Impact of distortional hardening and the strength differential effect on the prediction of large deformation behavior of the Ti6Al4V alloy

Abstract

The ability of three plasticity models to predict the mechanical behavior of Ti6Al4V until fracture is presented. The first model is the orthotropic yield criterion CPB06 developed by Cazacu et al. (Int J Plast 22:1171–1194, 2006) with a distortional hardening, allowing for the description of material anisotropy and the strength differential effect. The second model is the anisotropic Hill’48 yield criterion with distortional hardening, describing the material anisotropy with quadratic functions but is unable to model the strength differential effect. Finally, the third model is the classical Hill’48 yield locus with isotropic hardening. Distortional hardening is modeled through five yield surfaces associated with five levels of plastic work. Each model is validated by comparing the finite element predictions with experimental results, such as the load and displacement field histories of specimens subjected to different stress triaxiality values. Tensile tests are performed on round bars with a V-notch, a through-hole, and two different radial notches; compression tests are performed on elliptical cross-section samples. The numerical results show that none of the models can perfectly predict both the measured load and the sample shape used for validation. However, the CPB06 yield criterion with distortional hardening minimizes the global error of the model predictions. The results provide a quantification of the influence of mechanical features such as hardening phenomenon, plastic anisotropy, and tension–compression asymmetry. The impact of these features on the prediction of the post-necking deformation behavior of the Ti6Al4V alloy is explored.

Appendix: Reduction of the CPB06 yield locus to Hill’48

The equivalent anisotropic stress associated to the Hill’48 yield criterion is defined as:

$$\bar{\sigma }_{\text{Hill'48}}^{2} = 1/2\,{\varvec{\upsigma}}^{T} :{\mathbf{H}}:{\varvec{\upsigma}},$$

(9)

where ${\varvec{\upsigma}} = \left( {\begin{array}{*{20}c} {\sigma_{11} } \\ {\sigma_{22} } \\ {\sigma_{33} } \\ {\sigma_{12} } \\ {\sigma_{13} } \\ {\sigma_{23} } \\ \end{array} } \right)$ and ${\mathbf{H}} = \left[ {\begin{array}{*{20}c} {G + H} & { - H} & { - G} & 0 & 0 & 0 \\ { - H} & {H + F} & { - F} & 0 & 0 & 0 \\ { - G} & { - F} & {F + G} & 0 & 0 & 0 \\ 0 & 0 & 0 & {2N} & 0 & 0 \\ 0 & 0 & 0 & 0 & {2L} & 0 \\ 0 & 0 & 0 & 0 & 0 & {2M} \\ \end{array} } \right]$

On the other hand, the equivalent stress for the CPB06 yield criterion is defined by:

$$\bar{\sigma }_{\text{CPB06}}^{a} = B\left\{ {\left( {\left| {\varSigma_{1} } \right| - k\,\varSigma_{1} } \right)^{\,a} + \left( {\left| {\varSigma_{2} } \right| - k\,\varSigma_{2} } \right)^{\,a} + \left( {\left| {\varSigma_{3} } \right| - k\,\varSigma_{3} } \right)^{a} } \right\} ,$$

(10)

where k is a parameter which takes into account the SD effect and a is the degree of homogeneity.$\varSigma_{1} ,$$\varSigma_{2} ,$$\varSigma_{3}$ are the principal values of the tensor ${\varvec{\Sigma}}$ as defined by:

$${\varvec{\Sigma}} = {\mathbf{C}}:{\mathbf{S}} ,$$

(11)

where C is a fourth-order orthotropic tensor that accounts for the plastic anisotropy of the material and S is the deviator of the Cauchy stress tensor defined by:

$${\mathbf{S}} = {\mathbf{L}}:{\varvec{\upsigma}} .$$

(12)

The tensor C represented in Voigt notations is defined as follows:

$${\mathbf{C}} = \left[ {\begin{array}{*{20}c} {C_{11} } & {C_{12} } & {C_{13} } & 0 & 0 & 0 \\ {C_{12} } & {C_{22} } & {C_{23} } & 0 & 0 & 0 \\ {C_{13} } & {C_{23} } & {C_{33} } & 0 & 0 & 0 \\ 0 & 0 & 0 & {C_{44} } & 0 & 0 \\ 0 & 0 & 0 & 0 & {C_{55} } & 0 \\ 0 & 0 & 0 & 0 & 0 & {C_{66} } \\ \end{array} } \right]$$

(13)

The tensor L is defined by:

$${\mathbf{L}} = \left[ {\begin{array}{*{20}c} {2/3} & { - 1/3} & { - 1/3} & 0 & 0 & 0 \\ { - 1/3} & {2/3} & { - 1/3} & 0 & 0 & 0 \\ { - 1/3} & { - 1/3} & {2/3} & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} } \right]$$

(14)

Rearranging the equations, the tensor ${\varvec{\Sigma}}$ can be written as follows

$${\varvec{\Sigma}} = {\mathbf{C}}:{\mathbf{L}}:{\varvec{\upsigma}} .$$

(15)

where:

$${\mathbf{C}}:{\mathbf{S}} = \left[ {\begin{array}{*{20}c} {\Phi_{1} } & {\Psi_{1} } & {\Omega_{1} } & 0 & 0 & 0 \\ {\Phi_{2} } & {\Psi_{2} } & {\Omega_{2} } & 0 & 0 & 0 \\ {\Phi_{3} } & {\Psi_{3} } & {\Omega_{3} } & 0 & 0 & 0 \\ 0 & 0 & 0 & {C_{44} } & 0 & 0 \\ 0 & 0 & 0 & 0 & {C_{55} } & 0 \\ 0 & 0 & 0 & 0 & 0 & {C_{66} } \\ \end{array} } \right]$$

(16)

$$\begin{aligned} \Phi_{1} = \frac{2}{3}C_{11} - \frac{1}{3}C_{12} - \frac{1}{3}C_{13} \hfill \\ \Phi_{2} = \frac{2}{3}C_{12} - \frac{1}{3}C_{22} - \frac{1}{3}C_{23} \hfill \\ \Phi_{3} = \frac{2}{3}C_{13} - \frac{1}{3}C_{23} - \frac{1}{3}C_{33} \hfill \\ \end{aligned}$$

$$\begin{aligned} \Psi_{1} = - \frac{1}{3}C_{11} + \frac{2}{3}C_{12} - \frac{1}{3}C_{13} \hfill \\ \Psi_{2} = - \frac{1}{3}C_{12} + \frac{2}{3}C_{22} - \frac{1}{3}C_{23} \hfill \\ \Psi_{3} = - \frac{1}{3}C_{13} + \frac{2}{3}C_{23} - \frac{1}{3}C_{33} \hfill \\ \end{aligned}$$

(17)

$$\begin{aligned} \Omega_{1} = - \frac{1}{3}C_{11} - \frac{1}{3}C_{12} + \frac{2}{3}C_{13} \hfill \\ \Omega_{2} = - \frac{1}{3}C_{12} - \frac{1}{3}C_{22} + \frac{2}{3}C_{23} \hfill \\ \Omega_{3} = - \frac{1}{3}C_{13} - \frac{1}{3}C_{23} + \frac{2}{3}C_{33} \hfill \\ \end{aligned}$$

$$B = \left\{ {\left( {\left| {\Phi_{1} } \right| - k\,\Phi_{1} } \right)^{\,a} + \left( {\left| {\Phi_{2} } \right| - k\,\Phi_{2} } \right)^{\,a} + \left( {\left| {\Phi_{3} } \right| - k\Phi_{3} } \right)^{a} } \right\}^{{\frac{ - 1}{a}}}$$

(18)

Consider the general form of CPB06 with $a = 2$. As the Hill’48 model does not account for the SD effect, the value of k is set equal to zero (with $k = 0$). Then, Eq. (10) becomes:

$$\bar{\sigma }_{\text{CPB06}}^{a} = B\left( {\varSigma_{1}^{2} + \varSigma_{2}^{2} + \varSigma_{3}^{2} } \right) .$$

(19)

Replacing Eq. (18) into (19), one can have:

$$\bar{\sigma }_{\text{CPB06}}^{a} = \frac{{\varSigma_{1}^{2} + \varSigma_{2}^{2} + \varSigma_{3}^{2} }}{{\Phi_{1}^{2} + \Phi_{2}^{2} + \Phi_{3}^{2} }} .$$

(20)

Moreover, $\varSigma_{1}^{2} + \varSigma_{2}^{2} + \varSigma_{3}^{2}$ can be written as follows:

$$\varSigma_{1}^{2} + \varSigma_{2}^{2} + \varSigma_{3}^{2} = tr\left( {{\varvec{\Sigma}}^{2} } \right) = \varSigma_{11}^{2} + \varSigma_{22}^{2} + \varSigma_{33}^{2} + 2\varSigma_{12}^{2} + 2\varSigma_{13}^{2} + 2\varSigma_{23}^{2}$$

$$\varSigma_{1}^{2} + \varSigma_{2}^{2} + \varSigma_{3}^{2} = {\varvec{\Sigma}}^{T} :{\mathbf{J}}:{\varvec{\Sigma}} ,$$

(21)

with ${\mathbf{J}} = \left[ {\begin{array}{*{20}c} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \\ \end{array} } \right]$

By considering Eq. (21), Eq. (19) becomes:

$$\bar{\sigma }_{\text{CPB06}}^{2} = B^{2} {\varvec{\Sigma}}^{T} {\mathbf{J}}{\varvec{\Sigma}} = B^{2} {\varvec{\upsigma}}^{T} \left( {{\mathbf{CL}}} \right)^{T} {\mathbf{J}}\left( {{\mathbf{CL}}} \right){\varvec{\upsigma}}$$

$$\bar{\sigma }_{\text{CPB06}}^{2} = B^{2} {\varvec{\upsigma}}^{T} :{\mathbf{H}}^{*} :{\varvec{\upsigma}},$$

(22)

with ${\mathbf{H*}} = \left[ {\begin{array}{*{20}c} {\Phi_{1}^{2} + \Phi_{2}^{2} + \Phi_{3}^{2} } & {\Phi_{1} \Psi_{1} + \Phi_{2} \Psi_{2} + \Phi_{3} \Psi_{3} } & {\Phi_{1} \Omega_{1} + \Phi_{2} \Omega_{2} + \Phi_{3} \Omega_{3} } & 0 & 0 & 0 \\ 0 & {\Psi_{1}^{2} + \Psi_{2}^{2} + \Psi_{3}^{2} } & {\Psi_{1} \Omega_{1} + \Psi_{2} \Omega_{2} + \Psi_{3} \Omega_{3} } & 0 & 0 & 0 \\ 0 & 0 & {\Omega_{1}^{2} + \Omega_{2}^{2} + \Omega_{3}^{2} } & 0 & 0 & 0 \\ 0 & 0 & 0 & {2C_{44} } & 0 & 0 \\ 0 & 0 & 0 & 0 & {2C_{55} } & 0 \\ 0 & 0 & 0 & 0 & 0 & {2C_{66} } \\ \end{array} } \right]$

Then, equating the Hill’48 (Eq. 9) with the CPB06 yield criterion reduced to Eq. (22), we have:

$$\bar{\sigma }_{\text{HILL}}^{2} = \frac{1}{2}{\varvec{\upsigma}}^{T} :{\mathbf{H}}:{\varvec{\upsigma}} = \bar{\sigma }_{\text{CPB06}}^{2} = B^{2} {\varvec{\upsigma}}^{T} :{\mathbf{H}}^{*} :{\varvec{\upsigma}}$$

(23)

As $a = 2$ and $k = 0$, one can have the system of equations that links the parameters of the Hill’48 yield criterion with the parameters of the CPB06 yield criterion:

$$\frac{1}{2}\left( {G + H} \right) = B^{2} \left( {\Phi_{1}^{2} + \Phi_{2}^{2} + \Phi_{3}^{2} } \right)$$

$$\frac{1}{2}\left( {F + H} \right) = B^{2} \left( {\Psi_{1}^{2} + \Psi_{2}^{2} + \Psi_{3}^{2} } \right)$$

$$\frac{1}{2}\left( {F + G} \right) = B^{2} \left( {\Omega_{1}^{2} + \Omega_{2}^{2} + \Omega_{3}^{2} } \right)$$

$$- \frac{H}{2} = B^{2} \left( {\Phi_{1} \Psi_{1} + \Phi_{2} \Psi_{2} + \Phi_{3} \Psi_{3} } \right)$$

(24)

$$- \frac{G}{2} = B^{2} \left( {\Phi_{1} \Omega_{1} + \Phi_{2} \Omega_{2} + \Phi_{3} \Omega_{3} } \right)$$

$$- \frac{F}{2} = B^{2} \left( {\Psi_{1} \Omega_{1} + \Psi_{2} \Omega_{2} + \Phi_{3} \Omega_{3} } \right)$$

$$L = 2B^{2} C_{44}^{2}$$

$$M = 2B^{2} C_{55}^{2}$$

$$N = 2B^{2} C_{66}^{2}$$