Table of Contents
ISRN Mechanical Engineering
VolumeΒ 2011, Article IDΒ 570140, 10 pages
http://dx.doi.org/10.5402/2011/570140
Research Article

Equivalent Elastic Modulus of Asymmetrical Honeycomb

Department of Mechanical Engineering, Tokyo University of Science, Kagurazaka 1–3, Shinjuku-ku, Tokyo 162-8601, Japan

Received 19 March 2011; Accepted 9 April 2011

Academic Editors: J.Β Botsis, A.Β Tounsi, and X.Β Yang

Copyright Β© 2011 Dai-Heng Chen and Kenichi Masuda. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 compliance 𝐢𝑦𝑦|max and the minimum compliance 𝐢𝑦𝑦|min of elastic modulus 𝐢𝑦𝑦 in one rotation of the (π‘₯,𝑦) coordinate system vary as the shape of the hexagon is changed. However, 𝐢𝑦𝑦|max takes a minimum and 𝐢𝑦𝑦|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 𝐢11=𝐢22.

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 Figure 1, the analyzed model in the present study is honeycomb consisting of hexagonal cells, and the honeycomb core height is denoted by β„Ž. 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 𝑑1, 𝑑2, and 𝑑3, respectively, as shown in the figure; the internal angles formed by the hexagon edges are 𝛾12, 𝛾23, and 𝛾31𝛾12+𝛾23+𝛾31=2πœ‹.(1)

fig1
Figure 1: Geometry of honeycomb: (a) honeycomb plate and (b) hexagonal cell.

The cell wall material is homogeneous and isotropic with an elastic modulus of 𝐸𝑠 and Poisson’s ratio of πœπ‘ . 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 𝐢𝑖𝑗 (𝑖,𝑗=1,2,3), which is applicable to the relation of stress and strain in the π‘₯- and 𝑦-coordinate plane shown in Figure 1βŽ›βŽœβŽœβŽœβŽπœ€π‘₯πœ€π‘¦πœ€π‘₯π‘¦βŽžβŽŸβŽŸβŽŸβŽ =⎑⎒⎒⎒⎣𝐢11𝐢12𝐢13𝐢21𝐢22𝐢23𝐢31𝐢32𝐢33⎀βŽ₯βŽ₯βŽ₯βŽ¦βŽ›βŽœβŽœβŽœβŽπœŽπ‘₯πœŽπ‘¦πœπ‘₯π‘¦βŽžβŽŸβŽŸβŽŸβŽ .(2)

The shear strain πœ€π‘₯𝑦 used here is defined as a tensor, the engineering definition of which is 𝛾π‘₯𝑦 with πœ€π‘₯𝑦=𝛾π‘₯𝑦/2.

2. Analysis of Elastic Moduli 𝐢12, 𝐢22, and 𝐢32 for the (π‘₯,𝑦) Coordinates with 𝑦-Axis Parallel to Edge 𝑙1

Initially, the 𝑦-axis is parallel to cell wall 1 in the case of πœƒ1=0∘ in Figure 1, and only the 𝑦-direction stress πœŽπ‘¦ is considered to act. As shown in Figure 2, the angles πœƒ2 and πœƒ3 are taken as those between edges 𝑙2 and 𝑙1, and edges 𝑙3 and 𝑙1, respectively πœƒ2=πœ‹βˆ’π›Ύ12,πœƒ3=πœ‹βˆ’π›Ύ31.(3)

570140.fig.002
Figure 2: Geometry of hexagon with an edge parallel to 𝑦-axis (πœƒ1=0).
2.1. Force, Moment, and Displacement Acting on Each Cell Wall

Due to the stress πœŽπ‘¦ 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 𝑇1+𝑇2=πœŽπ‘¦β„Žξ€·π‘™2sinπœƒ2+2𝑙3sinπœƒ3ξ€Έ,𝑇1=πœŽπ‘¦β„Žξ€·π‘™2sinπœƒ2+𝑙3sinπœƒ3ξ€Έ,𝑇1=𝑇2+𝑇3,(4) from which, the forces 𝑇1, 𝑇2, and 𝑇3 are given by the following equations: 𝑇1=πœŽπ‘¦β„Žξ€·π‘™3sinπœƒ3+𝑙2sinπœƒ2ξ€Έ,𝑇2=πœŽπ‘¦β„Žπ‘™3sinπœƒ3,𝑇3=πœŽπ‘¦β„Žπ‘™2sinπœƒ2.(5) The moments due to 𝑇𝑖 are given as follows: 𝑀1𝑀=0,2=𝑇212𝑙2sinπœƒ2=12πœŽπ‘¦β„Žπ‘™2sinπœƒ2𝑙3sinπœƒ3,𝑀3=𝑇312𝑙3sinπœƒ3=12πœŽπ‘¦β„Žπ‘™2sinπœƒ2𝑙3sinπœƒ3.(6) Here, 𝑀2=𝑀3.

570140.fig.003
Figure 3: Mechanics of cell walls subjected to stress πœŽπ‘¦ in the y-direction.

The π‘₯-direction displacement 𝛿𝑖π‘₯ 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 π‘₯-direction of the ends of cell walls 1, 2, and 3 are given by the following equations: 𝛿1π‘₯𝛿=0,2π‘₯=𝑇2sinπœƒ2cosπœƒ2πΈπ‘ β„Žξ‚΅π‘™2𝑑2ξ‚Ά3ξ€·+21+πœπ‘ ξ€Έπ‘‡2𝑙2sinπœƒ2cosπœƒ2π‘˜πΈπ‘ β„Žπ‘‘2βˆ’π‘™2𝑇2cosπœƒ2sinπœƒ2πΈπ‘ β„Žπ‘‘2,𝛿3π‘₯𝑇=βˆ’3sinπœƒ3cosπœƒ3πΈπ‘ β„Žξ‚΅π‘™3𝑑3ξ‚Ά3ξ€·βˆ’21+πœπ‘ ξ€Έπ‘‡3𝑙3sinπœƒ3cosπœƒ3π‘˜πΈπ‘ β„Žπ‘‘3+𝑙3𝑇3cosπœƒ3sinπœƒ3πΈπ‘ β„Žπ‘‘3.(7) Similarly, the displacements in the 𝑦-direction of each cell wall are given as follows: 𝛿1𝑦𝑙=βˆ’1𝑇1πΈπ‘ β„Žπ‘‘1,𝛿2𝑦=𝑇2sin2πœƒ2πΈπ‘ β„Žξ‚΅π‘™2𝑑2ξ‚Ά3ξ€·+21+πœπ‘ ξ€Έπ‘‡2𝑙2sin2πœƒ2π‘˜πΈπ‘ β„Žπ‘‘2+𝑙2𝑇2cos2πœƒ2πΈπ‘ β„Žπ‘‘2,𝛿3𝑦=𝑇3sin2πœƒ3πΈπ‘ β„Žξ‚΅π‘™3𝑑3ξ‚Ά3ξ€·+21+πœπ‘ ξ€Έπ‘‡3𝑙3sin2πœƒ3π‘˜πΈπ‘ β„Žπ‘‘3+𝑙3𝑇3cos2πœƒ3πΈπ‘ β„Žπ‘‘3.(8) Here, the sign of the displacement follows the coordinates shown in Figure 4, and π‘˜ is a correction coefficient related to shear deformation, which is taken as π‘˜=1 in this work (previous research has shown that results for π‘˜=1 agree well with those of numerical analysis by the finite element method [4]).

570140.fig.004
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 𝛿𝑖π‘₯ and 𝛿𝑖𝑦, as well as the coordinates of each wall end (π‘₯𝑖,𝑦𝑖) (𝑖=1,2,3). The displacements in the π‘₯- and 𝑦-directions of each wall end 𝛿𝑖π‘₯ and 𝛿𝑖𝑦, which are found in (7) and (8), can be seen as those due to the rigid body displacements 𝑒0 and 𝑣0, the rigid body rotation πœ”π‘₯𝑦, and the uniform strain field in the unit πœ€π‘₯, πœ€π‘¦, and πœ€π‘₯𝑦 and are then described by the following equations:𝛿𝑖π‘₯=π‘₯π‘–πœ€π‘₯+π‘¦π‘–πœ€π‘₯𝑦+π‘¦π‘–πœ”π‘₯𝑦+𝑒0,𝛿𝑖𝑦=π‘¦π‘–πœ€π‘¦+π‘₯π‘–πœ€π‘₯π‘¦βˆ’π‘₯π‘–πœ”π‘₯𝑦+𝑣0.(𝑖=1,2,3)(9) Six unknowns, namely, πœ€π‘₯, πœ€π‘¦, πœ€π‘₯𝑦, πœ”π‘₯𝑦, 𝑒0, and 𝑣0, are determined by solving these equations. By using these strain fields obtained from (9), the equivalent elastic moduli 𝐢12, 𝐢22, and 𝐢32 can be found from the following equation. 𝐢12=πœ€π‘₯πœŽπ‘¦,𝐢22=πœ€π‘¦πœŽπ‘¦,𝐢32=πœ€π‘₯π‘¦πœŽπ‘¦.(10)

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 Figure 5, 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 π‘ˆ21 and 𝑉21, and π‘ˆ31 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: βŽ›βŽœβŽœβŽœβŽœβŽœβŽπ‘ˆ21𝑉21π‘ˆ31𝑉31⎞⎟⎟⎟⎟⎟⎠=⎑⎒⎒⎒⎒⎒⎣𝐿2π‘₯0βˆ’πΏ2π‘¦βˆ’πΏ2𝑦0βˆ’πΏ2𝑦𝐿2π‘₯βˆ’πΏ2π‘₯𝐿3π‘₯0𝐿3𝑦𝐿3𝑦0𝐿3𝑦𝐿3π‘₯βˆ’πΏ3π‘₯⎀βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ¦βŽ›βŽœβŽœβŽœβŽœβŽœβŽπœ€π‘₯πœ€π‘¦πœ€π‘₯π‘¦πœ”π‘₯π‘¦βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽ ,(11) where 𝐿2π‘₯ and 𝐿2𝑦 are the distances in the π‘₯- and 𝑦-directions between 𝑂1 and 𝑂2, while 𝐿3π‘₯ and 𝐿3𝑦 are the distances between 𝑂1 and 𝑂3𝐿2π‘₯=𝑙2sinπœƒ2+𝑙3sinπœƒ3,𝐿2𝑦=𝑙2cosπœƒ2βˆ’π‘™3cosπœƒ3,𝐿3π‘₯=𝑙3sinπœƒ3,𝐿3𝑦=𝑙1+𝑙3cosπœƒ3.(12) From (11), the strain field can be obtained as a function of the relative displacements between units as follows:πœ€π‘₯=𝐿3π‘¦π‘ˆ21+𝐿2π‘¦π‘ˆ31𝐿2𝑦𝐿3π‘₯+𝐿2π‘₯𝐿3𝑦,πœ€π‘¦=βˆ’πΏ3π‘₯𝑉21+𝐿2π‘₯𝑉31𝐿2𝑦𝐿3π‘₯+𝐿2π‘₯𝐿3𝑦,πœ€π‘₯𝑦=βˆ’πΏ3π‘₯π‘ˆ21+𝐿3𝑦𝑉21+𝐿2π‘₯π‘ˆ31+𝐿2𝑦𝑉312𝐿2𝑦𝐿3π‘₯+𝐿2π‘₯𝐿3𝑦,πœ”π‘₯𝑦=βˆ’πΏ3π‘₯π‘ˆ21βˆ’πΏ3𝑦𝑉21+𝐿2π‘₯π‘ˆ31βˆ’πΏ2𝑦𝑉312𝐿2𝑦𝐿3π‘₯+𝐿2π‘₯𝐿3𝑦.(13)

fig5
Figure 5: Condition of junctions between wall ends: (a) a part of honeycomb consisting of three units and (b) displacement of cell wall ends 𝐴, 𝐡, and 𝐢.

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, Figure 5(b) also shows the displacements of points 𝐴, 𝐡, and 𝐢. Point 𝐴 of unit 1 is on cell wall 3, and the displacement of points 𝐴 with respect to point 𝑂1 in the respective π‘₯- and 𝑦-direction, π‘ˆπ΄1 and 𝑉𝐴1, are equal to the cell wall deformation itselfπ‘ˆπ΄1=𝛿3π‘₯,𝑉𝐴1=𝛿3𝑦.(14) Point 𝐡 of unit 2 is on cell wall 2 and point 𝐢 of unit 3 is on cell wall 1. Therefore, the displacements of points 𝐡 and 𝐢 with respect to point 𝑂1 in the respective π‘₯- and 𝑦-direction, π‘ˆπ‘–1 and 𝑉𝑖1 (𝑖=𝐡,𝐢), are calculated by adding the relative displacements between the units to the displacements due to the cell wall deformation itselfπ‘ˆπ΅1=𝛿2π‘₯+π‘ˆ21,𝑉𝐡1=𝛿2𝑦+𝑉21,π‘ˆπΆ1=𝛿1π‘₯+π‘ˆ31,𝑉𝐢1=𝛿1𝑦+𝑉31.(15) Since points 𝐴, 𝐡, and 𝐢 are the same point prior to deformation, as shown in Figure 5(a), the displacement of points 𝐴, 𝐡, and 𝐢 after deformation must be the same and the condition that π‘ˆπ΄1=π‘ˆπ΅1=π‘ˆπΆ1 and 𝑉𝐴1=𝑉𝐡1=𝑉𝐢1 holds true. From this condition, the relative displacements π‘ˆ21 and 𝑉21, and π‘ˆ31 and 𝑉31 of 𝑂2 and 𝑂3 with respect to 𝑂1 are given as follows:π‘ˆ21=𝛿3π‘₯βˆ’π›Ώ2π‘₯,𝑉21=𝛿3π‘¦βˆ’π›Ώ2𝑦,π‘ˆ31=𝛿3π‘₯βˆ’π›Ώ1π‘₯,𝑉31=𝛿3π‘¦βˆ’π›Ώ1𝑦.(16)

By substituting (12) and (16) into (13), the strain field can be obtained, and then, each equivalent elastic modulus 𝐢12, 𝐢22, and 𝐢32 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 𝐢11, 𝐢12∼𝐢33 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 Figure 6, we take the (𝛼′,𝛽′) coordinates based on cell wall 1, the (π›Όξ…žξ…ž,π›½ξ…žξ…ž) coordinates based on cell wall 2 and the (π›Όξ…žξ…žξ…ž,π›½ξ…žξ…žξ…ž) coordinates based on cell wall 3, in which the 𝛽-axis is set to be parallel to the cell wall. Thus, πΆξ…ž12, πΆξ…ž22, and πΆξ…ž32 in the (𝛼′,π›½ξ…ž) coordinates, πΆξ…žξ…ž12, πΆξ…žξ…ž22, and πΆξ…žξ…ž32 in the (π›Όξ…žξ…ž,π›½ξ…žξ…ž) coordinates and πΆξ…žξ…žξ…ž12, πΆξ…žξ…žξ…ž22, and πΆξ…žξ…žξ…ž32 in the (π›Όξ…žξ…žξ…ž,π›½ξ…žξ…žξ…ž) 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 𝐢11, 𝐢12∼𝐢33 to be found are attached to the (π‘₯,𝑦) coordinates of Figure 6. Thus, we transform coordinates from the (π‘₯,𝑦) coordinate system to the (𝛼,𝛽) coordinate system. Here, we suppose an angle πœƒ between the (π‘₯,𝑦) coordinate system and the (𝛼,𝛽) coordinate system. In the (π›Όξ…ž,π›½ξ…ž) coordinate system, πœƒ is πœƒξ…ž=πœƒ1, for the (π›Όξ…žξ…ž,π›½ξ…žξ…ž) coordinate system, it is πœƒξ…žξ…ž=πœƒ1+(πœ‹βˆ’π›Ύ12), and for the (π›Όξ…žξ…žξ…ž,π›½ξ…žξ…žξ…ž) coordinate system, it is πœƒξ…žξ…žξ…ž=πœƒ1+(πœ‹βˆ’π›Ύ12)+(πœ‹βˆ’π›Ύ23). For example, by transforming the stress and strain in the (π‘₯,𝑦) coordinate system to the stress and strain in the (π›Όξ…ž,π›½ξ…ž) coordinate system, the following equation can be obtained from (2): βŽ›βŽœβŽœβŽœβŽπœ€π›Όβ€²πœ€π›½β€²πœ€π›Όβ€²π›½β€²βŽžβŽŸβŽŸβŽŸβŽ =[𝑇]βˆ’1⎑⎒⎒⎒⎣𝐢11𝐢12𝐢13𝐢21𝐢22𝐢23𝐢31𝐢32𝐢33⎀βŽ₯βŽ₯βŽ₯⎦[𝑇]βŽ›βŽœβŽœβŽœβŽπœŽπ›Όβ€²πœŽπ›½β€²πœπ›Όβ€²π›½β€²βŽžβŽŸβŽŸβŽŸβŽ .(17) The coordinate transformation matrix [𝑇] is given below: [𝑇]=⎑⎒⎒⎒⎣cos2πœƒ1sin2πœƒ12sinπœƒ1cosπœƒ1sin2πœƒ1cos2πœƒ1βˆ’2sinπœƒ1cosπœƒ1βˆ’sinπœƒ1cosπœƒ1sinπœƒ1cosπœƒ1cos2πœƒ1βˆ’sin2πœƒ1⎀βŽ₯βŽ₯βŽ₯⎦.(18) However, the stress-strain equations in the (π›Όξ…ž,π›½ξ…ž) coordinate system are expressed by the following equation:βŽ›βŽœβŽœβŽœβŽπœ€π›Όβ€²π›½π›½β€²πœ€π›Όβ€²π›½β€²βŽžβŽŸβŽŸβŽŸβŽ =βŽ‘βŽ’βŽ’βŽ’βŽ£πΆξ…ž11πΆξ…ž12πΆξ…ž13πΆξ…ž21πΆξ…ž22πΆξ…ž23πΆξ…ž31πΆξ…ž32πΆξ…ž33⎀βŽ₯βŽ₯βŽ₯βŽ¦βŽ›βŽœβŽœβŽœβŽπœŽπ›Όβ€²πœŽπ›½β€²πœπ›Όβ€²π›½β€²βŽžβŽŸβŽŸβŽŸβŽ .(19) Since both (17) and (19) are the same, the following equation is obtained:βŽ‘βŽ’βŽ’βŽ’βŽ£πΆξ…ž11πΆξ…ž12πΆξ…ž13πΆξ…ž21πΆξ…ž22πΆξ…ž23πΆξ…ž31πΆξ…ž32πΆξ…ž33⎀βŽ₯βŽ₯βŽ₯⎦=[𝑇]βˆ’1⎑⎒⎒⎒⎣𝐢11𝐢12𝐢13𝐢21𝐢22𝐢23𝐢31𝐢32𝐢33⎀βŽ₯βŽ₯βŽ₯⎦[𝑇].(20) As stated above, πΆξ…ž12, πΆξ…ž22, and πΆξ…ž32 are known and from (20), they can be expressed as functions of the components 𝐢11∼𝐢33, which are to be foundπΆξ…ž12=𝐢12cos4πœƒ1βˆ’ξ€·βˆ’πΆ13+2𝐢32ξ€Έcos3πœƒ1sinπœƒ1+𝐢11+𝐢22βˆ’2𝐢33ξ€Έcos2πœƒ1sin2πœƒ1+𝐢23βˆ’2𝐢31ξ€Έcosπœƒ1sin3πœƒ1+𝐢21sin4πœƒ1,πΆξ…ž22=𝐢22cos4πœƒ1+𝐢23+2𝐢32ξ€Έcos3πœƒ1sinπœƒ1+𝐢12+𝐢21+2𝐢33ξ€Έcos2πœƒ1sin2πœƒ1+𝐢13+2𝐢31ξ€Έcosπœƒ1sin3πœƒ1+𝐢11sin4πœƒ1,πΆξ…ž32=𝐢32cos4πœƒ1+𝐢12βˆ’πΆ22+𝐢33ξ€Έcos3πœƒ1sinπœƒ1+ξ€·βˆ’πΆ23+𝐢31βˆ’πΆ32+𝐢13ξ€Έcos2πœƒ1sin2πœƒ1+𝐢11βˆ’πΆ21βˆ’πΆ33ξ€Έcosπœƒ1sin3πœƒ1βˆ’πΆ31sin4πœƒ1.(21) Similarly, by transforming the (π‘₯,𝑦) coordinate system to the (π›Όξ…žξ…ž,π›½ξ…žξ…ž) and (π›Όξ…žξ…žξ…ž,π›½ξ…žξ…žξ…ž) coordinate systems, πΆξ…žξ…ž12, πΆξ…žξ…ž22, and πΆξ…žξ…ž32 and πΆξ…žξ…žξ…ž12, πΆξ…žξ…žξ…ž22, and πΆξ…žξ…žξ…ž32 can be expressed as functions of 𝐢11∼𝐢33. Therefore, by solving these nine simultaneous equations, the nine components, 𝐢11∼𝐢33, can be determined

570140.fig.006
Figure 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 of 𝑑1=𝑑2=𝑑3=0.05𝑙3, Poisson’s ratio of πœπ‘ =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[𝐢]=⎑⎒⎒⎒⎣⎀βŽ₯βŽ₯βŽ₯βŽ¦Γ—4.076βˆ’2.4581.505βˆ’2.4582.6143.3260.7531.6638.568103𝐸𝑠.(22) 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: [𝐢]=⎑⎒⎒⎒⎣⎀βŽ₯βŽ₯βŽ₯βŽ¦Γ—11.01βˆ’0.510420.83βˆ’0.51040.27501.70910.420.85435.72103𝐸𝑠.(23) It can be seen from these results that the symmetry of the elastic moduli holds𝐢21=𝐢12,2𝐢31=𝐢13,2𝐢32=𝐢23.(24)

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, 𝑑1=𝑑2=𝑑3=0.0866𝑙. Here, 𝑙 is the length of one edge of the regular hexagon.

Figure 7(a) shows hexagon π΄ξ…žπ΅πΆπ·ξ…žπΈπΉ (geometry 1), which is formed from the regular hexagon ABCDEF by fixing edges 𝐡𝐢 and 𝐸𝐹 and moving only points 𝐴 and 𝐷 in the π‘₯-direction by βˆ’Ξ” and Ξ”, 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 𝐢22|regular taken as the standard, the elastic modulus 𝐢𝑖𝑗 along the vertical axis is compared with 𝐢22|regular. In the figure, the following is observed.

fig7
Figure 7: Geometry 1 of hexagonal cell: (a) horizontal movement of point 𝐴 and 𝐷 and (b) parallelogram √(Ξ”=3𝑙/2).
570140.fig.008
Figure 8: Equivalent elastic moduli for geometry 1 shown in Figure 7.
(1)The elastic modulus 𝐢22, which expresses the magnitude of the 𝑦-direction strain due to the stress in the 𝑦-direction, is a maximum for Ξ”=0, that is, the regular hexagon, and 𝐢22 decreases with increasing Ξ”. When βˆšΞ”/(3𝑙/2)β†’1, 𝐢22 is not 0 but converges to 𝐢22/𝐢22|regular=0.0288, because hexagon π΄ξ…žπ΅πΆπ·ξ…žπΈπΉ becomes parallelogram π΄ξ…žπ΅π·ξ…žπΈ when βˆšΞ”/(3𝑙/2)=1, as shown in Figure 7(b). For the parallelogram π΄ξ…žπ΅π·ξ…žπΈ, the elastic modulus 𝐢22 is 𝐢22=√3𝑙/(𝑑𝐸𝑠).(2)The elastic modulus 𝐢11, which expresses the magnitude of the π‘₯-direction strain due to the stress in the π‘₯-direction, appears not to be strongly influenced by the change in geometry due to the movement of points 𝐴 and 𝐷 in the π‘₯-direction; for each Ξ”,𝐢11 remains nearly constant.(3)The elastic modulus 𝐢32, which expresses the magnitude of the shear strain due to the stress in the 𝑦-direction, for the case of Ξ”=0, that is, for the regular hexagon, is zero due to symmetry. As Ξ” increases and the geometry deviates from that of a regular hexagon, |𝐢32| increases; however, in the vicinity of about βˆšΞ”/(3𝑙/2)β‰…0.6, |𝐢32| decreases, because shear deformation due to the stress πœŽπ‘¦ decreases, as the geometry approaches that of a parallelogram.

In order to investigate the resultant deformation due to 𝐢32 and 𝐢22, we consider the displacement of the upper end π‘ˆπ‘₯ and π‘ˆπ‘¦ of a honeycomb plate in geometry 1 under a tensile stress πœŽπ‘¦, as shown in Figure 9(a). For a plate length of 𝐿 under πœŽπ‘¦, the displacement is given as follows:π‘ˆπ‘₯=πœŽπ‘¦πΆ23𝐿,π‘ˆπ‘¦=πœŽπ‘¦πΆ22𝐿.(25) Figure 9(b) shows the ratio of the compliance π‘ˆ/(πœŽπ‘¦πΏ) of the plate to the compliance 𝐢22|regular of the regular hexagon. Here, ξ”π‘ˆ=π‘ˆ2π‘₯+π‘ˆ2𝑦 is the displacement of the upper end of the plate. In Figure 9(b), it can be seen that the comprehensive compliance due to 𝐢32 and 𝐢22 increases as Ξ” increases and reaches a maximum in the vicinity of βˆšΞ”/(3𝑙/2)=0.5. As Ξ” further increases, when geometry 1 deviates from the regular hexagon greatly, the compliance conversely becomes smaller.

fig9
Figure 9: Resultant deformation due to 𝐢32 and 𝐢22: (a) displacement of plate subjected to stress πœŽπ‘¦ and (b) compliance π‘ˆ/(πœŽπ‘¦πΏ).
(4)The elastic modulus 𝐢12, which expresses the magnitude of the π‘₯-direction strain due to stress in the 𝑦-direction, is always 𝐢12<0. For Ξ”=0, |𝐢12| is a maximum and becomes zero when βˆšΞ”/(3𝑙/2)=1.

In addition, the ratio of 𝐢12 and 𝐢22 is Poisson’s ratio 𝜐21, 𝜐21=βˆ’πΆ12/𝐢22. Figure 10 shows the change in Poisson’s ratio 𝜐21 with changing Ξ”. Poisson’s ratio 𝜐21 for Ξ”=0 is 𝜐21β‰…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, 𝜐21β‰…0.971; however, as indicated by Gibson et al. [3], 𝜐21=1 when only bending deformation of the cell wall is considered). Near βˆšΞ”/(3𝑙/2)=0.9 Poisson’s ratio reaches its maximum value of about 3.1.

570140.fig.0010
Figure 10: Change in Poisson’s ratio 𝜐21 with changing Ξ” for geometry 1.
(5)For geometry 1 of the hexagonal cell, we also investigate the maximum value 𝐢𝑦𝑦|max and the minimum value 𝐢𝑦𝑦|min of the elastic moduli 𝐢𝑦𝑦 in one rotation of the (π‘₯,𝑦) coordinate axes, which are shown in Figure 11. When Ξ”=0; that is, when geometry 1 is a regular hexagon, 𝐢𝑦𝑦|max is at a minimum, and 𝐢𝑦𝑦|min is at a maximum; both equal the elastic modulus of regular hexagonal cell 𝐢22|regular. When Ξ”β‰ 0, the compliance 𝐢𝑦𝑦|min for a certain direction becomes small, however, the compliance 𝐢𝑦𝑦|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:
570140.fig.0011
Figure 11: Change in 𝐢𝑦𝑦|max and 𝐢𝑦𝑦|min with changing Ξ” for geometry 1.

𝐢11=𝐢22=𝐢22||regular𝐢for𝑦-axisparalleltoacellwall,12=𝐢12||regular𝐢for𝑦-axisparalleltoacellwall,13=𝐢23=𝐢31=𝐢32𝐢=0,33=𝐢22βˆ’πΆ12.(26)

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 πœƒξ…ž=0,πœƒξ…žξ…ž=πœ‹/3,πœƒξ…žξ…žξ…žπΆ=2πœ‹/3,ξ…žπ‘–2=πΆξ…žξ…žπ‘–2=πΆξ…žξ…žξ…žπ‘–2(𝑖=1,2,3).(27) By substituting (27) into (21), (26) can be obtained.

(6)The elastic modulus 𝐢33, which expresses the magnitude of the shear strain due to the shear stress, is a minimum when Ξ”=0; however, as the geometry approaches that of a parallelogram, 𝐢33 becomes larger, since shear deformation is generated easily.

Next, we consider the hexagonal cell π΄π΅ξ…žπΆξ…žπ·πΈξ…žπΉξ…ž, which is referred to as geometry 2 here and is formed from the regular hexagon by fixing points 𝐴 and 𝐷, and moving points 𝐡, 𝐢, 𝐸, and 𝐹 in the 𝑦-direction, as shown in Figure 12(a). The nonzero elastic moduli for geometry 2 (𝐢32=𝐢13=0 from left-right symmetry) are shown in Figure 13. For geometry 2, points π΅ξ…ž and πΆξ…ž, as well as points πΈξ…ž and πΉξ…ž, converge when Ξ”=βˆ’π‘™/2, transforming the hexagon into rhomboid π΄π΅ξ…žπ·πΈξ…ž. However, when Ξ”=𝑙/2, the three points 𝐴, π΅ξ…ž, and πΉξ…ž and the three points πΆξ…ž, 𝐷, and πΈξ…ž form straight lines, transforming the hexagon to rectangle π΅ξ…žπΆξ…žπΈξ…žπΉξ…ž. In Figure 13, when Ξ” changes from the rhomboid to the rectangle, the following is observed. (1) 𝐢33 becomes large; that is, 𝐢33 increases from the value of 𝐢33/𝐢22|regular=0.022 for the rhomboid to 𝐢33/𝐢22|regular=9.42 for the rectangle. (2) The elastic moduli 𝐢11,𝐢22, and 𝐢12 (absolute values) each become smaller. Namely, these elastic moduli decrease from the values of 𝐢11/𝐢22|regular=3.43,𝐢22/𝐢22|regular=1.94 and 𝐢12/𝐢22|regular=βˆ’2.56 for the rhomboid to 𝐢11/𝐢22|regular=0.033,𝐢22/𝐢22|regular=0.67, and 𝐢12=0 for the rectangle.

fig12
Figure 12: (a) Geometry 2 of hexagonal cell; (b) geometry 3 of hexagonal cell.
570140.fig.0013
Figure 13: Equivalent elastic moduli for geometry 2 shown in Figure 12(a).

The maximum value 𝐢𝑦𝑦|max and the minimum value 𝐢𝑦𝑦|min of the elastic modulus 𝐢𝑦𝑦 in one rotation of the (π‘₯,𝑦) coordinate axes are shown in Figure 14. 𝐢𝑦𝑦|max and 𝐢𝑦𝑦|min take a minimum and a maximum, respectively, when Ξ”=0, that is, when geometry 2 is a regular hexagon, which is similar to the case of geometry 1.

570140.fig.0014
Figure 14: Change in 𝐢𝑦𝑦|max and 𝐢𝑦𝑦|min with changing Ξ” for geometry 2.

Lastly, we consider a hexagonal cell π΄ξ…žπ΅ξ…žπΆξ…žπ·ξ…žπΈξ…žπΉξ…ž, referred to as geometry 3 here, which is formed from the regular hexagon by moving the upper edge 𝐹𝐴𝐡 and lower edge CDE to the upper and lower sides by Ξ” in the 𝑦-direction, respectively, as shown in Figure 12(b). The nonzero elastic moduli for geometry 3 (𝐢32=𝐢13=0 from left-right symmetry) are shown in Figure 15. As shown in the figure, 𝐢22 decreases. This is because the length of the cell walls parallel to the 𝑦-axis increases with increasing Ξ”; 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 𝜎π‘₯ increases, thus increasing 𝐢11. Furthermore, even if Ξ” changes, the elastic modulus 𝐢12 remains constant, maintaining a value of 𝐢12/𝐢22|regular=βˆ’0.971. When Ξ” changes, the force acting at the sloping cell walls due to the stress πœŽπ‘¦ does not change because the width of one unit does not change in the π‘₯-direction. Therefore, the deformation of the sloping cell walls is the same and the equivalent strain in the π‘₯-direction, πœ€π‘₯, also remains the same.

570140.fig.0015
Figure 15: Equivalent elastic moduli for geometry 3 shown in Figure 12(b).

The maximum value 𝐢𝑦𝑦|max and the minimum value 𝐢𝑦𝑦|min of the elastic modulus 𝐢𝑦𝑦 in one rotation of the (π‘₯,𝑦) coordinate axes are shown in Figure 16, from which it is seen that 𝐢𝑦𝑦|max is at a minimum and 𝐢𝑦𝑦|min is at a maximum when Ξ”=0, that is, when geometry 3 is a regular hexagon, as in the cases of geometry 1 and geometry 2.

570140.fig.0016
Figure 16: Change in 𝐢𝑦𝑦|max and 𝐢𝑦𝑦|min with changing Ξ” 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 modulus 𝐢11, 𝐢12∼𝐢33 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 𝐢𝑦𝑦|max and the minimum 𝐢𝑦𝑦|min of elastic modulus 𝐢𝑦𝑦 in one rotation of the (π‘₯,𝑦) coordinate system vary as the shape of the hexagon is changed. However, 𝐢𝑦𝑦|max takes the minimum and 𝐢𝑦𝑦|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 𝐢11=𝐢22 and 𝐢33=𝐢22βˆ’πΆ12.

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