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 $$\mathbf{C}$$.