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

Problem 1: Part 3

The rate of deformation is defined as

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

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

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

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

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

 Figure 2. Motion of continuum-based beam.

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

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

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

Expressed in terms of the local basis vectors ${\displaystyle \mathbf {e} _{x}}$ , ${\displaystyle \mathbf {e} _{y}}$ , and ${\displaystyle \mathbf {e} _{z}}$ , the angular velocity and the radial vector are

{\displaystyle {\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,

${\displaystyle \mathbf {v} _{\text{PC}}=(\omega ~\mathbf {e} _{z})\times (y~\mathbf {e} _{y})=y~\omega ~\mathbf {e} _{z}\times \mathbf {e} _{y}=-y~\omega ~\mathbf {e} _{x}~.}$

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

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

Now, in terms of the local basis vectors

${\displaystyle \mathbf {v} ^{M}=v_{x}^{M}~\mathbf {e} _{x}+v_{y}^{M}~\mathbf {e} _{y}+0~\mathbf {e} _{z}~.}$

Therefore,

${\displaystyle \mathbf {v} =v_{x}^{M}~\mathbf {e} _{x}+v_{y}^{M}~\mathbf {e} _{y}-y~\omega ~\mathbf {e} _{x}=(v_{x}^{M}-y~\omega )\mathbf {e} _{x}+v_{y}^{M}\mathbf {e} _{y}~.}$

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

{\displaystyle {\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

${\displaystyle D_{ij}={\frac {1}{2}}\left({\frac {\partial v_{i}}{\partial x_{j}}}+{\frac {\partial v_{j}}{\partial x_{i}}}\right),\qquad i,j=1,2,3~.}$

In terms of the local basis, these components are

{\displaystyle {\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 ${\displaystyle \theta }$  be the rotation of the normal. Then, the rotation is given by (see Figure 3)

${\displaystyle \theta ={\frac {\partial u_{y}^{M}}{\partial x}}}$

where ${\displaystyle u_{y}^{M}}$  is the displacement in the local ${\displaystyle y}$ -direction at a point on the reference line.

 Figure 3. Euler-Bernoulli beam kinematics.

The angular velocity of the normal is given by

${\displaystyle \omega ={\frac {\partial \theta }{\partial t}}={\frac {\partial v_{y}^{M}}{\partial x}}~.}$

Hence,

{\displaystyle {\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}}}