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

$\begin{align} F_{ij} & = \frac{\partial x_i }{\partial X_j} \\ & = \begin{bmatrix} A & \kappa & 0 \\ \kappa & A & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{align}$

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

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

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 $\mathbf{C}$.