# Equivalent Elastic Modulus of Asymmetrical Honeycomb

Authors: Dai-Heng Chen; Kenichi Masuda
Journal: ISRN Mechanical Engineering
 (2011)

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## Abstract

The equivalent elastic moduli of asymmetrical hexagonal honeycomb are studied by using a theoretical approach. The deformation of honeycomb consists of two types of deformations. The first is deformation inside the unit, which is caused by bending, stretching, and shearing of cell walls and rigid rotation of the unit; the second is relative displacement between units. The equivalent elastic modulus related to a direction parallel to one cell wall of the honeycomb is determined from the relative deformation between units. In addition, a method for calculating other elastic moduli by coordinate transformation is described, and the elastic moduli for various shapes of hexagon, which are obtained by systematically altering the regular hexagon, are investigated. It is found that the maximum complianceCyy|max and the minimum compliance Cyy|min of elastic modulus Cyy in one rotation of the (x,y) coordinate system vary as the shape of the hexagon is changed. However, Cyy|max takes a minimum and Cyy|min takes a maximum when the honeycomb cell is a regular hexagon, for which the equivalent elastic moduli are unrelated to the selected coordinate system, and are constant with C11=C22.


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## Body

## 1. Introduction

To date, honeycomb materials consisting of regular hexagonal cells or symmetrical hexagonal cells [1–4] have been the subject of extensive research. In the present study, a general method is proposed for finding the equivalent elastic moduli for the two-dimensional (2D) problem of honeycomb consisting of an array of hexagonal cells, including asymmetrical hexagonal cells. Moreover, the equivalent elastic moduli for several hexagonal geometries are found using the proposed method, and a systematic investigation is carried out into the effects of changes in geometry of hexagonal cells on the equivalent elastic moduli of honeycomb.Research into the equivalent elastic moduli of asymmetrical honeycombs has been carried out by Overaker et al. [5], who proposed a method for finding the equivalent elastic moduli of an asymmetrical honeycomb by fitting an equivalent strain field to satisfy the displacements of each end of cell wall in one unit. In the present analysis, the method of Overaker et al. [5] is used although we simultaneously attempt to find the strain field via a different approach. Specifically, by treating the deformation of the honeycomb as the sum of the deformation of each cell wall in one unit and the relative displacement between each unit, the equivalent elastic modulus of the honeycomb can be found from the relative displacement between units. The relative displacement between units is determined by the condition of the junctions between the cell wall ends of each adjacent unit after the deformation.As shown in Figure1, the analyzed model in the present study is honeycomb consisting of hexagonal cells, and the honeycomb core height is denoted by h. In order to form a honeycomb by periodically arraying hexagonal cells, two opposing edges of the hexagon must have the same length and be parallel. Here, the length and thickness are l1, l2, and l3, and t1, t2, and t3, respectively, as shown in the figure; the internal angles formed by the hexagon edges are γ12, γ23, and γ31(1)γ12+γ23+γ31=2π.Geometry of honeycomb: (a) honeycomb plate and (b) hexagonal cell.
(a)(b)The cell wall material is homogeneous and isotropic with an elastic modulus ofEs and Poisson’s ratio of υs. The aim of this analysis is to find the equivalent elastic modulus for the honeycomb plane problem. Specifically, we want to find the equivalent elastic modulus Cij (i,j=1,2,3), which is applicable to the relation of stress and strain in the x- and y-coordinate plane shown in Figure 1(2)(εxεyεxy)=[C11C12C13C21C22C23C31C32C33](σxσyτxy).The shear strainεxy used here is defined as a tensor, the engineering definition of which is γxy with εxy=γxy/2.


## 2. Analysis of Elastic ModuliC12, C22, and C32 for the (x,y) Coordinates with y-Axis Parallel to Edge l1

Initially, they-axis is parallel to cell wall 1 in the case of θ1=0∘ in Figure 1, and only the y-direction stress σy is considered to act. As shown in Figure 2, the angles θ2 and θ3 are taken as those between edges l2 and l1, and edges l3 and l1, respectively
(3)θ2=π-γ12,θ3=π-γ31.Figure 2
Geometry of hexagon with an edge parallel toy-axis (θ1=0).


### 2.1. Force, Moment, and Displacement Acting on Each Cell Wall

Due to the stressσy acting in the y-direction, the force Ti (i=1~3) and moment Mi (i=1~3) act on each cell wall, as shown in Figure 3. From the equilibrium of forces, we can obtain (4)T1+T2=σyh(l2sin⁡θ2+2l3sin⁡θ3),T1=σyh(l2sin⁡θ2+l3sin⁡θ3),T1=T2+T3,
from which, the forces T1, T2, and T3 are given by the following equations:
(5)T1=σyh(l3sin⁡θ3+l2sin⁡θ2),T2=σyhl3sin⁡θ3,T3=σyhl2sin⁡θ2.
The moments due to Ti are given as follows:
(6)M1=0,M2=T212l2sin⁡θ2=12σyhl2sin⁡θ2l3sin⁡θ3,M3=T312l3sin⁡θ3=12σyhl2sin⁡θ2l3sin⁡θ3.
Here, M2=M3.Figure 3
Mechanics of cell walls subjected to stressσy in the y-direction.Thex-direction displacement δix and y-direction displacement δiy are found for each cell wall of the unit shown in Figure 4. The bold lines in Figure 4 denote the cell walls, the thin lines denote the boundary of one unit; by arrangement of these units the honeycomb is formed. The displacements of the cell walls are caused by bending deformation, shear deformation and tensile deformation of the cell walls, generated by each force and moment. Taking the junction of the three walls as the origin, the displacements in the x-direction of the ends of cell walls 1, 2, and 3 are given by the following equations:
(7)δ1x=0,δ2x=T2sin⁡θ2cos⁡⁡θ2Esh(l2t2)3+2(1+υs)T2l2sin⁡θ2cos⁡⁡θ2kEsht2-l2T2cos⁡⁡θ2sin⁡θ2Esht2,δ3x=-T3sin⁡θ3cos⁡⁡θ3Esh(l3t3)3-2(1+υs)T3l3sin⁡θ3cos⁡⁡θ3kEsht3+l3T3cos⁡⁡θ3sin⁡θ3Esht3.
Similarly, the displacements in the y-direction of each cell wall are given as follows:
(8)δ1y=-l1T1Esht1,δ2y=T2sin2θ2Esh(l2t2)3+2(1+υs)T2l2sin2θ2kEsht2+l2T2co⁡s2θ2Esht2,δ3y=T3sin2θ3Esh(l3t3)3+2(1+υs)T3l3sin2θ3kEsht3+l3T3co⁡s2θ3Esht3.
Here, the sign of the displacement follows the coordinates shown in Figure 4, and k is a correction coefficient related to shear deformation, which is taken as k=1 in this work (previous research has shown that results for k=1 agree well with those of numerical analysis by the finite element method [4]).Figure 4
Displacements of ends of cell walls.


### 2.2. Analysis of Equivalent Elastic Moduli

Overaker et al. [5] proposed an elegant method for fixing the equivalent strain field, which satisfies the displacements of each cell wall found in the previous section. Figure 4 shows the displacements of each wall end in a unit δix and δiy, as well as the coordinates of each wall end (xi,yi)  (i=1,2,3). The displacements in the x- and y-directions of each wall end δix and δiy, which are found in (7) and (8), can be seen as those due to the rigid body displacements u0 and v0, the rigid body rotation ωxy, and the uniform strain field in the unit εx, εy, and εxy and are then described by the following equations:(9)δix=xiεx+yiεxy+yiωxy+u0,δiy=yiεy+xiεxy-xiωxy+v0.(i=1,2,3)
Six unknowns, namely, εx, εy, εxy, ωxy, u0, and v0, are determined by solving these equations. By using these strain fields obtained from (9), the equivalent elastic moduli C12, C22, and C32 can be found from the following equation. 								(10)C12=εxσy,C22=εyσy,C32=εxyσy.The strain field produced in the honeycomb can also be determined from the relative displacements between units. In fact, the deformation of the whole honeycomb is performed by the relative displacements between each unit. Here, we consider a part of honeycomb consisting of three units, as shown in Figure5, in which the three units are denoted as units 1, 2, and 3, counterclockwise from the lower left, and the cell wall joints of each unit are O1, O2, and O3. Denote the relative displacements of O2 and O3 with respect to O1 by U21 and V21, and U31 and V31, respectively, as shown in Figure 5(b). Thus, the following equations can be obtained from the relation between the strain field produced in the honeycomb and the displacements of O2 and O3 with respect to O1:
(11)(U21V21U31V31)=[L2x0-L2y-L2y0-L2yL2x-L2xL3x0L3yL3y0L3yL3x-L3x](εxεyεxyωxy),
where L2x and L2y are the distances in the x- and y-directions between O1 and O2, while L3x and L3y are the distances between O1 and O3(12)L2x=l2sin⁡θ2+l3sin⁡θ3,L2y=l2cos⁡⁡θ2-l3cos⁡⁡θ3,L3x=l3sin⁡θ3,L3y=l1+l3cos⁡⁡θ3.
From (11), the strain field can be obtained as a function of the relative displacements between units as follows:(13)εx=L3yU21+L2yU31(L2yL3x+L2xL3y),εy=-L3xV21+L2xV31(L2yL3x+L2xL3y),εxy=-L3xU21+L3yV21+L2xU31+L2yV312(L2yL3x+L2xL3y),ωxy=-L3xU21-L3yV21+L2xU31-L2yV312(L2yL3x+L2xL3y).Condition of junctions between wall ends: (a) a part of honeycomb consisting of three units and (b) displacement of cell wall endsA, B, and C.
(a)(b)The relative displacements between units are determined by the condition of the junctions between the wall ends of each adjacent unit. In order to consider the connection between each cell wall after deformation, Figure5(b) also shows the displacements of points A, B, and C. Point A of unit 1 is on cell wall 3, and the displacement of points A with respect to point O1 in the respective x- and y-direction, UA1 and VA1, are equal to the cell wall deformation itself(14)UA1=δ3x,VA1=δ3y.
Point B of unit 2 is on cell wall 2 and point C of unit 3 is on cell wall 1. Therefore, the displacements of points B and C with respect to point O1 in the respective x- and y-direction, Ui1 and Vi1 (i=B,C), are calculated by adding the relative displacements between the units to the displacements due to the cell wall deformation itself(15)UB1=δ2x+U21,VB1=δ2y+V21,UC1=δ1x+U31,VC1=δ1y+V31.
Since points A, B, and C are the same point prior to deformation, as shown in Figure 5(a), the displacement of points A, B, and C after deformation must be the same and the condition that UA1=UB1=UC1 and VA1=VB1=VC1 holds true. From this condition, the relative displacements U21 and V21, and U31 and V31 of O2 and O3 with respect to O1 are given as follows:(16)U21=δ3x-δ2x,V21=δ3y-δ2y,U31=δ3x-δ1x,V31=δ3y-δ1y.By substituting (12) and (16) into (13), the strain field can be obtained, and then, each equivalent elastic modulus C12, C22, and C32 can be determined from (10).


### 2.3. Calculation of Elastic Modulus Matrix

In the previous section, we presented a method for finding the three equivalent elastic moduli for the directions parallel to a cell wall constituting the hexagon cell; however, these are only three of the nine components of the elastic modulus described in (2). To express the elastic characteristics of a hexagonal honeycomb, it is necessary to know all nine components. In this section, using the three equivalent elastic moduli relating to the directions parallel to a cell wall, the nine components of the honeycomb equivalent elastic modulus C11, C12~C33 are derived.Since the approach described above allows the three equivalent elastic moduli for the direction parallel to any cell wall to be found, the three elastic moduli can be found for each direction of cell walls 1, 2, and 3, respectively. Specifically, as shown in Figure6, we take the (α′,β′) coordinates based on cell wall 1, the (α′′,β′′) coordinates based on cell wall 2 and the (α′′′,β′′′) coordinates based on cell wall 3, in which the β-axis is set to be parallel to the cell wall. Thus, C12′, C22′, and C32′ in the (α′,β′) coordinates, C12′′, C22′′, and C32′′ in the (α′′,β′′) coordinates and C12′′′, C22′′′, and C32′′′ in the (α′′′,β′′′) coordinates can be found for each coordinate system (the prime superscripts of the coordinate system correspond with those of the elastic moduli). However, the nine components of the elastic modulus C11, C12~C33 to be found are attached to the (x,y) coordinates of Figure 6. Thus, we transform coordinates from the (x,y) coordinate system to the (α,β) coordinate system. Here, we suppose an angle θ between the (x,y) coordinate system and the (α,β) coordinate system. In the (α′,β′) coordinate system, θ is θ′=θ1, for the (α′′,β′′) coordinate system, it is θ′′=θ1+(π-γ12), and for the (α′′′,β′′′) coordinate system, it is θ′′′=θ1+(π-γ12)+(π-γ23). For example, by transforming the stress and strain in the (x,y) coordinate system to the stress and strain in the (α′,β′) coordinate system, the following equation can be obtained from (2):
(17)(εα′εβ′εα′β′)=[T]-1[C11C12C13C21C22C23C31C32C33][T](σα′σβ′τα′β′).
The coordinate transformation matrix [T] is given below:
(18)[T]=[cos⁡2θ1sin2θ12sin⁡θ1cos⁡⁡θ1sin2θ1cos⁡2θ1-2sin⁡θ1cos⁡⁡θ1-sin⁡θ1cos⁡⁡θ1sin⁡θ1cos⁡⁡θ1cos⁡2θ1-sin2θ1].
However, the stress-strain equations in the (α′,β′) coordinate system are expressed by the following equation:(19)(εα′ββ′εα′β′)=[C11′C12′C13′C21′C22′C23′C31′C32′C33′](σα′σβ′τα′β′).
Since both (17) and (19) are the same, the following equation is obtained:(20)[C11′C12′C13′C21′C22′C23′C31′C32′C33′]=[T]-1[C11C12C13C21C22C23C31C32C33][T].
As stated above, C12′, C22′, and C32′ are known and from (20), they can be expressed as functions of the components C11~C33, which are to be found(21)C12′=C12cos⁡4θ1-(-C13+2C32)cos⁡3θ1sin⁡θ1+(C11+C22-2C33)cos⁡2θ1sin2θ1+(C23-2C31)cos⁡θ1sin3θ1+C21sin4θ1,C22′=C22cos⁡4θ1+(C23+2C32)cos⁡3θ1sin⁡θ1+(C12+C21+2C33)cos⁡2θ1sin2θ1+(C13+2C31)cos⁡⁡θ1sin3θ1+C11sin4θ1,C32′=C32cos⁡4θ1+(C12-C22+C33)cos⁡3θ1sin⁡θ1+(-C23+C31-C32+C13)cos⁡2θ1sin2θ1+(C11-C21-C33)cos⁡θ1sin3θ1-C31sin4θ1.
Similarly, by transforming the (x,y) coordinate system to the (α′′,β′′) and (α′′′,β′′′) coordinate systems, C12′′, C22′′, and C32′′ and C12′′′, C22′′′, and C32′′′ can be expressed as functions of C11~C33. Therefore, by solving these nine simultaneous equations, the nine components, C11~C33, can be determinedFigure 6
(α,β) coordinates based on each cell wall.By using this method, the honeycomb equivalent elastic components are found. For example, for a cell thickness oft1=t2=t3=0.05l3, Poisson’s ratio of υs=0.3, the honeycomb equivalent elastic moduli for a hexagon with parameters of l1/l3=l2/l3=1, γ12=140°, γ23=120°, and γ31=100°can be determined(22)[C]=[4.076-2.4581.505-2.4582.6143.3260.7531.6638.568]×103Es.
Moreover, for a hexagon with parameters of l1/l3=2.45, l2/l3=0.6, γ12=153.7°, γ23=115°, and γ31=91.3°, the following honeycomb equivalent elastic moduli are calculated:
(23)[C]=[11.01-0.510420.83-0.51040.27501.70910.420.85435.72]×103Es.
It can be seen from these results that the symmetry of the elastic moduli holds(24)C21=C12,2C31=C13,2C32=C23.


## 2.1. Force, Moment, and Displacement Acting on Each Cell Wall

Due to the stressσy acting in the y-direction, the force Ti (i=1~3) and moment Mi (i=1~3) act on each cell wall, as shown in Figure 3. From the equilibrium of forces, we can obtain (4)T1+T2=σyh(l2sin⁡θ2+2l3sin⁡θ3),T1=σyh(l2sin⁡θ2+l3sin⁡θ3),T1=T2+T3,
from which, the forces T1, T2, and T3 are given by the following equations:
(5)T1=σyh(l3sin⁡θ3+l2sin⁡θ2),T2=σyhl3sin⁡θ3,T3=σyhl2sin⁡θ2.
The moments due to Ti are given as follows:
(6)M1=0,M2=T212l2sin⁡θ2=12σyhl2sin⁡θ2l3sin⁡θ3,M3=T312l3sin⁡θ3=12σyhl2sin⁡θ2l3sin⁡θ3.
Here, M2=M3.Figure 3
Mechanics of cell walls subjected to stressσy in the y-direction.Thex-direction displacement δix and y-direction displacement δiy are found for each cell wall of the unit shown in Figure 4. The bold lines in Figure 4 denote the cell walls, the thin lines denote the boundary of one unit; by arrangement of these units the honeycomb is formed. The displacements of the cell walls are caused by bending deformation, shear deformation and tensile deformation of the cell walls, generated by each force and moment. Taking the junction of the three walls as the origin, the displacements in the x-direction of the ends of cell walls 1, 2, and 3 are given by the following equations:
(7)δ1x=0,δ2x=T2sin⁡θ2cos⁡⁡θ2Esh(l2t2)3+2(1+υs)T2l2sin⁡θ2cos⁡⁡θ2kEsht2-l2T2cos⁡⁡θ2sin⁡θ2Esht2,δ3x=-T3sin⁡θ3cos⁡⁡θ3Esh(l3t3)3-2(1+υs)T3l3sin⁡θ3cos⁡⁡θ3kEsht3+l3T3cos⁡⁡θ3sin⁡θ3Esht3.
Similarly, the displacements in the y-direction of each cell wall are given as follows:
(8)δ1y=-l1T1Esht1,δ2y=T2sin2θ2Esh(l2t2)3+2(1+υs)T2l2sin2θ2kEsht2+l2T2co⁡s2θ2Esht2,δ3y=T3sin2θ3Esh(l3t3)3+2(1+υs)T3l3sin2θ3kEsht3+l3T3co⁡s2θ3Esht3.
Here, the sign of the displacement follows the coordinates shown in Figure 4, and k is a correction coefficient related to shear deformation, which is taken as k=1 in this work (previous research has shown that results for k=1 agree well with those of numerical analysis by the finite element method [4]).Figure 4
Displacements of ends of cell walls.


## 2.2. Analysis of Equivalent Elastic Moduli

Overaker et al. [5] proposed an elegant method for fixing the equivalent strain field, which satisfies the displacements of each cell wall found in the previous section. Figure 4 shows the displacements of each wall end in a unit δix and δiy, as well as the coordinates of each wall end (xi,yi)  (i=1,2,3). The displacements in the x- and y-directions of each wall end δix and δiy, which are found in (7) and (8), can be seen as those due to the rigid body displacements u0 and v0, the rigid body rotation ωxy, and the uniform strain field in the unit εx, εy, and εxy and are then described by the following equations:(9)δix=xiεx+yiεxy+yiωxy+u0,δiy=yiεy+xiεxy-xiωxy+v0.(i=1,2,3)
Six unknowns, namely, εx, εy, εxy, ωxy, u0, and v0, are determined by solving these equations. By using these strain fields obtained from (9), the equivalent elastic moduli C12, C22, and C32 can be found from the following equation. 								(10)C12=εxσy,C22=εyσy,C32=εxyσy.The strain field produced in the honeycomb can also be determined from the relative displacements between units. In fact, the deformation of the whole honeycomb is performed by the relative displacements between each unit. Here, we consider a part of honeycomb consisting of three units, as shown in Figure5, in which the three units are denoted as units 1, 2, and 3, counterclockwise from the lower left, and the cell wall joints of each unit are O1, O2, and O3. Denote the relative displacements of O2 and O3 with respect to O1 by U21 and V21, and U31 and V31, respectively, as shown in Figure 5(b). Thus, the following equations can be obtained from the relation between the strain field produced in the honeycomb and the displacements of O2 and O3 with respect to O1:
(11)(U21V21U31V31)=[L2x0-L2y-L2y0-L2yL2x-L2xL3x0L3yL3y0L3yL3x-L3x](εxεyεxyωxy),
where L2x and L2y are the distances in the x- and y-directions between O1 and O2, while L3x and L3y are the distances between O1 and O3(12)L2x=l2sin⁡θ2+l3sin⁡θ3,L2y=l2cos⁡⁡θ2-l3cos⁡⁡θ3,L3x=l3sin⁡θ3,L3y=l1+l3cos⁡⁡θ3.
From (11), the strain field can be obtained as a function of the relative displacements between units as follows:(13)εx=L3yU21+L2yU31(L2yL3x+L2xL3y),εy=-L3xV21+L2xV31(L2yL3x+L2xL3y),εxy=-L3xU21+L3yV21+L2xU31+L2yV312(L2yL3x+L2xL3y),ωxy=-L3xU21-L3yV21+L2xU31-L2yV312(L2yL3x+L2xL3y).Condition of junctions between wall ends: (a) a part of honeycomb consisting of three units and (b) displacement of cell wall endsA, B, and C.
(a)(b)The relative displacements between units are determined by the condition of the junctions between the wall ends of each adjacent unit. In order to consider the connection between each cell wall after deformation, Figure5(b) also shows the displacements of points A, B, and C. Point A of unit 1 is on cell wall 3, and the displacement of points A with respect to point O1 in the respective x- and y-direction, UA1 and VA1, are equal to the cell wall deformation itself(14)UA1=δ3x,VA1=δ3y.
Point B of unit 2 is on cell wall 2 and point C of unit 3 is on cell wall 1. Therefore, the displacements of points B and C with respect to point O1 in the respective x- and y-direction, Ui1 and Vi1 (i=B,C), are calculated by adding the relative displacements between the units to the displacements due to the cell wall deformation itself(15)UB1=δ2x+U21,VB1=δ2y+V21,UC1=δ1x+U31,VC1=δ1y+V31.
Since points A, B, and C are the same point prior to deformation, as shown in Figure 5(a), the displacement of points A, B, and C after deformation must be the same and the condition that UA1=UB1=UC1 and VA1=VB1=VC1 holds true. From this condition, the relative displacements U21 and V21, and U31 and V31 of O2 and O3 with respect to O1 are given as follows:(16)U21=δ3x-δ2x,V21=δ3y-δ2y,U31=δ3x-δ1x,V31=δ3y-δ1y.By substituting (12) and (16) into (13), the strain field can be obtained, and then, each equivalent elastic modulus C12, C22, and C32 can be determined from (10).


## 2.3. Calculation of Elastic Modulus Matrix

In the previous section, we presented a method for finding the three equivalent elastic moduli for the directions parallel to a cell wall constituting the hexagon cell; however, these are only three of the nine components of the elastic modulus described in (2). To express the elastic characteristics of a hexagonal honeycomb, it is necessary to know all nine components. In this section, using the three equivalent elastic moduli relating to the directions parallel to a cell wall, the nine components of the honeycomb equivalent elastic modulus C11, C12~C33 are derived.Since the approach described above allows the three equivalent elastic moduli for the direction parallel to any cell wall to be found, the three elastic moduli can be found for each direction of cell walls 1, 2, and 3, respectively. Specifically, as shown in Figure6, we take the (α′,β′) coordinates based on cell wall 1, the (α′′,β′′) coordinates based on cell wall 2 and the (α′′′,β′′′) coordinates based on cell wall 3, in which the β-axis is set to be parallel to the cell wall. Thus, C12′, C22′, and C32′ in the (α′,β′) coordinates, C12′′, C22′′, and C32′′ in the (α′′,β′′) coordinates and C12′′′, C22′′′, and C32′′′ in the (α′′′,β′′′) coordinates can be found for each coordinate system (the prime superscripts of the coordinate system correspond with those of the elastic moduli). However, the nine components of the elastic modulus C11, C12~C33 to be found are attached to the (x,y) coordinates of Figure 6. Thus, we transform coordinates from the (x,y) coordinate system to the (α,β) coordinate system. Here, we suppose an angle θ between the (x,y) coordinate system and the (α,β) coordinate system. In the (α′,β′) coordinate system, θ is θ′=θ1, for the (α′′,β′′) coordinate system, it is θ′′=θ1+(π-γ12), and for the (α′′′,β′′′) coordinate system, it is θ′′′=θ1+(π-γ12)+(π-γ23). For example, by transforming the stress and strain in the (x,y) coordinate system to the stress and strain in the (α′,β′) coordinate system, the following equation can be obtained from (2):
(17)(εα′εβ′εα′β′)=[T]-1[C11C12C13C21C22C23C31C32C33][T](σα′σβ′τα′β′).
The coordinate transformation matrix [T] is given below:
(18)[T]=[cos⁡2θ1sin2θ12sin⁡θ1cos⁡⁡θ1sin2θ1cos⁡2θ1-2sin⁡θ1cos⁡⁡θ1-sin⁡θ1cos⁡⁡θ1sin⁡θ1cos⁡⁡θ1cos⁡2θ1-sin2θ1].
However, the stress-strain equations in the (α′,β′) coordinate system are expressed by the following equation:(19)(εα′ββ′εα′β′)=[C11′C12′C13′C21′C22′C23′C31′C32′C33′](σα′σβ′τα′β′).
Since both (17) and (19) are the same, the following equation is obtained:(20)[C11′C12′C13′C21′C22′C23′C31′C32′C33′]=[T]-1[C11C12C13C21C22C23C31C32C33][T].
As stated above, C12′, C22′, and C32′ are known and from (20), they can be expressed as functions of the components C11~C33, which are to be found(21)C12′=C12cos⁡4θ1-(-C13+2C32)cos⁡3θ1sin⁡θ1+(C11+C22-2C33)cos⁡2θ1sin2θ1+(C23-2C31)cos⁡θ1sin3θ1+C21sin4θ1,C22′=C22cos⁡4θ1+(C23+2C32)cos⁡3θ1sin⁡θ1+(C12+C21+2C33)cos⁡2θ1sin2θ1+(C13+2C31)cos⁡⁡θ1sin3θ1+C11sin4θ1,C32′=C32cos⁡4θ1+(C12-C22+C33)cos⁡3θ1sin⁡θ1+(-C23+C31-C32+C13)cos⁡2θ1sin2θ1+(C11-C21-C33)cos⁡θ1sin3θ1-C31sin4θ1.
Similarly, by transforming the (x,y) coordinate system to the (α′′,β′′) and (α′′′,β′′′) coordinate systems, C12′′, C22′′, and C32′′ and C12′′′, C22′′′, and C32′′′ can be expressed as functions of C11~C33. Therefore, by solving these nine simultaneous equations, the nine components, C11~C33, can be determinedFigure 6
(α,β) coordinates based on each cell wall.By using this method, the honeycomb equivalent elastic components are found. For example, for a cell thickness oft1=t2=t3=0.05l3, Poisson’s ratio of υs=0.3, the honeycomb equivalent elastic moduli for a hexagon with parameters of l1/l3=l2/l3=1, γ12=140°, γ23=120°, and γ31=100°can be determined(22)[C]=[4.076-2.4581.505-2.4582.6143.3260.7531.6638.568]×103Es.
Moreover, for a hexagon with parameters of l1/l3=2.45, l2/l3=0.6, γ12=153.7°, γ23=115°, and γ31=91.3°, the following honeycomb equivalent elastic moduli are calculated:
(23)[C]=[11.01-0.510420.83-0.51040.27501.70910.420.85435.72]×103Es.
It can be seen from these results that the symmetry of the elastic moduli holds(24)C21=C12,2C31=C13,2C32=C23.


## 3. Effects of Geometry on Elastic Moduli

In order to investigate whether the geometry of hexagonal cell affects each of the equivalent elastic moduli, the equivalent elastic moduli are found for various hexagons that deviates from the regular hexagon, which is taken as a basic geometry here. For the following investigation, in order to observe the effects due to changes in the cell geometry, each cell wall thickness of the basic regular hexagon is taken to be the same,t1=t2=t3=0.0866l. Here, l is the length of one edge of the regular hexagon.Figure7(a) shows hexagon A′BCD′EF (geometry 1), which is formed from the regular hexagon ABCDEF by fixing edges BC and EF and moving only points A and D in the x-direction by -Δ and Δ, respectively. Each equivalent elastic modulus corresponding to the hexagonal cell of geometry 1 shown in Figure 7(a) is shown in Figure 8. Here, with the elastic modulus of a regular hexagonal cell C22|regular taken as the standard, the elastic modulus Cij along the vertical axis is compared with C22|regular. In the figure, the following is observed.Geometry 1 of hexagonal cell: (a) horizontal movement of pointA and D and (b) parallelogram (Δ=3l/2).
(a)(b)Figure 8
Equivalent elastic moduli for geometry 1 shown in Figure7.(1)
The elastic modulusC22, which expresses the magnitude of the y-direction strain due to the stress in the y-direction, is a maximum for Δ=0, that is, the regular hexagon, and C22 decreases with increasing Δ. When Δ/(3l/2)→1, C22 is not 0 but converges to C22/C22|regular=0.0288, because hexagon A′BCD′EF becomes parallelogram A′BD′E when Δ/(3l/2)=1, as shown in Figure 7(b). For the parallelogram A′BD′E, the elastic modulus C22 is C22=3l/(tEs).(2)
The elastic modulusC11, which expresses the magnitude of the x-direction strain due to the stress in the x-direction, appears not to be strongly influenced by the change in geometry due to the movement of points A and D in the x-direction; for each Δ,C11 remains nearly constant.(3)
The elastic modulusC32, which expresses the magnitude of the shear strain due to the stress in the y-direction, for the case of Δ=0, that is, for the regular hexagon, is zero due to symmetry. As Δ increases and the geometry deviates from that of a regular hexagon, |C32| increases; however, in the vicinity of about Δ/(3l/2)≅0.6, |C32| decreases, because shear deformation due to the stress σy decreases, as the geometry approaches that of a parallelogram.In order to investigate the resultant deformation due toC32 and C22, we consider the displacement of the upper end Ux and Uy of a honeycomb plate in geometry 1 under a tensile stress σy, as shown in Figure 9(a). For a plate length of L under σy, the displacement is given as follows:(25)Ux=σyC23L,Uy=σyC22L.
Figure 9(b) shows the ratio of the compliance U/(σyL) of the plate to the compliance C22|regular of the regular hexagon. Here, U=Ux2+Uy2 is the displacement of the upper end of the plate. In Figure 9(b), it can be seen that the comprehensive compliance due to C32 and C22 increases as Δ increases and reaches a maximum in the vicinity of Δ/(3l/2)=0.5. As Δ further increases, when geometry 1 deviates from the regular hexagon greatly, the compliance conversely becomes smaller.Resultant deformation due toC32 and C22: (a) displacement of plate subjected to stress σy and (b) compliance U/(σyL).
(a)(b)(4)
The elastic modulusC12, which expresses the magnitude of the x-direction strain due to stress in the y-direction, is always C12<0. For Δ=0, |C12| is a maximum and becomes zero when Δ/(3l/2)=1.In addition, the ratio ofC12 and C22 is Poisson’s ratio υ21, υ21=-C12/C22. Figure 10 shows the change in Poisson’s ratio υ21 with changing Δ. Poisson’s ratio υ21 for Δ=0 is υ21≅0.971 (as the tensile deformation and the shear deformation of the cell wall are also taken into consideration in the present research, in addition to the bending deformation of the cell wall, υ21≅0.971; however, as indicated by Gibson et al. [3], υ21=1 when only bending deformation of the  cell wall is considered). Near Δ/(3l/2)=0.9 Poisson’s ratio reaches its maximum value of about 3.1.Figure 10
Change in Poisson’s ratioυ21 with changing Δ for geometry 1.(5)
For geometry 1 of the hexagonal cell, we also investigate the maximum valueCyy|max and the minimum value Cyy|min of the elastic moduli Cyy in one rotation of the (x,y) coordinate axes, which are shown in Figure 11. When Δ=0; that is, when geometry 1 is a regular hexagon, Cyy|max is at a minimum, and Cyy|min is at a maximum; both equal the elastic modulus of regular hexagonal cell C22|regular. When Δ≠0, the compliance Cyy|min for a certain direction becomes small, however, the compliance Cyy|max for other direction becomes large. That is, when deviating the cell form from a regular hexagonal cell, the rigidity of the honeycomb can increase for a certain specific direction; however, direction for which the rigidity becomes small also exists. For the regular hexagon cell, it is found that the equivalent elastic moduli are unrelated to the selected coordinate system, and the compliance of arbitrary direction is always the same as follows:Figure 11
Change inCyy|max and Cyy|min with changing Δ for geometry 1.(26)C11=C22=C22|regularfory-axisparalleltoacellwall,C12=C12|regularfory-axisparalleltoacellwall,C13=C23=C31=C32=0,C33=C22-C12.It is not dependent on whether the tensile or the shear deformation is taken into the analysis of equivalent elastic modulus that (26) holds. Equation (26) is based on the characteristic symmetry of the regular hexagon. That is, using the symbols shown in Figure 6, for a regular hexagon, we have (27)θ′=0,θ′′=π/3,θ′′′=2π/3,Ci2′=Ci2′′=Ci2′′′(i=1,2,3).
By substituting (27) into (21), (26) can be obtained.(6)
The elastic modulusC33, which expresses the magnitude of the shear strain due to the shear stress, is a minimum when Δ=0; however, as the geometry approaches that of a parallelogram, C33 becomes larger, since shear deformation is generated easily.Next, we consider the hexagonal cellAB′C′DE′F′, which is referred to as geometry 2 here and is formed from the regular hexagon by fixing points A and D, and moving points B, C, E, and F in the y-direction, as shown in Figure 12(a). The nonzero elastic moduli for geometry 2 (C32=C13=0 from left-right symmetry) are shown in Figure 13. For geometry 2, points B′ and C′, as well as points E′ and F′, converge when Δ=-l/2, transforming the hexagon into rhomboid AB′DE′. However, when Δ=l/2, the three points A, B′, and F′ and the three points C′, D, and E′ form straight lines, transforming the hexagon to rectangle B′C′E′F′. In Figure 13, when Δ changes from the rhomboid to the rectangle, the following is observed. (1) C33 becomes large; that is, C33 increases from the value of C33/C22|regular=0.022 for the rhomboid to C33/C22|regular=9.42 for the rectangle. (2) The elastic moduli C11,C22, and C12 (absolute values) each become smaller. Namely, these elastic moduli decrease from the values of C11/C22|regular=3.43,C22/C22|regular=1.94 and C12/C22|regular=-2.56 for the rhomboid to C11/C22|regular=0.033,C22/C22|regular=0.67, and C12=0 for the rectangle.(a) Geometry 2 of hexagonal cell; (b) geometry 3 of hexagonal cell.
(a)(b)Figure 13
Equivalent elastic moduli for geometry 2 shown in Figure12(a).The maximum valueCyy|max and the minimum value Cyy|min of the elastic modulus Cyy in one rotation of the (x,y) coordinate axes are shown in Figure 14. Cyy|max and Cyy|min take a minimum and a maximum, respectively, when Δ=0, that is, when geometry 2 is a regular hexagon, which is similar to the case of geometry 1.Figure 14
Change inCyy|max and Cyy|min with changing Δ for geometry 2.Lastly, we consider a hexagonal cellA′B′C′D′E′F′, referred to as geometry 3 here, which is formed from the regular hexagon by moving the upper edge FAB and lower edge CDE to the upper and lower sides by Δ in the y-direction, respectively, as shown in Figure 12(b). The nonzero elastic moduli for geometry 3 (C32=C13=0 from left-right symmetry) are shown in Figure 15. As shown in the figure, C22 decreases. This is because the length of the cell walls parallel to the y-axis increases with increasing Δ; these cell walls only undergo tensile deformation, and the amount of deformation is small compared to the bending deformation. Moreover, due to the increase in the length of one unit in the y-direction, the force acting at the sloping cell walls due to stress σx increases, thus increasing C11. Furthermore, even if Δ changes, the elastic modulus C12 remains constant, maintaining a value of C12/C22|regular=-0.971. When Δ changes, the force acting at the sloping cell walls due to the stress σy does not change because the width of one unit does not change in the x-direction. Therefore, the deformation of the sloping cell walls is the same and the equivalent strain in the x-direction, εx, also remains the same.Figure 15
Equivalent elastic moduli for geometry 3 shown in Figure12(b).The maximum valueCyy|max and the minimum value Cyy|min of the elastic modulus Cyy in one rotation of the (x,y) coordinate axes are shown in Figure 16, from which it is seen that Cyy|max is at a minimum and Cyy|min is at a maximum when Δ=0, that is, when geometry 3 is a regular hexagon, as in the cases of geometry 1 and geometry 2.Figure 16
Change inCyy|max and Cyy|min with changing Δ for geometry 3.


## 4. Conclusions

In this research, the equivalent elastic moduli of asymmetrical hexagonal honeycomb are studied by using a theoretical approach. The deformation of honeycomb consists of two types of deformations. The first is deformation inside the unit, which is caused by bending, stretching, and shearing of cell walls and rigid rotation of the unit; the second is relative displacement between units. The relative displacements between units are determined by condition of the junctions between wall ends of each adjacent unit, and the equivalent elastic modulus related to a direction parallel to one cell wall of the honeycomb is determined from the relative deformation between units. In addition, using the three equivalent elastic moduli relating to the directions parallel to the cell wall, the nine components of the honeycomb equivalent elastic modulusC11, C12~C33 are derived by coordinate transformation. Using the proposed calculation equation, the elastic moduli for various shapes of hexagon, which are obtained by systematically altering the regular hexagon, are investigated. It is found that the maximum compliance Cyy|max and the minimum Cyy|min of elastic modulus Cyy in one rotation of the (x,y) coordinate system vary as the shape of the hexagon is changed. However, Cyy|max takes the minimum and Cyy|min takes the maximum when the honeycomb cell is a regular hexagon, for which the equivalent elastic moduli are unrelated to the selected coordinate system and are constant with C11=C22 and C33=C22-C12.



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*Source: 570140-2011-06-14.xml*