Mechanical Properties of Semi-Regular Lattices

(cid:1) The in-plane mechanical properties of 7 semi-regular lattices were derived analytically. (cid:1) We found a stretching-dominated topology with an elastic buckling strength 43% higher than a regular triangular lattice. (cid:1) One bending-dominated semi-regular tessellation is 85% stiffer and 11% stronger than a regular hexagonal lattice.


Introduction
Micro-architectured and lattice materials have a huge advantage over conventional fully-dense solids: their topology can be designed to achieve specific, and often unique, properties [1,2].For example, architectures have been created to reach the theoretical limit on stiffness [3][4][5], achieve a high fracture toughness [6][7][8], increase energy absorption capacities [9,10], have a negative Poisson's ratio [11,12], avoid localised deformation [13,14], or have unusual elastic properties such as a ratio of bulk to shear modulus as high as 10 3 [15,16].These lattice materials can be categorised as either spatial (3D) or planar (2D) topologies [17].Even though spatial lattices are potentially more efficient, planar topologies have remained attractive as they are easier to manufacture on a large scale.
Planar lattices can be classified into three categories: regular, semi-regular, and other tessellations [17].Regular lattices are made by tessellating a single regular polygon.There are only three regular tessellations [18]: hexagonal, square and triangular lattices, see Fig. 1.In contrast, semi-regular lattices are assembled by tessellating two or more regular polygons, with the same arrangement at each vertex.This arrangement generates eight semi-regular lattices [18], and these are shown in Fig. 2. Lastly, other lattices can be created using one or more polygons (regular or irregular) and/or by changing the nodal connectivity (defined as the number of bars meeting at each joint).
The mechanical behaviour of any lattice can be categorised in two groups: bending-or stretching-dominated [19].Stretchingdominated lattices are significantly stiffer and stronger than bending-dominated topologies; however, the latter offer superior energy absorption capacities.A key parameter to determine if a lattice is bending-or stretching-dominated is the nodal connectivity [20].Planar lattices with a nodal connectivity Z 6 3, such as the hexagonal lattice (Fig. 1a), are bending-dominated.Otherwise, topologies with Z P 6, like the triangular lattice (Fig. 1c), are stretching-dominated.Finally, lattices with Z ¼ 4 or 5 can be either bending-or stretching-dominated.For example, a square lattice (Fig. 1b) has a nodal connectivity Z ¼ 4 and it is stretchingdominated when loading is aligned with the struts, but bendingdominated otherwise.
Semi-regular tessellations, however, have received considerably less attention; the trihexagonal lattice (also referred to as kagome, see Fig. 2a) is the only topology to have been studied extensively.This architecture has impressive properties: it is stretchingdominated and as stiff and strong as the triangular lattice despite having a lower nodal connectivity [28,17].In addition, the tri-hexagonal lattice has a remarkably high fracture toughness [29] and a great potential for actuation [30,31].Reports on other semi-regular topologies are scarce: the stiffness and strength of the snub-square lattice were reported by Their and St-Pierre [32]; whereas the elastic and shear moduli of the snubtrihexagonal and rhombi-trihexagonal were obtained numerically by Elsayed and Pasini [33] and Pronk et al. [34], respectively.At the moment, it is impossible to evaluate the potential of most semi-regular tessellations since many of their mechanical properties remain unknown.Therefore, the aim of this study is to provide all missing properties for semi-regular lattices and compare their performances to those of regular topologies.
This article is structured as follows.Analytical equations for the elastic modulus, compressive strength, as well as shear modulus and strength, are derived in Section 2. These analytical expressions are validated using Finite Element (FE) simulations, and a description of the modelling approach is given in Section 3. A comparison between analytical and numerical results is presented in Section 4 along with a discussion on the potential of semi-regular lattices.

Analytical modelling
In this section, the modulus and strength of all semi-regular lattices are derived analytically for in-plane uniaxial compression and shear.Only the trihexagonal lattice (Fig. 2a) is excluded from this analysis since its properties are already available in the literature [28,23,17].Here, we assume that all lattices are made from an isotropic linear elastic, perfectly plastic solid, characterized by a Young's modulus E s and a yield strength r ys .All cell walls are considered to have a thickness t, a length ', and an out-of-plane depth b.The cell walls are considered to behave (i) as pin-jointed trusses in the case of stretching-dominated lattices or (ii) as Euler-Bernoulli beams for bending-dominated topologies.Two failure modes are considered when predicting the strength of a lattice: plastic collapse/yielding and elastic buckling.Bending-dominated topologies fail by plastic collapse when the maximum bending moment in a cell wall reaches the fully plastic moment: Otherwise, stretching-dominated lattices fail by yielding when the axial stress in a cell wall (in tension or compression) reaches the yield strength r ys of the parent material.In contrast, elastic buckling occurs when the axial compressive force in a cell wall reaches the Euler buckling load [35]: where I ¼ bt 3 =12 is the second moment of area and n is the end constraint factor.The operative failure mode is the one associated with the lowest load.Note that these assumptions are identical to those used in previous studies [21,28,23,17], which will enable a fair comparison between our results and the properties of regular lattices.Our analysis of bending-dominated lattices will rely heavily on the stiffness matrix of a single beam.Consider a beam of length ' subjected to a transverse displacement D, and rotations h i and h j at ends i and j, respectively, see Fig. 3.These rotations and displacement will give rise to a transverse force V ij , and bending moments M ij and M ji at ends i and j, respectively.These quantities are related by [36]: This equation can be used to model the deformation of lattices as follows.First, the bending moments and transverse force for each bar can be expressed using Eq. ( 3).Second, equilibrium conditions at each vertex, and the boundary conditions, are used to form a system of equations from which the deflections and rotations can be solved.This procedure is demonstrated next.

Truncated-hexagonal lattice
The truncated-hexagonal tessellation has two dodecagons and a triangle meeting at each vertex, see Fig. 4. It has a nodal connectivity Z ¼ 3 and therefore, its behaviour is bending-dominated [20].This topology has 6-fold rotational symmetry and consequently, its in-plane elastic properties are isotropic [37].The relative density of the truncated-hexagonal lattice is given by: 2.1.1.Compression Consider the truncated-hexagonal lattice subjected to a uniaxial compressive stress r in x 2 , see Fig. 4a.A representative unit cell is shown in Fig. 4a, where the nominal stress r is replaced by an equivalent force: In compression, the truncated-hexagonal lattice deforms primarily by bending bar bb.This strut will have a transverse deflection D and a rotation h at both ends, see Fig. 4a.The component of F perpendicular to bar bb corresponds to the transverse force V bb , which can also be expressed as a function of D and h using the stiffness matrix introduced in Eq. (3).This gives: In addition, equilibrium of moments at b requires that: Again, these moments were obtained using the stiffness matrix given in Eq. (3).Eqs. ( 6) and ( 7) can be used to solve for D and h, which gives: The component of D along x 2 is used to compute the compressive strain: The elastic modulus of the lattice is E ¼ r=e 2 , which returns: Otherwise, the component of D in x 1 gives a deformation: and therefore the Poisson's ratio is: Next, we turn our attention to the compressive strength of the truncated-hexagonal lattice.This topology is bending-dominated and the maximum bending moment in a cell wall is: Equating this to the fully plastic moment (Eq.( 1)) gives the plastic collapse strength of the lattice: Otherwise, bar aa is loaded in compression and may buckle elastically.The axial force in this bar is N aa ¼ F, and equating this to the Euler buckling load (Eq.( 2)) returns the elastic buckling strength of a truncated-hexagonal lattice: where the end constraint factor n ¼ 0:394 is derived analytically in Appendix A.1.The same procedure can be followed to derive the compressive strength in x 1 .This analysis is not detailed here for the sake of brevity, but it leads to ðr pl Þ 1 ¼ ðr pl Þ 2 .Note that elastic buckling does not occur for uniaxial compression in x 1 .

Shear
Consider the truncated-hexagonal lattice loaded in pure inplane shear as shown in Fig. 4b.A representative unit cell in Fig. 3. Deformation and reaction loads for a beam in bending.The end rotations h i and h j and the deflection D create the bending moments M ij and M ji , and a transverse force included in Fig. 4b, where the nominal shear stress s is replaced by an equivalent force: In shear, the truncated-hexagonal lattice deforms by bending bars aa and bb.Their deformation is characterised by four parameters: the transverse deflections D 1 and D 2 , and rotations h 1 and h 2 , see Fig. 4b.Using the stiffness matrix in Eq. ( 3), the transverse force in bar aa can be expressed as: Next, the nominal shear strain can be expressed as: and the shear modulus This result respects the relationship G ¼ E=ð2ð1 þ mÞÞ characteristic of isotropic materials (see Eq. ( 11) and ( 13) for E and m, respectively).
The truncated-hexagonal lattice fails by plastic collapse in shear.The maximum bending moment is located at the ends of bar aa and is: Setting M aa equal to the fully plastic moment (Eq.( 1)) returns the shear strength:

. Rhombi-trihexagonal lattice
The rhombi-trihexagonal lattice has a triangle, two squares, and a hexagon meeting at each vertex, see Fig. 5.The bars forming the squares are anticipated to bend when subjected to compression or shear and therefore, the lattice is expected to be bendingdominated despite its nodal connectivity Z ¼ 4.This pattern has a 6-fold rotational symmetry and consequently, it is in-plane elastically isotropic [37].Its relative density is given by:

Compression
Consider the rhombi-trihexagonal lattice subjected to an inplane compressive stress r in the x 2 direction, see Fig. 5a.The unit cell, given in Fig. 5a, includes an equivalent force F, which is related to the compressive stress r by: In compression, the lattice is expected to deform by bending bars ac and bd.Rotations at vertices b and b0 are equal (but in opposite directions) due to symmetry, and there are no rotations at a, see Fig. 5a.Using Eq. (3), the transverse force in bars ac and bd can be written as a function of the displacement field, which gives: Otherwise, the sum of moments at vertex b should be zero, and this gives: Using Eq. ( 32) and (33) to solve for D and h returns: The compressive strain is calculated with the component of D in the x 2 direction, which gives: and the elastic modulus E ¼ r=e 2 becomes: In addition, the strain in x 1 direction is: and therefore, the Poisson's ratio is: The rhombi-trihexagonal lattice is bending-dominated and the maximum bending moment is: Equating this to the fully plastic moment (Eq.( 1)) returns the plastic collapse strength of the lattice: This topology may also fail by elastic buckling of bar be.The axial compressive force in this bar is N be ¼ F, and equating this to the Euler buckling load (Eq.( 2)) gives us the elastic buckling strength: where the end constraint factor n ¼ 0:871 is obtained in Appendix A.2. Finally, note that when the rhombi-trihexagonal lattice is compressed in the x 1 direction, plastic collapse is the only failure mode and we find

Shear
The rhombi-trihexagonal lattice subjected to pure shear is shown in Fig. 5b along with its unit cell, where the nominal shear stress s has been replaced by an equivalent shear force F given by: The deformation of the lattice in shear is characterised by four variables: two deflections, D 1 and D 2 , and two rotations, h 1 and h 2 , see Fig. 5b.Using Eq. ( 3), the transverse force in bar be can be expressed as: whereas that in bar ac (or bd) is: Also, equilibrium of moments at vertex a returns: and at vertex b gives: The displacement field can be solved with Eqs.(44)-(47), and this yields: Next, we can write the shear strain: and the shear modulus G ¼ s=c becomes: The rhombi-trihexagonal lattice is in-plane isotropic and consequently, G ¼ E=ð2ð1 þ mÞÞ.
This lattice is bending-dominated in shear and the maximum bending moment in a cell wall is: Setting this equal to the fully plastic moment (Eq.( 1)), returns the plastic collapse strength: The truncated-square lattice, shown in Fig. 6, has a square and two octagons meeting at each vertex.It has a nodal connectivity Z ¼ 3 and consequently, its behaviour is bending-dominated [20].This tessellation has a 4-fold rotational symmetry, and therefore, its elastic properties are not isotropic [37].Its relative density is given by: The deformation of a truncated-square lattice in compression can be analysed with the unit cell shown in Fig. 6a, where the nominal compressive stress r is replaced by an equivalent force: This tessellation deforms by bending bar ab.This strut has a transverse deflection D, but rotations are prevented at both extremities due to symmetry.Therefore, the transverse force in beam ab is: from which it is straightforward to find the deflection: Next, the compressive strain is: and the elastic modulus becomes: where the subscript 2 is used to emphasise that elastic properties are not isotropic for this topology.Otherwise, the strain in x 1 is: and the Poisson's ratio is: The maximum bending moment in a truncated-square lattice is: ð64Þ and equating this to the fully plastic moment (Eq.( 1)) returns the plastic collapse strength: A truncated-square lattice can also fail by elastic buckling of bar aa.
The axial compressive force in this strut is N aa ¼ F, and setting this equal to the Euler buckling load (Eq.( 2)) yields the elastic buckling strength: where the end constraint factor n ¼ 0:734 is derived in Appendix A.3.Finally, note that all properties are the same for compression in x 1 because of the 4-fold rotational symmetry of the truncatedsquare lattice.

Shear
Consider the truncated-square lattice subjected to pure shear.The deformation of its unit cell is given in Fig. 6b, where the nominal shear stress s has been replaced by an equivalent force: In shear, the truncated-square lattice deforms by bending the vertical and horizontal bars, and the displacement field is characterised by their transverse deflection D and rotation h.The transverse force in bar aa is: and equilibrium of moments at vertex a gives: Using the last two equations to find D and h returns: Next, the shear strain can be expressed as: and the shear modulus G 12 ¼ s=c becomes: Otherwise, the maximum bending moment in a truncated-square lattice is at vertex b and is given by: Equating this to the fully plastic moment (Eq.( 1)) gives us the plastic collapse strength in shear:

Truncated-trihexagonal lattice
The truncated-trihexagonal lattice has a square, a hexagon and a dodecagon meeting at each vertex, see Fig. 7. Its behaviour is bending-dominated, since it has a nodal connectivity Z ¼ 3 [20].This topology has a 6-fold rotational symmetry; therefore, its inplane elastic properties are isotropic.Its relative density is given by: 2.4.1.Compression When a truncated-trihexagonal lattice is loaded in compression, the deformation of its unit cell is characterised by two displacements (D 1 ; D 2 ) and three rotations (h 1 ; h 2 ; h 3 ), see Fig. 7a.Here, the compressive stress r is replaced by an equivalent force: Bars ac and bd experience the same transverse force, which is given by: Next, equilibrium of forces, in the x 2 direction, at vertex b gives: Finally, equilibrium of moments at vertices a; b and e yields: respectively.Solving the displacement field using Eqs.( 78)-(82) returns: The vertical components of D 1 and D 2 are used to compute the compressive strain: and the elastic modulus becomes: Moreover, the strain in the x 1 direction is: and therefore the Poisson's ratio is:  Next, we turn our attention to the compressive strength of the lattice.The maximum bending moment is: and equating this to the fully plastic moment (Eq.( 1)) gives us the plastic collapse strength: A truncated-trihexagonal lattice can also fail by elastic buckling of bar ef.The axial compressive force in this bar is N ef ¼ F and setting equal to the Euler buckling load (Eq.( 2)) returns the elastic buckling strength: where the end constraint factor n ¼ 0:682 is detailed in Appendix A. 4. Following the same procedure, it is possible to show that ðr pl Þ 1 ¼ ðr pl Þ 2 ; however, the elastic buckling strength is different and given by: where the end constraint factor is n ¼ 0:793 for compression along x 1 , please refer to Appendix A.4 for more details.

Shear
The deformation of the truncated-trihexagonal lattice in shear is characterised by three rotations and four displacements, see the unit cell in Fig. 7b.Here, the shear stress s is replaced by an equivalent force: The transverse force in bar ef is simply: and that in bar ac and bd is given by: Also, the transverse forces in bars aa and ee have to respect: Moreover, equilibrium of moments at vertices a; b, and e return: respectively.Finally, compatibility of displacements in x 2 requires that: The displacement field is solved using the previous seven equations and this yields: With the above displacement field, we can compute the shear strain: and the shear modulus G ¼ s=c becomes: The truncated-trihexagonal lattice is in-plane isotropic and we find that G ¼ E=ð2ð1 þ mÞÞ.
The maximum bending moment in a truncated-trihexagonal lattice is: and equating this to the fully plastic moment (Eq.( 1)) returns the plastic collapse strength in shear, which is: The snub-square lattice has three triangles and two squares meeting at each vertex, see Fig. 8, and therefore, it has a nodal connectivity Z ¼ 5. Their and St-Pierre [32] have shown that its behaviour is stretching-dominated in compression, but it is bendingdominated in shear, as we will show below.This tessellation has a 4-fold rotational symmetry and consequently, its in-plane elastic properties are not isotropic [37].Its relative density is given by:

. Compression
The uniaxial compressive behaviour of the snub-square lattice was investigated previously by Their and St-Pierre [32].Their analytical work revealed that the snub-square lattice is stretchingdominated in compression with an elastic modulus: Using the internal loads provided in Their and St-Pierre [32], we find that the Poisson's ratio is: In compression, the snub-square lattice can fail by either elastic buckling or yielding, depending on its relative density.The yield strength of the snub-square lattice is [32]: ð118Þ whereas the elastic buckling strength is given by [32]: where the end constraint factor n ¼ 1:693 is derived analytically in Appendix A.5.Finally, note that

Shear
In shear, the snub-square lattice deforms by bending, and the deformation of its unit cell is characterised by a deflection D and a rotation h, see Fig. 8.The shear stress s is replaced by an equivalent force: Equilibrium of forces, along x 1 , at vertex b requires that: where N ab ¼ N ba0 is the axial load in bar ab.Combining these two equations to remove N ab yields: Otherwise, equilibrium of moments at vertex a gives: Using the last two equations to find h and D returns: Next, the shear strain can be expressed as: and the shear modulus becomes: The maximum bending moment in a snub-square lattice subjected to shear is: and equating this to the fully plastic moment (Eq.( 1)) returns the plastic collapse strength in shear:

Elongated-triangular lattice
The elongated-triangular lattice has three triangles and two squares meeting at each vertex, see Fig. 9; these are the same polygons as in a snub-square tessellation, but their arrangement is different.Therefore, the elongated-triangular lattice also has a nodal connectivity Z ¼ 5, but it is clear, upon inspection, that its behaviour is stretching-dominated when compressed in x 1 or x 2 , and bending-dominated in shear.Its elastic properties are not isotropic, and its relative density is: which is the same as a snub-square lattice, see Eq. (115).

Compression
Consider the elongated-triangular lattice loaded in compression along the x 2 direction, see Fig. 9a.A unit cell is shown in Fig. 9a, where the nominal compressive stress r is replaced by an equivalent force: For this loading scenario, all bars are carrying axial forces only.Using the method of sections, it is straightforward to find the axial load in each bar and this gives: where a negative sign indicates a tensile force.The shortening u of the unit cell is obtained by equating the work done by external forces to the internal energy.This gives: where the shortening of bar ij is Next, the nominal compressive strain in x 2 is: and the elastic modulus E 2 ¼ r=e 2 becomes: Otherwise, the nominal strain in x 1 is: which gives a Poisson's ratio: In compression, the elongated-triangular lattice fails by either elastic buckling or yielding depending on its relative density.For loading in x 2 , the highest compressive force is in bar aa and setting N aa ¼ r ys bt gives us the yield strength of the elongated-triangular lattice: Otherwise, equating N aa to the Euler buckling load (Eq.( 2)) returns the elastic buckling strength of the lattice: where the end constraint factor n ¼ 0:881 is derived in Appendix A.6.
The same procedure can be followed to derive the properties for compression in x 1 .In this case, the internal loads are: and the elastic modulus becomes: which is almost equal to E 2 .Otherwise, the yield strength is: ð143Þ whereas the elastic buckling strength is given by: where the end constraint factor n ¼ 1:515 is derived in Appendix A.6.

Shear
In shear, the elongated-triangular lattice deforms by bending bar aa, see Fig. 9b.The deformation of the unit cell is characterised by only two variables: the transverse deflection D and rotation h of beam aa.The shear stress s can be replaced by an equivalent force: and the transverse load in beam aa can be expressed as: Moreover, equilibrium of moments at vertex a gives: Using the last two equations to solve for h and D returns: Next, the shear strain is given by: and the shear modulus becomes: In shear, the maximum bending moment in an elongated-triangular lattice is: and equating this to the fully plastic moment (Eq.( 1)) gives us the plastic collapse strength of the lattice in shear:

Snub-trihexagonal lattice
A snub-trihexagonal lattice has four triangles and a hexagon meeting at each vertex, see Fig. 10.It has a nodal connectivity Z ¼ 5, and our analysis below will show that its behaviour is stretching-dominated in both compression and shear.Its elastic properties are isotropic since this pattern has a 6-fold rotational symmetry [37].Its relative density is given by:

Compression
In compression, the cell walls of the snub-trihexagonal lattice are carrying axial forces only.The unit cell, shown in Fig. 10a, includes nine independent bars and their axial loads are labelled N 1 ; N 2 ; . . .; N 9 .Equilibrium equations at vertices a; b and c give: Next, the macroscopic stress field is r 11 ¼ 0; r 12 ¼ 0, and r 22 ¼ r, and these give: respectively.The previous nine equations are used to solve for the axial forces, and this returns: where a negative sign tension.Next, the nominal compressive strain can be obtained by equating the work done by the external load to the internal strain energy, and this gives: Finally, the elastic modulus becomes: Using the principle of virtual work, we find that the strain along x 1 is: where D i ¼ N i '=ðbtE s Þ is the extension/shortening of bar i.With this result, the Poisson's ratio becomes: The snub-trihexagonal lattice fails by elastic buckling or yielding depending on the relative density.For compression in x 2 , bar 4 carries the highest compressive force.Setting N 4 ¼ r ys bt gives us the yield strength of the lattice: Otherwise, equating N 4 to the Euler buckling load (Eq.( 2)) returns the elastic buckling strength: where the end constraint factor n ¼ 1:541 is derived analytically in Appendix A.7.
The same procedure can be used for uniaxial compression in x 1 .For this loading scenario, the internal loads are: Of course, we find the same elastic modulus in x 1 since this topology is isotropic.Here, the highest compressive load is in bar 7, and setting N 7 ¼ r ys bt returns the yield strength: Finally, equating N 7 to the Euler buckling load (Eq.( 2)) yields the elastic buckling strength: where the end constraint factor n ¼ 1:661 is derived in Appendix A.7.

Shear
In shear, the cell walls of the snub-trihexagonal lattice also carry axial forces only.The unit cell includes nine independent bars and these are identified in Fig. 10b.The six equilibrium equations in (155) are also valid for shear.Otherwise, the macroscopic stress field is r 12 ¼ s; r 22 ¼ 0, and r 11 ¼ 0, and these give: respectively.With these three equations and those in (155), we can solve for the nine axial forces, which returns: where a negative sign denotes tension.Then, the shear strain is obtained by equating the work done by external forces to the strain energy, and this yields: Finally, the shear modulus becomes: The relationship G ¼ E=ð2ð1 þ mÞÞ is respected since the snubtrihexagonal lattice is in-plane isotropic.
In shear, bar 9 carries the highest (compressive) load.Setting N 9 ¼ r ys bt gives us the yield strength of the lattice: ð175Þ whereas equating N 9 to the Euler buckling load (Eq.( 2)) returns the elastic buckling strength: where the end constraint factor n ¼ 1:741 is derived analytically in Appendix A.7.

Finite Element modelling
Finite Element simulations were conducted to validate the analytical expressions derived in Section 2. All simulations were done using the implicit solver of the commercial software Abaqus 6.18.The parent material was modelled as an isotropic linear elastic, perfectly-plastic solid with a Young's modulus E s ¼ 200 GPa, a Poisson's ratio m ¼ 0:3 and a yield strength r ys ¼ 200 MPa.For each topology, the relative density was varied from 0.01 to 0.3 by changing the thickness t of the cell walls, while keeping their length fixed at ' ¼ 10 mm.The cell walls were discretised using shear-flexible Timoshenko beam elements (B21 in Abaqus notation), and a mesh convergence study showed that ten elements per bar (corresponding to a mesh size of 1 mm) offers accurate predictions, see Supplementary material.All simulations included a small geometric imperfection, which had the shape of the first eigenmode and an amplitude of 0:05t.This small imperfection was necessary to trigger buckling, but had a negligible effect on the stiffness of the lattice.
Each topology was modelled using a periodic unit cell (see Fig. S1 in Supplementary material).Periodicity was enforced with the following equations [26]: where Du i and Dh are the differences in displacement and rotation, respectively, between two corresponding points on either sides of the unit cell; Dx j is the vector connecting these two corresponding points; and ij is the macroscopic strain tensor.The compressive response along x 2 was simulated by prescribing 22 and letting Abaqus calculate r 22 provided that r 11 ¼ r 12 ¼ 0. Similarly, the compressive response in x 1 was obtained by setting 11 and calculating r 11 with r 22 ¼ 0. Otherwise, the response in shear was obtained by imposing 12 and computing r 12 while ensuring that r 11 ¼ r 22 ¼ 0.More details on how these periodic boundary conditions were implemented in Abaqus are given in Supplementary material.

Comparison between analytical and Finite Element results
Analytical results are compared to FE simulations in Fig. 11, where four properties (E 2 ; r 2 ; G 12 ; and s 12 ) are plotted as a function of relative density.In each plot, results are shown for all eight semi-regular lattices (including, for completeness, the trihexagonal tessellation even though its properties were derived analytically by Wang and McDowell [25], Fan et al. [23]).For each topology, there is an excellent agreement between analytical and FE predictions, and this holds true for the four properties plotted in Fig. 11.Note that there is also an excellent agreement between analytical and FE results for E 1 and r 1 , but these results are not shown here for the sake of brevity and because E 1 ¼ E 2 and r 1 ¼ r 2 for many topologies.
Our analytical work predicts that the failure mode of stretchingdominated lattices will switch from yielding to elastic buckling as the relative density decreases.This transition is clearly visible in Fig. 11b,d and for both failure modes, there is an excellent agreement between analytical and FE predictions.Based on our analytical modelling, the compressive strength r 2 of bending-dominated lattices should display a similar transition (from plastic collapse to elastic buckling) but this switch is not visible in Fig. 11b because it occurs at q < 0:01 for this choice of material properties where r ys =E s ¼ 0:001.The relative density at which the failure mode changes to elastic buckling is sensitive to r ys =E s [21]; therefore additional FE simulations were performed with r ys =E s ¼ 0:01 to capture this transition for bending-dominated lattices.These results are provided in Appendix B and show a very good agreement between analytical and FE predictions.
Note that the FE simulations presented here are done using Timoshenko beam elements, which account for axial, bending and shear deformations.In contrast, our analytical model neglects (i) axial and shear deformations inside the cell walls of bendingdominated lattices and (ii) bending and shear deformations for the bars of stretching-dominated topologies (see Section 2).These assumptions are justified based on the excellent agreement between analytical and FE results, see Fig. 11 and Appendix B. Therefore, we conclude that the analytical expressions derived in Section 2 are validated for relative densities ranging from 0.01 to 0.3.

Comparison between regular and semi-regular lattices
The mechanical properties of all eight semi-regular lattices are compared to those of regular tessellations in Table 1.Topologies are divided in three groups depending on their behaviour: (i) bending-dominated lattices; (ii) topologies that are stretchingdominated in compression, but bending-dominated in shear; and (iii) stretching-dominated lattices.Below, we discuss each group in turn.
Four of the eight semi-regular lattices have a bendingdominated behaviour, and their properties are compared to those of a regular hexagonal tessellation in Table 1.Note that all bending-dominated lattices are in-plane elastically isotropic, except for the truncated-square topology.Amongst all bendingdominated lattices, the truncated-hexagonal tessellation clearly offers the best performances: it is 85% stiffer than a hexagonal lattice and has a slightly higher plastic collapse strength.This is true for both compression and shear.The truncated-hexagonal has a lower elastic buckling strength than a hexagonal lattice, but this failure mode occurs only at low values of relative densities.Otherwise, the truncated-square and truncated-trihexagonal lattices have performances that are similar to those of the hexagonal topology.Finally, the rhombi-trihexagonal tessellation has by far the lowest properties: it is three times more compliant and about two times weaker than the hexagonal lattice.
Two semi-regular topologies, the elongated-triangular and the snub-square lattices, are stretching-dominated in compression along x 1 or x 2 , but bending-dominated in shear.The elongatedtriangular topology has similar properties to a regular square lattice: they both have a high elastic modulus and compressive strength in x 1 or x 2 , but they are very compliant in shear.In contrast, the snub-square lattice has a shear modulus about five times higher than that of the elongated-triangular and square topologies, but its elastic modulus and compressive strength are lower.
Finally, only two semi-regular tessellations have a stretchingdominated behaviour: trihexagonal and snub-trihexagonal lattices.Their properties are compared to those of a regular triangular lattice in Table 1.Note that these three lattices are inplane elastically isotropic.The snub-trihexagonal lattice is 30% more compliant and has a lower yield strength than the triangular and trihexagonal topologies (which have identical values for E 2 ; G 12 ; ðr pl Þ 1 and ðr pl Þ 2 ).The three lattices differ in their resistance to elastic buckling: the snub-trihexagonal lattice is 43% stronger than a triangular lattice, but weaker than a trihexagonal topology.

Conclusion
The in-plane mechanical properties of seven semi-regular lattices were presented in this study.For each topology, analytical expressions were derived for the elastic modulus and strength under uniaxial compression and shear.These analytical equations were then verified with finite element simulations.
The analysis allowed us to classify the behaviour of these seven tessellations: four were found to be bending-dominated; one was stretching-dominated; and two were stretching-dominated in compression but bending-dominated in shear.The properties of these seven semi-regular tessellations were also compared to those of regular lattices, and this revealed the potential of the truncated-hexagonal lattice.The truncated-hexagonal tessellation is elastically isotropic and it is 85% stiffer and 11% stronger than its hexagonal counterpart.With such combination of properties, the truncated-hexagonal lattice is a promising topology for automotive, rail and aerospace applications where a high stiffness and strength are required.Finally, work is underway to manufacture these semi-regular lattices and measure the properties derived in this study.

Data availability
The raw/processed data required to reproduce these findings cannot be shared at this time as the data also forms part of an ongoing study.

Declaration of Competing Interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

Table 1
The mechanical properties of all eight semi-regular lattices are compared to those of regular tessellations (hexagonal, square and triangular).A reference frame is included in Fig. 1 for regular lattices.Expressions for regular topologies and for the trihexagonal lattice are taken from Gibson and Ashby [21], Wang and McDowell [23], Fan et al. [25].

A.1. Truncated-hexagonal lattice
When compressed in x 2 , the truncated-hexagonal lattice is anticipated to buckle in the swaying mode shown Fig. 12. Equilibrium of moments at vertices a and b, and the transverse forces in bars aa and ab give the following set of equations: where each moment and transverse force can be expressed as a function of h 1 ; h 2 ; D 1 and D 2 using Eq. ( 178).This gives: where s ij ; s Ã ij ; s ij and c ij are all functions of k ij as defined in Eqs. ( 179)-(186).Substituting Eq. ( 189) in (188) returns a linear system of equations: A½h 1 ; h 2 ; D 1 ; D 2 T ¼ 0, where the determinant of A should be zero for a non-trivial solution to exists.Setting detðAÞ ¼ 0 returns a lengthy expression that includes k aa ; k ab ; k bb , and k bc .Each k ij is proportional to the axial load in bar ij (see Eq. ( 183)); therefore, using structural analysis, we find that: Above, each k ij is expressed as a function of k aa , since bar aa is the most loaded strut and the one expected to buckle.Substituting Eq. (190) in detðAÞ ¼ 0 yields an expression where k aa is the only unknown.Solving this numerically gives k aa ¼ 1:240, and consequently, the end constrain factor for a truncated-hexagonal lattice compressed in x 2 is:

. Rhombi-trihexagonal lattice
A rhombi-trihexagonal lattice is expected to buckle elastically in the swaying mode shown Fig. 13 when compressed in x 2 .The buckling shape is characterised by rotations h 1 and h 2 , and a lateral displacement D. Equilibrium of moments at vertices a and b, as well as the transverse load in bar ae return the following set of equations: FE eigenvalue buckling predictions, that the anticipated buckling patterns corresponds to the first eigenmode for each topology.
where the moments and transverse load are obtained using Eq. ( 178) and expressed as: The next steps are the same as those detailed in Appendix A. 1. First,substitute (193) in (192) to form a linear system of equations and set the determinant of the coefficient matrix equal zero.Second, express the resulting equation as a function of k ae only (since this is the bar expected to buckle) using the following relations: Solving the resultant expression returns k ae ¼ 2:736.Therefore, the end constraint factor for a rhombi-trihexagonal lattice compressed in x 2 is: A.3.Truncated-square lattice A truncated-square tessellation is anticipated to buckle in a periodic swaying mode as shown in Fig. 14 when compressed in x 2 .The analysis below is for compression in x 2 , but, due to symmetry, the result is exactly the same when the lattice in compressed in x 1 .The buckling shape in Fig. 14 is characterised by two rotations, h 1 and h 2 , and two transverse displacements, D 1 and D 2 .Equilibrium of moments at vertices a and b, as well as the transverse loads in bars aa and bb return the following set of equations: where M ij and V ij are obtained using Eq. ( 178) and given by: Note that bar bb does not carry any axial load and therefore, s bb ¼ 4 and c bb ¼ 0:5.Following the same procedure used for the two previous topologies, and the fact that k 2 ab ¼ 0:353 k 2 aa , we find that k aa ¼ 2:307.Thus, the end constraint factor for a truncated-square lattice is:

. Truncated-trihexagonal lattice
A truncated-trihexagonal lattice may fail by elastic buckling when compressed in either x 1 or x 2 .The periodic buckling shapes, for both loading directions, are given in Fig. 15.First, consider compression in x 1 : the buckling shape is a swaying mode characterised by h 1 ; h 2 ; h 3 ; D 1 and D 2 .Equilibrium of moments at vertices a; c, and d, and transverse loads of bars aa and dd give the following set of equations: where  M Note that bar ab, does not carry any axial load and therefore, s ab ¼ 4 and c ab ¼ 0:5.Substituting Eq. ( 200) in (199), then setting the determinant of the coefficient matrix to zero, and making use of the following relations: returns an expression where k dd is the only unknown (bar dd is the bar expected to buckle for compression in x 1 ).Solving this expression numerically returns k dd ¼ 2:491; therefore, the end constrain factor for a truncated-trihexagonal lattice compressed in x 1 is: Next, consider a truncated-trihexagonal lattice compressed in x 2 , see Fig. 15b.We have: where the moments and transverse forces are: For compression in x 2 , bar ab is expected to buckle and the proportionality between axial forces implies that: Following the same procedure as in x 1 , we find that k ab ¼ 2:143 and the end constrain factor for compression in x 2 is: A.5. Snub-square lattice A snub-square lattice may fail by elastic buckling when compressed in x 1 or x 2 .The analysis is the same for both loading directions due to symmetry and therefore, we will consider only compression in x 2 here.The periodic buckling mode for this loading scenario is shown in Fig. 16.Equilibrium of moments at vertex a requires that: where the moments are: Note that the above results are based on the internal loads derived by Their and St-Pierre [32].They showed that N ab ¼ 0, whereas bars ad and ac are both in compression with N ad ¼ 0:577N ac .Substituting these moments in Eq. (207) gives: and, with the definitions in Eq. ( 181) and (182), this becomes: This equation can be expressed as a function of k ac only since k 2 ad ¼ 0:577k 2 ac .Doing that and solving numerically returns k ac ¼ 5:319 and consequently, the end constraint factor for a snub-square lattice is: A.6.Elongated-triangular lattice An elongated-triangular lattice is anticipated to buckle in the pattern shown Fig. 17a when compressed in x 1 .Equilibrium of moments at vertex a requires that: where The above expressions are based on the fact that bar ab is in compression whereas bars ac; ad and ae do not carry any axial load for compression in x 1 , see Eq. (141).Substituting Eq. ( 213) in (212) gives: Solving this expression numerically returns k ab ¼ 4:761 and consequently, the end constraint factor for compression in x 1 is: Otherwise, an elongated-triangular lattice is expected to buckle in the swaying pattern shown in Fig. 17b when compressed in x 2 .
The vertical bar ae has a transverse displacement D, and all vertices have the same rotation h.Again, the sum of moments should be zero at vertex a and therefore, Eq. ( 212) remains valid, but for this loading direction the moments are: where the subscript 1 appear in M ab because this bar is tension.In addition, M ac ¼ M ad since they have the same rotations and carry the same axial compressive load, see Eq. ( 133).Substituting Eq. ( 216) in (212) yields: and the transverse force in bar ae gives: The last two expressions form a system of linear equations and setting the determinant of the coefficient matrix equal to zero, along with the fact that the proportionality between internal axial loads (see Eq. ( 133)) implies that: The snub-trihexagonal lattice is stretching-dominated and may fail by elastic buckling when loaded in compression or shear.First, consider the lattice under uniaxial compression in the x 1 direction.The periodic buckling shape for this loading scenario is shown in Fig. 18a, and equilibrium of moments at vertices b; c, and d return: respectively.Using Eq. ( 178) to express each moment as a function of the rotations, the above three equations become:  s bc ðh 3 À c bc h 2 Þ þ s ab ðh 3 À c ab h 1 Þþ respectively.Note that s cd ¼ s bj ¼ 4 and c cd ¼ c bj ¼ 0:5 since these bars are not carrying any axial load, see Eq. (166).The three expressions above form a linear system of equations Ah ¼ 0, where the determinant of A should be zero to ensure that a non-trivial solution exists.Setting detðAÞ ¼ 0 returns a lengthy expression with coefficients s mn and c mn , which can be expressed as a function of k mn using Eqs.(181)-(185).Next, we use the internal loads given in Eq. (166) to express each term as a function of k ch (this is the bar expected to buckle for compression in x 1 ) and this yields: This returns a lengthy expression where k ch is the only unknown, and solving numerically yields k ch ¼ 5:219.Therefore, the end constraint factor n for compression in x 1 is: Next, consider the snub-trihexagonal lattice compressed in the x 2 direction.The anticipated buckling pattern for this case is shown in Fig. 18b.The three expressions in Eq. ( 221) are still valid, but the moments are different for this loading scenario, which gives: Otherwise, the internal loads given in Eq. ( 159) implies that: Following the same procedure detailed above (for compression in x 1 ), we obtain a constitutive equation as a function of k df only, since this is the most loaded bar and the one expected to buckle.Solving this expression returns k df ¼ 4:841, and therefore, the end constraint factor for compression in x 2 is: Finally, a snub-trihexagonal lattice can also buckle elastically under shear according to the pattern shown in Fig. 18c.For this buckling mode, the three expressions in Eq. (221) become: In addition, the internal loads in Eq. (172) implies that: Following the same procedure employed for compression, we obtain a constitutive equation as a function of k bj and find k bj ¼ 5:469.Consequently, the end constraint factor in shear is:

Fig. 11 .
Fig. 11.Comparison between analytical (lines) and FE (symbols) results for the normalised mechanical properties of semi-regular lattices: (a) elastic modulus E2, (b) compressive strength r2, (c) shear modulus G12, and (d) shear strength s12, all plotted as a function of relative density q.The properties of the parent material are Es ¼ 200 GPa and rys ¼ 200 MPa.
long expression where k ae is the only unknown.Solving this expression numerically returns k ae ¼ 2:767 and consequently, the end constraint factor n for compression in x 2 is: