We investigated the potential of demi-regular tessellations to outperform existing prismatic lattice materials, such as the triangular topology. The in-plane mechanical properties of three demi-regular tessellations were derived analytically, and then corroborated by both finite element simulations and experiments. Our analysis showed that these three topologies are elastically isotropic and have a stretching-dominated behaviour. The elastic modulus of these three demi-regular topologies was 5-20% lower than that of a triangular lattice. Demi- regular lattices, however, had a high elastic buckling strength, up to 42% higher than their triangular counterpart. This characteristic makes them ideal topologies for ultra-lightweight applications, where failure is usually governed by elastic buckling. Keywords: Lattice material, Demi-regular tessellation, Cellular solid, Elastic buckling, Analytical modelling, Finite element analysis 1. Introduction Prismatic (2D) lattices are particularly attractive for multi-functional applications where both structural efficiency and heat transfer capabilities are required [1, 2]. The most common prismatic lattices are the triangular, hexagonal and square lattices, see Fig. 1. Their elastic modulus and compressive strength have been documented using analytical and numerical methods [3–7] and then, corroborated by experiments [3, 8–11]. This work showed that lattices have either a bending- or a stretching-dominated behaviour depending upon the ∗Corresponding Author: luc.st-pierre@aalto.fi Preprint submitted to Elsevier April 5, 2023 Triangular Hexagonal Elongated - triangular Semi - regular lattices Regular lattices Square Kagome S nub - trihexagonal Truncated - hexagonal Truncated - square Truncated - t ri hexagonal Rhombi - trihexagonal Snub - squareFigure 1: The eleven 1-uniform tessellations include three regular and eight semi-regular lattices. Regular lattices are made from a single regular polygon, whereas semi-regular tessellations are assembled from multiple regular polygons while maintaining a unique vertex configuration. connectivity of the joints [12]. With three bars per joint, the hexagonal lattice is bending- dominated and consequently, its elastic modulus and yield strength scale as ¯ ρ3and ¯ρ2, respectively, where ¯ ρis the relative density of the lattice [3]. In contrast, the triangular lattice, with six bars per joints, has a stretching-dominated behaviour. Its elastic modulus and yield strength both scale linearly with ¯ ρ, which makes it significantly stiffer and stronger than a hexagonal lattice [5]. At low relative densities, however, the collapse mode of the triangular lattice switches from yielding to elastic buckling, and this is associated with a significant reduction in compressive strength since the buckling resistance scales with ¯ ρ3 [3, 5, 11]. The development of light, stiff and strong lattice materials would therefore benefit from discovering new topologies that have a stiffness comparable to the triangular lattice, but a higher resistance to elastic buckling. This is the motivation behind this study. It is insightful to introduce notions of Euclidean tilings to guide the search for new 2 Tessellation A Tessellation B Tessellation CFigure 2: The twenty 2-uniform or demi-regular tessellations. These are assembled from regular polygons and have two different vertex configurations, which are marked in blue. topologies. Triangular, hexagonal and square topologies are the only three regular lattices, defined as tessellations made from a single regular and convex polygon [13]. In contrast, semi-regular lattices are assembled from at least two different regular polygons, while main- taining a unique vertex configuration (the sequence of polygons meeting at each joint). There are eight semi-regular lattices [13] and these are shown in Fig. 1. Since both regular and semi-regular lattices have a single vertex configuration, they form a category known as 1-uniform tilings [14]. Tessellations with two different vertex configurations can also be cre- ated from regular polygons, and these are called 2-uniform or demi-regular lattices. There are 20 demi-regular tessellations [14], and they are given in Fig. 2. The properties of semi-regular lattices have been reported using analytical and numerical methods [15–21], and two topologies were found to be stretching-dominated: the Kagome and snub-trihexagonal lattices. Even though both tessellations have a higher resistance to 3 elastic buckling than a triangular lattice, their practical use remains limited because (i) the kagome lattice is highly sensitive to imperfections [22, 23], and (ii) the snub-trihexagonal lattice is significantly more compliant than a triangular lattice [21]. These reasons explain why the triangular lattice remains the optimal prismatic topology for high stiffness and strength. While the properties of semi-regular lattices have been documented, those of demi-regular tessellations remain unknown. Therefore, this study aims to quantify their properties and determine if they can outperform a regular triangular lattice. The scope of this study is limited to three topologies that we will refer to as demi-regular lattices A, B, and C, see Fig. 2. These three tessellations were selected for two reasons. First, they have a six-fold rotational symmetry and therefore, their in-plane elastic properties will be isotropic [24]. Second, they have a high nodal connectivity (4, 5 or 6 bars per joint) and consequently, we expect their behaviour to be stretching-dominated [12]. We will use analytical, numerical, and experimental methods to show that these three demi-regular lattices have a comparable stiffness but a higher resistance to elastic buckling than a triangular lattice. This article is structured as follows. Analytical equations for the elastic modulus and compressive strength of each demi-regular lattice are derived in Section 2. To validate these analytical expressions, Finite Element (FE) simulations and experiments are developed and described in Section 3. Finally, analytical, numerical, and experimental results are presented in Section 4, where the performances of demi-regular tessellations are compared to those of a regular triangular lattice. 2. Analytical modelling In this section, analytical expressions are derived for the in-plane elastic modulus, Pois- son’s ratio and compressive strength of demi-regular lattices A, B, and C, see Fig. 2. Our analysis assumes that all lattices are made from a linear elastic, perfectly plastic material, characterised by a Young’s modulus Esand a yield strength σys. All cell walls are consid- ered to have a length ℓ, thickness t, and out-of-plane width b. The cell walls are assumed to behave as trusses since all four lattices are anticipated to be stretching-dominated. 4 Two collapse mechanisms are considered when deriving the compressive strength of a lattice: yielding and elastic buckling. Yielding occurs when the (tensile or compressive) stress in a cell wall reaches the yield strength σys. Otherwise, elastic buckling takes place when the compressive load in a cell wall reaches the Euler buckling load [25]: Tcr=n2π2EsI ℓ2, (1) where I=bt3/12 is the second moment of area and nis the end constraint factor (for example, n= 1 for pinned joints, whereas n= 2 when both ends are fixed). The compressive strength of a lattice is governed by the collapse mechanism that has the lowest failure load. Finally, we emphasise that the assumptions detailed above are identical to those used in previous studies [3, 5, 11, 21], which will enable a fair comparison between the properties of these three demi-regular lattices and other tessellations. 2.1. Demi-regular lattice A Demi-regular lattice A is a tessellation composed of triangles and hexagons, see the unit cell in Fig. 3. Its relative density ¯ ρis given by: ¯ρ=14√ 3 9t ℓ = 2.694t ℓ . (2) 2.1.1. Loading in x1direction Consider demi-regular lattice A loaded in the x1direction by a uniform compressive stress σ1. Because of symmetry, there are only five independent bars and these are labelled on the unit cell shown in Fig. 3. Equilibrium equations for joint Agive: T1−T2+T3+ 2(T4−T5) = 0 , (3) T1+T2−T3= 0, (4) where Tiis the internal force in bar i. Otherwise, the macroscopic stress field is σ11=σ1 andσ22= 0, and this implies: T1+ 2(T2+T4+T5) = 3√ 3σ1bℓ, (5) 5 3ℓ33ℓ512 𝐴5 11 2 2 23 34 4 𝑥2 𝑥1Figure 3: Periodic unit cell of demi-regular lattice A. T2+T3= 0, (6) respectively. This indeterminate truss can be solved using the principle of minimum strain energy as follows. First, we use (3) to (6) to express T2,T3,T4andT5as functions of T1. Second, we use these expressions to write the strain energy Uas a function of T1only. Third, the minimum strain energy is found by setting ∂U/∂T 1= 0, and this allows to solve for T1. These steps are detailed in Appendix A.1.1, and return the following internal forces: T1=√ 3 10σ1bℓ, (7a) T2=−√ 3 20σ1bℓ, (7b) T3=√ 3 20σ1bℓ, (7c) T4=7√ 3 10σ1bℓ, (7d) T5=4√ 3 5σ1bℓ, (7e) 6 where a negative sign (-) indicates tension, whereas a positive value corresponds to com- pression. Next, the nominal compressive strain in x1is given by: ϵ1=2∆4+ ∆ 5 3ℓ=11√ 3 15σ1 Esℓ t, (8) where ∆ i=Tiℓ/(Esbt) is the extension/shortening of bar i. Therefore, the elastic modulus E=σ1/ϵ1becomes: E Es=5√ 3 11t ℓ = 0.292¯ρ. (9) Otherwise, we can find the nominal strain in x2using the principle of virtual work, which gives: ϵ2=1 15Esbt T1+ 12T2+22 3T3−4T4−4 3T5 =−4√ 3 15σ1 Esℓ t, (10) and therefore, the Poisson’s ratio is: ν=−ϵ2 ϵ1=4 11= 0.364. (11) Since this tessellation is isotropic, the shear modulus G=E/(2(1 + ν)) and this yields: G Es=√ 3 6t ℓ = 0.107¯ρ. (12) Next, we turn our attention to the compressive strength of demi-regular lattice A. For compression in x1, bar 5 carries the highest compressive force, see (7). The elastic buckling strength is obtained by equating T5to the Euler buckling load in (1), which returns: σ1,el Es=5√ 3 144n2 1π2t ℓ3 = 0.081¯ρ3, (13) where the end constraint factor n1= 1.633 is derived analytically in Appendix B.1. The collapse mode is anticipated to change from elastic buckling to yielding as the relative density increases. The yield strength of the lattice is obtained by setting T5=btσys, which gives: σ1,pl σys=5√ 3 12t ℓ = 0.268¯ρ. (14) 7 2.1.2. Loading in x2direction Now, consider demi-regular lattice A loaded in the x2direction by uniform compressive stress σ2. Expressions (3) and (4) remain valid since they represent the equilibrium of joint A, see Fig. 3. Otherwise, enforcing the macroscopic stress field σ22=σ2andσ11= 0 yields: T2+T3=√ 3σ2bℓ, (15) T1+ 2(T2+T4+T5) = 0 , (16) respectively. The additional equation needed to solve the system is obtained with the princi- ple of minimum strain energy, see Appendix A.1.2, and we find the following internal forces: T1=√ 3 10σ2bℓ, (17a) T2=9√ 3 20σ2bℓ, (17b) T3=11√ 3 20σ2bℓ, (17c) T4=−3√ 3 10σ2bℓ, (17d) T5=−√ 3 5σ2bℓ. (17e) For compression in x2, bar 3 carries the highest compressive force. The elastic buckling strength of the lattice is obtained by setting T3equal to the Euler buckling load in (1), and this yields: σ2,el Es=5√ 3 99n2 2π2t ℓ3 = 0.094¯ρ3, (18) where the end constraint factor n2= 1.455 is derived in Appendix B.1. Otherwise, the yield strength of the lattice is obtained by setting T3=btσys, which gives: σ2,pl σys=20√ 3 33t ℓ = 0.390¯ρ. (19) 8 3+3ℓ1+3ℓ 𝑥2 𝑥151 𝐴𝐵52 66 3 3 44 1 2 6Figure 4: Periodic unit cell of demi-regular lattice B. 2.2. Demi-regular lattice B Demi-regular lattice B is made from triangles and squares as shown on the unit cell in Fig. 4. The relative density of this tessellation is given by: ¯ρ=18 3 + 2√ 3t ℓ = 2.785t ℓ . (20) 2.2.1. Loading in x1direction First, consider demi-regular lattice B loaded in the x1direction by a compressive stress σ1. This tessellation has six independent bars, which are labelled in Fig. 4. Equilibrium equations for joints AandBgive: T2+T3−√ 3T4= 0, (21) T1−T3−√ 3T4+ 2T5= 0, (22) √ 3(T1+T3)−T4−2T6= 0. (23) Otherwise, the macroscopic stress field is σ11=σ1andσ22= 0 and this implies: T1+T2+ 2T5= (1 +√ 3)σ1bℓ, (24) √ 3(T1+T3) +T4+ 2T6= 0, (25) respectively. Again, demi-regular lattice B is an indeterminate truss and we can use the principle of minimum strain energy to solve the system. Intermediate steps are included in 9 Appendix A.2.1 and lead to the following internal forces: T1=−(1 +√ 3) 24σ1bℓ, (26a) T2=11(1 +√ 3) 24σ1bℓ, (26b) T3=1 +√ 3 24σ1bℓ, (26c) T4=3 +√ 3 6σ1bℓ, (26d) T5=7(1 +√ 3) 24σ1bℓ, (26e) T6=−(3 +√ 3) 12σ1bℓ. (26f) The nominal compressive strain ϵ1is obtained by equating the work done by external loads to the internal strain energy and this gives: ϵ1=2√ 3−3 6Esσ1tℓb2 8T2 1+ 4T2 2+ 8T2 3+ 8T2 4+ 4T2 5+ 4T2 6 =47√ 3 72σ1 Esℓ t. (27) Therefore, the elastic modulus, E=σ1/ϵ1, of demi-regular lattice B becomes: E Es=24√ 3 47t ℓ = 0.318¯ρ. (28) Otherwise, the nominal strain in x2can be calculated using the principle of virtual work, which gives: ϵ2=3−√ 3 18Esbt 7T1−5 2T2+ 5T3−1 2T5+ 3√ 3T6 =−17√ 3 72σ1 Esℓ t, (29) and the Poisson’s ratio is: ν=−ϵ2 ϵ1=17 47= 0.362. (30) Since this tessellation is isotropic, its shear modulus is G=E/(2(1 + ν)) and this yields: G Es=3√ 3 16t ℓ = 0.117¯ρ. (31) Next, we turn our attention to the compressive strength of demi-regular lattice B. For compression in x1, the highest compressive force is in bar 2, see (26). Therefore, the elastic 10 buckling strength is obtained by equating T2to the Euler critical load in (1), which returns: σ1,el Es=(√ 3−1) 11n2 1π2t ℓ3 = 0.083¯ρ3, (32) where the end constraint factor n1= 1.650 is derived analytically in Appendix B.2. On the other hand, the yield strength of the lattice is obtained by setting T2=btσys, which gives: σ1,pl σys=24 11(1 +√ 3)t ℓ = 0.287¯ρ. (33) 2.2.2. Loading in x2direction Now, consider demi-regular lattice B loaded in the x2direction by a compressive stress σ2. Equations (21), (22) and (23) remain valid since they represent the equilibrium of joints AandB, see Fig. 4. On the other hand, enforcing the macroscopic stress field σ22=σ2and σ11= 0 yields: √ 3(T1+T3) +T4+ 2T6= (3 +√ 3)σ2bℓ, (34) T1+T2+ 2T5= 0, (35) respectively. The additional equation needed to solve the system is obtained with the princi- ple of minimum strain energy, see Appendix A.2.2, and we find the following internal forces: T1=7(1 +√ 3) 24σ2bℓ, (36a) T2=−5(1 +√ 3) 24σ2bℓ, (36b) T3=5(1 +√ 3) 24σ2bℓ, (36c) T4= 0, (36d) T5=−(1 +√ 3) 24σ2bℓ, (36e) T6=(3 +√ 3) 4σ2bℓ. (36f) 11 3+23ℓ2+3ℓ 𝑥2 𝑥15 7𝐴𝐵5 6263 34 4 121 8 997 𝐶𝐷Figure 5: Periodic unit cell of demi-regular lattice C. Here, bar 6 carries the highest compressive load. Equating T6to the Euler critical load in (1) returns the elastic buckling strength of the lattice: σ2,el Es=1 3(3 +√ 3)n2 2π2t ℓ3 = 0.089¯ρ3, (37) where the end constraint factor n2= 1.659 is derived in Appendix B.2. Otherwise, the yield strength is obtained by setting T6=btσys, which gives: σ2,pl σys=4 (3 +√ 3)t ℓ = 0.304¯ρ. (38) 2.3. Demi-regular lattice C Demi-regular lattice C is made from three regular polygons: triangles, squares and hexagons, see its unit cell in Fig. 5. Its relative density is given by: ¯ρ=54 12 + 7√ 3t ℓ = 2.238t ℓ . (39) 2.3.1. Loading in x1direction Consider demi-regular lattice C loaded in the x1direction by a uniform compressive stress σ1. This tessellation has nine independent bars that are labelled in Fig. 5. Equilibrium equations for joints A,B,CandDgive: T1+T2−√ 3T3= 0, (40) 12 √ 3T4−T5−2T6= 0, (41) T4+√ 3T5−2T9= 0, (42) √ 3T3+T5−2T6−T8= 0, (43) −T3+√ 3(T5+T8)−2T7= 0, (44) √ 3T2−T4= 0. (45) Otherwise, the macroscopic stress field is σ11=σ1andσ22= 0, which implies that: T1+T2+√ 3T4= (2 +√ 3)σ1bℓ, (46) √ 3T2+T3+ 2(T7+T9) = 0 . (47) Again, the additional equation needed to solve the system is obtained using the principle of minimum strain energy. Intermediate steps are included in Appendix A.3.1 and lead to the following internal forces: T1=11(2 +√ 3) 24σ1bℓ, (48a) T2=13(2 +√ 3) 96σ1bℓ, (48b) T3=19(3 + 2√ 3) 96σ1bℓ, (48c) T4=13(3 + 2√ 3) 96σ1bℓ, (48d) T5=−11(2 +√ 3) 96σ1bℓ, (48e) T6=25(2 +√ 3) 96σ1bℓ, (48f) T7=−17(3 + 2√ 3) 96σ1bℓ, (48g) T8=−(2 +√ 3) 24σ1bℓ, (48h) T9=(3 + 2√ 3) 96σ1bℓ. (48i) The nominal compressive strain ϵ1is obtained by equating the work done by external loads to the internal strain energy, which gives: ϵ1=7√ 3−12 3Esσ1tℓb2 2T2 1+ 8 T2 2+T2 3+T2 4+T2 5+T2 6 + 4 T2 7+T2 8+T2 9 13 =143√ 3 144σ1 Esℓ t. (49) Therefore, the elastic modulus E=σ1/ϵ1becomes: E Es=48√ 3 143t ℓ = 0.260¯ρ. (50) Otherwise, the nominal strain along x2can be calculated using the principle of virtual work, which gives: ϵ2=2−√ 3 6Esbt −5√ 3 6(T1−T2)−5 2(T3−T4) +√ 3 6(29T5−7T6) +31 4T7+7√ 3 3T8+17 4T9! =−65√ 3 144σ1 Esℓ t,(51) and the Poisson’s ratio is: ν=−ϵ2 ϵ1=65 143= 0.455. (52) Since this tessellation is isotropic, the shear modulus is G=E/(2(1 + ν)) and this yields: G Es=3√ 3 26t ℓ = 0.089¯ρ. (53) Next, we turn our attention to the compressive strength of demi-regular lattice C. For compression in x1, the maximum compressive load is in bar 1, see (48). Equating T1to the Euler buckling load in (1) gives us the elastic buckling strength of the lattice: σ1,el Es=2(2−√ 3) 11n2 1π2t ℓ3 = 0.105¯ρ3, (54) where the end constraint factor n1= 1.567 is derived analytically in Appendix B.3. In contrast, the yield strength is obtained by setting T1=btσys, which gives: σ1,pl σys=24(2−√ 3) 11t ℓ = 0.261¯ρ. (55) 14 2.3.2. Loading in x2direction Now, consider demi-regular lattice C loaded in the x2direction by a uniform compressive stress σ2. Expressions (40) to (45) remain valid since they represent the equilibrium of joints A,B,CandD, see Fig. 5. Otherwise, enforcing the macroscopic stress field σ22=σ2and σ11= 0, yields: √ 3T2+T3+ 2(T7+T9) = (3 + 2√ 3)σ2bℓ, (56) T1+T2+√ 3T4= 0, (57) respectively. The additional equation needed to solve the system is obtained with the princi- ple of minimum strain energy, see Appendix A.3.2, and we find the following internal forces: T1=−5(2 +√ 3) 24σ2bℓ, (58a) T2=5(2 +√ 3) 96σ2bℓ, (58b) T3=−5(3 + 2√ 3) 96σ2bℓ, (58c) T4=5(3 + 2√ 3) 96σ2bℓ, (58d) T5=29(2 +√ 3) 96σ2bℓ, (58e) T6=−7(2 +√ 3) 96σ2bℓ, (58f) T7=31(3 + 2√ 3) 96σ2bℓ, (58g) T8=7(2 +√ 3) 24σ2bℓ, (58h) T9=17(3 + 2√ 3) 96σ2bℓ. (58i) Here, the highest compressive force is in bar 7. Therefore, the elastic buckling strength is obtained by equating T7to the Euler buckling load in (1), which gives: σ2,el Es=8(2√ 3−3) 93n2 2π2t ℓ3 = 0.098¯ρ3, (59) 15 where the end constraint factor n2= 1.667 is derived analytically in Appendix B.3. Oth- erwise, the yield strength of the lattice is obtained by setting T7=btσys, which yields: σ2,pl σys=32(2√ 3−3) 31t ℓ = 0.214¯ρ. (60) 3. Materials and methods Both Finite Element (FE) simulations and experiments were conducted to verify the analytical expressions derived in Section 2. Below, our FE modelling approach is presented first, followed by a description of our experimental protocol. 3.1. Finite element modelling All simulations were done with the commercial software Abaqus and using the standard solver (Static, Implicit step). Periodic unit cells, identical to those in Fig. 3 to 5, were used for each topology. In all cases, the bar length was kept fixed to ℓ= 10 mm, but the bar thickness twas varied to cover a range of relative densities ¯ ρfrom 0.01 to 0.3. Each bar was discretised using at least 70 Timoshenko beam elements (B21 in Abaqus notation) based on a mesh convergence study (not shown here). It was necessary to include a geometric imperfection in all simulations to trigger buckling. This imperfection had the shape of the first non-swaying eigenmode and its amplitude was set to 5% of the bar thickness, as used in previous studies on prismatic lattices [11, 21]. The parent material was modelled as an elastic perfectly plastic solid. The elastic regime was linear and isotropic, characterised by a Young’s modulus Es= 200 GPa and a Poisson’s ratio νs= 0.26, up to a yield strength σys= 250 MPa. These material properties, repre- sentative of steel, were selected to ensure that both collapse modes (elastic buckling and yielding) occur over the range of relative densities considered (0 .01≤¯ρ≤0.3). Periodic boundary conditions were imposed on each unit cell using the following con- straint equations [7, 11, 21]: ∆ui=ϵij∆xjand ∆ ϕ= 0, (61) 16 where ∆ uiand ∆ ϕare the difference in displacement and rotation, respectively, between corresponding points of the unit cell; ϵijis the macroscopic nominal strain tensor; and ∆xjis the displacement vector connecting two corresponding points of the unit cell. The compressive response of the lattice along x1was obtained by prescribing ϵ11, and letting Abaqus calculate σ11assuming σ22= 0. Likewise, the compressive response in x2was obtained by imposing ϵ22, and computing σ22provided σ11= 0. Finally, the response in shear was obtained by prescribing ϵ12and computing σ12while keeping σ11=σ22= 0. 3.2. Specimen manufacture and compression tests Experiments were designed to (i) corroborate the analytical expressions derived in Sec- tion 2 and (ii) compare the performances of demi-regular tessellations to those of a regular triangular lattice. All samples were produced by additive manufacturing using a Formlabs Form 3 machine, which relies on stereolithography to cure a photo-polymerising resin. The overall dimensions of each lattice are given in Fig. 6. All samples had a relative density ¯ρ= 0.17, and this was achieved by keeping the bar length fixed at ℓ= 10 mm and adjusting the strut thickness tas indicated in Table 1. In all cases, the specimens had a depth of 25 mm in the prismatic direction. The samples were compressed only in the x2direction since they are elastically isotropic and this is the strongest orientation for all topologies (except tessellation C where the elastic buckling strength is 7% higher in x1, see Section 2). The specimens were fabricated as follows. First, the geometry of the sample was created in Abaqus and a stl file was exported to the Form 3 printer. Second, the specimen was printed with a layer thickness of 0.05 mm and using Formlabs Clear resin. All samples were printed with their prismatic axis perpendicular to the printing bed. Finally, when the print was completed, the lattice was washed in an isopropyl alcohol (IPA) solution and post-cured under UV light at a temperature of 60◦C for 30 min, as recommended in the Formlabs documentation. The printed lattices had no visible imperfections, such as broken or wavy bars. All samples were compressed at rate of 0.05 mm/s using a MTS electromechanical testing machine with a capacity of 30 kN. Both the compressive force and displacement were recorded by the testing machine. 17 𝑥2 𝑥1(a) 80 69120 104(b) 142 109 (c)129 112(d)Figure 6: Dimensions, in mm, of the tested samples: demi-regular tessellations (a) A, (b) B, and (c) C as well as (d) a triangular lattice. All samples had a bar length ℓ= 10 mm and a depth of 25 mm in the prismatic direction. Table 1: Values of relative density ¯ ρ, bar thickness t, and bar length ℓused in the experiments. Topology ¯ ρ t [mm] ℓ[mm] t/ℓ A 0.17 0.63 10 0.063 B 0.17 0.61 10 0.061 C 0.17 0.76 10 0.076 Triangular 0.17 0.49 10 0.049 18 00.020.040.060.080.101020304050 Figure 7: Responses of three tensile tests done on the polymer used to manufacture all lattices. Tensile tests were conducted to characterise the behaviour of the Clear resin used to manufacture all lattices. Following the procedure detailed above, three tensile specimens were printed with a thickness of 0.6 mm and a gauge length of 15 mm (dimensions comparable to those of a cell wall inside the lattice samples). The tensile responses, measured at a strain rate of 5 ·10−4s−1, are shown in Fig. 7. The measurements have a small scatter with an average Young’s modulus Es= 2.1 GPa and a mean ultimate tensile strength of 54 MPa. 4. Results and discussion Our results are presented below in three parts. First, numerical and analytical results are compared to validate the expressions derived in Section 2. Second, we use our analytical results to compare the properties of demi-regular tessellations to those of a triangular lattice. Third, we present our experimental results to corroborate the performances of demi-regular tessellations. 4.1. Comparison between analytical and finite element results Finite Element (FE) simulations are compared to analytical predictions in Fig. 8a for the elastic modulus Eand in Fig. 8b for the shear modulus G. Both moduli are normalised by the Young’s modulus of the solid Es; hence, the results in Fig. 8 are insensitive to the choice 19 0.010 .10.010.1 0 .010 .110-310-2 Figure 8: Comparison between analytical and FE results: (a) elastic modulus Eand (b) shear modulus G, both plotted as a function of relative density ¯ ρand normalised by the Young’s modulus of the parent material Es= 200 GPa. Results for demi-regular tessellations A, B, and C are compared to those of a triangular lattice. of parent material. Results are given for all three demi-regular lattices. In all cases, there is an excellent agreement between FE and analytical results for both EandG. This is true over the entire range of relative density considered here (0 .01≤¯ρ≤0.3). Next, FE predictions of the compressive strength are compared to analytical results in Fig. 9a and b for compression in x1andx2, respectively. Here, the compressive strength is normalised by the yield strength σys= 250 MPa of the parent material. In all cases, the collapse mode changes from elastic buckling (at low relative densities) to yielding (at high values of ¯ ρ) and this transition is noticeable in Fig. 9 by the change in slope. For both collapse modes, there is an excellent agreement between FE and analytical results. This holds true for both loading directions, and for all three demi-regular lattices. Even though analytical and FE models share similar assumptions (both rely on periodic unit cells and consider an elastic, perfectly plastic parent material), there is an important difference. The simulations are done using Timoshenko beam elements, which account for axial, bending and shear deformations. In contrast, our analytical model assumes that 20 0.010 .110-410-310-210-1 0 .010 .110-410-310-210-1 Figure 9: Comparison between analytical and FE results for the normalised compressive strength in (a) x1 and (b) x2directions, both plotted as a function of relative density ¯ ρ. Results for demi-regular tessellations A, B, and C are compared to those of a triangular lattice. The properties of the parent material are Es= 200 GPa and σys= 250 MPa. each bar behaves as a pin-jointed truss, neglecting bending and shear deformations. This assumption is justified based on the excellent agreement between FE and analytical results in Fig. 8 and 9. Therefore, we conclude that the analytical expressions derived in Section 2 are validated by FE simulations. 4.2. Comparison between demi-regular and triangular lattices 4.2.1. Analytical results Now that our analytical equations are verified, we can use them to compare the perfor- mances of demi-regular tessellations to those of a triangular lattice. This is a fair comparison since these four topologies are stretching-dominated and in-plane elastically isotropic. The mechanical properties of demi-regular tessellations A, B, and C are compared to those of a triangular lattice in Fig. 8 and 9, as well as in Table 2, where all analytical expressions are summarised. Here, the properties of a triangular lattice are collected from previous studies [3–6, 11] and its orientation is defined in Fig. 6d. Demi-regular lattices have a comparable, yet slightly lower elastic modulus Ethan a 21 Table 2: The mechanical properties of demi-regular tessellations A, B, and C are compared to those of a triangular lattice. The properties of a triangular lattice are collected from [3–6, 11]. Topology ¯ ρ E/E s ν G/E sσ1,el/Esσ2,el/Esσ1,pl/σysσ2,pl/σys A 2 .694t/ℓ 0.292¯ρ0.364 0 .107¯ρ0.081¯ρ30.094¯ρ30.268¯ρ 0.390¯ρ B 2 .785t/ℓ 0.318¯ρ0.362 0 .117¯ρ0.083¯ρ30.089¯ρ30.287¯ρ 0.304¯ρ C 2 .238t/ℓ 0.260¯ρ0.455 0 .089¯ρ0.105¯ρ30.098¯ρ30.261¯ρ 0.214¯ρ Triangular 3 .464t/ℓ 0.333¯ρ0.333 0 .125¯ρ0.061¯ρ30.069¯ρ30.333¯ρ 0.500¯ρ triangular lattice, see Table 2 and Fig. 8a. Tessellation B has the highest elastic modulus, being only 5% more compliant than a triangular lattice, followed by topologies A and then C. These observations also hold true for the shear modulus G. The triangular lattice has a higher yield strength than any of the three demi-regular lattices, see Table 2 and Fig. 9. For compression in x1, all three demi-regular topologies have a similar yield strength, which is 15-20% lower than a triangular lattice, see Fig. 9a. When compressed in x2, demi-regular lattice A has the highest yield strength, yet it remains 20% lower than a triangular lattice. The main advantage of demi-regular lattices is their high resistance to elastic buckling, see Table 2 and Fig. 9. All demi-regular lattices have a significantly higher elastic buckling strength than a triangular lattice, and this holds true for both compression in x1andx2. The elastic buckling strength in x2is greater than that in x1, except for tessellation C. Nonetheless, demi-regular lattice C offers the highest elastic buckling strength in x2, being 42% stronger than a triangular lattice. Amongst demi-regular lattices, tessellation B has the lowest elastic buckling strength σ2,el, yet it remains about 30% stronger than a triangular lattice. In summary, our analytical results show that demi-regular tessellations are slightly more compliant but significantly stronger than a triangular lattice when failure is governed by elastic buckling. In the next section, experiments are presented to corroborate these findings. 4.2.2. Experimental results The measured compressive responses are plotted in Fig. 10 for demi-regular tessellations A, B, and C as well as for a triangular lattice. All tests have a linear elastic regime up to a peak stress, followed by a gradual softening response. All lattices failed by elastic buckling 22 00.0040.0080.0120.0160.0200.20.40.60.811.2 Figure 10: The measured compressive responses of demi-regular tessellations A, B, and C are compared to that of a triangular lattice. All samples have a relative density ¯ ρ= 0.17. and their deformation is shown in Fig. 11. Buckling took place in the diagonal bars for tessellation A and the triangular lattice (see Fig. 11a,d), whereas the vertical struts buckled in demi-regular lattices B and C (see Fig. 11b,c). These deformation modes are in good agreement with the periodic buckling patterns considered in our analytical model, which are included in Fig. 11 for comparison. Two tests were done for each topology and the elastic modulus and compressive strength measured are summarised in Fig. 12a and b, respectively. Here, the elastic modulus Eis nor- malised by Es= 2.1 GPa, whereas the compressive strength is normalised by σys= 54 MPa (see the tensile response of the parent material in Fig. 7). The experiments have a good repeatability, with variations inferior to 10%, except for tessellation A which shows a larger disparity between tests. Taking the average of both tests, the experimental results indicate that demi-regular lattice B is 5% more compliant than a triangular lattice, whereas this value increases to 13% for tessellation C. The tests also show that demi-regular lattice C is 47% stronger than a triangular lattice. These differences in elastic modulus and compressive strength are in-line with with our analytical predictions, see Table 2. 23 2=1% (d) 2=1.64% (c) 2=1.44% (b) 2=1.54% (a)Experiments Analytical Experiments AnalyticalFigure 11: Photographs showing the deformation modes of demi-regular tessellations (a) A, (b) B, and (c) C, as well as (d) a triangular lattice. All lattices have a relative density ¯ ρ= 0.17 and fail by elastic buckling. The instant when each photograph was taken is indicated on the responses in Fig. 10 with a circular symbol. For scale, all bars have a length of 10 mm. For each topology, the period buckling shape considered in our analytical model is also included for comparison. TriB A C 00.010.020.030.040.050.06 C A B T ri00.0050.010.0150.02 Figure 12: Comparison between experimental and analytical results: (a) normalised elastic modulus and (b) normalised compressive strength in x2. Results are given for demi-regular tessellations A, B, and C, as well as a triangular lattice. In all cases, the relative density is ¯ ρ= 0.17 and the properties of the parent material areEs= 2.1 GPa and σys= 54 MPa. 24 Our analytical predictions are also included in Fig. 12 for comparison. Considering the average of both tests, the largest difference between experimental and analytical values is 9% for the elastic modulus and 8% for the compressive strength. Therefore, the measured properties are in good agreement with our analytical results and confirm that demi-regular lattices are slightly more compliant than a triangular lattice, but have a higher elastic buckling strength. 5. Conclusion This study explored the potential of three demi-regular tessellations as stiff and strong lattice materials. Their in-plane mechanical properties were first derived analytically, then validated using Finite Element simulations, and corroborated by experiments performed on polymer lattices produced by additive manufacturing. All three demi-regular tessellations had an in-plane isotropic and stretching-dominated behaviour. Consequently, their properties were compared to those of a triangular lattice, which exhibits the same characteristics. This comparison showed that demi-regular topolo- gies are 5-20% more compliant, but up to 42% stronger than a triangular lattice when the failure mode is elastic buckling. With such combination of properties, demi-regular tessella- tions are attractive topologies for ultra-light and/or polymer lattices, which usually fail by elastic buckling of the constituent members. There is growing evidence that prismatic lattices are not only stiff and strong, but also have a high fracture toughness [26–30]. Work is currently underway to evaluate if these three demi-regular tessellations are tougher than regular and semi-regular lattices. Acknowledgements AM wishes to thank the Finnish Cultural Foundation for their financial support (grant 00220662). LS is grateful for the funding received from the Academy of Finland (grant 322007). 25 A. Principle of minimum strain energy A.1. Demi-regular lattice A A.1.1. Loading in x1 Using (3) to (6), we express T2,T3,T4, and T5as functions of T1, which yields: T2=−T1 2, (A.1a) T3=T1 2, (A.1b) T4=−T1 2+3√ 3 4σ1bℓ, (A.1c) T5=T1 2+3√ 3 4σ1bℓ. (A.1d) Next, the strain energy for the unit cell shown in Fig. 3 is: U=ℓ 2Esbt 6T2 1+ 16T2 2+ 8T2 3+ 8T2 4+ 4T2 5 , (A.2) and substituting (A.1) in (A.2) returns: U=ℓ 2Esbt 15T2 1−3√ 3T1σ1bℓ+81 4(σ1bℓ)2 . (A.3) The minimum strain energy is obtained by setting the derivative of U, with respect to T1, equal to zero, which gives: ∂U ∂T1= 0 = ⇒T1=√ 3 10σ1bℓ. (A.4) Finally, substituting (A.4) in (A.1) returns the internal forces in bars 2 to 5, and these are listed in (7). A.1.2. Loading in x2 First, we use (3), (4), (15) and (16) to express T2,T3,T4, and T5as functions of T1, which gives: T2=−T1 2+√ 3 2σ2bℓ, (A.5a) T3=T1 2+√ 3 2σ2bℓ, (A.5b) 26 T4=−T1 2−√ 3 4σ2bℓ, (A.5c) T5=T1 2−√ 3 4σ2bℓ. (A.5d) Second, we substitute these expressions in (A.2) to obtain the strain energy: U=ℓ 2Esbt 15T2 1−3√ 3T1σ2bℓ+81 4(σ2bℓ)2 . (A.6) Third, differentiating Uwith respect to T1to find the minimum strain energy yields: ∂U ∂T1= 0 = ⇒T1=√ 3 10σ2bℓ. (A.7) Finally, substituting (A.7) in (A.5) returns the internal forces, which are given in (17). A.2. Demi-regular lattice B A.2.1. Loading in x1 First, we express T2, T3, T4, T5andT6as functions of T1using (22) to (25), which gives: T2=T1+(1 +√ 3) 2σ1bℓ, (A.8a) T3=−T1, (A.8b) T4=(3 +√ 3) 6σ1bℓ, (A.8c) T5=−T1+(1 +√ 3) 4σ1bℓ, (A.8d) T6=−(3 +√ 3) 12σ1bℓ. (A.8e) The strain energy for the unit cell illustrated in Fig. 4 is: U=ℓ 2Esbt 8T2 1+ 4T2 2+ 8T2 3+ 8T2 4+ 4T2 5+ 4T2 6 , (A.9) and substituting (A.8) into (A.9) yields: U=ℓ 2Esbth 24T2 1+ 2(1 +√ 3)T1σ1bℓ+ 4(2 +√ 3)(σ1bℓ)2i . (A.10) Differentiating Uwith respect to T1and setting it equal to zero to find the minimum strain energy returns: ∂U ∂T1= 0 = ⇒T1=−(1 +√ 3) 24σ1bℓ. (A.11) Finally, substituting this result in (A.8) gives the internal forces in bars 2 to 6, see (26). 27 A.2.2. Loading in x2 Here, we use (21) to (23), (34) and (35) to express T2,T3,T4,T5, and T6as functions of T1, which yields: T2=T1−(1 +√ 3) 2σ2bℓ, (A.12a) T3=−T1+(1 +√ 3) 2σ2bℓ, (A.12b) T4= 0, (A.12c) T5=−T1+(1 +√ 3) 4σ2bℓ, (A.12d) T6=(3 +√ 3) 4σ2bℓ. (A.12e) Substituting these in (A.9) returns the strain energy: U=ℓ 2Esbth 24T2 1−14(1 +√ 3)T1σ2bℓ+ 8(2 +√ 3)(σ2bℓ)2i . (A.13) Next, we differentiate this with respect to T1to find the minimum strain energy, and this gives: ∂U ∂T1= 0 = ⇒T1=7(1 +√ 3) 24σ2bℓ. (A.14) Finally, substituting T1into (A.12) returns the internal forces given in (36). A.3. Demi-regular lattice C A.3.1. Loading in x1 Using (40) to (47), we write T2, . . . , T 9as functions of T1, which gives: T2=−1 4T1+(2 +√ 3) 4σ1bℓ, (A.15a) T3=√ 3 4T1+(3 + 2√ 3) 12σ1bℓ, (A.15b) T4=−√ 3 4T1+(3 + 2√ 3) 4σ1bℓ, (A.15c) T5=−1 4T1, (A.15d) T6=−1 4T1+3(2 +√ 3) 8σ1bℓ, (A.15e) 28 T7=√ 3 4T1−7(3 + 2√ 3) 24σ1bℓ, (A.15f) T8=T1−(2 +√ 3) 2σ1bℓ, (A.15g) T9=−√ 3 4T1+(3 + 2√ 3) 8σ1bℓ. (A.15h) Next, the strain energy for the unit cell in Fig. 5 is given by: U=ℓ 2Esbt 2T2 1+ 8(T2 2+T2 3+T2 4+T2 5+T2 6) + 4( T2 7+T2 8+T2 9) , (A.16) and substituting (A.15) in (A.16) yields: U=ℓ 2Esbt" 12T2 1−11(2 +√ 3)T1σ1bℓ+11(7 + 4√ 3) 2(σ1bℓ)2# . (A.17) Differentiating Uwith respect to T1and setting it equal to zero to find the minimum strain energy returns: ∂U ∂T1= 0 = ⇒T1=11(2 +√ 3) 24σ1bℓ. (A.18) Finally, substituting this result in (A.15) returns the internal forces given in (48). A.3.2. Loading in x2 Here, we use (40) to (45), (56) and (57) to express T2, . . . , T 9as functions of T1, which yields: T2=−1 4T1, (A.19a) T3=√ 3 4T1, (A.19b) T4=−√ 3 4T1, (A.19c) T5=−1 4T1+(2 +√ 3) 4σ2bℓ, (A.19d) T6=−1 4T1−(2 +√ 3) 8σ2bℓ, (A.19e) T7=√ 3 4T1+3(3 + 2√ 3) 8σ2bℓ, (A.19f) T8=T1+(2 +√ 3) 2σ2bℓ, (A.19g) 29 𝑃 𝑃 𝑎 𝑏 Δ 𝜃𝑎𝑃 𝑉𝑀𝑎𝑃 𝑉 𝑏𝜃𝑏(a) (b) 𝑀ℓFigure B.1: (a) A bar of length ℓsubjected to a compressive load P. (b) The end rotations θaandθb generate bending moments MaandMb, respectively, and the transverse displacement ∆ causes the shear force V. T9=−√ 3 4T1+(3 + 2√ 3) 8σ2bℓ. (A.19h) Next, we substitute (A.19) in (A.16), which gives: U=ℓ 2Esbt" 12T2 1+ 5(2 +√ 3)σ2bℓT 1+7(7 + 4√ 3) 2(σ2bℓ)2# . (A.20) The minimum strain energy Uallows us to find T1, and this yields: ∂U ∂T1= 0 = ⇒T1=−5(2 +√ 3) 24σ2bℓ. (A.21) Substituting this back in (A.19) returns the internal forces given in (58). B. End constraint factor Lattices with low relative densities usually fail by elastic buckling, and analytical pre- dictions for this failure mode are sensitive to the end constraint factor n, see (1). The end constraint factor has been derived analytically for regular [6, 11] and semi-regular [21] lattices, and here, we extend this work to demi-regular lattices A, B, and C. The method used to calculate the end constraint factor relies on the stiffness matrix of a single bar. Consider a bar of length ℓloaded by an axial compressive force P, as shown in Fig. B.1a. Next, assume that small rotations θaandθbare imposed at ends a andb, respectively, and a small transverse displacement ∆ is applied, see Fig. B.1b. The corresponding bending moments MaandMb, and shear force Vare given by [31]: 30   Ma Mb V  =EsI ℓ s sc ¯s/ℓ sc s ¯s/ℓ ¯s/ℓ ¯s/ℓ s∗/ℓ2   θa θb ∆  , (B.1) where Esis the elastic modulus and Iis the second moment of area of the bar. In addition, ¯sands∗are defined as: ¯s=s(1 +c) and s∗= 2¯s−Pℓ2 EsI, (B.2) in which parameters sandcvary depending on the axial load P. When the bar is loaded in compression ( P >0),sandcare given by: s=λ(sinλ−λcosλ) 2−2 cosλ−λsinλ, (B.3) and c=λ−sinλ sinλ−λcosλ, (B.4) where λ=r P EsIℓ. (B.5) Otherwise, when the bar is in tension ( P <0),sandchave the form: s=s1=λ1(λ1coshλ1−sinhλ1) 2−2 cosh λ1+λ1sinhλ1, (B.6) c=c1=sinhλ1−λ1 λ1coshλ1−sinhλ1, (B.7) λ1=r -P EsIℓ, (B.8) where the subscript 1 is used simply to differentiate these expressions from those in (B.3) to (B.5). Finally, when the bar carries no axial load ( P= 0), the parameters sandcare: s= 4 and c= 0.5. (B.9) 31 The approach to find the end constraint factor nis as follows. First, the periodic buck- ling shape with the longest wavelength is identified for each topology and for each loading direction. Then, equilibrium conditions and (B.1) are combined to form a constitutive equa- tion from which λcan be solved. Once λis known, it is straightforward to compute the end constraint factor nsince λ=πn. B.1. Demi-regular lattice A First, consider demi-regular lattice A compressed in the x1direction. For this loading scenario, the periodic buckling shape illustrated in Fig. B.2a is the one associated with the lowest load. This pattern has two rotations, denoted as θ1andθ2, whereas there are no rotations at vertices bandd. By equilibrium, the bending moments at vertices aandfare zero, and this gives: Mab+Mac+Mad+Mae+Mak= 0, (B.10a) 2Mfb+ 2Mfc+Mfg= 0, (B.10b) respectively. Here, the notation Mijdenotes the bending moment in bar ijat end i. Each term in (B.10) is expressed as a function of θ1andθ2using (B.1), and this returns: Mab=EsI ℓ(s1abθ2), (B.11a) Mac=EsI ℓ(sacθ2−saccacθ2), (B.11b) Mad=EsI ℓ(sadθ2), (B.11c) Mae=EsI ℓ(saeθ2−saecaeθ2), (B.11d) Mak=EsI ℓ(sakθ2−sakcakθ1), (B.11e) Mfb=EsI ℓ(s1fbθ1), (B.11f) Mfc=EsI ℓ(sfcθ1−sfccfcθ2), (B.11g) Mfg=EsI ℓ(sfgθ1−sfgcfgθ1), (B.11h) 32 (a) (b) 2 11 -1 2 2 2 -12 12-2 -21 -1 -1Figure B.2: Periodic buckling modes of demi-regular lattice A compressed in (a) x1and (b) x2. 33 where sijandcijare functions of λijas defined in relations (B.3) to (B.9). Each λijis proportional to the axial force in bar ij(see (B.5)); therefore, with the internal loads in (7), we find that: λad= 0.9354λae, (B.12a) λac=λfg= 0.3536λae, (B.12b) λ1ab=λ1fb= 0.25λae, (B.12c) λak=λfc= 0.25λae. (B.12d) Above, each λijis expressed as a function of λaesince bar aeis the most loaded strut and the one expected to buckle. Substituting (B.11) in (B.10) returns a linear system of equations: A[θ1, θ2]T=0, where the determinant of Ashould be zero for a non-trivial solution to exists. With the expressions in (B.12), setting det( A) = 0 returns a lengthy expression where λaeis the only unknown. Solving numerically yields λae= 5.131 and therefore, the end constraint factor for compression in x1is: n1=λae π= 1.633. (B.13) Next, consider demi-regular lattice A compressed in the x2direction. In this case, the lattice collapses according to the periodic buckling shape shown in Fig. B.2b. Equation (B.10) remain valid, but the bending moments are now given by: Mab=EsI ℓ(sabθ2), (B.14a) Mac=EsI ℓ(sacθ2−saccacθ2), (B.14b) Mad=EsI ℓ(s1adθ2), (B.14c) Mae=EsI ℓ(s1aeθ2−s1aec1aeθ2), (B.14d) Mak=EsI ℓ(sakθ2−sakcakθ1), (B.14e) Mfb=EsI ℓ(sfbθ1), (B.14f) 34 Mfc=EsI ℓ(sfcθ1−sfccfcθ2), (B.14g) Mfg=EsI ℓ(sfgθ1−sfgcfgθ1), (B.14h) where the parameters sijandcijare again functions of λij. Next, we use the internal loads in (17) to express each λijas a function of λfcsince this is the bar sustaining the highest force and expected to buckle. This gives: λak=λfc, (B.15a) λac=λfg= 0.4264λfc, (B.15b) λab=λfb= 0.9045λfc, (B.15c) λ1ad= 0.7385λfc, (B.15d) λ1ae= 0.6030λfc. (B.15e) Similarly to the analysis in x1, substituting (B.14) in (B.10) returns a linear system of equations: A[θ1, θ2]T=0, where the determinant of Ashould be zero for a non-trivial solu- tion to exists. Setting det( A) = 0, and using the relations in (B.15), returns an expression where λfcis the only unknown. Solving numerically gives λfc= 4.5697 and therefore, the end constraint factor in x2is: n2=λfc π= 1.4546. (B.16) B.2. Demi-regular lattice B When compressed in x1, demi-regular lattice B buckles according to the periodic pattern shown in Fig. B.3a. This mode is characterised by three rotations denoted θ1,θ2andθ3. Equilibrium of moments at vertices a,bandkgives: Mab+Mac+Mad+Mae+Maf= 0, (B.17a) Mba+Mbc+Mbk+Mbg+Mbh= 0, (B.17b) Mkb+Mkg+Mkh= 0, (B.17c) 35 (a) (b) 2 1 −3 3 −2 2 −3 ℎ −1 3 3 1 − − Figure B.3: Periodic buckling modes of demi-regular lattice B compressed in (a) x1and (b) x2. respectively, where each bending moment is expressed as a function of θ1,θ2andθ3using (B.1), which yields: Mab=EsI ℓ(−sabθ3+sabcabθ2), (B.18a) Mac=EsI ℓ(−s1acθ3−s1acc1acθ3), (B.18b) Mad=EsI ℓ(−sadθ3−sadcadθ2), (B.18c) Mae=EsI ℓ(−saeθ3−saecaeθ3), (B.18d) Maf=EsI ℓ(−s1afθ3+s1afc1afθ1), (B.18e) Mba=EsI ℓ(sbaθ2−sbacbaθ3), (B.18f) Mbc=EsI ℓ(sbcθ2−sbccbcθ3), (B.18g) Mbk=EsI ℓ(sbkθ2−sbkcbkθ1), (B.18h) 36 Mbg=EsI ℓ(sbgθ2+sbgcbgθ3), (B.18i) Mbh=EsI ℓ(sbhθ2+sbhcbhθ3), (B.18j) Mkb=EsI ℓ(−skbθ1+skbckbθ2), (B.18k) Mkg=EsI ℓ(−s1kgθ1+s1kgc1kgθ3), (B.18l) Mkh=EsI ℓ(−s1khθ1+s1khc1khθ3). (B.18m) Above, the parameters sijandcijare functions of λijas defined in (B.3) to (B.9). Each λij is proportional to the axial force in bar ijand, with the internal loads in (26), we find that: λbg=λbh=λad= 0.3015λkb, (B.19a) λab=λbc= 0.7936λkb, (B.19b) λae= 0.7977λkb, (B.19c) λ1ac= 0.5612λkb, (B.19d) λ1af=λ1kg=λ1kh= 0.3015λkb, (B.19e) where all λijare expressed as functions of λkbsince this is the most loaded strut and the one expected to buckle. To find λkb, we first substitute (B.18) in (B.17) to form a system of equations: A[θ1, θ2, θ3]T=0. Next, we set det( A) = 0 and use (B.19) to obtain an expression as a function of λkbonly. Solving this expression numerically returns λkb= 5.1839 and consequently, the end constraint factor in x1is: n1=λkb π= 1.650. (B.20) Next, consider demi-regular lattice B compressed in the x2direction. In this case, the lattice is expected to deform according to the periodic buckling shape shown in Fig. B.3b. Equilibrium of vertex agives: Mab+Mac+Mad+Mae+Maf= 0, (B.21) 37 where each bending moment is: Mab=EsI ℓ(−sabθ), (B.22a) Mac=EsI ℓ(−sacθ+saccacθ), (B.22b) Mad=EsI ℓ(−sadθ), (B.22c) Mae=EsI ℓ(−s1aeθ+s1aec1aeθ), (B.22d) Maf=EsI ℓ(−safθ). (B.22e) Note that sab= 4 because bar abcarries no load, see (36) and (B.9). Otherwise, each λij is expressed as a function λac(since this is the most loaded bar and the one expected to buckle) using the internal loads in (36), which gives: λad= 0.6936λac, (B.23a) λ1ae= 0.3102λac, (B.23b) λaf= 0.8207λac. (B.23c) Substituting (B.22) in (B.21), and then making use of (B.23), returns a lengthy expression where λacis the only unknown. Solving numerically returns λac= 5.2149 and therefore, the end constraint factor in x2is: n2=λac π= 1.659. (B.24) B.3. Demi-regular lattice C When compressed in the x1direction, demi-regular lattice C buckles in the periodic pattern shown in Fig. B.4a. This deformation mode is characterised by two rotations ( θ1 andθ2), whereas there is no rotation at vertices dandh. Equilibrium of vertices aandc gives: 2Mad+ 2Mae+Maf= 0, (B.25a) Mcg+Mch+Mck+Mcl+Mcm= 0, (B.25b) 38 (a) (b) 2 1−2ℎ1−2 −1 −2 −21 −2 1ℎ−1 21 −1 2Figure B.4: Periodic buckling modes of demi-regular lattice C compressed in (a) x1and (b) x2. respectively, where the bending moments are: Mad=EsI ℓ(sadθ1), (B.26a) Mae=EsI ℓ(saeθ1−saecaeθ2), (B.26b) Maf=EsI ℓ(safθ1−safcafθ1), (B.26c) Mcg=EsI ℓ(−s1cgθ2−s1cgc1cgθ2), (B.26d) Mch=EsI ℓ(−schθ2), (B.26e) Mck=EsI ℓ(−sckθ2+sckcckθ1), (B.26f) Mcl=EsI ℓ(−s1clθ2), (B.26g) Mcm=EsI ℓ(−s1cmθ2−s1cmc1cmθ2). (B.26h) 39 Again, substituting (B.26) in (B.25), then setting the determinant of the coefficient matrix to zero, and making use of the following relations: λad= 0.5436λaf, (B.27a) λae=λck= 0.8648λaf, (B.27b) λ1cg= 0.8180λaf, (B.27c) λch= 0.7538λaf, (B.27d) λ1cl= 0.5λaf, (B.27e) λ1cm= 0.3015λaf, (B.27f) returns an expression where λafis the only unknown. Solving numerically, we find λaf= 4.922 and hence, the end constraint factor in x1is: n1=λaf π= 1.567. (B.28) Finally, consider demi-regular lattice C compressed in the x2direction. In this case, the lattice collapses according to the periodic buckling shape shown in Fig. B.4b. Equilibrium of vertices bandcimplies: Mbd+Mbe+Mbg+Mbh= 0, (B.29a) Mcg+Mch+Mck+Mcl+Mcm= 0, (B.29b) respectively, where the bending moments are: Mbd=EsI ℓ(sbdθ2), (B.30a) Mbe=EsI ℓ(sbeθ2−sbecbeθ1), (B.30b) Mbg=EsI ℓ(s1bgθ2+s1bgc1bgθ1), (B.30c) Mbh=EsI ℓ(sbhθ2−sbhcbhθ2), (B.30d) Mcg=EsI ℓ(−scgθ1+scgccgθ1), (B.30e) Mch=EsI ℓ(−s1chθ1−s1chc1chθ2), (B.30f) 40 Mck=EsI ℓ(−s1ckθ1), (B.30g) Mcl=EsI ℓ(−sclθ1+sclcclθ2), (B.30h) Mcm=EsI ℓ(−scmθ1+scmccmθ1), (B.30i) where sijandcijare defined in (B.3) to (B.9). Substituting (B.30) in (B.29) gives a system of equations: A[θ1, θ2]T=0. Next, setting det( A) = 0 and making use of the following relations: λbd=λ1ck= 0.4016λcg, (B.31a) λbe=λcl= 0.7349λcg, (B.31b) λ1bg=λ1ch= 0.3611λcg, (B.31c) λbh= 0.7405λcg, (B.31d) λcm= 0.7221λcg, (B.31e) returns an expression where λcgis the only unknown. Solving numerically gives λcg= 5.237 and we find that the end constraint factor for compression in x2is: n2=λcg π= 1.667. (B.32)