Figure 3 shows irregular honeycomb structures loaded in two different directions, where \(\sigma_{{1}}\) and \(\sigma_{{2}}\) are uniformly distributed loads in two mutually perpendicular directions.

### 3.1 Elastic Modulus in the \(\sigma_{{1}}\) Direction

Figure 4 shows the structural parameters and force analysis of rod *AB*, where *b* is the thickness of the unit structure, *t* is the depth of the unit structure, \(l_{i}\) and *h* are the lengths of the inclined cell walls with inclination angle \(\theta\) and the length of the vertical rod, respectively. The moment \(M_{{1}}\) of rod *AB* can be expressed as

$$ M_{{1}} { = }\frac{{P_{1} l_{1} \sin \theta_{1} }}{2}, $$

(1)

where \(P_{1} = \sigma (h + l_{1} \sin \theta_{1} )b\) is the force in the \(\sigma_{{1}}\) direction.

From the standard beam theory [33], the deflection \(\delta_{AB}\) of rod *AB* can be expressed as

$$ \delta_{AB} = \frac{{p_{1} l_{{1}}^{3} \sin \theta_{1} }}{12EI}. $$

(2)

Axial force \(F^{\prime}\) along rod *AB* is

$$ F^{\prime} = p_{1} \cos \theta_{1} . $$

(3)

Axial deformation \(\Delta l\) can be expressed as

$$ \Delta l = \frac{{F^{\prime}L}}{EA} = \frac{{p_{1} \cos \theta_{1} l_{1} }}{Ebt}, $$

(4)

where *E* is the elastic moduli of the original materials, and the total deformation \(\delta_{1}\) of rod *AB* along the \(\sigma_{{1}}\) direction is

$$ \delta_{1} = \delta_{AB} \sin \theta_{1} + \Delta l\cos \theta_{1} = \frac{{p_{1} l_{{1}}^{3} \sin^{2} \theta_{1} }}{12EI} + \frac{{p_{1} \cos^{2} \theta_{1} l_{1} }}{Ebt}. $$

(5)

Similarly, the total deformation \(\delta_{{2}}\) of rod *BC* along the \(\sigma_{{1}}\) direction can be expressed as follows:

$$ \delta_{2} = \delta_{BC} \sin \theta_{2} + \Delta l\cos \theta_{2} = \frac{{p_{1} l_{2}^{3} \sin^{2} \theta_{2} }}{12EI} + \frac{{p_{1} \cos^{2} \theta_{2} l_{2} }}{Ebt}. $$

(6)

Combing Eqs. (2)–(6), the strain \(\varepsilon_{1}\) parallel to the \(\sigma_{{1}}\) direction is given by

$$ \varepsilon_{1} = \frac{{\delta_{1} + \delta_{2} }}{{l_{1} \cos \theta_{1} + l_{2} \cos \theta_{2} }} = \frac{\sigma \gamma }{{l_{1} \cos \theta_{1} + l_{2} \cos \theta_{2} }}, $$

(7)

where

$$ \gamma = b\left[ {(h + l_{1} \sin \theta_{1} )\left( {\frac{{l_{{1}}^{3} \sin^{2} \theta_{1} }}{12EI} + \frac{{\cos^{2} \theta_{1} l_{1} }}{Ebt}} \right) + (h + l_{2} \sin \theta_{2} )\left( {\frac{{l_{2}^{3} \sin^{2} \theta_{2} }}{12EI} + \frac{{\cos^{2} \theta_{2} l_{2} }}{Ebt}} \right)} \right]. $$

(8)

Thus, the elastic modulus \(E_{1U}\) in the \(\sigma_{{1}}\) direction can be expressed as follows:

$$ E_{1U} = \frac{{(l_{1} \cos \theta_{1} + l_{2} \cos \theta_{2} )}}{\gamma }. $$

(9)

### 3.2 Elastic Modulus in the \(\sigma_{2}\) Direction

To derive the expression of the transverse elastic modulus, stress \(\sigma_{{2}}\) is applied, as shown in Figure 3(b). Figure 5 shows that the deflection of rod *BD* consists of two parts: bending deformation and rotational deformation. The bending deformation \(\delta_{{{2}vb}}\) caused by the moment \(M_{{1}}\) in the \(\sigma_{{2}}\) direction can be expressed as

$$ \delta_{2vb} = \left( {\frac{{w\cos \alpha \left( {\frac{{s_{1} }}{\sin \alpha }} \right)^{3} }}{3EI}} \right)\cos \alpha , $$

(10)

where

$$ \left\{ \begin{aligned} &w = \sigma_{2} (l_{1} \cos \theta_{1} + l_{2} \cos \theta_{2} )b, \hfill \\ &I = {{bt^{3} } \mathord{\left/ {\vphantom {{bt^{3} } {12,}}} \right. \kern-\nulldelimiterspace} {12,}} \hfill \\ & M_{1} = ws_{1} \cot \alpha . \hfill \\ \end{aligned} \right. $$

(11)

Because the rotation angles of the three rods connected to point *B* are identical, the rotation angle \(\phi\) of joint *B* can be written as

$$ \phi = \frac{{M_{1} l_{1} }}{{l_{1} + l_{2} }}\frac{{l_{1} }}{6EI}. $$

(12)

Thus, the deformation \(\delta_{{{2}vr}}\) of the cell wall with an inclination angle \(\alpha\) in the \(\sigma_{{2}}\) direction is given by:

$$ \delta_{2vr} = \phi \left( {\frac{{s_{1} }}{\sin \alpha }} \right)\cos \alpha . $$

(13)

Therefore, the total deformations of rod *BD* and rod *FH* in the \(\sigma_{{2}}\) direction are

$$ \delta_{2BD} = \left( {\frac{{w\cos \alpha \left( {\frac{{s_{1} }}{\sin \alpha }} \right)^{3} }}{3EI}} \right)\cos \alpha + \frac{{M_{1} l_{1} }}{{l_{1} + l_{2} }}\frac{{l_{1} }}{6EI}\left( {\frac{{s_{1} }}{\sin \alpha }} \right)\cos \alpha , $$

(14)

$$ \delta_{v2FH} = \left( {\frac{{w\cos \beta \left( {\frac{{s_{2} }}{\sin \beta }} \right)^{3} }}{3EI}} \right)\cos \beta + \frac{{M_{1} l_{4} }}{{l_{3} + l_{4} }}\frac{{l_{4} }}{6EI}\left( {\frac{{s_{2} }}{\sin \beta }} \right)\cos \beta . $$

(15)

Now, the axial deformations of rod *BD* and rod *FH* in the \(\sigma_{{2}}\) direction can be expressed as

$$ \Delta l_{2BD} = \frac{{ws_{1} }}{Ebt}\sin \alpha , $$

(16)

$$ \Delta l_{2FH} = \frac{{ws_{2} }}{Ebt}\sin \beta . $$

(17)

The deflections \(\delta_{vBD} {\kern 1pt}\) and \(\delta_{vGF} {\kern 1pt}\) of rods *AB* and *GF* (Figure 6) in the \(\sigma_{{2}}\) direction can be expressed as

$$ \delta_{vAB} = \frac{{\left( {\frac{{l_{1} w}}{{l_{1} + l_{2} }}\cos \theta_{1} } \right)l_{1}^{3} }}{12EI}\cos \theta_{1} , $$

(18)

$$ \delta_{vGF} = \frac{{\left( {\frac{{l_{4} w}}{{l_{3} + l_{4} }}\cos \theta_{4} } \right)l_{4}^{3} }}{12EI}\cos \theta_{4} . $$

(19)

The axial deformation of rods *AB* and *GF* in the \(\sigma_{{2}}\) direction can be expressed as

$$ \Delta l^{\prime}_{2} = \frac{{\frac{{l_{1} w}}{{l_{1} + l_{2} }}\sin \theta_{1} l_{1} }}{Ebt}\sin \theta_{1} , $$

(20)

$$ \Delta l^{\prime}_{3} = \frac{{\frac{{l_{4} w}}{{l_{3} + l_{4} }}\sin \theta_{4} l_{4} }}{Ebt}\sin \theta_{4} . $$

(21)

Thus, the total deformation \(\delta_{2} {\kern 1pt}\) of the structure in the \(\sigma_{{2}}\) direction can be expressed as

$$ \delta_{{2}} = \delta_{v2BD} + \delta_{v2FH} + \Delta l_{2BD} + \Delta l_{2FH} + \delta_{vAB} + \delta_{vGF} + \Delta l^{\prime}_{2} + \Delta l^{\prime}_{3} , $$

(22)

while Eq. (22) can be rewritten as follows:

$$ \delta_{{2}} = w\varphi , $$

(23)

where

$$ \begin{aligned} \varphi &= \left[ {\left( {\frac{{\cos \alpha \left( {\frac{{s_{1} }}{\sin \alpha }} \right)^{3} }}{3EI}} \right)\cos \alpha + \frac{{{\text{s}}_{1} \cot \alpha l_{1} }}{{l_{1} + l_{2} }}\frac{{l_{1} }}{6EI}\left( {\frac{{s_{1} }}{\sin \alpha }} \right)\cos \alpha } \right. \hfill \\ & \quad + \left( {\frac{{\cos \beta \left( {\frac{{s_{2} }}{\sin \beta }} \right)^{3} }}{3EI}} \right)\cos \beta + \frac{{{\text{s}}_{1} \cot \alpha l_{4} }}{{l_{3} + l_{4} }}\frac{{l_{4} }}{6EI}\left( {\frac{{s_{2} }}{\sin \beta }} \right)\cos \beta \hfill \\ & \quad+ \frac{{s_{1} }}{Ebt}\sin \alpha + \frac{{s_{2} }}{Ebt}\sin \beta + \frac{{\left( {\frac{{l_{1} }}{{l_{1} + l_{2} }}\cos \theta_{1} } \right)l_{1}^{3} }}{12EI}\cos \theta_{1} \hfill \\ & \quad \left. { + \frac{{\left( {\frac{{l_{4} }}{{l_{3} + l_{4} }}\cos \theta_{4} } \right)l_{4}^{3} }}{12EI}\cos \theta_{4} + \frac{{\frac{{l_{1} }}{{l_{1} + l_{2} }}\sin \theta_{1} l_{1} }}{Ebt}\sin \theta_{1} + \frac{{\frac{{l_{4} }}{{l_{3} + l_{4} }}\sin \theta_{4} l_{4} }}{Ebt}\sin \theta_{4} } \right], \hfill \\ \end{aligned} $$

(24)

Strain \(\varepsilon_{2} {\kern 1pt}\) in the \(\sigma_{{2}}\) direction can be obtained as

$$ \varepsilon_{2} = \frac{{\delta_{2} }}{{h + s_{1} + s_{2} + l_{1} \sin \theta_{1} + l_{4} \sin \theta_{4} }} = \frac{w\varphi }{{h + s_{1} + s_{2} + l_{1} \sin \theta_{1} + l_{4} \sin \theta_{4} }}, $$

(25)

where \(w = \sigma_{2} (l_{1} \cos \theta_{1} + l_{2} \cos \theta_{2} )b{\kern 1pt}\); thus, the elastic modulus in the \(\sigma_{{2}}\) direction of the structure can be expressed as

$$ E_{2U} = \frac{{\sigma_{2} }}{{\varepsilon_{2} }} = \frac{{h + s_{1} + s_{2} + l_{1} \sin \theta_{1} + l_{4} \sin \theta_{4} }}{{(l_{1} \cos \theta_{1} + l_{2} \cos \theta_{2} )b\varphi }}. $$

(26)

### 3.3 Poisson’s Ratio \(v_{{{12}}}\)

Poisson’s ratios were calculated by taking the negative ratios of strains normal to and parallel to the loading direction. Poisson’s ratio \(v_{{{12}}}\) of the unit structure can be defined as

$$ v_{12} = - \frac{{\varepsilon_{2} }}{{\varepsilon_{1} }}, $$

(27)

where \(\varepsilon_{{1}}\) and \(\varepsilon_{{2}}\) are strains in the \(\sigma_{{1}}\) and \(\sigma_{2}\) directions, respectively, due to the load in the \(\sigma_{{1}}\) direction. In addition, \(\varepsilon_{{1}}\) can be obtained from Eq. (7) and \(\varepsilon_{{2}}\) can be expressed as

$$ \varepsilon_{2} = - \frac{{\delta^{\prime}_{1} + \delta^{\prime}_{2} }}{{h + s_{1} + s_{2} + l_{1} \sin \theta_{1} + l_{4} \sin \theta_{4} }}, $$

(28)

where

$$ \left\{ {\begin{array}{*{20}l} {\delta^{\prime}_{1} = \delta_{AB} \cos \theta_{1} + \Delta l\sin \theta_{1} = \frac{{p_{1} l_{{1}}^{3} \sin \theta_{1} \cos \theta_{1} }}{12EI} + \frac{{p_{1} l_{1} \sin \theta_{1} \cos \theta_{1} }}{Ebt},} \hfill \\ {\delta^{\prime}_{2} = \delta_{GF} \cos \theta_{4} + \Delta l\sin \theta_{4} = \frac{{p_{2} l_{4}^{3} \sin \theta_{4} \cos \theta_{4} }}{12EI} + \frac{{p_{2} l_{4} \sin \theta_{4} \cos \theta_{4} }}{Ebt}.} \hfill \\ \end{array} {\kern 1pt} } \right. $$

(29)

The Poisson’s ratio of a structure in the \(\sigma_{{1}}\) direction can be expressed as

$$ v_{12} = \frac{{\left( {\delta^{\prime}_{1} + \delta^{\prime}_{2} } \right)\left( {l_{1} \cos \theta_{1} + l_{2} \cos \theta_{2} } \right)}}{{\left( {\delta_{1} + \delta_{2} } \right)\left( {h + s_{1} + s_{2} + l_{2} \sin \theta_{2} + l_{3} \sin \theta_{3} } \right)}}. $$

(30)

### 3.4 Poisson’s Ratio \(v_{21}\)

Poisson’s ratio of a structure for loading in the \(\sigma_{{2}}\) direction can be expressed as

$$ v_{21} = - \frac{{\varepsilon^{\prime}_{1} }}{{\varepsilon^{\prime}_{2} }}, $$

(31)

where \(\varepsilon_{{1}}^{\prime }\) and \(\varepsilon_{{2}}^{\prime }\) are the strains in the \(\sigma_{{1}}\) and \(\sigma_{{2}}\) directions, respectively. \(\varepsilon_{{2}}^{\prime }\) can be obtained from Eq. (23) as

$$ \varepsilon^{\prime}_{2} = \frac{w\varphi }{{h + s_{1} + s_{2} + l_{1} \sin \theta_{1} + l_{4} \sin \theta_{4} }}, $$

(32)

where \(w = \sigma_{2} (l_{1} \cos \theta_{1} + l_{2} \cos \theta_{2} )b{\kern 1pt}\), and \(\varepsilon_{{1}}^{\prime }\) can be obtained as

$$ \varepsilon^{\prime}_{1} = - \frac{{\delta_{vAB1} + \delta_{vBC1} }}{{l_{1} \cos \theta_{1} + l_{2} \cos \theta_{2} }} = - \frac{{w\varphi^{\prime}}}{{l_{1} \cos \theta_{1} + l_{2} \cos \theta_{2} }}, $$

(33)

where \(\delta_{vAB1}\) and \(\delta_{vBC1} {\kern 1pt}\) are the deformations in the \(\sigma_{{1}}\) direction due to the load in the \(\sigma_{{2}}\) direction

$$ \begin{gathered} \varphi^{\prime} = \frac{{\left( {\frac{{l_{1} }}{{l_{1} + l_{2} }}\cos \theta_{1} } \right)l_{1}^{3} }}{12EI}\sin \theta_{1} + \frac{{\left( {\frac{{l_{2} }}{{l_{1} + l_{2} }}\cos \theta_{2} } \right)l_{2}^{3} }}{12EI}\sin \theta_{2} \hfill \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} + \frac{{\frac{{l_{1} }}{{l_{1} + l_{2} }}\sin \theta_{1} }}{{E{\text{bt}}}}\cos \theta_{1} + \frac{{\frac{{l_{2} }}{{l_{1} + l_{2} }}\sin \theta_{2} }}{{E{\text{bt}}}}\cos \theta_{2} . \hfill \\ \end{gathered} $$

(34)

Thus, Poisson’s ratio \(v_{{{21}}} {\kern 1pt}\) of a structure in the \(\sigma_{{2}}\) direction can be expressed as

$$ v_{21} = \frac{{\varphi^{\prime}\left( {h + s_{1} + s_{2} + l_{2} \sin \theta_{2} + l_{3} \sin \theta_{3} } \right)}}{{\varphi \left( {l_{1} \cos \theta_{1} + l_{2} \cos \theta_{2} } \right)}}. $$

(35)