A Generalized Ogden-Type Elastically Isotropic Hyperelastic Model Including Elastic-Viscoplastic Response

Abstract

The objective of this paper is to generalize an Ogden-type model for elastically isotropic response to include elastic-viscoplastic response. The proposed model uses a strain energy function that depends on the total dilatation and the maximum and minimum elastic distortional stretches. A novel feature of the model is that these elastic distortional stretches are expressed in terms of two independent invariants of an elastic distortional deformation tensor that is determined by an evolution equation. The Cauchy stress is determined by derivatives of the strain energy function, the dilatation and the elastic distortional deformation tensor without the need for determining its principal directions. Examples demonstrate the response of the model for hyperelastic response but the proposed formulation can also model a smooth elastic-plastic transition with rate-independent or rate-dependent response with hardening.

Appendices

Appendix A: Bounds

Consider the function

$$ \begin{aligned} & f(x_{1},x_{3}) = x_{1} + \frac{1}{x_{1} x_{3}} + x_{3} \quad \text{for} \quad x_{1} \ge 1 \,,\quad 0 < x_{3} \le 1 \,, \end{aligned} $$

(40)

for which

$$\begin{aligned} &\frac{{\partial }f}{{\partial }x_{1}} = 1 - \frac{1}{x_{1}^{2}x_{3}} \,, \quad \frac{{\partial }f}{{\partial }x_{3}} = 1 - \frac{1}{x_{1}x_{3}^{2}} \,, \\ &\frac{{\partial }^{2} f}{{\partial }x_{1}^{2}} = \frac{2}{x_{1}^{3} x_{3}} > 0 \,, \quad \frac{{\partial }^{2} f}{{\partial }x_{3}^{2}} = \frac{2}{x_{1} x_{3}^{3}} > 0 \,, \\ &\frac{{\partial }^{2} f}{{\partial }x_{1} {\partial }x_{3}} = \frac{{\partial }^{2} f}{{\partial }x_{3} {\partial }x_{1}} = \frac{1}{x_{1}^{2} x_{3}^{2}} > 0 \,. \end{aligned}$$

(41)

These results can be used to determine that $x_{1}=x_{3}=1$ yields the minimum value of $f$ with

$$ \begin{aligned} &f \ge 3 \,. \end{aligned} $$

(42)

Next, using the spectral form of ${\mathbf {B}_{e}^{\prime}}$, it follows from (4) that

$$ \begin{aligned} &\alpha _{1} = {\mathbf {B}_{e}^{\prime}}\cdot {\mathbf {I}}= \lambda _{e1}^{{\prime }2} + \frac{1}{\lambda _{e1}^{{\prime }2} \lambda _{e3}^{{\prime }2}} + \lambda _{e3}^{ {\prime }2} \ge 3 \,, \\ &\alpha _{2} = {\mathbf {B}_{e}^{\prime}}\cdot {\mathbf {B}_{e}^{\prime}}= \lambda _{e1}^{{\prime }4} + \frac{1}{\lambda _{e1}^{{\prime }4} \lambda _{e3}^{{\prime }4}} + \lambda _{e3}^{ {\prime }4} \ge 3 \,, \\ &\alpha _{2} = \big(\frac{\alpha _{1}}{3} {\mathbf {I}}+ {\mathbf {B}_{e}^{\prime \prime }}\big) \cdot \big(\frac{\alpha _{1}}{3} {\mathbf {I}}+ {\mathbf {B}_{e}^{\prime \prime }}\big) = \frac{\alpha _{1}^{2}}{3} + {\mathbf {B}_{e}^{\prime \prime }}\cdot {\mathbf {B}_{e}^{\prime \prime }}\,, \\ &\alpha _{2} \ge \frac{\alpha _{1}^{2}}{3} \ge \alpha _{1} \ge 3 \,, \\ &{\mathbf {B}_{e}^{\prime}}^{-1} \cdot {\mathbf {I}}= \frac{1}{\lambda _{e1}^{{\prime }2}} + \lambda _{e1}^{ {\prime }2} \lambda _{e3}^{{\prime }2} + \frac{1}{\lambda _{e3}^{{\prime }2}} \ge 3 \,. \end{aligned} $$

(43)

Then, using (12), it can be shown that

$$ \begin{aligned} &{\mathbf {A}}_{p} \cdot {\mathbf {I}}= 3 \big(\frac{\alpha _{1}}{3} - \frac{3}{{\mathbf {B}_{e}^{\prime}}^{-1} \cdot {\mathbf {I}}} \big) \ge 0 \,, \\ &{\mathbf {A}}_{p} \cdot {\mathbf {B}_{e}^{\prime}}= \alpha _{2} - \frac{3 \alpha _{1}}{{\mathbf {B}_{e}^{\prime}}^{-1} \cdot {\mathbf {I}}} \ge \alpha _{1} \big(1 - \frac{3}{{\mathbf {B}_{e}^{\prime}}^{-1} \cdot {\mathbf {I}}}\big) \ge 0 \,. \end{aligned} $$

(44)

Furthermore, it can be numerically shown that

$$ \begin{aligned} &{\mathbf {A}}_{p} \cdot {\mathbf {B}_{e}^{\prime}}\ge {\mathbf {A}}_{p} \cdot {\mathbf {I}}\,, \end{aligned} $$

(45)

which prove the results (18).

Appendix B: Details of Some Derivatives

With the help of (28) it can be shown that

$$\begin{aligned} & \frac{{\partial }\lambda _{e1}^{{\prime }}}{{\partial }\alpha _{1}} = \frac{1}{2 \lambda _{e1}^{{\prime }}} \Big[ \frac{1}{3} + \frac{2}{3} \cos \big( \frac{\pi}{6} + \beta \big) \frac{{\partial }B}{{\partial }\alpha _{1}} - \frac{2B}{3} \sin \big(\frac{\pi}{6} + \beta \big) \frac{{\partial }\beta}{ {\partial }\alpha _{1}} \Big]\,, \\ & \frac{{\partial }\lambda _{e1}^{{\prime }}}{{\partial }\alpha _{2}} = \frac{1}{2 \lambda _{e1}^{{\prime }}} \Big[ \frac{2}{3} \cos \big( \frac{\pi}{6} + \beta \big) \frac{{\partial }B}{{\partial }\alpha _{2}} - \frac{2B}{3} \sin \big(\frac{\pi}{6} + \beta \big) \frac{{\partial }\beta}{ {\partial }\alpha _{2}} \Big]\,, \\ & \frac{{\partial }\lambda _{e2}^{{\prime }}}{{\partial }\alpha _{1}} = \frac{1}{2 \lambda _{e2}^{{\prime }}} \Big[ \frac{1}{3} + \frac{2}{3} \sin \big( \beta \big) \frac{{\partial }B}{{\partial }\alpha _{1}} + \frac{2B}{3} \cos \big(\beta \big) \frac{{\partial }\beta}{ {\partial }\alpha _{1}} \Big]\,, \\ & \frac{{\partial }\lambda _{e2}^{{\prime }}}{{\partial }\alpha _{2}} = \frac{1}{2 \lambda _{e2}^{{\prime }}} \Big[\frac{2}{3} \sin \big( \beta \big) \frac{{\partial }B}{{\partial }\alpha _{2}} + \frac{2B}{3} \cos \big(\beta \big) \frac{{\partial }\beta}{ {\partial }\alpha _{2}} \Big]\,, \\ & \frac{{\partial }\lambda _{e3}^{{\prime }}}{{\partial }\alpha _{1}} = \frac{1}{2 \lambda _{e3}^{{\prime }}} \Big[ \frac{1}{3} - \frac{2}{3} \cos \big( \frac{\pi}{6} - \beta \big) \frac{{\partial }B}{{\partial }\alpha _{1}} - \frac{2B}{3} \sin \big(\frac{\pi}{6} - \beta \big) \frac{{\partial }\beta}{ {\partial }\alpha _{1}} \Big]\,, \\ & \frac{{\partial }\lambda _{e3}^{{\prime }}}{{\partial }\alpha _{2}} = \frac{1}{2 \lambda _{e3}^{{\prime }}} \Big[- \frac{2}{3} \cos \big( \frac{\pi}{6} - \beta \big) \frac{{\partial }B}{{\partial }\alpha _{2}} - \frac{2B}{3} \sin \big(\frac{\pi}{6} - \beta \big) \frac{{\partial }\beta}{ {\partial }\alpha _{2}} \Big]\,. \end{aligned}$$

(46)

Moreover, using (31), it follows that

$$ \begin{aligned} &\frac{{\partial }B}{{\partial }\alpha _{1}} = - \frac{\alpha _{1}}{2B} \,, \quad \frac{{\partial }B}{{\partial }\alpha _{2}} =\frac{3}{4B} \,, \\ &C = \frac{27-\alpha _{1}^{3}+3 \alpha _{1} B^{2}}{2B^{3}} = -\sin (3 \beta ) \,, \\ & \frac{{\partial }C}{{\partial }\alpha _{1}} = \frac{3\big[\alpha _{1}(27-\alpha _{1}^{3})- \alpha _{1}^{2} B^{2} +2 B^{4}\big]}{4B^{5}} \,, \quad \frac{{\partial }C}{{\partial }\alpha _{2}} = - \frac{9\big(27-\alpha _{1}^{3}+\alpha _{1} B^{2}\big)}{8B^{5}} \,, \\ &\frac{{\partial }\beta}{ {\partial }\alpha _{1}} =- \frac{\alpha _{1}(27-\alpha _{1}^{3})- \alpha _{1}^{2} B^{2} +2 B^{4}}{4B^{5} \cos (3\beta )} \,, \quad \frac{{\partial }\beta}{ {\partial }\alpha _{2}} = \frac{3\big(27-\alpha _{1}^{3}+\alpha _{1} B^{2}\big)}{8B^{5} \cos (3\beta )} \,. \end{aligned} $$

(47)

Again, using (31), it follows that

$$ \begin{aligned} &27-\alpha _{1}^{3} = -3 \alpha _{1} B^{2}- 2B^{3} \sin (3 \beta ) \,, \end{aligned} $$

(48)

which can be used to deduce that

$$ \begin{aligned} &\frac{{\partial }\beta}{ {\partial }\alpha _{1}} = \frac{\alpha _{1} B \sin (3 \beta )+2 \alpha _{1}^{2}-B^{2}}{2B^{3} \cos (3 \beta )} \,, \quad \frac{{\partial }\beta}{ {\partial }\alpha _{2}} = - \frac{3 [B \sin (3 \beta ) + \alpha _{1}]}{4 B^{3} \cos (3 \beta )} \,. \end{aligned} $$

(49)