# Elasticity/Sample midterm5
## Sample Midterm Problem 5

Suppose that, under the action of external forces, a material point $\mathbf {p} =(X_{1},X_{2},X_{3})$ in a body is displaced to a new location $\mathbf {q} =(x_{1},x_{2},x_{3})$ where
    $x_{1}=A~X_{1}+\kappa ~X_{2}~;~~x_{2}=A~X_{2}+\kappa ~X_{1}~;~~x_{3}=X_{3}$
and $A$ and $\kappa$ are constants.

### Part (a)

A displacement field is called proper and admissible if the Jacobian ($J$) is greater than zero. If a displacement field is proper and admissible, then the deformation of the body is continuous.
Indicate the restrictions that must be imposed upon $A$ so that the deformation represented by the above displacement is continuous.

#### Solution

The deformation gradient $(F)$ is given by
    ${\begin{aligned}F_{ij}&={\frac {\partial x_{i}}{\partial X_{j}}}\\&={\begin{bmatrix}A&\kappa &0\\\kappa &A&0\\0&0&1\end{bmatrix}}\end{aligned}}$
Therefore, the requirement is that $J={\text{det}}(F)>0$ where
    $J=A^{2}-\kappa ^{2}$
The restriction is
    ${|A|>|\kappa |}$

### Part (b)

Suppose that $A=0$. Calculate the components of the infinitesimal strain tensor ${\boldsymbol {\varepsilon }}$ for the above displacement field.

#### Solution

The displacement is given by $\mathbf {u} =\mathbf {x} -\mathbf {X}$. Therefore,
    $\mathbf {u} ={\begin{bmatrix}\kappa ~X_{2}-X_{1}\\\kappa ~X_{1}-X_{2}\\0\end{bmatrix}}$
The infinitesimal strain tensor is given by
    ${\boldsymbol {\varepsilon }}={\frac {1}{2}}({\boldsymbol {\nabla }}u+{\boldsymbol {\nabla }}u^{T})$
The gradient of $\mathbf {u}$ is given by
    ${\boldsymbol {\nabla }}u={\begin{bmatrix}-1&\kappa &0\\\kappa &-1&0\\0&0&0\end{bmatrix}}$
Therefore,
    ${{\boldsymbol {\varepsilon }}={\begin{bmatrix}-1&\kappa &0\\\kappa &-1&0\\0&0&0\end{bmatrix}}}$

### Part (c)

Calculate the components of the infinitesimal rotation tensor $\mathbf {W}$ for the above displacement field and find the rotation vector ${\boldsymbol {\omega }}$.

#### Solution

The infinitesimal rotation tensor is given by
    $\mathbf {W} ={\frac {1}{2}}({\boldsymbol {\nabla }}u-{\boldsymbol {\nabla }}u^{T})$
Therefore,
    ${\mathbf {W} ={\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}}}$
The rotation vector ${\boldsymbol {\omega }}$ is
    ${{\boldsymbol {\omega }}={\begin{bmatrix}0\\0\\0\end{bmatrix}}}$

### Part (d)

Do the strains satisfy compatibility ?

#### Solution

The compatibility equations are
    ${\begin{aligned}\varepsilon _{11,22}+\varepsilon _{22,11}-2\varepsilon _{12,12}&=0\\\varepsilon _{22,33}+\varepsilon _{33,22}-2\varepsilon _{23,23}&=0\\\varepsilon _{33,11}+\varepsilon _{11,33}-2\varepsilon _{13,13}&=0\\(\varepsilon _{12,3}-\varepsilon _{23,1}+\varepsilon _{31,2})_{,1}-\varepsilon _{11,23}&=0\\(\varepsilon _{23,1}-\varepsilon _{31,2}+\varepsilon _{12,3})_{,2}-\varepsilon _{22,31}&=0\\(\varepsilon _{31,2}-\varepsilon _{12,3}+\varepsilon _{23,1})_{,3}-\varepsilon _{33,12}&=0\end{aligned}}$
All the equations are trivially satisfied because there is no dependence on $X_{1}$, $X_{2}$, and $X_{3}$.
    ${\text{Compatibility is satisfied.}}$

### Part (e)

Calculate the dilatation and the deviatoric strains from the strain tensor.

#### Solution

The dilatation is given by
    $e={\text{tr}}{\boldsymbol {\varepsilon }}$
Therefore,
    ${e=-2~~~~{\text{(Note: Looks like shear only but not really.)}}}$
The deviatoric strain is given by
    ${\boldsymbol {\varepsilon }}_{d}={\boldsymbol {\varepsilon }}-{\frac {{\text{tr}}~{\boldsymbol {\varepsilon }}}{3}}\mathbf {I}$
Hence,
    ${{\boldsymbol {\varepsilon }}_{d}={\begin{bmatrix}-{\cfrac {1}{3}}&\kappa &0\\\kappa &-{\cfrac {1}{3}}&0\\0&0&-{\cfrac {2}{3}}\\\end{bmatrix}}}$

### Part (f)

What is the difference between tensorial shear strain and engineering shear strain (for infinitesimal strains)?

#### Solution

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

### Part (g)

Briefly describe the process which you would use to calculate the principal stretches and their directions.

#### Solution

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