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

${\begin{aligned}N_{1}(\xi )&={\frac {1}{2}}\xi (\xi -1)\\N_{2}(\xi )&=1-\xi ^{2}\\N_{3}(\xi )&={\frac {1}{2}}\xi (\xi +1)\end{aligned}}$ In matrix form,

$\mathbf {N} (\xi )={\begin{bmatrix}{\frac {1}{2}}\xi (\xi -1)&1-\xi ^{2}&{\frac {1}{2}}\xi (\xi +1)\end{bmatrix}}~.$ The trial functions (in terms of the parent coordinates) are

${\begin{aligned}u_{h}(\xi ,t)&=\mathbf {N} (\xi )~\mathbf {u} (t)\\v_{h}(\xi ,t)&=\mathbf {N} (\xi )~\mathbf {v} (t)\\a_{h}(\xi ,t)&=\mathbf {N} (\xi )~\mathbf {a} (t)~.\end{aligned}}$ The mapping from the Eulerian coordinates to the parent element coordinates
is

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

$x(\xi ,t)=\mathbf {N} (\xi )~\mathbf {x} (t)~.$ Therefore, the derivative with respect to $\xi$ is

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

$x_{,\xi }=\mathbf {N} _{,\xi }~\mathbf {x} (t)={\begin{bmatrix}{\frac {1}{2}}(2\xi -1)&-2\xi &{\frac {1}{2}}(2\xi +1)\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}~.$ Now, the $\mathbf {B}$ matrix is given by

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

$\mathbf {B} ={\cfrac {1}{2x_{,\xi }}}{\begin{bmatrix}2\xi -1&-4\xi &2\xi +1\end{bmatrix}}~.$ The rate of deformation is then given by

${D=\mathbf {N} _{,x}~\mathbf {v} (t)=\mathbf {B} ~\mathbf {v} (t)={\cfrac {1}{2x_{,\xi }}}{\begin{bmatrix}2\xi -1&-4\xi &2\xi +1\end{bmatrix}}{\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}~.}$ The stress can be calculated using the relation

$\sigma =E^{\sigma D}~D~.$ The internal forces are given by

$\mathbf {f} _{\text{int}}=\int _{x_{1}}^{x_{3}}\mathbf {B} ^{T}~\sigma ~A~dx~.$ Plugging in the expression for $\mathbf {B}$ , and changing the limits of
integration, we get

${\mathbf {f} _{\text{int}}=\int _{-1}^{1}{\cfrac {1}{2x_{,\xi }}}{\begin{bmatrix}2\xi -1\\-4\xi \\2\xi +1\end{bmatrix}}~\sigma ~A~x_{,\xi }~d\xi ={\frac {1}{2}}\int _{-1}^{1}{\begin{bmatrix}2\xi -1\\-4\xi \\2\xi +1\end{bmatrix}}~\sigma ~A~d\xi ~.}$ If node $2$ is midway between node $1$ and node $3$ ,

$x_{2}={\frac {1}{2}}(x_{1}+x_{3})~.$ Then we have,

$x_{,\xi }={\frac {1}{2}}(2\xi -1)~x_{1}-\xi ~(x_{1}+x_{3})+{\frac {1}{2}}(2\xi +1)~x_{3}={\frac {1}{2}}(x_{3}-x_{1})~.$ The rate of deformation becomes

$D={\cfrac {1}{x_{3}-x_{1}}}{\begin{bmatrix}2\xi -1&-4\xi &2\xi +1\end{bmatrix}}{\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}$ which is a linear function of $\xi$ .

However, if node $2$ moves away from the middle during deformation, then
$x_{,\xi }$ 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 $x_{,\xi }$ is zero. In that case,
the Jacobian becomes

$J={\cfrac {A}{A_{0}}}x_{,X}={\cfrac {A}{A_{0}}}x_{,\xi }~(X_{,\xi })^{-1}=0~.$ Similarly, when $x_{,\xi }$ is negative, $J$ is negative. This implies that the conservation of mass is violated.

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

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

$x_{2}={\cfrac {x_{1}+3x_{3}}{4}}~.$ This means that as node $2$ gets closer and closer to node $3$ , the rate of
deformation become infinite at node $3$ and then negative.

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

$x_{2}={\cfrac {3x_{1}+x_{3}}{4}}~.$ This means that as node $2$ gets closer and closer to node $1$ , the rate of
deformation becomes infinite at node $1$ 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.