# 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
    ${\begin{aligned}{\boldsymbol {\omega }}&=0~\mathbf {e} _{x}+0~\mathbf {e} _{y}+\omega ~\mathbf {e} _{z}=\omega ~\mathbf {e} _{z}\\\mathbf {r} &=0~\mathbf {e} _{x}+y~\mathbf {e} _{y}+0~\mathbf {e} _{z}=y~\mathbf {e} _{y}~.\end{aligned}}$
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
    $\mathbf {v} ^{M}=v_{x}^{M}~\mathbf {e} _{x}+v_{y}^{M}~\mathbf {e} _{y}+0~\mathbf {e} _{z}~.$
Therefore,
    $\mathbf {v} =v_{x}^{M}~\mathbf {e} _{x}+v_{y}^{M}~\mathbf {e} _{y}-y~\omega ~\mathbf {e} _{x}=(v_{x}^{M}-y~\omega )\mathbf {e} _{x}+v_{y}^{M}\mathbf {e} _{y}~.$
Therefore, the velocity of any point $P$ in terms of the local basis at its orthogonal projection at the reference line is
    ${\begin{aligned}\mathbf {v} (x,y,z,t)&=v_{x}~\mathbf {e} _{x}&+&v_{y}~\mathbf {e} _{y}&+&v_{z}~\mathbf {e} _{z}\\&=[v_{x}^{M}(x,t)-y~\omega (x,t)]~\mathbf {e} _{x}&+&v_{y}^{M}(x,t)~\mathbf {e} _{y}&+&0~\mathbf {e} _{z}~.\end{aligned}}$
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
    ${\begin{aligned}D_{xx}&={\frac {\partial v_{x}}{\partial x}}={\frac {\partial v_{x}^{M}}{\partial x}}-y{\frac {\partial \omega }{\partial x}}\\D_{yy}&={\frac {\partial v_{y}}{\partial y}}=0\\D_{xy}&={\frac {1}{2}}\left({\frac {\partial v_{x}}{\partial y}}+{\frac {\partial v_{y}}{\partial x}}\right)={\frac {1}{2}}\left(-\omega +{\frac {\partial v_{y}^{M}}{\partial x}}\right)\\D_{zz}&={\frac {\partial v_{z}}{\partial z}}=0\\D_{yz}&={\frac {1}{2}}\left({\frac {\partial v_{y}}{\partial z}}+{\frac {\partial v_{z}}{\partial y}}\right)=0\\D_{zx}&={\frac {1}{2}}\left({\frac {\partial v_{z}}{\partial x}}+{\frac {\partial v_{x}}{\partial z}}\right)=0\end{aligned}}$
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)
    $\theta ={\frac {\partial u_{y}^{M}}{\partial x}}$
where $u_{y}^{M}$ 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
    $\omega ={\frac {\partial \theta }{\partial t}}={\frac {\partial v_{y}^{M}}{\partial x}}~.$
Hence,
    ${\begin{aligned}D_{xx}&={\frac {\partial v_{x}^{M}}{\partial x}}-y{\frac {\partial \omega }{\partial x}}={\frac {\partial v_{x}^{M}}{\partial x}}-y{\frac {\partial ^{2}v_{y}^{M}}{\partial x^{2}}}\\D_{yy}&=0\\D_{xy}&={\frac {1}{2}}\left(-\omega +{\frac {\partial v_{y}^{M}}{\partial x}}\right)=0\\D_{zz}&=0\\D_{yz}&=0\\D_{zx}&=0\end{aligned}}$