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Section 1.8 strain measures

Subsection 1.8.1 Cauchy-Green strain tensor

Let us consider the following product

\begin{equation*} \transp{\tens{F}}\,\tens{F}\,. \end{equation*}

By applying the right polar decomposition it can be obtained what follows

\begin{equation*} \transp{\tens{F}}\,\tens{F} = \transp{\left(\tens{R}\tens{U}\right)}\, \tens{R}\tens{U} = \transp{\tens{U}}\transp{\tens{R}} \, \tens{R}\tens{U} = \tens{U}\,\tens{U} = \tens{U}^2\,, \end{equation*}

result showinig how \(\transp{\tens{F}}\,\tens{F}\) contains only deformation contribution related to \(\tens{U}\text{.}\) Such a product

\begin{equation} \tens{C} = \transp{\tens{F}}\,\tens{F}\,,\label{strain_C_eq}\tag{1.8.1} \end{equation}

is called right Cauchy-Green tensor and, as \(\tens{U}\text{,}\) is symmetric and positive definite. \(\tens{C}\) constitutes an important measure of strain.

Remark 1.8.1.

One might think that, known \(\tens{U}^2 \text{,}\) the simple operation \(\tens{U} = \sqrt{\tens{C}}\) allows the evaluation of \(\tens{U}\text{.}\) The operation is possible but not trivial because it requires the evaluation of the eigenvalues and eigenvectors of \(\tens{C} \text{.}\) For this reason the tensor \(\tens{C} \) or other derived quantities are used directly in strain measurements.

In order to illustrate the procedure described above, however, consider the following deformation gradient

\begin{equation*} \mat{F}= \left[ \begin{array}{ccc} 4.4623 \amp -3.2249 \amp -0.4874 \\ 2.9623 \amp 1.7249 \amp -1.9874 \\ -1.7500 \amp -1.0607 \amp 5.2500 \end{array}\right]\,. \end{equation*}

Starting from this expression, the following MATLAB® instructions allow you to evaluate the eigenvectors and eigenvalues of \(\tens{U}^2\text{.}\)

F = [4.4623   -3.2249   -0.4874;
2.9623    1.7249   -1.9874;
-1.7500   -1.0607    5.2500];
U2 = transpose(F)*F;
[u, D2] = eig(U2)
Listing 1.8.2.

The following instructions instead, evaluating the principal stretches as

\begin{equation*} \lambda_1 = \sqrt{D2(1,1)} = 2\,,\quad \lambda_2 = \sqrt{D2(2,2)} = 5\,,\quad \lambda_3 = \sqrt{D2(3,3)} = 7\,, \end{equation*}

and the main directions by extracting them from the columns of the eigenvectors matrix returned by MATLAB®, allow to calulate tensor \(\tens{U}\) and tensor \(\tens{R}\) by using

\begin{equation*} \tens{R} = \tens{F}\tens{U}^{-1}\,. \end{equation*}

It is also possible to calculate \(\tens{V}\) by using

\begin{equation*} \tens{V} = \tens{F}\transp{\tens{R}}\,. \end{equation*} 

u1 = u(:,1);
u2 = u(:,2);
u3 = u(:,3);
lam1 = sqrt(D2(1,1));
lam2 = sqrt(D2(2,2));
lam3 = sqrt(D2(3,3));
U = lam1*u1*transpose(u1)+lam2*u2*transpose(u2)+lam3*u3*transpose(u3)
R = F*inv(U)
V = F*transpose(R)
Listing 1.8.3.

In a similar way the left Cauchy-Green tensor can be obtained, i.e.

\begin{equation*} \tens{F}\,\transp{\tens{F}} = \tens{V}\tens{R} \, \transp{\left(\tens{V}\tens{R}\right)} = \tens{V}\tens{R} \, \transp{\tens{R}} \transp{\tens{V}} = \tens{V}\,\tens{V} = \tens{V}^2\,, \end{equation*}

from which

\begin{equation} \tens{B} = \tens{F}\,\transp{\tens{F}}\,.\tag{1.8.2} \end{equation}
Remark 1.8.4.

Between the two Cauchy-Green tensors, \(\tens{C}\) is most used in solid mechanics because it acts on the position of the point expressed in the reference configuration.

Insight 1.8.5. meaning of the components of \(\tens{C}\).

Let's apply to the gradient of the transformation \(\tens{F}\) what has been said in Section 1.2 about the meaning of the matrices associated with linear transformations. That is, we express the matrix associated with \(\tens{F}\) as follows

\begin{equation*} \mat{F}=\left[\begin{array}{ccc} \tens{F}\vec{e}_1 \amp \tens{F}\vec{e}_2 \amp \tens{F}\vec{e}_3 \end{array}\right]\,, \end{equation*}

where the transformed basis vectors appear in the columns of \(\mat{F}\text{.}\) We then evaluate the matrix associated with the tensor \(\tens{C} \) on the basis of the definition (1.8.1)

\begin{equation*} \mat{C}=\transp{\mat{F}}\mat{F}= \left[\begin{array}{c} \transp{(\tens{F}\vec{e}_1)} \\ \transp{(\tens{F}\vec{e}_2)} \\ \transp{(\tens{F}\vec{e}_3)} \end{array}\right] \left[\begin{array}{ccc} \tens{F}\vec{e}_1 \amp \tens{F}\vec{e}_2 \amp \tens{F}\vec{e}_3 \end{array}\right]\,, \end{equation*}

which gives

\begin{equation*} \mat{C}=\left[\begin{array}{ccc} \transp{(\tens{F}\vec{e}_1)}\tens{F}\vec{e}_1 \amp \transp{(\tens{F}\vec{e}_1)}\tens{F}\vec{e}_2 \amp \transp{(\tens{F}\vec{e}_1)}\tens{F}\vec{e}_3 \\ \transp{(\tens{F}\vec{e}_2)}\tens{F}\vec{e}_1 \amp \transp{(\tens{F}\vec{e}_2)}\tens{F}\vec{e}_2 \amp \transp{(\tens{F}\vec{e}_2)}\tens{F}\vec{e}_3 \\ \transp{(\tens{F}\vec{e}_3)}\tens{F}\vec{e}_1 \amp \transp{(\tens{F}\vec{e}_3)}\tens{F}\vec{e}_2 \amp \transp{(\tens{F}\vec{e}_3)}\tens{F}\vec{e}_3 \end{array}\right]\,. \end{equation*}

This result allows us to establish the following for the coefficients of \(\mat{C}\)

  • For the generic coefficient of \(\mat{C}\) placed on the main diagonal, \(C_{ii}\text{,}\) we have
    \begin{equation} C_{ii} = \transp{(\tens{F}\vec{e}_i)}\tens{F}\vec{e}_i = \tens{F}\vec{e}_i \cdot \tens{F}\vec{e}_i = \| \tens{F}\vec{e}_i\|^2\,.\label{Cii_pre_eq}\tag{1.8.3} \end{equation}
    It can therefore be seen that the generic component \(ii\) on the main diagonal represents the square of the norm of the transformed vector \(\tens{F}\vec{e}_i \text{.}\) Since the vector \(\vec{e}_i\) is of unit length, the component \(C_{ii}\) is therefore the square of the stretch along direction \(\vec{e}_i\text{.}\) Then the following can be established
    \begin{equation} C_{ii} = \| \tens{F}\vec{e}_i\|^2 = \func{\lambda}{\vec{e}_i}^2\,,\label{Cii_eq}\tag{1.8.4} \end{equation}
    where \(\func{\lambda}{\vec{e}_i}\) is the stretch along the direction \(\vec{e}_i \). In the case of the diagonal form for the tensor \(\tens{C} \) this stretch coincides with the corresponding principal stretch or eigenvalue of \(\tens{U}\text{.}\)
  • For the components outside the main diagonal it is obtained instead
    \begin{equation*} C_{ij} = \transp{(\tens{F}\vec{e}_i)}\tens{F}\vec{e}_j = \tens{F}\vec{e}_i \cdot \tens{F}\vec{e}_j = \cos{\func{\theta}{\tens{F}\vec{e}_i,\tens{F}\vec{e}_j}} \| \tens{F}\vec{e}_i\|\, \| \tens{F}\vec{e}_j\|\,, \end{equation*}
    where \(\func{\theta}{\tens{F}\vec{e}_i,\tens{F}\vec{e}_j}\) is the angle between \(\tens{F}\vec{e}_i\) and \(\tens{F}\vec{e}_j\text{.}\) By using (1.8.4), the following result can be obtained
    \begin{equation*} C_{ij} = \cos{\func{\theta}{\tens{F}\vec{e}_i,\tens{F}\vec{e}_j}} \, \func{\lambda}{\vec{e}_i} \, \func{\lambda}{\vec{e}_j}\,, \end{equation*}
    such a result shows the the generic component \(C_{ij}\) depends on stretches along \(\vec{e}_i\) and \(\vec{e}_j\) but, mainly, on the angle between the trasformed basis vectors \(\tens{F}\vec{e}_i\) and \(\tens{F}\vec{e}_j\text{.}\) By introducing the angle
    \begin{equation} \func{\gamma}{\vec{e}_i,\vec{e}_j} = \func{\Theta}{\vec{e}_i,\vec{e}_j} - \func{\theta}{\tens{F}\vec{e}_i,\tens{F}\vec{e}_j} = \frac{\pi}{2} - \func{\theta}{\tens{F}\vec{e}_i,\tens{F}\vec{e}_j}\,,\tag{1.8.5} \end{equation}
    that is the angle given by the difference between the angles formed by \(\vec{e}_i\) and \(\vec{e}_j \) before and after the transformation, \(C_{ij}\) can be rewritten as follows
    \begin{align*} C_{ij} \amp = \cos{\left(\frac{\pi}{2} - \func{\gamma}{\vec{e}_i,\vec{e}_j}\right)} \, \func{\lambda}{\vec{e}_i} \, \func{\lambda}{\vec{e}_j}\,,\\ \amp = \sin{\func{\gamma}{\vec{e}_i,\vec{e}_j}} \, \func{\lambda}{\vec{e}_i} \, \func{\lambda}{\vec{e}_j}\,, \end{align*}
    and then
    \begin{equation} C_{ij} = \sin{\func{\gamma}{\vec{e}_i,\vec{e}_j}}\, \func{\lambda}{\vec{e}_i} \, \func{\lambda}{\vec{e}_j}\,.\label{Cij_eq}\tag{1.8.6} \end{equation}
    The quantity \(\func{\gamma}{\vec{e}_i,\vec{e}_j}\) is called shear angle or shear. When \(\func{\gamma}{\vec{e}_i,\vec{e}_j}\) is zero, i.e. when the angle between \(\vec{e}_i\) and \(\vec{e}_j\) directions does not change, the \(C_{ij}\) component is zero.

Subsection 1.8.2 Green-Lagrange strain tensor

A standard strain measure is based on the difference between the squared lengths in the current configuration and in the reference one. In particular what follows can be obtained

\begin{align*} \frac{1}{2}\left(\|d\vec{x}\|^2 - \|d\vec{X}\|^2\right) \amp = \frac{1}{2}\left(d\vec{x}\cdot d\vec{x} - d\vec{X}\cdot d\vec{X}\right)\,,\\ \amp = \frac{1}{2}\left( (\tens{F} d\vec{X})\cdot(\tens{F} d\vec{X}) - d\vec{X}\cdot d\vec{X}\right) \,,\\ \amp = \frac{1}{2}\left( d\vec{X}\cdot(\transp{\tens{F}}\,\tens{F} d\vec{X}) - d\vec{X}\cdot(\tens{I}\, d\vec{X})\right) \,,\\ \amp = d\vec{X}\cdot \frac{1}{2}\left( \transp{\tens{F}}\,\tens{F} - \tens{I}\right)d\vec{X}\,. \end{align*}

The following tensor is so defined

\begin{equation} \tens{E} = \frac{1}{2}\left( \transp{\tens{F}}\,\tens{F} - \tens{I}\right) = \frac{1}{2}\left(\tens{C} - \tens{I}\right)\,.\tag{1.8.7} \end{equation}

The name of the tensor is Green-Lagrange strain tensor (the use of the \(\frac{1}{2} \) factor will become clear in the following). \(\tens{E}\) is symmetric but is not positive definite.

Remark 1.8.6.

The tensor \(\tens{C} = \tens{U}^2 \) is positive definite as it is related to the square of the scale factor with which the length of the generic segment \(d\vec{X}\) is modified. In the field of admissible deformations, this scale factor, as for areas and volumes, cannot cancel. With the Green-Lagrange tensor a measure of the elongation of the segment \(d\vec{X} \) with respect to its initial size is introduced. This makes possible negative values of the Green-Lagrange measure when the segment shortens or null values when the segment length does not change.

Subsection 1.8.3 strain tensors and displacement gradient

The mechanical models that will be discussed during the course are typically formulated using the displacement field as a kinematic entity which, remember, is defined as \(\vec{u} = \vec{x} - \vec{X}\text{.}\) Hence \(\tens{F} = \tens{I} + \tens{\nabla u} \text{.}\) Therefore the Green-Lagrange tensor can be expressed as

\begin{align*} \tens{E} \amp = \frac{1}{2}\left( \transp{\tens{F}}\,\tens{F} - \tens{I}\right)\,,\\ \amp = \frac{1}{2}\left( \transp{(\tens{I}+\tens{\nabla u})}\,(\tens{I}+\tens{\nabla u}) - \tens{I}\right)\,,\\ \amp = \frac{1}{2}\left(\tens{I}+\tens{\nabla u} + \transp{\tens{\nabla u}} + \transp{\tens{\nabla u}}\,\tens{\nabla u} - \tens{I}\right)\,,\\ \amp = \frac{1}{2}\left(\tens{\nabla u} + \transp{\tens{\nabla u}}\right) + \frac{1}{2} \transp{\tens{\nabla u}}\,\tens{\nabla u} \,. \end{align*}

Therefore, with respect to the displacement gradient, the Green-Lagrange strain tensor becomes

\begin{equation} \tens{E} = \frac{1}{2}\left(\tens{\nabla u} + \transp{\tens{\nabla u}}\right) + \frac{1}{2} \transp{\tens{\nabla u}}\,\tens{\nabla u} \,.\tag{1.8.8} \end{equation}

Also \(\tens{C}\) can be reformulated as follows

\begin{equation*} \tens{C} = \transp{\tens{F}}\tens{F} = \transp{(\tens{I}+\tens{\nabla u})} (\tens{I}+\tens{\nabla u})\,, \end{equation*}

from which

\begin{equation} \tens{C} = \tens{I}+ \tens{\nabla u} + \transp{\tens{\nabla u}} + \transp{\tens{\nabla u}}\,\tens{\nabla u}\,.\tag{1.8.9} \end{equation}