strain measures

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}\,,\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.

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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.

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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 \(\tens{U}^2\text{.}\)

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Listing 1.8.2.

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$ 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)

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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*}

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Listing 1.8.3.

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$ 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)

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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}

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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.

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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\,.\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\,,\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 \(\tens{U}\text{.}\)

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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}\,.\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.

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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.

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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.

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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}

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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}

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