Minimum weight design of sandwich beams withhoneycomb core of arbitrary density

Annette MeidellNarvik University College, P.O. Box 385, N-8505 Narvik, Norway

Abstract

We consider minimum weight optimization of sandwich beams for agiven stiffness. The core consists of regular hexagonal honeycomb struc-ture of arbitrary density. We present a simple algorithm for estimatingall design parameters. Concrete examples are also given.

1 IntroductionEngineering and mathematical aspects of the homogenization theory have beenconsidered in number of paper, see e.g. [2], [3], [6] and the references therein.An elementary introduction to the homogenization method is given in the bookPersson et. al. [8].In this paper we consider minimum weight optimization of sandwich beams

with regular hexagonal honeycomb core structure of arbitrary density. Bothfacings consist of a material with Youngs modulus Ef and density ρf . The cell-wall material in the core is an isotropic material with plain strain bulk modulusK, shear modulus G and density ρ. The objective function (to be minimizedfor a given stiffness) is the total mass m, and the free variables are the face-thickness tf , the distance between the centroids of the faces d (≈ tc) and thevolume fraction of the cell-wall material v. In order to calculate the stiffness ofthe sandwich beam, we have to know the effective longitudinal shear modulusof the core G∗L. Moreover, if the core is relatively stiff, we also need to knowthe value of the effective transversal Young’s modulus of the core E∗T (i.e. theeffective Young’s modulus in the x1-direction and x2-direction). By using FEM-computations and homogenization techniques, an approximate formula for G∗Lwas obtained in [7]. Despite its simplicity it turns out that this formula is validwith error less than 0.25% for any volume fraction. Combining some recentapproximate formulae for the transversal bulk and shear moduli obtained in [1]with a general formula of Hill [4] we also obtain a relatively simple approximativeformula for E∗T with error less than 1% for any volume fraction. By playing withthese formulae we find an approximative formula for the stiffness of the beamfrom which we can deduce a simple algorithm for estimating the values of d, vand tf which minimizes the mass of the beam for a given stiffness. It turns out

1

Figure 1: Sandwich beam with regular hexagonal honeycomb core.

that the problem can be divided into 3 cases depending on the required stiffnessand the material properties. We also consider the case when constraints on dand v are added to the minimization problem.The paper is organized as follows. In Section 2 we establish all effective pa-

rameters of the effective stress-strain relation in the core. The general minimumweight problem is considered in Section 3. In Section 4 we simplify this prob-lem and present the solutions for the case when no constraints on d and v areassumed. The case of ”medium stiff” beams is studied in more detail in Section5. In Section 6 we use the general solution to consider the case when the facingsare made of carbon fiber reinforced plastic and the cell-wall material in the coreis polystyrene. Constrained problems are considered and exemplified in Section7. Finally, we have collected the proofs of our main results in an Appendix.

2

2 Stiffness properties of the coreIn the case when the effective stress-strain relation in the core is of the form⎡⎢⎢⎢⎢⎢⎢⎣

hσ11ihσ22ihσ33ihσ12ihσ23ihσ13i

⎤⎥⎥⎥⎥⎥⎥⎦ =⎡⎢⎢⎢⎢⎢⎢⎣

K∗T +G∗T K∗T −G∗T l∗ 0 0 0K∗T −G∗T K∗T +G∗T l∗ 0 0 0l∗ l∗ n∗ 0 0 00 0 0 G∗T,45 0 00 0 0 0 G∗L 00 0 0 0 0 G∗L

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣he11ihe22ihe33ihγ12ihγ23ihγ13i

⎤⎥⎥⎥⎥⎥⎥⎦(1)

(where h·i denotes the local average taken over a small representative volumeelement of the structure), we say that the effective stiffness matrix on the rightside satisfies square symmetry. The inverse of this matrix takes the form⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1E∗T

− ν∗TE∗T

− ν∗LE∗L

0 0 0

− ν∗TE∗T

1E∗T

− ν∗LE∗L

0 0 0

− ν∗LE∗L

− ν∗LE∗L

1E∗L

0 0 0

0 0 0 1G∗T,45

0 0

0 0 0 0 1G∗L

0

0 0 0 0 0 1G∗L

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦, (2)

described by the Young’s moduli E∗T , E∗L, the Poisson’s ratios ν

∗T , ν

∗L and the

shear moduli G∗T,45, G∗L. Using this, it is possible to obtain the relations

4

E∗T=

1

G∗T+

1

K∗+4 (ν∗L)

2

E∗L, (3)

4ν∗TE∗T

=1

G∗T− 1

K∗− 4 (ν

∗L)2

E∗L. (4)

Moreover, using the exact same approach as Hill [4] did for the the case G∗T =G∗T,45 and utilizing the fact that the core is built up by only one material, wefind that ν∗L = ν and E∗L = vE. For such composite structures it is also possibleto show that

1

K∗T=

1

K+A1

µ1

K+1

G

¶, (5)

1

G∗T=

1

G+A2

µ1

K+1

G

¶, (6)

1

G∗T,45=

1

G+A3

µ1

K+1

G

¶, (7)

where A1, A2 and A3 are constants which only depend on the geometry of thestructure (see Vigdergauz [9]). By (3), (5) and (6), we obtain that

4

E∗T= (A1 +A2 + 1)

µ1

K+1

G

¶+4ν2

vE.

3

Recalling that G and K are related to the Young’s modulus E and Poisson’sratio ν by the formulae

K =E

2 (1 + ν) (1− 2ν) , G =E

2 (1 + ν), (8)

we finally obtain that

E∗T =E

(A1 +A2 + 1) (1− ν2)− ν2

v

. (9)

For hexagonal honeycombs the following approximate formulae were presentedin [1]:

A1 ≈Ã√

3

2

µl

t

¶− 12

!Ã1− 0.099263

µt

l

¶− 0.27239

µt

l

¶2!, (10)

A2 ≈

⎧⎪⎪⎪⎨⎪⎪⎪⎩³√

34

³¡lt

¢3+¡lt

¢´− 12

´(1− 0.31343 ¡ tl ¢+

0.36676¡tl

¢2+ 0.22223

¡tl

¢3)

if 0 ≤ tl ≤ 0.957,

1.235¡tl − 1.5305

¢ ¡tl − 1.732

¢if 0.957 ≤ t

l ≤ 1.732,(11)

where the relative cell-wall thickness t/l (see Figure 1) is related to the volumefraction v by the formula

t

l=√3¡1−√1− v

¢. (12)

Both of these approximation formulae have proven to hold with maximumerror less than 1% for all volume fractions. This is verified by performing FEM-computations with very high accuracy for a large number of volume fractions.Moreover, arguing exactly as in [1] concerning the effective bulk and shear mod-uli we are able to show that we obtain almost the same accuracy for E∗T if wereplace A1 and A2 in (9) by (10) and (11). We may certainly also obtain anexplicit formula for K∗T , G

∗T and ν∗T /E

∗T with a similar accuracy by making use

of (4), (5), (6), (10) and (11).By using FEM-computations and homogenization techniques, the following

approximate formula for hexagonal honeycombs was obtained in [7]:

G∗L ≈52Gv

(2− v) (−4v2 + 4v + 52) . (13)

Despite its striking simplicity it turns out that this formula is valid with errorless than 0.25% for any volume fraction.Since the core structure in Figure 1 possesses hexagonal symmetry, the

effective elastic properties of the structure become transversely isotropic, i.e.G∗T,45 = G∗T in (1) (see e.g. [5]). We can therefore obtain explicit formulae forall elements in the compliance matrix (2) and certainly also for all elements ineffective stiffness matrix in (1).

4

3 The general minimum weight problemWe recall that the maximum deflection of the sandwich beam δ is the sum ofdeflection due to pure bending and pure shear deformation (see e.g. [10]), i.e.

δ = δb + δs, (14)

where

δb =PL3

B1 (EI)eqand δs =

PL

B2 (AG)eq.

Here, P is the fource, L is the lenght of the beam and B1 and B2 are positiveconstants depending on the loading- and boundary conditions. The equivalentflexural rigidity (EI)eq is given by

(EI)eq =Efbt

3f

6+

E∗T bt3c

12+

Efbtfd2

2(15)

and the equivalent shear rigidity is given by

(AG)eq =bd2G∗Ltc

. (16)

Expressed in terms of tf , tc and v we obtain that the compliance of the beam(the inverse of the stiffness) δ/P multiplied with b (the width) is given by

δb

P=

L3

B1

³Ef t3f6 +

E∗T (v)t3c12 +

Ef tf (tc+tf )2

2

´ + L

B2(tc+tf )

2G∗L(v)tc

, (17)

where the expressions for E∗T (v) and G∗L(v) are given in the previous section.The compliance δ/P and all other parameters in this expression are assumedto be fixed except for tc, tf and v. Our task is to minimize the mass m, orequivalently m/b (since b is fixed):

m

b= 2ρfLtf| {z }

facings

+ vρLtc| {z }core

(18)

with the free variables tc > 0, tf > 0 and 0 < v ≤ 1, under the constraint (17).This task is a matter of constraint nonlinear programming, and a numericalsolution can be found by choosing appropriate computational methods. Suchcomputations are not within the scope of this paper. Instead, we will considera simplified constraint minimization problem for which the solution (tc, tf , v)is obtained quite easily. Inserting this approximate solution into (17) and (18)enables us to estimate the accuracy of our calculations.

5

4 The simplified minimum weight problemIf tf << tc and tcEc << tfEf we have that (EI)eq can be reduced to

(EI)eq ≈Efbtfd

2

2. (19)

Moreover, if tf << tc, the shear rigidity clearly reduces to

(AG)eq ≈ bdG∗L. (20)

If we insert (19) and (20) into (14) we obtain that the compliance of the beamδ/P is given by

δ

P=

2L3

B1Efbtfd2+

L

B2bdG∗L, (21)

from which we can deduce that

tf =2L3

B1Efd2³b δP − L

B2dG∗L

´ . (22)

Moreover, the fact that ¯̄̄̄52

−4v2 + 4v + 52 − 1¯̄̄̄< 0.02

for 0 ≤ v ≤ 1 motivates the following approximation of G∗L given in (13):

G∗L = Gv

2− v

(which coincide with the Hashin-Strikman upper bound for the shear modulus).Thus, the weight per unit depth w = m/b is given by

w =m

b= 2ρfLtf + vρLd = 2ρfL

⎛⎜⎜⎝ 2L3

B1Efd2µb δP − L

B2(dG∗L)

¶⎞⎟⎟⎠+ vρLd =

1

d2B1Ef4ρfL

4

¡b δP¢− d

B1Ef4ρfL

3B2Gv

2−v

+ vρLd. (23)

The minimum weight solutions v, d (and hence also tf by using (22)) arefound as follows (for verification, see the Appendix).

Case 1 If

bδ

P<3

8

sB1Efρ

2ρfB32G

3

6

then d = Ld1, where d1 is the only solution to the fourth order equation

−2µbδ

P

¶d1 +

1

B2G+

B1Efρ

4ρf

µµbδ

P

¶d21 −

1

B2Gd1

¶2= 0 (24)

satisfying µbδ

P

¶d1 − 1

B2G> 0.

In this case v = 1.

Case 2 If3

8

sB1Efρ

2ρfB32G

3≤ b

δ

P<1

2

sB1Efρ

2ρfB32G

3(25)

then

d =Lq

B1Efρ2B2ρfG

− 2B2G¡b δP¢ ,

and

v = 4

Ã1− 2

s2B2ρfG

B1EfρB2G

µbδ

P

¶!. (26)

In this case, d ranges from

bd = 4LqB1Efρ2ρfB2G

(corresponding to v = 1) and ∞ (corresponding to v = 0).

Case 3 If1

2

sB1Efρ

2ρfB32G

3< b

δ

P

then the function wopt(·) defined bywopt(d) = min

0≤v≤1w(v, d),

decreases in its whole region of definition and converges to its limit

2ρL2

B2G¡b δP¢

as d→∞. Moreover, the volume-fraction

v =1¡

b δP¢dL +

1B2G

ÃL

d

s8ρf

B1EfρB2G+

2

B2G

!(27)

vanishes as d→∞. Consequently, there are no minimum-solutions in thiscase.

7

Remark 1 For Case 2, we have the simple relation

d =bdv

(28)

(see Appendix). This shows that the combination of d and v making the weightw minimal when bδ/P belongs to the interval (25) are those making mc/b equalto bdρL, where mc is the mass of the core material, i.e.

mc

b= ρLdv =

4L2ρqB1Efρ2ρfB2G

= 4L2

s2ρρfB2G

B1Ef.

It is interesting to note that this value is independent of d and v, and maybemore interesting, that it is independent of bδ/P , i.e. it is independent of therequired stiffness of the beam.

Remark 2 We may certainly also minimize the total cost of the sandwich byreplacing ρf and ρ with ρfcf and ρc, respectively, where cf and c are the costsper kg of the face material and the cell-wall material, respectively.

5 On Case 2Let us discuss some other aspects of Case 2. From (26) we have that

bδ

P=

sB1Efρ

2B2ρfG

1

2B2G

³1− v

4

´.

Hence, by (22) and the relation (28)

d =4L

vq

B1Efρ2ρfB2G

, (29)

we obtain that

tf =Lv

2q

B1Efρf2ρB2G

. (30)

Substituting v with (26) we also obtain

tf = 4L

ÃsρB2G

2B1Efρf− 2

¡b δP¢G2B2

2

B1Ef

!.

Moreover, by (29) and (30) we find the simple relation

d

tf=8

v2ρfρ. (31)

8

We now get that the mass of facings mf (divided with b) is given by

mf = 2tfLρf =2L2vρfq2B1EfρfρB2G

= L2v

s2ρρfB2G

B1Ef.

Recalling that the mass of the core is given by

mc

b= 4L2

s2ρρfB2G

B1Ef,

we obtain thatmf

mc=

v

4.

Let us now say some words about the validity of our simplified model. It iseasy to see that the first term of (15) is less than p % of the second term if

100

3

µtfd

¶2< p.

According to (31) this condition is satisfied if

100

3

µv2ρ

8ρf

¶2< p. (32)

The third term of (15) is less than p % of the second term if

E∗T t3c

Ef tfd2100

6< p. (33)

Using that tc = d− tf we obtain from (31) the relation

E∗T t3c

Ef tfd2100

6=

E∗T

µ³8ρfv2ρ − 1

´3t3f

¶Ef tf

³³8ρfv2ρ

´tf

´2 100

6=

E∗T³8ρfv2ρ − 1

´3Ef

³8ρfv2ρ

´2 100

6.

Thus, in order to obtain that the third term of (15) is less than p % of thesecond term, we must fulfill the following condition:

E∗T³8ρfv2ρ − 1

´3Ef

³8ρfv2ρ

´2 100

6< p. (34)

Finally, we have that the error obtained by using (15) is less than p % if

100tfd

< p,

i.e.

100v2

8

ρ

ρf< p. (35)

9

6 An exampleLet us consider the case when both ends of the beam are built in. For this caseB1 = 192 and B2 = 4. We assume that the facings are made of carbon fiberreinforced plastic CFRP with Youngs modulus Ef = 138 GN/m2 and densityρf = 1600 kg/m3. The cell-wall material in the core is assumed to be madeof Polystyrene PS with Youngs modulus E = 3.370 GN/m2, density ρ = 1330kg/m3 and Poissons ratio ν = 0.34; hence with corresponding shear modulus

G =E

2(1 + ν)= 1.2575 GN/m2.

Let us first calculate the dimensionless parametersB1Efρ

2ρfB2G=

s(192) (138) (1330)

2 (1600) (4) (1.2575)= 46.79.

By inserting the above parameters into (25) we obtain ”Case 2” if

3.488 m2/GN ≤ bδ

P≤ 4.651 m2/GN.

For this case we obtain from (26), (29) and (30) the relations

v = 4− 0.86µbδ

P

¶,

d =L

v (11.698),

and

tf =Lv

93.58.

For example, if bδ/P = 4 m2/GN,

v = 0.56, d =L

6.5509, tf =

L

167.11.

Using (12), (11), (10) and (9), in that order, we find that

E∗T = 0.95365 GN/m2.

For these values (32), (34) and (35) are satisfied for p = 0.05%, p = 2.6% andp = 3.9%, respectively.For the case

bδ

P≤ 3.488 m2/GN

we obtain by (24) that d = Ld1, where d1 is the solution of the equation

−2µbδ

P

¶d1 + 0.198 81 + 5506.2

µµbδ

P

¶d21 − 0.198 81d1

¶2= 0 (36)

10

Figure 2: The optimal value of d/L versus the compliance bδ/P in the case when thefacings are made of CFRP and the cell-wall material in the core is made of polystyrene.

satisfying µbδ

P

¶d1 − 0.198 81 > 0.

In Figure 2 we have plotted the curve of d/L as function of bδ/P. This is doneby solving (24) for a number of values of bδ/P ≤ 3.488 and by plotting thefunction

d

L=

1¡4− 0.86 ¡b δP ¢¢ (11.698) (37)

in the interval 3.488 ≤ bδ/P ≤ 4.651.

7 Adding constraintsWhen the compliance

bδ

P<1

2

sB1Efρ

2ρfB32G

3,

it appears that bd is the smallest obtainable value of dmaking the weight minimal(see the Appendix). If the relative thickness

bdL=

4qB1Efρ2ρfB2G

is too large, we must consider to use a different design than a sandwich beam,e.g. a truss structure. A different way is to reduces the relative thickness by

11

adding a constraint on d of the type

d ≤ dmax.

7.1 Constraints on d

From our discussion at the end of the Appendix we obtain the following mini-mum weight solutions v, d (and hence also tf by using (22)).

Case 1 Let

bδ

P<3

8

sB1Efρ

2ρfB32G

3.

If

−2µbδ

P

¶dmaxL

+1

B2G+

B1Efρ

4ρf

õbδ

P

¶µdmaxL

¶2− 1

B2G

µdmaxL

¶!2(38)

is positive and µbδ

P

¶dmaxL− 1

B2G> 0,

then d = Ld1, where d1 is the only solution to the fourth order equation

−2µbδ

P

¶d1 +

1

B2G+

B1Efρ

4ρf

µµbδ

P

¶d21 −

1

B2Gd1

¶2= 0 (39)

satisfying µbδ

P

¶d1 − 1

B2G> 0.

Otherwise, d = dmax. For both alternatives, v = 1.

Case 2 If3

8

sB1Efρ

2ρfB32G

3≤ b

δ

P<1

2

sB1Efρ

2ρfB32G

3,

then

d = min

⎧⎨⎩dmax,Lq

B1Efρ2B2ρfG

− 2B2G¡b δP¢⎫⎬⎭

and

v = min

(1,

1¡b δP¢dL +

1B2G

ÃL

d

s8ρf

B1EfρB2G+

2

B2G

!). (40)

12

Case 3 If1

2

sB1Efρ

2ρfB32G

3< b

δ

P,

then we put d = dmax (due to the fact that wopt(·) is decreasing in itsregion of definition) and

v = min

(1,

1¡b δP¢dL +

1B2G

ÃL

d

s8ρf

B1EfρB2G+

2

B2G

!).

7.2 Constraints on d and v

Due to manufacturing reasons, v can certainly not be chosen arbitrarily small.A thin walled honeycomb-core also gives rise to several failure mechanisms, likecell-wall buckling, face wrinkling etc. In practice, for Case 2 and 3 we thereforehave to make sure that the value of v which comes out of the above algorithmis larger than a safe value vmin > 0.By the discussion at the end of the Appendix we find that adding both

constraintsd ≤ dmax and v ≥ vmin,

gives a solution (d, v) to the minimum-mass problem which simply is the samesolution as that given in Subsection 7.1, provided vu ≥ vmin. Otherwise it is thepair (d0, vmin) where

d0 = min {dmax, Ld1}and d1 is the only solution to the fourth order equation

−2µbδ

P

¶d1 +

2− vminB2Gvmin

+B1Efρvmin

4ρf

µµbδ

P

¶d21 −

2− vminB2Gvmin

d1

¶2= 0 (41)

satisfying µbδ

P

¶d1 − 2− vmin

B2Gvmin> 0.

Note that dmax ≥ Ld1 if

−2µbδ

P

¶dmaxL+2− vminB2Gvmin

+B1Efρvmin

4ρf

õbδ

P

¶µdmaxL

¶2− 2− vmin

B2Gvmin

dmaxL

!2≥ 0

(42)and µ

bδ

P

¶dmaxL− 2− vmin

B2Gvmin> 0.

Otherwise, dmax < Ld1.

13

7.3 Example

We consider the same example as in Section 6, but add the constraints d ≤dmax = L(1/6.5509) and v ≥ vmin = 0.08. It is clear that d/L will be constantand equal to dmax/L = 1/6.5509 ≈ 0.15 in two intervals. There are two valuesof bδ/P where d/L change from this constant value, one in the interval 0 ≤bδ/P ≤ 3.488 and another in the interval 3.488 ≤ bδ/P ≤ 4.651 (see Figure 3).The first one is found by solving (36) with respect to bδ/P for d1 = 1/6.5509,i.e.

−2µbδ

P

¶µ1

6.5509

¶+0.199+(5506)

õbδ

P

¶µ1

6.5509

¶2− 0.199

µ1

6.5509

¶!2= 0,

(43)µbδ

P

¶1

6.5509− 0.199 > 0,

which gives bδ/P = 1.6163. The second one is found by solving (37) with respectto bδ/P for d/L = 1/6.5509, i.e.

1

6.5509=

1¡4− 0.86 ¡b δP ¢¢ (11.698) ,

which gives bδ/P = 4.The expression (40) is now written as follows:

v = min

(1,

Ld 8.497 8 + 8.283 6

103¡b δP¢dL + 4.141 8

). (44)

The value of bδ/P making v = 0.08 (in the region where d/L = 1/6.5509 ) isfound from the equation

(6.5509) 8.4978 + 8.283 6

103¡b δP¢ ¡

16.5509

¢+ 4.141 8

= 0.08,

i.e.63.952¡

b δP¢152.65 + 4.141 8

= 0.08,

which gives

bδ

P= 5.2096.

For bδ/P > 5.2096 we have that d/L = min {1/6.5509, d1} , where d1 is foundby solving (41), which now takes the form

−2µbδ

P

¶d1 + 4.7714 + 440.50

µµbδ

P

¶d21 − (4.7714) d1

¶2= 0, (45)µ

bδ

P

¶d1 − 4.771 4 > 0. (46)

14

Figure 3: The optimal value of d/L versus the compliance bδ/P when the constraintsd/L < 0.15 and v > 0.1 are added, the facings are made of CFRP and the cell-wallmaterial in the core is made of polystyrene.

Putting d1 = 1/6.5509 into (45) and (46) we find the value of bδ/P making theleft side of (45) change sign from negative to positive, namely bδ/P = 36.407.Adding the results obtained in Section 6 we obtain the following solution

(d, v) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(0.15, 1) if 0 ≤ bδ/P ≤ 1.61,(d0, 1) if 1.61 ≤ bδ/P ≤ 3.49,µ

1

(4−0.86(b δP ))(11.698), 4− 0.86 ¡b δP ¢¶ if 3.49 ≤ bδ/P ≤ 4.00,µ

0.15, 63.952

(b δP )152.65+4.141 8

¶if 4.00 ≤ bδ/P ≤ 5.21,

(0.15, 0.08) if 5.21 ≤ bδ/P ≤ 36.41,(d00, 0.08) if 36.41 ≤ bδ/P,

where d0(= d1) is the solution of (36) and d00(= d1) is the solution of (45).

Acknowledgements. I thank Professor Dag Lukkassen for some adviseswhich have improved the mathematical presentation of my results.

References[1] G. Beeri, D. Lukkassen, A. Meidell, Estimating the Vigdergauz constants

for regular hexagonal honeycombs. To appear in: Composites Part B.

[2] S. Berggren, D. Lukkassen, A. Meidell and L. Simula, On stiffness proper-ties of square honeycombs and other unidirectional composites, CompositesB. 32, 6, 503-511, 2001.

[3] J. Bystrøm, J. Helsing and A. Meidell, Some computational aspects of it-erated structures, Composites Part B., Vol 32/6, 485-490, 2001.

15

[4] R. Hill, Theory of mechanical properties of fibre-strengthened materials-I.Elastic behaviour, Journal of the Mechanics and Physics of Solids, Vol. 12,199-212, 1964.

[5] L.D. Landau and E.M. Lifshitz, Theory of elasticity. Pergamon Press, Ox-ford, 1986.

[6] D. Lukkassen, L.-E. Persson, P. Wall, Some engineering and mathematicalaspects on the homogenization method. Composites Engineering 5, 5, 519-531, 1995.

[7] A. Meidell, The out-of-plane shear modulus of two-component regular hon-eycombs with arbitrary thickness. In: Mechanics of Composite Materialsand Structures (eds. C.A. Mota Soares, C.M. Mota Soares and M.J.M.Freitas), NATO ASI, Troia, Portugal, Vol. III, 367-379, 1998.

[8] L.E. Persson, L. Persson, N. Svanstedt and J. Wyller, The homogenizationmethod: An introduction, Studentlitteratur, Lund, 1993.

[9] S. Vigdergauz, Complete elasticity solution to the stress problem in a pla-nar grained structure. Mathematics and Mechanics of Solids 4(4), 407-441,1999.

[10] D. Zenkert, An Introduction to Sandwich Construction, EMAS, 1995.

AppendixLet us now verify the formulae for the minimum values stated in Section 4. By(23) we have that

w =1

rd2 − sv

2−vd+ kvd, (47)

where

r =B1Ef

4ρfL4

µbδ

P

¶, s =

B1Ef

4ρfL3B2G

, k = ρL, (48)

i.e.w =

v

d (v (rd+ s)− 2s) + kvd. (49)

Noting that v/ (d (v (rd+ s)− 2s)) is the mass of the facings, we obtain thatd (v (rd+ s)− 2s) > 0, i.e. that 2s/ (rd+ s) < v. Therefore, by the expression

∂w

∂v=

−2dsd2 (v (rd+ s)− 2s)2 + kd,

we find that the only possible 0-value for ∂w/∂v is obtained when

d (v (rd+ s)− 2s) =r2s

k, (50)

16

i.e.

v =1

rd+ s

Ã1

d

r2s

k+ 2s

!. (51)

We therefore find that

∂w

∂v=

⎧⎪⎪⎪⎨⎪⎪⎪⎩< 0 if 0 < d (v (rd+ s)− 2s) <

q2sk ,

0 if d (v (rd+ s)− 2s) =q

2sk ,

> 0 ifq

2sk < d (v (rd+ s)− 2s) .

(52)

In addition, we must satisfy the condition v ≤ 1. Hence, using (49), (50) and(51) , we obtain that for a fixed d, the minimum value of w(·, d), denoted wopt(d),is given by

wopt(d) =

⎧⎨⎩1

d((rd−s)) + kd if 0 < d (rd− s) <q

2sk ,

1+2d√2sk+2skd2

d(rd+s) =(√2skd+1)

2

rd2+sd if d (rd− s) ≥q

2sk .

(53)

Differentiating we obtain that

w0opt(d) =

⎧⎨⎩−(2rd−s)d2((rd−s))2 + k if 0 < d (rd− s) <

q2sk ,

(√2skd+1)(d(s

√2sk−2r)−s)

(rd2+sd)2if d (rd− s) ≥

q2sk .

(54)

Due to the fact that

− (2rd− s)

d2 ((rd− s))2 =

−rd− (rd− s)

d2 ((rd− s))2 = −

Ãr

d ((rd− s))2 +

1

d2 (rd− s)

!, (55)

we clearly see that the first expression in the piecewise-defined function (53),

− (2rd− s)

d2 ((rd− s))2 + k,

is increasing in d. This shows that the first expression in (53),

1

d ((rd− s))+ kd,

has at most one local minimum and no local maxima in that interval.It is clear from the second expression in (54) that the second expression in

(53), ³√2skd+ 1

´2rd2 + sd

, (56)

17

is decreasing in d if s√2sk − 2r ≤ 0. Similarly, if s√2sk − 2r > 0, i.e.

1

2s√2sk > r, (57)

we obtain thatd =

s

s√2sk − 2r (58)

is the only positive local minimum-point of (56). On the other hand, (58) hasto be in the range of definition of (56) in order to be a local minimum of wopt ,i.e. we must have that d (rd− s) ≥p2s/k. This condition is satisfied for

r ≥ 3s√2sk

8. (59)

Indeed, if (59) is satisfied, then

d =s

s√2sk − 2r ≥

4√2sk

.

Hence,

d (rd− s) ≥ 4√2sk

(3s√2sk

8

4√2sk− s) =

r2s

k.

We may therefore conclude that (58) is the only local minimum of the secondexpression in (53) in its range of definition if (57) and (59) are satisfied, i.e.

1

2s√2sk ≥ r ≥ 3

8s√2sk, (60)

and non minimum-points in its range of definition if r is outside this interval.We also note that in this interval, we obtain from (51) that

v =1

rd+ s

Ã1

d

r2s

k+ 2s

!=

1

r³

ss√2sk−2r

´+ s

Ã1s

s√2sk−2r

r2s

k+ 2s

!=

s√2sk − 2r

s³s√2sk − r

´ Ã2s√2sk − 4r√2sk

+ 2s

!=

s√2sk − 2r

s³s√2sk − r

´ ³s√2sk − r´ 4√

2sk=

s√2sk − 2rs

4√2sk

,

i.e.

v =s√2sk − 2rs

4√2sk

.

18

Thus we find the simple relation

v =4

d√2sk

.

For the crucial case d (rd− s) =p2s/k, i.e. when

d =1

2r

⎛⎝s+

ss2 + 4r

r21

ks

⎞⎠ ,

the first expression in (54) takes the value

− (2rd− s)

d2 ((rd− s))2 + k =

− (2rd− s)2sk

+ k =

−Ã2r 12r

Ãs+

rs2 + 4r

q2 1ks

!− s

!2sk

+ k =−rs2 + 4r

q2 1ks

2sk

+ k.

Thus if r ≥ (3/8) s√2sk we obtain that

− (2rd− s)

d2 ((rd− s))2+ k ≤

−krs2 + 4

³38s√2sk

´q2 1ks

2s+ k = 0.

Combined with the fact that the first expression in (54) is increasing in d (seeabove), this shows that for r ≥ (3/8) s√2sk, the first expression in (53) has nominimum-values in its range of definition. Conversely, if r < (3/8) s

√2sk then

we obtain that − (2rd− s)

d2 ((rd− s))2+ k > 0 (61)

for d (rd− s) =p2s/k, and since the left side obviously is negative for small val-

ues of d (rd− s), it follows similarly that it has exactly one zero, and hence, thefirst expression in (53) has exactly one minimum-point in its range of definition.

Summing up, we obtain the following solutions depending on the value of r:If r < (3/8) s

√2sk then v = 1 and d is the only solution to the equation

− (2rd− s)

d2 ((rd− s))2+ k = 0 (62)

satisfying d (rd− s) > 0.If (3/8) s

√2sk ≤ r < (1/2) s

√2sk then

d =s

s√2sk − 2r . (63)

19

and

v =s√2sk − 2rs

4√2sk

.

If (1/2) s√2sk ≤ r then wopt(·) is decreasing in its whole region of definition

and converges to its limit 2sk/r as d→∞. Moreover, the volume-fraction

v =1

rd+ s

Ã1

d

r2s

k+ 2s

!(64)

vanishes as d→∞.Using (48) we find that (62) can be written as follows

−2µbδ

P

¶d+

L

B2G+ d2

B1Efρ

4ρfL3

µµbδ

P

¶d− L

B2G

¶2= 0.

It is easy to see that the solution d = d1L, where d1 is the solution correspondingto the case L = 1, i.e.

−2µbδ

P

¶d1 +

1

B2G+

B1Efρ

4ρf

µµbδ

P

¶d21 −

1

B2Gd1

¶2= 0,µ

bδ

P

¶d1 − 1

B2G> 0.

From (55) we note that if r < (3/8) s√2sk and the compliance bδ/P is

reduced, i.e. that

r =B1Ef

4ρfL4

µbδ

P

¶is reduced, then the minimum-point d has to increase (in order to make theleft side of (55) maintain its value −k). On the other hand, if (3/8) s√2sk ≤r < (1/2) s

√2sk and the compliance bδ/P reduces, then the minimum-point d,

given by the formula (63), reduces. This shows that for r < (1/2) s√2sk the

smallest obtainable value of d making the weight minimal is

bd = 4√2sk

,

which is obtained for r = (3/8) s√2sk.

If we add the constraint d ≤ dmax, where dmax is a fixed positive number,it is possible to show that there always exists a solution denoted (d0, v0) to theminimum-problem for all positive values of r. This solution is simply the pair(du, vu) obtained in the unconstrained case provided du ≤ dmax, otherwise it isthe pair (dmax, v0) where

v0 = min

(1,

1

rdmax + s

Ã1

dmax

r2s

k+ 2s

!).

20

This formula is easily seen by (52). Note that for the case r < (3/8) s√2sk we

do not need to solve the fourth other equation (62) in order to see whether thissolution is less than dmax or not. Since the left side of (62) increases with d, itis enough to evaluate the sign of

− (2rdmax − s)

d2max ((rdmax − s))2 + k.

If this value is negative, then dmax < du, and if this value is positive, thendmax > du.If we in addition add the constraint v ≥ vmin > 0, there also exists a solution

denoted (d00, v00) to the minimum-mass problem. This solution is simply the pair(d0, v0) obtained above provided vu ≥ vmin, otherwise it is the pair (d0, vmin)where

d0 = min {dmax, dα}and d = dα is the value making (47) minimal for v = vmin. This value is foundexactly as we found the solution of (62) except that we have to replace s with(2− v) s/v and k with kv.More precisely, dα is the only solution to the equation

− (2rdα − sα)

d2α ((rdα − sα))2 + kα = 0 (65)

satisfying dα (rdα − sα) > 0, where sα = (2− vmin) s/vmin and kα = kvmin.Similarly as above we do not need to solve (65) in order to see whether thissolution is less than dmax or not, and we find that dmax < dα if

− (2rdmax − sα)

d2max ((rdmax − sα))2 + kα

is negative and vice versa.

21