Elasticity/Sample midterm5

Sample Midterm Problem 5

Suppose that, under the action of external forces, a material point $𝐩=(X_1,X_2,X_3)$ in a body is displaced to a new location $𝐪=(x_1,x_2,x_3)$ 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

$\begin{array}{cc}{F}_{ij}& =\frac{\partial x_i}{\partial X_j}\\ & =\left[\begin{array}{ccc}A& \kappa & 0\\ \kappa & A& 0\\ 0& 0& 1\end{array}\right]\end{array}$

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 $\mathit{\epsilon }$ for the above displacement field.

Solution

The displacement is given by $𝐮=𝐱-𝐗$. Therefore,

$𝐮=\left[\begin{array}{c}\kappa X_2-X_1\\ \kappa X_1-X_2\\ 0\end{array}\right]$

The infinitesimal strain tensor is given by

$\mathit{\epsilon }=\frac{1}{2}(\mathrm{\nabla }u+\mathrm{\nabla }{u}^T)$

The gradient of $𝐮$ is given by

$\mathrm{\nabla }u=\left[\begin{array}{ccc}-1& \kappa & 0\\ \kappa & -1& 0\\ 0& 0& 0\end{array}\right]$

Therefore,

$\mathit{\epsilon }=\left[\begin{array}{ccc}-1& \kappa & 0\\ \kappa & -1& 0\\ 0& 0& 0\end{array}\right]$

Part (c)

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

Solution

The infinitesimal rotation tensor is given by

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

Therefore,

$𝐖=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

The rotation vector $\mathit{\omega }$ is

$\mathit{\omega }=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Part (d)

Do the strains satisfy compatibility ?

Solution

The compatibility equations are

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

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

$e=\text{tr}\mathit{\epsilon }$

Therefore,

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

The deviatoric strain is given by

${\mathit{\epsilon }}_d=\mathit{\epsilon }-\frac{\text{tr}\mathit{\epsilon }}{3}𝐈$

Hence,

${\mathit{\epsilon }}_d=\left[\begin{array}{ccc}-\frac{1}{3}& \kappa & 0\\ \kappa & -\frac{1}{3}& 0\\ 0& 0& -\frac{2}{3}\end{array}\right]$

Part (f)

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

Solution

The tensorial shear strains are ${\epsilon }_{12}$, ${\epsilon }_{23}$, ${\epsilon }_{31}$. 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 ($𝐅$).
• Compute the right Cauchy-Green deformation tensor ($𝐂={𝐅}^T\bullet 𝐅$).
• Calculate the eigenvalues and eigenvectors of $𝐂$.
• The principal stretches are the square roots of the eigenvalues of $𝐂$.
• The directions of the principal stretches are the eigenvectors of $𝐂$.