What is the proper method to scale deviatoric stresses back to the yield surface in plane stress 2D?
I cannot get the radial return method to produce the desired results under plain stress conditions in 2D. I'll show this using an example in both 3D and 2D in the hope that my mistake might become apparent.
Three Dimensions
Assume an arbitrary trial Cauchy Stress tensor of:
$\sigma_{tr} = \begin{bmatrix} 90 & 20 & 0 \\ 20 & 90& 0 \\ 0 & 0& 0 \end{bmatrix}$
Then the hydrostatic component is given by: $\sigma_{tr,hyd} = \begin{bmatrix} 50 & 0 & 0 \\ 0 & 50& 0 \\ 0 & 0& 50 \end{bmatrix}$
And the deviatoric component by: $\sigma_{tr,dev} = \begin{bmatrix} 40 & 20 & 0 \\ 20 & 10& 0 \\ 0 & 0& -50 \end{bmatrix}$
We will use the Von Mises criterion to determine whether the stresses are still in the yield surface. This calculated as (for general plane stress):
$\sigma_v = \sqrt{\sigma_{11}^2-\sigma_{11}\sigma_{22}+\sigma_{22}^2+3\sigma_{12}^2} = 86.6025$
For the sake of argument let's assume a yield stress of $\sigma_y=80$. In which case the deviatoric stresses need to be scaled back by a factor of $\alpha={\sigma_y}/{\sigma_v}=0.9238$. So the new deviatoric stress is given by:
$\sigma_{dev}=\alpha*\sigma_{tr,dev} = \begin{bmatrix} 36.95 & 18.48 & 0 \\ 18.48 & 9.24& 0 \\ 0 & 0& -46.19 \end{bmatrix}$
And the scaled back
$\sigma=\sigma_{tr,hyd}+\sigma_{dev} = \begin{bmatrix} 86.95& 18.48 & 0 \\ 18.48 & 59.24& 0 \\ 0 & 0& 3.81 \end{bmatrix}$
If we calculate the corresponding Von Mises stress we get $\sigma_{v}=80$ indicating that the scaling was correct.
Two Dimensions
I haven't repeated a lot of the text here for brevity. Check page 157 here for confirmation of hydrostatic stress in 2D, it's just $(\sigma_{11}+\sigma_{22})/2$. Assuming plane stress then:
$\sigma_{tr} = \begin{bmatrix} 90 & 20 \\ 20 & 90 \\ \end{bmatrix}$ $\sigma_{tr,hyd} = \begin{bmatrix} 75 & 0 \\ 0 & 75 \\ \end{bmatrix}$ $\sigma_{tr,dev} = \begin{bmatrix} 15 & 20 \\ 20 & -15 \\ \end{bmatrix}$ $\sigma_v = 86.6025$
The problem is that scaling the deviatoric stresses ($\alpha$ is the same as before) results in:
$\sigma_{dev}=\alpha*\sigma_{tr,dev} = \begin{bmatrix} 13.86 & 18.48 \\ 18.48 & -13.86 \\ \end{bmatrix}$
And then:
$\sigma=\sigma_{tr,hyd}+\sigma_{dev} = \begin{bmatrix} 88.86& 18.48 \\ 18.48 & 61.14 \\ \end{bmatrix}$
Calculating the Von Mises stress gives a value of $\sigma_{v}=85$, which is outside the yield surface. What is the problem here? What is the proper method to scale the deviatoric stresses back to the yield surface in 2D?
