*2.2. Stress Space in Cylindrical Coordinates*

Since plastic deformation in most metals does not depend on hydrostatic stress components, it is useful to transform principal stresses from their representation as a 3-dimensional (3D) Cartesian vector of principal stresses *σ* = (*σ*1, *σ*2, *σ*3) into a cylindrical coordinate system with *s* = (*σ*eq, *θ*, *p*), where the equivalent stress *σ*eq represents the norm of the stress deviator *σ* ′ , and the polar angle *θ* lies in the deviatoric plane normal to the hydrostatic axis *p*, which has already been used by Hill [2]. This coordinate transformation improves the efficiency of the training, because only two-dimensional data for the equivalent stress and the polar angle need to be used as training features, whereas the hydrostatic component is disregarded. Hence, by exploiting basic physical principles, we effectively reduce the dimensionality of the problem from 6 independent components of an arbitrary stress tensor to 2 degrees of freedom, without loosing the generality of the formulation. As the polar angle *θ* can be considered a generalized Lode angle [21], it is noted that the Lode angle, by definition, describes the axiality of a loading state in a way that uniaxial loads in different directions result in the same Lode angle. Since our formulation aims at describing anisotropy in the plastic deformation, uniaxial stresses in different directions must possess different angles. To achieve this, we introduce a complex-valued deviatoric stress

$$
\sigma'\_c = \boldsymbol{\sigma} \cdot \mathbf{a} + \mathbf{i}\,\boldsymbol{\sigma} \cdot \mathbf{b} = \sqrt{2/3} \sigma\_{\rm eq} \, e^{i\theta} \,, \tag{12}
$$

where *i* is the imaginary unit, such that the polar angle

$$\theta = \arg \sigma\_c' = -i \ln \frac{\boldsymbol{\sigma} \cdot \boldsymbol{\mathfrak{a}} + i \boldsymbol{\sigma} \cdot \boldsymbol{\mathfrak{b}}}{\sqrt{2/3} \sigma\_{\text{eq}}} \,. \tag{13}$$

with the unit vectors *a* = (2, −1, −1)/ √ 6 and *b* = (0, 1, −1)/ √ 2 that span the plane normal to the hydrostatic axis *c* = (1, 1, 1)/ √ 3.

To transform the gradient of the yield function from this cylindrical stress space back to the principle stress space, in which form it is used to calculate the direction of the plastic strain increments in the return mapping algorithm of the plasticity model, we introduce the Jacobian matrix for this coordinate transformation as

$$\mathbf{J} = \frac{\partial \mathbf{s}}{\partial \sigma} = \begin{pmatrix} \frac{\partial \sigma\_{\text{eq}}}{\partial \sigma\_1} & \frac{\partial \theta}{\partial \sigma\_1} & \frac{\partial p}{\partial \sigma\_1} \\ \frac{\partial \sigma\_{\text{eq}}}{\partial \sigma\_2} & \frac{\partial \theta}{\partial \sigma\_2} & \frac{\partial p}{\partial \sigma\_1} \\ \frac{\partial \sigma\_{\text{eq}}}{\partial \sigma\_3} & \frac{\partial \theta}{\partial \sigma\_3} & \frac{\partial p}{\partial \sigma\_1} \end{pmatrix} \tag{14}$$

where *∂σ*eq/*∂σ<sup>j</sup>* is given in Equation (10), *∂p*/*∂σ<sup>j</sup>* = 1/3 and

$$\frac{\partial \theta}{\partial \sigma} = -i \left( \frac{\mathbf{a} + i \mathbf{b}}{\sigma \cdot \mathbf{a} + i \sigma \cdot \mathbf{b}} - \frac{3 \sigma'}{\sigma\_{\text{eq}}^2} \right) . \tag{15}$$

With this Jacobian, the gradient can be calculated as

$$\frac{\partial f}{\partial \sigma} = \mathbf{J} \frac{\partial f}{\partial \mathbf{s}}.\tag{16}$$
