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 $\mathbf{C}$ .

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