# Nonlinear finite elements/Effect of mesh distortion

## An Example: Effect of Mesh Distortion

Consider the three-noded quadratic displacement element shown in Figure 1.

 Figure 1. Reference and Current Configurations of a 3-noded element.

The shape functions for the parent element are

{\displaystyle {\begin{aligned}N_{1}(\xi )&={\frac {1}{2}}\xi (\xi -1)\\N_{2}(\xi )&=1-\xi ^{2}\\N_{3}(\xi )&={\frac {1}{2}}\xi (\xi +1)\end{aligned}}}

In matrix form,

${\displaystyle \mathbf {N} (\xi )={\begin{bmatrix}{\frac {1}{2}}\xi (\xi -1)&1-\xi ^{2}&{\frac {1}{2}}\xi (\xi +1)\end{bmatrix}}~.}$

The trial functions (in terms of the parent coordinates) are

{\displaystyle {\begin{aligned}u_{h}(\xi ,t)&=\mathbf {N} (\xi )~\mathbf {u} (t)\\v_{h}(\xi ,t)&=\mathbf {N} (\xi )~\mathbf {v} (t)\\a_{h}(\xi ,t)&=\mathbf {N} (\xi )~\mathbf {a} (t)~.\end{aligned}}}

The mapping from the Eulerian coordinates to the parent element coordinates is

${\displaystyle x(\xi ,t)=N_{1}(\xi )~x_{1}(t)+N_{2}(\xi )~x_{2}(t)+N_{3}(\xi )~x_{3}(t)~.}$

In matrix form,

${\displaystyle x(\xi ,t)=\mathbf {N} (\xi )~\mathbf {x} (t)~.}$

Therefore, the derivative with respect to ${\displaystyle \xi }$ is

${\displaystyle x_{,\xi }=N_{1,\xi }~x_{1}(t)+N_{2,\xi }~x_{2}(t)+N_{3,\xi }~x_{3}(t)={\frac {1}{2}}(2\xi -1)~x_{1}-2\xi ~x_{2}+{\frac {1}{2}}(2\xi +1)~x_{3}~.}$

In matrix form,

${\displaystyle x_{,\xi }=\mathbf {N} _{,\xi }~\mathbf {x} (t)={\begin{bmatrix}{\frac {1}{2}}(2\xi -1)&-2\xi &{\frac {1}{2}}(2\xi +1)\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}~.}$

Now, the ${\displaystyle \mathbf {B} }$ matrix is given by

${\displaystyle \mathbf {B} =\mathbf {N} _{,\xi }(x_{,\xi })^{-1}~.}$

Therefore,

${\displaystyle \mathbf {B} ={\cfrac {1}{2x_{,\xi }}}{\begin{bmatrix}2\xi -1&-4\xi &2\xi +1\end{bmatrix}}~.}$

The rate of deformation is then given by

${\displaystyle {D=\mathbf {N} _{,x}~\mathbf {v} (t)=\mathbf {B} ~\mathbf {v} (t)={\cfrac {1}{2x_{,\xi }}}{\begin{bmatrix}2\xi -1&-4\xi &2\xi +1\end{bmatrix}}{\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}~.}}$

The stress can be calculated using the relation

${\displaystyle \sigma =E^{\sigma D}~D~.}$

The internal forces are given by

${\displaystyle \mathbf {f} _{\text{int}}=\int _{x_{1}}^{x_{3}}\mathbf {B} ^{T}~\sigma ~A~dx~.}$

Plugging in the expression for ${\displaystyle \mathbf {B} }$, and changing the limits of integration, we get

${\displaystyle {\mathbf {f} _{\text{int}}=\int _{-1}^{1}{\cfrac {1}{2x_{,\xi }}}{\begin{bmatrix}2\xi -1\\-4\xi \\2\xi +1\end{bmatrix}}~\sigma ~A~x_{,\xi }~d\xi ={\frac {1}{2}}\int _{-1}^{1}{\begin{bmatrix}2\xi -1\\-4\xi \\2\xi +1\end{bmatrix}}~\sigma ~A~d\xi ~.}}$

If node ${\displaystyle 2}$ is midway between node ${\displaystyle 1}$ and node ${\displaystyle 3}$,

${\displaystyle x_{2}={\frac {1}{2}}(x_{1}+x_{3})~.}$

Then we have,

${\displaystyle x_{,\xi }={\frac {1}{2}}(2\xi -1)~x_{1}-\xi ~(x_{1}+x_{3})+{\frac {1}{2}}(2\xi +1)~x_{3}={\frac {1}{2}}(x_{3}-x_{1})~.}$

The rate of deformation becomes

${\displaystyle D={\cfrac {1}{x_{3}-x_{1}}}{\begin{bmatrix}2\xi -1&-4\xi &2\xi +1\end{bmatrix}}{\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}}$

which is a linear function of ${\displaystyle \xi }$.

However, if node ${\displaystyle 2}$ moves away from the middle during deformation, then ${\displaystyle x_{,\xi }}$ is no longer constant and can become zero or negative. Under such situations the rate of deformation is either infinite or the element inverts upon itself since the isoparametric map is no longer one-to-one.

Let us consider the case where ${\displaystyle x_{,\xi }}$ is zero. In that case, the Jacobian becomes

${\displaystyle J={\cfrac {A}{A_{0}}}x_{,X}={\cfrac {A}{A_{0}}}x_{,\xi }~(X_{,\xi })^{-1}=0~.}$

Similarly, when ${\displaystyle x_{,\xi }}$ is negative, ${\displaystyle J}$ is negative. This implies that the conservation of mass is violated.

To find the location of ${\displaystyle x_{2}}$ when this happens, we set the relation for ${\displaystyle x_{,\xi }}$ to zero. Then we get,

${\displaystyle {\frac {1}{2}}(2\xi -1)~x_{1}-\xi ~(x_{1}+x_{3})+{\frac {1}{2}}(2\xi +1)~x_{3}=0\qquad \implies \qquad x_{2}={\cfrac {x_{1}+x_{3}}{2}}+{\cfrac {x_{3}-x_{1}}{4\xi }}~.}$

If ${\displaystyle x_{,\xi }=0}$ at ${\displaystyle \xi =1}$, then

${\displaystyle x_{2}={\cfrac {x_{1}+3x_{3}}{4}}~.}$

This means that as node ${\displaystyle 2}$ gets closer and closer to node ${\displaystyle 3}$, the rate of deformation become infinite at node ${\displaystyle 3}$ and then negative.

If ${\displaystyle x_{,\xi }=0}$ at ${\displaystyle \xi =-1}$, then

${\displaystyle x_{2}={\cfrac {3x_{1}+x_{3}}{4}}~.}$

This means that as node ${\displaystyle 2}$ gets closer and closer to node ${\displaystyle 1}$, the rate of deformation becomes infinite at node ${\displaystyle 1}$ and then negative.

These effects of mesh distortion can be severe in multiple dimensions. That is the reason that linear elements are preferred in large deformation simulations.