# Introduction to 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 $\mathbf {C}$ .