bending deformation of honeycomb consisting of regular hexagonal cells_chen2011

8/9/2019 Bending Deformation of Honeycomb Consisting of Regular Hexagonal Cells_chen2011

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Bending deformation of honeycomb consisting of regular hexagonal cells

D.H. Chen ⇑

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

a r t i c l e i n f o

Article history:Available online 14 August 2010

Keywords:HoneycombEquivalent elastic modulusFlexural rigidityTorsion of plateElasticity

a b s t r a c t

In this study, the exural rigidity of a honeycomb consisting of regular hexagonal cells is investigated. It isfound that the bending deformation of the honeycomb cannot be evaluated by using the equivalent elas-

tic moduli obtained from the in-plane deformation since the moments acting on inclined walls of honey-comb cell are different for the in-plane deformation and bending deformation. Based on the fact that theinclined wall of the honeycomb is twisted under the condition that the rotation angle in both connectionedges is zero in bending deformation, a theoretical technique for calculating the honeycomb exuralrigidity is proposed. In the theoretical analysis, a torsion problem of a thin plate was solved by usingthe generalized variational principle. The validity of the present analysis is demonstrated by numericalresults obtained by the nite element method.

2010 Elsevier Ltd. All rights reserved.

1. Introduction

In recent years, honeycombs have attracted considerable inter-est as advanced composite materials that satisfy the high perfor-

mance requirements of machine design. Honeycombs are widelyused in areas ranging from the aerospace industry to householdapplications. Thus, it is becoming ever more important to developmethods for evaluating the elastic properties of various honey-comb materials. Very many studies have been performed on thistopic to calculate the in-plane equivalent elastic moduli of honey-comb [1–9] , including a systematic review of various analyticalmethods by Hohe and Becker [10] . Moreover, the three-dimen-sional problem that incorporates the effect of honeycomb heighthas also been analyzed by Grediac [11] , Shi and Tong [12] , Becker[13] , Hohe and Becker [14] , Xu and Qiao [15] , Chen and Davalos[16] and Chen and Ozaki [17] to more accurately calculate the elas-tic moduli of honeycomb.

In the design process of honeycomb structures, nding the

bending deection when a load acts in the direction perpendicularto the plane of honeycomb structure is also necessary; however,this problem has been only partially addressed. Abd El-Sayedet al. [18] proposed a calculation method of honeycomb curvaturein bending deformation, however it is not highly precise as shownlater. Alderson et al. [19] investigated honeycomb curvature inbending deformation using simulation results of nite elementmodel. No highly precise theoretical analysis of this problem hasbeen reported to date.

In the deformation analysis of honeycomb structures, the hon-eycomb is usually considered to be a homogeneous plate withequivalent elastic modulus due to their cyclic cell structure. For ahoneycomb formed from hexagonal cells shown in Fig. 1, the

in-plane equivalent elastic modulus, namely the equivalent elasticmoduli along the x-axis and y-axis E x and E y and Poisson’s ratios m xyand m yx have been considered in the work of Gibson et al. [3] . Theyobtained

E x ¼ cos h

sin 2 hð1 þ sin hÞt l

3

E s;

E y ¼ ð1 þ sin hÞ

cos 3 ht l

3

E s;

m xy ¼ cos 2 h

sin hð1 þ sin hÞ ;

m yx ¼ sin hð1 þ sin hÞ

cos 2 h :

ð1 Þ

In the present research, for the honeycomb cell wall the materialYoung’s modulus and Poisson’s ratio are denoted by E s and ms,respectively.

In the analysis of a plate bending deformation, the necessaryelastic moduli are the exural rigidities D x, D y and D1 [20] whichrelate moments M x and M y with curvatures of the deformed plate1/ q x and 1/ q y as

M x ¼ D x1

q xþ D1

1

q y; M y ¼ D y

1

q yþ D1

1

q x: ð2 Þ

Based on the Kirchhoff hypothesis, the exural rigidity of a plate isgiven from its in-plane equivalent elastic modulus by the followingequation:

0263-8223/$ - see front matter 2010 Elsevier Ltd. All rights reserved.doi: 10.1016/j.compstruct.2010.08.006

⇑ Tel.: +81 3 5228 8367; fax: +81 3 5213 0977.E-mail address: [email protected]

Composite Structures 93 (2011) 736–746

Contents lists available at ScienceDirect

Composite Structures

j o u rn a l home page : www.e l s ev i e r. c om/ loca t e / co mps t ruc t

http://dx.doi.org/10.1016/j.compstruct.2010.08.006mailto:[email protected]://dx.doi.org/10.1016/j.compstruct.2010.08.006http://www.sciencedirect.com/science/journal/02638223http://www.elsevier.com/locate/compstructhttp://www.elsevier.com/locate/compstructhttp://www.sciencedirect.com/science/journal/02638223http://dx.doi.org/10.1016/j.compstruct.2010.08.006mailto:[email protected]://dx.doi.org/10.1016/j.compstruct.2010.08.006


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D x ¼ E xh

3

12 ð1 m xym yxÞ ; D y ¼

E yh3

12 ð1 m xym yxÞ ; D1 ¼

E xm yxh3

12 ð1 m xym yxÞ :

ð3 Þ

It is desirable that Eq. (3) is also applicable to the honeycomb bend-ing problem. However, for a regular hexagonal cell ( h = 30 ) Eq. (1)yields m xy m yx = 1 for Poisson’s ratio, and the exural rigidities D x, D yand D1 dened by Eq. (3) become innite, thus making analysis

impossible.Such a problem of m xy m yx = 1 is caused by the approximationtechnique used in deriving Eq. (1) , in which only the bendingdeformation of cell walls is taken into account. Masters and Evans[5] extended the analysis of the in-plane elastic modulus of honey-comb by taking the tensile deformation and shear deformation, be-sides the bending deformation, of the cell walls into consideration,which leads to

E x ¼ E scos h

sin 2 hð1 þ sin hÞt l

3 1

1 þ 2 ð1 þ mS Þ þ 1tan 2 hh i t l 2 ;

E y ¼ E sð1 þ sin hÞ

cos 3 ht l

3 1

1 þ 2 ð1 þ mS Þ þ tan 2 h þ 2cos 2 h

t l

2 ;

m xy ¼ cos2

hsin hð1 þ sin hÞ

1 þ ½2 ð1 þ mS Þ 1 t l

2

1 þ 2 ð1 þ mS Þ þ 1tan 2 hh i t l 2 ;

m yx ¼ sin hð1 þ sin hÞ

cos 2 h

1 þ ½2 ð1 þ mS Þ 1 t l 2

1 þ 2 ð1 þ mS Þ þ tan 2 h þ 2cos 2 h t l

2 ;

ð4 Þ

which is slightly different to Eq. (1) . Using Eq. (4) to determine Pois-son’s ratio of a regular hexagon yields m xy m yx – 1. Then, the bend-ing rigidities D x, D y and D1 dened by Eq. (3) become nite values,thereby making the analysis possible. It seems that the bendingdeformation of honeycomb can be analyzed by using the exuralrigidities obtained by substituting Eq. (4) into Eq. (3). However,our investigation, the details of which will be shown later, showsa regrettable fact that the exural rigidity obtained in this way can-

not be accurately applied to the honeycomb bending analysis.The rst motive to begin this research is looking for the reasonwhy Eq. (3) is inapplicable to the honeycomb bending problem,and trying to propose an effective technique for analyzing honey-comb bend. In this study a detailed investigation of the bendingmechanics of honeycomb is carried out, and a theoretical tech-nique for determining the exural rigidity of honeycomb is pro-posed. The validity of the present analysis is demonstrated bynumerical results obtained by the nite element method (FEM).

2. Problem used for examination of honeycomb exural rigidity

In the present research, as an example of honeycomb structure,we consider the 7-row, 8-column honeycomb model shown in

Fig. 2, which is constituted of periodically arranged regular hexag-onal cells, shown in Fig. 1. The length, thickness and height of the

cell wall are l = 10 mm, t = 0.5 mm and h = 2.5 mm, respectively. In

this model, the honeycomb is supported at points E and F shown inthe gure (points at y = 50 mm and y = 50 mm on the y-axis), amoment M x is applied uniformly to edges BC and DA, and amoment M y is applied uniformly to edges AB and CD. Here, in orderto distinguish the honeycomb equivalent global stress andmoment from those for each cell wall, a bar is placed above thesymbol for global stress and moment. Furthermore, the Young’smodulus is E s = 10 7 N/mm 2 and Poisson’s ratio is ms = 0.3 for thehoneycomb cell wall material.

In order to examine the effectiveness of honeycomb exuralrigidity investigated in this research, a thin plate is used to modelthe honeycomb. The thin plate imitating the honeycomb shown inFig. 2 has the same size of a b (a = 16 lcos h = 138.56 mm,b = 8 l(1 + sin h) = 120 mm) and height h = 2.5 mm, as shown inFig. 3. This plate model is used as the object of the theoretical anal-ysis. The load and boundary conditions for the plate model are thesame as in the honeycomb model, that is, the plate is supported atpoints E and F shown in the gure, the moment M x is applied toedges BC and DA and the moment M y is applied to edges AB andCD.

Based on the theory of thin plate, the theoretical solution of bending deection w to the problem shown in Fig. 3 is given as

w ¼ C 0 þ C 1 x2 þ C 2 y2 : ð5 Þ

Fig. 1. Geometry of regular hexagonal cell.

Fig. 2. Analyzed honeycomb model.

Fig. 3. Analyzed plate model, which imitates the honeycomb model shown in Fig. 2.

D.H. Chen / Composite Structures 93 (2011) 736–746 737


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Here, coefcients C 1 and C 2 are determined from the condition thatthe bending moments equal M x and M y in x- and y-directions,respectively. Then

D xC 1 þ D1 C 2 ¼ 1

2M x; D1 C 1 þ D yC 2 ¼

1

2M y; ð6 Þ

from which

C 1 ¼ D1 M y D yM x2 D xD y D

21

; C 2 ¼ D1 M x D xM y2 D xD y D

21

: ð7 Þ

Then the condition of the deection at point E being zero gives

C 0 ¼ C 2 y2E ; ð8 Þ

where yE is the y-coordinate of point E , yE = 50.Moreover, the honeycomb bending is also analyzed by FEM,

which is directly applied to the honeycomb model shown inFig. 2. Numerical analysis is carried out with the commercialFEM analysis package MSC.Marc. These results are then comparedwith the theoretical solution given by Eq. (5) .

3. Differences of honeycomb in-plane deformation and out-of-plane bending deformation

Using the honeycomb exural rigidity obtained by substitutingEq. (4) into Eq. (3) , deection of the thin plate shown in Fig. 3 is ob-tained from Eq. (5) . This result is then compared with the numer-ical results obtained by FEM analysis for the honeycomb modelshown in Fig. 2.

Table 1 shows the deection of point G7 (point at x = 60.63 mmon the x-axis) found from the honeycomb model using FEM analy-sis and from the thin plate model using Eq. (5) for the followingthree load cases: (1) M x ¼ 28 :33 N (distribution density of the mo-ment per unit length; same for M y), M y ¼ 0; (2) M x ¼ 0,M y ¼ 28 :33 N; (3) M x ¼ M y ¼ 28 :33 N. From the results shown inTable 1 , when M x ¼ M y as in the case of load (3), the theoretical

solution of the plate model is in good agreement with the numer-ical result of the honeycomb model; however, when the acting loadis only M x or only M y as in the case of load (1) or (2), the amount of deection for point G7 in both models differs by a factor of morethan 20. Therefore, it is apparent that the exural rigidity basedon honeycomb in-plane elastic modulus, as expressed in Eq. (3) ,is inapplicable to analysis of the honeycomb bending problem.

In order to propose an applicable calculation technique for hon-eycomb exural rigidity, we have to investigate the reason why Eq.(3) is inapplicable to the honeycomb bending problem.

Here, we compare the bending deformation caused by uniformmoments M x and M y with the in-plane deformation caused by uni-form stresses r x and r y to investigate the bending mechanics of honeycomb.

The deformation of honeycomb is analyzed using one unitshown in Fig. 4, because the honeycomb is constituted by period-ical arraying of these units.

In the following investigation, excluding Section 4, the globalcoordinate system ( x, y, z ) is used for the honeycomb as a whole,as shown in Figs. 1–4 ; however, a local coordinate system ( n,g,f )is used for each cell plate, as shown in Fig. 5. The coordinate plane

( x, y) of the global coordinate system and the plane ( n, f ) of the localcoordinate system are both located at the center of the height of the cell wall, and the z and g components are equal, z = g.

Assuming the honeycomb to be an equivalent homogeneousmaterial, the bending stress caused by the bending moments M xand M y is given as follows :

r x ¼ M x z

I h; r y ¼

M y z I h

: ð9 Þ

Here, I h ¼ h3

12 . In this case, as shown in Fig. 6, the force acting on thevertical cell wall is

qnjv ¼ 2 r yl cos h; qf jv ¼ 0 ; ð10 Þ

and forces acting on the inclined wall in the x and y directions are p xjl and p yjl.

p xjl ¼ r xlð1 þ sin hÞ; p yjl ¼ r yl cos h: ð11 Þ

Here, the subscripts v and l are used to denote the vertical wall andinclined wall, respectively, and due to the left–right symmetry weonly show the forces acting on the left inclined wall.

The components of force for the inclined wall in the n-directionand f-direction (out-of-plane direction normal to the wall plane)

can be obtained from forces p xjl and p yjl and are expressed asqnjl ¼ p xjl cos h þ p yjl sin h; qf jl ¼ p yjl cos h p xjl sin h: ð12 Þ

In the case of the in-plane deformation the force acting on cell wallsdue to an in-plane load r x and r y can also be expressed by Eqs.(10)–(12), however the force for honeycomb in-plane deformationis equal along the height direction ( g direction). Therefore, fromthe equilibrium conditions of moment around the g axis relatingto the inclined wall, moments M gjlare necessary to be generated onboth edges of the plate, as shown in Fig. 7a.

M gjl ¼ qf2 ð

13 Þ

However, for honeycomb bending deformation, the force acting onthe inclined wall are distributed linearly with the resultant forcebeing zero, as shown in Fig. 7b. With such a force distribution, themoment equilibrium around the g-axis is established. This differsfrom the in-plane deformation, as there is no need for a moment

Table 1

Deection at point G7 in honeycomb model and plate model for three cases of loadingconditions.

Load case (1) (2) (3)

FEM analysis for Fig. 2 0.91 0.87 0.045Eqs. (5) and (3) for Fig. 3 23.51 23.46 0.044

A

B

DC

A’

C’

B’

D’

x z

y

Fig. 4. One unit of honeycomb.

Fig. 5. Local coordinate system used for each cell wall.

738 D.H. Chen / Composite Structures 93 (2011) 736–746


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on the edge of the inclined wall to satisfy the moment equilibrium.Therefore, Eq. (13) for calculating the moment M gjl in the in-planeproblem does not hold for the honeycomb bending problem. In or-der to conrm this fact, moments acting on an inclined wall in thehoneycomb in-plane deformation and bending deformation areinvestigated based on numerical results of FEM analysis, as shownin Fig. 8.

The hexagon shown in Fig. 8a is located at the center of the hon-eycomb model shown in Fig. 2, and the nodes on the inclined cellwall in g = h/2 that are used in the FEM analysis are denoted bypoints E to M . For the honeycomb bending deformation causedby a moment M y and for the honeycomb in-plane deformationcaused a stress r y acting uniformly along the height direction, mo-

ments M i(i = E M ) generated at nodes E to M in g = h/2 are shownin Fig. 8b. The moments M i at each node shown on the vertical axisof the gure are divided by the moment M eq (13) dened by Eq. (13) .As shown in the gure, the moments acting on the cell wall for in-plane deformation can be evaluated by Eq. (13) (since node E isabout 0.1 l from the edge, M E =M eq:ð13 Þ ¼ 0 :40:5 ¼ 0:8). However, the mo-ments acting on the cell wall in the case of bending deformation,

excluding both edges, are almost zero and thus cannot be evalu-ated by Eq. (13) . Furthermore, in order to see the features of distri-bution of the moments along the length of the cell wall due tobending deformation, the ratio M i/M E of the moment of each pointto the moment of point E are also shown in Fig. 8b. As shown in thisgure, moments acting on the cell wall during bending deforma-tion are concentrated at the connection edges, and this distributionis conrmed to greatly differ from in-plane deformation. From thisfact, we can understand why Eq. (3) is inapplicable to the honey-comb bending problem.

The cause of moments forming in the inclined cell wall duringbending deformation is for maintaining the corner form of cell wall junction. The cell wall junction is more difcult to deform than thecentral part and can be assumed to be xed ends. This can be con-rmed in the numerical results obtained by the nite elementmethod. For the honeycomb bending deformation investigated inFig. 8b, the displacement in f-direction, namely, the deection of the inclined cell wall w is shown in Fig. 9. From this gure, theinclination in the vicinity of the junction ( n = ±l/2) decreases and

Fig. 6. Forces acting on each cell wall of a honeycomb.

(a) (b)

Fig. 7. Forces acting in f -direction on the inclined cell wall (a) in-plane deformation (b) bending deformation.

−1

0

1

M i

/ M

e q . (

1 3 ) ,

M i /

M E

MLKJIHGFEE

Mi/ME for bending deformation

Mi/Meq.(13) for bending deformation

Mi/Meq.(13) for in−plane

deformation

(a) (b)

Fig. 8. Distribution of moments acting on the inclined wall (a) positions of point E M (b) distribution of moments.



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the inclination can be assumed to approach zero at n = ±l/2. Thusfor the deection w of the cell wall we have the following bound-ary condition at the connection edges :

@ w@ n n¼ l=2

¼ 0 : ð14 Þ

Such restraint for the cell wall edges is also expected for the in-planedeformation as well as the bending deformation of honeycomb. Forthe in-plane deformation it is conrmed easily that the deection wsatises the condition of Eq. (14) at the connection edges due to ac-tion of the moment M g dened by Eq. (13) . This means that, for thedeection analysis of cell wall, the condition of giving force qf withthe moment M g = qf /2 is equivalent to a condition giving the force q f

with the boundary condition of @ w /@ njn=±l/2 = 0.Accordingly, in honeycomb bending deformation, moments arealso necessary to be generated at the connection edges of the in-clined cell wall in order to satisfy Eq. (14) ; however, since the mo-ment equilibrium is established around the n axis and g axis foronly the qf force, the values of the moment cannot be found usingEq. (13) . Therefore, in order to analyze the honeycomb bendingmachanics based on deformation of each cell wall, we have toknow the deection of a thin plate subjected to force qf underthe boundary condition of Eq. (14) .

4. Theoretical analysis of plate torsion

In this section, we analyze the deection w of the rectangular

thin plate ABCD shown in Fig. 10, for which the plate length, heightand thickness are l, h and t , respectively, and the material Young’smodulus and Poisson’s ratio are E s and ms, respectively. The plate issubjected to a linearly distributed force Q in the z -direction,

Q ¼ Q 0 y

h=2; ð15 Þ

which acts along edge AD ( x = l/2) and edge BC ( x = l/2), and as aboundary condition for the deection w the rotation angle aroundthe y-axis at both edges AD and BC is zero,

@ w@ x x¼ l=2

¼ 0 : ð16 Þ

The purpose of our analysis is to determine the rotation angle

around the x-axis of the edge BC with respect to edge AD. Here thistwisting angle is denoted by 2 ctwist .

4.1. Equilibrium equations and boundary conditions

Based on the equilibrium of force and moment, the plate deec-tion w satises the following differential equation:

@ 4 w@ x4

þ 2 @ 4 w@ x2 @ y2

þ @ 4 w@ y4

¼ 0 : ð17 Þ

Considering the symmetry of the problem, only one-fourth of theregion, namely, the BEOF region ( x P 0, y P 0 in Fig. 10), is consid-ered. The boundary conditions for this region are listed below.

(1) From the left-right and up-down symmetry, the deectionand moment at x = 0 (edge OF) and y = 0 (edge OE) are zero.

w ¼ 0 ; ð x ¼ 0 or y ¼ 0 Þ; ð18 ÞM x ¼ 0 ; ð x ¼ 0 Þ; ð19 ÞM y ¼ 0 ; ð y ¼ 0 Þ: ð20 Þ

(2) The shear force and moment at y = h/2 (edge BF) are zero.

Q y þ @ M xy

@ x ¼ 0 ; ð y ¼ h=2 Þ; ð21 Þ

M y ¼ 0 ; ð y ¼ h=2 Þ: ð22 Þ

(3) At x = l/2 (edge BE), the shear force given by Eq. (15) isapplied and the rotation angle around the y-axis is zero.

Q x þ @ M xy

@ y ¼

Q 0 yh=2

; ð x ¼ l=2 Þ: ð23 Þ

@ w@ x

¼ 0 ; ð x ¼ l=2 Þ: ð24 Þ

(4) No concentrated force acts at corner B.

M xy ¼ 0 ; ð x ¼ l=2 ; y ¼ h=2 Þ: ð25 Þ

4.2. Conguration of the solution

The deection of a rectangular plate can be determined withoutdifculty using a trigonometric series if the boundary condition issuch that the plate is simply supported along two opposite edges.Therefore, the solution to the plate shown in Fig. 10 is assumed tobe a sum of four solutions, each of which satises the equilibriumequation, as follows:

w ¼ Kxy þ w1 þ w2 þ w3 : ð26 Þ

Here, w 1, w 2 and w 3 are solutions to the following three Problems

1–3 . In all three problems, among the four edges there exist twoopposite edges being simply supported.

−0.5 0 0.50

0.002

0.004

0.006

0.008

0.01

w / t

ξ /l

Fig. 9. Deection of inclined cell wall.

x

y

z

h

t

Q

Q−

A B

D

E

F

O

C

Fig. 10. Plate subjected to twisting moment.



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Z l=2

0

@ 3 w@ y3

þ ð2 msÞ @ 3 w@ x2 @ y" #

y¼h=2

sin ðam xÞdx ¼ 0 ;

ðm ¼ 1 ; 3 ; . . .Þ: ð46 Þ

(2) By assuming only bm(m = 1,3, . . .) is non-zero, bm – 0, thevirtual deection becomes

dw ¼ ½ X m sin ðbm yÞdbm ; ð47 Þand from Eq. (44) we obtain

Z h=2

0

@ w@ x x¼l=2 sin ðbm yÞdy ¼ 0 ; ðm ¼ 1 ; 3 ; . . .Þ: ð48 Þ

(3) By assuming only c m(m = 1,3, . . .) is non-zero, c m – 0, the vir-tual deection becomes

dw ¼ ½ Z m sin ðbm yÞdc m ; ð49 Þ

and from Eq. (44) we obtain

Z h=2

0

@ 3 w@ x3

þ ð2 msÞ @ 3 w@ x@ y2" #

x¼l=2

q0 yh=28<

:9=;sin ðbm yÞdy ¼ 0 ; ðm ¼ 1 ; 3 ; . . .Þ: ð50 ÞHere,

q0 ¼ Q 0Ds

; Ds ¼ E st

3

12 1 m2s :

(4) By assuming only K is non-zero, K – 0, the virtual deectionbecomes

dw ¼ xydK ; ð51 Þ

and from Eq. (44) we obtain

Z

l=2

0

@ 3 w@ y3

þ ð2 msÞ @ 3 w@ x2 @ y" #

y¼h=2

h2

x

dx

þ Z h=2

0

@ 3 w@ x3

þ ð2 msÞ @ 3 w@ x@ y2" #

x¼l=2

q0 yh=28


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For the deection w expressed by Eq. (26) , only the solution of Kxyand w3 of Problem 3 have non-zero deection at edge BE. Fig. 11shows the distribution of deection w j x=l/2 along edge BE found withN = 49 for a plate with l = 10 mm, h = 10 mm and t = 1 mm. The ver-tical axis shows the deection wj x=l/2 divided by Khl/4, where valuesof K are obtained from the solution of the simultaneous equations.As shown in the gure, for deection wj x=l/2 , compared with Kxy, theinuence of w3 is extremely small, and along edge BE, the deectionw j x=l/2 is almost equal to ( Kl/2) y and is distributed linearly. The va-lue of ctwist calculated from Eq. (57) equals Kl/2 to three signicantgures.

4.4. Values of ctwist

It is seen from the simultaneous Eqs. (46), (48), (50) and (52)that the twisting angle ctwist can be expressed in the followingform:

ctwist ¼ qs ðh=l; msÞ Q 0E sl

l

t 3

; ð58 Þ

where coefcient qs (h/l,m) is a function of h / l and ms only.Table 2 shows the twisting coefcient qs obtained from FEM

numerical analysis using the MSC.Marc package and from the pres-

ent theoretical analysis. As shown in the table, the results of thepresent analysis agree well with the numerical results of FEManalysis.

5. Theoretical analysis of honeycomb exural rigidity

In this section, the deformation of one unit, as shown in Fig. 4, isanalyzed and the honeycomb exural rigidity is found. The bend-ing deformation of one unit is evaluated by the rotation of eachwall edge of the unit (AA 0, CC0 and DD 0) around the x-axis and y-axis.

Under the action of the moments M x and M y, there may existstwo kind of forces linearly distributed along cell wall edges. The

one is the force Q , as given in Eq. (15) , which acts in the f -direction.In this case the cell wall deformation is evaluated by the twistingangle 2 ctwist and is given by Eq. (58) shown in Section 4. The otheris an in-plane force component T , which acts in n-direction and isexpressed as

T ¼ T 0g

h=2: ð59 Þ

The in-plane bending deformation of a plate due to the in-planeforce T can be analyzed as a problem of pure bending of a beamin material mechanics. The rotation angle hT of edge n = l/2 with re-spect to edge n = l/2 is calculated as follows:

hT ¼ T 0 lE sth =2 : ð60 Þ

5.1. Case of only moment M y acting

Fig. 12 shows the rotation angle for each cell wall when onlymoment M y is acting. The sign of each rotation angle shown inthe gure is labeled as positive. The same labeling is used inFig. 13 below.

5.1.1. Vertical wall rotation angleThe vertical wall is subjected to an in-plane force T . From Eq.(10) we obtain

T 0 ¼ r yjg¼h=2 2 l cos h ¼ 2 l cos hM yh=2

I h: ð61 Þ

Substituting this equation into Eq. (60) , the rotation angle hvx of edge AA 0 with respect to edge BB 0 of the vertical wall becomes

hv x ¼ 2M yl

2 cos hEI h t

: ð62 Þ

In this case, since the out-of-plane force is Q = 0, the following holdstrue:

hv y ¼ 0 : ð63 Þ

5.1.2. Rotation angle due to in-plane bending of the inclined wallThe in-plane force T acting on the inclined wall is obtained from

Eq. (12) .

T 0 ¼ r yjg¼h=2 l cos h sin h ¼ l cos h sin hM yh=2

I h: ð64 Þ

Substituting this equation into Eq. (60) , the rotation angle hlr of theinclined wall is obtained as follows:

hlr ¼ M yl

2 cos h sin hEI h t

: ð65 Þ

Table 2

Torsion coefcients qs.

h/ l ms FEM Present theoret ical analysis

0.25 0.0 0.112 0.1100.3 0.143 0.138

0.50 0.0 0.198 0.1960.3 0.246 0.238

1.00 0.0 0.293 0.3090.3 0.344 0.344

Fig. 12. Rotation angle of each cell wall when only moment M y is acting.

Fig. 13. Rotation angle of each cell wall when only moment M x is acting.



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value of twisting coefcient qs is obtained from FEM analysis, andthe other is result for which qs is obtained from theoretical analysisdemonstrated in Section 4.

Calculations were then carried out using the D x, D y and D1 val-ues listed in Table 3 for the cases where the plate model shown inFig. 3 was acted upon by three types of load: (1) M x ¼ 28 :33N; M y ¼ 0; (2) M x ¼ 0; M y ¼ 28 :33 N; and (3) M x ¼ M y ¼ 28 :33 N.The deection for each point G1 G7 on the plate was found byusing Eq. (5) . These results are listed in Tables 4–6 . The tables alsolist the numerical results of FEM analysis of the honeycomb modelshown in Fig. 2 for comparison. As shown in the tables, the resultsof the deection analysis based on honeycomb exural rigidity thatwere determined by the proposed method in Section 5 are in goodagreement with the numerical results of FEM analysis. However,the results of the plate deection analysis based on the honeycombin-plane elastic modulus in the cases of loads (1) and (2) do not to-tally agree with the numerical results of FEM analysis. In the caseof load (3) shown in Table 6 , that is, for load conditions of M x ¼ M y,the analysis results based on the in-plane elastic modulus are inexact agreement with the analysis results of this study and agreewell with the numerical results of the FEM analysis. WhenM x ¼ M y and h = 30 , the force q f in direction normal to wall planebecomes zero, as calculated with Eq. (12) .

qf jl ¼ M yg

I hl½cos 2 h sin h sin

2h ¼ 0 ð84 Þ

If the force qf is zero, then the twisting deformation of the inclinedcell wall of the honeycomb must also be zero.

In Tables 4–6 are shown also results of the plate deection anal-ysis using the calculation method of curvature proposed by Abd El-Sayed et al. [18] . Their results are correct in general as comparedwith the results obtained from FEM analysis. In their analysis,the twisting deformation of the inclined cell wall is taken into ac-count. However, since the boundary condition at the connectionedges, namely Eq. (14) is not satised in the analysis, the resultsare always larger than the results of FEM about ten percent, asshown in Tables 4 and 5 .

7. Conclusion

In this research, the exural rigidity of a honeycomb consistingof regular hexagonal cells was investigated and the following re-sults were obtained:

(1) The honeycomb bending deformation can not be treated as aplate bending problem using the equivalent in-plane elasticmodulus.

(2) The moments acting on the honeycomb inclined cell walldiffer for honeycomb bending deformation and in-planedeformation.

(3) The honeycomb inclined cell wall undergoes twisting defor-mation for the bending deformation; however, the rotationangle at both connection edges is zero.

(4) A method for calculating the honeycomb exural rigiditybased on the bending and twisting deformation of each plate

forming the honeycomb was proposed. In the theoreticalanalysis, the torsion problem of a thin plate was solved by

Table 4

Deection at points G1 G7 in honeycomb model and plate model when M x ¼ 28 :33 N is acting.

Point FEM for Fig. 2 Theoretical analysis using Eq. (5) Abd El-Sayed et al. [18]

Eq. (83) with qs obtained from FEM Eq. (83) with qs obtained in Section 4 Eqs. (3) and (4)

G1 0.11 0.11 0.11 3.41 0.13G2 0.27 0.27 0.25 7.95 0.30G3 0.30 0.30 0.29 9.08 0.34G4 0.33 0.33 0.31 9.75 0.37G5 0.42 0.43 0.41 12.04 0.48G6 0.62 0.62 0.60 16.63 0.70G7 0.91 0.92 0.88 23.51 1.03

Table 5

Deection at points G1 G7 in honeycomb model and plate model when M y ¼ 28 :33 N is acting



G1 0.14 0.15 0.14 3.44 0.16G2 0.34 0.34 0.33 8.03 0.38G3 0.39 0.39 0.38 9.17 0.44G4 0.41 0.42 0.41 9.84 0.46G5 0.49 0.50 0.48 12.11 0.55

G6 0.64 0.65 0.62 16.65 0.72G7 0.87 0.87 0.84 23.46 0.98

Table 6

Deection at points G1 G7 in honeycomb model and plate model when M x ¼ 28 :33N and M y ¼ 28 :33N are acting.



G1 0.032 0.034 0.034 0.034 0.034G2 0.074 0.079 0.079 0.079 0.079G3 0.084 0.090 0.090 0.090 0.090G4 0.085 0.091 0.091 0.091 0.091G5 0.063 0.069 0.069 0.069 0.069G6 0.020 0.024 0.024 0.024 0.024G7 0.045 0.044 0.044 0.044 0.044



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using the generalized variational principle. Based on thenumerical results of FEM analysis, the effectiveness of theproposed method was demonstrated.

Furthermore, in order to carry out analysis of bending deforma-tion by treating a honeycomb as a thin equivalent plate, it is alsonecessary to determine the exural rigidity modulus D xy; however,

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