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

Problem 1: Part 3

The rate of deformation is defined as

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

where $𝐯$ is the velocity. In index notation, we write

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

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.

File:ContinuumBeamMotion.png Figure 2. Motion of continuum-based beam.

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

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

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

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

$\begin{array}{cc}\mathit{\omega }& =0{𝐞}_x+0{𝐞}_y+\omega {𝐞}_z=\omega {𝐞}_z\\ 𝐫& =0{𝐞}_x+y{𝐞}_y+0{𝐞}_z=y{𝐞}_y.\end{array}$

Therefore,

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

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

$𝐯={𝐯}^M+{𝐯}_{\text{PC}}.$

Now, in terms of the local basis vectors

${𝐯}^M=v_x^M{𝐞}_x+v_y^M{𝐞}_y+0{𝐞}_z.$

Therefore,

$𝐯=v_x^M{𝐞}_x+v_y^M{𝐞}_y-y\omega {𝐞}_x=(v_x^M-y\omega ){𝐞}_x+v_y^M{𝐞}_y.$

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

$\begin{array}{cccccc}𝐯(x,y,z,t)& =v_x{𝐞}_x& +& v_y{𝐞}_y& +& v_z{𝐞}_z\\ & =[v_x^M(x,t)-y\omega (x,t)]{𝐞}_x& +& v_y^M(x,t){𝐞}_y& +& 0{𝐞}_z.\end{array}$

The components of the rate of deformation tensor are

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

In terms of the local basis, these components are

$\begin{array}{cc}{D}_{xx}& =\frac{\partial v_x}{\partial x}=\frac{\partial v_x^M}{\partial x}-y\frac{\partial \omega }{\partial x}\\ D_{yy}& =\frac{\partial v_y}{\partial y}=0\\ D_{xy}& =\frac{1}{2}(\frac{\partial v_x}{\partial y}+\frac{\partial v_y}{\partial x})=\frac{1}{2}(-\omega +\frac{\partial v_y^M}{\partial x})\\ D_{zz}& =\frac{\partial v_z}{\partial z}=0\\ D_{yz}& =\frac{1}{2}(\frac{\partial v_y}{\partial z}+\frac{\partial v_z}{\partial y})=0\\ D_{zx}& =\frac{1}{2}(\frac{\partial v_z}{\partial x}+\frac{\partial v_x}{\partial z})=0\end{array}$

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.

File:EulerBernoulliBeam1.png 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,

$\begin{array}{cc}{D}_{xx}& =\frac{\partial v_x^M}{\partial x}-y\frac{\partial \omega }{\partial x}=\frac{\partial v_x^M}{\partial x}-y\frac{{\partial }^2{v}_y^M}{\partial x^2}\\ D_{yy}& =0\\ D_{xy}& =\frac{1}{2}(-\omega +\frac{\partial v_y^M}{\partial x})=0\\ D_{zz}& =0\\ D_{yz}& =0\\ D_{zx}& =0\end{array}$