## QEPH (Quadrilateral ElastoPlastic Physical Hourglass Control) Element

With one-point integration formulation, if the non-constant part follows exactly the state of constant part for the case of elasto-plastic calculation, the plasticity will be under-estimated due to the fact that the constant equivalent stress is often the smallest one in the element and element will be stiffer. Therefore, defining a yield criterion for the non-constant part seems to be a good idea to overcome this drawback.

From In-plane Strain-rate Construction, Equation 4 and Equation 9, you have the rate of stresses of non-constant part:

${\left\{{\stackrel{˙}{\sigma }}_{i}\right\}}_{}^{H}=\left[C\right]{\left\{\stackrel{˙}{\epsilon }\right\}}^{H}=\left[C\right]\left\{\begin{array}{c}{\varphi }_{,x}{\stackrel{˙}{q}}_{x}^{\alpha }\\ {\varphi }_{,y}{\stackrel{˙}{q}}_{y}^{\alpha }\\ 0\end{array}\right\}=\eta \left\{\begin{array}{c}{\stackrel{˙}{\sigma }}_{x\eta }^{\alpha }\\ {\stackrel{˙}{\sigma }}_{y\eta }^{\alpha }\\ 0\end{array}\right\}+\xi \left\{\begin{array}{c}{\stackrel{˙}{\sigma }}_{x\xi }^{\alpha }\\ {\stackrel{˙}{\sigma }}_{y\xi }^{\alpha }\\ 0\end{array}\right\}$

Where, $\alpha =m,b$ corresponds to the membrane and bending terms respectively.
Note: The shear terms are eliminated to avoid shear locking.

The transverse shear terms can also be written as the same way:

${\left\{{\stackrel{˙}{\tau }}_{i}\right\}}_{}^{H}=\eta \left\{\begin{array}{c}{\stackrel{˙}{\tau }}_{x\eta }^{}\\ {\stackrel{˙}{\tau }}_{y\eta }^{}\end{array}\right\}+\xi \left\{\begin{array}{c}{\stackrel{˙}{\tau }}_{x\xi }^{}\\ {\stackrel{˙}{\tau }}_{y\xi }^{}\end{array}\right\}$

You can now redefine 12 generalized hourglass stresses by integrating their rate ones, and the stress field can be expressed by:
Membrane, bending
$\left\{{\sigma }_{i}^{\alpha }\right\}={\left\{{\sigma }_{i}^{\alpha }\right\}}^{0}+{\left\{{\sigma }_{i}^{\alpha }\right\}}^{H}={\left\{{\sigma }_{i}^{\alpha }\right\}}^{0}+\eta \left\{\begin{array}{c}{\sigma }_{x\eta }^{\alpha }\\ {\sigma }_{y\eta }^{\alpha }\\ 0\end{array}\right\}+\xi \left\{\begin{array}{c}{\sigma }_{x\xi }^{\alpha }\\ {\sigma }_{y\xi }^{\alpha }\\ 0\end{array}\right\}$
Shear
$\left\{{\tau }_{i}\right\}={\left\{{\tau }_{i}\right\}}^{0}+{\left\{{\tau }_{i}\right\}}_{}^{H}={\left\{{\tau }_{i}\right\}}^{0}+\eta \left\{\begin{array}{c}{\stackrel{˙}{\tau }}_{x\eta }^{}\\ {\stackrel{˙}{\tau }}_{y\eta }^{}\end{array}\right\}+\xi \left\{\begin{array}{c}{\stackrel{˙}{\tau }}_{x\xi }^{}\\ {\stackrel{˙}{\tau }}_{y\xi }^{}\end{array}\right\}$

Even the redefinition for shear is not necessary as it is not included in the plastic yield criterion, but the same stress calculation as the constant part with the updated Lagrangian formulation is always useful when large strain is involved.

## Plastic Yield Criterion

The von Mises type of criterion for any point in the solid element is written by:

$f={\sigma }_{eq}^{2}\left(\xi ,\eta ,\zeta \right)-{\sigma }_{y}^{2}=0$

Where, ${\sigma }_{y}^{}$ is evaluated at the quadrature point.

As only one criterion is used for the non-constant part, two choices are possible:
• taking the mean value, that is: $f=f\left({\overline{\sigma }}_{eq}^{}\right);{\overline{\sigma }}_{eq}^{}=\frac{1}{\Omega }\underset{\Omega }{\int }{\sigma }_{eq}^{}d\Omega$
• taking the value by some representative points, such as eight Gauss points

The second choice has been used in this element.

## Elasto-plastic Hourglass Stress Calculation

The incremental hourglass stress is computed by:
• Elastic increment

${\left({\sigma }_{i}\right)}_{n+1}^{trH}={\left({\sigma }_{i}\right)}_{n}^{H}+\left[C\right]{\left\{\stackrel{˙}{\epsilon }\right\}}^{H}\text{Δ}t$

• Check the yield criterion
• If $f\ge 0$ , the hourglass stress correction will be done by unradial return

${\left({\sigma }_{i}\right)}_{n+1}^{H}=P\left({\left({\sigma }_{i}\right)}_{n+1}^{trH},f\right)$