A nonlocal damage model for plain concrete consistent with cohesive fracture

Abstract

A rate-independent damage constitutive law is proposed to describe the fracture of plain concrete under tensile loading. Here, the target scale is the individual crack. In order to deal with localised damage, the model is inherently nonlocal: the gradient of the damage field is explicitly involved in the constitutive equations; it is parameterised by a nonlocal length scale which is interpreted as the width of the process zone. The model is defined so that its predictions are close to those of a cohesive law for vanishing nonlocal length scales. Therefore, the current model is plainly consistent with cohesive zone model analyses: the nonlocal length scale appears as a small parameter which does not need any specific identification. And four parameters—among which the tensile strength and the fracture energy—enable to adjust the softening cohesive response. Besides, a special attention has been paid to the shape of the initial damage surface and to the relation between damage and stiffness. The damage surface takes into account not only the contrast between tensile and compressive strengths but also experimental evidences regarding its shape in multiaxial tension. And the damage–stiffness relation is defined so as to describe important phenomena such as the stiffness recovery with crack closure and the sustainability of compressive loads by damaged structures. Finally, several comparisons with experimental data (global force/opening responses, size dependency, curved crack paths, crack opening profiles) enable to validate qualitatively and quantitatively the pertinence of the constitutive law in 2D and 3D.

Appendices

Appendix: Introduction of a distorted norm relative to a closed set

Consider a normed vector space E and a closed set ${\mathcal {D}}\subset E$ which contains the point 0. For any $x\in E$, we introduce the following subset of $\mathbb {R}^{+}$: $I\left( x \right) =\left\{ {t\in \mathbb {R}^{+}\;;\;t x\in {\mathcal {D}}} \right\}$. As $0\in {\mathcal {D}}$, $I\left( x \right)$ is nonempty since $0\in I\left( x \right)$ and we denote $t^{{*}}\left( x \right) \in \bar{{\mathbb {R}}}$ the supremum of $I\left( x \right)$, which is evidently positive. Then, we define the following function $\upchi$ that we named the distorted norm relative to ${\mathcal {D}}$:

Definition

$\upchi$ is the function from E to $\mathbb {R}^{+}$ defined as $\upchi \left( x \right) =t^{{*}}\left( x \right) ^{\;-\;1}$. In particular, it can be noticed that $\upchi \left( 0 \right) =0$.

At this stage, $\upchi$ is not necessarily defined everywhere in E since $t^{{*}}\left( x \right)$ may be equal to 0. Therefore, we introduce the condition that 0 is an interior point of ${\mathcal {D}}$ (there exists an open set containing 0 and contained in ${\mathcal {D}})$.

Property

Under the condition that $0\in \mathrm{int}\left( {\mathcal {D}} \right)$, the domain of the function $\upchi$ denoted dom$\left( \upchi \right)$ is equal to E and there exists a constant $C_U >0$ such that:

$$\begin{aligned} \forall \;x\in E\quad 0\le \upchi \left( x \right) \le C_U \left\| x \right\| \end{aligned}$$

Demonstration

Since $0\in \mathrm{int}\left( {\mathcal {D}} \right)$, there exists $\upeta >0$ such that $\forall \;x\in E\;\;\left\| x \right\| \le \upeta \Rightarrow x\in {\mathcal {D}}$. Therefore, $\forall \;x\in E\backslash \{0\}$ one has $\upeta x/{\left\| x \right\| }\in {\mathcal {D}}$ so that $\upeta /{\left\| x \right\| }\in I\left( x \right)$ and, by definition of $t^{{*}}$, $t^{{*}}\left( x \right) \ge \upeta /{\left\| x \right\| }>0$. Therefore, $\upchi \left( x \right) =1/{t^{{*}}(x)}$ is well defined and $x\in \mathrm{dom}\left( \upchi \right)$. In addition $\upchi \left( 0 \right) =0$. The conclusion follows: $\forall \;x\in E\;\;\upchi \left( x \right) \le C_U \left\| x \right\|$ with $C_U =1/\upeta$.

Property

The function $\upchi$ is positive homogeneous of degree 1:

$$\begin{aligned}&\forall \;x\in \mathrm{dom}\left( \upchi \right) \quad \forall \;\lambda \in \mathbb {R}^{+}\quad \lambda x\in \mathrm{dom}\left( \upchi \right) \quad \hbox {and}\nonumber \\&\quad \upchi \left( {\lambda x} \right) =\lambda \upchi \left( x \right) . \end{aligned}$$

Demonstration

The property holds for $\lambda =0$ since $\upchi \left( 0 \right) =0$. Now, consider any $\lambda >0$ and any $x\in \mathrm{dom}\left( \upchi \right)$. One has:

$$\begin{aligned} t\in I\left( {\lambda x} \right)&\Leftrightarrow \left[ {t\ge 0\;\hbox {and}\;t \lambda x\in {\mathcal {D}}} \right] \\&\Leftrightarrow \left[ {\lambda t\ge 0\;\hbox {and}\;\lambda t x\in {\mathcal {D}}} \right] \Leftrightarrow \lambda t\in I\left( x \right) \end{aligned}$$

Therefore, the relation on the supremum reads $t^{{*}}\left( x \right) =\lambda t^{{*}}\left( {\lambda x} \right)$ so that $\lambda x\in \mathrm{dom}\left( \upchi \right)$ and $\upchi \left( {\lambda x} \right) =\lambda \upchi \left( x \right)$.

In order to exhibit a relation between the set ${\mathcal {D}}$ and the lower level sets $D_\upalpha$ of $\upchi$, defined as $D_\upalpha =\left\{ {x\in \mathrm{dom}(\upchi )\;;\;\;\upchi \left( x \right) \le \upalpha } \right\}$, an additional condition should be introduced: ${\mathcal {D}}$ is star convex with respect to the point 0, that is for any $x\in {\mathcal {D}}$ the segment $\left[ {0\;x} \right]$ is included in ${\mathcal {D}}$.

Property

Under the condition that ${\mathcal {D}}$ is star convex with respect to 0, the lower level sets of $\upchi$ are homothetic to ${\mathcal {D}}$: $\forall \;\upalpha >0\;\;D_\upalpha =\left\{ {\upalpha x\;;\;x\in {\mathcal {D}}} \right\}$. In particular, the set ${\mathcal {D}}$ is equal to the lower level set $D_1$.

Demonstration

First, let us demonstrate that ${\mathcal {D}}=D_1$. If $x\in {\mathcal {D}}$, then $1\in I\left( x \right)$ and hence $t^{{*}}\left( x \right) \ge 1$. Therefore, $x\in \mathrm{dom}\left( \upchi \right)$ and $\upchi \left( x \right) =1/{t^{{*}}\left( x \right) }\le 1$ so that $x\in D_1$. Reciprocally, consider that $x\in D_1$. Then $\upchi \left( x \right) \le 1$ and $t^{{*}}\left( x \right) \ge 1$. We distinguish two cases: $t^{{*}}\left( x \right) =1$ and $t^{{*}}\left( x \right) >1$. In the first case, there exists a sequence $\left( {t_n } \right) \in I\left( x \right) ^{\mathbb {N}}$ which converges towards 1. The sequence $\left( {t_n x} \right)$ hence belongs to ${\mathcal {D}}^{\mathbb {N}}$ and converges towards x. As ${\mathcal {D}}$ is a closed set, $x\in {\mathcal {D}}$. In the second case, there exists $t_1 \in I\left( x \right)$ with $t_1 \ge 1$. Thanks to the condition that ${\mathcal {D}}$ is star convex with respect to 0, $\left[ {0\;\;t_1 x} \right] \subset {\mathcal {D}}$ and in particular $x\in {\mathcal {D}}$ which concludes the demonstration that ${\mathcal {D}}=D_1$.

Now, consider $\upalpha >0$. Thanks to the property that $\upchi$ is positive homogeneous of degree one, one has:

$$\begin{aligned} D_\upalpha= & {} \left\{ {x\in \mathrm{dom}\upchi \;;\;{\upchi (x)}/\upalpha \le 1} \right\} \\= & {} \left\{ {x\in \mathrm{dom}\upchi \;;\;\upchi (x/\upalpha )\le 1} \right\} =\left\{ {\upalpha y\;;\;y\in D_1 } \right\} \end{aligned}$$

As ${\mathcal {D}}=D_1$, this concludes the demonstration.

Property

Under the condition that ${\mathcal {D}}$ is star convex with respect to 0, one has $x/{\upchi \left( x \right) }\;\in \;\mathrm{bd}\left( {\mathcal {D}} \right)$ the boundary of ${\mathcal {D}}$ for any $x\in \mathrm{dom}\left( \upchi \right)$ such that $\upchi \left( x \right) \ne 0$. In particular, if $\upchi \left( x \right) =1$ then $x\in \mathrm{bd}\left( {\mathcal {D}} \right)$.

Demonstration

First, consider the case $\upchi \left( x \right) =1$. Then $t^{{*}}\left( x \right) =1$ which implies that $x\in D$ as demonstrated previously. Moreover, for any ${\upvarepsilon } >0$, $1+{\upvarepsilon } \notin I\left( x \right)$ by definition of the supremum, so that $\left( {1+{\upvarepsilon } } \right) \;x\notin {\mathcal {D}}$. In any neighbourhood of x, one can find a point that belongs to ${\mathcal {D}}$ (the point x itself) and a point that does not belong to ${\mathcal {D}}$. This is a characterisation of the boundary of ${\mathcal {D}}$: $x\in \mathrm{bd}\left( {\mathcal {D}} \right)$.

Consider now the general case $\upchi \left( x \right) \ne 0$. Then $\upchi \left( x \right) >0$ and $\upchi \left( {x/{\upchi (x)}} \right) =1$, thanks to the positive homogeneity of degree one of $\upchi$. We can apply the first part of the demonstration and conclude that $x/{\upchi (x)}\in \mathrm{bd}\left( {\mathcal {D}} \right)$.

Unfortunately, the converse of the previous property does not hold in general: a point $x\in \mathrm{bd}\left( {\mathcal {D}} \right)$ is not necessarily characterised by $\upchi \left( x \right) =1$ so that $x/{\upchi (x)}\in \mathrm{bd}\left( {\mathcal {D}} \right)$ does not necessarily characterises the function $\upchi$. Some additional regularity is required on ${\mathcal {D}}$. A practical case of interest is a convex set ${\mathcal {D}}$, which encompasses the condition of star convexity with respect to zero.

Property

If the set ${\mathcal {D}}$ is convex and if $0\in \mathrm{int}\left( {\mathcal {D}} \right)$, then the boundary of ${\mathcal {D}}$ is characterised by $\upchi \left( x \right) =1$, i.e. $\upchi \left( x \right) =1\Leftrightarrow x\in \mathrm{bd}\left( {\mathcal {D}} \right)$.

Demonstration

As ${\mathcal {D}}$ is convex, the condition of star convexity with respect to 0 is fulfilled and the following implication has been proven above: $\upchi \left( x \right) =1\Rightarrow x\in \mathrm{bd}\left( {\mathcal {D}} \right)$.

Now, let us demonstrate the converse. If $x\in \mathrm{bd}\left( {\mathcal {D}} \right)$ then $x\in {\mathcal {D}}$ (since ${\mathcal {D}}$ is a closed set) and $t^{{*}}\left( x \right) \ge 1$ . Now assume (wrongly) that $t^{{*}}\left( x \right) >1$ so that there exists $\uprho >0$ with $\left( {1+\uprho } \right) x\in {\mathcal {D}}$. In this case, we will show that there exists a neighbourhood of x included in ${\mathcal {D}}$, in contradiction with the assumption that $x\in \mathrm{bd}\left( {\mathcal {D}} \right)$.

Indeed, if $0\in \mathrm{int}\left( {\mathcal {D}} \right)$, there exists $\upeta >0$ such that $\forall \;z\in E\;\;\left\| z \right\| \le \upeta \;\;\Rightarrow \;\;z\in {\mathcal {D}}$. Besides, consider the ball ${\mathcal {B}}$ centred on x with radius $\upeta \uprho /{(1+\uprho )}$. For any point $y\in {\mathcal {B}}$, let us define:

$$\begin{aligned} z=\frac{1+\uprho }{\uprho }\left( {y-x} \right) \end{aligned}$$

Then $\left\| z \right\| \le \upeta$ so that $z\in {\mathcal {D}}$. Moreover, $x,\;y\;\hbox {and}\;z$ are also related through the following equation:

$$\begin{aligned} y=\frac{\uprho }{1+\uprho }z+\frac{1}{1+\uprho }\left( {1+\uprho } \right) x \end{aligned}$$

As $z\in {\mathcal {D}}$ and $\left( {1+\uprho } \right) x\in {\mathcal {D}}$ then $y\in {\mathcal {D}}$ thanks to the convexity of ${\mathcal {D}}$. Thus the neighbourhood ${\mathcal {B}}$ of x is included in ${\mathcal {D}}$, in contradiction with $x\in \mathrm{bd}\left( {\mathcal {D}} \right)$. Therefore, the assumption $t^{{*}}\left( x \right) >1$ is false, which demonstrates that $t^{{*}}\left( x \right) =1$ and $\upchi \left( x \right) =1$.

Corollary

Assume that the set ${\mathcal {D}}$ is convex and contains a neighbourhood of 0. Now consider the following problem for any $x\in E$:

(P) find $\uprho >0$ so that $x/\uprho \in \mathrm{bd}\left( {\mathcal {D}} \right)$

Then the problem (P) admits at most one solution:

• If $\upchi \left( x \right) >0$, the solution exists and is equal to $\upchi \left( x \right)$;

• If $\upchi \left( x \right) =0$, no solution exists.

Demonstration

As ${\mathcal {D}}$ is convex and $0\in \mathrm{int}\left( {\mathcal {D}} \right)$, $x/\uprho \in \mathrm{bd}\left( {\mathcal {D}} \right) \Leftrightarrow \upchi \left( {x/\uprho } \right) =1\Leftrightarrow \uprho =\upchi \left( x \right)$, thanks to the previous property, the fact that the domain of $\upchi$ is E and the positive homogeneity of degree one of $\upchi$. As $\uprho$ should be strictly positive, one has to distinguish two cases: if $\upchi \left( x \right) >0$ then $\uprho =\upchi \left( x \right)$; else there is no solution to the problem (P).

The former corollary provides an alternative definition of the function $\upchi$ for a convex, set ${\mathcal {D}}$ which contains a neighbourhood of the point 0. Indeed, for a given point $x\in E$, $\upchi \left( x \right)$ is equal to the unique solution to the problem (P) when it exists; else, $\upchi \left( x \right) =0$. Moreover, it can be shown that the function $\upchi$ is also convex and continuous in this case.

Property

If the set ${\mathcal {D}}$ is convex and such that $0\in \mathrm{int}\left( {\mathcal {D}} \right)$, then the function $\upchi$ is convex.

Demonstration

First, as $0\in \mathrm{int}\left( {\mathcal {D}} \right)$, $\upchi$ is defined everywhere in E. Then, consider two points $x\in E$ and $y\in E$ and a real $0<\upgamma <1$. We should prove that $\upchi \left( {\upgamma x+\left( {1-\upgamma } \right) y} \right) \le \upgamma \upchi \left( x \right) +\left( {1-\upgamma } \right) \upchi \left( y \right)$. Three cases should be distinguished:

• (a) $\upchi \left( x \right) \ne 0$ and $\upchi \left( y \right) \ne 0$

Let us introduce two additional points:

$$\begin{aligned} {y}'= & {} \frac{\upchi \left( x \right) }{\upchi \left( y \right) }y \\ z= & {} \frac{\upgamma \upchi \left( x \right) }{\upgamma \upchi \left( x \right) +\left( {1-\upgamma } \right) \upchi \left( y \right) }x\nonumber \\&+\,\frac{\left( {1-\upgamma } \right) \upchi \left( y \right) }{\upgamma \upchi \left( x \right) +\left( {1-\upgamma } \right) \upchi \left( y \right) }{y}'\\= & {} \frac{\upchi \left( x \right) \;\;\left[ {\upgamma x+\left( {1-\upgamma } \right) y} \right] }{\upgamma \upchi \left( x \right) +\left( {1-\upgamma } \right) \upchi \left( y \right) } \end{aligned}$$

The definition is licit since $\upchi \left( x \right) \ne 0$ and $\upchi \left( y \right) \ne 0$. Geometrically, the point ${y}'$ is collinear to y and belongs to the same level set of $\upchi$ as x since $\upchi \left( {{y}'} \right) =\upchi \left( x \right)$. And the point z is colinear to $\upgamma x+(1-\upgamma )y$ and belongs to the segment $\left[ {x\;\;{y}'} \right]$. Thanks to the convexity of ${\mathcal {D}}$, the lower level set $D_{\upchi \left( x \right) }$ is also convex because it is homothetic to ${\mathcal {D}}$. And as $x\in D_{\upchi \left( x \right) }$ (by definition) and ${y}'\in D_{\upchi \left( x \right) }$, then $z\in D_{\upchi \left( x \right) }$ which implies that $\upchi \left( z \right) \le \upchi \left( x \right) \ne 0$. One can now conclude:

$$\begin{aligned} \upchi \left( {\upgamma x+\left( {1-\upgamma } \right) y} \right)= & {} \frac{\upgamma \upchi \left( x \right) +\left( {1-\upgamma } \right) \upchi \left( y \right) }{\upchi \left( x \right) }\upchi \left( z \right) \\\le & {} \upgamma \upchi \left( x \right) +\left( {1-\upgamma } \right) \upchi \left( y \right) \end{aligned}$$

• (b) $\upchi \left( x \right) =0$ and $\upchi \left( y \right) \ne 0$

This covers also the case where $\upchi \left( x \right) \ne 0$ and $\upchi \left( y \right) =0$ by permuting x with y. The inequality to be demonstrated then reads:

$$\begin{aligned}&\upchi \left( {\upgamma x+\left( {1-\upgamma } \right) y} \right) \le \left( {1-\upgamma } \right) \upchi \left( y \right) \quad \Leftrightarrow \\&\quad \upchi \left( {\frac{\upgamma }{\left( {1-\upgamma } \right) \upchi \left( y \right) } x+\frac{1}{\upchi \left( y \right) }y} \right) \le 1 \end{aligned}$$

Let us denote:

$$\begin{aligned} \uprho =\frac{\upgamma }{\left( {1-\upgamma } \right) \upchi \left( y \right) }\quad \hbox {and}\quad y_1 =\frac{1}{\upchi \left( y \right) }y \end{aligned}$$

As ${\mathcal {D}}$ is convex (hence star convex with respect to 0) and $\upchi \left( y \right) \ne 0$, it has been showed above that $y_1 \in \mathrm{bd}\left( {\mathcal {D}} \right)$. Moreover, the lower level set $D_1$ is equal to ${\mathcal {D}}$ so that the former inequality is equivalent to $z=\uprho x+y_1 \in {\mathcal {D}}$ which should be demonstrated for any $\uprho >0$, $y_1 \in \mathrm{bd}\left( {\mathcal {D}} \right)$ and x such that $\upchi \left( x \right) =0$.

The demonstration is led in two steps. First, let us introduce a sequence $\left( {t_n } \right) \in ] {0\;\;1} ]^{\mathrm{N}}$ which converges toward 1 and consider the sequence of points $z_n =\uprho x+t_n y_1$ which converges towards z. The expression of $z_n$ can be reformulated as follows:

$$\begin{aligned} z_n =\left( {1-t_n } \right) \frac{\uprho x}{1-t_n }+t_n y_1 \end{aligned}$$

As $\upchi \left( x \right) =0$, $I\left( x \right) ={\mathbb {R}}^{+}$ and hence ${\uprho x}{/}{\left( {1-t_n } \right) }\in {\mathcal {D}}$. Moreover, $y_1 \in {\mathcal {D}}$ since ${\mathcal {D}}$ is closed. Therefore $z_n \in {\mathcal {D}}$ thanks to the convexity of ${\mathcal {D}}$.The next step leads to the conclusion: as $z_n$ converges towards z, $z_n \in {\mathcal {D}}$ and ${\mathcal {D}}$ is closed, $z\in {\mathcal {D}}$, which demonstrates the former inequality.

• (c) $\upchi \left( x \right) =0$ and $\upchi \left( y \right) =0$

In this case, we need to show that $\upchi \left( {\upgamma x+\left( {1-\upgamma } \right) y} \right) =0$, which is equivalent to $I\left( {\upgamma x+\left( {1-\upgamma } \right) y} \right) ={\mathbb {R}}^{+}$. To that purpose, consider any $t>0$. Then:

$$\begin{aligned} t\left( {\upgamma x+\left( {1-\upgamma } \right) y} \right) =\frac{1}{2}\left[ {2\upgamma t x} \right] +\frac{1}{2}\left[ {2\left( {1-\upgamma } \right) t y} \right] \end{aligned}$$

As both terms of the right hand side belong to ${\mathcal {D}}$ since $I\left( x \right) ={\mathbb {R}}^{+}$ and $I\left( y \right) ={\mathbb {R}}^{+}$, the convexity of ${\mathcal {D}}$ enables to conclude that the left hand side also belongs to ${\mathcal {D}}$ for any $t>0$, so that $I\left( {\upgamma x+\left( {1-\upgamma } \right) y} \right) ={\mathbb {R}}^{+}$.

Gathering the three cases (a), (b) and (c) finally shows that the function $\upchi$ is convex.

Corollary

If the set ${\mathcal {D}}$ is convex and such that $0\in \mathrm{int}\left( {\mathcal {D}} \right)$, then the function $\upchi$ fulfils the triangular inequality: $\forall \;x,y\in E\;\;\upchi \left( {x+y} \right) \le \upchi \left( x \right) +\upchi \left( y \right)$ .

Demonstration

The triangular inequality is a straightforward consequence of the convexity and the positive homogeneity:

$$\begin{aligned} \upchi \left( {x+y} \right)= & {} 2 \upchi \left( {\frac{1}{2}x+\frac{1}{2}y} \right) \\\le & {} 2\left( {\frac{1}{2}\upchi (x)+\frac{1}{2}\upchi (y)} \right) =\upchi \left( x \right) +\upchi \left( y \right) \end{aligned}$$

Property

If the set ${\mathcal {D}}$ is convex and such that $0\in \mathrm{int}\left( {\mathcal {D}} \right)$, then the function $\upchi$ is Lipschitz continuous.

Demonstration

Thanks to the triangular inequality, one has for any $x\in E$ and $y\in E$:

Moreover, as $0\in \mathrm{int}\left( {\mathcal {D}} \right)$, it has been shown above that there exists a constant $C_U >0$ such that $0\le \upchi \left( z \right) \le C_U \left\| z \right\|$ for any $z\in E$. In particular, $\upchi \left( {x-y} \right) \le C_U \left\| {y-x} \right\|$ and $\upchi \left( {y-x} \right) \le C_U \left\| {y-x} \right\|$. The previous inequalities hence imply:

$$\begin{aligned} \exists \;C_U >0\quad \forall x,y\in E\quad \left| {\upchi \left( y \right) -\upchi \left( x \right) } \right| \le C_U \left\| {y-x} \right\| \end{aligned}$$

At last, one can explained why $\upchi$ is initially referred to as a norm, thanks to the following properties:

Property

If ${\mathcal {D}}$ is a bounded set, then $\upchi \left( x \right) >0$ for any $x\ne 0$ and there exists a constant $C_L >0$ such that $\forall \;x\in E\quad \upchi \left( x \right) \ge C_L \left\| x \right\|$.

Demonstration

As ${\mathcal {D}}$ is bounded, there exists $M>0$ such that $\forall \;x\in D \quad \left\| x \right\| \le M$. Then, by definition of $I\left( x \right)$, one has for any $x\in E\backslash \left\{ 0 \right\}$ and $t\in I\left( x \right)$: $t\ge 0$ and $t x\in {\mathcal {D}}$. Therefore, $\left\| {t x} \right\| =t\left\| x \right\| \le M$, hence $t\le M/{\left\| x \right\| }$. The latter inequality holds also true for the supremum: $t^{{*}}\left( x \right) \le M/{\left\| x \right\| }$. And finally, by definition of $\upchi$, one has $\upchi \left( x \right) \ge C_L \left\| x \right\|$ with $C_L =1/M$ and in particular $\upchi \left( x \right) >0$. At last, the fact that $\upchi \left( 0 \right) =0$ completes the demonstration.

Property

Assume that the set ${\mathcal {D}}$ is convex, bounded and contains a neighbourhood of 0. Assume in addition that it is symmetric with respect to the point 0, that is $x\in D\Leftrightarrow -x\in D$. Then the function $\upchi$ is a norm.

Demonstration

As $0\in \mathrm{int}\left( {\mathcal {D}} \right)$, $\upchi$ is a mapping from E to ${\mathbb {R}}^{+}$.

As ${\mathcal {D}}$ is bounded, $\upchi \left( x \right) =0\;\;\Leftrightarrow \;\;x=0$.

As ${\mathcal {D}}$ is convex and contains a neighbourhood of 0, the function $\upchi$ fulfils the triangular inequality.

Finally, for any $x\in E$ and $\lambda \ge 0$, one has $\upchi \left( {\lambda x} \right) =\lambda \upchi \left( x \right)$. Moreover, $\upchi \left( {-x} \right) =\upchi \left( x \right)$ thanks to the symmetry of ${\mathcal {D}}$ with respect to 0. Hence $\upchi \left( {-\lambda x} \right) =\lambda \upchi \left( x \right)$. Both equalities can be gathered so that for any $x\in E$ and any $\uprho \in \mathbb {R}$, $\upchi \left( {\uprho x} \right) =\left| \uprho \right| \upchi \left( x \right)$.

This concludes the demonstration that $\upchi$ is a norm.

Note that if ${\mathcal {D}}$ is not bounded, then the function $\upchi$ is only a semi-norm. Besides, if ${\mathcal {D}}$ is not a symmetric set, $\upchi$ is not a norm because $\upchi \left( {-\;x} \right) \ne \upchi \left( x \right)$ a priori, which refers to Minkowski norms (but without the required regularity).

Appendix: Properties of conewise elasticity potentials

Consider a strain energy density ${\varvec{{\upvarepsilon }}}\mapsto \hbox {w}\left( {\varvec{{\upvarepsilon }}} \right)$ which is positive, continuously differentiable, positive homogeneous of degree 2 and convex. Now assume that there exists a set of strain tensors ${\mathcal {C}}$ (not reduced to the point ${\varvec{{\upvarepsilon }}}=0$) where the potential $\hbox {w}$ is equal to zero; this corresponds to a strain domain with zero stiffness.

Property

${\mathcal {C}}$ is a closed convex cone.

Demonstration

Since the function $\hbox {w}$ is continuous and ${\mathcal {C}}$ is the inverse image of the closed set $\left\{ 0 \right\}$ under w, ${\mathcal {C}}$ is closed. Besides, it is a cone thanks to the property of positive homogeneity of $\hbox {w}$. And finally, as ${\mathcal {C}}$ corresponds to the minimisers of a convex function, it is convex.

As the stiffness can be equal to zero, one can expect that some stress states are not admissible. This corresponds to stress tensors for which the dual potential is infinite, where the latter is defined as:

$$\begin{aligned} {\varvec{{\upsigma }}}\mapsto \hbox {w}^{{*}}\left( {\varvec{{\upsigma }}} \right) =\mathop {\sup }\limits _{\varvec{{\upvarepsilon }}} \;\left[ {\varvec{\upsigma }} \mathbf{:}{\varvec{{\upvarepsilon }}} -\hbox {w}\left( {\varvec{{\upvarepsilon }}} \right) \right] \end{aligned}$$

Property

The set of admissible stress tensors corresponding to finite values of the dual potential is the dual cone of ${\mathcal {C}}$, i.e. ${\mathcal {C}}^{{*}}=\left\{ {{\varvec{{\upsigma }}}\;\;\hbox {s.t.}\;\;\forall \;{\varvec{{\upvarepsilon }}}\in {\mathcal {C}}\;\;{{\varvec{\upsigma }} \mathbf{:}{\varvec{\upvarepsilon }} }\le 0} \right\}$. Actually, an additional technical assumption is necessary to show that stress tensors lying on the boundary of ${\mathcal {C}}^{ {*}}$ are admissible: it is detailed in the demonstration and it refers to a nonzero stiffness in any direction normal to C.

Demonstration

• (a) First, consider the case where ${\varvec{{\upsigma }}}$ is such that there exists ${\varvec{{\upvarepsilon }}}\in {\mathcal {C}}$ so that ${ {\varvec{\upsigma }} \mathbf{:}{\varvec{\upvarepsilon }} }>0$. Then the same inequality holds for any $\lambda {\varvec{{\upvarepsilon }}}$ with $\lambda >0$ leading to:

  $$\begin{aligned} \hbox {w}^{{*}}\left( \upsigma \right)\ge & {} \mathop {\sup }\limits _{\lambda >0} \left[ {\lambda { {\varvec{\upsigma }} \mathbf{:}{\varvec{\upvarepsilon }} }-\hbox {w}\left( {\lambda {\varvec{{\upvarepsilon }}}} \right) } \right] \\= & {} +\infty \quad \hbox {since}\quad \hbox {w}\left( {\lambda {\varvec{{\upvarepsilon }}}} \right) =0 \end{aligned}$$

• (b) Conversely, consider the case where ${\varvec{{\upsigma }}}$ is such that ${ {\varvec{\upsigma }} \mathbf{:}{\varvec{\upvarepsilon }} }<0$ for any ${\varvec{{\upvarepsilon }}}\in {\mathcal {C}}$ and decompose ${\varvec{{\upvarepsilon }}}$ as ${\varvec{{\upvarepsilon }}}=\lambda \mathbf{d}$ where $\lambda =\left\| {\varvec{{\upvarepsilon }}} \right\|$ and $\left\| \mathbf{d} \right\| =1$ (i.e. $\mathbf{d}\in {\mathcal {S}}_1$ the unit sphere which is compact since it is closed and bounded in a space of finite dimension). Then, one has:

  $$\begin{aligned}&\hbox {w}^{{*}}\left( {\varvec{{\upsigma }}} \right) =\mathop {\sup }\limits _{\lambda \ge 0} \;\left[ {\lambda {{\varvec{\upsigma }} \mathbf{:d}}_\lambda -\lambda ^{2}\hbox {w}\left( {\mathbf{d}_\lambda } \right) } \right] \quad \hbox {where}\\&\quad \mathbf{d}_\lambda \in \mathop {\arg \max }\limits _{\mathbf{d}\in {\mathcal {S}}_1 } \left[ {\lambda {\varvec{\upsigma }} \mathbf{:d}-\lambda ^{2}\hbox {w}\left( \mathbf{d} \right) } \right] \end{aligned}$$

  $\mathbf{d}_\lambda$ is indeed defined because the function $\mathbf{d}\mapsto \lambda {\varvec{\upsigma }} \mathbf{:d}-\lambda ^{2}\hbox {w}\left( \mathbf{d} \right)$ is continuous and the unit sphere is compact. Moreover, $\mathbf{d}_\lambda \in {\mathcal {S}}_1$ so that one can extract a convergent sequence when $\lambda \rightarrow \infty$, still named $\mathbf{d}_\lambda$ for the sake of simplicity; its limit is denoted $\mathbf{d}_\infty$. If $\mathbf{d}_\infty \notin {\mathcal {C}}$ then $\hbox {w}\left( {\mathbf{d}_\infty } \right) >0$ and by continuity of w, there exist $m>0$ and $\lambda _0$ so that for any $\lambda \ge \lambda _0$, $\hbox {w}\left( {\mathbf{d}_\lambda } \right) >m$. Then:

  $$\begin{aligned} \;\forall \;\lambda\ge & {} \lambda _0 \quad \lambda {{\varvec{\upsigma }} \mathbf{:d}}_\lambda -\lambda ^{2}\hbox {w}\left( {\mathbf{d}_\lambda } \right) \le \lambda \left\| {\varvec{{\upsigma }}} \right\| -\lambda ^{2}m\\\le & {} \frac{\left\| {\varvec{{\upsigma }}} \right\| ^{2}}{4m}\quad \Rightarrow \hbox {w}^{{*}}\left( {\varvec{{\upsigma }}} \right) <+\infty \end{aligned}$$

  And if $\mathbf{d}_\infty \in \mathcal {C}$, then ${ {\varvec{\upsigma }} \mathbf{:d}}_\infty <0$ by assumption. Hence there exist $m>0$ and $\lambda _0$ so that for any $\lambda \ge \lambda _0$, ${{\varvec{\upsigma }} \mathbf{:d}}_\lambda <-\;m$. Then:

  $$\begin{aligned} \;\forall \;\lambda\ge & {} \lambda _0 \quad \lambda {{\varvec{\upsigma }} \mathbf{:d}}_\lambda -\lambda ^{2}\hbox {w}\left( {\mathbf{d}_\lambda } \right) \le \lambda {{\varvec{\upsigma }} \mathbf{:d}}_\lambda \\\le & {} -\;\lambda _0 m\quad \Rightarrow \hbox {w}^{{*}}\left( {\varvec{{\upsigma }}} \right) <+\infty \end{aligned}$$

• (c) The case where the stress tensor belongs to the boundary of the dual cone should now be addressed. With the previous notations, the potentially problematic situation consists of (i) $\mathbf{d}_\infty \in \mathcal {C}$ with (ii) ${{\varvec{\upsigma }} \mathbf{:d}}_\infty =0$ (${\varvec{{\upsigma }}}$ is thus an outer normal of $\mathcal {C}$ in $\mathbf{d}_\infty )$ and (iii) there exists a subsequence (still denoted $\mathbf{d}_\lambda )$ such that ${{\varvec{\upsigma }} \mathbf{:d}}_\lambda >0$ (hence $\mathbf{d}_\lambda \notin {\mathcal {C}})$. In that case, an additional hypothesis of transverse stiffness is required to conclude:

  $$\begin{aligned}&\forall \;\mathbf{d}\in \partial C\cap S_1 \quad \exists \;K>0\quad \exists \;V\in {\mathcal {V}}\left( \mathbf{d} \right) \\&\forall \;\mathbf{{d}'}\in V\cap S_1 \backslash C\quad w\left( {\mathbf{{d}'}} \right) \ge K\left\| {\mathbf{{d}'}-\mathbf{d}} \right\| ^{2} \end{aligned}$$

  where $\partial C$ is the boundary of C and ${\mathcal {V}}\left( d \right)$ denotes the set of neighbourhoods of $\mathbf{d}$; we also recall that $\hbox {w}\left( \mathbf{d} \right) =0$ and $\hbox {w}'\left( \mathbf{d} \right) =0$. The condition refers to a nonzero stiffness in directions normal to C, see Curnier et al. (1995) for its admissibility.

This additional condition is applied to the sequence $\mathbf{d}_\lambda$:

$$\begin{aligned}&\exists \;K>0\quad \exists \;\lambda _0 \ge 0\quad \forall \;\lambda \ge \lambda _0 \quad w\left( {\mathbf{d}_\lambda } \right) \\&\quad \ge K\left\| {\mathbf{d}_\lambda -\mathbf{d}_\infty } \right\| ^{2} \end{aligned}$$

It leads to the following upper bound for any $\lambda \ge \lambda _0$ :

$$\begin{aligned} \lambda {{\varvec{\upsigma }} \mathbf{:d}}_\lambda -\lambda ^{2}\hbox {w}\left( {\mathbf{d}_\lambda } \right)\le & {} \lambda { {\varvec{\upsigma }} \mathbf{:d}}_\lambda -K \lambda ^{2}\left\| {\mathbf{d}_\lambda -\mathbf{d}_\infty } \right\| ^{2}\quad \\= & {} \lambda {\varvec{\upsigma }} \mathbf{:}\left( {\mathbf{d}_\lambda -\mathbf{d}_\infty } \right) -K \lambda ^{2}\left\| {\mathbf{d}_\lambda -\mathbf{d}_\infty } \right\| ^{2} \\\le & {} \lambda \left\| {\varvec{{\upsigma }}} \right\| \left\| {\mathbf{d}_\lambda -\mathbf{d}_\infty } \right\| -K \lambda ^{2}\left\| {\mathbf{d}_\lambda -\mathbf{d}_\infty } \right\| ^{2} \\\le & {} \mathop {\sup \;}\limits _{x\ge 0} \left( {\left\| {\varvec{{\upsigma }}} \right\| x-K x^{2}} \right) \quad =\quad {\left\| {\varvec{{\upsigma }}} \right\| ^{2}}{/}{4K} \\ \end{aligned}$$

It proves that $\hbox {w}^{{*}}\left( {\varvec{{\upsigma }}} \right)$ is finite and concludes the demonstration.