7 Appendix

7.1 Derivation of the deformation gradient tensor \(\tilde {F}\) in normal Cartesian coordinates system

In what follows the expression for the deformation gradient tensor \(\mathbf {\tilde {F}}\) is derived. This tensor transform one vector into another vector.

For simplicity it is assumed that the deformed and the undeformed states are described using the same coordinates system. In addition, it is assumed that this coordinates system is the normal Cartesian system with basis vectors \(\mathbf {i,j,k}\). Later these expression will be written in the more general case where the coordinate systems are different and use curvilinear coordinate. Other than using different notation, the derivation is the same in both cases.

Considering a point \(P\) in the undeformed state. This point will have coordinates \(\left ( X_{1},X_{2},X_{3}\right )\). When the body undergoes deformation, this point will be displaced to a new location. The image of this point in the deformed state is called the point \(p\). The coordinates of the the point \(p\) is \(\left ( x_{1},x_{2},x_{3}\right )\).

The coordinates \(x_{i}\) is function of the coordinates \(X_{j}\). These functions constitute the mapping between the undeformed shape and the deformed shape. These functions can be written in general as

\begin{align*} x_{1} & =f_{1}\left ( X_{1},X_{2},X_{3}\right ) \\ x_{2} & =f_{2}\left ( X_{1},X_{2},X_{3}\right ) \\ x_{3} & =f3\left ( X_{1},X_{2},X_{3}\right ) \end{align*}

Therefore by knowing the functions \(f_{i}\) the position of any point in the deformed state can be located if its position in the undeformed state is known. It is more customary to write the function \(f_{i}\) using the name of the coordinate itself. For example writing \(x_{1}=x_{1}\left ( X_{1},X_{2},X_{3}\right ) \) instead of \(x_{1}=f_{1}\left ( X_{1},X_{2},X_{3}\right ) \) as was done above.

However this can be a little confusing since it uses the letter \(x_{i}\) as function when on the RHS and a variable on the LHS. Hence here the choice was to use a new name for the mapping function.

From the above we the expression for a differential change in each of the 3 coordinates using the differentiation chain rule is determined as follows

\begin{align} dx_{1} & =\frac {\partial f_{1}}{\partial X_{1}}dX_{1}+\frac {\partial f_{1}}{\partial X_{2}}dX_{2}+\frac {\partial f_{1}}{\partial X_{3}}dX_{3}\nonumber \\ dx_{2} & =\frac {\partial f_{2}}{\partial X_{1}}dX_{1}+\frac {\partial f_{2}}{\partial X_{2}}dX_{2}+\frac {\partial f_{2}}{\partial X_{3}}dX_{3}\nonumber \\ dx_{3} & =\frac {\partial f_{3}}{\partial X_{1}}dX_{1}+\frac {\partial f_{3}}{\partial X_{2}}dX_{2}+\frac {\partial f_{3}}{\partial X_{3}}dX_{3} \tag {1}\end{align}

Considering now a differential vector element \(\mathbf {dr}\) in the deformed state. This vector can be written as

\begin{equation} \mathbf {dr}=\mathbf {i\ }dx_{1}+\mathbf {j\ }dx_{2}+\mathbf {k\ }dx_{3} \tag {2}\end{equation}

Combining equations (1) and (2) gives

\begin{align*} \mathbf {dr} & =\mathbf {i\ }\left ( \frac {\partial f_{1}}{\partial X_{1}}dX_{1}+\frac {\partial f_{1}}{\partial X_{2}}dX_{2}+\frac {\partial f_{1}}{\partial X_{3}}dX_{3}\right ) \\ & +\mathbf {j\ }\left ( \frac {\partial f_{2}}{\partial X_{1}}dX_{1}+\frac {\partial f_{2}}{\partial X_{2}}dX_{2}+\frac {\partial f_{2}}{\partial X_{3}}dX_{3}\right ) \\ & +\mathbf {k\ }\left ( \frac {\partial f_{3}}{\partial X_{1}}dX_{1}+\frac {\partial f_{3}}{\partial X_{2}}dX_{2}+\frac {\partial f_{3}}{\partial X_{3}}dX_{3}\right ) \end{align*}

The above equation can be written in matrix form as follows

\begin{equation}\begin {pmatrix} dx_{1}\\ dx_{2}\\ dx_{3}\end {pmatrix} =\begin {pmatrix} \frac {\partial f_{1}}{\partial X_{1}} & \frac {\partial f_{1}}{\partial X_{2}} & \frac {\partial f_{1}}{\partial X_{3}}\\ \frac {\partial f_{2}}{\partial X_{1}} & \frac {\partial f_{2}}{\partial X_{2}} & \frac {\partial f_{2}}{\partial X_{3}}\\ \frac {\partial f_{3}}{\partial X_{1}} & \frac {\partial f_{3}}{\partial X_{2}} & \frac {\partial f_{3}}{\partial X_{3}}\end {pmatrix}\begin {pmatrix} dX_{1}\\ dX_{2}\\ dX_{3}\end {pmatrix} \tag {3}\end{equation}

It is seen that the components of \(\mathbf {dr}\) can be obtained from the components \(\mathbf {dR}\) by pre-multiplying the components of \(\mathbf {dR}\) by the above \(3\times 3\) matrix. Hence this matrix acts as a transformation rule which maps one vector to another, it is a second order tensor, which is called the deformation gradient tensor \(\mathbf {\tilde {F}}\)

\begin{equation} \mathbf {dr=\tilde {F}\cdot dR} \tag {4}\end{equation}

This relation can be written also in dyadic form as follows

\begin{multline} \mathbf {i\ }dx_{1}+\mathbf {j\ }dx_{2}+\mathbf {k\ }dx_{3}=\nonumber \\ \left ( \mathbf {ii}\frac {\partial f_{1}}{\partial X_{1}}+\mathbf {ij}\frac {\partial f_{1}}{\partial X_{2}}+\mathbf {ik}\frac {\partial f_{1}}{\partial X_{3}}+\mathbf {ji}\frac {\partial f_{2}}{\partial X_{1}}+\mathbf {jj}\frac {\partial f_{2}}{\partial X_{2}}+\mathbf {jk}\frac {\partial f_{2}}{\partial X_{3}}+\mathbf {ki}\frac {\partial f_{3}}{\partial X_{1}}+\mathbf {kj}\frac {\partial f_{3}}{\partial X_{2}}+\mathbf {kk}\frac {\partial f_{3}}{\partial X_{3}}\right ) \ \nonumber \\ \mathbf {\cdot }\left ( \mathbf {i\ }dX_{1}+\mathbf {j\ }dX_{2}+\mathbf {k\ }dX_{3}\right ) \tag {5}\end{multline}

To carry the multiplication on the RHS in the above equation, the normal dot product convention is followed using the following rules. \begin{align*} \mathbf {ii\cdot i} & \mathbf {=i}\left ( \mathbf {i\cdot i}\right ) =1\\ \mathbf {ij\cdot i} & \mathbf {=i}\left ( \mathbf {j\cdot i}\right ) =0\\ \mathbf {ik\cdot i} & \mathbf {=i}\left ( \mathbf {k\cdot i}\right ) =0\\ \mathbf {ji\cdot i} & \mathbf {=j}\left ( \mathbf {i\cdot i}\right ) =1\\ & etc\cdots \end{align*}

Performing the multiplication gives

\begin{multline*} \mathbf {i\ }dx_{1}+\mathbf {j\ }dx_{2}+\mathbf {k\ }dx_{3}=\\ \left ( \mathbf {ii}\frac {\partial f_{1}}{\partial X_{1}}+\mathbf {ij}\frac {\partial f_{1}}{\partial X_{2}}+\mathbf {ik}\frac {\partial f_{1}}{\partial X_{3}}+\mathbf {ji}\frac {\partial f_{2}}{\partial X_{1}}+\mathbf {jj}\frac {\partial f_{2}}{\partial X_{2}}+\mathbf {jk}\frac {\partial f_{2}}{\partial X_{3}}+\mathbf {ki}\frac {\partial f_{3}}{\partial X_{1}}+\mathbf {kj}\frac {\partial f_{3}}{\partial X_{2}}+\mathbf {kk}\frac {\partial f_{3}}{\partial X_{3}}\right ) \cdot \mathbf {i\ }dX_{1}\\ +\left ( \mathbf {ii}\frac {\partial f_{1}}{\partial X_{1}}+\mathbf {ij}\frac {\partial f_{1}}{\partial X_{2}}+\mathbf {ik}\frac {\partial f_{1}}{\partial X_{3}}+\mathbf {ji}\frac {\partial f_{2}}{\partial X_{1}}+\mathbf {jj}\frac {\partial f_{2}}{\partial X_{2}}+\mathbf {jk}\frac {\partial f_{2}}{\partial X_{3}}+\mathbf {ki}\frac {\partial f_{3}}{\partial X_{1}}+\mathbf {kj}\frac {\partial f_{3}}{\partial X_{2}}+\mathbf {kk}\frac {\partial f_{3}}{\partial X_{3}}\right ) \cdot \mathbf {j\ }dX_{2}\\ +\left ( \mathbf {ii}\frac {\partial f_{1}}{\partial X_{1}}+\mathbf {ij}\frac {\partial f_{1}}{\partial X_{2}}+\mathbf {ik}\frac {\partial f_{1}}{\partial X_{3}}+\mathbf {ji}\frac {\partial f_{2}}{\partial X_{1}}+\mathbf {jj}\frac {\partial f_{2}}{\partial X_{2}}+\mathbf {jk}\frac {\partial f_{2}}{\partial X_{3}}+\mathbf {ki}\frac {\partial f_{3}}{\partial X_{1}}+\mathbf {kj}\frac {\partial f_{3}}{\partial X_{2}}+\mathbf {kk}\frac {\partial f_{3}}{\partial X_{3}}\right ) \cdot \mathbf {k\ }dX_{3}\end{multline*}

The dot multiplication is simplified using the above mentioned rules to obtain

\begin{multline*} \mathbf {i\ }dx_{1}+\mathbf {j\ }dx_{2}+\mathbf {k\ }dx_{3}=\\ \left ( \mathbf {i}\frac {\partial f_{1}}{\partial X_{1}}dX_{1}+\mathbf {0}+\mathbf {0}+\mathbf {j}\frac {\partial f_{2}}{\partial X_{1}}dX_{1}+\mathbf {0}+\mathbf {0}+\mathbf {k}\frac {\partial f_{3}}{\partial X_{1}}dX_{1}+\mathbf {0}+\mathbf {0}\right ) \\ +\left ( \mathbf {0}+\mathbf {ij}\frac {\partial f_{1}}{\partial X_{2}}dX_{2}+\mathbf {0}+\mathbf {0}+\mathbf {jj}\frac {\partial f_{2}}{\partial X_{2}}dX_{2}+\mathbf {0}+\mathbf {0}+\mathbf {kj}\frac {\partial f_{3}}{\partial X_{2}}dX_{2}+\mathbf {0}\right ) \\ +\left ( \mathbf {0}+\mathbf {0}+\mathbf {ik}\frac {\partial f_{1}}{\partial X_{3}}dX_{3}+\mathbf {0}+\mathbf {0}+\mathbf {jk}\frac {\partial f_{2}}{\partial X_{3}}dX_{3}+\mathbf {0}+\mathbf {0}+\mathbf {kk}\frac {\partial f_{3}}{\partial X_{3}}dX_{3}\right ) \end{multline*}

Simplifying gives

\begin{multline*} \mathbf {i\ }dx_{1}+\mathbf {j\ }dx_{2}+\mathbf {k\ }dx_{3}=\\ \left ( \mathbf {i}\frac {\partial f_{1}}{\partial X_{1}}dX_{1}+\mathbf {j}\frac {\partial f_{2}}{\partial X_{1}}dX_{1}+\mathbf {k}\frac {\partial f_{3}}{\partial X_{1}}dX_{1}\right ) +\left ( \mathbf {i}\frac {\partial f_{1}}{\partial X_{2}}dX_{2}+\mathbf {j}\frac {\partial f_{2}}{\partial X_{2}}dX_{2}+\mathbf {k}\frac {\partial f_{3}}{\partial X_{2}}dX_{2}\right ) \\ +\left ( \mathbf {i}\frac {\partial f_{1}}{\partial X_{3}}dX_{3}+\mathbf {j}\frac {\partial f_{2}}{\partial X_{3}}dX_{3}+\mathbf {k}\frac {\partial f_{3}}{\partial X_{3}}dX_{3}\right ) \end{multline*}

Collecting similar terms on the RHS gives

\begin{multline*} \mathbf {i\ }dx_{1}+\mathbf {j\ }dx_{2}+\mathbf {k\ }dx_{3}=\\ \mathbf {i}\left ( \frac {\partial f_{1}}{\partial X_{1}}dX_{1}+\frac {\partial f_{1}}{\partial X_{2}}dX_{2}+\frac {\partial f_{1}}{\partial X_{3}}dX_{3}\right ) +\mathbf {j}\left ( \frac {\partial f_{2}}{\partial X_{1}}dX_{1}+\frac {\partial f_{2}}{\partial X_{2}}dX_{2}+\frac {\partial f_{2}}{\partial X_{3}}dX_{3}\right ) \\ +\mathbf {k}\left ( \frac {\partial f_{3}}{\partial X_{1}}dX_{1}+\frac {\partial f_{3}}{\partial X_{2}}dX_{2}+\frac {\partial f_{3}}{\partial X_{3}}dX_{3}\right ) \end{multline*}

comparing the components of the vector on the LHS with those component of the vector on the RHS gives equation (1) as expected.

In addition to the matrix form and the dyadic form, the transformation from \(\mathbf {dR}\) to \(\mathbf {dr}\) can be expressed using indices notation as follows

\[ dx_{i}=\frac {\partial f_{i}}{\partial X_{j}}dX_{j}\]

7.2 Useful identities and formulas

A matrix \(\mathbf {A}\) is orthogonal if \(AA^{T}=I\) where \(I\) is the identity matrix.

If a matrix/tensor \(\mathbf {A}\) is orthogonal then \(\mathbf {A}^{-1}=\mathbf {A}^{T}\). In component form, \(a_{ij}^{-1}=a_{ji}\)

\(\left ( \mathbf {\tilde {A}\cdot \tilde {B}}\right ) ^{-1}=\mathbf {\tilde {B}}^{-1}\cdot \mathbf {\tilde {A}}^{-1}\)

\(\left ( \mathbf {\tilde {A}\cdot \tilde {B}}\right ) ^{T}=\mathbf {\tilde {B}}^{T}\cdot \mathbf {\tilde {A}}^{T}\)

\(\left ( \mathbf {\tilde {A}\cdot \tilde {B}}\right ) ^{-T}=\left ( \left ( \mathbf {\tilde {A}\cdot \tilde {B}}\right ) ^{T}\right ) ^{-1}=\left ( \mathbf {\tilde {B}}^{T}\cdot \mathbf {\tilde {A}}^{T}\right ) ^{-1}\)

\(\mathbf {\tilde {F}}=\mathbf {\tilde {R}\cdot \tilde {U}}\)

\(\mathbf {\tilde {F}}=\mathbf {\tilde {V}\cdot \tilde {R}}\)

\(\mathbf {\tilde {U}=\tilde {R}}^{T}\cdot \mathbf {\tilde {V}\cdot \tilde {R}}\)

\(\mathbf {\tilde {V}=\tilde {R}}\cdot \mathbf {\tilde {U}\cdot \tilde {R}}^{T}\)

\(\mathbf {\tilde {U}=}\left ( \mathbf {\tilde {F}}^{T}\cdot \mathbf {\tilde {F}}\right ) ^{\frac {1}{2}}\)

\(\mathbf {\tilde {F}}_{q}=\mathbf {\tilde {Q}}\cdot \mathbf {\tilde {F}}\)