Compute the components of [F] for the motions associated with Eqs. (2.50), (2.52), and (2.55), and then calculate the associated values of [E]and [\varepsilon]. u_{x}=(\Lambda -1)X, u_{y}=0,(2.50) u_{x}=\kappa Y, u_{y}=0,(2.52) u_{x}=(\cos \phi -1)X+\sin \phi Y, u_{y}=(-\sin \phi )X+(\cos \phi -1)Y. (2.55)
Question 6.1: Compute the components of [F] for the motions associated wit...
First, consider the mathematically simple 1-D stretching motion given by x=\Lambda X, y=Y, and z=Z, where \Lambda is a stretch ratio (i.e., just a number for each equilibrium stretch). Clearly,
\left[F\right]=\left[\begin{matrix} \frac{\partial x}{\partial X}\frac{\partial x}{\partial Y}\frac{\partial x}{\partial Z} \\ \frac{\partial y}{\partial X}\frac{\partial y}{\partial Y}\frac{\partial y}{\partial Z} \\ \frac{\partial z}{\partial X}\frac{\partial z}{\partial Y} \frac{\partial z}{\partial Z} \end{matrix} \right]=\left[\begin{matrix} \Lambda 0 0 \\ 0 1 0 \\ 0 0 1 \end{matrix} \right]and therefore
\left[E\right]=\frac{1}{2}\left(\left[F\right]^{T}\left[F\right]-\left[I\right] \right)=\left[\begin{matrix} \frac{1}{2}(\Lambda ^{2}-1) & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right],whereas
\left[\varepsilon \right]=\frac{1} {2}\left(\left[F\right]+\left[F\right]^{T}-2\left[I\right] \right)\left[\begin{matrix} (\Lambda-1) & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right],as we found earlier. Again, for \Lambda \sim 1 (small strain), the numerical values of \left[E\right] and \left[\varepsilon\right] differ little, but for larger values typically experienced by soft tissues (stretches often on the order of 10–100 %), the difference becomes pronounced. For example, if \Lambda=1.5, a 50% extension, then E_{11}=0.625 (exact) and \varepsilon_{11}=0.5 (approximate), thus revealing a 20 % error in the computation of the strain. Question: Why would this motion be difficult to achieve in the lab? Some Nonlinear Problems Second, consider a simple shear motion given by x=X + \kappa Y, y=Y, and z=Z, where \kappa is just a number for each equilibrium motion. Hence
\left[F \right]=\left[\begin{matrix} 1 & \kappa & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]and, therefore,
\left[E \right]=\frac{1}{2}\left(\left[\begin{matrix} 1 & 0 & 0 \\ \kappa & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]\left[\begin{matrix} 1 & \kappa & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]-\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \right) =\frac{1}{2}\left[\begin{matrix} 0 & \kappa & 0 \\ \kappa & \kappa ^{2} & 0 \\ 0 & 0 & 0 \end{matrix} \right],whereas
\left[\varepsilon \right]=\frac{1}{2}\left(\left[\begin{matrix} 1 & \kappa & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]+\left[\begin{matrix} 1 & 0 & 0 \\ \kappa & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]-\left[\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix} \right] \right)=\frac{1}{2}\left[\begin{matrix} 0 & \kappa & 0 \\ \kappa & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right].This comparison reveals a significant conceptual difference between \left[E \right] and \left[\varepsilon \right]. Note that the extensional strain in the Y direction E_{YY}=\kappa ^{2}/2 whereas \varepsilon_{yy}=0; that is, shear and extension are coupled in the (exact) nonlinear theory, whereas the linearization of \left[\varepsilon \right] loses this coupling. Although \kappa ^{2} /2 will be negligible in comparison to \kappa/2 if κ\ll 1, this will not be the case for large shears, as experienced by the heart during each cardiac cycle. Again, therefore, the exact (nonlinear) theory must be used when the deformations or rigid rotations are large. The latter is revealed by considering the third case, the rigid-body motion associated with Eq. (2.55):
u_{x}=(\cos \phi -1)X+\sin \phi Y,
u_{y}=(-\sin \phi )X+(\cos \phi -1)Y. (2.55)
x=X\cos \phi +Y\sin \phi , y=-X\sin \phi +Y\cos \phi , z=Z.
In this case,
\left[E \right]=\frac{1}{2}\left(\left[\begin{matrix} \cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{matrix} \right]\left[\begin{matrix} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{matrix} \right]-\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \right)=\left[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right],
as it should, for \left[E\right] is insensitive to rigid-body motion, but
\left[\varepsilon \right]=\frac{1}{2}\left(\left[\begin{matrix} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{matrix} \right]+\left[\begin{matrix} \cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{matrix} \right]-\left[\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix} \right] \right)=\left[\begin{matrix} \cos \phi -1 & 0 & 0 \\ 0 & \cos \phi -1 & 0 \\ 0 & 0 & 0 \end{matrix} \right],
as found in Eq. (2.56), which reveals that \left[\varepsilon \right] is inappropriately sensitive to a rigid-body rotation unless the rotation is small (i.e., as \phi\rightarrow 0, cos \phi\rightarrow 1). Although these three motions are very simple, they serve to illustrate the use of \left[F\right] as a fundamental measure of the motion.
\varepsilon _{xx}=\cos \phi -1, \varepsilon _{yy}=\cos \phi -1, \varepsilon _{xy}=\frac{1}{2}(\sin \phi -\sin \phi )=0. (2.56)