law for a homogenous and isotropic material is \[\begin{aligned} \label{eqA.146} \begin{aligned} &\varepsilon_{11}=\frac{1}{E}[\sigma_{11}-v \sigma_{22}-v \sigma_{33}]+\alpha(T-T_{0}) \\ &\varepsilon_{22}=\frac{1}{E}[-v \sigma_{11}+\sigma_{22}-v \sigma_{33}]+\alpha(T-T_{0}) \\ &\varepsilon_{33}=\frac{1}{E}[-v \sigma_{11}-v \sigma_{22}+\sigma_{33}]+\alpha(T-T_{0}) \end{aligned}\mbox{, and } \begin{aligned} &\gamma_{23}=\sigma_{23} / G \\ &\gamma_{31}=\sigma_{31} / G \\ &\gamma_{12}=\sigma_{12} / G \end{aligned}.\end{aligned}\] The strain energy for an isotropic material given by eq. ([eqA.140]) is positive for any state of strain if \[\begin{aligned} \label{eqA.147} \frac{E}{1+\nu}>0 \quad \text{and} \quad \frac{E}{1-2 \nu}>0.\end{aligned}\] To satisfy eq. ([eqA.147]) the range of Poisson’s ratio is \(-1<\nu<1 / 2\), and modulus \(E>0\). However, materials with a negative Poisson’s ratio are theoretically possible but they are rare.²