# Nonlinear finite elements/Homework 6/Solutions/Problem 1/Part 3

## Problem 1: Part 3

The rate of deformation is defined as

$\boldsymbol{D} = \text{sym}(\boldsymbol{\nabla} \mathbf{v}) = \frac{1}{2}\left[\boldsymbol{\nabla} \mathbf{v} + (\boldsymbol{\nabla} \mathbf{v})^T\right]$

where $\mathbf{v}$ is the velocity. In index notation, we write

$D_{ij} = \text{sym}(v_{i,j}) = \frac{1}{2}\left(v_{i,j} + v_{j,i}\right) = \frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right), \qquad i,j=1,2,3~.$

Given the above definition, derive equations (9.2.1) through (9.2.7) of the book chapter.

The motion of a point $P$ on the beam with respect to a point on the reference line $C$ is shown in Figure 2.

 Figure 2. Motion of continuum-based beam.

Since the normal ($\mathbf{n}$) rotates as a rigid body, the velocity of point $P$ with respect to $C$ is given by

$\mathbf{v}_{\text{PC}} = \boldsymbol{\omega}\times\mathbf{r}$

where $\boldsymbol{\omega}$ is the angular velocity of the normal, and $\mathbf{r}$ is the vector from $C$ to $P$.

Expressed in terms of the local basis vectors $\mathbf{e}_x$, $\mathbf{e}_y$, and $\mathbf{e}_z$, the angular velocity and the radial vector are

\begin{align} \boldsymbol{\omega} & = 0~\mathbf{e}_x + 0~\mathbf{e}_y + \omega~\mathbf{e}_z = \omega~\mathbf{e}_z \\ \mathbf{r} & = 0~\mathbf{e}_x + y~\mathbf{e}_y + 0~\mathbf{e}_z = y~\mathbf{e}_y~. \end{align}

Therefore,

$\mathbf{v}_{\text{PC}} = (\omega~\mathbf{e}_z)\times(y~\mathbf{e}_y) = y~\omega~\mathbf{e}_z\times\mathbf{e}_y = -y~\omega~\mathbf{e}_x ~.$

Let $\mathbf{v}^M(\mathbf{x},t)$ be the velocity of the point $C$ at time $t$. Then the actual velocity of point $P$ is

$\mathbf{v} = \mathbf{v}^M + \mathbf{v}_{\text{PC}} ~.$

Now, in terms of the local basis vectors

$\mathbf{v}^M = v^M_x~\mathbf{e}_x + v^M_y~\mathbf{e}_y + 0~\mathbf{e}_z~.$

Therefore,

$\mathbf{v} =v^M_x~\mathbf{e}_x + v^M_y~\mathbf{e}_y- y~\omega~\mathbf{e}_x =(v^M_x - y~\omega)\mathbf{e}_x + v^M_y \mathbf{e}_y ~.$

Therefore, the velocity of any point $P$ in terms of the local basis at its orthogonal projection at the reference line is

\begin{align} \mathbf{v}(x,y,z,t) & = v_x~\mathbf{e}_x & + & v_y~\mathbf{e}_y & + &v_z~\mathbf{e}_z \\ & = [v^M_x(x,t) - y~\omega(x,t)]~\mathbf{e}_x & + &v^M_y(x,t)~\mathbf{e}_y & + &0~\mathbf{e}_z~. \end{align}

The components of the rate of deformation tensor are

$D_{ij} = \frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right), \qquad i,j=1,2,3~.$

In terms of the local basis, these components are

\begin{align} D_{xx} & = \frac{\partial v_x}{\partial x} = \frac{\partial v^M_x}{\partial x} - y\frac{\partial \omega}{\partial x} \\ D_{yy} & = \frac{\partial v_y}{\partial y} = 0 \\ D_{xy} & = \frac{1}{2}\left(\frac{\partial v_x}{\partial y} + \frac{\partial v_y}{\partial x}\right) = \frac{1}{2}\left(-\omega + \frac{\partial v^M_y}{\partial x}\right)\\ D_{zz} & = \frac{\partial v_z}{\partial z} = 0 \\ D_{yz} & = \frac{1}{2}\left(\frac{\partial v_y}{\partial z} + \frac{\partial v_z}{\partial y}\right) = 0 \\ D_{zx} & = \frac{1}{2}\left(\frac{\partial v_z}{\partial x} + \frac{\partial v_x}{\partial z}\right) = 0 \end{align}

For the Euler-Bernoulli beam theory, the normals remain normal to the reference line. Let $\theta$ be the rotation of the normal. Then, the rotation is given by (see Figure 3)

$\theta = \frac{\partial u^M_y}{\partial x}$

where $u^M_y$ is the displacement in the local $y$-direction at a point on the reference line.

 Figure 3. Euler-Bernoulli beam kinematics.

The angular velocity of the normal is given by

$\omega = \frac{\partial \theta}{\partial t} = \frac{\partial v^M_y}{\partial x}~.$

Hence,

\begin{align} D_{xx} & = \frac{\partial v^M_x}{\partial x} - y\frac{\partial \omega}{\partial x} = \frac{\partial v^M_x}{\partial x} - y\frac{\partial^2 v^M_y}{\partial x^2} \\ D_{yy} & = 0 \\ D_{xy} & = \frac{1}{2}\left(-\omega + \frac{\partial v^M_y}{\partial x}\right) = 0 \\ D_{zz} & = 0 \\ D_{yz} & = 0 \\ D_{zx} & = 0 \end{align}