## Appendix C New Analytical Relationships for Hexagonal Packing Geometry  
In this appendix, the derivation of new analytical relationships for the hexagonal packing geometry is presented. Hexagonal closed-packing (HCP) is known as the mostly efficient way a space can be filled by an arrangement of spheres (Figure 11a). Sphere constructing an HCP structure fill up 74% of space. For an open-cell lattice structure from the concept of HCP, the spheres can be visualized to inflate from all direction until they create a second lattice structure made of polyhedrons (demonstrated by black solid lines in Figure 11a). To find the unit cell of the new polyhedral, first we select two rows of spheres on top of each other and the corresponding polyhedral lattice structure surrounding them (Figure 11b). After removal of the spheres we are left with six polyhedral cells, in which we can find a unit cell whichafter tessellation in space can create the HCP-based polyhedral lattice structure (the red unit cell highlighted in Figure A1c). Dimensions and loads acting on a hexagonal packing unit cell are shown in Figure A1d. The initial spheres demonstrated in Figure A1a create an angle \(\theta\) for which we have \(\cos\theta=\frac{1}{3}\) and \(\sin\theta=\frac{2\sqrt{2}}{3}\). The cross-sectional area of the hexagonal packing unit cell is therefore \(A_{UC}=\frac{3\sqrt{3}}{2}l^{2}\sin^{2}\theta\). This structure is composed of two types of vertical struts OA (type A) and BC (type C) and one type of inclined strut OB (type B). In each unit cell, there are two type A struts and \(6\times 1/3=2\) type B struts. The term \(1/3\) is due to the fact that each type B strut is shared by three adjacent unit cells. Displacement of each of the three types of struts are as follows:  
(i) Type-A strut: Under external load \(P\), the change in the length of strut OA is (Equation (2) in [35])  
\[\delta_{A,z}=-\frac{Pl}{2 AE}(1+\cos\theta)\] (A30)  
(ii) Type-C strut: Under external load \(P^{\prime}\), the change in the length of strut BC is (Equation (3) in [35])  
\[\delta_{C,z}=-\frac{P^{\prime}l}{2 AE}(1-\cos\theta)\] (A31)  
(iii) Type-B strut OB: Displacements of B with respect to O in directions respectively parallel and perpendicular to strut OB are (see Figure 2a,c in the main paper)  
\[\delta_{B}=\frac{P^{\prime}l}{3 AE}\cos\theta\delta^{\prime}{}_{B}=\frac{Ml^{2 }}{6 EI}=\frac{P^{\prime}\sin\theta l^{3}}{36 EI}\] (A32)  
Total displacement of \(C\) with respect to O in the \(z\) and \(x\) directions are respectively:  
\[\Delta_{C,z}=\delta_{C,z}+\delta_{B,z}+\delta^{\prime}{}_{B,z}=-\frac{P^{ \prime}l}{2 AE}(1-\cos\theta)-\frac{P^{\prime}l}{3 AE}\cos^{2}\theta-\frac{p^{ \prime}l}{36 EI}\sin^{2}\theta\] (A33)  
\[\Delta_{C,x}=\delta_{B,x}+\delta^{\prime}{}_{B,x}=-\frac{p^{\prime}l}{3 AE} \sin\theta\cos\theta+\frac{p^{\prime}(l)}{36 EI}\sin\theta\cos\theta\]  
Therefore, the strains in the \(z\) and \(x\) directions are:  
\[\varepsilon_{z}=\frac{2\delta_{A,z}}{l(1+\cos\theta)}=-\frac{P}{AE}\; \varepsilon_{x}=\frac{\Delta_{C,x}}{l\sin\theta}=\frac{-\frac{P^{\prime}l}{3 AE }\sin\theta\cos\theta+\frac{p^{\prime}(l)}{36 EI}\sin\theta\cos\theta}{l\sin \theta}=-\frac{P^{\prime}}{3 AE}\cos\theta+\frac{P^{\prime}l^{2}}{36 EI}\cos\theta\] (A34)  
On the other hand, we know that due to continuity of the material and symmetry of each unit cell with respect to the neighbouring cell: \(\delta_{A,z}=\Delta_{C,z}\).