A Linear Elastic Model for the In-Plane Behavior of a Honeycomb Structure

A linear elastic model for the in-plane behavior of a honeycomb structure

G.J. Oosterbaan (ID: 0543487)

MT 08.28

Bachelor Thesis Report

Supervisor: dr.ir. J.A.W. van Dommelen

Eindhoven University of Technology Department of Mechanical Engineering Section Mechanics of Materials

Eindhoven (Netherlands), July 2008 Abstract

The in-plane behavior of regular hexagonal honeycomb structures will be discussed in this report. In chapter 2 a homogenization procedure is used to determine the plane stress compliance matrix which gives a linear relation between strain and stress. For the determination of the compliance matrix basic beam bending equations are used. These basic beam bending equations use the assumptions of Bernoulli/Euler/Kirchhoﬀ. The compliance matrix is only valid in the linear elastic regime. The linear elastic regime is given by the initial yield surface shown in section 3.1. Initial yielding will occur if the ﬁrst material point reaches the yield stress. If the total cross section of a cell wall reaches the yield stress plastic collapse will occur. The plastic collapse yield surface is given in section 3.2. The initial yield surface has the shape of a parallelogram the plastic collapse yield surface has an ellipse shape. These shapes are due the bending dominated character for uniaxial loading and the tensile character for equi-biaxial loading. For validation purposes and visualization of the Von-Misses stress distribution there are done some numerical simulations in chapter 4. From these numerical experiments it can be concluded that for high slenderness ratios the compliance matrix is accurate. Furthermore it can be seen that the locations with the highest Von-Misses stresses are near the cell wall intersections.

Preface/Acknowledgements

The demand for a material model of a hexagonal honeycomb structure is initialized by the University Racing Eindhoven Team (URE). The URE uses honeycomb sandwich panels as a body for their race car. Because the URE has switched to a carbon ﬁber body in 2008 there was no demand from their side anymore. Because there was no more demand from the URE this report has a more scientiﬁc and theoretical approach.

Furthermore I would like to thank my coaches Varvara Kouznetsova and Hans van Dom- melen from the Mechanics of Materials Department of the TU/e for their inspiration and patience with this research.

Contents

1 Introduction 7

2 Linear elastic model for in-plane loading 9 2.1 Homogenization procedure ...... 9 2.1.1 Constitutive relations ...... 9 2.1.2 Applying constitutive relations to unit cell ...... 11

3 Yielding under biaxial loading 15 3.1 Initial Yielding ...... 16 3.2 Full plasticity collapse ...... 17

4 Numerical simulations 21 4.1 Modeling in Marc-Mentat ...... 21 4.2 Results ...... 27 4.2.1 Subproblem 1: loading in 11-direction ...... 27 4.2.2 Subproblem 2: loading in 22-direction ...... 28 4.2.3 Subproblem 3: shear loading ...... 28 4.3 Von Mises Stress distribution ...... 30

5 Conclusion 35

A Constitutive relations 37 A.1 Derivation of energy density ...... 37 A.2 Derivation of strain tensor ...... 38 A.3 Derivation of stress tensor ...... 38

B Point symmetry and displacement relations 41 B.1 Subproblem 1 ...... 41 B.2 Subproblem 2 ...... 41 B.3 Subproblem 3 ...... 42

C Displacements as a function of the stress σ 45 C.1 Subproblem 1 ...... 46 C.2 Subproblem 2 ...... 47 C.3 Subproblem 3 ...... 49

3 4 List of Symbols

Scalars symbol description dimension c distance neutral bending line of cell wall [m] d size of a cell [m] fi force in ii-direction [N] h cell wall height [m] I second moment of area [m4] L size RVE [m] l cell wall length [m] M bending moment [Nm] Mp moment of full plasticity [Nm] ds(k) unit cell surface area belonging to point (k) [m2] t cell wall thickness [m] (k) ui displacement point (k) in ii-direction [m] W energy density [J/m2] V unit cell volume [m3] εij macroscopic strain element in ij-direction [-] ν Poisson’s ratio honeycomb [-] νs Poisson’s ratio solid material [-] σij macroscopic stress element in ij-direction [Pa] σa axial stress in cell wall [Pa] σb,max maximum bending stress in cell wall [Pa] σys yield stress material [Pa] ∗ σiy uniaxial initial macroscopic yield stress [Pa] ∗ σpl uniaxial full plasticity macroscopic yield stress [Pa]

Columns symbol description dimension ~e orthonormal base vector column [-] f˜ force column [N] u˜ displacement column [m] ˜ε strain column [-] σ˜ stress column [Pa] ˜ 5 Vectors symbol description dimension f~(k) force vector of point (k) [N] ~n(k) normal vector of point (k) [m] ~p(k) stress vector of point (k) [Pa] ~u(k) displacement vector of point (k) [m] ~x(k) coordinate vector from the origin O to point (k) [m] ~0 zero vector [-]

Tensors symbol description dimension ε strain tensor [-] σ stress tensor [Pa]

Matrices symbol description dimension C elasticity matrix [Pa] R rotation matrix [-] S compliance matrix [1/Pa]

Mathematical Operators symbol description Pm k=1 sum over the elements k=1,2,3,...,m · inner product × crossproduct : double inner product aT transpose of a ˜ ˜

6 Chapter 1

Introduction

In this report a linear elastic in-plane honeycomb material model will be given. The model will be for a regular hexagonal honeycomb structure. In figure 1.1.a a beehive honeycomb is shown and in figure 1.1.b a man made aluminium honeycomb is shown. Because nature has a tendency for mechanically efficient materials the honeycomb structure can be found in a lot of natural cellular materials as can be seen in figure 1.2. The man made honeycomb structures are commonly used as the core material of honeycomb sandwich panels or as kinetic energy absorbers. In structural applications the material should remain in the elastic regime, whereas for energy absorbtion plastic deformation is needed. Honeycomb structures are not just used in sandwich panels or as energy absorbers but have many other applications such as air directionalization, thermal panels, acoustic panels, light diffusion and radio frequency shielding. The main advantage of honeycomb panels is its high stiffness and bending strength combined with a low weight. The in-plane behavior of Honeycomb structures has to be studied to know the contribution of the honeycomb core to the total panel strength. Also cellular solids are often approximated by this linear elastic honeycomb model. This report will focus on the linear elastic behavior for structural applications. The main parameters of interest are:

(1) a relationship between the applied stress on the honeycomb and the resulting macroscopic strains; (2) the maximal stress that can be applied on the honeycomb before yielding occurs; (3) the maximal stress that can be applied on the honeycomb before total collapse occurs.

Chapter 2 will be devoted to (1). A homogenization technique is used to derive the compliance matrix S which gives a relation between the macroscopic strain and macroscopic stress of the honeycomb continuum. Also the macroscopic Poisson’s ratio will follow from this homogenization procedure. In chapter 3 the maximal stress that can be applied before yielding occurs and before the plastic collapse of the honeycomb structure occurs are considered. This results in two yield surfaces: one for initial yielding and one for plastic collapse of the honeycomb. In chapter 4 some numerical analyses are done with Marc-Mentat for validation of the analytical model and visualization of the deformations and the Von Mises stress distribution. In the last chapter a conclusion will be drawn and some recommendations for further research will be given.

7 (a) Beehive honeycomb. (b) Aluminium hexagonal honeycomb.

Figure 1.1: Regular honeycomb structures.

Figure 1.2: Natural cellular materials: (a) cork (b) balsa (c) sponge (d) trabecular bone (e) coral (f) cuttleﬁsh bone (g) iris leaf (h) plant stalk.

8 Chapter 2

Linear elastic model for in-plane loading

2.1 Homogenization procedure

In this chapter the linear elastic model for the honeycomb core will be set up. The model for the honeycomb core is largely based on the models used by Onck [1] and Gibson and Ashby [2]. The honeycomb can be seen as a continuum with properties that are obtained by homogenization over a Representative Volume Element (RVE). This homogenization is valid when the micro structural cell size d is much smaller as the length scale L of the RVE and the micro structural geometry is randomly distributed or if the microstructural geometry is equal and periodically distributed. In the case of the regular honeycomb the structure is equal and periodically distributed. First of all some basic linear elastic constitutive relations will be set up. Thereafter these constitutive relations are applied to the honeycomb to derive the compliance matrix S and the elasticity matrix C. The starting relations are force and moment equilibrium and energy conservation. The honeycomb material is characterized by Young’s modulus Es, the yield stress σY , the cell wall thickness t, the cell wall height h and the length l of the cell wall. It will be assumed that a uniform stress (static boundary conditions) or a uniform strain (kinematic boundary conditions) will be applied. The procedure for static boundary conditions is visualized in ﬁgure 2.1. The honeycomb structure used is a regular hexagonal cell structure. Because this cell structure is equal and periodically distributed the smallest RVE is in size equal to a unit cell. The geometric structure and the used unit cell are shown in ﬁgure 2.2.

2.1.1 Constitutive relations For the unit cell, force equilibrium should be satisﬁed,

m m m X X X f~(k) = ~p(k)ds(k) = σ · ~n(k)ds(k) = ~0 (2.1) k=1 k=1 k=1 where f~(k) is the force on a cell wall (k), m is the number of cell walls in the RVE. The stress vector ~p(k) is acting on surface ds(k) and can be calculated by taking the inner product between the uniform stress tensor σ and the normal vector ~n(k) on surface ds(k).

9 r e2 r r e1 e3 s ji e ij

Figure 2.1: Homogenization with static boundary conditions, source [1].

A moment equilibrium should be satisﬁed, m m X X ~x(k) × f~(k) = ~x(k) × (σ · ~n(k)ds(k)) = ~0 (2.2) k=1 k=1 The vector ~x(k) is a Eulerian coordinate vector from the origin O to the boundary point of cell wall (k) on which the force f~(k) is acting. The conservation of energy principle dictates that the internal work has to be the same as the external work. Because uniform stress is assumed the energy density is equal to the work done by the applied forces averaged over the RVE, m 1 1 X 1 W = σ : ε = f~(k) · ~u(k) (2.3) 2 V 2 k=1 where the ﬁrst term represents the internal stored energy W and the second term the external

10 r r (2) 60° ds f f (1) l r r n(2) n(1) (2) (1)

c 120°

3l l O

c' (3) (4) r r r n(3) n(4) e2 r r r (3) (4) e2 f f r e1 r e1 (a) Regular hexagonal cell structure. (b) Unit cell.

Figure 2.2: Unit cell structure.

work We averaged over the volume V of the RVE. In the ﬁrst term ε represents the uniform strain tensor. The derivation of the energy density function can be found in appendix A.1. Inserting the relation f~(k) = σ·~n(k)ds(k) in equation (2.3) and making use of the symmetry of the stress tensor σ yields a relation for the strain tensor ε,

m 1 X 1 ε = ~u(k)~n(k) + ~n(k)~u(k) ds(k) (2.4) V 2 k=1 the derivation of the strain tensor can be found in appendix A.2. The stress tensor σ is derived by substituting the displacement vector ~u(k) = ε · ~x(k) in the energy density equation (2.3),

m 1 X 1 σ = ~x(k)f~(k) + f~(k)~x(k) (2.5) V 2 k=1

For the derivation of equation (2.5) see appendix A.3.

2.1.2 Applying constitutive relations to unit cell The center point O can be seen as a point of symmetry, see appendix B for a visualization. This means that the forces at beam (1) and beam (2) are related to beam (3) and beam (4) respectively. Thus, it suﬃces to take only the upper half part of the unit cell with beams (1) and (2) into account.

11 √ Applying equation (2.4) and ﬁlling in for the normal vectors ~n(1) = 3 1 ~e, ~n(2) = 2 2 √ √ ˜ − 3 1 ~e and for the surface area ds = 3l, see ﬁgure 2.2.b, leads to the following strain 2 2 ˜ tensor ε which is dependent on the displacement vectors ~u(1) and ~u(2),  √ √  2 3 u(1) − u(2) 1 u(1) + u(2) + 3 u(1) − u(2) T 3lh 1 1 3lh 1 1 2 2 ε = ~e  √ ~e ˜ 1 (1) (2) (1) (2) 2 (1) (2) ˜ 3lh u1 + u1 + 3 u2 − u2 3lh u2 + u2 (2.6)

Now only a relation for the displacement vectors ~u(k) as a function of the stress tensor σ remains to be determined. This relation can be found by treating the cell walls as cantilever beams which are clamped in their intersection points and thus there are no hinges in the intersection points. The principle of superposition allows to handle the problem (total loading) as three diﬀerent subproblems (load cases) and sum these afterwards. The diﬀerent load cases are one for simple tension in 1-direction, one for simple tension in 2-direction, and one for shear in 12-direction. The results for the displacement vectors ~u(1) and ~u(2) as a function of the stress tensor σ are given below, for the derivation see appendix C.

Displacements subproblem 1 (tension in 11-direction)

2 4 9l σ11 3l σ11 ! + 3 (1) T √16Esht 16√Esht ~u = ~e 2 4 (2.7) 3 3l σ11 3 3l σ11 ˜ − 3 16Esht 16Esht

2 4 9l σ11 3l σ11 ! − − 3 (2) T √16Esht 16√Esht ~u = ~e 2 4 (2.8) 3 3l σ11 3 3l σ11 ˜ − 3 16Esht 16Esht

Displacements subproblem 2 (tension in 22-direction)

2 4 3l σ22 3l σ22 ! − 3 (1) T √16Esht 16√Esht ~u = ~e 2 4 (2.9) 9 3l σ22 3 3l σ22 ˜ + 3 16Esht 16Esht

2 4 3l σ22 3l σ22 ! − + 3 (2) T √16Esht 16√Ehst ~u = ~e 2 4 (2.10) 9 3l σ22 3 3l σ22 ˜ + 3 16Esht 16Esht

Displacements subproblem 3 (shear in 12-direction) √ √ 2 4 3 3l σ12 3 3l σ12 ! − 3 (1) T 8Esht 4Esht ~u = ~e 2 4 (2.11) 3l σ12 3l σ12 ˜ + 3 8Esht 2Esht

√ √ 2 4 3 3l σ12 3 3l σ12 ! − 3 (2) T 8Esht 4Esht ~u = ~e 2 4 (2.12) 3l σ12 3l σ12 ˜ − − 3 8Esht 2Esht By substitution of the expressions above in the strain tensor equation (2.6) the compliance matrix can be determined. The compliance matrix gives a relation for the strain column

12 T T ε = (ε11 ε22 ε12) as a function of the stress column σ = (σ11 σ22 σ12) . The compliance ˜matrix is given by ˜

√  l 2 l 2  3 + t 1 − t 0 3l 2 2 S =  1 − l 3 + l 0  (2.13) 4E ht  t t  s l 2 0 0 2 + 2 t

l Using the fact that the dimension l is much larger than dimension t, thus t 1, it is l n possible to discard the terms t where n < 2. This yields the reduced compliance matrix: √  1 −1 0  3l3 S = −1 1 0 (2.14) 4E ht3   s 0 0 2

Comparing the full compliance matrix and the reduced compliance matrix with the plane stress compliance matrix for linear elastic isotropic materials, given by

 1 ν  E − E 0 ν 1 S =  − E E 0  (2.15) 1+ν 0 0 E leads to the conclusion that the elastic modulus E and the Poisson’s ratio ν for the full compliance matrix of the honeycomb-continuum is equal to, 4E h E = s √ 2 3 l 3 + l t t (2.16) l 2 ( t ) −1 ν = l 2 ( t ) +3 and for the reduced compliance matrix these are equal to,

3 E = 4√Esht 3l3 (2.17) ν = 1

The elasticity matrix C is given by the inverse of the full compliance matrix S−1,

 l 2 l 2  3+( t ) ( t ) −1 l 2 l 2 0  8+8( t ) 8+8( t )   2 2  4Esht ( l ) −1 3+( l ) C = √  t t 0  (2.18) 3l  l 2 l 2   8+8( t ) 8+8( t )   1  0 0 l 2 2+2( t ) The elasticity matrix above is equal to the plane stress isotropic elasticity matrix (see Geers [4]) with the appropriate values for E and ν, given by

 1 ν 0  E C = ν 1 0 (2.19) (1 − ν2)   0 0 1 − ν

13 It has to be noticed that for an isotropic linear elastic material the Poisson’s ratio normally 1 has to be between −1 ≤ ν ≤ 2 . Because ν = 1 for the reduced compliance matrix this matrix becomes singular. Thus the elasticity matrix has to be determined as the inverse of the full compliance matrix and can not be taken from the reduced form. If the reduced values of the Young’s modulus E and the Poisson’s ratio ν are substituted in the plane stress elasticity matrix for isotropic materials it can be seen that the values of this matrix go to inﬁnity and thus not give appropriate values. For the elasticity matrix always the full form has to be taken.

14 Chapter 3

Yielding under biaxial loading

When the honeycomb is loaded and material points reach the yield stress hinges form near the intersections. If it is assumed that the material behavior is ideal plastic, such that there is a plastic plateau border with a constant yield stress σys, two different yield surfaces can be identified, one for initial yielding and one for collapse plasticity. Initial yielding takes place when the outer fiber on the cell wall starts yielding due bending. Collapse plasticity will occur when the hinges have reached full plasticity. When this happens the total honeycomb will collapse. For both yield surface the bending theory of Benoulli/Euler/Kirchof is assumed, see Brekelmans [5], this means that shear deformation is not taken into account. The biaxial loading and the location of the hinges is shown in figure 3.1.

s 22

s11 s11

s 22

Figure 3.1: Location of hinges, source [2].

15 3.1 Initial Yielding

Initial yielding is the case where the stress in the first material point reaches the yield stress σys. Because the bending stress is distributed linearly over the cell wall thickness initial yielding will start at the most outer fibers on the cell wall boundary. The tensile stress can just be superpositioned on the bending stress like is shown in figure 3.2. As can be seen in

s s a t a

M M

s a

t c y 1 x 2

s a -s -s s s s ys b,max 0 b,max ys

Figure 3.2: Stresses in a cell wall when initial yielding is reached, 1=central axis and 2=neutral axis.

ﬁgure 3.2 the neutral axis will move to a new position c from the cell wall, due to the axial stress. This distance c can be calculated as, σ + σ 1 σ c = b,max a t = t + a t (3.1) 2σb,max 2 2σb,max

If the axial stress σa is positive, the most outer fiber that reaches the yield stress will be in the part with positive maximum bending stress σb,max. On the other hand if the axial stress σa is negative yielding will first occur in the most outer fiber with negative bending stress σb,max. If no axial stress is applied the yield stress will be reached at both outer fibers at the same time and if no bending stress is present the total cross section will reach the yielding stress when initial yielding occurs. This means that four cases can be identified for the combination of positive and negative axial stress σa and positive and negative bending moment M,

1. σys = σa + σb,max for σa ≥ 0 and M ≥ 0 2. σ = σ − σ for σ ≥ 0 and M ≤ 0 ys a b,max a (3.2) 3. σys = −σa + σb,max for σa ≤ 0 and M ≥ 0 4. σys = −σa − σb,max for σa ≤ 0 and M ≤ 0 with, √ 3 l σ = (3σ + σ ) (3.3) a 4 11 22 t

16 and Mt 1 l 2 σ = = 2 (σ − σ ) (3.4) b,max 2I 4 22 11 t

In equation (3.4), for the bending moment M, the following is substituted 3 M = (σ − σ ) hl2 (3.5) 8 22 11 and for the second moment of area,

ht3 I = (3.6) 12

If the honeycomb is uniaxially loaded, so σ11 = 0 or σ22 = 0, the deformation is bending ∗ dominated. This leads to an initial yielding stress limit of σiy,

4 t2 σ∗ = σ (3.7) iy 9 l ys

Substitution of equation (3.3) and (3.4) in equation (3.2) and normalizing by the initial yielding stress limit, equation (3.7) leads to the following yield conditions: 1. 1 = σ11 √1 t − 1 + σ22 √1 t + 1 for σ ≥ 0 ∧ M ≥ 0 σ∗ 3 l σ∗ 3 3 l a iy iy 2. 1 = σ11 √1 t + 1 + σ22 √1 t − 1 for σ ≥ 0 ∧ M ≤ 0 σ∗ 3 l σ∗ 3 3 l a iy iy (3.8) 3. 1 = − σ11 √1 t + 1 + σ22 − √1 t + 1 for σ ≤ 0 ∧ M ≥ 0 σ∗ 3 l σ∗ 3 3 l a iy iy σ11 1 t σ22 1 t 4. 1 = ∗ − √ + 1 − ∗ √ + 1 for σa ≤ 0 ∧ M ≤ 0 σiy 3 l σiy 3 3 l With equation (3.8), the normalized initial yield surface for biaxial loading can be calculated l l for diﬀerent values of t , in ﬁgure 3.3 this is done for t = 5, 10 and 20. In the yield surface it can be seen that under equi-biaxial loading the honeycomb is the strongest. This is due to the fact that if σ11 = σ22 the bending moment vanishes and only the stronger tensile deformation remains. Furthermore it can be seen that the shape is a parallelogram, for initial yielding the relations are linear. It has to be noticed that under compression, elastic buckling will occur which limits the yield surface, see literature [2]. Because buckling will only occur if the cell walls are loaded in their length direction there will be no buckling if uniaxial compression in 11-direction is applied.

3.2 Full plasticity collapse

The honeycomb will collapse when the hinges reach full plasticity. This will happen if the bending moment M equals the moment that causes full plasticity Mp. It wil be assumed that the stress-strain curve of the material has a plateau border both in compression and in tension at the yield stress, the material behavior is thus ideal plastically. The cell wall is loaded by a moment M and an axial stress σa, see ﬁgure 3.4. In this ﬁgure the state of full plasticity is reached. This means that the stress σ in a hinge location is equal to the yield stress σys over the total thickness of the cell wall. Because of the axial stress the neutral bending line will

17 Initial yield surface 30 (l/t)=20 (l/t)=10 20 (l/t)=5

10 * iy σ / 0 22 σ −10

−20

−30 −30 −20 −10 0 10 20 30 σ /σ* 11 iy

l Figure 3.3: Initial yield surface for t = 5, 10 and 20.

18 s s a t a

M M

t c 1

-s 0 s s ys ys

Figure 3.4: Stresses in a cell wall when full plasticity is reached, 1=central axis and 2=neutral axis.

not overlap with the central line of the cell wall. It will move to a position c of the cell wall. This distance c can be calculated trough the following force equilibrium

σysch − σys(t − c)h = σath (3.9) which gives 1 σ c = t + a t (3.10) 2 2σys With this distance c, the moment of full plasticity can be calculated as

2! 1 2 σa Mp = σysht 1 − (3.11) 4 σys

Equating the plastic moment Mp to the maximal moment M equation (3.5) and ﬁlling in equation (3.3) for the axial stress σa leads to the following relation for the yield surface,

 √ 2 σ11 1 σ22 ! 2 3 3 + 1. σ22−σ11 = 2 t 1 − σys 3 σys for M ≥ 0 σys 3 l  t  4( l )  √ 2 (3.12) σ11 1 σ22 ! 2 3 3 + 2. σ11−σ22 = 2 t 1 − σys 3 σys for M ≤ 0 σys 3 l  t  4( l )

∗ Equation (3.12) can be normalized by the uniaxial plastic collapse stress σpl. If loaded uniaxial the axial stress σa can be discarded because the deformation is bending dominated. The

19 plastic collapse stress is the given by

2 t2 σ∗ = σ (3.13) pl 3 ys l

Normalizing the relation for the yield surface with the plastic collapse stress yields

√ 2 σ22−σ11 3 σ11 1 σ22 t 1. ∗ = 1 − ∗ + ∗ for M ≥ 0 σpl 2 σpl 3 σpl l √ 2 (3.14) σ11−σ22 3 σ11 1 σ22 t 2. ∗ = 1 − ∗ + ∗ for M ≤ 0 σpl 2 σpl 3 σpl l

l In ﬁgure 3.5 the collapse yield surface is drawn for the slenderness ratios t = 5, 10 and 20. The yield surface has its ellipse form due to the diﬀerence between uniaxial loading, which

Collapse yield surface 20 (l/t)=20 (l/t)=10 15 (l/t)=5

5 * pl σ / 0 22 σ

−5

−10

−15

−20 −20 −15 −10 −5 0 5 10 15 20 σ /σ* 11 pl

l Figure 3.5: Collapse yield surface for t = 5, 10 and 20. corresponds to almost pure bending, and equi-biaxial loading, which leads to pure tension of the cell walls. Also, here, it has to be noticed that elastic buckling will occur if the cell walls are loaded in their length direction which limits the yield surface under compression.

20 Chapter 4

Numerical simulations

The analytical displacement relations will be verified by numerical simulations in this chapter. The numerical simulations are done in the finite element method (FEM) software package MARC and Mentat. For every subproblem, uniaxial loading in 11-direction and 22-direction and shear loading, the displacement relations will be verified with a beam structure and a l l plane stress model. This will be done for slenderness ratios of t = 1000 and t = 20. The slenderness ratio of 1000 is taken as an extreme case and the slenderness ratio of 20 as a more practical value. The extreme case is taken because the Bernoulli/Euler/Kirchhoff beam theory is only valid for slender beams and in the beam theory no shear deformations are taken into account, i.e. only pure bending. In section 4.1 the modeling in Marc-Mentat will be explained. In section 4.2 the results will be shown for the three subproblems. At the end of this chapter, the Von Mises stress (see Geers [4]) distribution in a plane stress model for uniaxial loading, shear loading and equi-biaxial loading will be shown.

4.1 Modeling in Marc-Mentat

As mentioned above, two types of models are made for all the load cases, a beam model and a plane stress model. In a beam model only the basic beam theory is used, see Brekelmans [5]. A plane stress model assumes that there only exists in-plane stress. This is a valid assumption if the honeycomb displacements in the height direction are not suppressed. For ease of modeling, the length l is chosen equal to the height h which is taken as 1 m. The material properties that 9 are used are for normal aluminium with an E-modulus of Es = 70 ∗ 10 Pa and a Poisson’s ratio of νs = 0.34, see Fenner [3]. The uniform stress applied to the models with a slenderness 6 ratio of 1000 and 20 are σ11 ∨σ22 ∨σ12 = 10 Pa and σ11 ∨σ22 ∨σ12 = 10 Pa respectively. The resulting force of this stress will be applied as a point load in the nodes belonging to points (1) and (2). In the plane stress model this force is a distributed force over all the nodes of the edge belonging to the points (1) and (2), in the corner nodes of the edge, this force equals a 3 half of the distributed force. The force for subproblem 1 is equal to f1 = 2 lhσ11. This force is in the 11-direction for point (1) and√ in the opposite direction for point (2). For subproblem 2 the applied force equals f = 3 lhσ . This force is in the 22-direction for point (1) and 2 2 22 √ 3 3 (2). The shear loading subproblem 3 has forces equal to f1,s = 2 lhσ12 and f2,s = 2 lhσ12. Force f1,s is in the 11-direction for both points and f2,s is in the 22-direction for point (1) and in the opposite direction for point (2). For the equi-axial load case, force f1 and f2 will be applied at the same time. Because of the symmetry mentioned in appendix B, only

21 cell wall (1) will be modeled in the plane stress model of subproblem 1. In the beam model and the other plane stress models, the full upper half of the unit cell, see figure 2.2, will be modeled. The full plane stress model consists of three beams equal to subproblem 1 which are connected by a triangle where on each edge the same number of elements are used as in the thickness direction of the beam. The rotations and displacements of the plane stress model of subproblem 1 are suppressed in the nodes belonging to point c. The other models are fixed in the nodes belonging to the origin O. In the beam model, the rotations and displacements out of plane are suppressed for all the nodes. For the plane stress model this is not necessary because here the assumption of plane stress is used. The beam model is modeled with element type ”thin elastic beam 52”, for this model see figure 4.1. The element type used in the plane stress model is a full integration 4-node quadrilateral plane stress element. The plane stress model for subproblem 1 is shown in figure 4.2 and the total plane stress model is shown in figure 4.3. For the plane stress model the calculated displacements are taken from the most middle node belonging to the edge of point 1, the displacements of point 2 follow from the symmetry relations.

Figure 4.1: Beam model.

Before the definite analyzes is done an optimum element distribution for the plane stress model is determined. This is done by calculating the displacements for the beam of subproblem 1 with different element meshes and comparing these displacements with the analytical displacements and the beam model displacement. With a finer mesh (more elements) the truncation error will be less but this also needs more computing capacity. Thus, the best mesh is a mesh with a minimal number of elements when the truncation error is not changing significantly. First of all the minimal number of elements in length direction is determined. For the number of elements in length direction, it suffices to look at the plane stress model with a slenderness ratio of 1000. The length of the models with a different slenderness ratio is the same. For the number of elements in thickness direction, both models with a slenderness

22 Figure 4.2: Plane stress model beam 1.

Figure 4.3: Plane stress model.

23 ratio of 1000 and 20 have to be taken into account. In figure 4.4 and figure 4.5 the displacements of the plane stress model versus the number of elements in length direction and the displacement of the analytical and beam model are given in 11-direction and 22-direction respectively. The number of elements in thickness direction in these figures equals 16 and the different number of elements in length direction are [500 1000 2000 4000]. It can be seen that there is only a slight change between 2000 and 4000 elements. The number of elements that are thus chosen for the length direction are 4000 elements, if more elements are chosen this will not leed to a significant change anymore.

16 elements in thickness direction and l/t=1000 0.03 analytical beam model plane stress model 0.025

0.02

0.015

0.01 displacement in 11−direction [m] 0.005

0 0 500 1000 1500 2000 2500 3000 3500 4000 number of elements in length direction

Figure 4.4: Displacement in 11-direction as function of elements in length direction, sub- l problem 1, t = 1000 and σ11 = 10 Pa.

The displacements versus the number of elements in thickness direction for a thickness of l 0.001 m ( t = 1000) are shown in figure 4.6 and figure 4.7 for the 11-direction and 22-direction l respectively. For a thickness of 0.05 m ( t = 20) the figures are 4.8 and 4.9 for the 11-direction and 22-direction respectively. The number of elements in length direction is taken as 4000 and in thickness direction the numbers of elements are [2 4 8 16 32 64]. In all the four figures it can be seen that there is no significant change in the displacements if more than 16 elements are used. Thus the number of elements in thickness direction is taken equal to 16. The plane stress model of subproblem 1 has thus 64000 elements and the full plane stress models have 192256 elements.

24 16 elements in thickness direction and l/t=1000 0 analytical beam model −0.005 plane stress model

−0.01

−0.015

−0.02

−0.025

−0.03

−0.035 displacement in 22−direction [m] −0.04

−0.045

−0.05 0 500 1000 1500 2000 2500 3000 3500 4000 number of elements in length direction

Figure 4.5: Displacement in 22-direction as function of elements in length direction, sub- l problem 1, t = 1000 and σ22 = 10 Pa.

4000 elements in length direction and l/t=1000

analytical 0.0268 beam model plane stress model

0.0266

0.0264

0.0262

0.026 displacement in 11−direction [m]

0.0258

0.0256 0 10 20 30 40 50 60 70 number of elements in thickness direction

Figure 4.6: Displacement in 11-direction as function of elements in thickness direction, l subproblem 1, t = 1000 and σ11 = 10 Pa.

25 4000 elements in length direction and l/t=1000

analytical beam model −0.0446 plane stress model −0.0448

−0.045

−0.0452

−0.0454

−0.0456

−0.0458

−0.046 displacement in 22−direction [m] −0.0462

−0.0464

−0.0466 0 10 20 30 40 50 60 70 number of elements in thickness direction

Figure 4.7: Displacement in 22-direction as function of elements in length direction, sub- l problem 1, t = 1000 and σ22 = 10 Pa.

4000 elements in length direction and l/t=20 0.0218 analytical beam model 0.0217 plane stress model

0.0216

0.0215

0.0214

0.0213

0.0212 displacement in 11−direction [m]

0.0211

0.021 0 10 20 30 40 50 60 70 number of elements in thickness direction

Figure 4.8: Displacement in 11-direction as function of elements in thickness direction, l 6 subproblem 1, t = 20 and σ11 = 10 Pa.

26 4000 elements in length direction and l/t=20 −0.036 analytical beam model plane stress model −0.0362

−0.0364

−0.0366

−0.0368

−0.037 displacement in 22−direction [m]

−0.0372

−0.0374 0 10 20 30 40 50 60 70 number of elements in thickness direction

Figure 4.9: Displacement in 22-direction as function of elements in thickness direction, l 6 subproblem 1, t = 20 and σ22 = 10 Pa.

4.2 Results

After the displacements are calculated the strains which follow from the displacements by equation (2.6) will be compared with the analytical strains. The graphs in the following subsections are drawn for a stress range from -10 Pa to 10 Pa for a slenderness ratio of 1000 and a stress range from −106 Pa to 106 Pa for a slenderness ratio of 20.

4.2.1 Subproblem 1: loading in 11-direction The strains are calculated out of the displacements and plotted in the following figures 4.10 l l for t = 1000 and in 4.11 for t = 20. In these graphs it can be seen that the beam model and the plane stress model follow the analytical strains. Figures 4.6, 4.7, 4.8 and 4.9 show that the displacements from the beam model are exactly the same as those from the analytical approach. In these graphs it can also be seen that the displacements from the plane stress model have a slight offset from the analytical values. Discarding the assumption of Bernoulli/Euler/Kirchhoff introduces shear displacements thus the displacements should be greater than the analytical and beam model displacements. In the case of slender beams the shear displacements are negligible. Thus for the extreme case of a slenderness ratio of 1000 the offset is mostly due to a finite mesh distribution. On the other hand, in the figures for the slenderness ratio of 20, figures 4.8 and 4.9, it can be noticed that if more elements are used the interpolated plane stress model line will cross the analytical and beam model value of the displacements. This means that the offset, which is very small, of the plane stress model with a slenderness ratio of 20 could be due shear deformation. The plots of the strains versus the

27 stress, ﬁgures 4.10 and 4.11, show that ε11 = −ε22 thus the Poisson’s ratio of the continuum equals ν = 1, this is conform the reduced compliance matrix, equation (2.14). Furthermore it is satisfying that all the shear strains are equal to zero. This also conﬁrms the assumptions of the analytical model.

Strains as function of σ , l/t=1000 11 0.08 ε analytical 11 ε analytical 22 0.06 ε analytical 12 ε beam model 11 0.04 ε beam model 22 ε beam model 12 0.02 ε plane stress model 11 ε plane stress model 22

ε 0 ε plane stress model 12

−0.02

−0.04

−0.06

−0.08 −10 −5 0 5 10 σ [Pa] 11

l Figure 4.10: Strains as a function of σ11, subproblem 1 and t = 1000.

4.2.2 Subproblem 2: loading in 22-direction In the figures 4.12 and 4.13, the strain versus the stress is shown for a slenderness ratio of 1000 and 20 respectively. In these figures it can be seen that the beam model gives the same results as the analytical results. Also it can be seen that the strains of subproblem 2 are the strains of subproblem 1 multiplied by -1, this means that the E-modulus for uniaxial tension in 11-direction is equal to the E-modulus for uniaxial tension in 22-direction. That the E- modulus is the same for tension in 11-direction and 22-direction is in congruence with the isotropic character of the hexagonal honeycomb structure which is shown with the compliance matrix equation (2.13) in chapter 2. Taking a look at the strains of the plane stress model it can be noticed that the strains of the model with a slenderness ratio of 20 have a higher difference with the beam model and the analytical results as the one with a slenderness ratio of 1000. This could be due the shear deformations or truncation errors. In the case of shear deformations this means that the plane stress model is more accurate.

4.2.3 Subproblem 3: shear loading In ﬁgure 4.14 and 4.15, the strains are plotted against the stress for a slenderness ratio of 1000 and 20 respectively. Also for the case of shear loading, the beam model gives the exact

28 Strains as function of σ , l/t=20 11 0.06 ε analytical 11 ε analytical 22 ε analytical 0.04 12 ε beam model 11 ε beam model 22 ε beam model 0.02 12 ε plane stress model 11 ε plane stress model 22

ε 0 ε plane stress model 12

−0.02

−0.04

−0.06 −1 −0.5 0 0.5 1 σ [Pa] 6 11 x 10

l Figure 4.11: Strains as a function of σ11, subproblem 1 and t = 20.

Strains as function of σ , l/t=1000 22 0.08 ε analytical 11 ε analytical 22 0.06 ε analytical 12 ε beam model 11 0.04 ε beam model 22 ε beam model 12 0.02 ε plane stress model 11 ε plane stress model 22

ε 0 ε plane stress model 12

−0.02

−0.04

−0.06

−0.08 −10 −5 0 5 10 σ [Pa] 22

l Figure 4.12: Strains as a function of σ22, subproblem 2 and t = 1000.

29 Strains as function of σ , l/t=20 22 0.06 ε analytical 11 ε analytical 22 ε analytical 0.04 12 ε beam model 11 ε beam model 22 ε beam model 0.02 12 ε plane stress model 11 ε plane stress model 22

ε 0 ε plane stress model 12

−0.02

−0.04

−0.06 −1 −0.5 0 0.5 1 σ [Pa] 6 22 x 10

l Figure 4.13: Strains as a function of σ22, subproblem 2 and t = 20.

analytical results. The plane stress model of the high slenderness ratio is almost exact and the plane stress model with the slenderness ratio of 20 has a slight diﬀerence with the analytical results. This diﬀerence has to be sought in the explanation given in the subsection 4.2.2 above. In the graphs it can be noticed that the shear strains are two times higher than the strains in the case of uniaxial loading, this can also be seen in the reduced compliance matrix 2.14.

4.3 Von Mises Stress distribution

To localize the initial yielding locations in the geometry, the Von Mises stress distributions will be shown for uniaxial tension, shear loading and equi-axial tension. The Von Mises stress distribution for uniaxial tension in 11-direction is shown in figure 4.16. Because the Young’s modulus is the same for 11-direction or 22-direction, only uniaxial tension in the 11-direction will be shown. Thus the stress distribution will be similar for the 22-direction. Only bending occurs in the opposite direction. As can be seen in the figure, the locations with the highest Von Mises stresses are in the most outer fiber near the intersection (c) in beams (1) and (2). In figure 4.17, the Von Mises stress distribution is shown for shear loading, the deformation 1 is scaled with a factor 2 . It can be seen that for shear loading the highest Von Mises stress can be found just below the intersection point (c). This can be explained through the fact that the bending moment in cell wall (1) and cell wall (2) is in the same direction and thus is the bending moment just below the intersection point two times the bending moment of cell wall (1) or (2). As is explained in chapter 3, uniaxial tension and shear loading is bending dominated.

30 Strains as function of σ , l/t=1000 12 0.15 ε analytical 11 ε analytical 22 0.1 ε analytical 12 ε beam model 11 ε beam model 0.05 22 ε beam model 12 ε plane stress model 11 0 ε plane stress model 22

ε ε plane stress model 12 −0.05

−0.1

−0.15

−0.2 −10 −5 0 5 10 σ [Pa] 12

l Figure 4.14: Strains as a function of σ12, subproblem 3 and t = 1000.

Strains as function of σ , l/t=20 12 0.15 ε analytical 11 ε analytical 22 0.1 ε analytical 12 ε beam model 11 ε beam model 0.05 22 ε beam model 12 ε plane stress model 11 0 ε plane stress model 22

ε ε plane stress model 12 −0.05

−0.1

−0.15

−0.2 −1 −0.5 0 0.5 1 σ [Pa] 6 12 x 10

l Figure 4.15: Strains as a function of σ12, subproblem 3 and t = 20.

31 Figure 4.16: Von-Mises stress distribution (Pa) for uniaxial tension with σ = 106 Pa and l t = 20.

32 Figure 4.17: Von-Mises stress distribution (Pa) for shear loading with σ = 106 Pa and l 1 t = 20, deformation scaled with a factor 2 .

33 On the other hand if equi-biaxial loading is applied the bending moment vanishes and pure tension will remain. This can be seen in ﬁgure 4.18 where a homogenous Von Mises stress distribution is shown. In this ﬁgure the deformation is scaled with an factor 100. This means that initial yielding will occur over the total cell wall for perfect structures. Of course now initial yielding and plastic collapse occur at the same time (σys = σiy = σpl). The macroscopic stress before yielding occurs is of course much lower than the yield stress. This macroscopic yield stress can be calculated from chapter 3 and equals √1 t σ . 3 l ys

Figure 4.18: Von-Mise stress distribution (Pa) for equi-biaxial tension with σ = 106 Pa and l t = 20, deformation scaled with a factor 100.

34 Chapter 5

Conclusion

A regular honeycomb structure can be represented as a continuum by a homogenization procedure used in chapter 2. This homogenization procedure makes use of a Representative Volume Element (RVE). Because the honeycomb is regular the RVE can be chosen as a unit cell. The linear elastic behavior makes it possible to split the loading on a unit cell in three subproblems. These subproblems are: tension in 11-direction, tension in 22-direction and shear loading. By the bending equations of Bernoulli/Euler/Kirchhoff the displacements for a unit cell are determined. With these displacements the strain tensor can be calculated which leads to the compliance matrix equation (2.13). The beam bending equations assume pure bending and axial elongation, this means that no shear deformations are taken into account. l This theory only holds for slender beams where t 1. Because the cell walls are slender, the deformation mechanism is bending dominated for uniaxial loading and shear loading. This l 3 l is due to the fact that bending is related to ( t ) and axial elongation is related to t in the compliance matrix. On the other hand, if equi-biaxial loading is applied (σ11 = σ22) the bending moment vanishes and the deformation mechanism is pure axial elongation. Taking the limit for slender beams of the compliance matrix leads to the reduced compliance matrix equation (2.14). In this reduced compliance matrix only the bending deformation is taken into account. Both the compliance matrices can be described by the isotropic plane stress compliance matrix for the reduced form this yields in a Poisson’s ratio for the honeycomb structure of ν = 1. If ν = 1 the compliance matrix will be singular, this means that the elasticity matrix always has to be determined from the full compliance matrix. The yield surfaces found for initial yielding and plastic collapse of the honeycomb structure have the shape of a parallelogram and an ellipse respectively. These forms of the yield surfaces in chapter 3 find their origin in the fact that bending is weak and axial elongation is much stronger. The initial yield surface gives the stress range where all the material points stay in the elastic regime. The plastic collapse surface gives the stress range before total collapse of the structure occurs. It has to be noticed that under compression, the structure tends to buckle which limits the yield surface drastically. Buckling will only occur if compressed in the length direction of the cell walls, this means that uniaxial loading in 11-direction will not lead to buckling. The numerical results of chapter 4 lead to the conclusion that the analytical derived compliance matrix is exact with the beam model, whereas the displacements of the plane stress model differ approximately by 0.5-1% with the analytical results. This could be due to the fact that in the plane stress model shear deformations are taken into account or because

35 of the used element mesh, the used elements perform less good in bending. In the ﬁrst case, the plane stress model would yield better results in the second case the beam model would yield better results, but this needs some more research. In further research a more extensive numerical approach can be used where a plane strain model is introduced besides the plane stress model. Also it needs to be investigated if the assumed strain option in Marc-Mentat for the plane stress and plane strain model will give better results. With these new numerical simulations for a range of slenderness ratios the Young’s modulus and Poisson’s ratio can be calculated and compared with the analytical results for the Young’s modulus and Poisson’s ratio. With this numerical approach not only can be seen that the model is quite accurate but also from which slenderness ratio the model is valid. Normally the bending equations of Bernoulli/Euler/Kirchhoﬀ are valid for a slenderness ratio higher than 10. Also more research has to be done on the buckling behavior under compression. Now there are several buckling limiting equations, see [2], but this are ruogh approximations. Finally physical tests could be done for validation of the analytical and numerical results. It has to be noticed that the analytical and numerical models assume perfect structures, this of course will not be the case if physical tests are done.

36 Appendix A

Constitutive relations

A.1 Derivation of energy density

In linear elastic theory no energy dissipation occurs, thus the change in external work δWe on a body must equal the change in internal work δWi. The change in external work is given by, ZZZ ZZ δWe = ρ~q · δ~udV + ~p · δ~udS (A.1) V S Where the ﬁrst term represents change in external work related to body forces such as gravity and the second term relates the change to boundary forces on the surface. The stress tensor σ is a lineair function σ = F (ε(~x)) of the strain tensor ε. Because of this linearity we can write the change of internal work,

ZZZ ZZZ 1 δWi = δ WdV = δ σ : εdV (A.2) V V 2 The term for the energy density W in equation (A.2) can be derived from (A.1) by using the appropriate relations and assumptions, RRR RR 1. δWe = ρ~q · δ~udV + ~p · δ~udS RRRV RRS 2. = V ρ~q · δ~udV + S(σ · ~n) · δ~udS RRR RRR ~ 3. = V ρ~q · δ~udV + V ∇ · (σ · δ~u)dV RRR RRR ~ ~ T 4. = V ρ~q · δ~udV + V [(∇ · σ) · δ~u + σ :(∇δ~u) ]dV RRR ~ RRR ~ T (A.3) 5. = V (∇ · σ) + ρ~q) · δ~udV + V σ :(∇δ~u) dV 6. = RRR σ :(∇~ δ~u)T dV RRRV 7. = V σ : δεdV RRR 1 8. = δ V 2 σ : εdV 9. = δWi

In line 1 the stress vector ~p = σ · ~n is substituted, see equation (2.1). The divergence RR RRR ~ theorem S ~n · σdS = V ∇ · σdV is used in line 2, in line 5 the static equilibrium equation ∇~ · σ + ρ~q = ~0 is substituted. The symmetry of the stress tensor σ and the deﬁnition of the ~ T ~ 1 ~ ~ T strain tensor permits to write σ :(∇δ~u) = σ :(∇δ~u) = σ : 2 [∇δ~u + (∇δ~u) ] = σ : δε. This is substituted in line 6. Because the stress tensor σ is a lineair function σ = F (ε(~x)) of the

37 strain tensor ε the change in energy density can be written in rate form as δW = σ : δε = hR ˙ i R 1 δ t Wdt = δ t σ : ε˙dt = δ 2 σ : εdt and is substituted in line 7. This derivation can be found in Geers [4]. The strain tensor ε and the stress tensor σ are assumed to be uniform over the volume of the RVE, thus it follows that the energy density W is equal to the external work averaged over the volume as, m 1 X 1 W = f~(k) · ~u(k) (A.4) V 2 k=1

A.2 Derivation of strain tensor

A relation for the strain tensor ε is obtained by inserting f~(k) = σ · ~n(k)ds(k) in the equation for energy density (2.3) and making use of the symmetry of the stress tensor. This is done by ﬁrst inserting the relation for f~(k), m 1 1 X 1 W = σ : ε = (σ · ~n(k)ds(k)) · ~u(k) (A.5) 2 V 2 k=1 and second by using the identity (σ · ~n(k)) · ~u(k) = σ : ~n(k)~u(k) where ~n(k)~u(k) is a dyad,

m 1 1 X 1 W = σ : ε = σ : ~n(k)~u(k) ds(k) (A.6) 2 V 2 k=1 Now the symmetry of the stress tensor σ permits to write, m 1 1 X 1 1 W = σ : ε = σ : ~n(k)~u(k) + ~u(k)~n(k) ds(k) (A.7) 2 V 2 2 k=1 Finally solving for the strain tensor ε yields equation (2.4),

m 1 X 1 ε = ~u(k)~n(k) + ~n(k)~u(k) ds(k) (A.8) V 2 k=1

A.3 Derivation of stress tensor

The stress tensor σ is derived by inserting ~u(k) = ε · ~x(k) for the displacement vector ~u(k) in the energy density equation (2.3) where ~x(k) is the Eulerian coordinate vector of point (k). Also, the symmetry of the strain tensor ε is used. Thus inserting the relation for vector ~u yields, m 1 1 X 1 W = σ : ε = f~(k) · ε · ~x(k) (A.9) 2 V 2 k=1 using the identity f~(k) · (ε · ~x(k)) = ~x(k)f~(k) : ε yields,

m 1 1 X 1 W = σ : ε = ~x(k)f~(k) : ε (A.10) 2 V 2 k=1

38 the symmetry of the strain tensor ε leads to,

m 1 1 X 1 1 W = σ : ε = ~x(k)f~(k) + f~(k)~x(k) : ε (A.11) 2 V 2 2 k=1 now solving for the stress tensor σ yields equation (2.5)

m 1 X 1 σ = ~x(k)f~(k) + f~(k)~x(k) (A.12) V 2 k=1

39 40 Appendix B

Point symmetry and displacement relations

To visualize the point symmetry, the principle of superposition can be used to divide the loading in three different load cases, one load case for the 11-direction, one load case for the 22-direction and a shear load in the 12-direction. Thus if a certain symmetry holds for all the subproblems, the total problem possesses this certain symmetry. Taking a look at the different subproblems below, see figures B.1, B.2 and B.3, the following point symmetry holds,

~u(1) = −~u(3) ∧ ~u(2) = −~u(4)

B.1 Subproblem 1

In ﬁgure B.1.a, the loading in 11-direction is shown and in ﬁgure B.1.b, the displacements resulting from this loading are shown. According to the displacements this subproblem has a point symmetry (rotation with 180◦ with O point of rotation),

~u(1) = −~u(3) ∧ ~u(2) = −~u(4)

a line symmetry where the vertical midline through the unit cell is the symmetry line,

(1) (2) (1) (2) (3) (4) (3) (4) u1 = −u1 ∧ u2 = u2 ∧ u1 = −u1 ∧ u2 = u2

and a line symmetry where the horizontal midline through the unit cell is the symmetry line,

(1) (4) (1) (4) (2) (3) (2) (3) u1 = u1 ∧ u2 = −u2 ∧ u1 = u1 ∧ u2 = −u2

B.2 Subproblem 2

The loading in 22-direction is shown in figure B.2.a and the resulting displacements in figure B.2.b. From the figure for the displacements it can be seen that the symmetry relations are similar to that of subproblem 1. These symmetries are a point symmetry,

~u(1) = −~u(3) ∧ ~u(2) = −~u(4)

41 (2) (1) f1 (2) (1) f1 u1 u1

u2 u2

f1 f1 u1 u1 (3) (4) (3) (4) r r e2 e2

r r e1 e1 (a) Unit cell with (b) Unit cell with dis- forces in 11-direction. placements for loading in 11-direction.

Figure B.1: Load case subproblem 1. a line symmetry with the vertical line as mirror line, (1) (2) (1) (2) (3) (4) (3) (4) u1 = −u1 ∧ u2 = u2 ∧ u1 = −u1 ∧ u2 = u2 and a line symmetry with the horizontal line as mirror line, (1) (4) (1) (4) (2) (3) (2) (3) u1 = u1 ∧ u2 = −u2 ∧ u1 = u1 ∧ u2 = −u2

f2 f2

u2 u2

(2) (1) (2) u1 u1 (1)

c c

O O

c' c'

(3) (4) (3) (4) u1 u1

r r u2 u2 e2 e2 f2 f2 r r e1 e1 (a) Unit cell with (b) Unit cell with dis- forces in 22-direction. placements for loading in 22-direction.

Figure B.2: Load case subproblem 2.

B.3 Subproblem 3

The shear load case, subproblem 3, has forces acting as in ﬁgure B.3.a and resulting displacements as in ﬁgure B.3.b. This subproblem has only a point symmetry, ~u(1) = −~u(3) ∧ ~u(2) = −~u(4)

42 and the following relation between ~u(1) and ~u(2) can be identiﬁed,

(1) (2) (1) (2) u1 = u1 ∧ u2 = −u2

f 2 u2 (2) f (1) f (2) (1) 1 1 u1 u1

c c u2 f2

O O

f2 u2 c' c'

u u (4) (3) (4) 1 (3) 1 f1 f1 r r u2 e2 e2 f2

r r e1 e1 (a) Unit cell with (b) Unit cell with dis- shear forces in 12- placements for loading direction. in 12-direction.

Figure B.3: Load case subproblem 3.

43 44 Appendix C

Displacements as a function of the stress σ

To derive a relation for the displacements ~u(k) as a function of the stress tensor σ basic beam bending theory will be used. To make the problem less cumbersome it will also be divided in three subproblems by the superposition principle, subproblem 1 and subproblem 2 for loading in the 11-direction and 22-direction respectively, and subproblem 3 for shear loading in the 12-direction. Furthermore we can make use of the displacement relations determined in appendix B, if one relation for the displacement ~u(k) is given the others follow from the displacement relations. In the following, the displacement vector ~u(1) will be derived and ~u(2) ∗ will follow out of the displacement relations. The displacement uaxial in length direction of the beam can be calculated with Hooke’s Law, L P u∗ = beam (C.1) axial Eht

Where Lbeam is the beam length, P is the axial force, Es is the elastic modulus of the solid material, h is the height of a cell wall and t is the cell wall thickness. The basic beam bending theory is based on the hypothesis of Bernoulli/Euler/Kirchhoff, see Brekelmans [5]. The beam deflection is given by, ML2 DL3 qL4 u∗ = beam + beam + beam (C.2) bending 2EI 3EI 8EI And the angle of rotation is given by, ML DL2 qL3 ϕ∗ = beam + beam + beam (C.3) bending EI 2EI 6EI ∗ ∗ Where ubending is the deflection perpendicular to the length direction of the beam, ϕbending is the angle of rotation between the original and the deformed beam, M is a moment, D is a 1 3 shear force, q is a distributed loading and I is the second moment of area given by I = 12 ht . In this analysis there is only a shear force D and a bending moment M which reduces equation (C.2) and (C.3) respectively to, ML2 DL3 u∗ = beam + beam (C.4) bending 2EI 3EI ML DL2 ϕ∗ = beam + beam (C.5) bending EI 2EI

45 C.1 Subproblem 1

For pure loading in 1-direction the beam c-c’, see ﬁgure 2.2, is not subjected to bending and it has a much longer length in 2-direction than in 1-direction (l t), thus it’s elongation in 1-direction is much smaller than in 2-direction and it is only necessary to look at beam (1). It is assumed that beam (1) is clamped in the intersection point c. r r * r r* e2 e2 e2 e r 2 r u2 60º e1 * e f 60º 1 * r* 2 r* u2 e1 e1 f u (1) 60º 1 (1) 60º 1 f * * c 1 c u1

(a) Forces subproblem 1. (b) Displacements subproblem1.

Figure C.1: Subproblem 1 loading in 1-direction.

In figure C.1.a the force on beam (1) is shown and in figure C.1.b the resulting displacements are shown. In these figures, a new vector basis ~e∗ is introduced besides the old vector ∗ ˜ basis ~e. The new vector basis has one component ~e1 perpendicular to the length direction of ˜ ∗ the beam and one ~e2 in the length direction. It can be seen that the shear force D is equal ∗ ∗ to f1 and the axial force P is equal to f2 . With equations (C.1) and (C.4), the displacement vector ~u in the new basis ~e∗ can be written as, ˜

∗ ∗ L3f ∗ ! L3 T T u T T 1 T ∗ u1 ∗ bending ∗ ∗ ∗ 3EI ∗ 3EI 0 ∗ ~u = ~e ∗ = ~e ∗ = ~e Af = ~e Lf ∗ = ~e L f ˜ u ˜ u ˜ ˜ 2 ˜ 0 2 axial ˜ Eht Eht ˜ (C.6)

The vector basis ~e∗ is a rotated vector basis ~e by ~e∗ = R~e, where R is the rotation matrix given by, ˜ ˜ ˜ ˜ √ ! 1 − 3 R = √2 2 (C.7) 3 1 2 2 With this rotation matrix, the vector ~u can be written with respect to the basis ~e which leads to, ˜ √ L3 3L ! ~u = ~eT u = ~eT RT u∗ = ~eT RT Af ∗ = ~eT 6√EI 2Eht f ∗ (C.8) 3L3 L ˜ ˜ ˜ ˜ ˜ ˜ ˜ − 6EI 2Eht ˜

46 The force vector f~ can be written as,

T T T f~ = ~e∗ f ∗ = ~e∗ Rf = ~e∗ R~e · ~eT σ~e ·~eT n ds (C.9) ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ Substituting the force vector column f ∗ in equation (C.8) and ﬁlling in for the stress ˜ √ ! σ 0 3 matrix σ = 11 , for the normal vector kolom n = 2 and for the surface area 0 0 ˜ 1 √ 2 ds = 3l leads to the following displacement vector for the boundary of the cell wall (1),

2 4 ! 9l σ11 + l σ11 (1) T √16Eht √64EI ~u = ~e 2 4 3 3l σ11 3l σ11 ˜ 16Eht − 64EI 2 4 ! (C.10) 9l σ11 + 3l σ11 T √16Eht 16√Eht3 = ~e 2 4 3 3l σ11 3 3l σ11 ˜ 16Eht − 16Eht3 where has been made use of the fact that the length Lbeam of the beam is half of the cell wall 1 1 3 length l (Lbeam = 2 l) and the second moment of area I = 12 ht is substituted. The displacement vector ~u(2) follows from the line symmetry of subproblem 1,

! ! 2 4 ! (2) (1) − 9l σ11 − 3l σ11 (2) T u1 T −u1 T √ 16Eht 16√Eht3 ~u = ~e = ~e = ~e 2 4 (C.11) (2) (1) 3 3l σ11 3 3l σ11 ˜ u2 ˜ u2 ˜ 16Eht − 16Eht3

C.2 Subproblem 2

When loading in 2-direction the beam c-c’ will be loaded axially with a force 2f2 so we have to take the full upper part of the unit cell into account, see ﬁgure C.2. Beam c-c’ can be clamped in point O because of the line symmetric loading (no bending in this beam). The

* * f2 f2 * * * f2 f2 * u2 u2 * u2 u2 * f1 f1 60º 60º u1 60º 60º u1 u1 u1 (2) (1) (2) (1)

r r r * r* e e2 r* c e 2 120° e2 2 r* c e1 e 120° 60º r 60º r1 e1 e1 O O

(a) Forces subproblem 2. (b) Displacements subproblem 2.

Figure C.2: Subproblem 2 loading in 2-direction.

displacement vector ~u for this subproblem can thus be divided in two displacement vectors, one displacement vector ~u(c) for point (c) and one displacement vector ~u(1/c) for the boundary point of cell wall (1) relative to point (c) this can be written as,

(1/c) ~u(1) = ~u(c) + ~u(1/c) = ~eT u(c) +~eT RT u∗ (C.12) ˜ ˜ ˜ ˜ 47 Where has been made use of the fact that the relative displacement ~u(1/c) can be written as ~eT RT u∗(1/c) , RT is the transposed skew-symmetric rotation matrix. This rotation matrix R ˜gives a˜ basis transformation ~e∗ = R~e, see ﬁgure C.2 and is given by, √ ˜ ˜ 3 1 ! R = 2 √2 (C.13) −1 3 2 2

(c) (c) Because of the line symmetry of subproblem 2 displacement u1 = 0, the displacement u2 can be calculated by Hooke’s Law, equation (C.1) where the axial loading P is equal to 2f2. Vector ~u(c) can thus be written as, ! u(c) 0 0 ~u(c) = ~eT 1 = ~eT A(c)~e ·~eT f = ~eT ~e ·~eT f (C.14) (c) 0 2Lbeam ˜ u2 ˜ ˜ ˜ ˜ ˜ Eht ˜ ˜ ˜ The displacement vector ~u(1/c) can be written as,

∗ Lbeam ! (1/c) ∗T uaxial ∗T (1/c) ∗ ∗ ∗ ∗T Eht 0 ∗ ∗ ∗ ~u = ~e ∗ = ~e A ~e ·~e f = ~e L3 ~e ·~e f (C.15) u beam ˜ bending ˜ ˜ ˜ ˜ ˜ 0 3EI ˜ ˜ ˜ The force vector f~ can be written as in equation (2.1) and the force vector kolom f ∗ is the same as in equation (C.9), substituting these force vectors in equations (C.14) and˜ (C.15) respectively yields, ~u(c) = ~eT A(c)~e ·~eT f = ~eT A(c)~e · ~eT σ~e ·~eT n ds (C.16) ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜

(1/c) T T T T ~u∗ = ~e∗ A(1/c)~e∗ ·~e∗ = ~e∗ A(1/c)~e∗ ·~e∗ R~e · ~eT σ~e ·~eT n ds (C.17) ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ Thus the ﬁnal equation for the displacement vector ~u(1) is,

~u(1) = ~u(c) +~eT RT~e∗ · ~u∗(1/c) T (c)˜ T˜ T = ~e A ~e · ~e σ~e ·~e n ds (C.18) ˜ ˜ ˜ ˜ ˜ ˜ T T +~eT RT~e∗ ·~e∗ A(1/c)~e∗ ·~e∗ R~e · ~eT σ~e ·~eT n ds ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ √ ! 0 0 3 2 Filling in for the stress matrix σ = , for the normal vector kolom n = 1 , 0 σ22 ˜ √ 2 1 for the surface area ds = 3l, for the beam length Lbeam = 2 l and for the second moment of 1 3 area I = 12 ht yields,

! 2 4 ! 2 4 ! 0 3l σ22 − l σ22 3l σ22 − 3l σ22 (1) T √ T √ 16Eht √64EI T √16Eht 16√Eht3 ~u = ~e 2 +~e 2 4 = ~e 2 4 (C.19) 3l σ22 3l σ22 3l σ22 9 3l σ22 3 3l σ22 ˜ 2Eht ˜ 16Eht + 64EI ˜ 16Eht + 16Eht3 The displacement vector ~u(2) for boundary point of cell wall (2) follows out of the line symmetry from sub problem 2,

! 2 4 ! (1) − 3l σ22 + 3l σ22 (2) T −u1 T √ 16Eht 16√Eht3 ~u = ~e = ~e 2 4 (C.20) (1) 9 3l σ22 3 3l σ22 ˜ u2 ˜ 16Eht + 16Eht3

48 C.3 Subproblem 3

For the shear loading subproblem 3 point O remains fixed, there is no translation or rotation of this point. Thus we can model this problem by clamping point O, see figure C.3. In this ∗ ∗ figure it can be seen that beam (1) is subjected to an axial force f1 and a shear force f2 , beam O − c is subjected to a shear force D and a bending moment M.

* f2 * u2 f2 u f * f * * 2 * 1 60º 1 u1 60º u1 f1 f1 u1 u1 60º (2) (1) 60º (2) (1) f * * 2 u2 u f2 c 2 c r 120° r 120° r* r* e2 e2 r* e2 e2 r* e1 e 60º r 60º r1 e 1 O e1 O

(a) Forces subproblem 3. (b) Displacements subproblem 3.

Figure C.3: Subproblem 3 shear loading in 12-direction.

The displacement for the boundary point of cell wall (1) can thus be divided in one displacement for point c and one for the boundary point of cell wall (1) relative to point c, this can be written as,

(1/c) ~u(1) = ~u(c) + ~u(1/c) = ~eT u(c) +~eT RT u∗ (C.21) ˜ ˜ ˜ ˜ ∗ The point c rotates with ϕbending due to the bending of beam O-c, because it is assumed that ∗ 1 ∗ the angle ϕbending is small, only the displacement 2 lϕbending perpendicular on the beam can be taken into account. The rotation matrix R remains the same as in subproblem 2.

The shear force acting in point√ c is equal to D = 2f1 and the bending moment is given P2 (k) ~(k) 3 1 (k) by M = ~e3 · k=1 ~x × f = 2 lf2 − 2 lf1 where the vector ~x is the vector from point (c) to the boundary point of cell wall (k). With the help of equation (C.4) and the deﬁnition for the force vector f~ the displacement vector kolom u(c) gets, ˜

√ 3 3 ! 7l − 3l u(c) = A(c)~e · f~ = 48EI 16EI ~e · ~eT σ~e ·~eT n ds (C.22) ˜ ˜ 0 0 ˜ ˜ ˜ ˜ ˜

The displacement vector column u∗(1/c) can be written as, ˜ ∗(1/c) (1/c) ∗ ∗T ∗ 0 u = A ~e ·~e f + 1 ∗ ˜ ˜ ˜ ˜ 2 lϕbending l ! (C.23) T 0 2Eht 0 ∗ ∗ T T √ = 3 ~e ·~e R~e · ~e σ~e ·~e n ds + 4 l 3l σ12 0 24EI ˜ ˜ ˜ ˜ ˜ ˜ ˜ 16EI

49 1 In both equations C.22 and C.23 has been made use of the fact that the length L = 2 l. Substituting the vector columns of equation C.22 and equation C.23 in equation C.21 and √ ! 0 σ 3 12 2 ﬁlling in for the stress matrix σ = , for the normal vector column n = 1 , σ12 0 ˜ √ 2 1 3 for the surface area ds = 3l and for the second moment of area I = 12 ht yields for the displacement vector of the boundary point from cell wall (1),

~u(1) = ~eT u(c) +~eT RT u∗(1/c) ˜ ˜ ˜ ˜ = ~eT A(c)~e · ~eT σ~e ·~eT n ds ˜ ˜ ˜ ˜ ˜ ˜ !! T 0 T T (1/c) ∗ ∗ T T √ +~e R A ~e ·~e R~e · ~e σ~e ·~e n ds + 4 3 3l σ12 ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ 4Eht3 √ √ √ (C.24) 4 ! 2 4 ! 3l σ12 3 3l σ12 3l σ12 T − 3 T 8Eht − 2Eht3 = ~e 4Eht +~e 2 4 3l σ12 3l σ12 ˜ 0 ˜ + 3 √ √ 8Eht 2Eht 2 4 ! 3 3l σ12 3 3l σ12 T 8Eht − 4Eht3 = ~e 2 4 3l σ12 3l σ12 ˜ 8Eht + 2Eht3 The displacement vector for the boundary point of cell wall (2) can be obtained by the displacement relations for subproblem 3 given in appendix B, √ √ ! 2 4 ! (1) 3 3l σ12 3 3l σ12 (2) T u1 T 8Eht − 4Eht3 ~u = ~e = ~e 2 4 (C.25) (1) 3l σ12 3l σ12 ˜ −u2 ˜ − 8Eht − 2Eht3

50 Bibliography

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[2] Gibson, L.J., Ashby, M.F. (1988). Cellular Solids: Structure and Properties, ﬁrst edition. Oxford: Pergamon press.

[3] Fenner, R.T. (1999). Mechanics of Solids. Boca Raton: CRC Press.

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[5] Brekelmans, W.A.M. (1996). Platen en schalen, deel 1: balken en platen. Eindhoven: Eindhoven University of Technology.