Elasticity/Sample midterm5

The deformation gradient ${\displaystyle (F)}$ is given by

${\displaystyle {\begin{aligned}F_{ij}&={\frac {\partial x_{i}}{\partial X_{j}}}\\&={\begin{bmatrix}A&\kappa &0\\\kappa &A&0\\0&0&1\end{bmatrix}}\end{aligned}}}$

Therefore, the requirement is that ${\displaystyle J={\text{det}}(F)>0}$ where

${\displaystyle J=A^{2}-\kappa ^{2}}$

The restriction is

${\displaystyle {|A|>|\kappa |}}$

The displacement is given by ${\displaystyle \mathbf {u} =\mathbf {x} -\mathbf {X} }$. Therefore,

${\displaystyle \mathbf {u} ={\begin{bmatrix}\kappa ~X_{2}-X_{1}\\\kappa ~X_{1}-X_{2}\\0\end{bmatrix}}}$

The infinitesimal strain tensor is given by

${\displaystyle {\boldsymbol {\varepsilon }}={\frac {1}{2}}({\boldsymbol {\nabla }}u+{\boldsymbol {\nabla }}u^{T})}$

The gradient of ${\displaystyle \mathbf {u} }$ is given by

${\displaystyle {\boldsymbol {\nabla }}u={\begin{bmatrix}-1&\kappa &0\\\kappa &-1&0\\0&0&0\end{bmatrix}}}$

Therefore,

${\displaystyle {{\boldsymbol {\varepsilon }}={\begin{bmatrix}-1&\kappa &0\\\kappa &-1&0\\0&0&0\end{bmatrix}}}}$

The infinitesimal rotation tensor is given by

${\displaystyle \mathbf {W} ={\frac {1}{2}}({\boldsymbol {\nabla }}u-{\boldsymbol {\nabla }}u^{T})}$

Therefore,

${\displaystyle {\mathbf {W} ={\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}}}}$

The rotation vector ${\displaystyle {\boldsymbol {\omega }}}$ is

${\displaystyle {{\boldsymbol {\omega }}={\begin{bmatrix}0\\0\\0\end{bmatrix}}}}$

The compatibility equations are

${\displaystyle {\begin{aligned}\varepsilon _{11,22}+\varepsilon _{22,11}-2\varepsilon _{12,12}&=0\\\varepsilon _{22,33}+\varepsilon _{33,22}-2\varepsilon _{23,23}&=0\\\varepsilon _{33,11}+\varepsilon _{11,33}-2\varepsilon _{13,13}&=0\\(\varepsilon _{12,3}-\varepsilon _{23,1}+\varepsilon _{31,2})_{,1}-\varepsilon _{11,23}&=0\\(\varepsilon _{23,1}-\varepsilon _{31,2}+\varepsilon _{12,3})_{,2}-\varepsilon _{22,31}&=0\\(\varepsilon _{31,2}-\varepsilon _{12,3}+\varepsilon _{23,1})_{,3}-\varepsilon _{33,12}&=0\end{aligned}}}$

All the equations are trivially satisfied because there is no dependence on ${\displaystyle X_{1}}$, ${\displaystyle X_{2}}$, and ${\displaystyle X_{3}}$.

${\displaystyle {\text{Compatibility is satisfied.}}}$

The dilatation is given by

${\displaystyle e={\text{tr}}{\boldsymbol {\varepsilon }}}$

Therefore,

${\displaystyle {e=-2~~~~{\text{(Note: Looks like shear only but not really.)}}}}$

The deviatoric strain is given by

${\displaystyle {\boldsymbol {\varepsilon }}_{d}={\boldsymbol {\varepsilon }}-{\frac {{\text{tr}}~{\boldsymbol {\varepsilon }}}{3}}\mathbf {I} }$

Hence,

${\displaystyle {{\boldsymbol {\varepsilon }}_{d}={\begin{bmatrix}-{\cfrac {1}{3}}&\kappa &0\\\kappa &-{\cfrac {1}{3}}&0\\0&0&-{\cfrac {2}{3}}\\\end{bmatrix}}}}$

The tensorial shear strains are ${\displaystyle \varepsilon _{12}}$, ${\displaystyle \varepsilon _{23}}$, ${\displaystyle \varepsilon _{31}}$. The engineering shear strains are ${\displaystyle \gamma _{12}}$, ${\displaystyle \gamma _{23}}$, ${\displaystyle \gamma _{31}}$.

The engineering shear strains are twice the tensorial shear strains.

• Compute the deformation gradient (${\displaystyle \mathbf {F} }$).
• Compute the right Cauchy-Green deformation tensor (${\displaystyle \mathbf {C} =\mathbf {F} ^{T}\bullet \mathbf {F} }$).
• Calculate the eigenvalues and eigenvectors of ${\displaystyle \mathbf {C} }$.
• The principal stretches are the square roots of the eigenvalues of ${\displaystyle \mathbf {C} }$.
• The directions of the principal stretches are the eigenvectors of ${\displaystyle \mathbf {C} }$.