Effect of stress triaxiality and normalized Lode angle on ductile fracture of aluminum 2139-T8

Abstract

This paper presents experimental ductile fracture data for 2139-T8 aluminum alloy under various stress states for the ultimate purpose of probing the fracture envelope as a function of stress triaxiality and normalized Lode angle. A new ductile fracture model is also proposed and validated by experimental data, along with the evaluation of existing ductile fracture models. The data extracted from published literature on 2024-T351 aluminum alloy was also used as an additional experimental data set for validation. An extensive experimental program was implemented to produce data in a wide range of stress states, including tensile tests (with round smooth, round notched, and plate specimens), torsion, compression (with round smooth and round notched specimens), and shear-compression experiments (two different sizes). The combined effects of stress triaxiality and normalized Lode angle were used to define a 3D fracture envelope for fracture strain. A parallel finite element simulation (fine-tuned by the experimental results) has been performed for each experiment to evaluate the evolution of stress triaxiality and normalized Lode angle in the critical section of the specimens with complex geometries. Finally, these results were used to develop two fracture models whose predictions were compared with some of the existing models.

Appendix

Calculating stress triaxiality and normalized Lode angle for shear compression specimens with $\alpha =35.26^\circ$ in the elastic region based on Vural et al. [26].

• (a) Shear Compression, Large Gage Width ($w=t$)

The stress field of shear compression specimen with $w=t$ has the following form

$${\sigma }_{A}=\left[\begin{array}{ccc}{\sigma }_{xx}& {\sigma }_{xy}& 0\\ {\sigma }_{xy}& {\sigma }_{yy}& 0\\ 0& 0& {\sigma }_{zz}\end{array}\right]$$

where ${\sigma }_{zz}=0$ and ${\sigma }_{xx}={\sigma }_{yy}/2$, and since there is no traction along the sides of the gauge section

$$ \sigma_{xy} \cos \alpha - \sigma_{yy} \sin \alpha = 0 \Rightarrow \sigma_{xy} = \sigma_{yy} \tan \alpha = 2\sigma_{xx} \tan \alpha $$

Therefore,

$$ \begin{aligned} \sigma_{A} & = \left[ {\begin{array}{*{20}c} {\sigma_{xx} } & {\sigma_{xy} } & 0 \\ {\sigma_{xy} } & {\sigma_{yy} } & 0 \\ 0 & 0 & {\sigma_{zz} } \\ \end{array} } \right] = \sigma_{xx} \left[ {\begin{array}{*{20}c} 1 & {2\tan \alpha } & 0 \\ {2\tan \alpha } & 2 & 0 \\ 0 & 0 & 0 \\ \end{array} } \right] \Rightarrow \left\{ {\begin{array}{*{20}l} {\sigma_{m} = \sigma_{xx} } \hfill \\ {\overline{\sigma } = \left| {\sigma_{xx} } \right|\sqrt {3\left( {1 + 4{\text{ tan}}^{2} \alpha } \right)} } \hfill \\ \end{array} } \right. \\ & \quad \Rightarrow \eta = \frac{{\sigma_{m} }}{{\overline{\sigma }}} = - 0.33 \\ \end{aligned} $$

Also, the deviatoric stress tensor,

$$ \begin{aligned} S_{A} & = \sigma_{xx} \left[ {\begin{array}{*{20}c} 0 & {2\tan \alpha } & 0 \\ {2\tan \alpha } & 1 & 0 \\ 0 & 0 & { - 1} \\ \end{array} } \right] \Rightarrow J_{3} = S_{1} S_{2} S_{3} = 4 \sigma_{xx}^{3} {\text{tan}}^{2} \alpha \\ & \quad \Rightarrow \overline{\theta } = 1 - \frac{2}{\pi }\cos^{ - 1} \left( {\frac{27}{2}\frac{{J_{3} }}{{\overline{\sigma }^{3} }}} \right) = - 1 \\ \end{aligned} $$

• (b) Shear Compression, Small Gage Width ($w<<t$)

In the gauge section of the shear compression specimen with $w<<t$, the strain tensor has the state of simple shear

$$\varepsilon =\left[\begin{array}{ccc}0& {\varepsilon }_{xy}& 0\\ {\varepsilon }_{xy}& 0& 0\\ 0& 0& 0\end{array}\right]$$

And like the previous case, there is no traction along the sides of the gauge section

$$ \sigma_{xy} \cos \alpha - \sigma_{yy} \sin \alpha = 0 \Rightarrow \sigma_{xy} = \sigma_{yy} \tan \alpha = \sigma_{xx} \tan \alpha $$

Therefore,

$$ \begin{aligned} \sigma_{A} & = \left[ {\begin{array}{*{20}c} {\sigma_{xx} } & {\sigma_{xy} } & 0 \\ {\sigma_{xy} } & {\sigma_{yy} } & 0 \\ 0 & 0 & {\sigma_{zz} } \\ \end{array} } \right] = \sigma_{xx} \left[ {\begin{array}{*{20}c} 1 & {\tan \alpha } & 0 \\ {\tan \alpha } & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} } \right] \Rightarrow \left\{ {\begin{array}{*{20}l} {\sigma_{m} = \sigma_{xx} } \hfill \\ {\overline{\sigma } = \sqrt 3 \left| {\sigma_{xx} } \right|\tan \alpha } \hfill \\ \end{array} } \right. \\ & \quad \Rightarrow \eta = \frac{{\sigma_{m} }}{{\overline{\sigma }}} = - 0.82 \\ \end{aligned} $$

Also, the deviatoric stress tensor,

$$ \begin{aligned} S_{A} & = \sigma_{xx} \left[ {\begin{array}{*{20}c} 0 & {\tan \alpha } & 0 \\ {\tan \alpha } & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} } \right] \Rightarrow J_{3} = S_{1} S_{2} S_{3} = 4 \sigma_{xx}^{3} {\text{tan}}^{2} \alpha \\ & \quad \Rightarrow \overline{\theta } = 1 - \frac{2}{\pi }\cos^{ - 1} \left( {\frac{27}{2}\frac{{J_{3} }}{{\overline{\sigma }^{3} }}} \right) = 0 \\ \end{aligned} $$