Anisotropic auxetic composite laminates: a polar approach

Let us consider an anisotropic layer in a plane-stress state, with $\mathbb{S}$ and $\mathbb{Q}=\mathbb{S}^{-1}$ respectively the compliance and the reduced stiffness tensors. The in-plane Poisson’s ratio $\nu_{12}(\theta)$ at the direction inclined of $\theta$ on the $x_{1}-$axis is defined as[12, 27, 28, 29] (we adopt in this study the Kelvin’s notation[30, 31] for the elasticity tensors)

$$\nu_{12}(\theta):=-\frac{\mathbb{S}_{12}(\theta)}{\mathbb{S}_{11}(\theta)}.$$

(1)

Because $\mathbb{S}_{11}(\theta)>0\ \forall\theta$,

$$\nu_{12}(\theta)<0\iff\mathbb{S}_{12}(\theta)>0.$$

(2)

In the polar formalism, the above condition for an orthotropic layer becomes[25]

$$\mathbb{S}_{12}(\theta)=-t_{0}+2t_{1}-r_{0}\cos 4(\varphi_{0}-\theta)>0,$$

(3)

with $t_{0},t_{1},r_{0}$ non-negative tensor invariants depending upon the components of $\mathbb{S}$ and $\varphi_{0}$ a polar angle. In view of the forthcoming developments, it is worth to transform the previous relation. To this end, we express the polar parameters of the compliance $\mathbb{S}$ by those of the stiffness $\mathbb{Q}$:

$$\begin{array}[]{l}t_{0}=2\dfrac{T_{0}T_{1}-R_{1}^{2}}{\varDelta},\vskip 6.0pt plus 2.0pt minus 2.0pt\\ t_{1}=\dfrac{T_{0}^{2}-R_{0}^{2}}{2\varDelta},\vskip 6.0pt plus 2.0pt minus 2.0pt\\ r_{0}\mathrm{e}^{4\mathrm{i}\varphi_{0}}=\dfrac{2}{\varDelta}(R_{1}^{2}\mathrm{e}^{4\mathrm{i}\varPhi_{1}}-T_{1}R_{0}\mathrm{e}^{4\mathrm{i}\varPhi_{0}}),\end{array}$$

(4)

where $T_{0},T_{1},R_{0},R_{1}$ are the polar invariant (positive) moduli of $\mathbb{Q}$ and $\varPhi_{0},\varPhi_{1}$ the two polar angles of $\mathbb{Q}$, whose difference is the fifth invariant. $\varDelta$ is the invariant given by

$$\varDelta=4T_{1}(T_{0}^{2}-R_{0}^{2})-8R_{1}^{2}[T_{0}-R_{0}\cos 4(\varPhi_{0}-\varPhi_{1})].$$

(5)

Some simple passages give

$$\begin{split}r_{0}\cos 4(\varphi_{0}-\theta)&=\frac{2}{\varDelta}[R_{1}^{2}\cos 4(\varPhi_{1}-\theta)-\\ &-T_{1}R_{0}\cos 4(\varPhi_{0}-\theta)].\end{split}$$

(6)

As, actually, $\varDelta=\det\mathbb{Q}>0$, it can be ignored in the following, because it does not change the sign of $\mathbb{S}_{12}$. So, in terms of the polar parameters of $\mathbb{Q}$ the auxeticity condition is

$$\begin{split}2(T_{0}T_{1}-R_{1}^{2})-T_{0}^{2}+R_{0}^{2}+2[R_{1}^{2}\cos 4(\varPhi_{1}-\theta)-&\\ -T_{1}R_{0}\cos 4(\varPhi_{0}-\theta)]&<0.\end{split}$$

(7)

A rotation $\alpha$ of the frame corresponds to subtract $\alpha$ from the two polar angles. If $\alpha=\varPhi_{1}$ or, equivalently, if the reference frame is chosen in such a way that $\varPhi_{1}=0$, which corresponds, for unidirectional (UD) plies, to put the $x_{1}-$axis aligned with the fibres, the above relation becomes

$$\begin{split}2(T_{0}T_{1}-R_{1}^{2})-T_{0}^{2}+R_{0}^{2}+2[R_{1}^{2}\cos 4\theta-&\\ -T_{1}R_{0}\cos 4(\varPhi-\theta)]&<0,\end{split}$$

(8)

with $\varPhi=\varPhi_{0}-\varPhi_{1}$, the angular invariant of $\mathbb{Q}$. Common orthotropy corresponds to the condition[25]

$$\varPhi=K\frac{\pi}{4},\ \ K\in\{0,1\},$$

(9)

the value of $K$ determining two different types of orthotropic materials sharing the same polar moduli of $\mathbb{Q}$. As said in the Introduction, we consider in this paper only UD plies (i.e. layers reinforced by fabrics are not considered). It can be shown that such plies can be only ordinary orthotropic hence $R_{0}\neq 0,R_{1}\neq 0$.

So finally, for an UD layer, the auxeticity condition resumes to

$$\begin{split}\lambda(\theta):&=2(T_{0}T_{1}-R_{1}^{2})-T_{0}^{2}+R_{0}^{2}+\\ &+2[R_{1}^{2}-(-1)^{K}T_{1}R_{0}]\cos 4\theta<0,\end{split}$$

(10)

Let us now consider a laminate composed of identical UD plies all sharing the same reduced stiffness tensor $\mathbb{Q}$. We focus on the extension response, described by tensor[28, 25]

$$\mathbb{A}=\frac{1}{h}\sum_{j=1}^{n}({z_{j}}-{z_{j-1}})\mathbb{Q}_{(}\delta_{j}),\\$$

(11)

with $\delta_{j}$ the orientation of the $j-$th layer among the $n$ composing the laminate and $h$ the plate’s thickness. We further assume that the laminate is extension-bending uncoupled. This assumption is necessary, because otherwise the Poisson’s ratio for the extension, or also for the bending, behavior, should be practically impossible to be analyzed (in fact, for coupled laminates the compliances, in extension and in bending, depend in a very complicate manner upon $\mathbb{A},\mathbb{B}$ and $\mathbb{D}$, respectively the stiffness tensors in extension, coupling and bending[32, 33, 34]). It is worth to recall that uncoupling ($\mathbb{B}=\mathbb{O}$) can be obtained by suitable stacking sequences, but, contrarily to what commonly believed, not necessarily symmetric. Actually, asymmetric uncoupled laminates are much more numerous than the symmetric ones[35, 36]. So, uncoupling can be obtained rather easily and to assume it does not constitute a true limitation.

The auxeticity condition for an orthotropic tensor $\mathbb{A}$ is the same of that for $\mathbb{Q}$, provided that the polar parameters of $\mathbb{A}$ (denoted in the following by a superscript $A$) replace those of the layer:

$$\begin{split}\lambda^{A}(\theta):&=2(T_{0}^{A}T_{1}^{A}-{R_{1}^{A}}^{2})-{T_{0}^{A}}^{2}+{R_{0}^{A}}^{2}+\\ &+2[{R_{1}^{A}}^{2}-(-1)^{K^{A}}T_{1}^{A}R_{0}^{A}]\cos 4\theta<0.\end{split}$$

(12)

Because the plies are identical, the polar parameters of the extension stiffness tensor $\mathbb{A}$ can be put in the form[25]

$$\begin{array}[]{l}T_{0}^{A}=T_{0},\vskip 6.0pt plus 2.0pt minus 2.0pt\\ T_{1}^{A}=T_{1},\vskip 6.0pt plus 2.0pt minus 2.0pt\\ R_{0}^{A}\mathrm{e}^{4\mathrm{i}\varPhi_{0}^{A}}={R_{0}\mathrm{e}^{4\mathrm{i}\varPhi_{0}}}{\left(\xi_{1}+\mathrm{i}\xi_{2}\right)},\vskip 6.0pt plus 2.0pt minus 2.0pt\\ R_{1}^{A}\mathrm{e}^{2\mathrm{i}\varPhi_{1}^{A}}={R_{1}\mathrm{e}^{2\mathrm{i}\varPhi_{1}}}{\left(\xi_{3}+\mathrm{i}\xi_{4}\right)}.\end{array}$$

(13)

The quantities $\xi_{i},i=1,...,4$ are the lamination parameters[26] for $\mathbb{A}$:

$$\xi_{1}+\mathrm{i}\xi_{2}=\frac{1}{n}\sum_{j=1}^{n}\mathrm{e}^{4\mathrm{i}\delta_{j}},\ \ \xi_{3}+\mathrm{i}\xi_{4}=\frac{1}{n}\sum_{j=1}^{n}\mathrm{e}^{2\mathrm{i}\delta_{j}},$$

(14)

Eq. (13) shows, on the one hand, that the isotropic part of $\mathbb{A}$ is identical to that of the layer, $T_{0}$ and $T_{1}$, and, on the other hand, because the layer is UD, if we consider, as said above, an orthotropic extension behavior, choosing $\varPhi^{A}_{1}=0$ to fix the reference frame for the laminate, we get easily

$$\begin{array}[]{c}\xi_{2}=\xi_{4}=0,\vskip 6.0pt plus 2.0pt minus 2.0pt\\ (-1)^{K^{A}}R_{0}^{A}=(-1)^{K}R_{0}\xi_{1},\vskip 6.0pt plus 2.0pt minus 2.0pt\\ R_{1}^{A}=R_{1}\xi_{3}.\end{array}$$

(15)

So, the polar auxeticity condition for $\mathbb{A}$ becomes

$$\begin{split}\lambda^{A}(\theta)&=2(T_{0}T_{1}-R_{1}^{2}\xi_{3}^{2})-T_{0}^{2}+R_{0}^{2}\xi_{1}^{2}+\\ &+2[R_{1}^{2}\xi_{3}^{2}-(-1)^{K}T_{1}R_{0}\xi_{1}]\cos 4\theta<0.\end{split}$$

(16)

In the expression of $\lambda^{A}(\theta)$, there are three kind of variables/parameters: the independent variable $\theta$, determining the direction, the two lamination parameters $\xi_{1}$ and $\xi_{3}$, accounting for the geometry of the stack (i.e., the sequence of the orientation angles $\delta_{j}$), and the polar parameters $T_{0},T_{1},(-1)^{K}R_{0}$ and $R_{1}$ of the layer, representing the material part of $\lambda^{A}(\theta)$. It is worth to introduce new parameters for the material part, dimensionless like $\xi_{1}$ and $\xi_{3}$. This will reduce the number of independent material parameters and allow an easier study of the relation between geometry and material. To this end, we introduce the following ratios[37]:

$$\tau_{0}=\frac{T_{0}}{R_{1}},\ \ \tau_{1}=\frac{T_{1}}{R_{1}},\ \ \rho=\frac{R_{0}}{R_{1}}.$$

(17)

It is worth noting that these ratios can be introduced because, for a UD ply, $R_{1}\neq 0$. The auxeticity condition (16) can be rewritten in the equivalent form

$$\begin{split}\lambda^{A}(\theta)&=2(\tau_{0}\tau_{1}-\xi_{3}^{2})-\tau_{0}^{2}+\rho^{2}\xi_{1}^{2}+\\ &+2[\xi_{3}^{2}-(-1)^{K}\tau_{1}\rho\ \xi_{1}]\cos 4\theta<0,\end{split}$$

(18)

depending on five dimensionless quantities: three for the ply, material part, and two for the stack, geometric part.

We present in Tab. 1 the characteristics of twenty different UD plies commonly used for the fabrication of composite laminates.

The polar moduli, like any other set of parameters representing elasticity, must satisfy some thermodynamic conditions, stating the positivity of the elastic potential[42, 29, 25, 43]. In[37] it has been shown that using the dimensionless parameters (17), for an orthotropic ply these conditions are

$$\begin{split}&\tau_{0}-\rho>0,\\ &\tau_{1}\left[\tau_{0}+(-1)^{K}\rho\right]-2>0.\end{split}$$

(19)

They are necessary for establishing the set, in the space of the polar parameters, of materials giving rise to auxetic laminates, see below.

An interesting interpretation can be given to the auxeticity condition: Eq. (3) can equivalently be rewritten as

$$\frac{1}{4[t_{0}+r_{0}\cos 4(\varphi_{0}-\theta)]}>\frac{1}{8t_{1}}.$$

(20)

The two members of the above inequality have a direct mechanical interpretation[25]: the shear modulus $G_{12}(\theta)$ is

$$\begin{split}G_{12}(\theta)&=\frac{1}{4[t_{0}-r_{0}\cos 4(\varphi_{0}-\theta)]}=\\ &=\frac{1}{4[t_{0}+r_{0}\cos 4(\varphi_{0}-\theta-\frac{\pi}{4})]}.\end{split}$$

(21)

The bulk modulus $\kappa$ is

$$\kappa=\frac{1}{8t_{1}}.$$

(22)

So, the auxeticity condition for the direction $\theta+\frac{\pi}{4}$ actually corresponds to state that

$$G_{12}(\theta)>\kappa.$$

(23)

This condition also includes and generalizes the already discussed auxeticity condition[9, 10] for isotropy ($r_{0}=0$), that in 2D elasticity reads like:

$$\frac{1}{4t_{0}}>\frac{1}{8t_{1}}\Rightarrow G>\kappa.$$

(24)

Miki[44, 45, 28] has shown that the lamination domain of an orthotropic $\mathbb{A}$, i.e. the set of points of the plane $(\xi_{3},\xi_{1})$, that can correspond to an orthotropic behavior in extension, is the sector of parabola $\Omega$ in Fig. 2 defined by the bounds

$$2\xi_{3}^{2}-1\leq\xi_{1}\leq 1,\ \ -1\leq\xi_{1}\leq 1.$$

(25)

Each lamination point $(\xi_{3},\xi_{1})$ corresponds to an extension response, i.e. to a tensor $\mathbb{A}$, that, in general, can be realized by more that one stacking sequence.

We just recall that the lamination point P${}_{1}=(1,1)$ corresponds to a unidirectional laminate with all the plies with $\delta_{j}=0\ \forall j$, P${}_{2}=(-1,1)$ to a unidirectional laminate with $\delta_{j}=\frac{\pi}{2}\forall j$, P${}_{3}=(0,-1)$ to a balanced angle ply with $\delta_{j}=\pm\frac{\pi}{4}$, P${}_{4}=(0,0)$ to an isotropic laminate. Moreover, any angle-ply laminate is represented by a point of the parabolic boundary and any cross-ply laminate by a point of the line P${}_{1}-$P₂.

Miki[46] has also shown that a tensor $\mathbb{D}$ describing the bending behavior has exactly the same lamination domain, although the definition of the relevant lamination parameters is different. That is why all the results of this paper, concerning extension, can be exported identically to bending, the stacking sequence apart.

For an anisotropic laminate to be totally auxetic, equation (18) must be satisfied $\forall\theta\in\left[0,\frac{\pi}{2}\right]$. But $\lambda^{A}(\theta)\leq 0\ \forall\theta\iff\max\lambda^{A}(\theta)\leq 0$ which gives the condition

$$\begin{split}\varPsi(\xi_{3},\xi_{1}):&=2(\tau_{0}\tau_{1}-\xi_{3}^{2})-\tau_{0}^{2}+\rho^{2}\xi_{1}^{2}+\\ &+2|\xi_{3}^{2}-(-1)^{K}\tau_{1}\rho\ \xi_{1}|<0.\end{split}$$

(26)

For a chosen ply, the function $\varPsi(\xi_{3},\xi_{1})$ is defined on $\Omega$; it is shown in Fig. 3 for materials 2 and 4 in Tab. 1.

Then:

• •

  an UD ply can fabricate a TAAL $\forall(\xi_{3},\xi_{1})\in\Omega\iff\max\ \varPsi(\xi_{3},\xi_{1})<0$;

• •

  an UD ply can produce a TAAL for some lamination points $(\xi_{3},\xi_{1})\in\Omega\iff\min\ \varPsi(\xi_{3},\xi_{1})<0$.

The maximum and minimum of $\varPsi(\xi_{3},\xi_{1})$ depend on the material parameters $\tau_{0},\tau_{1}$ and $\rho$ and are rather cumbersome though not difficult to be found. The maximum has the same expression regardless the type of orthotropy, $K=0$ or $K=1$:

$$\varPsi_{\max}=\varPsi(0,\pm 1)=(\tau_{0}+\rho)(2\tau_{1}-\tau_{0}+\rho).$$

(27)

So, because the polar moduli are non-negative quantities, the condition for obtaining a TAAL $\forall(\xi_{3},\xi_{1})$ is

$$\varPsi_{0}:=2\tau_{1}-\tau_{0}+\rho<0.$$

(28)

The minimum can take different values and depends also on the type of orthotropy. If $K=0$,

$$\begin{array}[]{l}\varPsi_{\min}=-(\tau_{0}-\tau_{1})^{2}:=\varPsi_{1}\ \ \ \ \mathrm{if}\ \ \tau_{1}<\rho<\dfrac{\tau_{1}}{2\tau_{1}^{2}-1},\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \varPsi_{\min}=\left(\dfrac{\rho}{2\tau_{1}\rho-1}\right)^{2}-\dfrac{2\tau_{1}\rho}{2\tau_{1}\rho-1}-\tau_{0}^{2}+2\tau_{0}\tau_{1}:=\varPsi_{2}\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \hskip 25.60747pt\mathrm{if}\ \ \ \rho>\dfrac{1}{\tau_{1}}\ \ \mathrm{and}\ \ \dfrac{\rho(2\tau_{1}^{2}-1)-\tau_{1}}{2\tau_{1}\rho-1}>0,\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \varPsi_{\min}=(\tau_{0}-\rho)(2\tau_{1}-\tau_{0}-\rho):=\varPsi_{3}\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \hskip 25.60747pt\mathrm{if}\ \ \ \rho<\tau_{1}\ \ \mathrm{and}\ \ \rho<\dfrac{1}{\tau_{1}}.\end{array}$$

(29)

The condition for obtaining a TAAL for some points $(\xi_{3},\xi_{1})$ is hence, for $K=0$ orthotropic materials,

$$\varPsi_{\min}\in\{\varPsi_{1},\varPsi_{2},\varPsi_{3}\}<0.$$

(30)

Moreover, referring to Fig. 4: $\varPsi_{1}$ is get on the segment $q_{1}q_{2}$ and on its symmetric with respect to the $\xi_{1}-$axis; $\varPsi_{2}$ is get on the point $q_{3}$ and on its symmetric; $\varPsi_{3}$ is get on the segment $q_{4}q_{5}$ and on its symmetric.

The coordinates of the points $q_{i}$ are:

$$\begin{array}[]{l}q_{1}=\left(\tau_{1},\dfrac{\tau_{1}}{\rho}\right),\ q_{2}=\left(\sqrt{\dfrac{\rho+\tau_{1}}{2\rho}},\dfrac{\tau_{1}}{\rho}\right),\vskip 6.0pt plus 2.0pt minus 2.0pt\\ q_{3}=\left(\sqrt{\dfrac{\tau_{1}\rho}{2\tau_{1}\rho-1}},\dfrac{1}{2\tau_{1}\rho-1}\right),\vskip 6.0pt plus 2.0pt minus 2.0pt\\ q_{4}=\left(\sqrt{\tau_{1}\rho},1\right),\ q_{5}=(1,1).\end{array}$$

(31)

A similar analysis can be done for materials with $K=1$:

$$\begin{array}[]{l}\varPsi_{\min}=\varPsi_{1}\ \ \ \ \mathrm{if}\ \ \ \ \rho(2\tau_{1}^{2}-1)+\tau_{1}<0,\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \varPsi_{\min}=\left(\dfrac{\rho}{2\tau_{1}\rho+1}\right)^{2}-\dfrac{2\tau_{1}\rho}{2\tau_{1}\rho+1}-\tau_{0}^{2}+2\tau_{0}\tau_{1}:=\varPsi_{4}\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \hskip 25.60747pt\mathrm{if}\ \ \ \rho(2\tau_{1}^{2}-1)+\tau_{1}>0,\end{array}$$

(32)

so the condition for obtaining a TAAL for some points $(\xi_{3},\xi_{1})$ using a $K=1$ orthotropic material is

$$\varPsi_{\min}\in\{\varPsi_{1},\varPsi_{4}\}<0.$$

(33)

Still referring to Fig. 4, $\varPsi_{1}$ is get on the segment $q_{6}q_{7}$ and on its symmetric with respect to the $\xi_{1}-$axis, while $\varPsi_{4}$ is get on the point $q_{8}$ and on its symmetric and the coordinates of these points are:

$$\begin{array}[]{l}q_{6}=\left(\tau_{1},-\dfrac{\tau_{1}}{\rho}\right),\ q_{7}=\left(\sqrt{\dfrac{\rho-\tau_{1}}{2\rho}},-\dfrac{\tau_{1}}{\rho}\right),\vskip 6.0pt plus 2.0pt minus 2.0pt\\ q_{8}=\left(\sqrt{\dfrac{\tau_{1}\rho}{2\tau_{1}\rho+1}},-\dfrac{1}{2\tau_{1}\rho+1}\right).\end{array}$$

(34)

It is interesting to remark that, while $\varPsi_{\min}$ depends on all the material parameters, i.e., in the end, on the whole $\mathbb{Q}$, its position on $\Omega$ as well as the formula for its calculation depend only on $\tau_{1}$ and $\rho$. The zones of the different $\varPsi_{i},\ i=1,...,4,$ in the plane $(\tau_{1},\rho)$ are depicted in Fig. 5. On the same diagram, the positions of the materials in Tab. 1 are indicated by a small circle. It is apparent that all of them, of the type $K=0$, apart material 1, pine wood, belong to the zone where $\varPsi_{\min}=\varPsi_{2}$. Hence, for all of them, the minimum is get at the point $q_{3}$, on the boundary of $\Omega$. Such a lamination point can always be obtained by an angle-ply sequence, but not exclusively.

An anisotropic laminate is partially auxetic if Eq. (18) has at least one solution in $\Omega$. Hence, a PAAL can be fabricated with a UD material if and only if $\min\lambda^{A}(\theta)\leq 0$ which gives the condition

$$\begin{split}\eta(\xi_{3},\xi_{1}):&=2(\tau_{0}\tau_{1}-\xi_{3}^{2})-\tau_{0}^{2}+\rho^{2}\xi_{1}^{2}-\\ &-2|\xi_{3}^{2}-(-1)^{K}\tau_{1}\rho\ \xi_{1}|<0.\end{split}$$

(35)

This case is rather similar to that of TAALs and can be treated in the same way. The function $\eta(\xi_{3},\xi_{1})$, defined on $\Omega$, is shown in Fig. 6 for the same materials 2 and 4 in Tab. 1.

It is evident that:

• •

  an UD ply can fabricate a PAAL $\forall(\xi_{3},\xi_{1})\in\Omega\iff\max\ \eta(\xi_{3},\xi_{1})<0$;

• •

  an UD ply can produce a PAAL for some lamination points $(\xi_{3},\xi_{1})\in\Omega\iff\min\ \eta(\xi_{3},\xi_{1})<0$.

Like for function $\varPsi(\xi_{3},\xi_{1})$, the maximum and minimum of $\eta(\xi_{3},\xi_{1})$ depend on the material parameters $\tau_{0},\tau_{1}$ and $\rho$ and can be found by standard, though slightly articulated, differential calculus. Like for $\varPsi_{\max}$, also $\eta_{\max}$ is the same for $K=0$ and $K=1$ orthotropy:

$$\begin{array}[]{l}\eta_{\max}=(\tau_{0}-\rho)(2\tau_{1}-\tau_{0}-\rho):=\eta_{1}\ \mathrm{if}\ \rho\geq 2\tau_{1},\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \eta_{\max}=\tau_{0}(2\tau_{1}-\tau_{0}):=\eta_{2}\ \mathrm{if}\ \rho<2\tau_{1}.\end{array}$$

(36)

To remark that $\eta_{1}=\varPsi_{3}$; $\eta_{1}$ is get at $(0,-1)$ if $K=0$, at $(0,1)$ if $K=1$ i.e., in the two cases, for cross-ply balanced laminates, while $\eta_{2}$ is get at $(0,0)$ in both the cases, i.e. for isotropic laminates. The zones of $\eta_{1},\eta_{2}$ in the plane $(\tau_{1},\rho)$ are shown in Fig. 7, where the positions of the materials in Tab. 1 are again denoted by a small circle.

The condition for obtaining a PAAL $\forall(\xi_{3},\xi_{1})\in\Omega$ is hence

$$\eta_{\max}\in\{\eta_{1},\eta_{2}\}<0.$$

(37)

Also $\eta_{\min}$ can take different values and depends on the type of orthotropy. For the case $K=0$,

$$\begin{array}[]{l}\eta_{\min}=\eta_{1}\ \ \mathrm{if}\ \ \rho^{2}-\tau_{1}\rho+1<0,\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \eta_{\min}=(\tau_{0}+\rho)(2\tau_{1}-\tau_{0}+\rho)-4:=\eta_{3}\\ \hskip 82.51299pt\mathrm{if}\ \ \rho^{2}+\tau_{1}\rho-1<0,\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \eta_{\min}=-\left(\dfrac{\tau_{1}\rho-1}{\rho}\right)^{2}-\tau_{0}^{2}-2+2\tau_{0}\tau_{1}:=\eta_{4}\\ \hskip 19.91692pt\mathrm{if}\rho^{2}-\tau_{1}\rho+1>0\ \mathrm{and}\ \rho^{2}+\tau_{1}\rho-1>0.\end{array}$$

(38)

The condition for obtaining a PAAL for some points $(\xi_{3},\xi_{1})$ is hence, for $K=0$ orthotropic materials,

$$\eta_{\min}\in\{\eta_{1},\eta_{3},\eta_{4}\}<0.$$

(39)

The minimum $\eta_{1}$ is get in $\Omega$ at the point $(0,-1)$ and on the line $([-1,1],1)$, i.e. for any cross-ply combination and also for a laminate with all the layer orientations $\delta_{j}=0$ or $\delta_{j}=\dfrac{\pi}{2}$; $\eta_{3}$ is given by points $(\pm 1,1)$, which is the case of a laminate with the orientations of all the plies at the angle $\delta_{j}=0$ or $\delta_{j}=\dfrac{\pi}{2}$ and $\eta_{4}$ by the two symmetric points, on the boundary of $\Omega$, $\left(\pm\sqrt{\dfrac{\rho^{2}-\tau_{1}\rho+1}{2\rho^{2}}},\dfrac{1-\tau_{1}\rho}{\rho^{2}}\right)$, that can be obtained by angle-ply sequences, but not only.

For $K=1$ orthotropic materials, it is

$$\begin{array}[]{l}\eta_{\min}=(\tau_{0}-\rho)(2\tau_{1}-\tau_{0}-\rho)-4:=\eta_{5}\\ \hskip 82.51299pt\mathrm{if}\ \ \rho^{2}-\tau_{1}\rho-1<0,\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \eta_{\min}=-\left(\dfrac{\tau_{1}\rho+1}{\rho}\right)^{2}-\tau_{0}^{2}-2+2\tau_{0}\tau_{1}:=\eta_{6}\\ \hskip 108.12047pt\mathrm{if}\ \ \rho^{2}-\tau_{1}\rho-1>0.\end{array}$$

(40)

The condition for obtaining a PAAL for some points $(\xi_{3},\xi_{1})$ is hence, for $K=1$ orthotropic materials,

$$\eta_{\min}\in\{\eta_{5},\eta_{6}\}<0.$$

(41)

The minimum $\eta_{5}$ is get at the points $(\pm 1,1)$ i.e. by laminates with all the layer orientations $\delta_{j}=0$ or $\delta_{j}=\dfrac{\pi}{2}$; $\eta_{6}$ is given by the two symmetric points, on the boundary of $\Omega$, $\left(\pm\sqrt{\dfrac{\rho^{2}+\tau_{1}\rho+1}{2\rho^{2}}},\dfrac{1+\tau_{1}\rho}{\rho^{2}}\right)$, also these lamination points can be obtained, but not only, by angle-ply sequences. Just like $\varPsi_{\min}$, also $\eta_{\min}$ depends on all the material parameters, but its position on $\Omega$ as well as the formula for its calculation depend only on $\tau_{1}$ and $\rho$. The zones of the different $\eta_{i},\ i=1,...,4,$ in the plane $(\tau_{1},\rho)$ are shown in Fig. 8, where, once more, the positions of the materials in Tab. 1 are also indicated by a small circle. Apart materials 7, a glass-epoxy ply, and 14, a composite boron-aluminium, all the layers belong to the zone where $\eta_{\min}=\eta_{4}$. For all of them, the minimum is get at a lamination point on the boundary of $\Omega$, that can always be obtained by an angle-ply sequence, but not exclusively.