17.2.3 Rankine-Hill - Anisotropic

A plane stress continuum model, which can capture different strengths and softening characteristics in orthogonal directions, was formulated by Lourenço [61], Lourenço, Rots & Blaauwendraad [64], and Lourenço, De Borst & Rots [62]. It is based on multi-surface plasticity, comprising of an anisotropic Rankine yield criterion combined with an anisotropic Hill criterion for compression [Fig.17.23].

Figure 17.23: Rankine-Hill yield condition

Masonry is an example of a material for which this criterion applies, having different strengths parallel and perpendicular to the bed joints. The formulation of the model is an extension of the Rankine Principal Stress model [§17.1.5].

Rankine.

The yield criterion for the Rankine model reads

$$\sqrt{{ \tfrac{1}{2}\, \boldsymbol{\xi}^{\mathrm{\scriptscriptstyle{T}}}\mathbf{P\!}_{\mathrm{t}} \: \boldsymbol{\xi} }} {\tfrac{{1}}{{2}}} \boldsymbol\pi \boldsymbol\xi$$ (17.214)

with the projection matrix P_t and vector $\boldsymbol\pi$ given by

$${\tfrac{{1}}{{2}}} \left[\vphantom{ \negthickspace \begin{array}{ccc} 1 & -1 & \cdot... ...\\ [0.8ex] \cdot & \cdot & 4 \alpha_{\tau} \end{array} \negthickspace }\right. \begin{array}{ccc} 1 & -1 & \cdot\\ [0.8ex] -1 & 1 & \cdot\\ [0.8ex] \cdot & \cdot & 4 \alpha_{\tau} \end{array} \left.\vphantom{ \negthickspace \begin{array}{ccc} 1 & -1 & \cdot... ...\\ [0.8ex] \cdot & \cdot & 4 \alpha_{\tau} \end{array} \negthickspace }\right] \boldsymbol\pi \left\{\vphantom{ \negthickspace \begin{array}{c} 1 \\ [0.8ex] 1 \\ [0.8ex] 0 \end{array} \negthickspace }\right. \begin{array}{c} 1 \\ [0.8ex] 1 \\ [0.8ex] 0 \end{array} \left.\vphantom{ \negthickspace \begin{array}{c} 1 \\ [0.8ex] 1 \\ [0.8ex] 0 \end{array} \negthickspace }\right\}$$ (17.215)

The parameter $\alpha_{{\tau}}^{}$ controls the shear stress contribution to failure and can be expressed as

$$\alpha_{{\tau}}^{} {\frac{{f_{\mathrm{t}.x} \; f_{\mathrm{t}.y}}}{{\tau_{\mathrm{u}}^{2}}}}$$ (17.216)

where f_t.x and f_t.y are the tensile strengths in the x and y direction respectively, and $\tau_{{\mathrm{u}}}^{}$ is the shear strength at zero normal stress. The normal Rankine value is $\alpha_{{\tau}}^{}$ = 1 . The reduced stress vector reads

$$\boldsymbol\xi \boldsymbol\sigma \boldsymbol\Gamma$$ (17.217)

with

$$\boldsymbol\sigma \left\{\vphantom{ \negthickspace \begin{array}{c} \sigma _{x} \\ \sigma _{y} \\ \tau_{xy} \end{array} \negthickspace }\right. \begin{array}{c} \sigma _{x} \\ \sigma _{y} \\ \tau_{xy} \end{array} \left.\vphantom{ \negthickspace \begin{array}{c} \sigma _{x} \\ \sigma _{y} \\ \tau_{xy} \end{array} \negthickspace }\right\} \boldsymbol\Gamma \left\{\vphantom{ \negthickspace \begin{array}{c} \sigma _{\mathrm{t}.x} \\ \sigma _{\mathrm{t}.y} \\ 0 \end{array} \negthickspace }\right. \begin{array}{c} \sigma _{\mathrm{t}.x} \\ \sigma _{\mathrm{t}.y} \\ 0 \end{array} \left.\vphantom{ \negthickspace \begin{array}{c} \sigma _{\mathrm{t}.x} \\ \sigma _{\mathrm{t}.y} \\ 0 \end{array} \negthickspace }\right\}$$ (17.218)

Exponential softening is assumed in the orthogonal directions, described by

$$\sigma_{{\mathrm{t}.x}}^{} {\frac{{ f_{\mathrm{t}.x} }}{{ g_{fx} }}} \kappa \sigma_{{\mathrm{t}.y}}^{} {\frac{{ f_{\mathrm{t}.y} }}{{ g_{fy} }}} \kappa$$ (17.219)

where g_fx and g_fy depict the inelastic work in the orthogonal directions. Regularization of the energy dissipation is achieved by assuming that the inelastic work is uniformly distributed over the equivalent length h as follows

$${\frac{{ G_{\mathrm{f}} }}{{ h }}}$$ (17.220)

where h is related to the area of a finite element A_e (Feenstra [27])

$$\alpha_{{h}}^{} \sqrt{{ A_{e} }}$$ (17.221)

and $\alpha_{{h}}^{}$ is one for quadratic elements and $\sqrt{{2}}$ for linear elements (Rots [87]). Note that, in order to avoid snap-back at constitutive level, the element size must be limited to

$$\leq {\frac{{ G_{\mathrm{f}} \: E }}{{ f^{2}_{\mathrm{t}} }}}$$ (17.222)

Alternatively, regularization is achieved by adding a rate term to the cracking stress (Van Zijl [106]). A simple viscous cracking model is given by

$$\sigma_{{\mathrm{t}.x}}^{} \left(\vphantom{ f_{\mathrm{t}.x} + m_{x} \, \dot{\kappa} }\right. \dot{{\kappa}} \left.\vphantom{ f_{\mathrm{t}.x} + m_{x} \, \dot{\kappa} }\right) {\frac{{ f_{\mathrm{t}.x} }}{{ g_{fx} }}} \kappa \sigma_{{\mathrm{t}.y}}^{} \left(\vphantom{ f_{\mathrm{t}.y} + m_{y} \, \dot{\kappa} }\right. \dot{{\kappa}} \left.\vphantom{ f_{\mathrm{t}.y} + m_{y} \, \dot{\kappa} }\right) {\frac{{ f_{\mathrm{t}.y} }}{{ g_{fy} }}} \kappa$$ (17.223)

where m is the cracking viscosity, to be determined by inverse analysis. An alternative cracking rate model has been derived from the activation energy concept (Wu & Bazant [114]). This enhances the cracking stress as follows

$$\begin{split}\sigma _{\mathrm{t}.x} &= f_{\mathrm{t}.x} \: e^... ... \dot{\kappa} }{ \dot{\kappa}_{\mathrm{r}} }\right) \end{split}$$ (17.224)

where $\dot{{\kappa}}_{{\mathrm{r}}}^{}$ is a constant, low reference crack velocity. The other two model parameters k₀ , k₁ have no physical meaning and must be found by inverse analysis. The softening is governed by the maximum principal plastic strain as follows

$$\dot{{\kappa}} \dot{{\varepsilon }}_{{1.\mathrm{p}}}^{} \sqrt{{ \tfrac{1}{2}\, \dot{\boldsymbol{\varepsilon }}_{\mathrm{p... ...athrm{\scriptscriptstyle{T}}}\, \dot{\boldsymbol{\varepsilon }}_{\mathrm{p}} }}$$ (17.225)

with

$${\tfrac{{1}}{{2}}} \left[\vphantom{ \negthickspace \begin{array}{ccc} 1 & -1 & \cdot... ...dot\\ [\smallskipamount] \cdot & \cdot & 1 \end{array} \negthickspace }\right. \begin{array}{ccc} 1 & -1 & \cdot\\ [\smallskipamount] -1 & 1 & \cdot\\ [\smallskipamount] \cdot & \cdot & 1 \end{array} \left.\vphantom{ \negthickspace \begin{array}{ccc} 1 & -1 & \cdot... ...dot\\ [\smallskipamount] \cdot & \cdot & 1 \end{array} \negthickspace }\right]$$ (17.226)

The plastic strain follows from the flow rule

$$\dot{{\boldsymbol{\varepsilon }}}_{{\mathrm{p}}}^{} \dot{{\lambda}} {\frac{{ \partial{g} }}{{ \partial{\boldsymbol{\sigma }} }}}$$ (17.227)

with the plastic potential given by

$$\sqrt{{ \tfrac{1}{2}\, \boldsymbol{\xi}^{\mathrm{\scriptscriptstyle{T}}}\, \mathbf{P\!}_{g} \: \boldsymbol{\xi}}} {\tfrac{{1}}{{2}}} \boldsymbol\pi \boldsymbol\xi$$ (17.228)

with

$${\tfrac{{1}}{{2}}} \left[\vphantom{ \negthickspace \begin{array}{ccc} 1 & -1 & \cdot... ...dot\\ [\smallskipamount] \cdot & \cdot & 4 \end{array} \negthickspace }\right. \begin{array}{ccc} 1 & -1 & \cdot\\ [\smallskipamount] -1 & 1 & \cdot\\ [\smallskipamount] \cdot & \cdot & 4 \end{array} \left.\vphantom{ \negthickspace \begin{array}{ccc} 1 & -1 & \cdot... ...dot\\ [\smallskipamount] \cdot & \cdot & 4 \end{array} \negthickspace }\right]$$ (17.229)

By substitution of the plastic strains into (17.225) the equivalent strain increment reduces to

$$\dot{{\kappa}} \dot{{\lambda}}$$ (17.230)

Hill.

The Hill yield criterion is a rotated centered ellipsoid in the stress space, expressed as

$$\sqrt{{ \tfrac{1}{2}\, \boldsymbol{\sigma }^{\mathrm{\scriptscriptstyle{T}}}\, \mathbf{P\!}_{\mathrm{c}} \: \boldsymbol{\sigma }}} \bar{{\sigma }}_{{\mathrm{c}}}^{} \kappa_{{\mathrm{c}}}^{}$$ (17.231)

with the projection matrix

$$\left[\vphantom{ \negthickspace \begin{array}{ccc} 2 \, \displays... ... [\bigskipamount] \cdot & \cdot & 2 \gamma \end{array} \negthickspace }\right. \begin{array}{ccc} 2 \, \displaystyle{ \frac{ \bar{\sigma }_{\mat... ...athrm{c}} } } & \cdot \\ [\bigskipamount] \cdot & \cdot & 2 \gamma \end{array} \left.\vphantom{ \negthickspace \begin{array}{ccc} 2 \, \displays... ... [\bigskipamount] \cdot & \cdot & 2 \gamma \end{array} \negthickspace }\right]$$ (17.232)

The yield value $\bar{{\sigma }}_{{\mathrm{c}}}^{}$ is given by

$$\bar{{\sigma }}_{{\mathrm{c}}}^{} \sqrt{{ \bar{\sigma }_{\mathrm{c}.x} \: \bar{\sigma }_{\mathrm{c}.y} }}$$ (17.233)

where $\bar{{\sigma }}_{{\mathrm{c}.x}}^{}$ and $\bar{{\sigma }}_{{\mathrm{c}.y}}^{}$ are the yield values along the x direction and y direction respectively and $\kappa_{{\mathrm{c}}}^{}$ , which controls the hardening and softening behavior, is given by the hardening hypothesis

$$\dot{{\kappa}}_{{\mathrm{c}}}^{} {\frac{{ 1 }}{{ \bar{\sigma }_{\mathrm{c}} }}} \boldsymbol\sigma \dot{{\boldsymbol{\varepsilon }}}_{{\mathrm{p}}}^{} \dot{{\lambda}}_{{\mathrm{c}}}^{}$$ (17.234)

The parameters $\beta$ and $\gamma$ determine the shape of the yield surface: $\beta$ rotates the surface around the shear stress axis and can be determined from biaxial compression tests (Lourenço [61]), $\gamma$ controls the shear stress contribution to failure and is calculated from

$$\gamma {\frac{{ f_{\mathrm{c}.x} \: f_{\mathrm{c}.y} }}{{ \tau_{\mathrm{\mathrm{u}}^{2}} }}}$$ (17.235)

where $\tau_{{\mathrm{u}}}^{}$ is the pure (material) shear strength.

The yield surface hardens, as described by a parabolic hardening rule, followed by parabolic/exponential softening [Fig.17.24].

Figure 17.24: Hardening-softening law for Hill yield criterion

The peak strengths f_c.x and f_c.y are assumed to be reached simultaneously, i.e., isotropic hardening up to the plastic strain $\kappa_{{\mathrm{p}}}^{}$ . However, the subsequent softening is orthotropic, governed by different fracture energies in the orthogonal directions ( G_fc.x, G_fc.y ). For practical reasons, all stress values in Figure 17.24 are related to the peak strengths f_c.i (i indicating either of the directions x or y ) as follows: $\bar{{\sigma }}_{{\mathrm{i}.i}}^{}$ = ${\tfrac{{1}}{{3}}}$f_c.i , $\bar{{\sigma }}_{{\mathrm{m}.i}}^{}$ = ${\tfrac{{1}}{{2}}}$f_c.i and $\bar{{\sigma }}_{{\mathrm{r}.i}}^{}$ = ${\frac{{1}}{{7}}}$f_c.i . The three regions of this hardening-softening rule are given by

$$\begin{split}\bar{\sigma }_{1i}(\kappa_{\mathrm{c}}) &= \bar{... ...i} - \bar{\sigma }_{\mathrm{r}.i} } \right) \right) \end{split}$$ (17.236)

The intermediate plastic strain value $\kappa_{{\mathrm{m}.i}}^{}$ is given by

$$\kappa_{{\mathrm{m}.i}}^{} {\frac{{75}}{{67}}} {\frac{{ G_{f_{\mathrm{c}.i}} }}{{ h\:f_{\mathrm{c}.i} }}} \kappa_{{\mathrm{p}}}^{}$$ (17.237)

where h is the equivalent mesh length of (17.221). To avoid snap-back at constitutive level, it is required that

$$\kappa_{{\mathrm{m}.i}}^{} \ge {\frac{{ f_{\mathrm{c}.i} }}{{ E_{i} }}} \kappa_{{\mathrm{p}}}^{}$$ (17.238)