# Elasticity/Kinematics example 2
## Example 2

Given: A body occupies the unit cube $X_{i}\in [0,1]$ in the reference configuration. The mapping between the current and the reference configuration given by $x_{1}=X_{1}+\kappa ~X_{2}$, $x_{2}=X_{2}\,$, $x_{3}=X_{3}\,$.
Find:
* Sketch current configuration.
* Show that motion is isochoric.
* Find stretches in the directions ${\widehat {\mathbf {e} }}_{1}$, ${\widehat {\mathbf {e} }}_{2}$, $(1/{\sqrt {2}})({\widehat {\mathbf {e} }}_{1}+{\widehat {\mathbf {e} }}_{2})$, and $(1/{\sqrt {2}})({\widehat {\mathbf {e} }}_{1}-{\widehat {\mathbf {e} }}_{2})$.
* Find the orthogonal shear strain between the reference material directions ${\widehat {\mathbf {e} }}_{1}$ and ${\widehat {\mathbf {e} }}_{2}$. Also between directions $(1/{\sqrt {2}})({\widehat {\mathbf {e} }}_{1}+{\widehat {\mathbf {e} }}_{2})$ and $(1/{\sqrt {2}})({\widehat {\mathbf {e} }}_{1}-{\widehat {\mathbf {e} }}_{2})$.
* Find principal stretches and principal directions of stretch ($\kappa =0.4\,$).

### Solution:

* Deformed shape.
* All parallelograms that have the equal heights and the same base have equal areas. Hence, there is no volume change in this deformation. Hence isochoric.
* Stretches in the material direction ${\widehat {\mathbf {G} }}$ are given by
    $\lambda ({\widehat {\mathbf {G} }})={\sqrt {{\widehat {\mathbf {G} }}:{\boldsymbol {C}}:{\widehat {\mathbf {G} }}}}$
where ${\boldsymbol {C}}\,$ is the Cauchy-Green deformation tensor
    ${\boldsymbol {C}}={\boldsymbol {F}}^{T}\bullet {\boldsymbol {F}}$
We will use Maple to calculate the stretches in the four directions.
    $F:={\begin{bmatrix}1&k&0\\0&1&0\\0&0&1\end{bmatrix}}$
    $C:={\begin{bmatrix}1&k&0\\k&k^{2}+1&0\\0&0&1\end{bmatrix}}$
    $\lambda _{1}:=1\,$
    $\lambda _{2}:={\sqrt {k^{2}+1}}$
    $\lambda _{3}:={\frac {\sqrt {4+4k+2k^{2}}}{2}}$
    $\lambda _{4}:={\frac {\sqrt {4-4k+2k^{2}}}{2}}$
The following figure shows that the calculated stretches are correct.
* Stretches
* The orthogonal shear strain between two orthogonal units vectors ${\widehat {\mathbf {G} }}_{1}$ and ${\widehat {\mathbf {G} }}_{2}$ in the reference material co-ordinate system is given by
    $\gamma ({\widehat {\mathbf {G} }}_{1},{\widehat {\mathbf {G} }}_{2})=\sin ^{-1}\left({\frac {{\widehat {\mathbf {G} }}_{1}:\mathbf {C} :{\widehat {\mathbf {G} }}_{2}}{\lambda ({\widehat {\mathbf {G} }}_{1})\lambda ({\widehat {\mathbf {G} }}_{2})}}\right)$
Once again, we will use Maple to calculate these strains. The steps are the same upto the calculation of the stretches.
    $\gamma _{1}:=\sin ^{-1}({\frac {k}{\sqrt {k^{2}+1}}})$
    $\gamma _{2}:=-\sin ^{-1}({\frac {2k^{2}}{{\sqrt {4+4k+2k^{2}}}{\sqrt {4-4k+2k^{2}}}}})$
The following figure shows that the calculated orthogonal shear strains are correct.
* Orthogonal shear strains
The value of $\gamma _{1}\,$ can easily be verified using the definition of $\sin {\gamma }\,$. For the verification of the value of $\gamma _{2}\,$, the cosine law
    $\cos(A)={\frac {b^{2}+c^{2}-a^{2}}{2bc}}\,$
has to be used. Note that the actual length of the sides $b\,$ and $c\,$ of the triangle is obtained after multiplying the values shown in the figure by the length of the diagonal (${\sqrt {2}}$). Hence, the calculated values are correct.
    $A:={\frac {2k^{2}}{{\sqrt {4+4k+2k^{2}}}{\sqrt {4-4k+2k^{2}}}}}$
* The principal stretches are given by the square roots of the eigenvalues of ${\boldsymbol {C}}$. The principal directions are the eigenvectors of ${\boldsymbol {C}}$.
Once again, we use Maple for our calculations.
    $CC:={\begin{bmatrix}1&0.4&0\\0.4&1.16&0\\0&0&1\end{bmatrix}}$
    ${\text{eigvals}}:=0.67,1.0,1.49$
    ${\text{PrinStretch}}[1]:=0.82~,~~{\text{PrinStretch}}[2]:=1.0~,~~{\text{PrinStretch}}[3]:=1.22$
    ${\text{PrinDir}}[1]:=[0.77,-0.63,0.]~;~~{\text{PrinDir}}[2]:=[0,0,1]~;~~{\text{PrinDir}}[3]:=[0.63,0.77,0.]$