Introduction to Sandwich Construction from 1995 but Is Essentially the Same

This is a free download copy of An Introduction to Sandwich Structures. This text in not entirely the same as the published book An Introduction to Sandwich Construction from 1995 but is essentially the same. This version has been used as a text in a course on sandwich structures at KTH for some time and has been through several minor updates since 1995.

I have decided to make it publicly available since the publisher is no longer in business and therefore the book cannot be purchased anywhere.

Every effort has been made to ensure that the information given herein is accurate and appropriate to the topic covered. No responsibility is accepted by the publishers, the editor or the authors, or any companies or individuals mentioned, for direct or consequential damages claimed to result from the use of the material for any particular application. No legal responsibility is accepted for any errors, omissions or misleading statements in the information caused by negligence or otherwise.

AN INTRODUCTION TO

SANDWICH STRUCTURES STUDENT EDITION

DAN ZENKERT

CONTENTS i

PREFACE v

LIST OF SYMBOLS vii

CHAPTER 1. INTRODUCTION 1

CHAPTER 2. MATERIALS AND MATERIAL PROPERTIES 2.1 2.1 Face Materials 2.1 2.2 Estimation of Face Material Properties 2.3 2.2.1 Rule-of-mixtures 2.3 2.2.2 "Practical" rule-of-mixtures 2.4 2.2.3 Conversion weight fraction/volume fraction 2.5 2.2.4 Thickness prediction 2.5 2.2.5 Stiffness properties of the lamina 2.6 2.2.6 Stiffness properties of the laminate 2.7 2.2.7 Strength of composite laminates 2.10 2.3 Experimental Determination of Face Material Properties 2.10 2.4 Core Materials 2.13 2.4.1 Honeycomb cores 2.14 2.4.2 Balsa wood 2.17 2.4.3 Cellular foams 2.18 2.5 Fatigue Properties of Sandwich Core Materials 2.20 2.6 Estimations of Core Material Properties 2.22 2.7 Experimental Determination of Core Material Properties 2.24 2.8 Adhesives – Description and Properties 2.25 2.8.1 Requirements on the adhesive 2.25 2.8.2 Adhesives and their properties 2.28 2.9 Experimental Determination of the Adhesive Interface Properties 2.30 2.10 Estimation of Thermal Insulation 2.31 References 2.34 Exercises 2.37

CHAPTER 3. FUNDAMENTALS 3.1 3.1 Flexural Rigidity 3.1 3.2 Approximations in the Flexural Rigidity 3.2 3.3 Stresses in the Sandwich Beam 3.3 3.4 Shear Stresses 3.3 3.5 Approximation in the Shear Stress 3.4 3.6 Summary of Approximations 3.5 3.7 “The Sandwich Effect” 3.5 3.8 Sandwich with Dissimilar Faces 3.6 3.9 Equivalent Width 3.8 References 3.9 Exercises 3.9

i AN INTRODUCTION TO SANDWICH CONSTRUCTION

CHAPTER 4. BEAM THEORY 4.1 4.1 Shear Deformations 4.2 4.2 Shear Stiffness 4.3 4.3 Equations in Terms of the Displacement Field 4.3 4.4 Governing Beam Equations 4.6 4.5 Effect of Thick Faces 4.7 4.6 Rigid Core 4.11 4.7 Energy Relations 4.12 4.8 General Solution of Beam Problems 4.14 4.9 Examples of Beam Calculations 4.15 4.9.1 Cantilever beam 4.15 4.9.2 Shear beam 4.18 4.9.3 Design example 4.19 4.9.4 Beam subjected to point load 4.21 4.9.5 Beam subjected to uniform pressure 4.26 4.9.6 Beam subjected to hydrostatic pressure 4.27 4.9.7 Hyperstatic beam example 4.29 4.10 Torsion 4.30 4.11 Testing of Sandwich Beams 4.31 4.11.1 The three-point bend (TPB) specimen 4.31 4.11.2 The four-point bend (FPB) specimen 4.32 References 4.33 Exercises 4.34

CHAPTER 5 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS 5.1 5.1 Governing equations 5.1 5.2 Boundary Conditions for Sandwich Beams 5.5 5.3 Buckling of Simply Supported Column – Simple Solution 5.8 5.4 Rigorous Solution to Beam Buckling 5.9 5.4.1 Clamped edges 5.9 5.4.2 Simply supported edges 5.14 5.4.3 One edge clamped, the other free (cantilever beam) 5.16 5.4.4 One edge clamped, the other simply supported 5.18 5.5 Examples of Sandwich Beam Buckling 5.21 5.6 Buckling of Sandwich Columns with Thick Faces 5.22 5.7 Buckling Stress Exceeding the Elastic Limit 5.23 5.8 Free Vibration of Sandwich Beams 5.24 5.8.1 Simply supported edges 5.26 5.8.2 Clamped edges 5.29 5.8.3 One edge clamped, the other simply supported 5.30 5.8.4 One edge clamped, the other free (cantilever beam) 5.31 5.9 Examples of Sandwich Beam Free Vibration 5.33 5.10 Estimation of Elastic Properties on Free-Free Sandwich Beam 5.34 5.11 Approximate Solutions to Beam Buckling and Free Vibration Problems 5.38 5.11.1 Buckling of clamped sandwich beam 5.38 5.11.2 Free vibration of a cantilever sandwich beam 5.40

ii 5.11.3 Free vibration of a clamped sandwich beam 5.41 References 5.42 Exercises 5.42

CHAPTER 6 FACE WRINKLING 6.1 6.1 Winkler Foundation Approach 6.2 6.2 Hoff's Approach 6.3 6.3 Exponential Decay 6.6 6.4 Differential Equation Method 6.7 6.5 Wrinkling under Biaxial Load 6.9 6.6 Wrinkling under Multi-Axial Load 6.11 6.7 Intercellular Buckling 6.15 6.8 Imperfection Induced Wrinkling 6.16 6.9 Summary of Buckling Phenomena 6.21 References 6.22

CHAPTER 7 FAILURE MODES AND DESIGN CRITERIA 7.1 7.1 Formulae for Failure Loads 7.2 7.2 Failure Mode Maps 7.5 7.3 Design Criteria 7.7 7.4 Determination of Thicknesses 7.9 7.5 Single Parameter Optimum 7.10 7.6 Minimum Weight for Given Stiffness 7.12 7.7 Minimum Weight for Given Strength 7.14 References 7.16

CHAPTER 8 SANDWICH PLATES – FUNDAMENTAL EQUATIONS 8.1 8.1 Governing Equations 8.2 8.2 Partial Deflections 8.8 8.3 Equation of Motion 8.10 8.4 Governing Buckling Equation 8.12 8.5 Isotropic Sandwich Plates 8.13 8.6 Isotropic Sandwich Plate with Thick Faces 8.14 8.7 Cross-Section Properties 8.15 8.8 Energy Relations 8.17 8.9 Stresses and Strains 8.19 8.10 Thermal Stresses and Deformations 8.21 8.11 General Sandwich Theory for Anisotropic Plates 8.23 8.12 Boundary Conditions 8.29 References 8.32

CHAPTER 9 SOLUTIONS TO PLATE PROBLEMS 9.1 9.1 Rotationally-Symmetric Plates 9.1 9.2 Bending of a Rectangular, Simply Supported, Isotropic Sandwich Plate 9.5 9.3 Rectangular, Simply Supported, Orthotropic Sandwich Plate 9.11

iii AN INTRODUCTION TO SANDWICH CONSTRUCTION

9.4 Solution by Energy Method - Ritz's Method 9.21 9.5 Approximate Solutions for Bending of Orthotropic Sandwich Plates 9.22 9.5.1 Simply supported plate 9.22 9.5.2 Clamped plate 9.25 9.5.3 Two sides clamped, the other two simply supported 9.30 9.6 Buckling of a Simply Supported, Isotropic Sandwich Plate 9.32 9.7 Buckling of a Simply Supported, Orthotropic Sandwich Plate with Thin Faces 9.35 9.8 Approximate Buckling Formulae for Orthotropic Sandwich Plates with Various Edge Conditions 9.38 9.8.1 Simply supported plate 9.38 9.8.2 Loaded edges simply supported, the other edges clamped 9.40 9.8.3 Loaded edges simply supported, the other edges clamped 9.41 9.8.4 All edges clamped 9.45 9.9 Shear Buckling 9.48 9.10 Combined Buckling and Transverse Load 9.49 9.11 Free Vibration of a Simply Supported Sandwich Plate 9.50 9.12 Conclusions 9.53 References 9.53 Exercises 9.55

CHAPTER 10 SINGLE CURVED SANDWICH SHELLS 10.1 10.1 Fundamental Equations 10.1 10.2 Governing Buckling Equation 10.3 10.3 Buckling of Simply Supported Isotropic Plate Subjected to Uniaxial Load 10.6 10.4 Buckling due to External Lateral Pressure 10.10 10.5 Local Buckling of Curved Sandwich Panels 10.11 10.6 Bending of Single Curved Sandwich Beams - a Practical Approach 10.12 References 10.15

CHAPTER 11 LOCALISED LOADS - by OLE T. THOMSEN 11.1 11.1 Elastic Foundation Analogy 11.2 11.1.1 Classical Winkler foundation model 11.2 11.1.2 Two-parameter elastic foundation model 11.6 11.1.3 Specification of boundary conditions 11.12 11.1.4 Superposition with classical sandwich beam theory 11.14 11.1.5 Range of applicability – Winkler vs. two-parameter foundation model 11.16 11.2 Discussion: Application, Results and Parametric Effects 11.19 11.2.1 Application of method for solving engineering design problems 11.19 11.2.2 Example 11.20 11.2.3 Parametric Effects 11.23 11.3 Concluding Remarks 11.27 References 11.28

CHAPTER 12 SANDWICH AND FEM 12.1 12.1 General Remarks on FEM 12.1 12.2 Special Considerations for Sandwich Structures 12.2 12.3 Beam Analysis 12.3 12.4 Sandwich Beam Finite Element 12.4 12.4.1 Derivation of stiffness matrix by beam calculations 12.5 12.4.2 Governing equations 12.7 12.4.3 Weak form of DE using virtual work method 12.9 12.4.4 Displacement finite element formulation 12.10 12.4.5 Two-node shear deformable beam element 12.12 12.4.6. Shear Locking 12.13 12.4.7 Three-node Shear Deformable Beam 12.14 12.4.8 Stiffness matrix assembly 12.17 12.4.9 Boundary conditions 12.18 12.4.10 Solution of equation systems 12.21 12.4.11 Post-processing 12.23 12.4.12 Example of FEM-Calculation for a Cantilever Sandwich Beam 12.25 12.4.13 Example of FEM-Calculation for a Simply Supported Sandwich Beam 12.27 12.4.14 More Examples 12.30 12.4.15 Mixed Formulation for Two-Node Beam 12.30 12.5 Plate Analysis 12.31 12.6 Shear Deformable Plate and Shell Elements 12.32 12.6.1 Governing equations 12.32 12.6.2 Loads on an arbitrary plane 12.37 12.6.3 Element construction 12.40 12.7 Boundary Conditions 12.46 12.8 Spurious Zero Energy Modes 12.48 12.9 Effects of Reduced Integration 12.51 12.10 Point Loads in Plate Formulations 12.52 12.11 Shell Elements 12.53 12.12 Alternative Modelling of Sandwich Structures 12.53 References 12.55

CHAPTER 13 JOINTS AND LOAD INTRODUCTIONS 13.1 13.1 Inserts 13.1 13.1.1 The elements involved 13.3 13.1.2 Face sheet/insert interface 13.4 13.1.3 Core/insert interface 13.5 13.1.4 Stress concentrations due inserts 13.5 13.1.5 Inserts in panels subjected to shear 13.6 13.1.6 Summary 13.6 13.2 Insert Calculation Examples 13.7 13.2.1 Self-tapping screw or rivet 13.9

v AN INTRODUCTION TO SANDWICH CONSTRUCTION

13.2.2 Partial insert 13.11 13.2.3 Through-thickness insert 13.14 13.2.4 Through-the-thickness insert with flared ends 13.15 13.3 Joints 13.16 13.3.1 Basic types 13.16 13.3.2 T-joints 13.17 13.3.3 L-joints 13.18 13.3.4 V-joints 13.20 13.3.5 Localised deflection 13.21 13.3.6 Calculation example, T-joint 13.23 13.3.7 Calculation example, L-joint 13.25 13.3.8 T-joints - tests and bbservations 13.26 Bibliography 13.28

CHAPTER 14 MANUFACTURING- BY KARLSSON AND ÅSTRÖM 14.1 14.1 Face Materials 14.2 14.2 Core Materials 14.2 14.3 Wet Lay-up 14.3 14.3.1 Procedure 14.3 14.3.2 Characteristics 14.6 14.3.3 Applications 14.7 14.4 Pre-preg Lay-up 14.8 14.4.1 Procedure 14.8 14.4.2 Characteristics 14.10 14.4.3 Applications 14.11 14.5 Adhesive Bonding 14.11 14.5.1 Procedure 14.12 14.5.2 Characteristics 14.13 14.5.3 Applications 14.13 14.6 Liquid Moulding 14.13 14.6.1 Procedure 14.14 14.6.2 Characteristics 14.16 14.6.3 Applications 14.17 14.7 Continuous Lamination 14.19 14.8 Other Processes 14.20 14.9 Outlook 14.23 References 14.24

vi PREFACE

This text is intended to cover some of the most important aspects of the theory on load carrying sandwich panels. In that respect it is very similar to the now almost classical texts from the 1960s by Allen and Plantema. I have tried, however, to be more practical and to emphasise problem solving. Thus, this text covers less of the purely theoretical side of the topic, fewer special cases, but on the other hand contains more examples and solved problems. The problems are there not only to provide a useful engineering tool, but are also sometimes included for the purpose that they clarify a physical behaviour. In that respect it is important not only to study the theory but to do so along with solving the problems.

This book has fourteen separate chapters in which the theory of structural sandwich construction is evolved. The chapters should preferably be read in the sequence they appear and although some chapters are more or less stand-alone much of the theory in each new chapter is based on the preceding chapters.

The first chapter is only a brief introduction to the concept; it gives some of the historical background and then mentions some of the pioneer work and applications in the field. The materials used and their characteristics are covered in chapter 2 along with some practical aspects of bonding face sheets to different kinds of core materials. Special emphasis is given to core materials, their special features and how to predict their characteristics. For each material category, some information is given on how to extract relevant properties, either by theoretical estimations or from testing.

Basic definitions, stress and strain calculations and important approximations are given in chapter 3. This is followed by a comprehensive beam theory development in chapter 4. A cross-section shear stiffness is introduced, governing differential equations including transverse shear and rotary inertia effects are derived and expressions for the strain energy given. Some special features like the effect of a rigid core and thick faces are also included. The chapter also contains quite a large number of examples and solutions to standard beam problems.

Buckling of sandwich columns with thin and thick faces are treated in chapter 5 and the so typical instability mode for sandwich construction called wrinkling or local buckling is treated in chapter 6. A summary of failure modes and how to predict their respective failure loads are given in chapter 7. Design procedures including the stiffness and the failure modes of the preceding chapter are outlined in chapter 8. The evolution of design criteria leads to some simple formulae for minimum weight design which can easily be included in an engineering design process.

In chapter 9 the theory for sandwich plates is derived, in which plates with generally anisotropic, orthotropic or isotropic behaviour as well as thin and thick faces are included. Energy expressions, boundary conditions and cross-section properties are also presented. Chapter 10 contains a large number of solutions to bending, buckling and free vibration of sandwich plate problems. Solutions are derived using direct integration of the governing equation and approximate solutions based on the Ritz energy approach. The theory of flat

vii AN INTRODUCTION TO SANDWICH CONSTRUCTION plates is extended to single curved sandwich shells in chapter 11, which is followed by solutions to some fundamental problems. A more practical approach to the design of curved sandwich shells is also presented.

The last three chapters in the book are slightly different. Chapter 12 treats the very important problem of localised loads. The theory given is based on two foundation models and the chapter is concluded with its application to engineering design problems. In chapter 13 the finite element method (FEM) is discussed in conjunction with sandwich construction. The basis of FEM and the formulation for sandwich elements is not given as that lies outside the scope of this text, but some important considerations when using FEM for sandwich structures are given. The final chapter has no theoretical background but simply gives some practical aspects on the joining and load introduction problems encountered when designing sandwich structures. The chapter is included merely for the sake of completeness rather than being any basis for the foundations of the concept.

This book started out as an attempt to tidy up some old lecture notes, add to them and prepare material for an extended course on the topic of structural sandwich constructions. I guess that is how most books start out. I had the opportunity to read calmly through a lot of collected material and properly learn the topic myself. There is no teacher like having to prepare material for teaching a subject. Most parts of this text have been written and compiled during my stay at the Department of Mechanical Engineering, The University of Auckland, New Zealand, between May 1991 and July 1992.

I wish to acknowledge Professor Jan Bäcklund for his encouragement and interest in the progress of my writing. I also want to take the opportunity to express my appreciation to all the staff at the Department of Mechanical Engineering, The University of Auckland, and especially Professor John Duncan for giving me opportunity to spend a wonderful year in New Zealand. I am also greatly in debt to Mr. Lars Falk for the many and long hours of editing and proof-reading. Thanks also to Mr. Mats Roslund for help with some of the drawings.

Chapter 12 – Localised Loads, has been written and compiled by Dr. Ole Thybo Thomsen, Department of Mechanical Engineering, Aalborg University Centre, Aalborg, Denmark.

Chapter 13 - Sandwich and FEM is co-authored by Prof. Jan Bäcklund, Department of Lightweight Structures, Royal Institute of Technology, Stockholm, Sweden.

Since this is the first issue I expect there are still several mistakes, misprints or even errors present. Certainly some parts need further explanation while others may be redundant. I am therefore more than happy to receive comments, corrections and suggestions for improvements.

Stockholm, March 1995 © Dan Zenkert

viii PREFACE, 2:ND EDITION

The main reason for compiling this second edition is that this version is updated to comply with current educational activities at KTH and is therefore to be used in only that context. It is merely an update of the original text. However, there are some major changes. The first chapters, 1-4, are basically unaltered albeit with some minor amendments here and there. Chapter 5 has been revised substantially and is now much more complete with stricter derivations, complete solutions to some fundamental beam cases now also including comparisons with FE-calculations.

Chapter 6 has been going through some updates, though not very extensive ones apart from the inclusion of some more recent theories on wrinkling that has appeared from research in the last few years.

Chapters 7 and 8 have been merged but are otherwise more or less unchanged. Chapters 8-11 and 14 are basically unchanged except for some minor updates.

Chapter 12 on the FEM has gone through quite a revision now including the basic theory of finite element construction for shear deformable beams and plates.

I have added a chapter on manufacturing which is based on a paper by Karlsson and Åström. This is added for completeness of the book contents rather for inclusion in the present course syllabus.

Another major change is that some chapters are followed by a section of examples. These are examples coming from exams in the undergraduate courses taught at KTH since around 1993. The examples have more or less worked through solutions although these are rather brief in most cases.

In its present form, this text is only to be used for teaching activities at KTH.

Stockholm, December 2005 © Dan Zenkert

ix AN INTRODUCTION TO SANDWICH CONSTRUCTION

LIST OF SYMBOLS

The following is a list of the notation and symbols used throughout the text. Some of these may have duplicate definitions in parts of the text and are in those parts locally defined. Other symbols not mentioned are defined in the text when appearing. Mostly millimetres (mm) are used in the problems used, but metres (m) could equally well be used.

Latin symbol Description Unit A Extensional stiffness N/mm B First moment of area (function of z) N C Compliance mm/N

CE, CG, Cτ Constants defining core material properties see eq.(8.3) D Flexural rigidity (bending stiffness) Nmm

D0, Df, Dc Flexural rigidity of components in a sandwich Nmm E Young's modulus of elasticity N/mm2 (MPa) G Shear modulus N/mm2 K Buckling coefficient – L Length of beam mm M Bending moment Nmm N Normal force N/mm P Load N/mm Q Point load N R Reaction force or initial radius of shell (chapter 11) N/mm or mm R Rotary inertia in chapters 4, 9 and 10 kg S Shear stiffness N/mm T Transverse force N/mm

T1, T2, Tm Temperatures K U Strain energy, potential energy Nmm W Weight kg Z Curvature parameter – A, B, D Extension, coupling and bending stiffness matrices N/mm, N, Nmm N Vector or matrix of element shape functions - B Vector or matrix of element shape function derivatives - F Load vector N K Stiffness matrix N/mm Kσ Geometric stiffness matrix N/mm M Mass matrix kg a, b Sides of rectangular panel (width of beam) mm d Distance between centroids of the sandwich mm d Vector of nodal degrees-of-freedom mm e Distance defining position of neutral axis (see Fig.2.7) mm k Thermal transmittance W/m2K kb, ks Deformation coefficients for beams and panels –

x kT, kM Loading and geometry constants – m, n Integers – summation, modes, or exponents – px, py Portions of buckling load (px P = Px) – 2 2 q, qT Distributed load (load/surface area) or heat flux N/mm or W/m t Component thicknesses mm u, v, w Deformation components in x, y, and z-directions mm x, y, z Cartesian coordinate system mm zf Local z-axis for a single face mm r, ϕ, z Cylindrical coordinate system mm, –, mm wb, ws Deformations due to bending and shear, respectively mm

Greek symbol Description Unit Φ Spatial displacement function mm Ψ Time displacement function –

Γ Ratio of shear modulus of orthotropic core, Gcx /Gcy – α Coefficient of thermal expansion K-1 ε Strain – φ Shear factor – κ Curvature mm-1 ψ Cross-section rotation – γ Transverse shear strain –

γ0 In-plane shear strain – λ Thermal conductivity in chapter 2 W/mK λ Buckling coefficient in chapter 3 – ν Poisson's ratio – ω Natural frequency s-1 θ Cross-section or nodal rotation – ρ Density (mass/unit volume) kg/m3 ρ* Mass per unit surface area kg/m2 σ Direct stress N/mm2 (MPa) τ Shear stress N/mm2 (MPa)

Subscript Description x, y, z Property refers to Cartesian direction r, ϕ, z Property refers to cylindrical direction c Property refers to the core f Property refers to the face 1,2 or f1,f2 Property refers to dissimilar faces L, R Left and right, respectively tot Total b Refers to a pure bending case max Maximum value min Minimum value s Refers to a pure shear case r Short for reduced modulus

xi AN INTRODUCTION TO SANDWICH CONSTRUCTION v Index used for virtual displacements and rotations tan Short for tangent modulus y Yield point

Superscript Description ^ Ultimate value, e.g., σ^ is ultimate or allowable stress Maximum value, e.g., w is maximum deflection

Operators Description ddx/ i Ordinary differential with respect to variable xi

∂ / ∂xi Partial differential with respect to variable xi Δ Laplace operator

xii CHAPTER 1

INTRODUCTION

John Montagu (1718-1792), 4th Earl of Sandwich and British First Lord of the Admiralty during the American revolution, gave his name to – either due to lack of time or a healthy appetite – a lunch sandwiched between two meetings, referring to the act of making time or room for something [1]. Another tale states that John Montagu, who was a devoted billiard- player, invented sandwiches to eat at the billiard-table [2] – in 1762 he spent 24 hours at the table without any other food than his sandwiches [3]. The archetypal sandwich consists of two slices of bread with meat in between. The sandwich construction discussed in this text is built in a manner similar to the edible sandwich but instead of bread two thin sheets of high- performing material, and in between these the filling is replaced with a low density material.

Historically, the advantages of the concept of using two co-operating faces separated by a distance is thought to have been first discussed by a Frenchman, Duleau, in 1820, and later by Fairbairn [4], although it was not until 100 years later that the concept was first applied commercially. In World War One sandwich panels of asbestos faces with a fibreboard core were used and prior to World War Two some use was made of sandwich panels in small planes. However, it was the invention and widespread acceptance of structural adhesives in England and the United States in the 1930s that allowed the application of bonded sandwich composites. The pre-war four-engined air-liner de Havilland Albatross designed for an experimental transatlantic service had a sandwich fuselage. The Mosquito aircraft, produced in England during World War Two, saw mass production of sandwich panels for the first time, utilising veneer faces with a balsa core [5]. This was mainly due to a shortage of other materials as well as an appreciation of the structural efficiency of the concept. It was towards the end of World War Two, in the late 1940s, that some of the first theoretical works on sandwich constructions were published. Real pioneer work on structural sandwich theory worth noticing are that of Gough, Elam and de Bruyne [6] and that of Williams, Leggett and Hopkins [7]. Since these early days the technology of sandwich panels has progressed significantly and today far more comprehensive use of the advantages of sandwich panels is being made.

Development of core materials has continued from the 1940s through to today in an effort to reduce the weight of sandwich panels. Balsa, the first core material to be used, is still in use where weight is not critical such as in cruising yachts and launches. Although heavy it still generally offers advantages over single skin designs. The late 1940s and 1950s saw the advent of honeycomb core materials, developed primarily for the aerospace industry. Honeycomb

1.1 AN INTRODUCTION TO SANDWICH STRUCTURES cores currently offer the greatest shear strength and stiffness to weight ratios but require care in ensuring adequate bonding to the faces. The core materials have been produced in various forms and developed for a range of applications, generally utilising a hexagonal cell shape for optimum efficiency. The continued high cost of honeycomb cores has restricted their application predominantly to the aerospace industry. The late 1950s and early 1960s brought about the advent of the polyvinyl chloride (PVC) and polyurethane (PUR) core materials commonly used today in low and medium cost applications. Although PVC foams were developed in Germany in the early 1940s they were not utilised commercially until 15 years later due to the softness of these early cores. In the last twenty years few new cores have been developed with research oriented around face materials and core bonding techniques. The next generation of core materials under development are cellular thermoplastic cores where properties can be tailored by orienting the cell structure.

Research into the theoretical analysis of sandwich constructions began following World War Two with several papers being published between 1945 and 1955 on the strength and stability of sandwich beams, columns and plates. An extensive and very complete listing of these early works, through to 1960, is published by Plantema [8] in his book on the topic. The theoretical research has generally preceded the practical application of sandwich construction, from the 1940s right through to modern times. In the early years this was primarily due to the practical difficulties such as in bonding the face sheets to the core, providing cores of sufficient stiffness and establishing reliable repair and inspection procedures. More recently, similar problems are still restricting the use of sandwich composites in areas such as primary aircraft structure. A considerable portion of the early research into sandwich construction was undertaken by the Forest Product Laboratories of the United States Forest Service, e.g. [9]. These works are of both experimental and theoretical nature and this laboratory has produced several significant works. These papers present analytical solutions to various beam and panel bending problems, using some degree of simplification to solve the various problems, and are referred to in several places in this text. Also during this period Reissner [10] published his well known theory on sandwich plates which derives the differential equation for deflection of a sandwich panel. Other significant early works on sandwich panels are those of Libove and Batdorf [11], Hoff [12] and Mindlin [13]. Libove and Batdorf [11] derived differential equations for the deflection and shear forces in orthotropic panels with thin faces, Hoff [12] considers the strain energy of a sandwich panel in terms of the transverse and in-plane deflections and derives the governing equation for an isotropic sandwich panel but with respect to thick faces, and Mindlin [13] derived the governing equation of motion for an isotropic plate accounting for both transverse shear deflections and rotary inertia. These theories form the basis of the two important texts on sandwich constructions published in the 1960s by Plantema [8] and Allen [14].

In the last twenty years the emphasis in theoretical research has shifted to optimisation of laminates, with finite element analysis used as a design tool for the panel analysis problems. As a result little further work has been conducted into the theoretical analysis of sandwich panels, principally due to the difficulty of obtaining more exact solutions in deriving and solving the differential equations for the deflection of sandwich panels. The errors in the current approximations used are often negligible for practical composite laminates but require consideration. Finite element techniques utilising especially designed sandwich elements also

1.2 INTRODUCTION allow accurate analysis of sandwich design problems. These are generally more accurate than many of the existing analytical solutions which require several approximations and the use of finite difference methods to solve the differential equation. Research into sandwich construction over the last two decades has revolved primarily around the areas of impact resistance, fatigue and fracture analysis with these areas being of major concern to the aerospace industry. This research is now allowing the introduction of composite materials in aircraft primary structure. The theoretical analysis of sandwich beams and plates has received little attention with the aerospace industry content to use finite element analysis for design purposes.

The ASTM defines a sandwich structure as follows:

A structural sandwich is a special form of a laminated composite comprising of a combination of different materials that are bonded to each other so as to utilise the properties of each separate component to the structural advantage of the whole assembly.

A sandwich consists of three main parts as illustrated in Fig.1.1.

Adhesive joint Core material Adhesive joint Face material

Figure 1.1 Schematic of a structural sandwich panel (drawing courtesy of Lars Falk). Two thin, stiff and strong faces are separated by a thick, light and weaker core. The faces are adhesively bonded to the core to obtain a load transfer between the components. The modus operandi of a sandwich is much the same as that of an I-beam, which is an efficient structural shape because as much as possible of the material is placed in the flanges situated farthest from the centre of bending or neutral axis. Only enough material is left in the connecting web to make the flanges work together and to resist shear and buckling. In a sandwich the faces take the place of the flanges and the core takes the place of the web. The difference is that the core of a sandwich is of a different material from the faces and it is spread out as a continuous support for the faces rather than concentrated in a narrow web. The faces will act together to

1.3 AN INTRODUCTION TO SANDWICH STRUCTURES form an efficient stress couple counteracting the external bending moment. The core resists shear and stabilises the faces against buckling or wrinkling. The bond between the faces and the core must be strong enough to resist the shear and tensile stresses set up between them. The adhesive that bonds the faces to the core is thus of critical importance.

The faces usually consist of thin and, in a classical meaning, high performing material while the core material is a thick, light but relatively low performing material. The choice of constituents depends mainly on the specific application and the design criteria set up by it. The design of a structural sandwich will not be one of geometry only but an integrated process of sizing and materials selection.

The advantages given by this design may be summarised: high stiffness and strength to weight ratios, integration of functions such as thermal and acoustic insulation, high energy absorption capability, buoyancy and few details. The way that a sandwich enhances the flexural rigidity of a structure without adding substantial weight has made the concept even more advantageous since the introduction of composite materials. These materials generally offer at least the same or even higher strengths as metals such as aluminium alloy or steel but their moduli are often much lower giving poor stiffness performance. By using sandwiched composites this problem can easily be overcome. At present, the main drawbacks are: production methods in infancy, complicated quality control, load introduction and joining difficulties, and lack of knowledge concerning the effect of damage.

The sandwich concept has had a very strong development during the past years in increasing the knowledge of its mechanical behaviour, which has led to it being used in an increasing number of applications. Future research and development will, step by step, solve the above mentioned drawbacks and prove the sandwich concept even stronger.

There are many successful applications of sandwich constructions. The use of the sandwich concept is usually surrounded by so many obstacles for the designer and the industry (s)he works for that only advantage of chaning a component or structural part from one type of design to a sandwich design is not incentive enough. Therefore, in most application so far several integrated functions of the sandwich concept have together provided enough advantages to enforce a change in a structural concept. Below are pictures of different applications of sandwich structures, where they come from and what advantages are gained by using a sandwich design. The set of applications is by no means complete and new interesting ones appear by the day. Thus,

1.4 INTRODUCTION

Figure 1.1 The Corvette Visby, built entirely in carbond fibre composite sandwich. Courtesy of Kockums AB and FMV.

Figure 1.2 A car bonnet in GRP-sandwich for weight-savings, high stiffness and potentially lower productions costs. Courtesy of Daimler-Benz AG.

Figure 1.3 Steel-sandwich body for sub-way train. Courtesy of Bombardier Transportation.

1.5 AN INTRODUCTION TO SANDWICH STRUCTURES

Figure 1.5 Terraign vehicle in GRP-sandwich built in RTM giving few details, lightweight body and integrated thermal insulation. Courtesy of Hägglunds AB.

Figure 1.6 Example of self-supported truck structure where weight-saving is the main merit but also flat panels with smooth high surface finish and integrated thermal insulation.

Figure 1.8 Helicopter parts such as fuselage, flooring, tail-boom and main rotors. Advantages gained are low weight, fewer parts and therefore lower production costs and easier maintenance. Courtesy of Eurocopter Deutschland GmbH, Germany.

1.6 INTRODUCTION

Figure 1.8 The world’s largest sailing yacht – The Mirabelle being manufactured in a GFRP-sandwich structure. Courtesy of Vosper Thornycroft.

Figure 1.9 Large wind-mill blade are mainly made of single skin composites but some parts have core making up a sandwich structures. Courtesy of LM Glassfibre.

Figure 1.10 Skies are typically manufactured as a sandwich having foam, honeycomb or wood as the core.

1.7 AN INTRODUCTION TO SANDWICH STRUCTURES

Figure 1.11 Structural and semi-structural parts for commercial air-liners such as stabilisers, flaps and doors. The use of sandwich construction ensures control surfaces to remain flat even under load and providing a high stiffness.

Figure 1.12 Solar cell satellite panels giving ultra-high stiffness and strength per unit weight. Courtesy of ESA.

References [1] The American Heritage Dictionary, Second Collage Edition, Houghton Miffin Company.

[2] The American Peoples Encyclopaedia, Grolier Inc., 1964.

[3] Encyclopædia Britannica, vol. 10, 15th edition, Encyclopædia Britannica Inc., 1985.

1.8 INTRODUCTION

[4] W. Fairbairn, An Account of the Construction of the Britannia and Conway Tubular Bridges, John Weale et al., London, 1849.

[5] Flight, 17 Nov., 1938.

[6] Gough G.S., Elam C.F. and de Bruyne N.D., "The Stabilization of a Thin Sheet by a Continuous Supporting Medium", J. Roy. Aero. Soc., Vol 44, Jan. 1940, pp 12-43. [7] Williams D., Leggett D.M.A. and Hopkins H.G., "Flat Sandwich Panels under Compressive End Loads", A.R.C., R & M 1987, 1941.

[8] Plantema F.J., Sandwich Construction, John Wiley & Sons, New York, 1966.

[7] March H.W., “Effects of Shear Deformation in the Core of a Flat Rectangular Sandwich Panel – 1. Buckling under Compressive End Load, 2. Deflection under Uniform Transverse Load”, U.S. Forest Products Laboratory Report 1583, 1948.

[8] Reissner E., “The Effect of Transverse Shear Deformation on the Bending of Elastic Plates”, Journal of Applied Mechanics, Transactions of the ASME, Vol. 12, 1945, pp A69-A77.

[9] Libove C. and Batdorf S.B., “A General Small – Deflection Theory for Flat Sandwich Plates”, NACA TN 1526, 1948, also in NACA report 899.

[10] Hoff N.J., “Bending and Buckling of Rectangular Sandwich Plates”, NACA TN 2225, 1950.

[11] Mindlin R.D., “The Influence of Rotary Inertia and Shear on Flexural Motions of Isotropic, Elastic Plates”, Journal of Applied Mechanics, Transactions of the ASME, Vol. 18, 1951, pp 31-38.

[12] Allen H.G., Analysis and Design of Structural Sandwich Panels, Pergamon Press, Oxford, 1969.

1.9 CHAPTER 2

MATERIALS AND MATERIAL PROPERTIES

As outlined in the previous section, a sandwich consists of three or more constituents, the faces, the core and the adhesive joints. In general, the faces may be of different materials and even the two adhesive joints may be made of different adhesives, all depending on the requirements of the structure or the manufacturing process. The choice of materials is vast and since the introduction of fibre composites the choice of face materials has increased to an almost infinite number of different materials, all with different properties. Even the number of available cores has increased dramatically in recent years since the introduction of more and more competitive cellular plastics. Hence, the design of sandwich structures is just as much a materials selection problem as a sizing problem. The vast number of material choices may appear as an additional complexity but is really one of the main features of using sandwich constructions; the materials best suited for a specific application may be utilised and some drawbacks can be overcome by geometrical sizing. For example, some reinforced plastics lack the advantage of the high stiffness of metals, but by increasing the core thickness a feasible rigidity may still be obtained. Materials are often chosen on grounds that are not purely mechanical but rather for reasons such as environmental resistance, surface finish, the use of a specific manufacturing method, cost, wear resistance etc.

The following section is not intended to be complete in any way but is rather an introduction to the materials commonly used in load carrying sandwich structures. The data presented are typical values collected from different sources and no guarantee is given for their correctness. Material data have a nasty habit of showing quite large scatter, especially data for fibre composites, since they are very dependent on the manufacturing process and the testing method used. However, the numbers given should be fairly typical for the materials described.

2.1 Face Materials Quoting Allen [1], “Almost any structural material which is available in the form of thin sheet may be used to form the faces of a sandwich panel”, gives a good view of the variety available in materials selection. This is a prime feature of the concept, the designer has the opportunity through efficient design to utilise each material component to its ultimate limit.

2.1 AN INTRODUCTION TO SANDWICH STRUCTURES

The properties of primary interest for the faces are;

• High stiffness giving high flexural rigidity • High tensile and compressive strength • Impact resistance • Surface finish • Environmental resistance (chemical, UV, heat, etc.) • Wear resistance

The commonly used face materials can be divided into two main groups; metallic and non- metallic materials. The former group contains steel, stainless steel and aluminium alloys. There is a vast choice of alloys with different strength properties whereas the stiffness variation is very limited. The larger of these two groups is the latter including materials such as plywood, cement, veneer, reinforced plastic, and fibre composites.

Material ρ E 1 σ^ 2 α (kg/m3) (GPa) (MPa) (/°C 10-6) Metals Mild Steel 7800 206 250-500 13 Stainless Steel 7900 196 200 18 Aluminium alloy 2024 2700 73 300 23 Titanium alloy 4500 108 980 9 Wood Pine 520 12 47/7 4 Plywood 580 12.4 21 – 3 Unidirectional fibre composites (vf ≈ 0.6-0.7) Carbon/Epoxy 1600 180/10 1500/40 – Glass/Epoxy 1800 39/8 1060/30 – Kevlar/Epoxy 1300 76/6 1400/12 – 4 Bi-directional fibre composites (vf ≈ 0.3-0.4) Kevlar/Polyester 1300 17.5 375 – Glass weave/Polyester 1700 16 250 – Glass WR (woven roving)/Polyester 1600 12 215 –

5 Random fibres (vf ≈ 0.15-0.25) Glass CSM (chopped strand mat) 1500 6.5 85 – SMC (sheet moulding compound) 1800 9 60 – 1Elastic modulus in/perpendicular to fibre direction, 2Tensile strength in/perpendicular to fibre direction, 3Prepregs with high fibre volume fraction, 4Hand lay-ups with medium fibre volume fraction and 5low fibre volume fraction. Table 2.1 Typical mechanical properties of some commonly used face materials [2,3,4].

The most important of the above groups of materials are fibre composites, which since their introduction have had a major impact on the use of sandwich construction. The reason for this is that most composites offer strength properties similar to or even higher than those of metals, but the stiffness is often magnitudes lower. Thus, in order to achieve high rigidity, composites are more often sandwiched with a light core. Another important reason is that the

2.2 MATERIALS AND MATERIAL PROPERTIES manufacturing of sandwich composites is much easier than the manufacturing of metal face sandwich structures. A feature of composites is their anisotropic behaviour, i.e., they have different properties in different directions. This is an initial complexity often viewed as an obstacle by engineers but is in reality an advantage as it offers the opportunity to tailor properties in conjunction with the applied loads. For example, one can place a sufficient amount of fibres in one direction to carry the load in that particular direction, and a different amount in another direction. Hence, not only the material components are stressed to their ultimate limit but the component itself may be utilised in a more efficient way.

2.2 Estimation of Face Material Properties This section will only deal with estimation of properties for composite materials, since properties of metals are found from testing or handbook data. It starts by showing how the properties of the composite can be approximately calculated by using the properties of the material constituents on a microscopic level. The main purpose is to find approximate properties of the lamina (one layer) as function of the materials building up the lamina. The properties of the lamina can then be used to calculate the properties of the laminate (a stack of lamina), which is to be used as a face sheet of the sandwich. For more information on the topic, refer to references [4] and [5].

A lamina is defined as a thin orthotropic layer of a composite material. It is defined as the smallest element on a macroscopic level to be used to build up a composite laminate. The reasons for using the orthotropic approximation for the lamina are several; the lamina often consists of a of unidirectional, bi-directional or random layer of reinforcement. These can with good approximation be treated as orthotropic layers. It is also a great simplification to use an orthotropic lamina instead of a general anisotropic one for computational purposes.

Figure 2.1 Example of a lamina - a unidirectional layer of fibres and matrix. 2.2.1 Rule-of-mixtures

(i) Determination of E1

The average modulus in the fibre direction of a uni-directional composite lamina E1 is according to the rule-of-mixtures

E1 = Efvf + Emvm (2.1) where vf and vm are the volume fractions (vi = Vi/V), Ef is the fibre modulus and Em the matrix modulus. In more general terms this equation can be written as

EEv1 = ∑ ii (2.2) i

2.3 AN INTRODUCTION TO SANDWICH STRUCTURES where the summation is taken over all constituents. This approximation is called the parallel model, and the result a Voigt material which constitutes an upper bound of the elastic modulus any composite can have. The stress in the composite is thus a volume weighted average.

σ11= ∑σ iiv (2.3) i

(ii) Determination of E2 The apparent Young's modulus in the direction transverse to the fibres can be assessed in a similar way. 1 v v 1 v =+f m or more generally = ∑ i (2.4) E2 E f Em E2 i Ei

This approximation is called the serial model, and the result a Reuss material which constitutes a lower bound of the elastic modulus any composite can have.

(iii) Determination of ν12

The Poisson ratio ν12 = – ε2/ε1 by definition (rule-of-thumb: index 1 - direction of load, index

2 - direction of deformation). If we denote νf the fibre Poisson ratio and νm the matrix Poisson ratio we have as for the serial model that,

ν12 = νfvf + νmvm or more generally ν12 = ∑ νivi (2.5) i

(iv) Determination of G12

The in-plane shear modulus G12 is found by assuming that the shear stress acting on the fibres and the matrix are equal. If Gf is the fibre shear modulus and Gm the matrix shear modulus, the average shear modulus of the lamina is 1 v v 1 v =+f m or more generally = ∑ i (2.6) G12 Gf Gm G12 i Gi

2.2.2 “Practical” rule-of-mixtures It is known from experience that the Voigt material model agrees well with experiments whereas the Ruess model shows very poor agreement, especially for unidirectional composites. The tensile modulus may be predicted for a unidirectional composite using the rules-of-mixture but in practice the composite laminate may be built up of fabrics, e.g., weaves, roving, mats and/or in combination with unidirectional layers. The rule-of-mixtures in the fibre direction may be modified to suit this purpose by simply introducing reinforcement efficiency factors. The mixture rule along the fibres may now take the form

EvE1 = ∑αi ii (2.7) i In practice, this means that we only consider the amount of fibres in the laminate acting in the studied direction. For example, a bi-directional symmetric weave has only half of its fibres in the 1-direction and the other half in its 2-direction. Thus, the efficiency factor should be 0.5.

2.4 MATERIALS AND MATERIAL PROPERTIES

Hence, the efficiency factor α is simply the fraction of the fibres acting in the studied direction. The values of α are for different reinforcements 1 for unidirectional fibres and the matrix 0.5 for bi-directional symmetric 0.375 for random in-plane 0.2 for random in space For any other configuration use the definition above, e.g., a weave with 70% of the fibres in its 1-direction and 30 % in its 2-direction, α = 0.7.

2.2.3 Conversion weight fraction/volume fraction Most calculations on composites are carried out on the basis of volume fractions. However, in production it is normal to work in terms of weight fractions. It is therefore necessary to convert from one to the other. The following formulae are often useful.

The total weight is: WW= ∑ i =∑ρii VVv= ∑ρii (2.8a) i i i

Wiiiρ V ρiiv The weight fraction is: wi == = with ∑ wi = 1 (2.8b) W Vv∑∑ρii ρiiv i i i

Wi wi The total volume is: VV==∑∑i =W ∑ (2.8c) i i ρρi i i

ViiiW / ρ wii/ ρ The volume fraction is: vi == = with ∑ vi = 1 (2.8d) V Ww∑∑ii/ ρ wii/ ρ i i i

2.2.4 Thickness prediction The simplest way to estimate a composite thickness is by using the rule-of-mixtures once again. The thickness of 1 m2 of a material weighing 1 kg and with a density ρ (kg/m3) is simply 1/ρ (m). If a laminate consists of Wi (kg) of material i then the total volume equals

ΣWi/ρι. Hence, the thickness can be predicted using W * t = ∑ i where W* is the weight per unit area (2.9) i ρi

* Since the weight fraction, wi, usually is known, the total weight of the components Wf (weight * of reinforcement per unit area) and Wm (weight of matrix per unit area), are readily available and the thickness may therefore commonly be estimated. In practice, composite laminates always contains voids of unwanted particles or air bubbles that will increase the thickness. Anyone who has ever made a laminate will also know that the thickness will vary over the laminate, especially if it is made by hand lay-up.

2.5 AN INTRODUCTION TO SANDWICH STRUCTURES

2.2.5 Stiffness properties of the lamina Since the lamina layer is thin a state of plane stress prevails and we have

σ3 = τ23 = τ31 = 0 Hooke's law for such a layer is then

⎛ ε 1 ⎞ ⎡ 10//EE1212−ν ⎤⎛ σ 1 ⎞ ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ ε 2 ⎟ = −ν 12//EE 110 2 ⎜σ 2 ⎟ (2.10) ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎝γ 12 ⎠ ⎣⎢ 001/ G12 ⎦⎥⎝τ 12 ⎠ The inverse of the compliance matrix is easily found as

⎛ σ ⎞ ⎡ EEν 0 ⎤⎛ ε ⎞ ⎜ 1 ⎟ 1 1211 ⎜ 1 ⎟ σ = ⎢ν EE 0 ⎥ ε (2.11) ⎜ 2 ⎟ ⎢ 12 2 2 ⎥⎜ 2 ⎟ ⎜ ⎟ 1− νν12 21 ⎜ ⎟ ⎝τ 12 ⎠ ⎣⎢ 00G12() 1− νν 12 21 ⎦⎥⎝γ 12 ⎠ or in a shorter form: σl = Ql εl where index l stands for lamina.

In the manufacturing of composite laminates several lamina are assembled on top of each other but with different orientation to the global coordinate system. Thus, in order to describe the behaviour of the laminate one must know the behaviour of each lamina described in the global coordinate system (x, y, and z) and not in its local coordinate system (1,2 and 3). The stiffness of the lamina in the global coordinate system x-y is found by using the transformation matrix. Assuming that the local coordinate system, 1-2, is rotated and angle θ to the global coordinate system, x-y. The transformation matrix is then written as

22 ⎛ σ 1 ⎞ ⎡ cs2 sc⎤⎛σ x ⎞ ⎜ ⎟ ⎢ 22 ⎥⎜ ⎟ σ =−sc2 scσ or = T-1 (2.12a) ⎜ 2 ⎟ ⎢ ⎥⎜ y ⎟ σl σ ⎜ ⎟ ⎢ 22⎥⎜ ⎟ ⎝τ 12 ⎠ ⎣−−sc sc c s ⎦⎝τ xy ⎠ where c = cosθ and s = sinθ. Alternatively we can write

22 ⎛σ x ⎞ ⎡cs−2 sc⎤⎛ σ 1 ⎞ ⎜ ⎟ ⎢ ⎥⎜ ⎟ σ = sc22 2 scσ or σ = Tσ (2.12b) ⎜ y ⎟ ⎢ ⎥⎜ 2 ⎟ l ⎜ ⎟ ⎢ 22⎥⎜ ⎟ ⎝τ xy ⎠ ⎣sc−− sc c s ⎦⎝τ 12 ⎠ where T is the transformation matrix. For the strains we can in the same way write

22 ⎛ ε 1 ⎞ ⎡ cs sc⎤⎛ ε x ⎞ ⎜ ⎟ ⎢ 22 ⎥⎜ ⎟ ε =−sc scε or ε = Ttε (2.12c) ⎜ 2 ⎟ ⎢ ⎥⎜ y ⎟ l ⎜ ⎟ ⎢ 22⎥⎜ ⎟ ⎝γ 12 ⎠ ⎣−−22sc sc c s ⎦⎝γ xy ⎠ and similarly

2.6 MATERIALS AND MATERIAL PROPERTIES

22 ⎛ ε x ⎞ ⎡ cs− sc⎤⎛ ε 1 ⎞ ⎜ ⎟ ⎢ 22 ⎥⎜ ⎟ ε = sc scε or = (T-1)t (2.12d) ⎜ y ⎟ ⎢ ⎥⎜ 2 ⎟ ε εl ⎜ ⎟ ⎢ 22⎥⎜ ⎟ ⎝γ xy ⎠ ⎣22sc−− sc c s ⎦⎝γ 12 ⎠

Now we can write the transformed stiffness matrix as

t QTQT= l (2.13)

Hence, the stress-strain relation in the global coordinate system can be written

⎛σ ⎞ ⎡QQQ⎤⎛ ε ⎞ ⎜ x ⎟ 11 12 16 ⎜ x ⎟ σ = ⎢QQQ⎥ ε or σ = Qε ⎜ y ⎟ ⎢ 12 22 26 ⎥⎜ y ⎟ ⎜ ⎟ ⎜ ⎟ ⎝τ xy ⎠ ⎣⎢QQQ16 26 66 ⎦⎥⎝γ xy ⎠ whose components written out in full reads

4 2 2 4 Q11 = Ql11 cos θ + 2(Ql12 + 2Ql66) cos θ sin θ + Ql22 sin θ 2 2 4 4 Q12 = (Ql11 + Ql22 – 4Ql66) cos θ sin θ + Ql12 (sin θ + cos θ) 4 2 2 4 Q22 = Ql11 sin θ + 2(Ql12 + 2Ql66) cos θ sin θ + Ql22 cos θ 3 3 Q16 = (Ql11 – Ql12 – 2Ql66) sinθ cos θ + (Ql12 – Ql22 + 2Ql66) sin θ cosθ 3 3 Q26 = (Ql11 – Ql12 – 2Ql66) sin θ cosθ + (Ql12 – Ql22 + 2Ql66) sinθ cos θ 2 2 4 4 Q66 = (Ql11 + Ql22 – 2Ql12 – 2Ql66) sin θ cos θ + Ql66 (sin θ + cos θ)

As an alternative to the foregoing we can write all this in compliances. Now, if

−1 t −1 S = ()T SlT (2.14) which is the transformed compliance matrix. Its components are

4 2 2 4 S11 = Sl11 cos θ + (2Sl12 + Sl66) cos θ sin θ + Sl22 sin θ 2 2 4 4 S12 = (Sl11 + Sl22 – Sl66) cos θ sin θ + Sl12 (sin θ + cos θ) 4 2 2 4 S22 = Sl11 sin θ + (2Sl12 + Sl66) cos θ sin θ + Sl22 cos θ 3 3 S16 = (2Sl11 – 2Sl12 – Sl66) sinθ cos θ – (2Sl22 – 2Sl12 – Sl66) sin θ cosθ 3 3 S26 = (2Sl11 – 2Sl12 – Sl66) sin θ cosθ – (2Sl22 – 2Sl12 – Sl66) sinθ cos θ 2 2 4 4 S66 = 2(2Sl11 + 2Sl22 – 4Sl12 – Sl66) sin θ cos θ + Sl66 (sin θ + cos θ)

2.2.6 Stiffness properties of the laminate A laminate is a stack of arbitrary oriented laminae assembled to form a plate or a shell. The lamina was considered being thin hence having rigidities in its plane only, i.e., no bending stiffness. The laminate on the other hand has a finite thickness and thus a flexural rigidity.

2.7 AN INTRODUCTION TO SANDWICH STRUCTURES

2 z 1

Figure 2.2 A stack of lamina - a laminate. Denote (see also Fig.8.1)

t t N = NNNxyxy,, and M = MMMxyxy,,

t t t σε00==σστxi,, yi xyi , εεxo ,, yo γ xyo and κ = κκκxyxy ,,

Assume that the thickness and position of each lamina in the stack are known, as shown in Fig.2.3.

z0 1 2 3

z i i-1 zi

Figure 2.3 Geometry of an n-layered laminate. (Note that the total thickness t = 2z0). The relation between forces and bending moments to strains and curvatures can be written

N n zi ⎛ ⎞ ⎛ QQiiz ⎞⎛ε 00⎞ ⎛ AB⎞⎛ε ⎞ ⎜ ⎟ = dz = (2.15) ⎜ ⎟ ∑ ∫ ⎜ 2 ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ i=1 ⎝QQiizz⎠⎝ κ ⎠ ⎝ BD⎠⎝ κ ⎠ ⎝ M⎠ zi −1 where

n 1 1 A, B, D=− Q ⎡ zz,, z2 − z23 z − z3 ⎤ (2.16) []∑ i ⎢()ii−−−1 () i i1 () i i1 ⎥ i=1 ⎣ 2 3 ⎦

A is called the extensional stiffness matrix, B the extension-bending coupling matrix, and D the bending stiffness matrix.

By computing the stiffness matrix components we can estimate the desired properties for the face sheet. Assume that the face sheet laminate is orthotropic (which is true in most practical cases), we can then get the Poisson's ratios as

2.8 MATERIALS AND MATERIAL PROPERTIES

A12 A12 ν21 = and ν12 = (2.17) A11 A22 The average moduli are then A ()1− ν ν A ()1− ν ν A E = 11 12 21 , E = 22 12 21 and G = 66 (2.18) x t y t xy t where t is the laminate thickness. For symmetrical laminates we have that B = 0 and for orthotropic ones also that D16 = D26 = 0. The local bending stiffness’ of the face sheet laminate are then simply (see section 3 and 9)

D11 D22 Dfx = , Dfy = and Dfxy = 2D66 (2.19) 1− νν12 21 1− νν12 21 For sandwich structures, we need further relations than just for the composite face sheets. The core in the sandwich will, as will be seen in the next chapter, deform in shear. To treat this in a general way, one also need to be able to calculate and transform shear stiffness’, which stem from the transformation of shear stresses and strains. The first problem is to transform the properties in shear to an arbitrary coordinate system x'-y'. The other interesting properties are the shear moduli. Assuming that the material is orthotropic in at least some coordinate system, which will be true for all engineering materials, especially sandwich cores like honeycombs, foams, etc. These can then in a local 1-2 system be described by the stress-strain relation from

⎛τ 23 ⎞ ⎡G23 0 ⎤⎛γ 23 ⎞ ⎜ ⎟ = ⎢ ⎥⎜ ⎟ or τl = Cl γl (2.20) ⎝τ 13 ⎠ ⎣ 0 G13 ⎦⎝γ 13 ⎠ since honeycombs can be treated as orthotropic in at least one co-ordinate system. Study the geometrical changes imposed by the shear on a small section of the panel, as shown in Fig.2.4. γ x xz

z,3 y 2 2 γ γ γ 13 yz 23 γ yz 1 1 γ xz

Figure 2.4 Geometrical illustration of shear angle transformation The shear angles can from Fig.2.4 be written as

γ13 = γxz cosθ γ13 = γyz sinθ

γ23 = −γxz sinθ γ23 = γyz cosθ so that the shear angle transformation becomes

2.9 AN INTRODUCTION TO SANDWICH STRUCTURES

⎛γ ⎞ ⎡c − s⎤⎛γ ⎞ ⎜ 23 ⎟ ⎜ yz ⎟ t ⎜ ⎟ = ⎢ ⎥⎜ ⎟ or γ l = Ts γ (2.21) ⎝γ 13 ⎠ ⎣s c ⎦⎝γ xz ⎠

The shear stresses are transformed through their equilibrium equation and the results is

⎛τ ⎞ ⎡c − s⎤⎛τ ⎞ ⎜ 23 ⎟ ⎜ yz ⎟ −1 ⎜ ⎟ = ⎢ ⎥⎜ ⎟ or τ l = Ts τ (2.22) ⎝τ 13 ⎠ ⎣s c ⎦⎝τ xz ⎠

We can then write

t t τ = Ts τ l = TsCl γ l = TsClTs γ which means that C = TsClTs = TsClTs which written out in full appears as

2 2 ⎡G23c + G13 s (G13 − G23 )sc ⎤ C= ⎢ 2 2 ⎥ (2.23) ⎣(G13 − G23 )sc G23 s + G13c ⎦

This relation is also simple to derive from tensor transformation. In the same manner as the transformation of shear moduli, we can transform shear stiffness’ as (with changed order of the transverse force components)

⎛T2 ⎞ ⎡S 2 0 ⎤⎛γ 23 ⎞ ⎜ ⎟ = ⎢ ⎥⎜ ⎟ or Τl = Slγl (2.24) ⎝T1 ⎠ ⎣ 0 S1 ⎦⎝γ 13 ⎠

t These can now be transformed to any other co-ordinate system by S = T Sl T . In this way we can get the shear stiffness’ in any co-ordinate system, generally written as

⎛T ⎞ ⎡ S S ⎤⎛γ ⎞ ⎜ y ⎟ = y xy ⎜ yz ⎟ or Τ = Sγ (2.24) ⎜ ⎟ ⎢S S ⎥⎜ ⎟ ⎝Tx ⎠ ⎣ xy x ⎦⎝γ xz ⎠ where

2 2 Sx = Sxs + Syc

Sxy = (Sy − Sx)sc (2.25)

2 2 Sy = Sys + Sxc 2.2.7 Strength of composite laminates There exists an almost infinite number of fracture criteria for composite laminates, ranging from very simple estimates based on the rule-of-mixtures, to quite complex multi-dimensional criteria aiming at incorporating every possible effect of different stresses and failure modes. These will not be covered in this text but some of the most common fracture criteria are given in references [3,4,5].

2.3 Experimental Determination of Face Material Properties By testing it is here understood that testing is performed for the reason of extracting reliable material properties that are to be used in design or analysis of sandwich structures. As seen

2.10 MATERIALS AND MATERIAL PROPERTIES above, it is possible to estimate the properties of composite laminates using different theoretical analyses, and this can be done with fairly good accuracy. For metals one is, on the other hand, forced to find reliable data from manufactures data sheets or handbooks. The most reliable way to obtain accurate material properties for all materials is to perform tests in order to extract the necessary data. For sandwich structures, the most important data for the face sheets are the in-plane tensile and compressive moduli and strengths in different directions, the in-plane shear properties and the interlaminar strength.

(i) Tensile properties Tensile testing of sheet metal and composite laminates are usually performed by using a strip- like test specimen which is mounted in a testing machine and tested under a prescribed cross- head displacement rate. The specimen may be a straight, strip with or without end-tabs, or shaped as some kind of dog-bone with a narrow test section. There are several different standards for the tensile test of composites, e.g., ASTM D638M, ASTM D3039, ISO 3268 or DIN EN61 (see Fig.2.5), and other standards for metals. In order to measure the elastic modulus of the specimen, strain gauges or an extensometer is mounted on the specimen to measure the displacement (or strain). Different standards can yield slightly different results depending on the type of specimen used and how the strains are measured and moduli extracted.

R60

15 50 150

Figure 2.5 Schematic of a tensile test specimens. Typical dimensions (not referring to any special standard). For composite laminates, the test must be performed in different directions since there may be significant differences between different directions. Another reason for performing tests in different directions is the in-plane shear modulus of the laminate can be determined that way. There exists a standard for this type of test described in ASTM D3518-76. Assume that tests are performed in 0°, 90° and 45° directions of the laminate so that values of the modulus are obtained, E1 = E0, E2 = E90 and E45. From eq.(2.14) we then have

2.11 AN INTRODUCTION TO SANDWICH STRUCTURES

11⎡ 1 ⎛ 111⎞ ⎤ =+⎢ 2⎜ +⎟ + ⎥ (2.26) EE45 4 ⎣ 1211122⎝ν EGE⎠ ⎦

If the off-axis measurement (E45) is made, then the in-plane shear modulus G12 can be calculated. The off-axis specimen is more often taken in only 5° or 10° since that yields a better result, but in principal the estimate of G12 is the same although eq.(2.13) takes a little different form for other angles than 45°.

Other important data extracted from the tensile test are, of course, the ultimate tensile strength of the material in x- and y-directions and perhaps also the ultimate strains to failure in different directions.

(ii) Compressive properties The compressive properties are required for the same reason as the tensile. For many materials, the compressive moduli and strengths are quite different from those in tension. Compression testing is far more difficult than that in tension due to the inherent problem of buckling. To obtain a good compressive failure can be very difficult. The specimens used for compression testing usually have the same geometry and dimension as the tensile ones, but the rigging is different and consists of a supporting device that prevents the specimen from buckling. Some available standards for composite laminates are the DIN 53454, ASTM D695 and ISO 604 (see Fig.2.6).

Figure 2.6 Schematic illustration of the DIN 53454 compression test The same type of data as in the tensile test are obtained, but for compressive loading.

(iii) In-plane shear properties

As described above, it is possible to obtain values for the in-plane shear modulus, G12, by using an off-axis tensile or compressive test. No value can, however, be obtained for the in-

2.12 MATERIALS AND MATERIAL PROPERTIES plane shear strength of the material using that test. The ultimate in-plane shear strength is generally less important in the design of sandwich constructions but might in special cases be an essential property. The shear test is according to DIN 53399 is schematically illustrated in Fig.2.7. A composite laminate is bolted to a steel frame with a test window of about 60 by 60 mm. At the upper and lower ends of the steel frame a special loading device is attached which is then itself attached to the testing machine. By applying either a tensile or a compressive loading the laminate will shear.

Figure 2.7 Schematic of shear test (iv) Interlaminar strength Delamination is a common failure mode for composite laminates. But for laminates used as face sheets in a sandwich it is usually of little importance since the core in most cases has a lower tensile strength than the delamination strength of the laminate. However, for sandwiches with high-density cores, delamination could be a critical failure mode. Delamination can also occur due to either a tensile load transverse to the laminate or due to an out-of-plane shear load.

The tensile delamination strength is measured using a tensile test where a circular part of a laminate is bonded between to cylindrical metal chucks that are mounted into a testing machine and pulled apart. In the DIN 53397 the metal chucks and test piece has a diameter of 56 mm.

Another relevant test in this context is the apparent interlaminar shear strength, i.e., delamination due to shear. A good method to measure this property is the so-called short beam shear test described in ASTM D2344. A very short beam of a composite laminate is tested in three point bending so that the transverse shear stress in the beam is high compared to the in-plane bending stress. The beam will then fail in shear, i.e., a delamination crack will start growing at the edge in the middle of the laminate.

2.4 Core Materials This material component is perhaps the most important of all even though it might not appear as such at first glance. It is also the material component of which the engineer commonly has the least knowledge. The cores used in load carrying sandwich constructions can be divided into four main groups; corrugated, honeycomb, balsa wood and cellular foams, schematically illustrated in Fig.2.8.

2.13 AN INTRODUCTION TO SANDWICH STRUCTURES

Corrugate Honeycomb Cellular or d balsa

Figure 2.8 Main groups of core materials. First of all the core should have low density in order to add as little as possible to the total weight of the sandwich. Furthermore, its Young's modulus perpendicular to the faces should be fairly high to prevent a decrease in the core thickness and therefore a rapid decrease in the flexural rigidity. Even though the transverse forces creating normal stresses perpendicular to the core are usually low, even a small decrease in core thickness would create a large decrease in the flexural rigidity. The core is mainly subjected to shear so that the core shear strains produce global deformations and core shear stresses. Thus, a core must be chosen that would not fail under the applied transverse load and with a shear modulus high enough to give the required shear stiffness. The critical wrinkling load depends on both the Young's modulus and the shear modulus of the core. Other functions of the sandwich such as thermal and acoustical insulation depend mainly on the core material and its thickness. The properties of primary interest for the core may then be summarised as;

• Low density • Shear modulus • Shear strength • Stiffness perpendicular to the faces • Thermal insulation

Following is a description of the most commonly used core materials in load carrying sandwich constructions and their typical mechanical and physical properties. Only a limited set of data is given intended to show only the major differences between different materials.

2.4.1 Honeycomb cores Core materials of honeycomb type have been developed and used primarily in aerospace applications. However, cheap honeycomb materials made from impregnated paper are also used in building applications. Honeycomb materials can be manufactured in a variety of cell shapes but the most commonly used shape is the hexagonal shape shown in Fig.2.9a. Others are the square (2.8b), the over-expanded hexagonal (2.8c) or the so-called “flex-core” (2.8d). The two latter configurations are primarily used when the core needs to be curved in the manufacturing of the sandwich element. Over-expanded hexagonal and flex-core shapes reduce the anticlastic bending and cell wall buckling when curved. There are other cell shapes used such as rectangular, and reinforced hexagonal.

2.14 MATERIALS AND MATERIAL PROPERTIES

(a) (b) W

Figure 2.9 Commonly used cell configurations for honeycomb core materials. (a) Hexagonal, (b) square, (c) over expanded hexagonal, and (d) flex-core. The manufacturing of metal honeycombs is performed in two different ways: the corrugating process implies that pre-corrugated metal sheets are bonded together and stacked into blocks. When the adhesive has cured, blocks with the required thickness can be cut from the stack (see Fig.2.10). This process is commonly used in the manufacture of high-density metal honeycombs. The expansion process begins with the stacking of thin plane sheets of the web material on which adhesive nodes have been printed. By stacking many thin layers in this way a block is made. Each block may then be cut into a desired thickness (T-direction). When the adhesive has cured it may be expanded by pulling in the W-direction until a desired cell shape has been achieved.

web

Corrugation

web

Expansion

Figure 2.10 Manufacture of honeycomb cores - corrugating and expansion process. Aluminium alloy honeycomb has been extensively used in aerospace applications during the past decades. They are commonly made of the aluminium alloys 5052, 5056 and 2024. 5052 is a general purpose alloy, 5056 a high strength version of 5052 and 2024 a heat treated aluminium alloy with good properties even at elevated temperature. The 5052 and 5056 alloy

2.15 AN INTRODUCTION TO SANDWICH STRUCTURES honeycombs can be used in environments up 180°C and the 2024 up to 210°C. The foil out of which the honeycombs is made can be perforated with small holes so that curing of the adhesive may be vented to reduce the risk for corrosion due to the water condensation.

Kraft paper honeycombs are manufactured by impregnating paper with resin to make it water resistant. This provides a cheap, but still mechanically very good sandwich core. Some manufacturers can even fill the cells of Kraft paper honeycomb with a lightweight foam (usually PUR or phenolic) for improved thermal insulation.

1 ^ 1 Density Gc τ λ kg/m3 MPa MPa W/m °C Paper honeycomb 23 33/10 0.24/0.09 0.066-0.114 2 56 141/38 1.3/0.48 '' Aluminium alloy honeycomb 32 180/98 0.83/0.48 3.9 50 310/150 1.4/0.89 5.4 70 460/200 2.2/1.5 5.4 92 620/260 3.1/2.0 8.8 110 780/320 4.0/2.5 14.4 130 930/370 5.0/3.1 14.4 Glass/phenolic honeycomb 64 96/48 1.9/1.1 0.066-0.114 2 88 131/75 3.0/1.6 '' 112 206/89 4.0/2.0 '' 192 330/190 6.4/4.2 '' Nomex® honeycomb 29 26/14 0.56/0.34 0.066-0.114 2 48 42/28 1.2/0.68 '' 80 69/44 2.2/1.0 '' 129 112/64 3.2/1.7 '' 1Property in length (L) / width (W) direction, 2Varying with cell size. Table 2.2 Typical mechanical and thermal properties of some commonly used honeycomb core materials [6-13]. Non-metallic honeycombs, like fibre-reinforced plastic honeycombs, are manufactured by impregnating a prefabricated cell-shaped fabric in a bath of resin. Different honeycombs are available with glass, aramid or even carbon fibre fabric reinforcement. The matrix which the fabric is impregnated with is usually phenolic, heat resistant phenolic, polyimide or polyester. The phenolic impregnated types have a maximum working temperature up to 180°C, the polyimide ones can be used up to 250°C and the polyester impregnated up to 80°C. A well- known type of fibre-impregnated honeycomb is made of Nomex® paper, which is an aramid fibre based fabric expanded in much the same way as aluminium alloy honeycomb before being coated with resin. It is widely used because of its high toughness and damage resistance and since it has almost as high mechanical properties as aluminium alloy honeycomb. Nomex

2.16 MATERIALS AND MATERIAL PROPERTIES honeycomb can be used up 180°C at which its strength still is about 75% of its room temperature value.

The most commonly used honeycombs are made of aluminium alloy or impregnated glass or aramid fibre mats, such as Nomex®. Due to the manufacturing methods involved most honeycombs have not only different out-of-plane properties but also the in-plane properties are different from each other as both the corrugation and the expansion process produces double cell walls in one direction and single walls in the other. Over-expanded cells also create additional anisotropy. The three principal directions to which material properties of honeycombs are referred are; width (W), length (L) and transverse (T) directions, as indicated in Fig.2.8.

Data for some commonly used honeycomb core materials are given in Table 2.2. Honeycombs have excellent mechanical properties, very high stiffness perpendicular to the faces and the highest shear stiffness and strength to weight ratios of all available core materials. The main drawback is high cost. They are fairly easily shaped and conform well to doubly curved shapes due to the low in-plane modulus. A pre-set curvature could still be quite difficult to achieve due to the Poisson effect. This can be overcome by using another cell shape [14]. An example of this is that by varying the hexagonal cell shape, a core can be made that has a negative Poisson ratio, implying that the secondary curvature has the same sign as the primary (see eq.(8.3)). Hence, a core can be manufactured to fit a specific double or even single curvature. If the Poisson ratio νxy or νyx equals zero this implies that a sheet of core can be bent into a cylinder applying a single bending moment. If this Poisson ratio is made positive, as for hexagonal shape honeycombs, the bending will be anticlastic (different signs on κx and κy) making the sheet difficult to fit the shape of a doubly curved mould, but if the Poisson ratio is negative, as for the shape in Fig.2.11, the reaction to a single bending moment will be synclastic (same signs on κx and κy) and the core sheet will be easier to handle.

Figure 2.11 Example of a honeycomb with negative Poisson ratio. 2.4.2 Balsa wood Balsa was the first material used as cores in load carrying sandwich structures. Balsa is a wood but under a microscope it can be seen as a high-aspect-ratio closed-cell structure. The fibres or grains are oriented in the direction of growth producing cells with a typical length of 0.5-1 mm and with a diameter of about 0.05 mm, thus giving the cell ratio of approximately 1:25. The properties of balsa are therefore high in direction of growth but much lower in the others. Balsa exists in different qualities with densities in the regime 100 to 300 kg/m3. Balsa

2.17 AN INTRODUCTION TO SANDWICH STRUCTURES is also very sensitive to humidity with the properties rapidly declining with the water content. To overcome the above problem balsa is most commonly utilised in its “end-grain” shape. This means that the balsa wood is cut up in cubic pieces and bonded together edge wise so that a block is produced where the fibre direction is located perpendicular to the plane of the block. With this procedure several advantages are gained; the fibres and hence the principal direction of stiffness is perpendicular to the faces, and humidity is primarily spread along the fibres and hence damage would only cause localised humidity damage. The drawback is that all the small balsa blocks have different densities and the design limit must hence be taken from the piece having the lowest properties. End-grain balsa is only available in a limited number of densities. The mechanical properties, even though restricted to the minimum density, appear quite good and higher than most cellular plastics.

2.4.3 Cellular foams The relatively recent developments of high density and high quality cellular foams has had a major impact on the use of the sandwich concept. Cellular foams do not offer the same high stiffness- and strength-to-weight ratios as honeycombs but have other very important advantages. Firstly, cellular foams are in general less expensive than honeycombs but more importantly, a foam is a solid on a macroscopic level making the manufacturing of sandwich elements easier; the foam surface is easy to bond to, surface preparation and shaping is simple and connections of core blocks are easily performed by adhesive bonding. In addition to this, cellular foams offer high thermal insulation, acoustical damping, and the closed cell structure of most foams ensure that the structure will become buoyant and resistant to water penetration. There exist a variety of foams, with different advantages and disadvantages. Some of these are briefly described below [6].

Polyurethane foam (PUR) The urethane polymer is formed through the reaction between iso-cyanate and polyol, and tri- chlor-fluor-methane or carbon dioxide is used as a blowing agent and is vaporised by the heat released by the exothermal reaction. PUR foams are produced in many variations from soft foam with more or less open cells to rigid types with predominantly closed cells and in a wide density range. They can be made fire resistant by using additives containing phosphorus. Because of the high molecular weight PUR foams have low thermal conductivity and diffusion coefficients giving them very good insulation properties. Rigid PUR foams generally have quite brittle cell walls and hence the PUR core has low toughness and low ultimate elongation. The mechanical properties are lower than most other cellular plastic cores but PUR foams are probably the cheapest of all available core materials. The primary use of PUR is for insulation purposes or in less critical load bearing elements. An advantage is that PUR foam can be produced in finite size blocks as well as being foamed in-situ thus giving an integrated manufacturing process in conjunction with the manufacturing of sandwich elements.

Polystyrene foam (PS) Polystyrene foam is produced either by extrusion or by expansion in closed moulds. In both cases the plastic is mixed with the blowing agent, which then expands at elevated temperature. A major obstacle was that CFC was used as a blowing agent, but recently PS foams have been expanded without use of environmentally dangerous CFC-gases. PS has

2.18 MATERIALS AND MATERIAL PROPERTIES closed cells and is available in densities ranging from 15 to 300 kg/m3. PS foam has quite good mechanical and thermal insulation properties, and it is cheap. A drawback is its sensitivity to solvents, particularly styrene, and hence ester-based matrices cannot be used as adhesives. PS is primarily used as a thermal insulation material but lately it has also been used in load carrying structures such as refrigerated tanks and containers.

^ Density Gc σ λ kg/m3 MPa MPa W/m °C Balsa Wood 96 108 1.85 0.0509 130 134 2.49 0.0588 180 188 3.46 0.0710 Polyurethane foam 30 3 0.2 0.025 40 4 0.25 0.025 Polystyrene foam 30 8 0.25 0.035 60 20 0.6 0.035 Polyvinyl chloride foam (linear) 80 18 1.1 0.034 Polyvinyl chloride foam (cross-linked) 45 18 0.5 0.024 80 31 1.0 0.028 100 40 1.4 0.030 130 52 2.0 0.034 200 85 3.3 0.043 Poly-metacryl-imide foam 52 19 0.8 – 75 29 1.3 – 110 50 2.4 – 205 150 5.0 –

Table 2.3 Typical mechanical and thermal properties of some core materials [6-13]. Polyvinyl chloride foam (PVC) PVC foam exists in two different forms; one purely thermoplastic also called linear PVC foam, and one cross-linked iso-cyanate modified type. The linear PVC has great ductility, quite good mechanical properties but softens at elevated temperatures. The cross-linked PVC is more rigid, has higher mechanical properties, is less heat sensitive, but more brittle. Still, even the cross-linked PVC has an ultimate elongation of about 10% in tension that is much higher than any PUR foam. PVC foam is available in finite size blocks with densities from 30 to 400 kg/m3. The mechanical properties of PVC are higher than those of both PUR and PS, but it is also more expensive than those. It is a non-flammable foam but when burnt a hydrochloric acid gas is released. PVC foams are used in almost every type of sandwich application varying from pure insulation applications to aerospace structures and are hence the most widely used of all foams and perhaps of all core materials. PVC has about 95%

2.19 AN INTRODUCTION TO SANDWICH STRUCTURES closed cells for the lower densities and almost entirely closed cells for the higher, which is much appreciated in applications where water absorption is a problem.

Poly-methacryl-imide foam (PMI) Acryl-imide cellular plastics are made from expanded imide-modified polyacrylates. The mechanical properties are good, perhaps the best of all commercially available cellular foams, but the price is also the highest. PMI is fairly brittle with an ultimate elongation of approximately 3% in tension. The main advantage is the temperature resistance making it possible to use PMI foam in conjunction with epoxy prepregs in autoclave manufacturing in up to 180°C environments. The cell structure is very fine with closed cells and the densities available are from 30 to 300 kg/m3.

In Table 2.3 data for some commonly used core materials are given. Poisson's ratios are not included due to lack of reliable data. However, some unpublished results indicate that the Poisson's ratios for most foams are in the regime 0.2 to 0.4, although varying in different directions due to anisotropy. Most foam cores are only moderately anisotropic but with fairly equal properties in the plane of the block (xy-plane), and usually slightly higher in the out-of- plane direction.

2.5 Fatigue Properties of Sandwich Core Materials Repetitive stresses or deformations often cause damage or failure even if these are well below their allowable values for static strength. This phenomenon is usually called fatigue and occurs in, for example, slamming of boat hulls, vibration of non-moving parts in vehicles and aircraft, or any other type of repetitive loading of structures. Fatigue is generally said to cause the major part of all structural failures, but for sandwich structures this is not true. Sandwich constructions have gained a reputation for being a very good concept in avoiding fatigue failures. One reason may be that the faces may fail in local instability at loads lower than their fatigue limit and the core is designed with a high margin of safety due to lack of knowledge about its fatigue properties. It is vital to realise that the constituents in a sandwich are subjected to different kinds of loading; the faces exhibit almost entirely membrane tension/compression and the core pure shear. Fatigue of the faces could then easily be included in the design process since fatigue properties commonly are known, especially for metals, and the loading situation is simple. The core, on the other hand, exhibits a more complex loading situation and fatigue data are almost non-existent. Repetitive shear stresses may also cause fatigue failure of the adhesive joint. Remembering that most adhesives and cellular foam cores are viscoelastic, this makes the problem of fatigue even more complex. Even so, it is generally considered to be an obstacle that the fatigue properties of the core and the adhesive cannot be included in the design process.

Some limited data are available which are obtained using the ASTM C273 shear test specimen, for fatigue loading using R = 0.1 (R = Pmin/Pmax) [15,16]. Tests were performed on balsa wood, some metallic and non-metallic honeycombs and also on a few foams. Since these tests were conducted several years ago, most of the materials used there are today seldom used but the results are however still adequate and anyone interested can find S-N curves in refs. [15,16].

2.20 MATERIALS AND MATERIAL PROPERTIES

A new approach to test sandwiches and sandwich core materials in fatigue with emphasis on the shear behaviour of the core is to use a four-point bending test. If designed properly, the shear stress in the core in the region between the outer and inner supports will be critical and a shear fatigue failure in the core can be obtained even for cyclic loading. If special rigs are used, tests with negative R-values can be performed. The results given here were performed using such an apparatus. They are non-dimension shear fatigue data for some sandwich core materials. Here τ or τmax means the maximum alternating stress, τcr the static shear strength of the core as given in previous sections, and R the ratio of the minimum alternating stress τmin to the maximum, i.e., R = τmin/τmax. Fig.2.12-2.14 shows S-N curves for some commonly used core materials. The dots are actual test results but the dashed lines are only included to roughly indicate the relation between stress amplitude and fatigue life.

τ /τcr 1

0.8

0.6

0.4

0.2

0 LOG N 0123456 78

Figure 2.12 S-N diagram for unnotched fatigue of Nomex honeycomb (48 kg/m3, 3.2 mm cell size, with 0.6 mm thick carbon/epoxy faces). In L-direction; square marks () for R = 0.1 and diamond marks (♦) for R = −1, and solid lines. In W-direction: square marks ( ) for R = 0.1 and diamond marks (◊) for R = −1, and solid lines. [17].

τ /τcr 1

0.8

0.6

0.4

0.2

0 LOG N 0123456 78

Figure 2.13 S-N diagram for unnotched fatigue of PMI foam Rohacell WF51l(0.96 mm thick glass/epoxy faces). Square marks () for R = 0.1 and diamond marks (♦) for R = −1 [17].

2.21 AN INTRODUCTION TO SANDWICH STRUCTURES

τ /τcr 1

0.8

0.6

0.4

0.2

0 LOG N 0123456 78

Figure 2.14 S-N diagram for unnotched fatigue of PVC foam Divinycell H100 (3.6 mm thick glass/vinylester faces). Square marks () for R = 0.1 and diamond marks (♦) for R = −1 [17].

2.6 Estimation of Core Material Properties There is an excellent book written by Gibson and Ashby [18] on the topic of the behaviour of cellular materials, covering both honeycombs and foams. The theoretical background is based on the fact that the primary mode of deformation of cellular materials originates from cell wall bending rather than tension or compression. Formulae for the prediction of almost any mechanical property are derived and compared with experimental results. The derived theories agree very well with the experimental results. However, the derivation is based on the properties of the unfoamed material being known and sometimes that the geometry such as cell wall thickness, cell shapes etc. is known. This is usually not the case and in any case the engineer is primarily interested in macro-mechanical properties. Hence, for details refer to reference [18]. The properties of primary interest are the out-of-plane properties. These are given in [18] from a theoretical reasoning and in [16] based on testing.

Honeycomb Cores Starting with honeycomb cores, the out-of-plane properties can be estimated as follows: the two Poisson's ratios νTL and νTW must equal that of the solid from which the cell walls are built up, that is

νTL = νTW = νS (2.27) where index S refers to the solid material, e.g., aluminium alloy, steel, or reinforced phenolic resin. By means of the reciprocity theorem, this implies that

EL EW ννLT =≈TL 0 and ννWT =≈TW 0 (2.28) ET ET since the Young's modulus is much higher in the transverse (T) direction than in the plane of the honeycomb. This modulus is easily estimated by [16] ρ EET = S (2.29) ρS

2.22 MATERIALS AND MATERIAL PROPERTIES

where ρ is the density of the honeycomb and ρS is the density of the cell wall material. Gibson and Ashby [18] derived upper and lower bounds for the out-of-plane shear moduli for honeycombs as a function of cell geometry [18], which for regular hexagons with all cell walls having equal thickness reduces to

⎛ t ⎞ GG==115. ⎜ ⎟G (2.30) LT WT⎝ s⎠ S where t is the cell wall thickness and s is the diameter of a circle inscribed in the hexagonal cell (see Fig.6.4). In practice, however, honeycombs have double cell walls in the L-direction (see Fig.2.9) due to their manufacturing, and the shear modulus estimation is then modified to [16] 4t 16t G = G and G = G (2.31) LT3s S WT30s S For square cells the out-of-plane shear moduli are simply [16]

⎛ t ⎞ GG==⎜ ⎟G (2.32) LT WT⎝ s⎠ S

The strength in the T-direction can be estimated by [16,18] ρ σ$$T = σ S (2.33) ρ S

Cellular Foams Most foams are available in a wide range of densities and the first design problem is to choose the right density. It is seen from most material data of foams and honeycombs that the mechanical properties vary in some way with the density of the material. Thus, in general terms we can write

k l m m ECcE= ρ , GCcG= ρ , σρ$ c = Cσ , and τρ$ c = Cτ (2.34)

The constant Cσ and the exponent m may of course be different for the ultimate strength in tension and compression, respectively. In [18] it is suggested that k = 2 for open cell foams. Curves can, if needed, easily be extracted from material data. Remember, however, that development of materials is a continuous effort by material manufacturers and changes in material data thus occur frequently. The formulae derived by Gibson and Ashby [18] assume knowledge about the properties of the unfoamed polymer. This is, unfortunately, not always the case. The analysis also assumes other, sometimes not accessible, knowledge such as the proportion of open and closed cells, cell wall thickness, and cell geometry. If some, or all, of these properties are known, most mechanical properties can be predicted by formulae given by Gibson and Ashby [18]. However, most serious foam core manufacturers support extensive material testing of their products and supply comprehensive data sheets in which almost all mechanical and physical properties may be found.

2.23 AN INTRODUCTION TO SANDWICH STRUCTURES

2.7 Experimental Determination of Core Material Properties As before, by testing it is here understood that testing is performed for the reason of extracting reliable material properties that are to be used in design or analysis of sandwich structures. Most core material manufacturers perform very thorough materials testing for approval and quality control of all their materials. These tests are usually performed using some standard method and the test results are often used in the product data sheet on the material. Thus, the data given by most manufacturers can be used in the design of sandwich structures. Sometimes it is, however, necessary to perform own tests or to obtain data not given in the data sheet.

The most important core material data for the design and calculation of sandwich structures ^ ^ ^ are the out-of-plane moduli and strengths, i.e., Ecz, Gcx, Gcy, σcz, τcxz and τcyz.

Apart from these mechanical properties there are other important properties to be known like the core material density, water absorption, ageing, impact resistance, etc. Standards for such tests are available.

(i) Tensile properties The tensile test is performed by two different methods for core materials since different method must be used to obtain the in-plane properties and the out-of-plane (z-direction) properties. For cellular core materials the in-plane moduli and strengths are obtained using the same kind of tests as described for the composite laminates, see Fig.2.6. These are manufactured by simply cutting thin strips (usually around 10 mm thick) of the core material in x- and y-directions, which then are machined to the correct dimensions according to the standard.

A more important property is the strength and modulus in tension of the core in the direction transverse to the faces, the z-direction. The standard ASTM D1623-78 - Tensile and tensile adhesion properties of rigid cellular plastics can then be used for that purpose, as illustrated in Fig.2.15. The test is performed in the direction perpendicular to the plane of the core block (z-direction), i.e. perpendicular to the plane of the faces in a sandwich panel. For honeycomb cores the standard specimen ASTM C297-61 should rather be used which is basically the same type of test but a slightly different geometry. By this procedure one can obtain values of modulus and strength in tension in the z-direction.

(ii) Compressive properties The compression test of core materials is usually conducted according to the standard ASTM D1621-73 - Compression properties of rigid cellular plastics, or any other equivalent testing procedure, like the ISO 844. The test is performed in the direction perpendicular to the plane of the core block (z-direction), i.e. perpendicular to the plane of the faces in a sandwich panel. A schematic of the test specimen is shown in Fig.2.16. By this procedure one can obtain values of modulus and strength in compression in the z-direction. Similar tests in compression as for the composite laminate, see Fig.2.6, can be used on cellular core materials to obtain moduli and strengths for in-plane compression

2.24 MATERIALS AND MATERIAL PROPERTIES

(iii) Shear properties The shear test is perhaps the most important to perform on the core material and the sandwich construction to be used in the structure. It is vital to have accurate data for the shear modulus and the shear strength of the core. The test used for this purpose is usually the standard ASTM C273-61 - Shear properties in flatwise plane of flat sandwich constructions or sandwich cores. The test is performed in the direction zx- or zy-plane, i.e. out-of-plane shear properties are determined. A schematic of the test specimen is shown in Fig.2.17. A special device is used for mounting a clip-gauge extensometer so that the in-plane shear strain can be measured during the test and hence the shear modulus can be calculated. By this procedure one can obtain values of modulus and strength in shear.

Figure 2.15 Schematic of the Figure 2.16 Schematic of the Figure 2.17 Schematic of the ASTM D1623-78 tensile test ASTM D1621-73 compression ASTM C273-61 shear test test

2.8 Adhesives – Description and Properties First a discussion of some basic features of the adhesive when utilised in sandwich structures is given. The requirements on the adhesive are somewhat different from normal use in that bonding sandwich structures involves bonding two very dissimilar constituents, one solid component to a softer cellular one. This fact implies that attention must be given to various aspects in choosing the adhesive [19,20].

2.8.1 Requirements on the adhesive Surface preparation Metal and composite face material surfaces are to be prepared before bonding in the same manner as when bonding, e.g., metal-to-metal or composite-to-composite. This usually

2.25 AN INTRODUCTION TO SANDWICH STRUCTURES involves cleaning, either mechanically or chemically, and sometimes priming. Bonding to metal surfaces is usually greatly improved by pre-treatment with a wash primer.

The core, on the other hand, may be more difficult to clean but the same requirements are valid for all cores; they should be clean from particles, grease, oil and other substances that may influence the bond. Dust can be removed by vacuum or blowing of oil-free compressed air. Grease and oil can be removed from metal honeycomb by liquid immersion. Foams and balsa are more difficult when it comes to removing grease and oil. It is best to ensure that the core is not exposed to such substances before bonding. In some cases the core has to be prepared by a sawing or sanding operation that removes the surface layer and hence leaves clean surfaces for bonding. However, when sawing and sanding foams, make sure the tooling is sharp, so that it cuts through all cell walls properly. Inadequate tools will damage the surface cells leaving partially attached cell walls like lids covering the cells. If so, the adhesive will bond to the lids rather than penetrating down into the surface cells, thus creating a poor bond.

Solvents Some core materials are highly sensitive to certain solvents. For example, polystyrene foams are sensitive to styrene, which means that polyester or vinylester resins cannot be used as adhesives since they contain styrene, while epoxies and polyurethanes may be used. There are other similar combinations, which need to be investigated prior to the choice of adhesive.

Curing vapours Some adhesives, like phenolics, give off vapour when curing. Since the bond line is situated in a closed space between the face and the core this could lead to several problems such as • Internal pressure build-up preventing the surfaces from attachment during curing, resulting in disbonds. • The pressure could damage the core. • The core material moves during curing, resulting in unusable parts. • Corrosion of faces or core due to chemical action of the vapour.

Bonding pressure Some adhesives require a bonding pressure to prevent creation of pores in the adhesive during cure. In these cases care must be taken so that the core itself will not fail due to compression when the bonding pressure is applied.

Adhesive viscosity When bonding to honeycomb cores the adhesive must have exactly the right combination of surface wetting and controlled flow so that the adhesive does not flow down the cell wall [19]. It is preferable that if the adhesive does flow to a certain extent down the cell wall thus increasing the area of contact with the core.

In the case of a foam or balsa core sandwich the viscosity must be low enough to enable the adhesive to properly fill the surface cells leaving as little trapped air as possible. Low viscosity has another advantage in that when bonding pressure is applied, the adhesive will flow from rich areas to drier, or even totally dry areas, smoothing out the bond line thickness.

2.26 MATERIALS AND MATERIAL PROPERTIES

On the other hand, if the viscosity is too low there is a possibility that the bonding pressure will squeeze out the adhesive from between the face and the core, once again leaving a too thin bond line.

Bond thickness The amount of adhesive applied must be large enough to ensure that both surfaces are properly wet and that no dry areas exist. Each type of core material will require different amounts depending on cell size, cell shapes, type of adhesive, etc. Core material manufacturers can normally provide such information.

Strength The adhesive joint must be able to transfer the loads set up by the design, i.e., have the desired tensile and shear strengths. Even if the bond has the required static strength it may still cause premature failure due to fatigue. Most adhesives have varying strength with temperature, the adhesive used must therefore have adequate strength in the temperature range of its final environment.

Thermal stresses A frequent cause of debonding failures are thermal stresses. If one face is heated, e.g., by sunlight or any other heat source, it will deform due to thermal expansion. Since most core materials are very good insulators there will be a very high thermal gradient over the bond line, with for example a high temperature in the face and a low temperature in the core. This means that very high shear stresses will develop at the interface (the bond line) between the face and the core, and these shear stresses may cause debonding. If the structure is to be used in such an environment, and the face and core materials have very different thermal expansions, it is necessary to use a very ductile adhesive, i.e., one with a high strain to failure.

Toughness When discussing adhesives, toughness usually refers to the resistance of the adhesive to interface crack formation and growth under impact loading [20]. The toughness in this context depends on several parameters such as adhesive ductility, bond line thickness, surface preparation, face material, core material, core cell size etc. There are, however, toughened adhesives on the market for improved impact resistance. These are often ordinary resins in which elastomer particles like rubber have been mixed.

Viscoelastic properties Most adhesives have a significant viscoelastic behaviour, that is, their strength and stiffness depend not only on temperature but also on loading rate. Commonly, adhesives lose stiffness and become more ductile as the loading rate decreases, which is mainly due to stress relaxation, or creep. Very viscoelastic adhesives can be advantageous for example when there are high thermal gradients. As the temperatures slowly change and the thermal stresses increase, the creep in the adhesive relaxes the stresses. Too much creep in the adhesive joints will eventually cause unwanted changes in structural geometry.

2.27 AN INTRODUCTION TO SANDWICH STRUCTURES

Curing shrinkage Some adhesive resins, like polyesters, exhibit a significant volume change when curing. In fact, as much as 7% decrease in volume from it’s uncured to its fully cured state is common. Problems occur when bonding to fairly stiff in-plane core materials like high-density foams, where the shrinkage will create fairly high interface shear stresses. The problem is even more significant when a wet laminate is laid up directly onto the core, in which case the resin itself acts as adhesive. Even though there are fibres in the laminate which to some degree prevent matrix shrinkage this thick layer of material exhibiting a volume change bonded to another thick layer of material exhibiting no volume change creates significant and fully measurable interface shear stresses. The implication of these is clearly seen if one face only is laid up on a core that causes the entire plate or beam to curve when cured. The magnitude of these interface shear stresses can be so high that they severely decrease the strength of adhesive joints.

Curing exotherm Most thermoset adhesives exhibit an exothermal curing, that is, the curing process gives off heat. This is seldom any problem since the adhesive layer is thin and spread over a large area. However, if an entire wet laminate is cured in one piece on the cores the exotherm can, since heat only can flow from the laminate in one direction due to the high thermal insulation of the core, create temperatures so high that they will damage not only the core but also the laminate.

2.8.2 Adhesives and their properties There exist a variety of adhesives, far to many to mention all of them within the scope of this text. Most are for special purposes, e.g., special PUR for bonding to stainless steel, a toughened epoxy for aluminium alloys in high temperature applications etc. This text will therefore give only an introduction to the most commonly used adhesives, and the main features of each group. The choice of adhesive is primarily set to find an adhesive that satisfies the mechanical requirements of the structure for providing a good bond between the material components in the environment where the structure is to work, and considerations like fatigue, heat resistance, strength, ageing and creep are of primary interest. Fortunately, there are a wide variety of adhesives on the market that satisfactorily meet the requirements of a mechanically good bond between almost every plausible combination of sandwich materials. Secondly, the adhesive must also meet the requirements of the environment in which it is supposed to be used. Thus, issues like health considerations, manufacturing technique, curing time, curing temperature, special tooling requirements etc., will just as much decide the choice of adhesive system for the particular application and manufacturing environment, and such requirements will commonly have the greatest influence on the choice.

Epoxy resin Epoxies are low temperature curing resins, normally between 20 and 90°C, but some formulations are made for high temperature curing (130-220°C). They have the advantage of being used without solvents and curing without creating volatile by-products and have thus a low volume shrinkage. The lack of solvents makes epoxies usable with almost every type of core material. Epoxies are available as paste, powder, films, or as solid adhesives. They generally have quite good mechanical properties with a shear strength at room temperature of

2.28 MATERIALS AND MATERIAL PROPERTIES about 20-25 MPa [16]. The bond to metals is greatly improved by pre-treating the metal surface with a primer. A major drawback with epoxy resin is that it is highly allergic when in contact with human skin.

Modified epoxies Toughened epoxies are the same as above but mixed with synthetic rubber, like polysulfide elastomers, which greatly improves the peel resistance. The greater the portion of elastomers the greater ductility but the creep tendencies increase at the same time and the heat resistance decreases. Other modifications are the inclusion of Nylon to improve filleting and controlled flow [20]. These types are, however, sensitive to humidity. By mixing the epoxy with nitrile instead of Nylon the same advantages are gained but with maintained resistance to humidity. These are the most common of the toughened thermosetting adhesives and are usually limited to approximately 150°C service temperature. The shear strength of toughened epoxies approaches values of about 35 MPa. Toughened epoxy adhesive films are the most common material used when bonding honeycomb sandwich parts.

Phenolics Phenolic adhesives have excellent strength, high-temperature mechanical properties, and durability. The main drawback is that they give off some water when curing making venting essential. The viscosity is also quite high and adhesive films must thus be used. These characteristics have limited the use of phenolics to mainly the process of making honeycombs, where venting is no obstacle and a high temperature bond is required. The out- gassing makes phenolic unsuitable for use in bonding sandwich constructions [20], apart from applications where venting is possible. Phenolic adhesives are often modified with synthetic rubbers to improve the toughness.

Polyurethanes Polyurethane (PUR) adhesives are probably the most widely used adhesives for bonding sandwich elements. This is because they provide excellent adhesion to most materials, can be used as paste or liquid in a wide range of viscosities, may have long or short cure times, and can be made fire-retardant and water resistant [21]. PUR adhesives contain virtually no solvents and are thus environmentally friendly and the least toxic of all resins. There exist two different types of PUR-adhesives, one-component moisture-cured and two-component systems.

One-component PUR adhesives are in fact pre-reacted two-component adhesives that continue to cure when exposed to moisture. Moisture necessary for curing is simply provided by spraying water on the surfaces prior to bonding. Onset of curing can vary between minutes and several hours depending on the choice of adhesive. Two-component PUR adhesives consist of various polyols, water scavengers, catalysts, fire retardants, fillers, etc. The curing agent is usually a polymeric methylene-di-phenyl-diisocyanate, which is the least volatile of all isocyanates. The pot-life when mixed can be made to vary between 5 minutes and several hours, and the consistency from liquid to paste.

PUR adhesives can be applied by spraying, rolling or even by brushes. Curing must take place under pressure, preferably mechanical pressure but vacuum tables are commonly used in the

2.29 AN INTRODUCTION TO SANDWICH STRUCTURES making of sandwich elements. Heat drastically increases the cure time. PUR adhesives are mainly used in bonding of foam or balsa core sandwich structures.

Urethane acrylates Urethane acrylate is a resin that is compatible with polyesters and vinylesters. In fact, acrylates are so compatible that they can be incorporated in, e.g., a wet polyester laminate. Urethane acrylates are very tough, and exhibit almost no curing volume shrinkage. A way to drastically increase the face-to-core bond in foam core GRP-sandwich structures is to use urethane acrylate resin for the first reinforcing layer [22], that closest to and bonding to the core. The rest of the laminate can then be laminated wet using, for example, polyester resin on top of the acrylate layer and still provide a perfect interlaminar bond.

Polyester and vinylester resins Polyesters and vinylesters are the most commonly used matrix materials used for reinforced plastic composites outside the aerospace industry. Prefabricated laminates can be bonded to, e.g., foam or balsa cores using the same resin as in the laminate, and it will in most cases prove to be an adequate bond. Usually, however, the laminate is built-up directly onto the core and hence the first layer of the laminate is laid wet onto the core and bonds to it. It is essential in the above procedure that enough resin is used to fill all surface cells otherwise dry areas are left in the bond line. A problem with these resins is their curing volume shrinkage creating sometimes very high interface shear stresses. A way of decreasing the effect of shrinkage is to prime the core surface by applying a thin layer of resin to it, which fills only the surface cells, and which is allowed to cure before the rest of the laminate is applied.

2.9 Experimental Determination of the Adhesive Interface Properties The requirement on the adhesive joint is often stated as “it must be stronger than the tensile strength of the core”. This requirement is not very quantitatively stated but actually works rather well. To deduct this, the test for the tensile strength of core materials, see Fig.2.15, is used, but now the entire sandwich is tested, i.e., the faces are attached to the specimen that are then bonded to the test rig. The failure of the specimen should then be a clear core tensile failure rather than an interface failure.

There exist test methods to determine the peel strength of the face/core interface that basically peels off the face from the core and measures the energy required to do that. In such tests it is, however, difficult to get any quantitative measure of how good the interface is. The test is therefore mainly used to compare different adhesives, adhesive systems, manufacturing methods and material systems with respect to the peel strength. P P a a

P Mode I Mode II Figure 2.18 Suggestions for mode I and mode II fracture toughness test specimens.

2.30 MATERIALS AND MATERIAL PROPERTIES

A more quantitative measure of the strength of the face/core interface strength is to perform fracture toughness test in mode I and mode II, KIc and KIIc (GIc and GIIc). There are no specific standards for such tests but in Fig.2.18 suggestions for such tests are proposed. The mode II specimen, the cracked sandwich beam (CSB), have been proposed and analysed by Carlsson et al [23], and there exists some results and an analytical model for the calculation of GII as a function of the applied load. For the mode I test specimen shown in Fig.2.18a KI (or GI) must be found using the FEM and some results are available in [24] and [25].

2.10 Estimation of Thermal Insulation An inherent feature of sandwich constructions is, except their very high stiffness and strength to weight ratios, their thermal properties - an integrated function which depending on the choice of materials often yields a structure with very good thermal insulation properties. This is a most important feature giving the sandwich concept a very strong position in the design of structures such as containers, tanks and other transportation structures where thermal insulation is necessary. As will be seen, the thermal insulation depends mainly on the core thermal properties and its thickness. Hence, for many applications the thermal insulation properties of the entire sandwich poses some constraints on the choice of materials and the sizing.

Start by studying the sandwich cross-section illustrated in Fig.2.19, having temperature T1 on the inside and T2 on the outside. t t t 1 c 2

T f2

T T1 2

Tf1

z,q

Figure 2.19 Schematic of the temperature field through a sandwich cross-section.

Assume that the faces 1 and 2 have the surface temperatures Tf1 and Tf2, and also that the temperature varies linearly within each material, see Fig.2.19. Since the thermal conductivity for almost every material varies with temperature, further assume that the average temperature in the faces are Tf1 and Tf2, respectively, since they are usually very thin, and that the average temperature in the core is T + T T = ff12 (2.35) cm 2

2.31 AN INTRODUCTION TO SANDWICH STRUCTURES

The heat loss per unit area through a cross-section with different temperatures on each side, T1 and T2, as shown in Fig.2.19 equals the thermal transmittance times the temperature difference, i.e.,

qkTT=−()12 (2.36) with q positive in the positive z-direction, as indicated in Fig.2.19. In order to continue and estimate the value of k we must know something about the physics of heat transfer. There are three main causes of heat transfer that must be included in the analysis: conduction, convection and radiation. The latter, radiation, is often of minor importance and is therefore usually omitted in structural analysis. In special cases, however, such as in space applications, it may be more important.

(i) Conduction For an isotropic body with varying temperature, heat flows in the direction of the gradient of the temperature field, that is, towards and in the direction of the lower temperature. This phenomena is associated with molecular movement, which increases with increasing temperature, and when molecules with high energies (high velocity, high "temperature") collide with neighbouring slower moving molecules, the energy is transferred. This occurs in solids, as well as fluid and gas phases of all materials.

In this case one can approximate the flow to one dimension, from one side of the panel to the other. If the z-coordinate is as defined in Fig.2.19 the one-dimensional heat flow through a material obeys Fourier's law as [26] dT qt q =−λ or if integrated TT−= (2.37) dz 12λ where λ is the thermal conductivity of the specific material at the studied temperature.

(ii) Convection Consider the example of the sandwich in Fig.2.19 where the face material and the surrounding air or fluid have different temperatures. In a region near the face there will exist a thermal boundary layer in which there must be a temperature gradient. The heating or cooling of the air or fluid in the boundary layer causes it to move due to the temperature difference to the surrounding fluid. This motion causes new fluid to enter the boundary layer which in turn is heated, or cooled, and a heat transfer process is going. This is called free convection. The heat flux depends on the temperature difference between the media and the convection heat transfer coefficient, a, between just those two media at the specific temperature. In certain applications, the first media may instead be water, a boat for example, and in such a case, an a must be used for that particular material combination. If the fluid or gas is moving along the surface of the solid by some other means, e.g., water flowing along a ship panel due to the velocity of the ship, or air is flowing along the side of a tank due to the wind, then the convection heat transfer process is even further accelerated. This is called forced convection. The heat flux is written [26] qa=ΔT (2.38)

2.32 MATERIALS AND MATERIAL PROPERTIES where ΔT is the difference in temperature between the face sheet and the surrounding air temperature. a it the convection heat transfer coefficient which usually depends on several parameters, such as the forced fluid or gas velocity, and may therefore be difficult to obtain.

(iii) Radiation Thermal radiation is energy emitted by matter that is at finite temperature. All materials, regardless of form, emits energy. The emission is due to changes in the electron configurations or the constituent atoms or molecules. Radiation transfers energy regardless of the surrounding medium, as conduction and convection, and is, in fact, most efficient in vacuum. The radiation heat flux is given by Stefan-Boltzmanns law

q = σT4 (2.39) where T is the absolute temperature (in Kelvin) and σ the so called Stefan-Boltzmanns constant (σ = 5.67×10-8 W/m2K4). Such a surface is called an ideal radiator or a black body. For any other type of surface the radiation is less than for a black body and may be written

q = ε σ T4 (2.40) where ε is a radiative property called the emissivity, and has values 0 < ε <1. Conversely, if a surface is exposed to radiation, a portion of the energy will be absorbed by the surface proportional to the radiation through

qabs = γqrad (2.41) where qrad is the radiated energy, qabs the absorbed energy and γ the so called absorbtivity, which has values 0 < γ <1.

Going back to the sandwich cross-section in Fig.2.19 and assembling the total heat flow assuming steady-state conditions and one-dimensional flow only and further omit the radiation part since it is negligible in most circumstances and considering conduction and convection only. If one writes the thermal resistance, 1/k, of each heat transfer mode as q⋅ΔT, one can express the total flux at the temperature difference divided by the total resistance. Hence, the heat flux will be resisted by each of the described modes. First, energy from the air on the inside, with temperature T1, is transferred to the surface of face 1 through convection by

q = a1(T1 - Tf1) and then the flux must overcome the resistance of face 1 in order to get to the inside of the core as

q = λf1(Tf1 - Tc1) then though the core

q = λc(Tc1 - Tc2) through the next face

q = λf2(Tc2 - Tf2)

2.33 AN INTRODUCTION TO SANDWICH STRUCTURES and to the surrounding air on the other side

q = a2(Tf2 - T2) If the heat flux is constant (steady-state) then the temperature difference can be found as

⎛ 11t1 tc t2 ⎞ TT12−= q⎜ + + + +⎟ (2.42) ⎝ a1 λλλ1 c 22a ⎠

Hence, the thermal transmittance of the entire cross-section is

−1 ⎡ 11t1 tc t2 ⎤ k =++++⎢ ⎥ (2.43) ⎣a1 λλλ1 c 22a ⎦ where indices 1 and 2 refer to the faces and c to the core. α has values in the order of 2 to 25 for free convection of gases and even higher for fluids. For forced convection, wind or fluid flow for example, the convection heat transfer coefficient increases even further. Common face materials have rather high thermal conductivity, around 40 - 200 W/m°C for metals and in the order of 0.2-0.4 for composites. Common core materials, on the other hand, have λ- values such as 0.07 for balsa wood, 0.1 for non-metallic honeycombs and 0.03 for most foams. Considering that the core layer is often much thicker than the faces we can simplify the thermal transmittance equation to include the core layer only as

()TT− λ λ q ≈ 12c or k ≈ c (2.44) tc tc

Now it is clearly seen that in order to obtain good thermal insulation, that is, to prevent any heat loss by minimising the heat flux, it is advantageous to choose a thick core with a low thermal conductivity.

References [1] Allen H.G., Analysis and Design of Structural Sandwich Panels, Pergamon Press, Oxford, 1969.

[2] Formelsamling i Hållfasthetslära, Department of Strength of Materials and Solid Mechanics, The Royal Institute of Technology, Stockholm, Sweden, Publikation 104 (in Swedish), 9th Edition, pp 206-217.

[3] Tsai S.W., and Hahn H.T., Introduction to Composite Materials, Technomic, Lancaster PA, USA.

[4] Design Data – Fibreglass Composites, Fibreglass, Reinforcements Division, St.Helens, Merseyside, England.

[5] Jones R.M., Mechanics of Composite Materials, Mc-Graw Hill, New York, 1975.

[6] Jansson J.-F., Olsson K.-A. and Sörelius S.-E., Fibre Reinforced Plastics 1, Swedish Tech Books, Solna, Sweden, 1979.

2.34 MATERIALS AND MATERIAL PROPERTIES

[7] Baltek Corporation Technical Data Sheet, 10 Fairway Court, P.O. Box 195, Northvale, NJ 07647, USA.

[8] Honeycomb Products Inc., Veritcel data sheet, 9700 Bell Range Drive, Santa Fe Springs, CA 90670-2981, USA.

[9] Mechanical Properties of Hexcel Honeycomb Materials, TSB 120, Hexcel Corporation, Dublin, California, 1982.

[10] ECA-honeycomb data sheet, EuroComposites S.A., P.B. 95, Zone Industrielle, L-6401 Echternach, Luxembourg.

[11] Airex, Technical Data Sheet, Airex Ltd, Speciality Foams, CH-5643 Sins, Switzerland.

[12] Divinycell, Design Manual, Divinycell International AB, Box 201, S-312 22 Laholm, Sweden.

[13] Rohacell and Rohacell WF, Technical Data Sheet, Röhm, Postfach 4242, Kirschenallee, D-6100 Darmstadt 1, Germany.

[14] Caddock B.D., Evans K.E. and Masters I.G., “Honeycomb Cores with Negative Poisson's Ratio for Use in Composite Sandwich Panels”, Proc. of the ICCM/VIII, 3-E, 1991, Eds. G.S. Springer and S.W. Tsai.

[15] Werren F., “Shear-Fatigue Properties of Various Sandwich Constructions”, U.S. Forest Product Laboratory Report 1837, 1952.

[16] MIL-HDBK-23A, Structural Sandwich Composites, Government Printing Office, Washington D.C., USA, Dec 1968.

[17] Damage Tolerant Design of FRP-Sandwich Structures - DAMTOS, Brite/EuRam II, CT92-0297, Technical Report TR301, not yet available.

[18] Gibson L.J. and Ashby M.F., Cellular Solids – Structure and Properties, Pergamon Press, Oxford, 1988.

[19] The Basics on Bonded Sandwich Constructions, TSB 124, Hexcel Corporation, Dublin, California, 1986.

[20] Marshall A, “Sandwich Construction”, in Handbook of Composites, ed. George Lubin, Van Nostrand Reinhold Company, New York 1982. pp 557-601.

[21] Ströbeck C., “One- and Two-Component Polyurethane for Bonding Sandwich Elements”, in Proceeding of the First International Conference on Sandwich

2.35 AN INTRODUCTION TO SANDWICH STRUCTURES

Constructions, Eds. K.-A. Olsson and R.P. Reichard, Stockholm, 1989, EMAS Ltd., UK, pp 261-277.

[22] Norwood L.S., “The Use of Tough Resin Systems for Improved Frame to Hull Bonding in GRP Ships”, in Proceeding of the First International Conference on Sandwich Constructions, Eds. K.-A. Olsson and R.P. Reichard, Stockholm, 1989, EMAS Ltd., UK, pp 279-290.

[23] Carlsson L.A., Sendlein L.S. and Merry S.L., “Characterisation of Sheet/Core Debonding of Composite Sandwich Beams”, Proc. First International Conference on Sandwich Constructions, Eds. K.-A. Olsson and R.P. Reichard, Stockholm, June 19- 21, 1989, EMAS Ltd., UK.

[24] Prasad S and Carlsson L.A., “Debonding and Crack Kinking in Foam Core Sandwich Beams - I. Analysis of Fracture Specimens”, Engng. Fract. Mech., Vol 47, No 6, 1994, pp 813-824.

[25] Prasad S and Carlsson L.A., “Debonding and Crack Kinking in Foam Core Sandwich Beams - II. Experimental Investigation”, Engng. Fract. Mech., Vol 47, No 6, 1994, pp 825-841.

[26] Incropera F.P. and de Witt D., Fundamentals of Heat and Mass Transfer, Third edition, 1990, John Wiley & Sons, New York.

2.36 MATERIALS AND MATERIAL PROPERTIES

Exercises Example 2.1 Honeycomb materials are often used as core in load carrying sandwich panels. These exhibit orthotropic properties so that Ex ≠ Ey and Gxz ≠ Gyz. Why?

Ans. See Chapter 2.4

Example 2.2 What are the three main groups of core materials used in load carrying sandwich constructions, and what are their main advantages and disadvantages?

Advantages Disadvantages Balsa Good mechanical properties Sensitive to humidity Anisotropic Few available densities Honeycomb Good mechanical properties Expensive (not the paper version) Anisotropic Difficult to handle Heat resistant Difficult to bond to Wide choice of materials Polymer foams Large variety Lower mechanical properties Cheap Low heat resistance Thermal insulation Sensitive to chemicals Closed cells Easy to bond to

2.37 CHAPTER 3

FUNDAMENTALS

In this section the fundamentals of sandwich beam theory will be derived. It is virtually the same as engineering beam theory with the exception that now one must account for transverse shear deformations. This is what is usually called Timoshenko beam theory. Another novelty is that different loads will be carried by different parts of the structure. The theory given here is a brief summary of what is thoroughly described by Allen [1] and Plantema [2]. For simplicity, all beams are assumed to have unit width, i.e., loads, stiffness etc. will be given per unit width. Indices f are used for the face and c for core.

3.1 Flexural Rigidity Recall the basic problem of a straight beam subjected to a constant bending moment giving the beam a curvature κx (inverse of radius of curvature) according to Fig.3.1. x

Figure 3.1 A beam subjected to a bending moment. The strain in a fibre situated a distance z from the neutral axis is now

ε xx= κ z (3.1) i.e., linearly varying with z. The applied bending moment needed to cause the curvature κx is then,

Ez 2 EI Mzdz==σκdz = Ez2 dz =, where EI== Ez2 dz D (3.2) xx∫∫∫ x ∫ Rx Rx

Now, EI is the flexural rigidity which normally is the product of the elastic modulus E and the moment of inertia I. From now on, EI will be designated D for the following reason; if Young's modulus E varies along the z-coordinate then it cannot be removed outside the integral in eq.(3.2). Hence, the definition of the moment of inertia I is lost. Therefore, for a general cross section, eq.(3.2) must be used as given and the flexural rigidity D will be the only property well defined. The general expression for the strain will then be

3.1 AN INTRODUCTION TO SANDWICH STRUCTURES

M z ε = x (3.3) x D Hence, the strain still varies linearly with z over the cross-section.

Now that the basic equations are established, one may commence calculating the cross- sectional properties and stresses in a sandwich beam. First, define the coordinate system and positive directions for the loads as in Fig.3.3.

tf Ef q(x) T M x x Mx d E ,G t x c c c Nx Nx

Tx E f t z,w f

Figure 3.2 Sign convention for sandwich beams. In order not to complicate the analysis at this stage, assume a symmetrical lay-up of the sandwich, i.e., the faces have the same thickness tf and are of the same material with elastic modulus Ef. The core has thickness tc and modulus Ec. The flexural rigidity D is then for a cross-section as in Fig.3.2

Et32 Etd Et3 DEzdz==++=++2 ff ff cc 2DDD (3.4) ∫ 6212fc0 where d = tf + tc, the distance between the centroids of the faces. The first term corresponds to the flexural rigidity of the faces alone bending about their individual neutral axes, the second represents the stiffness of the faces associated with bending about the centroidal axis of the entire sandwich and the third term is the flexural rigidity of the core. 3.2 Approximations in the Flexural Rigidity

The faces are usually thin compared with the core, i.e., tf << tc, and the first term of eq.(3.4) is therefore quite small and is less than 1 percent of the second if

2 ⎛ d ⎞ d 3 ⎜ ⎟ > 100 or >. 5 77 (3.5) ⎝ t ff⎠ t

As a result of materials selection, the core usually has a much lower modulus than that of the face, i.e., Ec << Ef. Hence, the third term in eq.(3.4) is less than 1 percent of the second if

2 6Etdff 3 > 100 (3.6) Etcc

Thus, for a sandwich with thin faces, tf << tc, and a weak core, Ec << Ef , the flexural rigidity can be written approximately as Etd2 D = ff (3.7) 2

3.2 FUNDAMENTALS

For ordinary engineering materials used in sandwich construction, the core/face thickness ratio is commonly in the regime 10 to 50 and the face/core modulus ratio between 50 and 1000. An important fact can now be noted; the dominating term in the expression for the flexural rigidity is that of the faces bending about the neutral axis of the entire sandwich. This part is the one originating from a direct tension-compression of the faces. Now, if the adhesive joints connecting the faces to the core were absent, this term would vanish leaving only the contributions from terms 1 and 3 in eq.(3.4), corresponding to the bending of two faces and one core about their individual neutral axes and hence independent of each other. This means that a major part, or almost all, of the flexural rigidity will be lost. The fact that the flexural rigidity of a sandwich consists mainly of the part originating from the interaction of the constituents is usually called the sandwich effect.

3.3 Stresses in the Sandwich Beam Using the strain definition in eq.(3.3), the stresses in the sandwich due to bending are readily found. The face and core stresses are M zE t t σ =xf for cc<|z |< +t (3.8) f D 22f

M zE t σ = x c ≈ 0 for ||

Nx Nx εσεσεx0 = = , and thus fxf =00EE and cxc = (3.9) Et11++ Et 2 2 Etcc Ax where εx0 is the strain at the neutral axis. The strains and stresses due to bending and in-plane loads can then be superimposed.

3.4 Shear Stresses In the same manner as outlined above, a more general definition must also be found for the shear stress. Consider an element dx of a beam (for example of the beam in Fig.3.2) as shown in Fig.3.3. The shear force must balance the change in the direct stress field (the same equation arises by integration of the equilibrium equation)

τ xz ∂σ σ + x dx x ∂ x σx τ x z xz

∂τ xz ∂σx τ dz σ σ + dx xz+ ∂ z x x ∂ x dx

Figure 3.3 Beam section dx defining equilibrium for a sub-area.

3.3 AN INTRODUCTION TO SANDWICH STRUCTURES

()/dt+ 2 dσ dτ f dσ xxz+=0 → τ ()z = x dz xz ∫ dx dz z dx when using the fact that τxz at (d + tf)/2 is zero. Now, from eq.(3.8) and using dMx/dx = Tx we get

()/dt+ 2 T f TBz() τ ==x ∫ Ezdz x (3.10) D z D where B(z) is the first moment of area. The formula reduces for a homogeneous cross-section to the more well-known formula TxJ(z)/I where J is the first moment of area by its most usual definition. Now, instead, the integral

()/dt+ f 2 Bz()= ∫ Ezdz (3.11) z is the new, and more general, definition for the first moment of area. Now, in the core material for |z| ≤ tc /2 the first moment of area is

E fftd Etcc⎛ ⎞⎛ t c ⎞ Bz()=+−⎜ z⎟⎜ + z⎟ 2222⎝ ⎠⎝ ⎠ yielding the shear stress in the core

2 Tx ⎡ Etdff Etcc⎛ 2 ⎞⎤ τ c ()z =+−⎢ ⎜ z ⎟⎥ (3.12) D ⎣ 224⎝ ⎠⎦ and similarly in the faces for tc /2 ≤ |z| ≤ tc /2 + tf

2 E f ⎛ tc ⎞⎛ tc ⎞ Tx E f ⎛ tc 22⎞ Bz()=+−⎜ tz⎟⎜ ++tz⎟ ⇒=τ ( z ) ⎜ ++−tt t z ⎟ (3.13) 22⎝ f ⎠⎝ 2ff⎠ D 24⎝ cf f ⎠

The maximum shear stress appears at the neutral axis, i.e., for z = 0

2 T ⎛ Etdff Et ⎞ τ (=)=z 0 x ⎜ + cc⎟ (3.14) c,max D ⎝ 2 8 ⎠ and the shear stress in the face/core interface will be

⎛ tTcx⎞ ⎛ Etdff ⎞ ττ=== τ⎜ ⎟ ⎜ ⎟ (3.15) c,min f ,max ⎝ 22⎠ D ⎝ ⎠ and as seen in eq.(3.13), the shear stress in the outer fibre of the faces is zero, which evidently must be the case for a free surface.

3.5 Approximation in the Shear Stress The ratio between the maximum and minimum shear stress in the core according to eqs.(3.14) and (3.16) will be quite small, and less than 1 percent if

3.4 FUNDAMENTALS

4E fftd 2 >100 (3.16) Etcc

3.6 Summary of Approximations The approximations can now be summarised through eqs.(3.6-3.16) as: if the core is weak,

Ec<

M xfzE σc ()z = 0 σ f ()z = (3.17) ()DD0 + 2 f

2 TxffE td T E f ⎛ t ⎞ τ ()z = τ ()z = x ⎜ c ++−tt t22 z ⎟ c f 24⎝ cf f ⎠ 22()DD0 + f ()DD0 + 2 f if the core is weak, Ec << Ef and the faces are thin, tf << tc , then the formula reduces to the simplest possible form

Mx Tx σc ()z = 0, σ f ()z =± , τc ()z = , and τ f ()z = 0 (3.18) tdf d

This simplifies the modus operandi or the principal load carrying and stress distributions in a structural sandwich construction to: the faces carry bending moments as tensile and compressive stresses and the core carries transverse forces as shear stresses.

The stress distributions for the different degrees of approximation can also be graphically represented by plotting the above equations as functions of z, as illustrated in Fig.3.4.

No approximations Ec << Ef Ec << Ef and tf << tc

Figure 3.4 Direct and shear stresses for different levels of approximations.

3.7 “The Sandwich Effect” One can now investigate the consequences of the above results. Consider a homogeneous plate of a material with a given Young's modulus, E, and a given strength. Subject a beam of this material to a bending moment. Calculate the weight, bending stiffness and strength of the beam and set them to unity. Suppose we now cut the beam in two halves and separate the parts with a core (we can do that without adding substantial weight). Given the above

3.5 AN INTRODUCTION TO SANDWICH STRUCTURES analysis, one can now calculate the corresponding stiffness and strengths of sandwich beams and the relative properties given in Fig.3.5 will be found. Weight Flexural Bending rigidity strength

111 t

2t ~1 12 6 t/2

4t ~1 48 12

Figure 3.5 Comparison between homogeneous and sandwich cross-sections. Hence, by using the sandwich concept, the flexural rigidity and flexural strength can be substantially increased in comparison with a single skin structure without adding much weight.

3.8 Sandwich with Dissimilar Faces The sandwiches studied so far have been symmetrical, but naturally there exist sandwiches with dissimilar faces, i.e., faces of different thickness and/or different materials. First, the position of the neutral axis must be found. It is given by the coordinate system for which the first moment of area is zero when integrated over the entire cross-section. Or more physically, it is given by Ez 1 σεdz=== E dz dz Ezdz =0 ∫∫xx ∫ ∫ RxxR

Since the location of the origin of the sought coordinate system is unknown, make a coordinate transformation from a known point in the cross-section, e.g., z*= z – e, according to Fig.3.6.

t1 E1

tc Ec

e z

t z* 2 E2

Figure 3.6 Definition of location of the neutral axis in a sandwich with dissimilar faces.

3.6 FUNDAMENTALS

The equation then becomes B() z==+=⇒−=∫∫ Ezdz E() z** e dz0∫ Ez *** dz e ∫ Edz (3.19)

For a general sandwich cross-section as shown in Fig.3.6, this reduces to

⎛ t12t ⎞ ⎛ tc t 2⎞ Et11⎜ ++tccc⎟ +Et⎜ +⎟ =eEt[]11 + Etcc + Et 2 2 (3.20a) ⎝ 22⎠ ⎝ 22⎠

For a sandwich with a weak core this can be written as Etd Etd e = 11 and de− = 22 (3.20b) Et11+ Et 2 2 Et11+ Et 2 2

The flexural rigidity will now be

3 3 3 2 Et11 Et22 Etcc 2 2 ⎛ ttc + 2 ⎞ D = + + + Et() d−+ e Ete + Et ⎜ − e⎟ (3.21a) 12 12 12 11 22 cc⎝ 2 ⎠ where d=t1/2+tc+t2/2 (distance between centroids of the faces). If the core is weak, E c <

Mdxxx()− eE1122MEEtd M σ 1 =− =− ≈− , and D DEt()11+ Et 2 2td 1

MxxeE2121M E E td M x σ2 =≈ ≈ (3.22) D DEt()11+ Et 2 2td 2 It now also possible to re-derive the expressions for the shear stresses and they will appear as t t In face 1:−+−de 11 ≤≤−++zde 22

⎡ 2 ⎤ Tx E1 ⎛ t1 ⎞ 2 τ1()z = ⎢⎜− d + e − ⎟ − z ⎥ (3.23a) D 2 ⎣⎢⎝ 2 ⎠ ⎦⎥

t t In face 2: e − 2 ≤ z ≤ e + 2 2 2

3.7 AN INTRODUCTION TO SANDWICH STRUCTURES

2 Tx E22⎡⎛ t ⎞ 2 ⎤ τ 2 ()z =+⎢⎜e ⎟ − z ⎥ (3.23b) D 22⎣⎝ ⎠ ⎦

t In the core for − d + e + 1 ≤ z ≤ 0 2

⎡ ⎧ 2 ⎫⎤ Tx Ec ⎪⎛ t1 ⎞ 2 ⎪ τ c ()z = ⎢E1t1()d − e + ⎨⎜− d + e + ⎟ − z ⎬⎥ (3.23c) D ⎣⎢ 2 ⎩⎪⎝ 2 ⎠ ⎭⎪⎦⎥

t In the core for 0 ≤ z ≤ e − 2 2

⎡ 2 ⎤ TxcE ⎧⎛ t2 ⎞ 2 ⎫ τ ()z =+−Ete ⎜e ⎟ − z (3.23d) c ⎢ 22 ⎨⎝ ⎠ ⎬⎥ D ⎣⎢ 22⎩ ⎭⎦⎥

When assuming a weak core the core shear stress is constant and can be written

Tx E11t E 2td 2 τc = (3.24a) D Et11+ Et 2 2 and when also assuming thin faces, it further simplifies to T τ = x (3.24b) c d which is the same result as for symmetrical sandwiches.

3.9 Equivalent Width Another way of analysing a sandwich would be to treat it as an I-beam and is done in the following way. Consider a sandwich of width b and an I-beam with flanges of width b.

t f bb

Figure 3.7 The equivalent web width of an I-beam.

Find the thickness of the web, be, in the I-beam so that the two cross-sections will have the same flexural rigidity. Or, imagine one can compress the core of the sandwich until it has the same modulus as the faces.

3.8 FUNDAMENTALS

3 3 Ebtcc Ebtf ec Ec =⇒=be b (3.25) 12 12 E f

In this way a fictitious moment of inertia can be computed as

3 3 t f d Ec tc I =+2 2t f + (3.26) 12 212E f

However, remember that this value of I is only valid for a given Ef , so that the product Ef I = D. Take great care before using a property like the moment of inertia since it is not well defined unless specifically given together with a modulus of elasticity, E.

References [1] Allen H.G., Analysis and Design of Structural Sandwich Panels, Pergamon Press, Oxford, 1969.

[2] Plantema F.J., Sandwich Construction, John Wiley & Sons, New York, 1966.

3.9 AN INTRODUCTION TO SANDWICH STRUCTURES

Exercises Example 3.1 A sandwich panel has honeycomb core and stainless steel faces. Derive an expression for the flexural rigidities

Dx and Dy. Which approximations can be done and why?

Faces: E = 205.000 MPa faces f tf = 1.2 mm adhesive joint νf = 0.3 tc core Core: Gcx = 62 MPa Gcy = 38 MPa t = 25 mm t c f 2 Etdff 2Df D Ans. D ≈=84,, 432 120Nmm2/mm, =<0.. 00070 0 01, c ≈ 0 (honeycomb) 2 D D

Example 3.2 A cantilever beam of a sandwich according to the figure is subjected to a point load P. The cross-section is unsymmetrical, i.e. the faces are of different materials and thickness.

P E1 t1

tc E c L t2 E 2 (a) Calculate the position of the neutral axis. (b) Derive an expression for the maximum direct stress in the faces assuming the core to be weak and the faces to be thin. Hint: Seek an expression for the bending stiffness and derive the stress via the definition of strain Ans. See eqs.(3.20)-(3.22)

Example 3.3 Sandwich panels are know to have great stiffness and strength compared to its weight. Compare with a homogenous cross-section and discuss the reason for the above statement. Use the stress variation over the cross-section with a linearly varying strain.

Example 3.4 A beam has a homogeneous rectangular cross-section with dimension b × h. How much material can be saved if the beam instead was made in sandwich with faces of the same material separated a distance 3h, if the same bending stiffness should be remained? The weight of the core may be discarded. Ebh3 Ans. Homogeneous beam: D = , material: b × h hom 12 2 3 Ebt f d Ebh h Sandwich: D ≈ = → t = since d = 3h sand 2 12 f 54 bh The weight of the sandwich is then 2 × b × h/54 = 27

3.10 FUNDAMENTALS

Example 3.5 Calculate the distance e to the neutral axis for the sandwich cross-section shown below.

Ef tf

Ec tc

e 3tf 3E f 16Et22++ 6 Ett 2 Ett + Et Ans. e = ff cfc ffc cc 10Etff+ 2 Et cc

Example 3.6 Show that the normal stress in the face approximately can be written as

M x M x σ f 1 ≈ − and σ f 2 ≈ t1d t2 d for a sandwich with thin, but dissimilar (t1 ≠ t2 and/or E1 ≠ E2) faces. Hint: The position of the neutral axis from the middle axis of the lower face (index 2) is given by

⎛ t1 t2 ⎞ ⎛ tc t2 ⎞ E1t1⎜ + tc + ⎟ + Ectc ⎜ + ⎟ = e[]E1t1 + Ectc + E2t2 ⎝ 2 2 ⎠ ⎝ 2 2 ⎠ Ans. See Chapter 3, pages 3.8-3.9

Example 3.7 In the April issue of Proffesional Boatbuilder there was an article on the choice of core material when building boats of sandwich construction. The following paragraph is cited from this article.

”In designing cored laminates (cored laminates here means sandwich panels), you should consider shear modulus. But, if you need to increase the stiffness of the panel, you will get better results – in most cases – by using a thicker core than by using a core with higher shear modulus. Back to beam theory: the strength of a beam increases by the square of its height, and its stiffness increases by the cube of its height. In other words, a panel’s stiffness will double by increasing its thickness 12%”

There are a number of errors in the above quotation. Find these errors but also show and explain how it really is!

Ans. First of all, the author is apparently not all that well acquainted with the sandwich beam theory, only with classical beam theory. The strength of a beam may be written as

M x z M x σ f = E f ≈ ± for a thin face sandwich D t f d The strength may be referred to the bending moment that may be carried. Assuming that the face material is the same when increasing the core thickness (as indicated in the article), this can then be written as

M x ≈ σˆ f t f d where M x is maximum bending moment and σˆ f the face sheet strength

Hence, the strength of the beam increases linearly with increasing core thickness, not with the square (as in the case of a homogeneous beam cross-section).

The bending stiffness for a sandwich beam equals 3 2 2 E t E t d E t 3 E t d D = f f + f f + c c ≈ f f 6 2 12 2

3.11 AN INTRODUCTION TO SANDWICH STRUCTURES which is proportional to the square of the core thickness, and not the cube of the core thickness as stated in the article (which is the case for a homogeneous beam cross-section). The shear stiffness on the other hand (for a thin face sandwich) equals 2 Gc d S ≈ ≈ Gc d tc and is hence proportional to the core thickness. Thus, a 12% increase in core thickness will give a 25% increase in D and a 12% increase in S.

3.12 CHAPTER 4

BENDING OF SANDWICH BEAMS

For sandwich beams it is necessary to account for transverse shear deformations. In classical engineering beam theory these are rightfully neglected as they add only marginally to the total deformation of beams with relatively high shear stiffness, such as beams with homogeneous cross-sections. The only case where transverse shear deformations must be accounted for even for shear stiff beams is if the beams are short, for which engineering beam theory is hardly valid anyway, and two- or three dimensional elasticity theory must be used. This section will cover the foundations of bending, buckling and free vibration of sandwich beams and the formulae derived and given below are again a summary from Allen [1] and Plantema [2].

For any type of structure, e.g., beams, plates or shells, the deformation always consists of two parts

(i) deformations due to bending moments − bending − wb

(ii) deformations due to transverse forces − shear − ws

Shear deformations are usually neglected in classical analysis of structures with homogeneous cross-sections unless the studied member has a very short span, because the shear part is usually only a small fraction of the bending part. But for short beams or cross-sections with low shear stiffness this deformation component must be included and for sandwich beams the latter is commonly true. For a sandwich with thin faces the two deformation parts may be superimposed as

w = wb + ws (4.1)

w''dxb

γ c w'dx w'b dx w's dx

(a) (b) (c)

Figure 4.1 A deformed sandwich element illustrating (a) the total, (b) bending and (c) shear deformation, respectively.

4.1 AN INTRODUCTION TO SANDWICH STRUCTURES

4.1 Shear Deformations When a structural element is subjected to shear forces it will deform, without volume change however, according to Fig.4.2. This deformation can be divided into two different parts, transverse (middle) and in-plane (far right) shear deformation. τ γ t γ t γ t dx

γ γ 0 τ 0

Figure 4.2 Deformation of a structural element subjected to shear forces. These are again schematically illustrated in Fig.4.3 but drawn with the assumption that the shear deformation only occurs in the core, i.e. with Gf = ∞, and that this deformation is linear, i.e., Ec << Ef giving a constant core shear stress (τxz = Tx/d) and a constant shear strain.

A'' dws/dx A' A

γ-γ0 γ0 γ0

γ t = + c d

Figure 4.3 Shear deformation of a sandwich element.

Denote the total shear in the core by γ and the in-plane core shear part by γ0. By studying Fig.4.3 the following geometrical relation is found which is valid for a general case where a part of the shear deformation also may occur in the plane of the beam. A straight line A-B prior to deformation has moved due to the shear deformation; first the point A has moved to position A' due to the out-of-plane shear and then another distance, from A' to A'', due to the in-plane shear. Hence, the distance between A and A' can then be written

2 dws dws γtc γ 00tc Tx γ tc dwsx1 dT dt=−()γγ0 c ⇒ = − = − → = (4.2) dx dx d d S d dx 2 S dx since γ approximately equals Tx/Gd from eq.(3.18). By integration we then get that

x ⎛ Txcγγ0t ⎞ M xc0 tx w =−⎜ ⎟dx =− +constant (4.3a) s ∫ ⎝ ⎠ 0 S d S d

or Sws = Mx + Ax + B (4.3b)

Solving for ws leaves two unknown constants to be found from the boundary conditions. As seen, γ0 or A corresponds to a rigid body rotation and B to a rigid body translation, which are both governed by boundary conditions.

4.2 BEAM THEORY

4.2 Shear Stiffness The shear stiffness of a cross-section is defined as the relation between some measure of the shear strain and the transverse force, γ = Tx/S. For a homogeneous cross-section the shear stiffness, S, is often written as Gh S = (4.4) k where G is shear modulus, h the height and k a shear factor, which for a rectangular homogeneous cross-section equals 1.2 [3]. For a general cross-section, the shear stiffness can be computed more accurately by using an energy balance equation, that is, so that the potential energy of the applied load equals the strain energy of the system. The shear stiffness, S, is found by calculating the average shear angle of the cross-section, γ. This then becomes 1 1 T Tzzdzγτγ= ( ) ( ) , where by definition γ = x (4.5) 2 xxzxz2 ∫ S

Using the approximations for a sandwich with thin faces, tf << tc , weak core, Ec << Ef , and that the shear modulus of the faces Gf are large, it is seen that τxz = Tx /d (see eq.(3.18) and Fig.3.5). Eq.(4.5) then becomes

1 1 tc 2 T T T 2t T 2 T γ = x x dz = x c = x 2 x 2 ∫ d G d 2G d 2 2S −tc 2 c c so that the shear stiffness is defined by

G d 2 S = c (4.6) tc which is the same expression as found by the geometrical relation in eq.(4.2).

4.3 Equations in Terms of the Displacement Field Next consider the beam in Fig. 3.3 again along with Fig.4.4 defining local co-ordinate systems for the faces and the core. First define the kinematic assumption, namely that the in- plane and out-of-plane deformations can be written as (see Fig.4.4)

R x,u x

wb ws

z,w −θx

x z u "pure bending" "pure shear" 0 = 0 θx ≠ θx

Figure 4.4 Definition of deflections, rotation, bending and shear of a sandwich cross-section.

4.3 AN INTRODUCTION TO SANDWICH STRUCTURES

u(z) = u0 + zθx and w = wb + ws (4.7) where the in-plane deformation u is thus a linear function in z, and θx is the cross-section rotation (defined as a positive rotation about the y-axis), which only depends on the bending deformation wb of the beam (since shear only causes sliding of the cross-section). Thus, dw du du dθ du d 2 w θ =− b and ε(z) = = 0 + z x = 0 − z b (4.8) x dx dx dx dx dx dx 2 The governing equation for the shear part has already been found in section 4.1. For bending we proceed as for ordinary beam bending theory. Assuming small deflection theory, the radius of curvature can be found in terms of the displacement field by a geometrical study x dx x dφ

z,w

−θx

⎛ ∂θ x ⎞ − ⎜θ x + ⎟dx ⎝ ∂x ⎠ dx

Figure 4.5 Definition of curvature.

From Fig.4.5 it is then seen that, dx = Rx dφ, and dφ – (θx + dθx/dx dx) = –θx¸ so that dφ = dθx/dx dx. This means that

2 1 dθ x d wb = κ x = = − 2 (4.9) Rx dx dx

There are now several ways to proceed with the beam theory. We see that we can write the average shear strain as dw γ = +θ (4.10) dx x or using partial deflections equivalently as

dw dw dw T s = − b = x (4.11) dx dx dx S

This can done since the integration constant A in eq.(4.3b) (or γ0 in (4.3a)) must shared by the bending deflection and therefore in the combined case can be found from boundary conditions for the of entire deflection. This is seen more clearly in section 4.8.

The bending moment can now also be written as

4.4 BEAM THEORY

Ez 2 dθ dθ d 2 w M = σ zdz = dz = x Ez 2 dz = D x = −D b (4.12) x ∫ x ∫∫ 2 Rx dx dx dx

We now see that there are several ways to describe the kinematics of a sandwich beam. In the general case, as for plates, one commonly use the displacement w and the rotation θ as variables, and as seen above, all other properties like bending moments, transverse forces, shear strains, etc, can be obtained as function of these variables. This approach is somewhat more straightforward when one shall, for example, derive finite elements, since then the displacement and rotation are equivalent to physical degrees-of-freedom in the finite element formulation. We can equivalently use the partial deflection wb and ws to describe the same properties. The final governing equations can still obtained in one single variable w irrespective of choice of variables. The choice of partial deflections is used here since it provides are more physical interpretation of the modes of deformation of a sandwich beam.

A second approach to the above derivation is that used by Plantema [2]. First consider the case where the faces are thin, or at least thin enough to behave like membranes when the shear deformation is studied, i.e. any transverse shear deformation may take place without being resisted by any bending of the faces about their individual neutral axes. This treatment is equivalent to the concept of partial deflections; bending causes in-plane stresses and transverse forces shear stresses and deformations. The study only includes stresses and strains arising from the bending moments and transverse forces of the beam, i.e. u0 = 0. If so, the in- plane stresses are obtained by studying the bending curvature only as

Mx1 Nx1

z1 Mx

d T e z x Mx2 Nx2

Figure 4.6 Sandwich cross-section with local z-coordinates for the faces.

2 2 2 dwb dwb dwb σ11=−Ez 2 , σ22=−Ez 2 and σcc=−Ez 2 (4.13) dx dx dx

Since z1=z+d−e for the upper face and z2=z–e for the lower face, the in-plane force in the upper face can be written as (see Fig.4.6)

t1 2 ⎡ dw2 ⎤ dw2 NEzde=−() − + b dz=− E t() d e b (4.14) x111∫ ⎢ dx 2 ⎥ 1 11 dx 2 −t1 2 ⎣ ⎦ where e is the distance between the lower face and the neutral axis. In the same way, for the lower face

t2 2 ⎡ dw2 ⎤ dw2 NEze=−() + b dz=− E t e b (4.15) x222∫ ⎢ dx 2 ⎥ 2 22 dx 2 −t2 2 ⎣ ⎦

4.5 AN INTRODUCTION TO SANDWICH STRUCTURES

2 2 For equal faces the normal forces simply become Nxf1 = −Nxf2 = −D0/d d wb/dx [2]. The local bending moments in the faces will be

t1 2 ⎡ dw2 ⎤ Et3 dw2 dw2 MEzzde=−() − + b dz =−11 bb =−D (4.16) x1111∫ ⎢ dx 2 ⎥ 1 12 dx 2 1 dx 2 −t1 2 ⎣ ⎦ and in the same way for the lower face. For the core, the same integration may be performed resulting in dw2 N = 0 and MD=− b (4.17) xc xc c dx 2 The total bending moment is hence

d 2 w d 2 w dθ M = −()D + D + D + D b = −D b = D x (4.18) x 0 1 2 c dx 2 dx 2 dx The response of the sandwich beam is thus described by two constants; the flexural rigidity D and the shear stiffness S. These are defined considering distortions of an element under the action of only one of the loads Mx and Tx at the time. By definition (see also eq.(4.2))

Mx dTx dx D =− 22 and S = 22 (4.19) dwb dx dws dx The contribution to the curvature amounts to dw2 M 1 dT =−xx + (4.20) dx 2 D S dx

4.4 Governing Beam Equations The equilibrium equations can be derived using Fig.4.7 which shows an element dx in its distorted condition with forces acting on it in positive directions. q dx

dM M x x +dxdx dw dx dTx Tx dx +dx

2 dx dw d w + 2 dx dx dx

Figure 4.7 Distorted beam element. The equilibrium equations for the beam element is dM dT T = x and q +=x 0 (4.21) x dx dx

4.6 BEAM THEORY

By combining eqs.(4.21) and (4.19) one can find a relation between the partial deflections as dw2 D dw4 sb=− (4.22a) dx 2 S dx 4 And since both deflection terms share the same rigid body rotation this may be simplified to

dw D d 3 w s = − b (4.22b) dx S dx 3 It is here seen that in the case of pure bending and small deformations, an equation is obtained in ws only, which can be integrated twice giving two unknown integration constants to be found from the boundary conditions. These are directly comparable to the two unknowns found from the geometrical exercise in eq.(4.2). Rewriting the above using eq.(4.22) gives the two governing equations

dw4 dw2 D b = q or S s =−q (4.23) dx 4 dx 2 which is a similar equation but in wb. This can again be rewritten using the relation between wb and ws in eq.(4.22) so that the governing equation can be written without the use of partial deflection as

dw4 dw4 d 4 w D d 2 q D −=D s q but D s = − from eq.(4.22) dx 4 dx 4 dx 4 S dx 2 giving that

d 4 w ⎛ D d 2 ⎞ D = ⎜1− ⎟q (4.24) 4 ⎜ 2 ⎟ dx ⎝ S dx ⎠ which is the one-dimensional form of the governing equation for plates derived by Mindlin [4]. Now, either the complete differential equation in (4.24) is solved, or eqs.(4.23) may be solved separately to obtain the partial deflections. Usually the latter can be adviced since it leads to a more straight forward way of introducing correct boundary conditions.

4.5 Effect of Thick Faces The previously derived formulae neglect the bending stiffness of the faces themselves and in fact, in order to superimpose the bending and shear deformations the faces must act as membranes. For example, the fixed end of the cantilever in the example of section 4.9.1 would not undertake such a deformation if the faces had a bending stiffness of their own. As will be seen, this has a local effect at, e.g., the fixed edge or under a point load, where the transverse force has a discontinuous jump and therefore the shear deformation shape has a jump in its derivative. In short, as a result of the core shear stress, the core undergoes a shear strain and a corresponding deformation. This deformation must be shared by the faces.

Generally, the shear force Tx will not be constant but varying with x, which means that there 2 2 exists a curvature d ws/dx . This curvature will not add to the rotation of the sandwich as a whole, but it will act on each component separately, hence introducing local bending moments in the faces but no resulting axial force. There is no problem if the transverse force

4.7 AN INTRODUCTION TO SANDWICH STRUCTURES

2 2 is constant, but if there is a curvature d ws/dx ≠ 0, the faces exhibit an additional curvature and hence additional stresses build up. Assuming the faces to be rigid in shear (a most reasonable assumption) and according to Fig.4.6, the stresses will now be given by

2 2 2 2 dwbsdw dwbsdw σ11=−Ez 2 −Ez11 2 and σ22=−Ez 2 −Ez22 2 (4.25) dx dx dx dx The in-plane forces in the faces of eqs.(4.14) and (4.15) are unaffected by this assumption but the local bending moments in the faces become

t1 2 ⎡ dw2 dw2 ⎤ Et3 dw2 dw2 MEzzde=−() − +b +Ez2 s dz =−11 =−D (4.26) x1111∫ ⎢ dx 2 11 dx 2 ⎥ 1 12 dx 2 1 dx 2 −t1 2 ⎣ ⎦

t2 2 ⎡ dw2 dw2 ⎤ Et3 dw2 dw2 MEzze=−() +b +Ez2 s dz =−22 =−D x2222∫ ⎢ dx 2 22 dx 2 ⎥ f 12 dx 2 2 dx 2 −t2 2 ⎣ ⎦

The normal force and bending moment in the core is still the same as in eq.(4.17). The total bending moment is now

d 2 w d 2 w d 2 w d 2 w M = −()D + D b − 2D = −D b − 2D s (4.27) x 0 c dx 2 f dx 2 dx 2 f dx 2 when D1 + D2 = 2Df. Take the equilibrium of bending moments in the infinitesimal element in Fig.4.7 again but now use the entire bending moment given in eq.(4.27)

dT dM2 dw4 dw4 −=−=xx2()D ++DD b =q dx dx 2 fcdx 4 0 dx 4

dw4 dw4 dw4 2D ++()DD =++qDD ()s fcdx 4 0 dx 4 0 cdx 4 Next we may use eq.(4.22) but first remember that this equation was derived for thin faces why it is realised than in eq.(4.22) D only contains the terms D0 + Dc. We then arrive at dw4 dw4 D dw6 2D ++()DD =−q 0 ()DD + b fcdx 4 0 dx 4 S 0 cdx 6

6 2 4 D0 d wb D0 d ⎡ d w⎤ but (D0 + Dc ) 6 = 2 ⎢q − 2D f 4 ⎥ S dx S dx ⎣ dx ⎦

This gives after some rearrangements

d 6 w DS d 4 w ⎛ d 2 S ⎞ 2D − = ⎜ − ⎟q (4.28) f 6 4 ⎜ 2 ⎟ dx D0 dx ⎝ dx D0 ⎠

The same equation was derived by Hoff and Mautner [7] using variational principles. This is actually the one-dimensional form of the plate equation derived by Hoff [8] which also

4.8 BEAM THEORY accounts for the bending stiffness of the faces alone. Eq.(4.28) replaces eq.(4.24) when the thickness of the faces are accounted for but when Df is small then it equals (4.24). By rewriting these equations one arrives at dM dw3 dw3 dw dw3 2D dT2 x ==−TDb −22D ss =S −D s =−T f x dx tot dx3 f dx3 dx f dx3 x S dx 2 where Ttot is the total shear force carried by both the faces and the core, and Tx is the shear force carried by the core alone. Rewritten

2 dTx 22 2S 2 −=−bTx bT tot with b = (4.29) dx 2Df

A more conceptual approach is used by both Allen [1] and Plantema [2]. The reasoning goes as follows. The faces must share the extra deflection of the core shear deflection, and in order to do so an additional load must act on them. Therefore, not all the shear force is used to produce the core shear strain but a part must be used to make the faces undergo the same deformation as the core. Consider for simplification a fixed edge, which is enlarged in Fig.4.8.

Mf0 Mf

Tf0 Tf x Mx z,w

Tx Mf Tf

Figure 4.8 Additional loads acting on the faces at a clamped edge. If the faces were rigid in shear the shear force would be completely carried by the faces at the section x = 0 since the shear angle γ is zero. Assume the global loads Mx, Tx and a distributed load q to act on the beam. These will induce a bending deformation, wb, due to Mx and a shear deformation, ws, due to Tx. But, at x = 0 the rotation dws /dx is prevented by the boundary 2 2 condition dw /dx = 0. Hence, there is a local bending moment Mf0 exerted on each face. At the free edge of the cantilever this moment, Mf, must equal zero. Therefore, shear forces, Tf, must act to build up these moments at the clamped edge. The total shear must therefore be divided over the cross-section into Tx, which is the normal shear force producing the shear angle dws/dx, and Tf which acts on the faces building up the edge moment. Hence, the total shear force equals

Ttot= T x+ 2T f (4.30)

Now, Mx and Tx produce the ordinary deflection wb and ws as given previously. The faces must now undergo the deflection ws and in order to do so, the parts of the total load Tf and Mf must be applied to the faces. Therefore, the extra deformation of the face must be

wwwbs11+= s (4.31)

From the previous analyses we have that

4.9 AN INTRODUCTION TO SANDWICH STRUCTURES

dw dw3 dM dw dw3 TS==−sbD and therefore that T ==f S s1 =−D b1 x dx dx3 f dx f dx f dx3 where Sf is the shear stiffness of the face alone. Along with eqs.(4.30) and (4.31) this gives

3 3 dws dws1 dwb1 SDf dwb1 dwb1 TStot =+22Sf =S −3 −Df 3 dx dx dx Sf dx dx which with some rearranging leads to the differential equation

dw3 dw T SS b1 −=−a2 b1 a2 tot , where a 2 = f (4.32) dx3 dx S ()SSD+ 2 ff

This equation can now be solved for the unknowns wb1, ws1, Tx and Tf, by applying the boundary conditions. This has been done for a cantilever beam in [2]. It can be shown by numerical examples that preventing the warping at a clamped edge is a local effect which has damped out at a distance from the edge of the order of the total thickness of the beam. It can also be shown the total deflection of the beam, providing it is not very short, will be virtually unaffected and could therefore be calculated using the equations given in the previous 2 2 analysis. If the shearing of the faces is neglected, i.e., Sf is very large, then a = S/2Df = b and eq.(4.32) can rewritten in the simpler form dT2 x −=−bT22 bT (4.33) dx2 x tot which is the same as eq.(4.29). Now, ws1 = 0, Tx(0) = 0 and Tf(0) = Ttot /2, which means that the entire shear force is carried by the faces at the clamped edge. The most important effect of this is that the maximum stress in the faces is considerably affected by the local bending moment Mf, and could increase quite a lot compared with that computed using the previous analysis. For examples of using the above equations, see [2].

Returning to the cantilever first discussed, this can now be solved using the above equations.

Taking e.g., eq.(4.33) with Ttot = P and solving for Tx gives

bx− bx TCx =+12ee C + P

Using the boundary conditions that Tx(0) = 0 (Sf is infinite) and Tx(L) = P (if L is large) yields

bx()−2 L −bx dws P ⎡ee− ⎤ = ⎢ −2bL + 1⎥ dx Se⎣ 1 − ⎦ and since ws(0) = 0 the shear deformation will take the form P Px w = eebx()−2 L +−−−bxe b2 L −+1 s −2bL [] bS()1− e S

To illustrate the influence of thick faces let us take an example. A cantilever beam with 5 mm thick aluminium alloy faces and a 20 mm core with a shear modulus of 40 MPa. The beam has a length of 500 mm and a point load P = 100 N is applied at its edge. Hence, the ratio d/tf = 5 < 5.77 according to eq.(3.5). One can now plot the shear deformation as a function of x as

4.10 BEAM THEORY shown in Fig.4.9. One can see in Fig.4.9 that there is a zone near the clamped edge where the bending stiffness of the faces prevent the core shear deformation, and that the deformations in this zone are governed by the bending of the two faces about their individual neutral axes.

The effect is rapidly damped out and after about x = 2tc the slope of the shear deformation is the same as that assuming thin faces, that is, Tx(x > 2tc) = P. It is also seen that after the effect has damped out there is a constant difference between the two solutions, thick faces or thin faces. The load-point deflection is therefore only affected very little in this case the correct solution gives ws(L) = 37.27 mm while the thin face approach in eq.(4.6) gives 40.0 mm. 0 20406080100 0 x (mm) 2 4 6 W (mm) s 8

Figure 4.9 Shear deformation near the edge of a cantilever beam with thick faces. Solid line represents the correct and the dashed line the thin face solution.

4.6 Rigid core The shear stiffness given in eq.(4.2) for a weak core and thin faces can no longer be used with good accuracy if the core has significant moduli. The energy equation eq.(4.5) will still be valid though. However, that equation is quite complex and requires a fair amount of time to be calculated. So, for engineering design problems where eq.(4.2) is not satisfied Allen [1] proposed the following approximate solution to that problem. The shear strain is defined by the derivative du/dz, where u is the displacement in the x-direction. This gives the shear stress τc = Gc du/dz, which integrated gives the displacement u, and by using eq.(3.12) for the shear stress in the core, becomes

2 3 Tx ⎡ Etdff Etzcc⎛ z ⎞⎤ u =+−⎢ z ⎜ ⎟⎥ (4.34) GDc ⎣ 2243⎝ ⎠⎦

Hence, the displacement is no longer linear as was the case when Ec is very small. The x- displacement at the face/core interface, e.g., for z = tc/2 equals

3 ⎛ tTcx⎞ ⎡ Ettdffc Etcc⎤ u⎜ ⎟ =+ (4.35) ⎝ ⎠ ⎢ ⎥ 2424GDc ⎣ ⎦

Suppose now that the real core can be replaced by another core having Ec small so that the linear displacement in the x-direction imposed by ignoring the contribution of the core is * retained. The foregoing analysis may be repeated except that Gc is replaced by Gc and Ec is set to zero. However, the real value of Ec must still be used when calculating the flexural rigidity. This new core would have a linear u-displacement with a value at the interface (following eq.(3.15)) of

⎛ tcc⎞ τ tTcxE ffcttd u⎜ ⎟ == (4.36) ⎝ ⎠ ** 22Gc GDc 4

4.11 AN INTRODUCTION TO SANDWICH STRUCTURES

* * Now, Gc must be chosen so that these displacement coincide meaning that the new value Gc gives the right displacement in the approximate formulae given in eqs.(4.5-4.6). Thus, comparing the above gives

−1 ⎡ 2 ⎤ * Etcc GGcc=+⎢1 ⎥ (4.37) ⎣⎢ 6Etdff ⎦⎥

This procedure yields the correct deflections and face stresses but not the correct core shear stress; however that can be found using eq.(3.12).

4.7 Energy Relations Let's for simplicity study a symmetric sandwich cross-section. The strain energy per unit volume is

dUse =σij dεij which for a linear elastic material equals 1 dU = σ ε se 2 ij ij This can be integrated over the thickness to obtain the strain energy per unit area as

t / 2+t 1 −tc / 2 1 tc / 2 1 c f dU = σ ε dz + σ ε +τ γ dz + σ ε dz se ∫ f f ∫[]c c c c ∫ f 2 f 2 −t / 2−t 2 −t / 2 2 t / 2 c f c c −tc / 2 2 tc / 2 2 2 tc / 2+t f 2 1 σ f 1 ⎡σ c τ c ⎤ 1 σ f 2 = dz + ⎢ + ⎥dz + dz 2 ∫ E 2 ∫ E G 2 ∫ E −tc / 2−t f f −tc / 2 ⎣ c c ⎦ tc / 2 f

For the sandwich cross-section with thin faces and a weak core, we know that

M x M x Tx σc = 0, σ 1 = − , σ 2 = and τ c = t1d t2 d d and thus,

−t / 2 t / 2 t / 2+t 1 c ⎛ M ⎞⎛ M ⎞ 1 c ⎛ T ⎞⎛ T ⎞ 1 c f ⎛ M ⎞⎛ M ⎞ dU = ⎜− x ⎟⎜− x ⎟dz + ⎜ x ⎟⎜ x ⎟dz + ⎜ x ⎟⎜ x ⎟dz se 2 ∫ ⎜ t d ⎟⎜ t dE ⎟ 2 ∫ d ⎜ d G ⎟ 2 ∫ ⎜ t d ⎟⎜ t dE ⎟ −tc / 2−t f ⎝ f ⎠⎝ f f ⎠ −tc / 2 ⎝ ⎠⎝ c ⎠ tc / 2 ⎝ f ⎠⎝ f f ⎠ t / 2+t 1 −tc / 2 M 2 1 tc / 2 T 2 1 c f M 2 = x dz + x dz + x dz 2 ∫ t 2 d 2 E 2 ∫ d 2G 2 ∫ t 2 d 2 E −tc / 2−t f f f −tc / 2 c tc / 2 f f

2 2 M 2 1 T 2 1 M 2 1 T 2 1 ⎡ ⎛ d 2 w ⎞ ⎛ dw ⎞ ⎤ = x t + x t = x + x = ⎢D⎜ b ⎟ + S⎜ s ⎟ ⎥ t 2 d 2 E f 2 d 2G c 2 D 2 S 2 ⎜ dx 2 ⎟ dx f f c ⎣⎢ ⎝ ⎠ ⎝ ⎠ ⎦⎥

The expression for strain energy can also be obtained by considering the work done by the external moments and forces acting on an element. If we this time also account for thick face sheets we get that the work done by Mx1 and Mx2 equals (by use of eq.(4.27))

4.12 BEAM THEORY

2 2 1 ⎛ d 2 w ⎞ 1 ⎛ d 2 w ⎞ 1 ⎛ d 2 w ⎞ 1 ⎛ d 2 w ⎞ M ⎜− ⎟ = D ⎜ ⎟ and M ⎜− ⎟ = D ⎜ ⎟ x1 ⎜ 2 ⎟ 1 ⎜ 2 ⎟ x2 ⎜ 2 ⎟ 2 ⎜ 2 ⎟ 2 ⎝ dx ⎠ 2 ⎝ dx ⎠ 2 ⎝ dx ⎠ 2 ⎝ dx ⎠

Similarly, the work done by Nx1 and Nx2 are similarly

1 1 dθ 1 d 2 w 1 d 2 w N ε(e − d) = N ()e − d x = N (d − e) b = E t (d − e) 2 b 2 x1 2 x1 dx 2 x1 dx 2 2 1 1 dx 2

1 1 dθ 1 d 2 w 1 d 2 w and N ε(e) = N e x = − N e b = E t e2 b 2 x1 2 x1 dx 2 x1 dx 2 2 2 2 dx 2

The work done by Mxc is simply

2 1 ⎛ d 2 w ⎞ 1 ⎛ d 2 w ⎞ M ⎜− b ⎟ = D ⎜ b ⎟ xc ⎜ 2 ⎟ c ⎜ 2 ⎟ 2 ⎝ dx ⎠ 2 ⎝ dx ⎠ and finally the work done by Tx becomes

2 1 ⎛ dws ⎞ 1 ⎛ dws ⎞ Tx ⎜ ⎟ = S⎜ ⎟ 2 ⎝ dx ⎠ 2 ⎝ dx ⎠

The total energy density is then

2 2 2 1 ⎛ d 2 w ⎞ 1 ⎛ d 2 w ⎞ 1 ⎛ dw ⎞ dU = (D + D )⎜ b ⎟ + (D + D )⎜ ⎟ + S⎜ s ⎟ (4.38) se 0 c ⎜ 2 ⎟ 1 2 ⎜ 2 ⎟ 2 ⎝ dx ⎠ 2 ⎝ dx ⎠ 2 ⎝ dx ⎠

The total strain energy is now found by an integration over the entire length of the beam, resulting in

2 2 2 1 ⎡ ⎛ d 2 w ⎞ ⎛ d 2 w ⎞ ⎛ dw ⎞ ⎤ U = ⎢(D + D )⎜ b ⎟ + (D + D )⎜ ⎟ + S⎜ s ⎟ ⎥dx se 2 ∫ 0 c ⎜ dx 2 ⎟ 1 2 ⎜ dx 2 ⎟ dx ⎣⎢ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦⎥ (4.39) 2 2 2 2 1 ⎡ ⎛ d 2 w ⎞ ⎛ dw ⎞ ⎤ 1 ⎡ ⎛ dθ ⎞ ⎛ dw ⎞ ⎤ ≈ ⎢D⎜ b ⎟ + S⎜ s ⎟ ⎥dx = ⎢D⎜ x ⎟ + S⎜ s ⎟ ⎥dx 2 ∫ ⎜ dx 2 ⎟ dx 2 ∫ dx dx ⎣⎢ ⎝ ⎠ ⎝ ⎠ ⎦⎥ ⎣⎢ ⎝ ⎠ ⎝ ⎠ ⎦⎥

The potential energy of the transverse force, q, equals

U = − q(x)wdx (4.40) q ∫

4.13 AN INTRODUCTION TO SANDWICH STRUCTURES

4.8 General Solution of Beam Problems There are several different ways of solving sandwich beam problems. The most straight forward way is by direct integration of the differential equations for the partial deflections given in eq.(4.23). Taking the beam part deflection first and then the shear part, these equations are integrate in general terms to

dw4 x 3 x 2 D b = q → Dw=++++ qdx4 A A Ax A (4.41) dx 4 b ∫∫ ∫∫ 1 622 34

dw2 S s =−q → Sw=− qdx2 − A x − A (4.42) dx 2 s ∫∫ 12 where the A’s are integration constants. The total deflection can thus be written as

wx()=+= wbs () x w () x 1 ⎧ x 3 x 2 ⎫ 1 (4.43) qdx4 ++++ A A Ax A −++qdx2 A x A ⎨∫∫∫∫ 1 2 34⎬ {} ∫∫ 12 D ⎩ 62 ⎭ S However, the rigid body translations given by A A A 42+=5 D S D may be replaced by a new integration constant. Now, there are four integration constants to be found, and these are given by the four boundary conditions at the edges of the beam (two at each edge). These are commonly;

Simply supported: wb + ws = 0, Mx = 0 (4.44a)

dwb Clamped: wb + ws = 0, = 0 (4.44b) dx

Free: Mx = 0, Tx = 0 (no kinematic restraints) If necessary, a given bending moment or transverse force may have to be invoked, e.g. a zero bending moment. That is done by

dw2 dw2 dw2 dw2 1 dT M = 0 → bs=− =− x =0 (4.44c) x dx2 dx2 dx 2 dx2 S dx or

1 dws Tx = 0 → = 0 (4.44d) S dx given at the location x where the boundaries are.

4.14 BEAM THEORY

A second method is by using already given functions for the bending deformation. There are several handbooks where wb is given explicitly as function of x, for a large number of beam cases. By using eq.(4.22) which gives the relation between bending and shear deformation, as dw2 D dw4 sb=− dx 2 S dx 4 Hence, by differentiating the known bending deformation four times, the shear deformation is given by twice integration of this and applying two boundary conditions. In many cases, the in-plane shear γ0 is known to be zero for symmetry reasons, then eq.(4.22) may be used in an even simpler form

dw D dw3 s =− b dx S dx 3 and the shear part of the solution is obtained even faster. In any case, when the shear deformation is found the total deformation is found adding the two solutions.

Another way to solve sandwich beam problems is, of course, to solve the Timoshenko beam equation of eq.(4.24) or (4.28). However, since these equations are in the total deflection, it might become difficult to apply proper boundary conditions using this approach.

4.9 Examples of Beam Calculations Following are a number of examples of design and calculations concerning sandwich beams. Several of them are for elementary loading cases and these may be used to solve other beam cases by means of superposition. In all cases the beam is of length L and the x-coordinate is measured from the left edge.

4.9.1 Cantilever beam P

Pure bending 3 wb = PL / 3D

Pure shear

ws = PL / S

w = wb + ws

Assume thin faces, tf << tc, and weak core, Ec << Ef, and that all other material data are known. The flexural rigidity and the shear stiffness is then according to eqs.(3.4) and (4.6)

Etd2 Gd2 D = ff and S = c 2 tc

4.15 AN INTRODUCTION TO SANDWICH STRUCTURES

The transverse force, Tx, equals P and the bending moment Mx = –P(L–x). The stress in the faces is then

Md x max PL σ f ==Ef 2D tdf

The bending deformation wb is most easily found from a handbook and is PL332PL wb == 2 3D 3Etdff and the shear deformation is found through the integration of eq.(4.2)

L T PL w ==x dx s ∫ 0 S S

The total deformation is then,

PL3 PL PL3 ⎛ 3D ⎞ PL3 w = + = ⎜1+ ⎟ = ()1+ 3φ 3D S 3D ⎝ SL2 ⎠ 3D where φ = D/L2S is the so called shear factor. If φ is zero, there is no shear and the problem is pure bending. As the shear increases, so does the displacement. Now observe that the bending deformation is a function of L3 whereas the shear deformation only varies linearly with L. This means that the proportion of shear deformation to the total deformation strongly depends on the length of the beam. Hence, transverse shear deformation has a strong influence on the deformation for short (generally valid) and shear weak (S small) beams. For slender and shear stiff beams it is on the other hand negligible.

In order to show this more clearly, let us use a numerical example, take a sandwich with tf = 1 3 mm, tc = 30 mm, L = 500 mm, Gc = 30 MPa (80 kg/m PVC foam), and Ef = 200,000 MPa (steel). This gives

wb= 0.43 P mm/N, ws= 0.52 P mm/N and hence a ratio wb/ws= 0.83

Note again that this ratio will change depending on beam length. Now compare this with, e.g., a homogeneous steel beam with a rectangular cross-section of height h = 30 mm and the same length. The deformation of this beam will be 4PL3 PL w ==0. 0926P mm/N, w ==1.. 2 0 0003P mm/N,, i.e., w /w = 310 b Eh3 s Gh b s This explains why engineering beam theory generally neglects the contribution of the transverse shear deformation. However, remember that it still must be included for short beams.

It is useful to use other load cases for cantilevers. Take for example the load cases illustrated below. The total deformations can be computed as above, but there is yet another alternative way to solve this problem and that is to use eq.(4.23) and integrating twice to obtain the shear

4.16 BEAM THEORY deformation. There are two boundary conditions for the unknown integration constants. The result is, of course, the same as the above.

(i) Uniform load q

x Q Tx()=− Q⎜⎛1 ⎟⎞ , Mx()=− (Lx − )2 x ⎝ L⎠ x 2L

432 qL4 ⎡⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎤ qx wxb ()= ⎢⎜ ⎟ − 46⎜ ⎟ + ⎜ ⎟ ⎥ , and wxs ()=− (2Lx ) 24D ⎣⎝ L⎠ ⎝ L⎠ ⎝ L⎠ ⎦ 2S where the total distributed load Q = qL.

An alternative way of analysing the same beam is to use eq.(4.22b). First use the expression for the deformation field in bending as given above. Its derivatives with respect to x of

dw qL43⎡41212x x 2 x ⎤ d 2 w qL4 ⎡12x 2 24x 12 ⎤ b =−+ , b = − + ⎢ 4 32⎥ 2 ⎢ 4 3 2 ⎥ dx 24D ⎣ L L L ⎦ dx 24D ⎣ L L L ⎦

d 3 w qL4 ⎡24x 24⎤ q b = − = (x − L) 3 ⎢ 4 3 ⎥ dx 24D ⎣ L L ⎦ D

According to eq.(4.23) the shear deformation is then

dw q qx 2 qLx qx s = − (x − L) → w = − + + B → wx()=− (2Lx ) dx S s 2S S s 2S since ws(0) = 0 and thus B = 0.

(ii) Hydrostatic pressure q(x)

2 2 3 qLmax ⎛ ⎛ x ⎞ ⎞ qLmax ⎡ ⎛ x ⎞ ⎛ x ⎞ ⎤ Txx ()=−⎜1 ⎜ ⎟ ⎟ , Mxx ()= ⎢32⎜ ⎟ −−⎜ ⎟ ⎥ 2 ⎝ ⎝ L⎠ ⎠ 6 ⎣ ⎝ L⎠ ⎝ L⎠ ⎦

4 53 2 2 qLmax ⎡⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎤ qLxmax ⎡ ⎛ x ⎞ ⎤ wxb ()= ⎢⎜ ⎟ −10⎜ ⎟ + 20⎜ ⎟ ⎥ , and wxs ()=−⎢3 ⎜ ⎟ ⎥ 120D ⎣⎝ L⎠ ⎝ L⎠ ⎝ L⎠ ⎦ 6S ⎣ ⎝ L⎠ ⎦

4.17 AN INTRODUCTION TO SANDWICH STRUCTURES

qL qx where the total load Q = max and the distributed load qx()= max 2 L

q(x)

qL x 2 qL2 x 3 Tx()=−max ⎜⎛1 ⎟⎞ , Mx()=−max ⎜⎛1 − ⎟⎞ x 2 ⎝ L⎠ x 6 ⎝ L⎠

4 2345 qLmax ⎡ ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎤ wxb ()= ⎢10⎜ ⎟ − 10⎜ ⎟ + 5⎜ ⎟ − ⎜ ⎟ ⎥ , and 120D ⎣ ⎝ L⎠ ⎝ L⎠ ⎝ L⎠ ⎝ L⎠ ⎦

2 qLxmax ⎡ ⎛ x ⎞ ⎛ x ⎞ ⎤ wxs ()=−⎢33⎜ ⎟ + ⎜ ⎟ ⎥ 6S ⎣ ⎝ L⎠ ⎝ L⎠ ⎦

qL qLx( − ) where the total load Q = max and the distributed load qx()= max 2 L 4.9.2 “Shear beam” P

x Ra Rb 2a a 2a The aim of the following example is to show that shear deflection may occur in ways that are not directly obvious unless one is well acquainted with their computation. The bending deformation of the above beam is easily calculated using engineering beam theory. Use the general form of the shear deformation given in eq.(4.3). We must use this expression this time since we need to consider rigid body rotation as well when analysing the pure shear case and this is not included when bending is omitted. From vertical and bending moment equilibrium the reaction loads at the supports are calculated as

Ra = –2P and Rb = 3P In a cross-section located at x ≤ 2a the loads and shear deformation are M γ xt γ xt T ==00, M , w =−xc00 +=−+C cC xxs S d d in the cross-section given by 2a ≤ x ≤ 3a M γ xt 22P()x − a γ xt T =−222P, M =− P(x − a ), w =−x 00cc +=−C −+C xx s S d S d

4.18 BEAM THEORY and finally in the cross-section 3a ≤ x ≤ 5a

Txx==−−+P, M 2233P(x a )P (x − a ),

M γ xt 2233P()()x − a P x − a γ xt w =−xc00 +=−C + −+ cC s S d S S d

There are now two unknowns, the in-plane shear γ0, and the constant C. To solve these there are two boundary conditions, w(2a) = w(3a) = 0. Using these in any of the above equations will give 42Pa Pd C =− and γ 0 =− S tSc The shear deformation is hence given by 2P xaw≤=−2 : ()xa2 s S

230a ≤≤x aw: s = 3P 35ax≤≤ aw: =(xa −3 ) s S This gives the following shape of the shear deformation

w w (0) (5 a )

4Pa 6Pa The displacements of the two free edges are then: w ()0 =− and wa()5 = . s S s S We could equally well have used eq.(4.3b) and determined the integration constants A and B instead and that will yield the same result.

4.9.3 Design example Consider a sandwich beam of length L which is simply supported at its edges and subjected to a uniformly distributed load Q per unit width. The faces are made of GRP with a modulus Ef ^ 3 and a strength σf.The core is a cross-linked PVC foam with density 100 kg/m and has a shear ^ modulus Gc and a shear strength τc. The maximum allowed deformation of the beam is L/50. Calculate the face thickness.

^ Data: Ef = 20,000 MPa, σf = 100 MPa ^ Gc = 40 MPa (Ec = 100 MPa), τc = 1.5 MPa

tc = 50 mm, L = 1 000 mm and Q = 100 N/mm width

Assume initially that Ec << Ef and tf << tc. Then

4.19 AN INTRODUCTION TO SANDWICH STRUCTURES

Etd2 Gd2 D = ff and S = c 2 tc The forces and bending moments acting on the beam are Q L Qx() L − x T =−⎜⎛ x⎟⎞ and M = x L ⎝ 2 ⎠ x 2L with the maximum values Q QL ||T = and ||M = x max 2 x max 8 (i) strength design: Find the face thickness for maximum allowable stress.

Md x max QL QL QL σ fcr, ==≈⇒Ef t f ≈=25mm. 288D tdffctt 8tcfcrσ ,

The maximum shear stress in the core is then

T Q τ ==≈x max 1 MPa (which is satisfactory) c d 2d The wrinkling stress is found in eq.(6.14)

3 σ ff==0. 5EEGcc 215 MPa (which is satisfactory)

(ii) stiffness design: Find the face thickness for allowable deformation. 5QL3 w = w + w , where w = b s b 384D The shear deformation is found from, e.g., eq.(4.2) and since the loading is symmetrical we have that γ0 = 0.

L 2T QL w ==x dx s ∫ 0 S 8S or from eq.(4.3) as M γ t x QL w =−x 0 c +C with w (0) = w (L) = 0 ⇒==⇒γ CwL02() = s S d s s 0 s 8S Then,

5QL3 QL Etd2 5QL3 L QL −1 w =+ ⇒=D ff =⎜⎛ − ⎟⎞ 384D 8S 2 384⎝ 50 8S ⎠

4.20 BEAM THEORY

by assuming that d ≈ tc and rearranging it is found that tf = 3.8 mm. Hence, a minimum of 3.8 mm thick faces are needed to fulfil the requirements. Check with the approximations in eqs.(3.5 and 3.6):

2 d 6Etdff eq.(3.6) ⇒=14.. 5 > 5 77 eq.(3.7) ⇒=>3 105. 6 100 (which is satisfactory) t f Etcc

4.9.4 Beam subjected to point load (i) Simply supported edges

aL P bL ML MR

RL RR This problem is solved by simply superimposing the deformations due to bending and shear. One can also arrive at the shear deformation by using eq.(4.22) on the bending deformation. The results are;

RL = Pb, RR = Pa and ML = MR = 0

3 3 PL ⎡ 2 ⎛ x ⎞ ⎛ x ⎞ ⎤ Pbx x ≤ aL wx()=+= wbs () x w () x bb⎢()1 −⎜ ⎟ − ⎜ ⎟ ⎥ + 6D ⎣ ⎝ L⎠ ⎝ L⎠ ⎦ S

Tx(x) = RL and Mx(x) = RLx

3 3 PL ⎡ 2 ⎛ Lx− ⎞ ⎛ Lx− ⎞ ⎤ Pa() L− x x ≥ aL wx()=+= wbs () x w () x aa⎢()1 −⎜ ⎟ − ⎜ ⎟ ⎥ + 6D ⎣ ⎝ L ⎠ ⎝ L ⎠ ⎦ S

Tx(x) = − RR and Mx(x) = RR(L − x) This simple superposition can be performed since the two extreme cases, S infinite or D infinite yield the same reaction forces, and hence the deformations occurring in the different modes are independent of each other.

(ii) One edge simply supported, the other clamped

aLP bL

This case is not as simple as it looks at first sight. The clamping moment at the edge will depend on the shear stiffness, S, giving the ordinary moment PL(3b–b2–2)/2b if S is infinite and MR = 0 if D is infinite. This fact yields an inconsistency; the transverse shear deformation will not be continuous under the load if the loads at the supports are calculated using beam bending theory and equilibrium equations since, due to a clamping moment MR ≠ 0, they will

4.21 AN INTRODUCTION TO SANDWICH STRUCTURES differ from the above (bP and aP, respectively). The shear deformation calculated by using support forces different from that would yield that the shear deformation ws(L) ≠ 0. The shear deformations following the transverse forces computed from the pure bending case are; the left reaction force times aL/S and the right reaction force times bL/S. It is now clearly seen that these two are unequal unless MR = 0, leading to an incompatibility at the edge.

The way to treat this is by neglecting that wb and ws equal zero at one of the edges and instead concentrating on the fact that their sum equals zero at the edges. In fact, the concept of partial deflections can still be used here but the clamping moment will depend on the shear stiffness which in consequence means that wb and ws are dependent, i.e., only their sum will yield the correct deformation field. Now, relax the boundary condition wb(0) = ws(0) = 0 and use instead wb(0) + ws(0) = 0, and compute the corresponding deformation fields due to bending and shear, respectively. Thus, treat the beam as a cantilever but ensure that the deflection of the free edge (at x = 0) is zero. Both partial deflections will depend on the reactions RR and RL which now can be solved using the boundary condition stated above. The equations required are the deformation fields as function of the reactions. These are

3 3 3 PL ⎡ 23⎛ x ⎞ ⎤ RLL ⎡⎛ x ⎞ ⎛ x ⎞ ⎤ x ≤ aL wxb ()=−⎢31b ⎜ ⎟ − b ⎥ − ⎢⎜ ⎟ − 32⎜ ⎟ + ⎥ 6D ⎣ ⎝ L⎠ ⎦ 6D ⎣⎝ L⎠ ⎝ L⎠ ⎦

Rx 1 wx()=+= w ()0 L RbL − R() aL − x and w(x) = w (x) + w (x) ssSS[]RL b s

Tx(x) = RL and Mx(x) = RLx

3 3 3 3 PL ⎡⎛ x ⎞ 23⎛ x ⎞ ⎤ RLL ⎡⎛ x ⎞ ⎛ x ⎞ ⎤ x ≥ aL wxb ()=−⎢⎜ ab⎟ −−32⎜ ab⎟ + ⎥ − ⎢⎜ ⎟ − 32⎜ ⎟ + ⎥ 6D ⎣⎝ L ⎠ ⎝ L ⎠ ⎦ 6D ⎣⎝ L⎠ ⎝ L⎠ ⎦

RaL RbL RbL RaL wL()=+− w ()00LR = ⇒=−w ()0 RL ssS S s S S

R which gives that wx()=−R ()Lx s S

Tx(x) = –RR and Mx(x) = RLx – P(x – aL) = PaL – RRx Using these equations along with two independent equations of equilibrium gives, after some manipulation and denoting the shear factor φ = D/L2S

Pb[(63φ +− b23 b )] Pa[(623φ +− b23 + b )] PbL()32 b−− b 2 R = , R = , M = L 21( +) 3φ R 21( +) 3φ R 21()+ 3φ

It is now seen that wb(L) = ws(L) = 0 but that wb(0) and ws(0) ≠ 0. It is also easy to verify that eq.(4.22) is satisfied. It is also seen that when S approaches infinity (φ = 0), the pure bending case, then

4.22 BEAM THEORY

P P PbL R =−()3bb23, R =−+()23bb23, M =−−()32bb2 L 2 R 2 R 2 and for the case of pure shear (φ = ∞)

RL = Pb, RR = Pa and MR = 0 which is the same as for the simply supported case above. It is easy to show that the total deformation at x = 0 actually equals zero, though the individual deflections do not.

An approximate approach would be superimposing the case infinite S with the case of infinite

D, i.e., wb + ws, that is, taking the bending deformation from a handbook and then adding the shear deformation derived in the case above with both edges simply supported. That is, use the set of reactions for S = ∞ when calculating the bending deformation and the reactions for

D = ∞ when calculating the shear deformations. This is the same thing as adding wb when φ =

0 with ws when φ = ∞ in the solution given above.

The exact and the approximate solutions can be compared in terms of deflections and reaction forces to determine the error made by using partial deflections. The parameter determining the degree of shear deformation is, as seen above, φ = D/L2S. Taking a = b = 0.5, L = 500 mm and a cross-section of 1 mm faces of modulus 106 MPa, and a 30 mm thick core, the following results are obtained for different values of the shear stiffness. In the approximate solution only the deflection is given since the reaction forces and bending moments are those of the pure bending case (column 2 in the Table 4.1). In the pure shear case, the clamping moment is zero, the reaction forces are those of the simply supported case above, the mid-point bending moment M(aL) = 125 Nmm and the corresponding deformations, wb + ws, are listed below. The solution below was obtained using the three-noded finite element described in chapter 12. Ten equal length three-nod elements were used.

φ 0 0.0620 0.620 2.07 62.0 Approximate solution w(aL) 0.00237 0.0064 0.0427 0.1368 4.03 Exact solution

RL 0.3125 0.3419 0.4344 0.4740 0.499

RR 0.6875 0.6581 0.5656 0.5258 0.501

MR –93.75 –79.05 –32.8 –13.0 –0.5 M (aL) 78.12 85.47 108.6 118.5 124.7 x w(aL) 0.00237 0.00688 0.0447 0.1393 4.04 FE -solution

RL 0.312 0.342 0.434 0.474 0.501

RR 0.688 0.658 0.566 0.526 0.499

MR –93.9 –79.1 –32.8 –13.0 –0.5 M (aL) 78.0 85.5 108.5 118.5 125.2 x w(aL) 0.00237 0.00688 0.0447 0.1394 4.04

Table 4.1 Comparison between analytical and FE calculations of a hyperstatic sandwich beam

4.23 AN INTRODUCTION TO SANDWICH STRUCTURES

As seen from Table 4.1, the approximate solution yields fairly accurate results in terms of mid-point deflection. However, the stress analysis will be severely wrong since the reaction forces are computed in an incorrect way. The fact that the reaction forces change with the shear factor φ means that any design stress also depends on φ. Any hyperstatic beam will exhibit this kind of behaviour. If, on the other hand, the reactions can be determined solely from equilibrium equations, as in case (i), the beam is isostatic.

(iii) Both edges clamped

aL P bL

The same thing as above must also be done here, that is, e.g., use the boundary condition wb(0) + ws(0) = 0 and dwb/dx = 0, along with the equations of equilibrium. The deformations due to bending and shear can directly be written as a function of the reaction forces and clamping moments and are

3 3 3 PL ⎡ 23⎛ x ⎞ ⎤ RLL ⎡⎛ x ⎞ ⎛ x ⎞ ⎤ x ≤ aL wx()=+= wbs() x w () x ⎢31b ⎜ −⎟ − b ⎥ − ⎢⎜ ⎟ − 32⎜ ⎟ + ⎥ 6D ⎣ ⎝ L⎠ ⎦ 6D ⎣⎝ L⎠ ⎝ L⎠ ⎦

2 2 MLL ⎡⎛ x ⎞ ⎛ x ⎞ ⎤ 1 − ⎢⎜ ⎟ − 21⎜ ⎟ + ⎥ +−−[]RbLRL R() aL x 2D ⎣⎝ L⎠ ⎝ LS⎠ ⎦

Tx(x) = RL and Mx(x) = ML + RLx

3 3 3 3 PL ⎡⎛ x ⎞ 23⎛ x ⎞ ⎤ RLL ⎡⎛ x ⎞ ⎛ x ⎞ ⎤ x ≥ aL wx()=−⎢⎜ ab⎟ −−32⎜ ab⎟ + ⎥ − ⎢⎜ ⎟ − 32⎜ ⎟ + ⎥ 6D ⎣⎝ L ⎠ ⎝ L ⎠ ⎦ 6D ⎣⎝ L⎠ ⎝ L⎠ ⎦

2 2 MLL ⎡⎛ x ⎞ ⎛ x ⎞ ⎤ 1 − ⎢⎜ ⎟ − 21⎜ ⎟ + ⎥ +−RLR () x 2D ⎣⎝ L⎠ ⎝ LS⎠ ⎦

Tx(x) = −RR and Mx(x) = MR + RR(L − x) The boundary conditions stated above yield two equations which along with two independent equations of equilibrium give the four unknowns as

Pb[(2 12++ a ) 12 bφ ] Pa[(2 12++ b ) 12 aφ ] R = , R = L 112+ φ R 112+ φ

PLab()6φ + b PLab()6φ + a M =− , M =− L 112+ φ R 112+ φ

We can now observe that the limits for these equations equal the special cases of infinite shear stiffness and infinite bending stiffness, respectively. Thus,

4.24 BEAM THEORY

2 2 2 2 Pure bending (φ = 0): RL = Pb (1+2a), RR = Pa (1+2b), ML = –PLab , MR = –PLa b

PL Pure shear (φ = ∞): R = Pb, R = Pa, MM==−ab L R LR 2

Hence, the clamping moments ML and MR will be equal if a = b and/or φ = ∞, and otherwise unequal. The same thing goes for the support forces RL and RR. If a = b, then all forces and moments will equal those of the pure bending case, independently of the shear stiffness. The interesting thing with this case is that the bending moments in the general case, a ≠ b, change from the ordinary clamping moments for φ = 0 to be equal as the shear stiffness decreases.

The same type of approximation can now be performed as in the case above, that is, to superimpose the deformation in the pure bending case with the pure shear case. That is, take wb from above with φ = 0 and superimpose the shear deformation given by the case φ = ∞. To illustrate the derived equations take the same example as above, but now assume e.g. a = 0.7 and b = 0.3 so that the change in the clamping moments can be seen. The results of the two forces and the two bending moments along with the deformation of the load point are given below. The solution below was obtained using the three-noded finite element described in chapter 12. Twenty equal length three-nod elements were used.

φ 0 0.0620 0.620 2.07 62.0 Approximate solution w(aL) 0.000803 0.00419 0.0347 0.114 3.39 Exact solution

RL 0.216 0.252 0.290 0.297 0.300

RR 0.784 0.748 0.710 0.703 0.700

ML –31.5 –40.5 –50.01 –51.7 –52.5

MR –73.5 –64.5 –54.99 –53.3 –52.5 M (aL) 44.1 47.7 51.51 52.2 52.5 x w(aL) 0.000803 0.00425 0.0348 0.114 3.39 FE -solution

RL 0.215 0.252 0.290 0.297 0.300

RR 0.784 0.748 0.710 0.703 0.700

ML –31.4 –40.5 –50.0 –51.7 –52.5

MR –73.6 –64.5 –55.0 –53.3 –52.5 M (aL) 43.9 47.7 51.5 52.3 52.5 x w(aL) 0.00092 0.00426 0.0348 0.114 3.39

Table 4.2 Comparison between analytical and finite element calculations of a clamped sandwich beam

Once again, the approximate solution can be used with good accuracy when calculating the deformations but the transverse forces and bending moments will be quite wrong unless the exact solution is used.

4.25 AN INTRODUCTION TO SANDWICH STRUCTURES

4.9.5 Beam subjected to uniform pressure (i) Simply supported edges q

This problem is once again solved by simply superimposing the deformations due to bending and shear. One can also arrive at the shear deformation by using eq.(4.22) on the bending deformation. The results are;

4 43 qL ⎡⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞⎤ q 2 wx()=+= wbs() x w () x ⎢⎜ ⎟ − 2⎜ ⎟ + ⎜ ⎟⎥ +−()Lx x 24D ⎣⎝ L⎠ ⎝ L⎠ ⎝ L⎠⎦ 2S

qx2 T (x) = R – qx and Mx()=− Rx x L xL2

QqL RR=== and M = M = 0 LR22 L R (ii) One edge simply supported, the other clamped q MR

RL RR

This problem is solved in the same way as outlined for point load above, that is, use the boundary condition wb(0) + ws(0) = 0 along with two independent equations of equilibrium to solve the three unknowns RL, RR and MR. The results are

4 4 3 3 qL ⎡⎛ x ⎞ ⎛ x ⎞ ⎤ RLL ⎡⎛ x ⎞ ⎛ x ⎞ ⎤ wx()=+= wbs() x w () x ⎢⎜ ⎟ − 43⎜ ⎟ + ⎥ − ⎢⎜ ⎟ − 32⎜ ⎟ + ⎥ 24D ⎣⎝ L⎠ ⎝ L⎠ ⎦ 6D ⎣⎝ L⎠ ⎝ L⎠ ⎦

1 ⎡ q 22⎤ +−+−RxL () L ()Lx S ⎣⎢ 2 ⎦⎥

qx2 T (x) = R – qx and Mx()=− Rx x L xL2 By these equations the reactions and the clamping moment can be derived as

4.26 BEAM THEORY

qL ⎛ 312+ φ ⎞ qL ⎛ 512+ φ ⎞ qL2 R = ⎜ ⎟ , R = ⎜ ⎟ , M =− L 8 ⎝ 13+ φ ⎠ R 8 ⎝ 13+ φ ⎠ R 81()+ 3φ

It is now seen that the reactions are the same as the ordinary bending case when φ = 0 and RL

= RR = qL/2 and MR= 0 when φ = ∞, which is similar to the point load case.

(iii) Both edges clamped q

It is quite quickly realised in this case that due to symmetry the reaction forces are equal and so are the clamping moments. By assuming otherwise and performing an analysis similar to the above, one reaches the following solution

4 432 qL ⎡⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎤ q 2 wx()=+= wbs() x w () x ⎢⎜ ⎟ − 2⎜ ⎟ + ⎜ ⎟ ⎥ +−()Lx x 24D ⎣⎝ L⎠ ⎝ L⎠ ⎝ L⎠ ⎦ 2S

q q ⎛ L2 ⎞ Tx()=− (Lx2 ) and Mx()=−−⎜ Lx x 2 ⎟ x 2 x 26⎝ ⎠ which is the same solution as given by superimposing the partial deflections. The reaction forces and clamping moments are qL qL2 RR==, MM==− LR2 LR12 4.9.6 Beam subjected to hydrostatic pressure These problems are solved in exactly the same manner as the above; therefore, the derivations are omitted here and only the result is given. The points of maximum deflection may be somewhat difficult to find in these cases as they do not appear in the middle of the beam or at any other obvious location. Both wb and ws will have their maxima found by differentiating and finding the stationary value but it is unlikely that they will appear at the same point. For practical purposes it may thus be more convenient simply to approach this by a trial-and-error procedure.

(i) Simply supported edges q(x)

Denote q(L) = qmax and thus total load Q = qmaxL/2 and q(x) = qmaxx/L.

22 qLmax⎡ ⎛ x ⎞ ⎤ qLx max ⎡ ⎛ x ⎞ ⎤ Txxx()=−⎢13⎜ ⎟ ⎥ and Mx ()=−⎢1 ⎜ ⎟ ⎥ 6 ⎣ ⎝ L⎠ ⎦ 6 ⎣ ⎝ L⎠ ⎦

4.27 AN INTRODUCTION TO SANDWICH STRUCTURES

4 53 2 qLmax⎡ ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞⎤ qLx max ⎡ ⎛ x ⎞ ⎤ wx()=+= wbs() x w () x ⎢3107⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟⎥ +−⎢1 ⎜ ⎟ ⎥ 360D ⎣ ⎝ L⎠ ⎝ L⎠ ⎝ L⎠⎦ 6S ⎣ ⎝ L⎠ ⎦

qL qL R ==max and R max LR63

(ii) One edge simply supported, the other clamped q(x)

qx23qx Tx()=− R max and Mx () = Rx − max xL26L xLL

4 5 3 3 qLmax ⎡⎛ x ⎞ ⎛ x ⎞ ⎤ RLL ⎡⎛ x ⎞ ⎛ x ⎞ ⎤ wx()=+= wbs () x w () x ⎢⎜ ⎟ − 54⎜ ⎟ + ⎥ − ⎢⎜ ⎟ − 32⎜ ⎟ + ⎥ 120D ⎣⎝ L⎠ ⎝ L⎠ ⎦ 6D ⎣⎝ L⎠ ⎝ L⎠ ⎦

1 ⎡ qmax 33⎤ +−−−Rx() L (xL ) S ⎣⎢ L 6L ⎦⎥

qL⎛15+ φ ⎞ qL⎛ 410+ φ ⎞ qL R = max ⎜ ⎟ , R = max ⎜ ⎟ , and M =− max L 10 ⎝13+ φ ⎠ R 10 ⎝ 13+ φ ⎠ R 15() 1+ 3φ

q(x)

qx x qx2 x Tx()=− R max⎜⎛2 −⎟⎞ and Mx ()=− Rx max ⎜⎛3 −⎟⎞ xL2 ⎝ L⎠ xL6 ⎝ L⎠

4 54 qLmax ⎡ ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎤ wx()=+= wbs () x w () x ⎢ −⎜ ⎟ + 51511⎜ ⎟ − ⎜ ⎟ + ⎥ 120D ⎣ ⎝ L⎠ ⎝ L⎠ ⎝ L⎠ ⎦

3 3 223 RLL ⎡⎛ x ⎞ ⎛ x ⎞ ⎤ 1 ⎡ qxmax qL max qx max ⎤ − ⎢⎜ ⎟ − 32⎜ ⎟ + ⎥ +−−++⎢RxL () L ⎥ 6D ⎣⎝ L⎠ ⎝ LS⎠ ⎦ ⎣ 236L ⎦

qL⎛11+ 40φ ⎞ qL⎛ 920+ φ ⎞ 7qL2 R = max ⎜ ⎟ , R = max ⎜ ⎟ , and M =− max L 40 ⎝ 13+ φ ⎠ R 40 ⎝ 13+ φ ⎠ R 120() 1+ 3φ

4.28 BEAM THEORY

(iii) Both edges clamped q(x)

qx23qx Tx()=− R max and Mx () = RxM + − max xL26L xLLL

2 2 ML⎡ x x ⎤ 1 ⎡ qmax 33⎤ − L ⎛ ⎞ − 21⎛ ⎞ + +−−−Rx() L (xL ) ⎢⎜ ⎟ ⎜ ⎟ ⎥ ⎢ L ⎥ 2D ⎣⎝ L⎠ ⎝ LS⎠ ⎦ ⎣ 6L ⎦

qL⎛ 340+ φ ⎞ qL⎛ 780+ φ ⎞ R = max ⎜ ⎟ , R = max ⎜ ⎟ , L 20 ⎝ 112+ φ ⎠ R 20 ⎝ 112+ φ ⎠

qL2 ⎛115+ φ ⎞ qL2 ⎛110+ φ ⎞ M =− max ⎜ ⎟ , and M =− max ⎜ ⎟ L 30 ⎝112+ φ ⎠ R 20 ⎝112+ φ ⎠

4.9.7 Hyperstatic beam example Consider a design case of a sandwich beam with one edge simply supported and the other clamped, subjected to a uniform pressure loading, as solved in example 4.9.5(ii). To illustrate the implication of hyperstatic beams assume that all materials and the cross-section geometry are given and the length of the beam is the sought design value. Take the following example: ^ tf = 0.43 mm, Ef = 210,000 MPa (steel), σf = 96 MPa ^ tc = 100 mm, Gc = 3 MPa (low density PUR), τc = 0.1 MPa uniform pressure q = 1.2 kPa (that is; a load 0.012 N/mm width of the beam) Maximum allowed deflection in the middle of the beam Δ = L/150.

The data gives that D = 451,500,000 Nmm and S = 300 N/mm. Hence, there are two strength requirements and one stiffness requirement on the beam. From example 4.9.5(ii) in conjunction with eq.(3.18) it follows that

qL4 131572++. φφ2 L I: wx(/)== L 2 < 192D 1+ 3φ 150

RR qL 512+ φ II: τ$ c >= d 8d 13+ φ

4.29 AN INTRODUCTION TO SANDWICH STRUCTURES

qL2 1 III: σ$ f > 8tdf 13+ φ

Since the shear factor φ is a function of beam length L, the above equations become fairly difficult to solve. However, by trial-and-error it is easy to find an approximate value of L for each design constraint. By inserting the values of D and S from above in these equations, rearranging the above equations for L as function of φ, L can be calculated for different choices of φ(L). Solving these equations for L we get

I: L = 6075 mm II: L = 13400 mm III: L = 5610 mm

Hence, the beam length should not exceed 5610 mm and the active constraint is that of face direct stress.. The same kind of approach has to take place whatever property is to be calculated, since thicknesses, materials and length all depend on φ. Hence, in whatever design situation, the calculation leads to a trial-and-error operation. We should note, however, that for other values of φ the maximum bending moment may not appear at the clamped edge and we will have to seek its maximum (which is a function φ) before we can state condition III.

4.10 Torsion

MTx

M Tx

Figure 4.10 Sandwich beam subjected to a torque M . Tx

The rate of twist of a beam subjected to a torque MTx is written dϕ M = Tx (4.45) dx GJ where GJ is the torsional stiffness and ϕ the twist. Providing the faces are isotropic and the beam is wide (b is much larger than the thickness of the sandwich) the torsional stiffness may be written [9]

33 8 ⎡⎛ t ⎞ ⎛ G ⎞⎛ t ⎞ ⎤ ⎢⎜ c ⎟ ⎜ cxy ⎟⎜ c ⎟ ⎥ 3 GJ =+⎜1 ⎟ +−⎜ 1⎟⎜ ⎟ Gbtff (4.46) 3 ⎢ 2t G 2t ⎥ ⎣⎝ f ⎠ ⎝ f ⎠⎝ f ⎠ ⎦ where b is the width of the beam (span in the y-direction). When the core shear modulus is sufficiently small compared to the shear modulus of the faces (a reasonable assumption in almost all practical cases) the beam acts as two faces twisting independently of each other reducing the torsional stiffness to the well-known relation

4.30 BEAM THEORY

33 8 ⎡⎛ t ⎞ ⎛ t ⎞ ⎤ 2Db GJ =+⎢⎜1 c ⎟ − ⎜ c ⎟ ⎥Gbt3 = 0 (4.47) 3 ⎢⎜ 22t ⎟ ⎜ t ⎟ ⎥ ff 1 + ν ⎣⎝ f ⎠ ⎝ f ⎠ ⎦ which is seen to be the same as the twisting stiffness Dxy (see chapter 8). 4.11 Testing of Sandwich Beams Sandwich beams are commonly used for testing since they can be used to obtain a variety of sandwich properties. It is particularly interesting to obtain flexural properties and, as will be shown below, both the bending stiffness D and the shear stiffness, S, can be measured. It is also possible, by appropriate design of the beam test, to obtain different failure modes and thus different fracture strengths can be found.

4.11.1 The three-point bend (TPB) test The three-point bend (TPB) specimen is schematically illustrated in Fig.4.11. The loads should be applied over some larger surface to prevent local indentation of the faces. A problem is that local load magnitude is P in the upper face and only P/2 in the lower giving rise to problems of local indentation in the upper face sheet at times. P

L P/2 P/2

Figure 4.11 The three-point bend test specimen. The maximum bending moment and transverse force appearing in the beam are PL P M = and T = (4.48) max 4 max 2 from which the maximum face stress and maximum core shear stress can be calculated. Depending on the span L the beam can be designed so that one can obtain a face tensile or compressive failure, or a core shear fracture. The deflection under the point load is PL3 PL w =+ (4.49) 48D 4S Strictly speaking, the bending stiffness D and the shear stiffness S could now be determined by performing two tests with different spans. A better approach is outlined in [1]; rewrite eq.(4.49) as w L2 1 w 1 1 1 =+ and =+ PL 48D 4S PL3248 D4 S L Now it is possible to perform a number of tests with the TPB using different spans L and then plotting the results in two different diagrams. The first diagram should be w/PL vs. L2, for which the asymptotic slope for large L-values will be equal to 1/48D, and a second diagram,

4.31 AN INTRODUCTION TO SANDWICH STRUCTURES w/PL3 vs. 1/L2, for which the asymptotic slope for small L-values equal 1/4S. This is illustrated in Fig.4.12. w w 1 3 PL PL 48D 1 4S

1 1 4S 2 1 L 48D 2 L

Figure 4.12 Determination of D and S using the three-point bend test.

4.11.2 The four-point bend (FPB) test The four-point bend (FPB) specimen is schematically illustrated in Fig.4.13. There is an inherent advantage by using the FPB rather than the TPB. The local load magnitude now has decreased since the applied load of the FPB can be 2P with the same local loading of P. This gives a lower risk of getting local failure due to indentation of the face sheet. A testing standard exist in the ASTM C393. P P L1

L PP2

Figure 4.13 The four-point bend test specimen. In the region between the inner and outer supports the transverse force is constant and equal to P, which in turn means that the shear stress in the core is constant over a fairly long part of the beam. Between the inner supports, over a length of L1, the bending moment is constant and equal to P(L2 − L1)/2, while in the same zone the transverse force is identically equal to zero. The deformation of the beam can easily be found by ordinary sandwich theory. The deformation of the load points, assuming the faces to be thin, is equal to the cross-head displacement of the testing machine and is given by

2 ⎛ LL21− ⎞ PLLL()()() 21−+212 L PL 21− L w⎜ ⎟ = + (4.50) ⎝ 2 ⎠ 24D 2S

The maximum deflection, in the middle of the beam can similarly be written

2 2 ⎛ LPLLLLLL2212⎞ ()(−+−2212 1 )()PL21− L w⎜ ⎟ = + (4.51) ⎝ 2 ⎠ 48D 2S

The maximum direct stress in the faces appear between the inner supports and the maximum transverse shear stress appear between the outer and inner support, and can in the case of thin equal faces, be written as

4.32 BEAM THEORY

M P()LL21− P σ f =± =± and τc = (4.52) tdff2td d

A key feature of the FPB specimen is the fact that in the mid-section the transverse forces is zero and the bending moment is constant. This means that the bending stiffness easily can be measured in this section since the curvature derives only from the bending of the beam and that curvature is constant over the entire mid-section.

A dial gauge is mounted on a rig which is placed on top of the FPB specimen in the region between the inner supports. The distance between the supports of the rig is denoted c, see Fig.4.14. When applying the load P the curvature will cause the mid-section to curve and thus the dial gauge will display a displacement as a function of the applied load. Since the bending moment is constant the bending stiffness is then given by c

L 1

Figure 4.14 Measurement of the bending stiffness in the four-point bend test specimen.

Mc2 PL()− L c2 w ==21 (4.53) 816D D

Another feature is that the spans L1 and L2 easily can be changed so the faces can become critical (will fail due to tensile or compressive failure when the load is increased) or the core will be critical in shear. Thus, by beam design, the FPB test can be used to extract a number of important sandwich characteristics. This design capability is utilised in the fatigue testing described in section 2.5 simply by choosing a FPB design which by margin makes the core critical in shear so that the core eventually will fail in shear as the fatigue loading is applied.

In [1] the effect of thick faces and the overhang of the FPB is considered leading to more accurate expressions for the deflection.

References [1] Allen H.G., Analysis and Design of Structural Sandwich Panels, Pergamon Press, Oxford, 1969.

[2] Plantema F.J., Sandwich Construction, John Wiley & Sons, New York, 1966.

[3] Roark R.J. and Young W.C., Formulas for Stress and Strain, McGraw-Hill, Fifth edition, London, 1976.

4.33 AN INTRODUCTION TO SANDWICH STRUCTURES

[4] Mindlin R.D., “The Influence of Rotary Inertia and Shear on Flexural Motions of Isotropic Elastic Plates”, Journal of Applied Mechanics, Transactions of the ASME, Vol. 18, 1951, pp 31-38.

[5] Timoshenko S.P., Vibration Problems in Engineering, Second Edition, D. Van Nostrand Company Inc., New York, N.Y., 1937, p 337.

[6] Clough R.W. and Penzien J., Dynamics of Structures, McGraw-Hill, New York, 1975, pp 301, 318.

[7] Hoff N.J. and Mautner S.E., “Bending and Buckling of Sandwich Beams”, Journal of the Aeronautical Sciences, Vol. 15, No 12, 1948, pp 707-720.

[8] Hoff N.J., “Bending and Buckling of Rectangular Sandwich Plates”, NACA TN 2225, 1950.

[9] Seide P., “On the Torsion of Rectangular Sandwich Plates”, Journal of Applied Mechanics, Vol. 23, no 2, 1956, pp 191-194

4.34 BEAM THEORY

Exercises Example 4.1 The shear stiffness S for a beam cross-section can be derived using the formula 1 1 T T γ = τ (z)γ (z)dz where by definition γ = x 2 x 2 ∫ xz xz A S Compute the shear stiffness for the sandwich panel in problem 3.1 assuming the faces to be thin and the core to be weak. 2 Gdc Ans. S ≈ → Sx = 1702.4 N and Sy = 1043.4 N tc Example 4.2 The shear stiffness for a cross-section depends on the shear strain over the section and is computed using the energy balance equation so that the work done by the transverse force T with respect to the average shear strain γ* equals the strain energy. In the general case one can write S = GA/k where k is a factor depending on the shape of the cross-section and the materials, G is the shear modulus and A the cross-section area. For homogeneous rectangular section k equals 1.2. What is k for a sandwich with thin faces tf<

Example 4.3 A ramp to a car ferry shall be designed. The ramp, as illustrated below is used for getting cars into the upper deck of the ferry. Only one car at the time is allowed on the ramp and the maximum weight of cars on the upper deck is 5 tons. The design load case is that when one car is in the middle of the ramp. The ramp is 10 m long, 2.5 m wide, must have a safety factor of 5 against fracture, and the maximum allowed deformation is 300 mm. Due to wear the faces is of a 5 mm thick stainless steel plate which has a yield point of 250 MPa. From a manufacturing point of view, a PVC core material should be chosen which is available in 4 different densities according to the table below. The ramp can be approximated with a simply supported beam subjected to two equally large point loads corresponding to the axles of the car. Which core material should be used and what thickness must it have? PP

3000

10000

b = 2 500 mm, tf = 5 mm, Ef = 205 000 MPa, Gf = 79 000 MPa and σf,cr = 250 MPa

Core/density Ec (MPa) Gc (MPa) τcr (MPa) H60 55 22 0.6 H100 95 38 1.2 H130 125 47 1.6 H200 195 75 3.0 Ans. By using tables of elementary beam cases one finds that ⎛ L ⎞ P(L − l)(2L2 + 2Ll − l 2 ) w⎜ ⎟ = and ⎝ 2 ⎠ 48D P w (L / 2) = (L − l) where P = 25,000 N, L = 10 m and l = 3 m s 2S MmaxdE f PL()− l P MPLlmax =−() → σmax =≈, Tmax = P → τmax = 22D tdf d 2 Approximate: S ≈ Gcd, D ≈ Eftfd /2, d = 2tf + tc (i) Stiffness: No safety factor (ii) Shear stress: P/b = 50 N/mm (iii) Face stress: P/b = 50 N/mm

(iv) Local buckling: P/b = 50 N/mm → σmax = 315 MPa for H60 - no problem!

4.35 AN INTRODUCTION TO SANDWICH STRUCTURES

The above gives the following minimum values of d. Core Stiffness Shear stress Face stress H60 51 84 70 H100 50 42 70 H130 50 31 70 H200 49 17 70 i.e., choose a 79 mm thick H60 core. (since a 79 mm thick H60 weighs less that a 65 mm thick H100)

Example 4.4 A sandwich beam with length L = 500 mm, is clamped at both ends and subjected to a uniformly distributed load q = 0.1 N/mm width (100 kPa for a unit width beam). It has faces of a unidirectional E-glass/epoxy laminate and a core of 75 kg/m3 PMI cellular plastic. Calculate the minimum face thickness required providing that none of the allowables are exceeded. The maximum allowed deformation is L/50. Present all stages in the calculation and show that any approximations done are adequately fulfilled.

q Faces: Ef = 27 GPa, σf,cr = 240 MPa Core: E = 105 MPa, G = 29 MPa x c c τc,cr = 1.3 MPa tc = 25 mm,

z Hint: The deformation for a clamped beam as illustrated above, without respect to transverse shear deformations, can be written 4 3 2 qL4 ⎡⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎤ w (x) = ⎢⎜ ⎟ − 2⎜ ⎟ + ⎜ ⎟ ⎥ b 24D L L L ⎣⎢⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦⎥ If the load is increased, what type of fracture would the beam exhibit? 2 qx q ⎛ 2 L ⎞ q Ans. wxs ()=− (Lx ), Mx()=−−⎜ Lx x ⎟ , and Tx()=− (Lx2 ) 2S 26⎝ ⎠ 2 qL42qL qL2 qL w =+, MM==−()0 , and TT==()0 max 384D 8S max 12 max 2 2 Assume Ec << Ef and tf << tc → D = Eftfd /2 and S = Gctc. (i) Stiffness: tf = 0.34 mm (ii) Shear stress: τc = 1.0 MPa - OK! (iii) Face stress: tf = 0.34 mm (iv) Wrinkling: σf = 217 MPa → tf = 0.38 mm

Example 4.5 Seek the maximum allowed length of the beam in the figure.

q wmax = L/200 q = 0.7·10-3 N/mm

L Faces (steel): tf = 0.43 mm Ef = 210,000 MPa σcr = 128 MPa Core (PUR): tc = 70 mm Gc = 3 MPa τcr = 0.1 MPa

Ans. D = 221,235,000 Nmm2/mm, and S = 210 N/mm (i) without respect to shear deformations LqL5 4 384D I: w == →=L 3 =4950 mm max 200 384D 1000q qL 2τ d II: τ =→=L max =20, 000 mm max 2d q 2 qL 8tdf σmax III: σmax =→=L = 6636 mm 8tdf q (ii) with respect to shear deformations

4.36 BEAM THEORY

LqL5 42qL 384D ⎛ 25qL⎞ I: w == +→=L 3 ⎜1 − ⎟ max 200 384D 8S 1000q ⎝ S ⎠ which after some trial-end-error gives L = 4280 mm. II and III remains unchanged - the beam is isostatic, i.e., the allowable length decreases when transverse shear is accounted for.

Example 4.6 Seek the maximum allowed length of the beam in the figure.

q wmax = L/150 q = 1.2·10-3 N/mm

LL Faces (steel): tf = 0.43 mm Ef = 210,000 MPa σcr = 96 MPa Core (PUR): tc = 100 mm Gc = 3 MPa τcr = 0.1 MPa

Ans. D = 451,500,000 Nmm2/mm, and S = 300 N/mm (i) without respect to shear deformations LqL2 4 384D I: w == →=L 3 =7840 mm max 150 384D 300q 5qL 8τ d II: τ =→=L max =13, 300 mm max 8d 5q 2 qL 8tdf σmax III: σmax =→=L = 5240 mm 8tdf q (ii) with respect to shear deformations (the beam is hyperstatic) L 17qL442⎡ 5qL qL ⎤⎛ 312+ θ ⎞ 3qL 2 I: wwLmax ≈==−+(/)2 ⎢ ⎥⎜ ⎟ + 150 384D ⎣⎢384D 16S ⎦⎥⎝ 13+ θ ⎠ 8S numerical trial-and-error gives L = 6130 mm qL 512+ θ 8τ d 13+ θ II: τ = →=L max → L = 13,390 mm max 8d 13+ θ 5q 512+ θ 2 qL 1 813tdf σθmax()+ III: σ max = →=L → L = 5610 mm 8tdf 13+ θ q i.e., the allowable length increases when transverse shear is accounted for. Compare with the previous example.

Example 4.7 Study a simply supported sandwich beam with properties D, S, and ρ*. Sketch the lowest eigen-frequency as function of beam length L. Draw in the same sketch the relation assuming pure bending and pure shear cases.

Faces (steel): tf = 0.43 mm, Ef = 210,000 MPa, 3 ρf = 7900 kg/m Core (PUR): tc = 70 mm, Gc = 3 MPa, L 3 ρc = 30 kg/m

Ans. D = 221,235 Nm2/m, S = 210,000 N/m, and ρ* = 8.9 kg/m2 The lowest frequency is given by 2 2 D π D ωπmn = = ρπθ**L42()11+ L2 ρπθ()+ 2 π2 D π S Pure bending: ω = , and pure shear: ω = mn L2*ρ mn L ρ*

4.37 AN INTRODUCTION TO SANDWICH STRUCTURES

Some values are (in s-1) Length (m) Sandwich (s-1) Pure bending (s-1) Pure shear (s-1) 0.5 953 6224 965 1 461 1556 482 2 205 389 241 4 76 97 121 8 23 24 60

Example 4.8 The sandwich beam (1) is subjected to a uniformly distributed load q, clamped in one end and supported by the sandwich beam (2) in the other end. The beam (2) is clamped in both ends. Calculate the clamping moment in beam (1) M1. Both beams have bending stiffness D, shear stiffness S, and a length L so that θ = 0.25. q (1)

(1) (2) (2) L L/2 L/2

Ans. Denote the force acting between the beams as R. We then have qL42qL RL 3RL RL3 RL wL()1 ()=+−−, wL()2 (/)2 =+, and w(1) = w(2) 823D S D S 48D 4S 6qL()1+ 4θ 3qL R = , with θ = 0.25 one finds that R = 17 + 60θ 8 3qL2 qL2 qL2 The clamping moment is then M = − = − 8 2 8

Example 4.9 Calculate the shear deformation for the beam in the figure. Draw the transverse and bending moment diagrams clearly. P P T(x)

2L L M(x)

Hint: The change in the transverse shear deformation can be written as dw T γ t s = x − 0 c dx S d Ans. The reaction loads at the supports are found by force and bending moment equilibrium. The left support load is zero and the one at x = 2L becomes 2P. From this one can find the transverse force and bending moment relations. The shear deformation is calculated by M()x γ 0xtc Pd wxs ()=−+C , ws(0) = ws(2L) = 0 →γ 0 =− and C =0 S d 2Stc

4.38 BEAM THEORY

P P T(x)

2L L M(x)

Example 4.10 Calculate the deformation of the sandwich beam in the figure below. It has bending stiffness D, shear stiffness S and length L. The load can be written as πx q(x) = q sin 0 L

Ans. The governing differential equation for the sandwich beam can in this be written as d 4 w D d 2q D + − q = 0 dx 4 S dx 2 Assume a deflection field of the same form as the loading, i.e. πx w(x) = w sin 0 L which satisfies the boundary conditions w(0) = w(L) = w’’(0) = w’’(L) = 0. Differentiate and insert into the D.E and the result is 4 2 ⎛ π ⎞ πx D ⎛ π ⎞ πx πx D⎜ ⎟ w0 sin − q0 ⎜ ⎟ sin − q0 sin = 0 ⎝ L ⎠ L S ⎝ L ⎠ L L

This expression must be valid for any x. Solve for w0 to become 2 2 ⎛ D ⎛ π ⎞ ⎞ ⎛ D ⎛ π ⎞ ⎞ q ⎜ ⎜ ⎟ +1⎟ q ⎜ ⎜ ⎟ +1⎟ 0 ⎜ ⎟ 0 ⎜ ⎟ S ⎝ L ⎠ S ⎝ L ⎠ πx w = ⎝ ⎠ and thus w = ⎝ ⎠ sin 0 4 4 ⎛ π ⎞ ⎛ π ⎞ L D⎜ ⎟ D⎜ ⎟ ⎝ L ⎠ ⎝ L ⎠ It is now seen that the deformation due to bending and due to shear can be written as L 2 q0 ⎛ L ⎞ πx q0 ⎛ L ⎞ πx wb = ⎜ ⎟ sin and ws = ⎜ ⎟ sin , respectively D ⎝ π ⎠ L S ⎝ π ⎠ L

Example 4.11 You shall design a floor panel to an aircraft which as a first approximation can be seen as a simply supported sandwich panel. However, since the side aspect ratio is large (a >> b) one can approximately calculate the stiffness of the panel by assuming a beam of length a. The dimensioning load case in this pre-study a uniform pressure q over the entire panel. The strength of the panel is governed by inserts and load introductions and is not considered in this preliminary design study. You shall now using given materials find the combination of face and core thickness, which minimises the weight of the floor panel. Assume that the face sheets are thin and the core is weak.

q Data: a = 500 mm

Ef = 52 000 MPa (carbon/epoxy) ρ = 1 700 kg/m3 f

4.39 AN INTRODUCTION TO SANDWICH STRUCTURES

Gc = 40 MPa (Nomex honeycomb) 3 ρf = 48 kg/m Stiffness: Max deformation ≤ a/100 = 5 mm Load: q = 200 kPa

4 ⎡ 4 3 ⎤ qL ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ q 2 w(x) = wb ()x + ws (x) = ⎢⎜ ⎟ − 2⎜ ⎟ + ⎜ ⎟⎥ + ()ax − x 24D ⎣⎢⎝ a ⎠ ⎝ a ⎠ ⎝ a ⎠⎦⎥ 2S

qa 4 ⎡ 1 1 1⎤ qa 2 ⎛ 1 1 ⎞ 5qa 4 qa 2 and thus, wˆ = ⎢ − 2 + ⎥ + ⎜ − ⎟ = + 24D ⎣16 8 2⎦ 2S ⎝ 2 4 ⎠ 384D 8S

2 2 E f t f d Gc d with D = and S = ≈ Gc d 2 tc

−1 a 5qa 4 qa 2 5qa 4 ⎛ qa 2 ⎞ One can then write wˆ = ≤ + or t ≥ ⎜ wˆ − ⎟ 2 f 2 ⎜ ⎟ 100 192E f t f d 8Gc d 192E f d ⎝ 8Gc d ⎠

4 −1 5qa ρ ⎛ qa 2 ⎞ The weight W = ρ t + 2ρ t ≈ κ + 2t ρ = f ⎜ wˆ − ⎟ + ρ d c c f f f f 2 ⎜ ⎟ c 96E f d ⎝ 8Gc d ⎠

Minimise this expression. The simplest way is to sketch W as function of d and then estimate the minimum value. Every way is good!! After some trial-and-error I go the answer

d ≈ 75 mm which then gives that tf = 0.38 mm.

Example 4.12 Calculate the first eigen-frequency approximately for a clamped sandwich beam as shown in the figure. Use Ritz' method with the deflection assumption given below. Length L = 1m Face thickness tf = 0.001m Core thickness tc = 0.025m 11 Face modulus Ef = 2×10 6 Core shear modulus Gc = 30×10 3 Face sheet density ρf = 8700 kg/m 3 Core density ρc = 100 kg/m πx Use w(x,t) = ()W +W sin 2 eiωt b s L Hint: a a πx a πx a sin 2 dx = , cos 2 dx = , ∫ a 2 ∫ a 2 0 0 a a a πx 3a πx 3a πx πx a sin 4 dx = , cos4 dx = , cos2 sin 2 dx = ∫ a 8 ∫ a 8 ∫ a a 8 0 0 0 Ans. By using πx w(x,t) = ()W + W sin 2 eiωt b s L we can write the kinetic energy as

L 2 L * 2 1 * ⎛ ∂w ⎞ 1 * 2 2 3Lρ ω 2 U k = ∫ ρ ⎜ ⎟ dx = − ∫ ρ ω w dx = − ()Wb + Ws 2 0 ⎝ ∂t ⎠ 2 0 16 The bending and shear strain energy is then also equal to

4.40 BEAM THEORY

2 2 1 LL⎛ ∂ 2 w ⎞ 1 ⎛ ∂w ⎞ DW 2π 4 SW 2π 2 U = D⎜ b ⎟ dx + S⎜ s ⎟ dx = b + s s 2 ∫∫⎜ 2 ⎟ 2 ∂x 3 L 00⎝ ∂x ⎠ ⎝ ⎠ L The total energy is then

2 4 2 2 * 2 DWb π SWs π 3Lρ ω 2 Π = + − ()Wb + Ws L3 L 16

Differentiation with respect to Wb and Ws yields

∂Π 2DW π 4 3Lρ *ω 2 = b − W + W = 0 3 ()b s ∂Wb L 8 and

2 * 2 ∂Π 2SWsπ 3Lρ ω = − ()Wb + Ws = 0 ∂Ws L 8 Solving for ω2 gives that

D ω = 4π 2 ⎛ 4π 2 D ⎞ 3ρ *L4 ⎜1+ ⎟ ⎜ 2 ⎟ ⎝ L S ⎠

Example 4.13

Your assignment is to extract the elastic modulus (Ef) of the laminate and the core shear modulus (Gc) for a sandwich panel. You have access to one small test piece in the form a beam and by testing this in the so called three-point bending set-up, with two different lengths (L1 = 800 mm och L2 = 600 mm), and measure the deflection under the load point using a dial gauge. Derive the necessary equations and calculate Ef and Gc. P

You know tf = 1.8 mm tc = 40 mm

The test results are for a load of 25 N/mm; Test 1: L = L1 = 800 mm → w = 13.43 mm Test21: L = L2 = 600 mm → w = 6.43 mm

You may assume tf << tc, Ec << Ef

Ans. P P

L L 1 2 3 3 PL1 PL1 PL1 PL2 PL2 PL2 w1 = + → S = w2 = + → S = 48D 4S ⎛ PL3 ⎞ 48D 4S ⎛ PL3 ⎞ 4⎜ w − 1 ⎟ 4⎜ w − 2 ⎟ ⎜ 1 ⎟ ⎜ 2 ⎟ ⎝ 48D ⎠ ⎝ 48D ⎠

4.41 AN INTRODUCTION TO SANDWICH STRUCTURES

This combined gives ⎛ 3 ⎞ ⎛ 3 ⎞ PL1 PL2 ⎜ PL1 ⎟ ⎜ PL2 ⎟ = or L2 w1 − = L1 w2 − ⎛ PL3 ⎞ ⎛ PL3 ⎞ ⎜ 48D ⎟ ⎜ 48D ⎟ 4⎜ w − 1 ⎟ 4⎜w − 2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎜ 1 ⎟ ⎜ 2 ⎟ ⎝ 48D ⎠ ⎝ 48D ⎠ Solve for D P P L L3 − L L3 2D L w − L w = L L3 − L L3 → D = 1 2 2 1 → E = 2 1 1 2 ()1 2 2 1 f 2 48D 48 L2 w1 − L1w2 t f d From above we can then derive

PL1 Stc S = → Gc = ⎛ PL3 ⎞ d 2 4⎜ w − 1 ⎟ ⎜ 1 ⎟ ⎝ 48D ⎠ With numbers, one arrives at Ef = 15.3 GPa and Gc = 48.8 MPa

Example 4.14 The so-called four-point bending (FPB) test is commonly used for sandwich beams. The reason for this is that several different properties can be measured in the same test and that the test easily can be modified to obtain different failure modes. The figure below shows schematically a typical testing set-up. What failure modes can plausibly be obtained in the test and where would they if so appear? Draw schematic figures of how the failure modes look and where they occur, use equations if necessary to explain. Even the bending stiffness D can be measured directly in the test by applying dial gauge. How can this be done?. If one type of failure mode occurs, how can the test be re-designed so another failure mode occurs? Load cell

Actuator L 2 Ans. The feature of the four-point bending test is that in the middle section the bending moment is constant (and has a maximum) and the shear force is zero. In this section one can assume failure occur only in the face sheets; either due to face tensile or compressive failure, or due to wrinkling. In the section between the inner and outer supports, the shear force is constant and here the specimen could exhibit core shear failure. If the load pads are small and/or the face sheets are thin, the specimen could also fail by local face indentation, under any of the load supports.

Since the bending moment is constant between the inner supports, and the shear force is zero, this section is subjected to pure bending. If one measures the deflection in for example the middle of the beam compared to the deflection of the inner supports, the bending stiffness could be derived as: Denote the relative displacement with δ (difference in displacement between the middle and the inner support). The deformation of a beam subjected to edge bending moments M is ML2 PL2 (L − L ) P(L − L ) δ = 1 = 1 2 1 since M = 2 1 8D 16D 2 where P is the support load. The measured value of δ thus gives the value of D. If for example, shear failure occurs on the specimen, one can move the inner supports inwards, thus increasing the bending moment for the same applied load P, thus moving closer to a face failure mode, and vice versa.

4.42 BEAM THEORY

Example 4.15 A cantilever sandwich beam is subjected to a hydrostatic pressure according to the figure below. Derive an expression for the shear deformation.

q(x) q(x) = qmaxx/L if x is defined from left to right.

The beam shear stiffness is S.

LLdw Tx() LqL⎛ x2 ⎞ qL2 Alternative 1: w ==s dx dx =max⎜1 −⎟dx ==..... max s ∫∫dx S ∫2S L2 3S 00 0 ⎝ ⎠ 3 532 qLmax ⎡⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎤ Alternative 2: wx()= ⎜ ⎟ −10⎜ ⎟ + 20⎜ ⎟ b ⎢⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎥ 120D ⎣⎢ L L L ⎦⎥ 3 2 2 dwbsS dw qLxmax ⎡ x ⎤ qLmax 3 =− gives wxs()=−⎢1 2 ⎥ → wLs ()= dx D dx 2S ⎣ 3L ⎦ 3S

4.43 CHAPTER 5

BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

In this section, formulae for the calculation of instability loads for the overall buckling mode will be derived. It follows much the same way as when deriving the ordinary Euler buckling load. The necessary extension here is to account for transverse shear deformations which will have the same influence on the behaviour for in-plane compressive loads as in bending since in both cases the behaviour depends on the stiffness of the beam.

5.1 Governing Equations In order to assess the motion of sandwich beams the governing equations must be derived assuming inertia forces and hence time is introduced as a variable. The following section will follow the work by Timoshenko [1] apart from that here a more general cross-section is considered. Timoshenko uses a form of partial deflections that is an agreement with what is used here, but another notation is used. In [1] ψ and β is used for ∂wb /∂x and ∂ws /∂x, respectively. Let us again study a differential element as in Fig.4.6 but now include a normal forces N. q dx

Nx dN x N + dx 2 x dx dw dw d w dM + dx M + x dx dx dx dx2 x dx dw dN dx x dTx N x + dx Tx + dx dx dx dx dx dw d 2w + dx dx dx2 Figure 5.1 Distorted beam element and projection of normal load Nx Then once again derive the equations of equilibrium. First define the two inertia forces: (i) Vertical inertia The body force acting on the element when subjected to an acceleration ∂2w/∂t2 is

5.1 AN INTRODUCTION TO SANDWICH STRUCTURES

∂ 2 w ∂ 2 w − ρ dz = −ρ * , with ρρ* = dz (5.1) ∫ ∂t 2 ∂t 2 ∫ where ρ* is the surface mass (mass per unit length per unit width) of the beam, which reduces to ρh for a homogeneous cross-section [1]. This mass may of course be a function of x if the beam has a varying cross-section. Hence, an acceleration in positive w-direction (downwards in Fig.5.1) creates a body force with opposite direction (upwards).

(ii) Rotary inertia

The cross-section rotates when bending, and this rotation equals ∂wb/∂x (see Fig.4.4). If one assumes the faces to be thin which implies that the shear strain is constant over the cross- section, we can write the x-displacement of a particle as +u

MR dw x b dx z − u

Figure 5.2 Definition of rotary inertia

∂w ∂θ ∂ 2 w u = zθ = −z b , since ε = z x = −z b x ∂x x ∂x ∂x 2 An angular acceleration creates an inertia bending moment in the opposite direction with magnitude

∂ 2u ∂ 2θ ∂ 3w ∂ 3w − ρz dz = − ρz 2 x dz = ρz 2 b dz =R b (5.2) ∫ ∂t 2 ∫ ∂t 2 ∫ ∂x∂t 2 ∂x∂t 2 and defined positive in the same direction as Mx. R is the rotary inertia, which reduces to ρI for a homogeneous cross-section [1].

The normal load Nx will affect the transverse load equilibrium. In the right of Fig.5.1 one can see that Nx will act in different planes once the beam will have a deflection. The resulting load in the z-direction due to the normal load is thus

dw ⎛ dN ⎞⎛ dw d 2 w ⎞ d 2 w dN dw dN d 2 w − N dx + N + x dx ⎜ + dx⎟ = N dx + x dx + x dx dx x ⎜ x ⎟⎜ 2 ⎟ x 2 2 dx ⎝ dx ⎠⎝ dx dx ⎠ dx dx dx dx dx

Neglecting terms of (dx)2 and realising that a normal load only will be considered providing it is applied and used for buckling analysis, we must have that dNx/dx must zero. Thus, only the first term in the above relation will give any contribution to the transverse load equilibrium equation.

Now, the equations of equilibrium take the following form ∂T ∂2w ∂2w Vertical: x ++qN −ρ* =0 ∂x x ∂x 2 ∂t 2

5.2 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

∂ 3 w ∂M Moment : − T + R b + x = 0 x ∂x∂t 2 ∂x This gives the two equations

∂ 2 w ∂ 2 w ∂ 2 w ρ * − q − N − S s = 0 (5.3a) ∂t 2 x ∂x 2 ∂x 2

∂ 3 w ∂w ∂ 3 w and D b + S s − R b = 0 (5.3b) ∂x 3 ∂x ∂x∂t 2 where the latter equation now constitutes the new relation between wb and ws to be used instead of eq.(4.22). The same set of governing equations can be found in the book by Reddy [2] but then written in terms of the displacement w and the rotation θ. By differentiation of eq.(5.3b) and combining that with eq.(5.3a) one can eliminate the partial deflections by

∂ 4 w ∂ 2 w ∂ 4 w ∂ 2 w ∂ 2 w ∂ 2 w D b + S s − R b = 0 and −=+−S s qN ρ * ∂x 4 ∂x 2 ∂x 2∂t 2 ∂x 2 x ∂x 2 ∂t 2

∂ 4 w ∂ 4 w ∂ 4 w ∂ 4 w R ∂ 2 ⎡ ∂ 2 w ∂ 2 w⎤ b - s * R 22==++−R 22 R 22 R 22 2 ⎢qNx 2 ρ 2 ⎥ ∂∂xt ∂∂xt ∂∂xt ∂∂xt St∂ ⎣ ∂x ∂t ⎦

4 4 4 4 2 2 2 ∂ wbs∂ w ∂ w ∂ w D ∂ ⎡ ∂ w * ∂ w⎤ and D 4 =−D 4 D 4 =+D 4 2 ⎢qN +x 2 −ρ 2 ⎥ ∂x ∂x ∂x ∂x Sx∂ ⎣ ∂x ∂t ⎦ gives the governing equation as

4 2 2 2 2 4 ∂ w ⎛ D ∂ R ∂ ⎞ ⎡ ∂ w * ∂ w⎤ ∂ w D 4 +−−⎜ 2 102 ⎟ ⎢qN+−x 2 ρ 2 ⎥ −=R 22 (5.4) ∂x ⎝ Sx∂ St∂ ⎠ ⎣ ∂x ∂t ⎦ ∂∂xt which is the so-called Timoshenko beam equation [1]. It is seen in the above expression that the first two terms represent the ordinary differential equation for a beam when neglecting the transverse shear deformation, the third term is the shear deformation which together with the first two terms equals eq.(4.20), the fourth term is a combination of shear and rotary inertia and lastly the fifth term represents the rotary inertia. In the case of free undamped vibration, the applied loads q and Nx are zero and eq.(5.3) will take the form ∂ 4 w ∂ 2 w ρ * ⎛ ∂ 4 w ∂ 4 w⎞ ∂ 4 w D +−ρ * ⎜ D −R ⎟ − R = 0 (5.5) ∂x 4 ∂tS2 ⎝ ∂∂xt22 ∂t 4 ⎠ ∂∂xt22

The cross-sectional properties ρ* and R are easily calculated for a sandwich as

* ρ =++ρ11tttρ 2 2 ρcc (5.6) where ρ is the density of the material component. Hence, ρ* can also be interpreted as the cross-section surface weight. The way to compute R is exactly the same as calculating the flexural rigidity D but with ρ substituted for E. Hence, from eq.(5.2) (see also eq.(3.21))

5.3 AN INTRODUCTION TO SANDWICH STRUCTURES

3 3 3 2 ρ1t1 ρ 2t2 ρctc 2 2 ⎛ tc + t2 ⎞ R = + + + ρ1t1 ()d − e + ρ 2t2e + ρctc ⎜ − e⎟ (5.7) 12 12 12 ⎝ 2 ⎠ Note that the contribution from the core (term 3 and 6) in this expression may be significant since the density ratio is usually lower than the modulus ratio. One important observation to make is that all units must be consistent, e.g., SI-units, or the coefficients in eq.(5.3) will have different dimensions. A suggestion is to use kilograms, metres, seconds and Newtons only, so that all terms in eq.(5.3) are in the dimension kg/ms2.

In the case of thick faces, the static beam bending equation of eq.(4.24) will take a slightly different form when including Nx and transverse inertia (but excluding rotary inertia)

d 6 w DS d 4 w ⎛ d 2 S ⎞⎛ ∂ 2 w ∂ 2 w ⎞ 2D − = ⎜ − ⎟⎜q + N − ρ * ⎟q (5.8) f 6 4 ⎜ 2 ⎟⎜ x 2 2 ⎟ dx D0 dx ⎝ dx D0 ⎠⎝ ∂x ∂t ⎠

In order to be able to solve beam buckling and free vibration problems using energy minimisation methods (Ritz method) we must have a complete set of energy relations to complement those derived in section 4.7 (eqs.(4.31) and (4.32)).

To assess the potential energy of the in-plane force, Nx,we do the following: Apply some lateral load producing bending of the beam and causing some deformation field w, as illustrated first in Fig.5.3. Assume that a normal force Nx is applied first, i.e., before any lateral loads, and that it remains constant during bending. The force induces normal strains in the beam and these are constant over the thickness. The work done by the constant load Nx can then be written as

L ε x L U = N dε dx = N ε dx = N δ p,N ∫∫ x x ∫ x x x x 00 0 since Nx is constant during the deformation. δx is the decrease in the projected length of the beam due to a lateral deflection w, as schematically illustrated in Fig.5.3. This will now equal the potential energy of this applied membrane load. In order to find this length δx study Fig.5.3. The length of the beam before bending is L. Since the strains produced by Nx are assumed to be small, the length of the beam remains constant. However, due to the lateral bending, the projected length L’ is smaller than L. By geometry we have that

2 ⎡ 1 ⎛ ∂w ⎞ ⎤ dL = dL'2 +dw2 and since ∂w/∂L is small dL ≈ dL'⎢1+ ⎜ ⎟ ⎥ ⎣⎢ 2 ⎝ ∂x ⎠ ⎦⎥

5.4 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

Nx Nx w w+dw

z dL δ L' x

Figure 5.3 Deformation of a beam element. The change in projected length of the beam element is then

2 1 ⎛ ∂w ⎞ dδ x = dL − dL'= ⎜ ⎟ 2 ⎝ ∂x ⎠

The potential energy of Nxdy is then hence

2 L ⎡1 ⎛ ∂w ⎞ ⎤ U = N ⎢ ⎜ ⎟ ⎥dx (5.9) N x ∫ 2 ∂x 0 ⎣⎢ ⎝ ⎠ ⎦⎥

The large deflection beam strain can now be written as

2 2 ∂u 1 ⎛ ∂w⎞ 1 ⎛ ∂w⎞ ε =+⎜ ⎟ =+εκz +⎜ ⎟ (5.10) xxx∂x 2 ⎝ ∂x ⎠ 0 2 ⎝ ∂x ⎠

The kinetic energy in the beam per unit volume is simply

2 2 ρ ⎡⎛ ∂u⎞ ⎛ ∂w⎞ ⎤ ⎢⎜ ⎟ + ⎜ ⎟ ⎥ 2 ⎣⎝ ∂t ⎠ ⎝ ∂t ⎠ ⎦ and by integrating this over the thickness we find the total kinetic energy to be

2 L ⎡ 2 2 ⎤ ⎛ ⎞ * 1 ⎢ ⎜ ∂ wb ⎟ ⎛ ∂w ⎞ ⎥ U k = R + ρ ⎜ ⎟ dx (5.11) 2 ∫ ⎢ ⎜ ∂x∂t ⎟ ⎝ ∂t ⎠ ⎥ 0 ⎣⎢ ⎝ ⎠ ⎦⎥

The total potential energy, which is a functional, may be written as

Π = Use + Uq + UN + Uk (5.12)

5.2 Boundary Conditions for Sandwich Beams The boundary conditions for sandwich beams become less obvious than for ordinary beams, due to the inclusion of shear deformations. In the following sections, all boundary conditions will again be revised and clarified.

The kinematic boundary conditions, however, are still trivial and are simply

5.5 AN INTRODUCTION TO SANDWICH STRUCTURES

Zero deflection: (for simply supported and clamped edges) w = 0 Zero rotation: (for clamped edges)

∂w θ = − b = 0 x ∂x This condition may be rewritten for cases of buckling and free vibration. This will be done explicitly when we solve buckling problems in section 5.4. In buckling, the boundary condition must thus be

dw dw dw dw T b = − s = − x = 0 dx dx dx dx S so that the transverse force Tx should be specified at the edge. This is more clearly shown below. In free vibration, to get the rotations, we use eq.(5.3a), now in the form

2 2 2 2 2 2 ∂ ws * ∂ w S ∂ ws S ⎡∂ w ∂ wb ⎤ * ∂ w S 2 = ρ 2 or * 2 = * ⎢ 2 − 2 ⎥ = ρ 2 ∂x ∂t ρ ∂x ρ ⎣ ∂x ∂x ⎦ ∂t from which it follows that

S ∂ 2 w ∂ 2 w S ∂ 2 w b = −ρ * + ρ * ∂x 2 ∂t 2 ρ * ∂x 2

The solution to the governing differential equation, expressed in terms of w, must then be integrated to obtain an expression for the rotation, which is set to zero at the boundary.

Zero bending moment: (for free and simply supported edges) We have that

∂ 2 w ∂ 2 w ∂ 2 w M = −D b = −D + D s = 0 x ∂x 2 ∂x 2 ∂x 2 But, from eq.(5.3a) we can rewrite

∂ 2 w ρ * ∂ 2 w N ∂ 2 w s = − x ∂x 2 S ∂t 2 S ∂x 2 and thus

⎛ ∂ 2 w ρ * ∂ 2 w N ∂ 2 w ⎞ M = −D⎜ − + x ⎟ = 0 x ⎜ 2 2 2 ⎟ ⎝ ∂x S ∂t S ∂x ⎠

Thus, for buckling cases, the boundary condition becomes

∂ 2 w ⎛ P ⎞ ∂ 2 w ⎜1− ⎟ = 0, i.e. either = 0 or P = S ∂x 2 ⎝ S ⎠ ∂x 2

5.6 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

In free vibration calculations, the boundary condition becomes

∂ 2 w ρ * ∂ 2 w − = 0 ∂x 2 S ∂t 2 Zero transverse force: (for free edges) Also here the boundary conditions are rather complicated to obtain. Let us first consider the buckling case. A free edge still has the edge load applied. For this edge to be in equilibrium we must have that dw T + N = 0 x x dx which can be rewritten to

3 2 d wb dw dM x d wb D + P = 0 through T = , M = −D and Nx = −P dx3 dx x dx x dx 2 but using that

d 3 w d 3 w d 3 w d 3 w 1 d 2T d 3 w P d 3 w d 3 w ⎛ P ⎞ b = − s = − x = − = ⎜1− ⎟ dx 3 dx 3 dx 3 dx 3 S dx 2 dx 3 S dx 3 dx 3 ⎝ S ⎠ we see that

⎛ P ⎞ d 3 w dw D⎜1− ⎟ + P = 0 ⎝ S ⎠ dx 3 dx which constitutes the correct boundary condition for a free edge under a normal compressive load P.

For the free vibration case, the condition must be sought after in a slightly different manner. Evidently we must now have a zero shear force Tx at the boundary (since Nx now is zero). We know that

∂w T = S s = 0 x ∂x which may rewritten to

∂w ∂ 3 w ∂ 3 w ⎛ ∂ 3 w ∂ 3 w ⎞ ∂ 2 ⎛ ∂w ∂w ⎞ S s = −D b + R b = −D⎜ − s ⎟ + R − s 3 2 ⎜ 3 3 ⎟ 2 ⎜ ⎟ ∂x ∂x ∂x∂t ⎝ ∂x ∂x ⎠ ∂t ⎝ ∂x ∂x ⎠ but from above ∂ws/∂x = 0, and from eq.(5.3a) we get that

∂3w ρ * ∂3w s = ∂x3 S ∂x∂t 2 Inserting this into the above yields that

5.7 AN INTRODUCTION TO SANDWICH STRUCTURES

∂ 3 w Dρ * ∂ 3 w ∂ 3 w D − − R = 0 ∂x3 S ∂x∂t 2 ∂x∂t 2 which constitutes the free edge boundary condition.

5.3 Buckling of Simply Supported Column – Simple Solution

x y x z P P L w

Figure 5.4 A simply supported sandwich beam subjected to in-plane compressive edge loads. Consider first a simply supported beam, as shown in Fig.5.4, and assume that bending and 2 2 shear deformation can be superimposed. It was previously shown that d wb /dx = –Mx/D, and 2 2 d ws /dx = 1/S dTx /dx. By using the partial deflections, w = wb + ws, the total curvature can be written as (see eq.(4.16)) dw2 dw2 dw2 M 1 dT =+=−+bs x x dx2 dx2 dx2 D S dx

According to Fig.5.4, Mx = Pw and hence Tx = P dw/dx. Rearranging eq.(5.8) then gives dw2 P Pw dw2 P S ⎜⎛10− ⎟⎞ += or +aw22 = 0 , with a =⎜⎛ ⎟⎞ (5.13) dx 2 ⎝ S ⎠ D dx 2 D ⎝ SP− ⎠

This equation can also be derived from eq.(5.4) with Nx = −P, q = 0. One solution is

w =+C12sinax C cosax (5.14)

Using the boundary conditions w(0) = w(L) = 0 gives that C2 = 0 and sin aL = 0. Taking the latter non-trivial solution, which also constitutes the stability criterion, gives

nππn 22 P S aL=⇒=⇒ nπ a =⎜⎛ ⎟⎞ L L2 D ⎝ SP− ⎠

Rearranging the expression, the critical buckling load P is solved as

5.8 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

2 2 n 2π 2 D n π D 2 (βL) 2 n 2 P P = L = = E with β = 1 (5.15a) cr n 2π 2 D n 2π 2 D 1+ n 2π 2φ / β 2 1+ 1+ L2 S (βL) 2 S

2 where φ again is the shear factor (φ = D/L S) and PE is the first Euler buckling load, i.e. the load for the case pure bending. Examine the validity of this expression; if S or the length of the beam approaches infinity the shear deformation can be neglected and eq.(5.15a) should equal the ordinary Euler load. By getting the limits of eq.(5.15a) this becomes obvious nD22π lim Pcr==P b (5.15b) SL→∞ and/or →∞ L2 where Pb is the ordinary pure bending Euler buckling load. If, on the other hand, the beam is very short or is weak in shear the limit will be

lim Pcr==S P S (5.15c) SL→→00 and/or which commonly is called the shear buckling load. The difference between taking account of shear deformation and not, as in ordinary Euler buckling, is that now there exists a finite buckling load even for short beams. For long and slender beams, the load will approach the Euler load. For the simple case of both ends being simply supported, eq.(5.15a) can hence be rewritten in the simplest possible form 111 =+ (5.16) PPPcr b s

5.4 Rigorous Solution to Beam Buckling

5.4.1 Clamped edges To get a more general solution to the buckling problem next consider a beam with both ends fixed. In order to include both symmetrical and anti-symmetrical buckling modes, introduce clamping moments and shear forces as illustrated in Fig.5.5. Let's here assume a coordinate system which originates from the centre of the beam so that we may later utilise symmetries in obtained equation system. x M M M 1 T1 1 T(x) 1 w w P P P T1 T1 M(x) L/2-x L

Figure 5.5 Schematic of a clamped beam in compression (note that x now starts in the middle of the beam). The shear force and bending moment at position x of the beam is readily found

⎛ L ⎞ dM x dw M=+− M Pw T ⎜ −x⎟ and T ==+P T x 11⎝ 2 ⎠ x dx dx 1

5.9 AN INTRODUCTION TO SANDWICH STRUCTURES

By rearranging and using the definition of a as in eq.(5.12) the following is obtained

dw2 Tx TL M +=−+aw22 a 1 a2 1 −a2 1 dx2 P 2P P

By direct integration of eq.(5.4) with Nx = −P and neglecting all time dependent terms, one arrives at the same equation dw2 +=+a2 w Ax B dx 2 In both cases the solution can be written as

w =+C12sinax C cosax +C 34x + C (5.17)

Substitution of w into the differential equation gives that C3 = –T1 /P and C4 = T1L/2P – M1 /P. Then the boundary condition can be used at the ends of the column.

There are two different ways of assessing the fixed end condition, either by assuming that the faces act as membranes which means that there may be a slope dws /dx at the end, or assuming that the entire slope must equal zero. The clamped boundary condition must now be given some special attention. The slope is not necessarily zero since we must allow for a slope due to shear deformation. The condition must thus be

dwb dw dws dw Tx dw 1 ⎛ dw ⎞ dw T1 = − = − = − ⎜ P + T1 ⎟ = 0 → = dx dx dx dx S dx S ⎝ dx ⎠ dx S − P

The boundary conditions are

dw dw T − C P w(−L/2) = w(L/2) =0, and = = 1 = 3 dx x=−L / 2 dx x=L / 2 S − P S − P

These four conditions give the following four equations 1 1 1 −+Csin aL C cos aL −+= C L C 0 (5.18a) 122 2 2 34

1 1 T aCcos aL++= aC sin aL C 1 (5.18b) 1232 2 ()SP−

1 1 1 Csin aL+++= C cos aL C L C 0 (5.18c) 122 2 2 34

1 1 T aCcos aL−+= aC sin aL C 1 (5.18d) 1232 2 ()SP−

There are two ways to solve this equation system; either one can write it as a matrix system and find the stability criterion by letting the determinant of the system matrix equal zero, or we can solve this explicitly by using symmetries in the equations. Since one often is referred to the former way, let's here proceed with the latter.

5.10 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

By adding and subtracting equations (5.18) the relations between the unknowns can be found as 21 1 C =−CaLCCaLsin, =− cos (5.19a,b) 31L 2 422

⎛ P⎞ 1 1 CaC=−⎜1 − ⎟ cosaL, 2 aC sin aL = 0 (5.19c,d) 31⎝ S ⎠ 2 22

(i) We can now formulate different solutions to this system of equations. One such is taking the non-trivial solution C2 ≠ 0 which leads to that sin aL/2 must be zero. This in turn also means that C4 ≠ 0 and that C1 = C3 = 0, from eq.(5.19b and c). sin aL/2 = 0 has the solution aL = 2nπ and by inserting this into the definition of a in eq.(5.13) we can solve for P, which then becomes

2 2 4n 2π 2 D n π D 2 (βL) 2 n 2 P P = L = = E with β = 0.5 (5.20) cr 4n 2π 2 D n 2π 2 D 1+ n 2π 2φ / β 2 1+ 1+ L2 S (βL) 2 S

2 2 Hence, eq.(5.16) holds even for this case but with PE = 4π D/L . From eq.(5.17b) it follows that C3 = 0, C1 = 0 (unless P = S) and that C4 = C2 for odd values of n and C4 = –C2 for even values of n. Hence, the buckling mode is

w(x) = C2(cos ax + 1) if n is odd and w(x) = C2(cos ax – 1) if n is even (5.21) This case is therefore symmetrical.

(ii) The other solution to the set of equation in (5.19d) is to take C2 = 0. From this it follows that C4 = 0 and the stability criterion is found from the relation between C1 and C3 in eq.(5.19c) as 1 1 P tan aL=− aL⎜⎛1 ⎟⎞ (5.22) 2 2 ⎝ S ⎠ and the corresponding buckling mode as 21x wC=−⎜⎛sin ax sin aL⎟⎞ (5.23) 1 ⎝ L 2 ⎠ and this is apparently the anti-symmetrical case. It can be shown for the n:th solution of eq.(5.22) that 2nπ < (aL)n < 2(n+1)π, meaning that the n:th anti-symmetrical buckling load is larger that the n:th, but smaller than the (n+1):th symmetrical buckling load. The stability criterion in eq.(5.22) may be rewritten using eq.(5.13) to

⎛ aD2 ⎞ 1 1 ⎜1 + ⎟ tanaL= aL (5.24) ⎝ S ⎠ 2 2

5.11 AN INTRODUCTION TO SANDWICH STRUCTURES

which then must solved for a number of roots an to find the value of anL/2 which gives the critical buckling load for anti-symmetrical buckling as

2 ⎛ aLn ⎞ 4 D ⎜ ⎟ ⎝ 2 ⎠ L2 Pcr = 2 (5.25) ⎛ aLn ⎞ 4 D 1 + ⎜ ⎟ ⎝ 2 ⎠ LS2

By using the second boundary condition, i.e., w = dw/dx = 0 at L/2 and –L/2, the non-zero right hand sides of eqs.(5.18b and d) will vanish but the symmetrical buckling load remains as given in eq.(5.20). However, the stability criterion of the anti-symmetrical case in eq.(5.24) changes to 1 aL tan aL = 22

This case, which is equivalent with a pure bending case, has the solutions for anL/2 equal to 4.493, 7.7253 and 10.9041 as the first three roots.

(iii) As mentioned previously, there is one more solution to the governing equations and that is for P = S. This provides a limiting value for the buckling which cannot be exceeded and as seen in eqs.(5.20) and (5.25) this value is obtained when an becomes large. Thus, all buckling modes, symmetric or unsymmetric, have the asymptotic value P = S when L and/or D becomes large. This buckling mode is often referred to as shear buckling or shear crimping. For very short columns, on the other hand, the effect of the stiffness of the face sheets must be accounted for and this covered later in this text.

Now, one may discuss the consequence of the derived formulae. The shear force in a cross- section x is equal to P dw/dx + T1, where T1 is the shear force at the end of the column. If the bending stiffness is assumed infinitely large it follows from eq.(5.3) that S d2w/dx2 = P 2 2 d w/dx or that P = S. Thus, the shear buckling load Ps is invariable with the boundary condition. Eq.(5.16) can be rearranged as 11111 =+≥+2 (5.26) PaDSPPcr b s

Comparing eq.(5.26) with eq.(5.16), where Pb is the buckling load of an “ordinary” column 2 with infinite shear rigidity, shows that eq.(5.16) holds for those cases where a = Pb/D. In general this will be true when a is independent of the shear stiffness S. This will be the case for all symmetrical buckling modes.

There is an alternative way to asses the governing equation for this case using the equilibrium equations. Eq.(5.3a) can be written as (using that Nx = −P)

∂ 2 w ∂ 2 w ∂ 2 w ⎛ ∂ 2 w ∂ 2 w ⎞ ∂ 2 w ∂ 2 w − P + S s = −P + S ⎜ − b ⎟ = 0 or S b − S − P = 0 2 2 2 ⎜ 2 2 ⎟ 2 ()2 ∂x ∂x ∂x ⎝ ∂x ∂x ⎠ ∂x ∂x which may be integrated to

5.12 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

∂w ∂w S b − ()S − P = A ∂x ∂x Eq.(5.3b) can be rewritten to

∂ 3 w ∂w ∂ 3 w ∂w ∂w D b + S s = D b + S − S b = 0 ∂x3 ∂x ∂x3 ∂x ∂x which combined gives that

∂ 3 w ∂w ∂w D b + S − ()S − P − A = 0 ∂x3 ∂x ∂x

∂ 3 w ∂w ∂ 2 w D b + P = A → D b + P w = Ax + B ∂x3 ∂x ∂x 2

The partial deflection wb can now be removed using the above as

∂ 2 w ∂w2 D ∂w2 S b = ()S − P → ()S − P + Pw = Ax + B ∂x 2 ∂x 2 S ∂x 2

∂w2 PS S or + w = (Ax + B) ∂x 2 D(S − P) D(S − P) which is the same as

∂w2 a 2 + a 2 w = (Ax + B) ∂x 2 P with a defined as in eq.(5.13). By inserting the solution of eq.(5.17) (which also must hold now) we get that the integration constants must be

A = C3P and B = C4P We can further from the above reasoning deduct the boundary conditions; A zero bending moment (simply supported edge) implies that

2 2 2 ∂ wb D ∂w ⎛ P ⎞ ∂w M x = −D = − ()S − P = −D⎜1− ⎟ = 0 ∂x 2 S ∂x 2 ⎝ S ⎠ ∂x 2 which means that

∂w2 = 0 unless P = S ∂x 2 A zero rotation (clamped edge) is found through

∂w ∂w ∂w ∂w C P S b = ()S − P + A = ()S − P + C P = 0 or + 3 = 0 ∂x ∂x ∂x 3 ∂x S − P

5.13 AN INTRODUCTION TO SANDWICH STRUCTURES

These results are exactly the same as derived previously. It may be worth noticing here that if we choose the coordinate system to start at the left edge instead we obtain another equation system, which will appear as

⎡ 0 1 0 1⎤ ⎢ ⎥⎛ C1 ⎞ ⎢ sin aL cos aL L 1⎥⎜ ⎟ ⎛ P ⎞ ⎜C2 ⎟ ⎢ ⎜1− ⎟a 0 1 0⎥ = 0 ⎢ ⎝ S ⎠ ⎥⎜C ⎟ ⎜ 3 ⎟ ⎢⎛ P ⎞ ⎛ P ⎞ ⎥⎜ ⎟ ⎢⎜1− ⎟a cos aL − ⎜1− ⎟asin aL 1 0⎥⎝C2 ⎠ ⎣⎝ S ⎠ ⎝ S ⎠ ⎦

The determinant to the coefficient matrix can be found as P − S aL (1− cos aL) + sin aL = 0 S 2

⎛ a 2 D ⎞ S a 2 D or 2(cos aL −1)⎜1+ ⎟ + aLsin aL = 0 since = 1+ from eq.(5.13) ⎝ S ⎠ S − P S which may be rewritten to the same criterion as in eq.(5.24). This is the solution given in [2].

5.4.1 Simply supported edges The simply supported case, see Fig.5.6, discussed first in this chapter, may be solved similarly, giving the same solution when applying the boundary conditions of zero bending moments at the edges of the beam. The solution to the problem may still obtained using eq.(5.17) although there are no clamping moments at the edges. And again, the values of the integration constants C are of no interest when obtaining the critical buckling load.

At the edges the boundary condition are obtained by applying

2 d wb Mx(0) = Mx(L) = 0 or D = 0 dx 2 x=0,L

d 2 w d 2 w d 2 w d 2 w 1 dT d 2 w P d 2 w d 2 w ⎛ P ⎞ but b = − s = − x = − = ⎜1− ⎟ = 0 at x = 0,L dx 2 dx 2 dx 2 dx 2 S dx dx 2 S dx 2 dx 2 ⎝ S ⎠ unless P = S, which then also provides us with a solution. We can also see this from eq.(5.3a) which when rewritten becomes

∂ 2 w ∂ 2 w ∂ 2 w ⎛ ∂ 2 w ∂ 2 w ⎞ N + S s = N + S⎜ − b ⎟ = 0 x 2 2 x 2 ⎜ 2 2 ⎟ ∂x ∂x ∂x ⎝ ∂x ∂x ⎠

∂ 2 w S + N ∂ 2 w ∂ 2 w P − S ∂ 2 w P − S ∂ 2 w or b = x or M = −D b = D = D = 0 at x = 0,L ∂x 2 S ∂x 2 x ∂x 2 S ∂x 2 S ∂x 2 The boundary conditions are in this case are then specified as

5.14 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

d 2 w d 2 w w(−L/2) = w(L/2) =0, and 2 = 2 = 0 dx x=−L / 2 dx x=L / 2

x T P 1 P w T1

Figure 5.6 Schematic of a simply supported beam in compression (note that x now starts in the middle of the beam). This gives the following four equations 1 1 1 −+Csin aL C cos aL −+= C L C 0 (5.27a) 122 2 2 34

1 1 a 2C sin aL − a 2C cos aL = 0 (5.27b) 1 2 2 2

1 1 1 C sin aL + C cos aL + C L + C = 0 (5.27c) 1 2 2 2 2 3 4

1 1 − a 2C sin aL − a 2C cos aL = 0 (5.27d) 1 2 2 2 By adding and subtracting these equations the relations between the unknowns are found as 2 1 1 C = − C sin aL, C = −C cos aL (5.28a,b) 3 L 1 2 4 2 2

1 1 − 2a 2C cos aL = 0, 2a 2C sin aL = 0 (5.28c,d) 2 2 1 2

(i) As above, this system of equations has two solutions, the first being C2 ≠ 0 which leads to that cos aL/2 must be zero. This in turn also means that C1 = C3 = C4 = 0, from eq.(5.28c). cos aL/2 = 0 has the solution aL = (2n − 1)π and by inserting this into the definition of a in eq.(5.13) we can solve for P, which then becomes

(2n −1) 2 π 2 D 2 (2n −1) 2 P P = L = E (5.29) cr (2n −1) 2 π 2 D 1+ (2n −1) 2 π 2φ 1+ L2 S This gives the symmetric buckling to ()21n − πx wx()= C cos (5.30) 2 L

5.15 AN INTRODUCTION TO SANDWICH STRUCTURES

(ii) The second solution comes from taking C2 = 0 which leads to that C4 = 0 and that sin aL/2 must be zero. This in turn also means that C1 ≠ 0 and that C3 again is zero. sin aL/2 = 0 has the solution aL = 2nπ and by inserting this into the definition of a in eq.(5.13) we can solve for P, which then becomes

(2n) 2 π 2 D 2 (2n) 2 P P = L = E (5.31) cr (2n) 2 π 2 D 1+ (2n) 2 π 2φ 1+ L2 S The anti-symmetric buckling mode is then 2nπx wx()= C sin (5.32) 1 L By combining eqs.(5.29) and (5.31) we cover all those buckling loads which are represented by eq.(5.14a) and the solutions are thus equivalent. This problem has a simpler solution by taking the origin of the x-coordinate at e.g. the left edge of the beam, resulting in one single solution w(x) = C sin ax and the buckling load of eq.(5.15a). This would, of course, have been much simpler, but was already done in section 5.2!

We can now also see that eq.(5.16) will hold for all buckling modes in this case. If we assume that (5.16) holds we can find the shear buckling load through eq.(5.3a) as

∂ 2 w ∂ 2 w P = S s ∂x 2 ∂x 2 and by letting wb = 0, we see that this only holds for P = Ps = S. Similarly, through eq.(5.3b) we have that

∂ 4 w ∂ 2 w ∂ 4 w ∂ 2 w D b + S s = D b + P = 0 ∂x 4 ∂x 2 ∂x 4 ∂x 2

For this case we take that ws = 0 so that wb = w. This then gives the buckling of an ordinary column, which has the solution

n 2π 2 D P = b L2 and then it is seen that eq.(5.16) holds for all buckling modes in the simply supported case.

5.4.3 One edge clamped, the other free (cantilever beam) There are two cases not considered so far ; the cantilever beam and a beam with one edge simply supported and the other clamped. It is now more appropriate to use a coordinate system that starts at the clamped edge of the beam, as illustrated in Fig.5.7.

5.16 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

M 1 T1 w T(x) dw P P dx x Nx Nx L

Figure 5.7 Schematic of a cantilever beam in compression (note that x now starts at the left edge of the beam). At the left edge the boundary conditions can be specified as before, i.e.,

dw T − C P w(0) = 0, and = 1 = 3 dx x=0 S − P S − P

However, the right edge requires some new attention. This edge is free of bending moment, and thus

d 2 w ⎛ P ⎞ d 2 w Mx(L) = 0 or 2 ⎜1− ⎟ = 0 → P = S or 2 = 0 dx ⎝ S ⎠ x=L dx x=L

At the right edge, the resulting transverse force should also be zero. However, from the definition of the in-plane load Nx acting along the beam when deriving the equilibrium equations we see that the free edge has several forces that must be in equilibrium in order to have a restraint free edge. From Fig.5.7 (to the right) we see that we can write dw dw dw dw T + N = 0 since sin ≈ and cos ≈ 1 x x dx dx dx dx which can be rewritten to

3 2 d wb dw dM x d wb D + P = 0 through T = , M = −D and Nx = −P dx3 dx x dx x dx 2 but using that

d 3 w d 3 w d 3 w d 3 w 1 d 2T d 3 w P d 3 w d 3 w ⎛ P ⎞ b = − s = − x = − = ⎜1− ⎟ dx3 dx 3 dx 3 dx 3 S dx 2 dx 3 S dx 3 dx 3 ⎝ S ⎠ we see that

⎛ P ⎞ d 3 w dw d 3 w dw D⎜1− ⎟ + P = 0 or + a 2 = 0 ⎝ S ⎠ dx 3 dx dx3 dx at a free edge subjected to an in-plane load P. The boundary conditions are now

dw T − C P d 2 w d 3 w dw w(0) = 0, = 1 = 3 , = 0 and + a 2 = 0 dx S − P S − P dx 2 dx 3 dx x=0 x=L x=L x=L

This gives the following four equations

C2 + C4 = 0 (5.33a)

5.17 AN INTRODUCTION TO SANDWICH STRUCTURES

⎛ P ⎞ a⎜1− ⎟C1 + C3 = 0 (5.33b) ⎝ S ⎠

2 2 −a C1 sin aL −a C2 cos aL = 0 (5.33c)

3 −a C1 = 0 (5.33d)

From this we see that C1 = C3 = 0. The third equation then gives that either C2 or cos aL is zero, the latter being the non-trivial solution constituting the stability criterion. The first equation then also gives that C4 = −C2. Thus, for cos aL to be zero aL = (2n−1)π/2. Using the definition of a gives the critical load to be

2 2 (2n −1) 2 π 2 D (2n −1) π D 2 (βL) 2 (2n −1) 2 P P = 4L = = E with β = 2 (5.34) cr (2n −1) 2 π 2 D (2n −1) 2 π 2 D 1+ (2n −1) 2 π 2φ / β 2 1+ 1+ 4L2 S (βL) 2 S

The buckling modes are then given by

⎛ (2n −1)πx ⎞ w(x) = C2 ⎜cos −1⎟ (5.35) ⎝ 2L ⎠

We could also of course have used symmetry, since this case must have the same buckling load as a twice as long simply supported beam, as illustrated in Fig.5.8, which is obvious by substituting L for 2L in eq.(5.15a). But that reasoning only gives us the value of the first mode!

Let's here also have a quick look at another way of solving the problem. Let's write eqs.(5.33) on the form

⎡ 0 1 0 1⎤⎛ C ⎞ ⎢ ⎛ P ⎞ ⎥⎜ 1 ⎟ a 1− 0 1 0 ⎢ ⎜ ⎟ ⎥⎜C2 ⎟ ⎝ S ⎠ = 0 ⎢ 2 2 ⎥⎜C ⎟ ⎢− a sin aL − a cos aL 0 0⎥⎜ 3 ⎟ ⎜ ⎟ ⎢ 3 ⎥ C ⎣ − a 0 0 0⎦⎝ 4 ⎠

The determinant of the coefficient matrix is zero if

a 5 cosaL = 0 i.e. cos aL = 0, as also found above.

5.4.4 One edge clamped, the other simply supported The simply supported/clamped beam will have the same buckling load as a twice as long clamped/clamped beam in its first unsymmetrical buckling mode, see Fig.5.8, i.e. by using eq.(5.25) and substituting L for 2L.

5.18 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

P P

Figure 5.8 Buckling of cantilever beam and beam with simply supported and clamped edges Higher buckling modes for both these cases must be pursued by solving the complete differential equation again with the correct boundary conditions. However, the first buckling mode is usually the sought after load since that constrains the load bearing capacity of the structure, and thus higher buckling modes are often of more academic interest. Anyhow, for the academic interest let’s perform this last of the four so called Euler buckling cases.

x M 1 T1 T2

P w P

Figure 5.9 Schematic of a simply supported/clamped beam in compression (note that x now starts at the left edge of the beam). Global equilibrium gives that

T1 = T2 and M1 = −T2L = −T1L The shear force and bending moment at position x of the beam is readily found

dM x dw Mx = M1 +T1x − Pw and T ==+P T x dx dx 1 By rearranging and using the definition of a as in eq.(5.13) the following is obtained

d 2 w T + a 2 w = a 2 1 (L − x ) dx 2 P The solution can again be written as in eq.(5.17) and substitution into the differential equation gives that C3 = –T1 /P and C4 = T1L/P

The boundary conditions are then

5.19 AN INTRODUCTION TO SANDWICH STRUCTURES

2 dw T1 d w w(0) = w (L) = 0, = and 2 = 0 dx x=0 S − P dx x=L

These four conditions give the following four equations

C2 + C4 = 0 (5.36a)

T1 − C3 P ⎛ P ⎞ aC1 + C3 = = or aC1 ⎜1− ⎟ + C3 = 0 (5.36b) (S − P) S − P ⎝ S ⎠

C1 sin aL + C2 cos aL + C3 L + C4 = 0 (5.36c)

2 2 − a C1 sin aL − a C2 cos aL = 0 (5.36d)

The first, second and fourth conditions results in

⎛ P ⎞ C4 = −C2, C3 = −aC1 ⎜1− ⎟ ⎝ S ⎠

C2 = −C1 tan aL which when inserted into the third condition gives that

⎛ P ⎞ tanaL = aL⎜1− ⎟ (5.37) ⎝ S ⎠ which then constitutes the stability criterion. This relation has roots an which must be calculated numerically from eq.(5.37). By using the definition of a in eq.(5.13) we can write the buckling load as

2 D(an L) 2 P = L (5.38) cr D(a L) 2 1+ n L2 S The mode buckling can be found as

⎛ ⎡ x ⎤⎞ w(x) = C1 ⎜sin ax + tan aL⎢1− − cos ax⎥⎟ (5.39) ⎝ ⎣ L ⎦⎠

Now, we can make a summary of the four Euler buckling cases: the fixed-fixed column studied above obviously correspond to the simply supported beam in Fig.5.4 but with only half its length and a column with one end fixed and the other free will correspond to the simply supported beam but with twice the length. Hence, eq.(5.15a) and (5.16) can be used in the three cases by just using the right length, as indicated in Fig.5.6. However, for the anti- symmetrical cases the buckling load can be approximated by the same formula so that a general expression as in eq.(5.36) may be used.

5.20 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

P P P P

β =2 β =1 β =0.5 β≈ 0.7

Figure 5.10 Euler's buckling cases.

111 π2 D =+ with Pb = 2 (5.40) PPScr b ()βL The buckling load for fixed-pinned end column (far right in Fig.5.10) is obtained by solving eq.(5.37) numerically. However, as shown in [3] only a slightly unconservative result will be obtained if eq.(5.15) is used for this case. Therefore, eqs.(5.15a) and (5.16) can for practical purposes also be used for the fixed-pinned end column by using L ≈ 0.7.

5.5 Examples of Sandwich Beam Buckling To exemplify this, let us use some real numbers. Take the beam in data as in section 4.10.3

Ef = 20 GPa, Gc = 40 MPa, tf = 2 mm, tc = 50 mm, L = 1000 mm, This gives

D = 54,080,000 Nmm and S = 2,163 N/mm → φ = 0.025 Now, calculating the buckling load for the first four modes gives the following results for four different beam configurations. The results are given in Table 5.1 and compared with the analytical formulae. The numerical results (FEM) where obtained using the three-nodes shear deformable sandwich element described in chapter 12. Twenty elements were used.

As seen in the example, the agreement between analytical and finite element calculation using a three-noded shear deformable beam finite element is excellent. However, such agreement is usually not found when performing stability calculations of sandwich panels with more complex geometry. Another interesting observation is that the shear deformation will have an increasingly larger effect on the buckling load for higher buckling modes. This is not at all surprising; higher buckling modes imply shorter wave length buckling and sandwich beam theory predicts more shear deformation for shorter beams. In the simply supported case for example, the first buckling mode corresponds to one deformation ”wave” for the beam with length L, the second corresponds to ”waves” each with length L/2. Thus, a half as long beam will have a four times higher shear factor.

5.21 AN INTRODUCTION TO SANDWICH STRUCTURES

mode 1 mode 2 mode 3 mode 4 CANTILEVER Analytical 125.7 772.2 1312.2 1625.3 ”pure bending, S large” 133.4 1200.9 3335.9 6538.4 FEM 125.7 772.3 1312.6 1626.0 SIMPLY SUPPORTED Analytical 428.1 1074.5 1491.5 1726.0 ”pure bending, S large” 533.7 2135.0 4803.7 8540.0 FEM 428.1 1074.7 1492.0 1726.7 SIMPLY SUPPORTED/CLAMPED Analytical 702.6 1270.2 1599.1 1785.0 ”pure bending, S large” 1091.9 3227.5 6430.1 10700 FEM 702.7 1270.6 1599.7 1785.9 CLAMPED/CLAMPED Analytical 1074.4 1364.2 1726.0 1808.9 ”pure bending, S large” 2135.0 4367.7 8540.0 12910 FEM 1074.7 1364.7 1726.7 1809.8

Table 5.1 Comparison between analytical and finite elements when calculating the buckling load for some different sandwich beams

5.6 Buckling of Sandwich Columns with Thick Faces The above equations are valid for faces that have no flexural rigidity of their own. This means that if the shear stiffness S is very low, so will be the overall buckling load. In the real case, the faces have a bending stiffness Df and even if S should be very low, the buckling load will still approach infinity for very short columns since that case would equal that of the two faces buckling independently of each other. Now use eq.(5.8) derived for the beam bending case, which is still valid since the buckling problem assumes a beam perturbed in bending. Consider the simply supported beam in Fig.5.4 again. Rewriting the governing equation using q = 0 and Nx = −P gives dw6 DS dw4 dw4 S dw2 20Dfxx6 −+−4 P 4 P 2 = dx D0 dx dx D0 dx Assuming the following deflected shape (the same as for the thin face case) nπx wa= sin (5.41) L we arrive at

6 442 ⎛ nππππ⎞ DS ⎛ n ⎞ ⎛⎛ n ⎞ S ⎛ n ⎞ ⎞ 2D ⎜ ⎟ + ⎜ ⎟ = P ⎜⎜ ⎟ + ⎜ ⎟ ⎟ f ⎝ ⎠ ⎝ ⎠ x ⎝ ⎠ ⎝ ⎠ L D0 L ⎝ L D0 L ⎠ which rewritten becomes

5.22 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

44 22 2nDDπ f 0 nDSπ 4 + 2 P = L L (5.42a) cr nD22π 0 + S L2

The limits for this equation are much the same as for eq.(5.15a), that is, that Pcr approaches the ordinary Euler load when S is large and/or L is large. If D is large and Df is small the load approaches the shear buckling load S and finally, if S is small the column will behave as two shear stiff beams, the faces, buckling independently of each other at a load equal to

2 2π Df lim Pcr = (5.42b) SL→→00 and/or L2 An alternative approach is taken in [3] and [4] and follows in brief. The total shear force in any section Ttot = P dw/dx. Using this in eq.(4.29) yields (assuming the faces to have infinite shear stiffness)

2 dT 22dw 2⎛ dw dw ⎞ x −=−=−+bT bP bP⎜ bs⎟ dx 2 x dx ⎝ dx dx ⎠

2 where b = S/2Df. From the beam bending analysis it follows that

3 3 dwb dwsxT T x D0 dwb TDx =− 0 3 and ==2 =−2 3 dx dx S 22bDffbD dx which yields that

5 3 3 dwbbb2 dw 22dw D dwb −+D0 5 bD0 3 =−+bP bP 2 3 dx dx dx 2bDf dx

dw5 ⎛ P ⎞ dw3 P dw b −−⎜b 2 ⎟ bb−=b 2 0 5 ⎜ ⎟ 3 dx ⎝ 2D f ⎠ dx D0 dx

The boundary conditions are w = M = 0 at x = 0 and x = L. These conditions are fulfilled by the sinusoidal displacement field given in eq.(5.41). Substitution into the differential equation and division by –(π/L)cos(πx/L) yields

⎡n 44ππ⎛ P ⎞ n 22 P ⎤ ⎢ +−⎜b 2 ⎟ − b 2 ⎥a = 0 L4 ⎜ 2D ⎟ L2 D ⎣⎢ ⎝ f ⎠ ⎦⎥

Solving for P, one arrives at eq.(5.42a) again.

5.7 Buckling Stress Exceeding the Elastic Limit If the buckling load is calculated with the above formulae and the corresponding face stress is found to exceed the elastic limit corrections to the buckling load must be done. This can be assessed in several ways but a simple and straight-forward way is the following [5].

5.23 AN INTRODUCTION TO SANDWICH STRUCTURES

Assume the face stresses already have exceeded their elastic limit prior to buckling. When the column starts to buckle, one face will, due to the superimposed bending moment, exhibit an increased compressive stress and the other will, for the same reason, unload. The rate at which the stresses change in these two faces are: the compressive increases with the rate of the tangent modulus at the specific strain point of the stress-strain curve, whereas the face that unloads does that with a slope equal to the elastic modulus. Hence, for each point on the stress-strain curve of the faces one can compute the tangent modulus, from that a neutral axis and thus a flexural rigidity D which depends on the strain. If this D is inserted in the buckling formula, the buckling load will be given, smaller than the elastic one, and also the corresponding face stress at this load. To find the buckling load one must now perform an iterative process to find the point on the stress-strain curve at which 2tfσf equals the computed buckling load P(ε). It is easy to show that this procedure is equivalent to replacing the moduli of the two faces with a reduced modulus

2E f Et Er = (5.43) EEft+ where Et is the tangent modulus. The application of the buckling equations e.g. (5.15a) etc., then requires a trial-and-error operation to find the correct buckling load. A more conservative approach would be to use Er = Et. 5.8 Free Vibration of Sandwich Beams The governing differential equation for free vibration of a sandwich beam was derived in eq.(5.4). By omitting the in-plane load Nx and the transverse load q, this is written as

∂ 4 w ∂ 2 w ρ * ⎛ ∂ 4 w ∂ 4 w ⎞ ∂ 4 w D + ρ * − ⎜ D − R ⎟ − R = 0 4 2 ⎜ 2 2 4 ⎟ 2 2 ∂x ∂t S ⎝ ∂x ∂t ∂t ⎠ ∂x ∂t or slightly rewritten

∂ 4 w ρ * ∂ 2 w ρ * ∂ 4 w ρ * R ∂ 4 w R ∂ 4 w + − + − = 0 (5.44) ∂x 4 D ∂t 2 S ∂x 2∂t 2 DS ∂t 4 D ∂x 2 ∂t 2 where ρ* [kg/m2] is the mass per unit length (also per unit width) of the beam, and R is the rotary inertia [kg]. In order to solve this, let’s first assume that the ratio R/D is very small and considerably smaller than ρ*/S (see section 5.6.1). We may then simplify eq.(5.44) to

∂ 4 w ρ * ∂ 2 w ρ * ∂ 4 w + − = 0 (5.45) ∂x 4 D ∂t 2 S ∂x 2∂t 2

The solution to this equation may be written in separable form by assuming w(x,t) = Φ(x)Ψ(t) (5.46)

By inserting this into eq.(5.45) one obtains

d 4Φ ρ * d 2 Ψ ρ * d 2Φ d 2 Ψ Ψ + Φ − = 0 dx 4 D dt 2 S dx 2 dt 2

5.24 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

By dividing each side with Φ and Ψ we get

1 d 4Φ ρ * 1 d 2 Ψ ρ * 1 d 2Φ 1 d 2 Ψ 1 d 4Φ 1 d 2 Ψ ⎛ ρ * ρ * 1 d 2Φ ⎞ + − = 0 → + ⎜ − ⎟ = 0 4 2 2 2 4 2 ⎜ 2 ⎟ Φ dx D Ψ dt S Φ dx Ψ dt Φ dx Ψ dt ⎝ D S Φ dx ⎠ so that

−1 1 d 4Φ ⎛ ρ * ρ * 1 d 2Φ ⎞ 1 d 2 Ψ ⎜ − ⎟ = − = λ 4 ⎜ 2 ⎟ 2 Φ dx ⎝ D S Φ dx ⎠ Ψ dt

The left hand side of this relation is clearly independent of t, and hence the right hand side must also be independent of t. Similarly, the right hand side is independent of x. Thus, since both side must be independent of x and t, each side must be constant. Now, this implies that

d 4Φ ⎛ ρ * ρ * 1 d 2Φ ⎞ d 4Φ ρ * d 2Φ ρ * − λ⎜ − ⎟Φ = + λ − λ Φ = 0 (5.47) 4 ⎜ 2 ⎟ 4 2 dx ⎝ D S Φ dx ⎠ dx S dx D and

d 2 Ψ + λΨ = 0 (5.48) dt 2 The latter of these equations has the solution

iωt −iωt 2 Ψ(t) = A1 sinωt + A2 cosωt = A3e + A4e with ω = λ (5.49)

The former equation (eq.(5.48)) can then be written

d 4Φ ρ * d 2Φ ρ * + ω 2 −ω 2 Φ = 0 (5.50) dx 4 S dx 2 D We now assume a solution to this differential equation on the form

an x Φ(x) = Bn e (5.51)

Inserting this into eq.(5.47) we get

⎡ * 2 * 2 ⎤ 4 ρ ω 2 ρ ω an x ⎢an + an − ⎥Bn e = 0 (5.52) ⎣ S D ⎦ which in turn must be valid for any x in order to fulfil the differential equation at every point of the beam. Thus, if the term within the bracket is zero the assumption made in eq.(5.52) can be made to satisfy the differential equation. This gives us the so called characteristic equation for the unknown coefficients an, which has the solution

2 1/ 2 ρ *ω 2 ⎡⎛ ρ *ω 2 ⎞ ρ *ω 2 ⎤ a 2 = − ± ⎢⎜ ⎟ + ⎥ = −a ± b (5.53) n 2S ⎜ 2S ⎟ D ⎣⎢⎝ ⎠ ⎦⎥ which has four roots. These roots must then be (symbolically and since a < b)

5.25 AN INTRODUCTION TO SANDWICH STRUCTURES

2 a1,2 = −a + b → a1 = − a + b = α and a2 = − − a + b = −α

2 a3,4 = −a − b → a3 = i a + b = iβ and a4 = −i a + b = −iβ and thus a2 = −a1 and a4 = −a3. This leads to the solution

αx −αx iβx −iβx Φ(x) = B1e + B2e + B3e + B4e (5.54) but since

e ±αx = coshαx ± sinhαx and e ±iβx = cos βx ± isin βαx this can be rewritten to

Φ(x) = C1 sinhαx + C2 coshαx + C3 sin βx + C4 cos βx (5.55)

The constants C are to be determined from the boundary conditions.

In the case of pure bending, i.e., when the shear stiffness is infinitely large, the roots of the characteristic equation becomes

* 2 1/ 2 * 2 1/ 4 * 2 1/ 4 2 ⎡ ρ ω ⎤ ⎡ ρ ω ⎤ ⎡ ρ ω ⎤ an = ±⎢ ⎥ → a1,2 = ±⎢ ⎥ = ±β and a3,4 = ±i⎢ ⎥ = ±iβ = ±iα ⎣ D ⎦ ⎣ D ⎦ ⎣ D ⎦ and hence, α = β. In the case of pure shear, i.e., when the bending stiffness is infinitely large, the roots of the characteristic equation becomes

1/ 2 ρ *ω 2 ⎛ ρ *ω 2 ⎞ ⎡ ρ *ω 2 ⎤ 2 ⎜ ⎟ an = − ± ⎜ ⎟ → a3,4 = 0 and a3,4 = −i⎢ ⎥ = ±iβ 2S ⎝ 2S ⎠ ⎣ S ⎦

This then has the solution

iβx −iβx Φ(x) = B3e + B4e = C3 sin βx + C4 cos βx (5.56)

The solution a3,4 = 0 means that the characteristic equation eq.(5.52) is fulfilled for any spatial deflection field and thus for any choice of the spatial constants Ci.

5.8.1 Simply supported edges The simply supported case, see Fig.5.6, is next studied but take the origin of the x-coordinate at the left edge of the beam instead. This will yield a much faster solution, although keeping the coordinate as in Fig.5.6 eventually will give the same result. The boundary conditions are in this case requiring zero deflection and zero bending moments at the edges. The first kinematic conditions are straight forward, but the natural conditions require some more elaboration. We have that

∂ 2 w ∂ 2 w ∂ 2 w M = −D b = −D + D s x ∂x 2 ∂x 2 ∂x 2 But, from eq.(5.3a) we can rewrite

5.26 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

∂ 2 w ρ * ∂ 2 w s = ∂x 2 S ∂t 2 and thus

∂ 2 w ∂ 2 w Dρ * ∂ 2 w ⎛ ∂ 2 w ρ *ω 2 ⎞ M = −D b = −D + = −D⎜ + w⎟ x 2 2 2 ⎜ 2 ⎟ ∂x ∂x S ∂t ⎝ ∂x S ⎠

However, at the simply supported edge, the displacement w is already prescribed to be zero. Thus, the boundary conditions in all can be written as

d 2 w d 2 w w(0) = w(L) =0, and 2 = 2 = 0 dx x=0 dx x=L

This gives the following four equations

C2 + C4 = 0 (5.57)

C1 sinhαL + C2 coshαL + C3 sin βL + C4 cos βL = 0

2 2 α C2 − β C4 = 0

2 2 2 2 α C1 sinhαL +α C2 coshαL − β C3 sin βL − β C4 cos βL = 0

The first equation gives that C4 = −C2. Inserting this into the third equation yields that

2 2 2 2 2 2 α C2 + β C2 = (α + β )C2 = 0 → C4 = C2 = 0 since (α + β ) = 2b ≠0 unless α = iβ, which happens when b = 0, i.e., when D becomes large. The second of eq.(5.57) now gives that

C1 sinhαL + C3 sin βL = 0 which inserting into the fourth equation results in

2 2 2 2 2 2 α C1 sinhαL − β C3 sin βL = −α C3 sin βL − β C3 sin βL = (α + β )C3 sin βL = 0

Thus, we have that either C3 = 0 (trivial solution) or that sin βL = 0 (non-trivial solution). Taking the latter of these implies that

2 1/ 2 m 2π 2 ρ *ω 2 ⎡⎛ ρ *ω 2 ⎞ ρ *ω 2 ⎤ βL = mπ → β 2 = = a + b = + ⎢⎜ ⎟ + ⎥ L2 2S ⎜ 2S ⎟ D ⎣⎢⎝ ⎠ ⎦⎥

Solving for ω gives

m 2π 2 D m 2π 2 D ωn = = (5.58) L2 ⎛ m 2π 2 D ⎞ L2 ρ * ()1+ m 2π 2φ ρ * ⎜1+ ⎟ ⎜ 2 ⎟ ⎝ L S ⎠

5.27 AN INTRODUCTION TO SANDWICH STRUCTURES with the spatial mode mπx Φ(x) = C sin (5.59) 3 L and the entire can thus be written as mπx w(x,t) = C sin eiωt (5.60) L There is a much quicker, but not so stringent way of solving this problem. Assuming harmonic motion, a displacement function that satisfies the boundary conditions is mπx w(x,t) = w sin eiωt = Φ(x)Ψ(t) L where Φ(x) is the spatial displacement function, Ψ(t) the time variation and ω the natural frequency. Inserting into the governing equation of eq.(5.4) yields

4 * 2 2 ⎡ ⎛ mπ ⎞ * 22ρ ⎡ ⎛ mπ ⎞ 42⎤ ⎛ mπ ⎞ ⎤ D⎜ ⎟ −−ρω Dω ⎜ ⎟ − RRωω− ⎜ ⎟ ΦΨ()xt ()= 0 ⎢ ⎝ ⎠ ⎢ ⎝ ⎠ ⎥ ⎝ ⎠ ⎥ ⎣ LS⎣ L ⎦ L ⎦

By introducing the quantity α = ρ*ω2/D and recognising that a non-trivial solution is for Φ(x), and Ψ(t) ≠ 0 the above can be rewritten as

4 2 ⎛ mπ ⎞ ⎛ mπ ⎞ ⎛ D R ⎞ α 2 DR −α −α ⎜ + ⎟ + = 0 ⎜ ⎟ ⎜ ⎟ ⎜ * ⎟ * ⎝ L ⎠ ⎝ L ⎠ ⎝ S ρ ⎠ ρ S

It can be seen from this expression that if the shear deformation (D/S = 0) and the rotary inertia (R = 0) are omitted, this expression equals that of the engineering beam theory giving the natural frequency

D ωπ= m22 m ρ* L4

For a homogeneous cross-section of isotropic material, the ratio D/S = 1.2(1+ν)h2/6 is approximately 3 times as large as R/ρ* = h2/12 and hence the effect of shear deformation is more important than the rotary inertia. In a sandwich, the ratio D/S is magnitudes larger than for a homogeneous cross-section and thus even more important. By omitting the effect of rotary inertia it is possible to simplify the above expression to

mπ 42mπ D ⎜⎛ ⎟⎞ −−αα⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ = 0 ⎝ L ⎠ ⎝ L ⎠ ⎝ S ⎠ or rewritten

D D ωπ= m 22 = m 22π m 2 * 422 * 4 ⎡ D ⎛ mπ ⎞ ⎤ ρπφLm()1 + ρ L ⎢1 + ⎜ ⎟ ⎥ ⎣ S ⎝ L ⎠ ⎦

5.28 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS with φ defined as above (φ = D/L2S). This can be seen as the engineering beam theory expression divided by a correction term (the denominator) that depends on the amount of shear deformation. It is important to notice that the correction depends on the mode of vibration, giving a small correction in the first eigen-frequency but larger and larger corrections for higher modes.

5.8.2 Clamped edges We can now use the same boundary conditions as specified previously for the buckling analysis. Let us use the same coordinate system as in Fig.5.5, i.e., with the x-coordinate having its origin in the middle of the beam. However, for this case the boundary conditions must be rewritten in order to obtain a solution. The clamped edges corresponds to that rotations are fixed at the edges, or

dw dw w(0) = w(L) = 0, and b (0) = b (L) = 0 dx dx The solution to the problem is once again given by eq.(5.55) and the first two boundary conditions again give that

C2 + C4 = 0 (5.61)

C1 sinhαL + C2 coshαL + C3 sin βL + C4 cos βL = 0

Now, to get the rotations, let us use eq.(5.3a), now in the form

∂ 2 w ∂ 2 w S s = ρ * = −ρ *ω 2 w(x) ∂x 2 ∂t 2 This gives that

2 2 2 S ∂ ws S ⎡∂ w ∂ wb ⎤ 2 * 2 = * ⎢ 2 − 2 ⎥ = −ω w(x) ρ ∂x ρ ⎣ ∂x ∂x ⎦ from which it follows that

S ∂ 2Φ S ∂ 2Φ b = ω 2Φ(x) + ρ * ∂x 2 ρ * ∂x 2 which can be integrated to

S ∂Φ ⎛ω 2 Sα ⎞ ⎛ω 2 Sα ⎞ b = C ⎜ + ⎟coshαx + C ⎜ + ⎟sinhαx * 1 ⎜ * ⎟ 2 ⎜ * ⎟ ρ ∂x ⎝ α ρ ⎠ ⎝ α ρ ⎠

⎛ω 2 Sβ ⎞ ⎛ω 2 Sβ ⎞ − C ⎜ − ⎟cosαx + C ⎜ − ⎟sinαx 3 ⎜ * ⎟ 4 ⎜ * ⎟ ⎝ β ρ ⎠ ⎝ β ρ ⎠

The boundary conditions are now

dw dw dΦ dΦ b (0) = b (L) = 0 or b (0) = b (L) = 0 dx dx dx dx

5.29 AN INTRODUCTION TO SANDWICH STRUCTURES which gives

⎛ω 2 Sα ⎞ ⎛ω 2 Sβ ⎞ C ⎜ + ⎟ − C ⎜ − ⎟ = 0 or C β ω 2 ρ * + Sα 2 − C α ω 2 ρ * − Sβ 2 = 0 1 ⎜ * ⎟ 3 ⎜ * ⎟ 1 ( ) 3 ( ) ⎝ α ρ ⎠ ⎝ β ρ ⎠ and C β (ω 2 ρ * + Sα 2 )coshαL + C β (ω 2 ρ * + Sα 2 )sinhαL 1 2 2 * 2 2 * 2 − C3α()ω ρ − Sβ cos βL + C4α ()ω ρ − Sβ sin βL = 0 which can be written as

C1S1 − C3S2 = 0

C1S1 coshαL + C2S1 sinhαL − C3S2 cosβL + C4S2 sinβL = 0 where

2 * 2 2 * 2 S1 = β (ω ρ + Sα ) and S 2 = α(ω ρ − Sβ ) This can now all be written in a matrix equation as

⎡ 0 1 0 1 ⎤⎛ C1 ⎞ ⎢ ⎥⎜ ⎟ sinhαL coshαL sin βL cos βL ⎜C2 ⎟ ⎢ ⎥ = 0 (5.62) ⎢ S 0 − S 0 ⎥⎜C ⎟ 1 2 ⎜ 3 ⎟ ⎢ ⎥⎜ ⎟ ⎣S1 coshαL S1 sinhαL − S2 cos βL S 2 sin βL⎦⎝C4 ⎠ for which there are two solutions; the trivial corresponding to all C terms being zero or that the determinant of the coefficient matrix is zero. The latter gives that

⎛ S2 S1 ⎞ ⎜ − ⎟sinhαLsin βL + 2 − 2coshαL cos βL = 0 (5.63) ⎝ S1 S 2 ⎠

This equation must now be solved numerically to obtain ω. The solution has been compared with finite element solutions and the results are given in table 5.2, and as seen the agreement is perfect. However, the finite element solution requires a fair amount of elements to give the exact results. In table 5.2, 40 three-noded sandwich beam elements has been used.

5.8.3 One edge clamped, the other simply supported This is the same problem as shown in Fig.5.9, but now free vibrations should be studied. The boundary conditions are a combination of the two above problems, namely

dw d 2 w w(0) = w(L) = 0, and b (0) = b (L) = 0 dx dx 2 which leads to the equations

C2 + C4 = 0 (5.64)

5.30 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

C1 sinhαL + C2 coshαL + C3 sin βL + C4 cos βL = 0

C1S1 − C3S2 = 0

2 2 2 2 α C1 sinhαL +α C2 coshαL − β C3 sin βL − β C4 cos βL = 0 or in matrix form

⎡ 0 1 0 1 ⎤⎛ C1 ⎞ ⎢ ⎥⎜ ⎟ sinhαL coshαL sin βL cos βL ⎜C2 ⎟ ⎢ ⎥ = 0 (5.65) ⎢ S 0 − S 0 ⎥⎜C ⎟ 1 2 ⎜ 3 ⎟ ⎢ 2 2 2 2 ⎥⎜ ⎟ ⎣α sinhαL α coshαL − β sin βL − β cos βL⎦⎝C4 ⎠ for which there are two solutions; the trivial corresponding to all C terms being zero or that the determinant of the coefficient matrix is zero. The latter gives that

2 2 2 2 (α + β )S 2 sinhαLcos βL + (α + β )S1 coshαLsin βL = 0 or

S 2 sinhαL cos βL + S1 coshαLsin βL = 0 (5.66)

This equation must now be solved numerically to obtain ω. The solution has been compared with finite element solutions and the results are given in Table 5.2, and as seen the agreement is perfect. However, the finite element solution requires a fair amount of elements to give the exact results. In Table 5.4, 40 three-noded sandwich beam elements has been used.

5.8.4 One edge clamped, the other free (cantilever beam) This is the same problem as shown in Fig.5.7, but now free vibrations should be studied. The boundary condition for the free edge needs first a little extra attention. The boundary conditions are

dw d 2 w dw w(0) = 0, b (0) = b (L) = s (L) = 0 dx dx 2 dx The first and second conditions have been treated already above. The third one is similar to the simply supported case but must now instead be written as

∂ 2 w ρ *ω 2 2 + w(L) = 0 ∂x x=L S since w at x = L is not necessarily zero in time. The fourth boundary condition must also be studied a little closer. We know from eq.(5.3a) that

∂w T = S s = 0 x ∂x which may rewritten to

5.31 AN INTRODUCTION TO SANDWICH STRUCTURES

∂w ∂ 3 w ∂ 3 w ⎛ ∂ 3 w ∂ 3 w ⎞ ∂ 2 ⎛ ∂w ∂w ⎞ S s = −D b + R b = −D⎜ − s ⎟ + R ⎜ − s ⎟ 3 2 ⎜ 3 3 ⎟ 2 ∂x ∂x ∂x∂t ⎝ ∂x ∂x ⎠ ∂t ⎝ ∂x ∂x ⎠ but from above ∂ws/∂x = 0, and from eq.(5.3a) we get that

∂3w ρ * ∂3w s = ∂x3 S ∂x∂t 2 Inserting this into the above yields that

∂ 3 w Dρ * ∂ 3 w ∂ 3 w ∂ 3 w Dρ *ω 2 ∂w ∂w D − − R = D + + Rω 2 = 0 ∂x3 S ∂x∂t 2 ∂x∂t 2 ∂x3 S ∂x ∂x which constitutes the free edge boundary condition. Since we already assumed R/D to be small, this relation becomes

∂ 3 w ρ *ω 2 ∂w + = 0 ∂x 3 S ∂x Thus, the boundary conditions in this case may be written

C2 + C4 = 0 (5.67)

⎛ω 2 Sα ⎞ ⎛ω 2 Sβ ⎞ C ⎜ + ⎟ − C ⎜ − ⎟ = 0 1 ⎜ * ⎟ 3 ⎜ * ⎟ ⎝ α ρ ⎠ ⎝ β ρ ⎠

⎛ ρ *ω 2 ⎞ ⎛ ρ *ω 2 ⎞ ⎜ 2 ⎟ ⎜ 2 ⎟ C1 ⎜α + ⎟sinhαL + C2 ⎜α + ⎟coshαL ⎝ S ⎠ ⎝ S ⎠

⎛ ρ *ω 2 ⎞ ⎛ ρ *ω 2 ⎞ ⎜ 2 ⎟ ⎜ 2 ⎟ + C3 ⎜− β + ⎟sin βL + C4 ⎜− β + ⎟cos βL = 0 ⎝ S ⎠ ⎝ S ⎠

⎛ αρ *ω 2 ⎞ ⎛ αρ *ω 2 ⎞ ⎜ 3 ⎟ ⎜ 3 ⎟ C1 ⎜α + ⎟coshαL + C2 ⎜α + ⎟sinhαL ⎝ S ⎠ ⎝ S ⎠

⎛ βρ *ω 2 ⎞ ⎛ βρ *ω 2 ⎞ ⎜ 3 ⎟ ⎜ 3 ⎟ + C3 ⎜− β + ⎟cos βL + C4 ⎜ β − ⎟sin βL = 0 ⎝ S ⎠ ⎝ S ⎠ with

2 * 2 2 * 2 S11 = ω ρ + Sα and S22 = ω ρ − Sβ the conditions can be rewritten into

C2 + C4 = 0

C1βS11 − C3αS 22 = 0

5.32 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

C1S11 sinhαL + C2 S11 coshαL + C3 S 22 sin βL + C4 S 22 cos βL = 0

2 2 2 2 C1βα S11 coshαL + C2 βα S11 sinhαL + C3αβ S 22 cos βL − C4αβ S 22 sin βL = 0 or in matrix form

⎡ 0 1 0 1 ⎤⎛ C1 ⎞ ⎢ ⎥⎜ ⎟ βS11 0 −αS 22 0 ⎜C2 ⎟ ⎢ ⎥ = 0 (5.68) ⎢ S sinhαL S coshαL S sin βL S cos βL ⎥⎜C ⎟ 11 11 22 22 ⎜ 3 ⎟ ⎢ 2 2 2 2 ⎥⎜ ⎟ ⎣α βS11 coshαL α βS11 sinhαL αβ S 22 cos βL −αβ S22 sin βL⎦⎝C4 ⎠ for which there are two solutions; the trivial corresponding to all C terms being zero or that the determinant of the coefficient matrix is zero. The latter gives that

⎛ 2 2 S11 ⎞ ⎛ 2 2 S11 ⎞ ⎛ S11 ⎞ ⎜ β −α ⎟ + ⎜α − β ⎟coshαLcos βL +αβ⎜1+ ⎟sinhαLsin βL = 0 (5.69) ⎝ S 22 ⎠ ⎝ S 22 ⎠ ⎝ S 22 ⎠

This equation must now be solved numerically to obtain ω. The solution has been compared with finite element solutions and the results are given in Table 5.2, and as seen the agreement is very good. However, the finite element solution requires a fair amount of elements to give the exact results. In Table 5.2, 20 three-noded sandwich beam elements has been used.

5.9 Examples of Sandwich Beam Free Vibration To exemplify this, let us use some real numbers. Take the beam in section 4.10.3 with data in SI-units

9 2 6 2 Ef = 20 10 N/m , Gc = 40 10 N/m , tf = 0.002 m, tc = 0.05 m, L = 1 m, 3 3 ρf = 1400 kg/m , and ρc = 100 kg/m

This gives

D = 54,080 Nm and S = 2,164,000 N/m ⇒ D/S = 0.025 m2

323 ρρffttd ff ρ t ρ* = 10.6 kg/m2 and R =+ +=cc 0.0048 kg ⇒ R/ρ* = 0.00045 m2 6212 This shows that the effect of shear deformation is about 55 times larger than the effect of rotary inertia. In calculating the rotary inertia it is seen that the term representing the core cannot be neglected. Hence, neglecting the rotary inertia term seems justified in this case. Some results are given in Table 5.2 for the first four vibration modes. The numerical results (FEM) where obtained using the three-nodes shear deformable sandwich element described in chapter 12. Twenty elements were used. With that many elements the FE solution is more or less exactly the same as the analytical one if rotary inertia is omitted (R = 0) in the FE- solution. The rotary inertia was included in the FEM solution but not in the analytical solutions.

5.33 AN INTRODUCTION TO SANDWICH STRUCTURES

mode 1 mode 2 mode 3 mode 4 CANTILEVER Analytical 238 1170 2616 4138 ”pure bending, S large” 251 1574 4407 8636 FEM 237 1165 2606 4122 SIMPLY SUPPORTED Analytical 631 2000 3536 5072 ”pure bending, S large” 705 2820 6345 11279 FEM 630 1996 3529 5065 SIMPLY SUPPORTED/CLAMPED Analytical 848 2154 3617 5115 ”pure bending, S large” 1101 3569 7446 12733 FEM 847 2150 3611 5108 CLAMPED/CLAMPED Analytical 1065 2286 3695 5158 ”pure bending, S large” 1598 4405 8636 14275 FEM 1064 2282 3689 5149

Table 5.2 Comparison between analytical and finite elements when calculating the free vibration frequencies (in s-1) for some different sandwich beams As seen in the example, transverse shear has a large effect even in the lowest eigen-frequency mode, and the effect increases drastically as the higher modes are considered. This is once again length dependent, just as in the bending or buckling case. If the length in the above example is increased to 3 m the first eigen-frequency in the simply supported cases neglecting -1 -1 shear deformation equals 78.3 s and if the shear deformation is included ωm= 77.3 s , which is a much smaller difference.

5.10 Estimation of Elastic Properties on Free-Free Sandwich Beam One way to performa a non-destructive test to extract the mechanical properties of a beam is to perform a vibration test and measure the natural frequencies. This data can then be used to calculate the properties. The beam is struck by an instrumented hammer which incorporates a load cell that registers the load during impact. An accelerometer that measures the vibrations caused by the impact is attached to the beam, see Fig.5.11. Data from the accelerometer and the load cell in the hammer is recorded and the so-called frequency response function (FRF) is then calculated. An example frequency response function can be seen in Fig.5.12. A peak in the frequency response function means that the structure is easily excited at this frequency, and such peaks represent the natural frequencies of the structure.

5.34 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

Figure 5.11 Principles of the test setup with the beam suspended using strings The most common way to perform this test is to suspend the beam in two lightweight strings and try to achieve so-called free - free boundary conditions, as indicated in Fig.5.11, since any kind of rigid support tends to influence the results. The result of such a test is presented as a frequency response plot in which the amplitude of the accelerometer signal is plotted versus frequency. A typical such plot is shown in Fig.5.12. The peaks in the frequency response plot identifies the eigenfrequencies.

Figure 5.12 Typical frequency response plot from sandwich beam vibration testing

5.35 AN INTRODUCTION TO SANDWICH STRUCTURES

For classical, homogeneous beams the Euler beam equation gives a simple relation between the bending stiffness and the natural frequencies.

2 D ω = ()β L n n ρ * L4

The factor (βL)2 is depending on the boundary conditions, and can be found tabulated in books on mechanical vibrations. For the free - free case of the test described above, the factors for the first three vibration modes are (βL)2 = 22.4, 61.7 and 121.0, respectively.

If the dimensions and density of the beam material is known (this is relatively easy to measure), the bending stiffness D (and thereby the E-modulus) can be calculated from if you measure one or more natural frequencies. In theory, one natural frequency would be enough to determine the E-modulus. In practice, the measurements are not always perfect, and by measuring more frequencies and calculating the corresponding E-modulus a least squares approximation can be made. There are however limits both to the theory and experiments as to how many natural frequencies can be accurately calculated/measured. Care should be taken when including higher frequencies so that you do not violate the theory or get unreliable results from the experiments. For sandwich beams there is a considerably more complicated relation between the natural frequencies and the material properties. First of all, there are two different materials involved in a sandwich beam, so two material properties are of interest, typically Ef and Gc. Thus you will have to measure at least two natural frequencies to be able to calculate the properties. In addition, the natural frequencies for the free - free sandwich beam is given in such a form that you must solve for them numerically.

The free-free beam thus has no kinematic boundary conditions, as illustrated in Fig.5.13.

Mx(0) = 0 Mx(L) = 0 Tx(0) = 0 Tx(L) = 0

Figure 5.13 Schematic of first vibration mode for a free-free sandwich beam The boundary conditions are already formulated and for this problem they become

∂ 2 w ρ *ω 2 2 + w x=0,L = 0 ∂x x=0,L S

∂ 3 w Dρ *ω 2 ∂w D + = 0 ∂x 3 x=0,L S ∂x x=0,L The four boundary conditions can now be written on the form

5.36 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

⎛ ρ *ω 2 ⎞ ⎛ ρ *ω 2 ⎞ ⎜ 2 ⎟ ⎜ 2 ⎟ C2 ⎜α + ⎟ + C4 ⎜− β + ⎟ = 0 (5.70) ⎝ S ⎠ ⎝ S ⎠

⎛ αρ *ω 2 ⎞ ⎛ βρ *ω 2 ⎞ ⎜ 3 ⎟ ⎜ 3 ⎟ C1 ⎜α + ⎟ + C3 ⎜− β + ⎟ = 0 ⎝ S ⎠ ⎝ S ⎠

⎛ ρ *ω 2 ⎞ ⎛ ρ *ω 2 ⎞ ⎜ 2 ⎟ ⎜ 2 ⎟ C1 ⎜α + ⎟sinhαL + C2 ⎜α + ⎟coshαL ⎝ S ⎠ ⎝ S ⎠

⎛ ρ *ω 2 ⎞ ⎛ ρ *ω 2 ⎞ ⎜ 2 ⎟ ⎜ 2 ⎟ + C3 ⎜− β + ⎟sin βL + C4 ⎜− β + ⎟cos βL = 0 ⎝ S ⎠ ⎝ S ⎠

⎛ αρ *ω 2 ⎞ ⎛ αρ *ω 2 ⎞ ⎜ 3 ⎟ ⎜ 3 ⎟ C1 ⎜α + ⎟coshαL + C2 ⎜α + ⎟sinhαL ⎝ S ⎠ ⎝ S ⎠

⎛ βρ *ω 2 ⎞ ⎛ βρ *ω 2 ⎞ ⎜ 3 ⎟ ⎜ 3 ⎟ + C3 ⎜− β + ⎟cos βL + C4 ⎜ β − ⎟sin βL = 0 ⎝ S ⎠ ⎝ S ⎠ with

2 * 2 2 * 2 S11 = ω ρ + Sα and S22 = ω ρ − Sβ the conditions can be rewritten into

C2 S11 + C4 S 22 = 0

C1αS11 + C3 βS 22 = 0

C1S11 sinhαL + C2 S11 coshαL + C3 S 22 sin βL + C4 S 22 cos βL = 0

2 2 2 2 C1βα S11 coshαL + C2 βα S11 sinhαL + C3αβ S 22 cos βL − C4αβ S 22 sin βL = 0 or in matrix form

⎡ 0 S11 0 S 22 ⎤⎛ C1 ⎞ ⎢ ⎥⎜ ⎟ αS11 0 βS 22 0 ⎜C2 ⎟ ⎢ ⎥ = 0 (5.71) ⎢ S sinhαL S coshαL S sin βL S cos βL ⎥⎜C ⎟ 11 11 22 22 ⎜ 3 ⎟ ⎢ ⎥⎜ ⎟ ⎣αS11 coshαL αS11 sinhαL βS22 cos βL − βS 22 sin βL⎦⎝C4 ⎠ for which there are two solutions; the trivial corresponding to all C terms being zero or that the determinant of the coefficient matrix is zero. The latter gives that 2αβ ()coshαLcos βL −1 + (β 2 −α 2 )sinhαLsin βL = 0 (5.72)

5.37 AN INTRODUCTION TO SANDWICH STRUCTURES

If we take the same numbers as used in section 5.9, the first three eigenfrequencies can be calculated analytically (actually by solving eq.(5.72) numerically) and compared with finite element analysis. The results are given in Table 5.3.

mode 1 mode 2 mode 3 mode 4 FREE-FREE Analytical (eq.(5.72) 1397 2995 4642 6226 FEM 1397 2995 4642 6228 ”pure bending, S large” 1598 4405 8636 14275 FEM 1600 4426 8725 14532

Table 5.3 Comparison between analytical and finite elements when calculating the free vibration frequencies for a free-free sandwich beam. As seen in Table 5.3, the analytical results compare very well with the FE-results. The reason for the discrepancy in the pure bending case (S very large) is that the elements do not perform all that well for such cases. Using more than twenty elements will however improve the correlation. The vibration modes for the first four eigenfrequencies are plotted in Fig.5.14.

Figure 5.14 The first four free vibration modes for a free-free sandwich beam

5.11 Approximate Solutions to Beam Buckling and Free Vibration Problems

5.11.1 Buckling of clamped sandwich beam An efficient way to solve various beam or plate problems is to use energy methods. This is done by assuming a deflection field, containing one or more unknowns, inserting this into the relation for the total potential energy and minimising with respect to the unknowns. This is usually called the Ritz’ method. The assumed deflection field, or ansatz, must at least satisfy the kinematic boundary conditions. If it also satisfies the natural boundary conditions, the obtained solution usually becomes better. In some sense, this method is similar to the finite element method but using only one element and some arbitrarily chosen deflection function. Even though the assumed deflection assumption does satisfy the boundary conditions, it does not necessarily satisfy the governing differential equation (actually it does not normally do so, for if it did one can solve the problem exactly). The solution is then not exact but should, given the deflection assumption, at least provide a solution in terms of deformation that is close to the correct value – a good estimation. Consequently, obtaining an approximate

5.38 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS deflection solution implies that any properties derived from derivatives of the displacements, e.g. strains and stresses, are usually quite poor estimates and cannot in general be trusted. In general, the better the deflection assumption, the better will the deflection solution be. As an example of this one can have a quick look at the simply supported sandwich beam subjected to a uniform pressure load. This is easily solved analytically. However, we could assume a deflection assumption of the kind πx w(x) = W sin L which is a function that satisfies all the boundary conditions (kinematic and natural). This function has one unknown, W. A solution can be found by differentiating the function, inserting into the total potential energy expressions (e.g. eqs.(4.39) and (4.40). Integration yields and expression for the total potential energy (e.g. eq.(5.12)) with only one unknown, W. The minimum of the energy if found by taking dΠ/dW = 0, from which W can be found. In this case, W will represent the deflection in the middle of the beam. The solution will be approximate, knowing that the exact solution is a fourth-order polynomial. However, we can choose a better deflection assumption, for example

m mπx w(x) = ∑Wm sin i=1 L

One can then proceed as above, but now we get m conditions dΠ/dWm = 0, giving us m equations for m unknown Wm. The more terms we include, m, the better the solution.

One can use Ritz’ method also to solve buckling and free vibration problems with complicated boundary conditions. The problem of finding an appropriate deflection assumption still remains. Below is one examples of how this can be done.

Let’s assume a deflection field of the type πx πx w(x) = w + w = W sin 2 +W sin b s b L s L Describing something that may look like the first buckling mode. This is seen to satisfy the boundary conditions, given as

wb(0) = wb(L) = 0

dw 2πW πx πx b = b sin cos = 0 for x = 0 and L dx L L L and for the shear component

ws(0) = ws(L) = 0 Inserting this into the energy equation yield the strain energy as

5.39 AN INTRODUCTION TO SANDWICH STRUCTURES

2 2 1 ⎡ ⎛ d 2 w ⎞ ⎛ dw ⎞ ⎤ DW 2π 4 SW 2π 2 U = ⎢D⎜ b ⎟ + S⎜ s ⎟ ⎥dx = b + b se 2 ∫ ⎜ dx 2 ⎟ dx L3 4L ⎣⎢ ⎝ ⎠ ⎝ ⎠ ⎦⎥

The potential energy of the applied load is given by

2 L ⎡1 ⎛ ∂w ⎞ ⎤ ⎛W 2π 2 W 2π 2 ⎞ U = −P ⎢ ⎜ ⎟ ⎥dx = −P⎜ b + b ⎟ N ∫ 2 ∂x ⎜ 4L 4L ⎟ 0 ⎣⎢ ⎝ ⎠ ⎦⎥ ⎝ ⎠

Minimising this expression then proceeds as follows;

4 2 2 ∂Π 2DWbπ 2PWbπ 4Dπ = 0 → 3 − = 0 → P = 2 ∂Wb L 4L L

∂Π 2SW π 2 2PW π 2 = 0 → s − s = 0 → P = S ∂Ws 4L 4L

Combing these using the principle for partial deflections gives

4Dπ 2 1 1 1 2 = + → P = L P P P cr 4Dπ 2 b s 1+ L2 S To obtain the unsymmetrical modes we must find a deflection assumption that is unsymmetrical and go over the analysis again.

5.11.2 Free vibration of a cantilever sandwich beam This problem can be solved in the same manner as for the simply supported beam. For the cantilever sandwich beam, as in 4.10.1, the assumed spatial/time displacement function is

⎡ (2m −1)πx ⎤ w(x,t) = w cos −1 eiωt ⎣⎢ 2L ⎦⎥ which is approximate, but satisfies the boundary conditions w = ∂w/∂x = 0 at x = 0, and that w(L,t) ≠ 0. The error made in this assumption is that really only the bending slope is supposed to be zero at the clamped edge. However, this fact can be shown to have little influence on the total deflection field away from the edge as long as the faces do not have significant thickness. By inserting this assumption in the governing differential equation and omitting the rotary inertia term, we get the following expressions for the energy terms one arrives at the natural frequency for a cantilever sandwich beam accounting for transverse shear deformation but not for rotary inertia as

2 2 D ()21m − π * 4 2 ρ L ⎛ 1⎞ 2 D ω m = =−⎜m ⎟ π 2 2 ⎝ ⎠ 2 ()21mD− π 2 * 42⎡ ⎛ 1⎞ ⎤ 41+ ρπφLm⎢1 +−⎜ ⎟ ⎥ 4LS2 ⎣ ⎝ 2⎠ ⎦

5.40 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

When also omitting the effect of transverse shear deformation, that is, assuming S is infinitely large, one arrives at the natural frequency for an ordinary beam as

1 2 D ωπ=−⎜⎛m ⎟⎞ 2 m ⎝ 2⎠ ρ * L4

This expression will yield good results, but due to the approximate and simplified deflection assumption, the first eigenfrequency will give somewhat erroneous results in the pure bending case. This is also see from the comparison with numerical results presented in Table 5.3 using the three-noded sandwich element described in chapter 12.

Mode -1 -1 FEM ωm {no shear} (s ) ωm{including shear} (s ) 1 176 171 238 2 1586 1271 1165 3 4406 2763 2605 4 8836 4306 4121 5 14275 5830 5657

Table 5.3 Comparison between analytical and finite element calculations of free undamped natural frequencies for a cantilever sandwich beam (rotary inertia omitted) As seen in Table 5.3, the solution for the first natural frequency shows very poor correlation with the numerical result. One reason is that the boundary condition at x = 0 assumes that the total slope is zero, but another error is that the transverse force is non-zero at x = L with the used approximate spatial assumption. Higher modes, on the other hand, show better correlation.

5.11.3 Free vibration of a clamped sandwich beam In this case one cannot find a simple displacement function that both satisfies the boundary conditions and is a solution to the governing equation. An approximate value for the first mode of vibration can however be found by using Ritz's method. A displacement function that satisfies the boundary conditions of the beam in 4.10.4(iii) is

2 πx iωt w(x,t) = (w + w )sin e b s L where once again the error is made that the slope of ws is also zero at the clamped edges. By inserting this assumption in the energy equations, integrating over the entire length of the beam, and letting the strain energy equal the kinetic energy one arrives at

2 4 2 2 wb π D ws π S 3L * 2 2 + = ρ (w + w ) ω L3 4L 16 b s for which a stationary value is sought. Differentiation with respect to w¯ b and w¯ s yields two equations that after substitution yield

D ωπ= 4 2 1 314ρπφ* L42()+

5.41 AN INTRODUCTION TO SANDWICH STRUCTURES

This value of the natural frequency will be slightly higher than the exact value. By comparing, it was found that the lowest natural frequency calculated by the above equation is about 2-15 percent higher than the value computed using the FE-method [1] for shear factors φ ranging from 0 to 1. This is also see from the comparison with numerical results presented in Table 5.4 using the three-noded sandwich element described in chapter 12.

-1 -1 ωm {no shear} (s ) ωm{including shear} (s ) Analytical 1628 1155 FEM 1606 1065 Table 5.4 Comparison between analytical and finite element calculations of the first free undamped natural frequency for a clamped sandwich beam

References [1] Timoshenko S.P., Vibration Problems in Engineering, Second Edition, D. Van Nostrand Company Inc., New York, N.Y., 1937, p 337.

[2] Reddy J.N. Mechanics of Laminated Composite Plates - Theory and Analysis, CRC Press Inc., Boca Raton, 1997.

[3] Plantema F.J., “Sandwich Construction”, John Wiley & Sons, New York, 1966.

[4] Allen H.G., “Analysis and Design of Structural Sandwich Panels”, Pergamon Press, Oxford, 1969.

[5] Zenkert D. and Olsson K.-A., “DP-Sandwich – the Utilization of Thin Steel Sheets in Compression”, Thin-Walled Structures, Vol. 7, 1989, pp 99-117.

5.42 BUCKLING AND FREE VIBRATION OF SANDWICH BEAMS

Exercises Example 5.1 Why can't the ordinary Euler load be used to predict buckling load of a sandwich beam? Ans. Transverse shear deflections are not accounted for in the Euler load.

Example 5.2 Draw schematically the buckling load as function of length for a sandwich column. Ans. See Fig.6.6

Example 5.3 Derive an expression for the buckling load of a simply supported sandwich column with length L, bending stiffness D, and shear stiffness S. Give the limiting values for L → 0, L → ∞, D → 0, and S → 0. Ans. See section 5.1

Example 5.4 A floor to a train car is made in a sandwich shall be designed. The dimensions of the floor is 26 by 3 metres and the analysis may hence be restricted to a simply supported beam with length L = 3 m, according to the figure. The floor must be able to carry a uniform pressure q = 6 kPa that corresponds to the weight of the passengers and equipment standing on the plus, multiplied with a safety factor. The eigen frequency of the floor must exceed 15 Hz, or 95 s-1. The core is PVC foam with a density 200 kg/m3 which has a modulus E = 200 MPa, a shear modulus Gc = 75 MPa and a shear strength τc,cr = 3 MPa. From sound and thermal insulation requirements the core must be 100 mm thick. The faces are made of carbon/vinylester laminates with a modulus E = 50 GPa, a Poisson ratio of 0.3 and a strength σf,cr = 600 MPa. Your job is to find the minimum required face thickness. You may assume that the rotary inertia can be neglected and that the moving mass (weight of panel plus the mass of the passengers) is approximately 250 kg/m2. Give all other made assumptions. q

L Hint: For the vibration analysis, assume mπx w(x,t) = W sin eωt L Ans. By inserting the deflection assumption into the Timoshenko beam equation one soon arrives at (the factor (1-ν2) equals 0.99 and may be omitted in D) 4 2 ⎛ mπ ⎞ ρ * Dω 2 ⎛ mπ ⎞ D⎜ ⎟ − ρ *ω 2 − ⎜ ⎟ = 0 ⎝ L ⎠ S ⎝ L ⎠ which when solved for ω becomes 2 D E f t f d ω = m2π 2 = m2π 2 m 2 E t d ⎡ D ⎛ mπ ⎞ ⎤ * 4 ⎛ 2 2 f f ⎞ ρ *L4 ⎢1+ ⎜ ⎟ ⎥ 2ρ L ⎜1+ m π ⎟ ⎜ 2G ⎟ ⎣⎢ S ⎝ L ⎠ ⎦⎥ ⎝ c ⎠ assuming this faces and a weak core. The minimum value for this is given by m = 1. By using the numbers in the * 2 problem (with ρ = 250 kg/m and making sure everything is in SI-units) and solving for tf gives that tf should be approximately 10 mm.

The local buckling strength is

σ w ≈ 0.5 E f EcGc = 450 MPa The maximum stress is

M max σ max = = 6.75 MPa for tf = 10 mm, hence no problem! t f d

5.43 AN INTRODUCTION TO SANDWICH STRUCTURES

The maximum shear stress in the core is T τ = max = 0.09 MPa, which is no problem either! c,max d Thus, the problem is solely one of stiffness!

Example 5.5 A simply supported sandwich beam is subjected to an in-plane compressive force P. Calculate the buckling load for the beam using Ritz' method (energy method). D,S

Use the deflection assumption mπx 1 1 1 w(x) = ()Wb +Ws sin and = + L P Pb Ps Hint: the integrals L mπx L L mπx L ∫ sin 2 dx = , ∫ cos 2 dx = 0 L 2 0 L 2 Ans. The strain energy of the beam can be written as 2 L ⎡ ⎛ 2 ⎞ 2 ⎤ 1 ⎢ ⎜ d wb ⎟ ⎛ dws ⎞ ⎥ U se = D + S⎜ ⎟ dxdy 2 ∫ ⎢ ⎜ dx 2 ⎟ ⎝ dx ⎠ ⎥ 0 ⎣ ⎝ ⎠ ⎦ L 4 L 2 1 2 ⎛ mπ ⎞ 2 mπx 1 2 ⎛ mπ ⎞ 2 mπx = ∫ DWb ⎜ ⎟ sin dx + ∫ SWs ⎜ ⎟ cos dx 2 0 ⎝ L ⎠ L 2 0 ⎝ L ⎠ L 4 2 DW 2 L ⎛ mπ ⎞ SW 2 L ⎛ mπ ⎞ = b ⎜ ⎟ + s ⎜ ⎟ 4 ⎝ L ⎠ 4 ⎝ L ⎠ The potential energy of the applied load is then L 2 L 2 ⎡1 ⎛ ∂w ⎞ ⎤ P mπx PL ⎛ mπ ⎞ U = N ⎢ ⎜ ⎟ ⎥dx = − ()W +W 2 cos2 dx = − ⎜ ⎟ ()W +W 2 N x ∫ 2 ∂x 2 ∫ b s L 4 L b s 0 ⎣⎢ ⎝ ⎠ ⎦⎥ 0 ⎝ ⎠ The total potential energy is now

Π = Use + UN Differentiate with respect to Wb and Ws 4 2 ∂Π DWb L ⎛ mπ ⎞ PL ⎛ mπ ⎞ = ⎜ ⎟ − ⎜ ⎟ ()Wb + Ws = 0 ∂Wb 2 ⎝ L ⎠ 2 ⎝ L ⎠ 2 2 ∂Π SWs L ⎛ mπ ⎞ PL ⎛ mπ ⎞ = ⎜ ⎟ − ⎜ ⎟ ()Wb + Ws = 0 ∂Ws 2 ⎝ L ⎠ 2 ⎝ L ⎠ which can be simplified into 2 ⎛ mπ ⎞ PWb DWb ⎜ ⎟ − P()Wb +Ws = 0 and SWs − P(Wb +Ws ) = 0 → Ws = ⎝ L ⎠ S − P which when inserted into the first equation yields that 2 ⎛ mπ ⎞ 2 D⎜ ⎟ ⎛ mπ ⎞ ⎛ P ⎞ L DW − PW 1+ = 0 giving that P = ⎝ ⎠ b ⎜ ⎟ b ⎜ ⎟ 2 ⎝ L ⎠ ⎝ S − P ⎠ D ⎛ mπ ⎞ 1+ ⎜ ⎟ S ⎝ L ⎠

5.44 CHAPTER 6

FACE WRINKLING

Several investigations have come to the conclusion that the effect of the core transverse deformability is negligible on the bending of sandwich beams, except for very extreme cases. The same goes for overall buckling of sandwich columns. However, there remains one case where the transverse normal stiffness of the core has an important influence and this is called face wrinkling or local buckling. The wrinkling phenomenon is a form of local instability of the faces associated with short waves of buckling of the faces. It may be viewed as buckling of a thin column (the faces) supported by a continuous elastic medium (the core). However, for sandwiches with corrugated or honeycomb cores, the core does not constitute a continuous support, which implies that, the buckling wave length must equal that of the length between the supports, e.g., the cell size of the honeycomb. This will be discussed later. Wrinkling may not only occur when the sandwich as whole is subjected to a compressive load, but also in ordinary bending as one of the faces will be in membrane compression. The wrinkling can occur in two ways, as illustrated in Fig.6.1. z

Anti-symmetrical Symmetrical

Figure 6.1 Wrinkling instability The displacements of the faces are transmitted to the core in which they damp out rapidly in the thickness direction. If the core thickness exceeds a certain value the mid-plane of the core remains flat for both the symmetrical and the anti-symmetrical case. Several different derivations of the wrinkling load will follow; first the simplest one based on a Winkler foundation model, then two improved ones based on an energy method and finally one based on solving the governing differential equation. Although the approaches are different they yield approximately the same result, apart from the first one, and the differences are mainly due to different assumptions on the decay of the transverse stress.

6.1 AN INTRODUCTION TO SANDWICH STRUCTURES

6.1 Winkler Foundation Approach The first way to approach the wrinkling problem is to use some foundation model, and the simplest one is the so called Winkler foundation model. This assumes that the support material (the core) consists of an array of continuously distributed linear springs, as shown in Fig.6.2.

Anti-symmetrical Symmetrical

Figure 6.2 Winkler foundation for wrinkling In the anti-symmetrical case it is seen that the springs remain unloaded even after wrinkling. It is further seen that the mode of deformation in the core is shear rather than tension/compression which the set of springs are unable to model and hence no solution can be derived in this case. In the symmetrical case, on the other hand, the model becomes more realistic since the mode of deformation in the core is both tension/compression and shear. In this case it assumes a core with some modulus perpendicular to the faces (Ecz) but with no out-of-plane shear stiffness, i.e., Gcxz = 0 The Winkler model then suggests that the core stress due to the wrinkling of the face is proportional to the face sheet deformation, and may be written as

σ zz(,)x 00=−K w (,)x (6.1) but since the springs are assumed to be linear, the core stress must be independent of z, and thus εz is constant from which is follows that w(x,z) can be written as A(x)z + B(x). Next use that w must be zero for z = tc/2 (see Fig.6.2) we get that w(x,z) = C(x)[z - tc/2] = wf(x)[1-2z/tc].

Assuming further that one may treat the set of springs as a continuous media so that εz = dw/dz and σz = Eczεz, we may then for z = 0 write

tc 2Ecz σ zz(,)xKwx0 =−0 () =Ewcz f () x so that Kz = (6.2) 2 tc The governing equation following this approach is that of a beam subjected to compressive end forces and a distributed load σz as shown in Fig.6.3.

dMx Mx+ M Tx dx x dN N N + x x x dx

σ dTx z Tx + dx

Figure 6.3 Forces, bending moments and stresses acting on the face sheet. By treating the face sheet as an ordinary beam or strut and neglecting transverse shear deformations (a most reasonable assumption for a thin composite or metal face), the governing equation for the beam is derived from the two equilibrium equations for the face dM dT dw2 x −=T 0 and x ++σ N =0 dx x dx zxdx 2

6.2 LOCAL BUCKLING OR WRINKLING PHENOMENA

2 2 leading to the differential equation (also using that d w/dx = −Mx/Df and taking the in-plane load as compressive, i.e., Nx = −P)

4 2 d w Pf d w K z 4 + 2 + w = 0 (6.3) dx D f dx D f

This is solved by assuming some appropriate function for the deflection field w that fulfils eq.(6.3). One such is simply πx wW= sin (6.4) l

Inserting this into eq.(6.3) yields an expression for Pf as

2 2 ⎛π ⎞ ⎛ l ⎞ Pf = D f ⎜ ⎟ + K z ⎜ ⎟ (6.5) ⎝ l ⎠ ⎝π ⎠

To find the critical load one must further minimise Pf with respect to the unknown wavelength l, which gives

2 4 dPf 2D f π 2K zl 4 D f π = − 3 + 2 = 0 → l = dl l π K z which inserted into eq.(6.5) gives the critical load as

3 E fxt f Ecz EtEfx f cz Pf ,cr = 2 D f K z = 2 or σcr = 0. 8165 (6.6) 6tc tc

This approach suffers from the drawback that the shear in the core is neglected. This is only reasonable either if the core truly has a very low shear modulus, or if the wavelength of the wrinkling is sufficiently long. In all other cases, there will be some shear lag effect smoothing out the stress σz so that it decreases with increasing z. The use of eq.(6.6) may therefore be somewhat unreliable.

6.2 Hoff's Method The next method to assess the wrinkling problem is that of Hoff [1]. In this case the shear stress in the core is accounted for and the shear lag is modelled by a linear decay function. Study Fig.6.4 showing an assumed shape of a perturbed face. l P Pf f

wf x

z h σz

Figure 6.4 Assumed shape of a symmetrical wrinkling. The deformation of the face induces tensile and compressive stresses in the core perpendicular to the face. If the wavelength, l, is short it would hardly effect the material in

6.3 AN INTRODUCTION TO SANDWICH STRUCTURES the middle of the core and hence it is assumed that the core is only affected in a small zone with depth h. Since the buckling is symmetrical the core deforms only in the z-direction. Assuming the face to undergo a sinusoidal displacement, and that the wave damps out linearly (linear decay) with coordinate z, we can then write W()h − z nπx W()h − z πx w = sin = sin (6.7) h L h l where L is the length of the beam or strut and l the wavelength of the wrinkles (see Fig.6.4) and that n is allowed to be a large number. Assuming the core to have low (or zero) in-plane modulus the tensile/compressive stress in the core can be written σz = Ec ∂w/∂z and if the deflections are in the z-direction only the core shear stress is τxz = Gc ∂w/∂x. Hence, the strain energy stored in the core over the length l equals

lh lh 2 2 2 1 2 1 2 E W l G π W h U = σ dxdz + τ dxdz = c + c se,c ∫∫ c ∫∫ c 2Ec 00 2Gc 00 4h 12l

The strain energy stored in the face due to bending is

2 D l ⎛ d 2w ⎞ W 2π 4t 3 E U = f ⎜ f ⎟ dx = f f se, f 2 ∫ ⎜ dx2 ⎟ 48l 3 0 ⎝ ⎠

The work done by the applied load is (see eq.(4.38))

2 1 l ⎛ dw ⎞ W 2π 2 W 2π 2t U = − P ⎜ f ⎟ dx = − P = − f σ p ∫ f ⎜ ⎟ f f 2 0 ⎝ dx ⎠ 4l 4l

The energy equation Π = Uf + Uc + Up can now be solved. The first minimisation yields

E W 2l G π 2W 2 h W 2π 4t 3 E W 2π 2t Π = c + c + f f − f σ 4h 12l 48l 3 4l f

∂Π E Wl G π 2Wh Wπ 4t 3 E Wπ 2t = c + c + f f − f σ = 0 dW 2h 6l 24l 3 2l f for σf which after some rearrangement results in

2 E l 2 hG π 2 E ⎛ t ⎞ σ = c + c + f ⎜ f ⎟ (6.8) f 2 ⎜ ⎟ π t f h 3t f 12 ⎝ l ⎠

The critical stress in this equation depends on the parameters l and h. The values of these are those that minimises the critical stress. Therefore, they are found by

2 2 2 dσ f Ec l Gc dσ f 2Ecl π E f t f = − 2 2 + = 0 and = 2 − 3 = 0 dh π t f h 3t f dl π t f h 6l

Solution of the two equations are

6.4 LOCAL BUCKLING OR WRINKLING PHENOMENA

2 h EEf c l E f 3 6 = 091.2 and = 165. (6.9) t f Gc t f EGcc

Substitution into eq.(6.8) yields the critical face stress as

3 σ f ,cr = 0.91 E f EcGc (6.10)

Now, this formula is only correct when the zone h, as given in eq.(6.9), is smaller than half the core thickness tc/2. If not, the assumed displacement field in eq.(6.7) must be changed so that tc/2 is substituted for h and the same derivation as outlined above is again performed. This will result in the following relations

⎛ ⎞ l Etfc E f Ect f tc = 142.4 and σ = 0.817 + 0.166G ⎜ ⎟ (6.11) f ,cr c ⎜ ⎟ t f Etcf tc ⎝ t f ⎠ As seen, eq.(6.11) gives the same result as the Winkler model in the previous section but with an extra term for the shear part. This is not surprising since even the Winkler model assumes some linear decay of the deformation and that w = 0 for z = tc/2. Hence, eq.(6.10) should be used for symmetrical wrinkling with thick cores and eq.(6.11) for thin cores.

By making a similar analysis of the anti-symmetrical case the following results are obtained

[1]. For h

2 h EEf c l E f 3 6 = 15.2 and = 215. (6.12) t f Gc t f EGcc

⎛ t ⎞ σ = 0.513 E E G + 0.33G ⎜ c ⎟ f ,cr f c c c ⎜ ⎟ ⎝ t f ⎠ and for h = tc/2 (for thin cores)

⎛ ⎞ l Etf c E f Ect f tc = 167.4 and σ = 0.59 + 0.387G ⎜ ⎟ (6.13) f ,cr c ⎜ ⎟ t f Etc f tc ⎝ t f ⎠ In order to illustrate the consequence of these formulae, let us take an example. Take a sandwich with aluminium alloy faces, i.e., Ef = 70,000 MPa, and a core with Ec = 100 MPa and Gc = 40 MPa. Now plot eqs.(6.10-6.13) as a function of tc /tf. First find the validity of these formulas using eqs.(6.9) and (6.12) to find that eq.(6.10) is valid for tc /tf > 30 and hence eq.(6.11) for tc /tf < 30 in the symmetrical case and eq.(6.12) when tc /tf > 49 making eq.(6.13) valid for tc/tf < 49 in the anti-symmetrical case.

6.5 AN INTRODUCTION TO SANDWICH STRUCTURES

2000 σf,cr [MPa] anti-symmetrical 1600

1200

800 symmetrical

400

0 ttc / f 0 20406080100

Figure 6.5 Critical wrinkling stress. The vertical lines corresponds to the limits tc /tf = 30 and 49 respectively.

As seen in Fig.6.5, anti-symmetrical buckling is more likely to occur when the ratio t c /tf <17 and the buckling mode would be symmetrical for t c /tf >17. From this it can also be concluded that the very simple formula eq.(6.10) is a conservative estimate of the buckling load for all cases. Hoff [1] found that the prediction of the buckling mode agreed very well with tests but suggested that in practical cases the load should be predicted using the conservative formula

3 σ f ,cr = 0.5 E f EcGc (6.14)

As a result of this, the wrinkling load will therefore be independent of the sandwich geometry and a function only of face and core properties. It also follows that the core will have the most influence on the wrinkling load with improving load-bearing capacity for cores with high elastic properties. The reason for using the conservative formula in eq.(6.14) is mainly due to the effect of initial irregularities. Derivations of this effect is performed in both [2] and [3] concluding that the effect has a maximum when the irregularities have a wavelength equal to 2L. In practical cases, initial irregularities are likely to reduce the wrinkling strength to about 80% of the theoretical [3]. Anyway, eq.(6.14) has proved to be one of the most useful formulas in the design of structural sandwich constructions since it shows very good agreement with both more sophisticated analytical formulae and also with experimental results.

6.3 Exponential Decay Shear lag problems are usually described by some exponential functions in hyperbolic sines or cosines. Therefore, another deflection function was used in [3] for the stress decay assuming the waves to damp out exponentially (exponential decay) as nπx πx wWe==−−kzsinWe kz sin (6.15) L l The following derivation based on this assumption also assumes symmetrical wrinkling, or rather that the buckling of one face has no effect on the face sheet on the other side. Using this function in the same way as outlined above, which is also the one-dimensional case of the procedure in section 6.5, results in the total energy as

6.6 LOCAL BUCKLING OR WRINKLING PHENOMENA

π 4W 2 D E W 2 kl G π 2W 2 P W 2π 2 Π = f + c + c − f 4l 3 8k 8kl 4l Minimising with respect to W leads to an expression for the load as

⎡ 4 2 2 ⎤ ∂Π π D f Ec kl Gcπ Pf π = W ⎢ 3 + + − ⎥ = 0 dW ⎣⎢ 2l 4k 4kl 2l ⎦⎥ which has two solutions; either W = 0 (trivial solution) or

π 2 D E kl 2 G P = f + c + c f l 2 2π 2 2k and by then letting ∂P/∂l = ∂P/∂k = 0 one soon arrives at

2 2 ∂Pf Ecl Gc 2 Gcπ = 2 − 2 = 0 → k = 2 ∂k 2π 2k Ecl

2 4 ∂Pf 2π D f Ec kl 2D f π = − 3 + 2 = 0 → k = 4 ∂l l π Ecl

This gives that

1/ 3 1/ 6 ⎛ G 2 ⎞ ⎛ 4D 2 ⎞ k = ⎜ c ⎟ and l = π ⎜ f ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2D f Ec ⎠ ⎝ EcGc ⎠

Inserting this into the expression for Pf yields that 3 P = 3 2D E G ⇒ σ = 0.853 E E G for ν = 0.3 (6.16) f ,cr 2 f c c f ,cr f c c f As seen, this is very close to the result obtained by Hoff [1] for the symmetrical case and thick core (see eq.(6.10)). In fact, the only difference is that another decay function has been used, otherwise the assumptions are the same. For anisotropic materials, the modulus of the face should be that which is included in the bending stiffness of the face, i.e., Efx and that the moduli of the core should be those corresponding to out-of-plane stresses, i.e., Ecz and Gcxz, respectively [4].

6.4 Differential Equation Method The following derivation assumes a thin beam (the face) under compressive end loads, continuously supported by an elastic medium that extends infinitely on one side of the beam (Fig.6.2 but with h very large). Hence, the face is unaffected by the opposite face. The differential equation for a homogeneous beam (the face) according to eq.(4.19) with Nx = –Pf is dw4 dw2 D +=P q (6.17) f dx 4 dx 2

6.7 AN INTRODUCTION TO SANDWICH STRUCTURES

Now, the transverse load q equals σz. Assume that the core is isotropic and the same shape of the face as above, i.e., wf = W sin(πx/l). By using Airy's stress function the stress, σc, necessary to deform the core in this manner may be written [2]

a πx 2πEc σc =− W sin , with a = . (6.18) l l ()()31−+ννcc In fact, the stress function used to derive this result contains an exponential decay. By substitution into eq.(6.17) and dividing by W sin(πx/l) one arrives at

4 2 ⎛π ⎞ ⎛π ⎞ ⎛ a ⎞ D f ⎜ ⎟ − Pf ⎜ ⎟ = −⎜ ⎟ ⎝ l ⎠ ⎝ l ⎠ ⎝ l ⎠

3 Now, with DEtfff= 12 and Pf = σf tf and after some rearrangement

2 2 π Etff⎛ ⎞ al⎛ ⎞ σ = + ⎜ ⎟ (6.19) f ⎜ ⎟ 2 ⎜ ⎟ 12 ⎝ l ⎠ π ⎝ t f ⎠

The critical stress is hence a function of l /tf and is found by taking dσf /d(l /tf) = 0. This gives

13/ 13/ l ⎛ ()()31−+ννcc⎞ ⎛ E f ⎞ = π⎜ ⎟ ⎜ ⎟ and ⎝ ⎠ t f 12 ⎝ Ec ⎠

1/ 3 ⎛ E E 2 ⎞ σ = 3⎜ f c ⎟ , with B = 12(3 −ν ) 2 1+ν 2 (6.20) f ,cr ⎜ ⎟ 1 []c ()c ⎝ B1 ⎠

This gives for νc = 0.3 and assuming the core to be isotropic with Gc = Ec /2(1+νc )

3 σ f ,cr = 0.78 E f EcGc (6.21) which is very close to the result in eq.(6.10) and (6.16) and also the proposed formula in eq.(6.14).

The above analyses have been derived for a state of plane stress, which is legitimate provided that the column is narrow. When it is wide, the column must be treated as a panel and a state of plane strain predominates. In an analysis similar to that of the column but for panels, Norris et al [5] derived the expression for the wrinkling stress as

E f Ec 3 σ f ,cr = Φ 2 2 Gc (6.22) (1−ν f )(1−ν c )

where the constant Φ equals 0.72 for νf = 0.3 and νc = 0.2. This indicates that the formula experimentally verified by Hoff [1], eq.(6.14), can be used even for panels, but with the E substituted by E/(1–ν2). In fact, the tests verifying this formula were performed on panels under uni-axial compressive loads.

6.8 LOCAL BUCKLING OR WRINKLING PHENOMENA

In practice, the actual wrinkling failure will occur as one buckle becomes unstable. Whether it buckles into or out of the core depends on the strength of the core and/or the adhesive joint. Buckling into the core occurs if the compressive strength of the core is lower than the tensile strength of the core or the adhesive joint and buckling outwards breaking the core or the adhesive joint if the other way. If the wrinkling stress exceeds the yield strength of the face it is customary to replace Ef by a reduced modulus just as in the overall buckling case in section 5. One such reduced modulus proposed by Norris et al. [5] is

4E ftE Er = 2 (6.23) ()EEf + t where Et is the tangent modulus. The application of eq.(6.14) then requires a trial-and-error operation to find the correct wrinkling stress. A more conservative approach would be to use

Er = Et.

The actual wrinkling load is very much affected by any initial waviness of the face. Some design guidelines to account for such effects are given in [6] where it is found that the factor proceeding the cube root expression in eqs.(6.10), (6.14), (6.16) and (6.21) decreases from between 0.5 - 0.9 down to such low numbers as 0.05 to 0.1 for very high waviness. However, waviness in the faces is usually carefully avoided, not only because of a reduction in the wrinkling stress, but mainly because a smooth surface is a sought after characteristic when using a sandwich design.

6.5 Wrinkling under Biaxial Load For a plate case one must consider deflections in both x- and y-directions. In this case, assume that the deflection of the face can be written as [3] πx πy wWf = sin sin (6.24) lxyl where lx and ly are the assumed wavelengths of the buckling in the x- and y-directions, respectively. The following analysis can now be performed as outlined in section 6.1, that is, assume a deformation field for the core, derive the energy of the system, and then find the critical values of lx, ly and the constant assumed for the deformation field. By then again -kz assuming an exponential decay as in eq.(6.15) [3], i.e., wc = wfe and that the core is infinitely thick (wc dies away so rapidly that this assumption usually is fulfilled). If the face undergoes vertical displacements only and the core has low in-plane moduli (e.g., honeycomb) we have that ∂w ∂w ∂w ∂u ∂v σ ==E ,, τ G τ =G , since == 0 cz cz∂z cxz cxz∂x cyz cyz ∂y ∂z ∂z

The strain energy in the core under the surface lx,ly equals (since σz= Ec ∂w/∂z, τxz= Gcx ∂w/∂x, and τyz= Gcy ∂w/∂y)

l 2 2 2 2 2 2 lx y ∞ ⎡ ⎤ ⎡ ⎤ 1 σ τ τ W 2 π L π L U = ⎢ z + xz + yz ⎥dxdydz = ⎢k l l E + y G + x G ⎥ se,c 2 ∫∫∫⎢ E G G ⎥ 16k ⎢ x y c l cx l cy ⎥ 000⎣ c cx cy ⎦ ⎣ x y ⎦

6.9 AN INTRODUCTION TO SANDWICH STRUCTURES

The strain energy in the square lx,ly of an orthotropic face is (see chapter 8.8)

2 l ly ⎡ 2 2 2 1 x D ⎛ ∂ w ⎞ ν D +ν D ⎛ ∂ w ⎞⎛ ∂ w ⎞ U = ⎢ fx ⎜ ⎟ − yx fx xy fy ⎜ ⎟⎜ ⎟ se, f ∫∫⎢ ⎜ 2 ⎟ ⎜ 2 ⎟⎜ 2 ⎟ 2 00 1−ν xyν yx ∂x 1−ν xyν yx ∂x ∂y ⎣⎢ ⎝ ⎠ ⎝ ⎠⎝ ⎠ 2 2 2 2 ⎤ D ⎛ ∂ w ⎞ ⎛ ∂ w ⎞ + fy ⎜ ⎟ + 2D ⎜ ⎟ ⎥dxdy ⎜ 2 ⎟ fxy ⎜ ⎟ ⎥ 1−ν xyν yx ∂y ∂x∂y ⎝ ⎠ ⎝ ⎠ ⎦⎥ 4 2 π W l l ⎡ D D 2D +ν D +ν D ⎤ x y ⎢ fx fy fxy yx fx xy fy ⎥ = 4 + 4 + 2 2 8(1−ν xyν yx ) ⎢ ⎥ ⎣ lx l y lx l y ⎦

3 where, e.g., Dfx = Efxtf /12, the flexural rigidity of the face alone in the x-direction and νxy and

νyx are the Poisson's ratios of the face. Finally, the potential energy of the external forces are (see chapter 8.8)

l 2 2 2 2 lx y ⎡ ⎤ 1 ⎛ dw ⎞ ⎛ dw ⎞ π W ⎛ Pxl y Pyl x ⎞ U = − ⎢P ⎜ ⎟ + P ⎜ ⎟ ⎥dxdy = − ⎜ + ⎟ p 2 ∫∫ x dx y ⎜ dy ⎟ 8k ⎜ l l ⎟ 00⎣⎢ ⎝ ⎠ ⎝ ⎠ ⎦⎥ ⎝ x y ⎠

* 2 Introducing the stiffness Df = [Dfx+Λ Dfy+Λ(2Dfxy+νyxDfx+νxyDfy)]/(1−νxyνyx) and Px = Pf, λ = 2 Py/Px, Λ = (lx/ly) , lx = l, Γ = Gcx/Gcy, the following equation can be derived from the condition ∂Π/∂W = 0.

π 2 kL2 E G P (1+ Λλ) = D* + c + cx (1+ ΛΓ) (6.25) f L2 f 2π 2 2k

The critical load is found by letting ∂Pf/∂l = ∂Pf/∂k = 0, which gives

16 13 ⎛ * 2 ⎞ 2 2 4Df ⎛ G ()1 + ΛΓ ⎞ L = π⎜ ⎟ and k = ⎜ cx ⎟ ⎜ ⎟ ⎜ * ⎟ ⎝ EGccx()1 + ΛΓ ⎠ ⎝ 2EDc f ⎠

By substituting these into eq.(6.25) and assuming λ < 1 the critical load can be found as

3 (1+ ΛΓ)1/ 3 P = 3 2D* E G f ,cr 2 f c c 1+ Λλ

* If the faces can be considered isotropic then the relation above reduces to (since Df = 2 Df(1−Λ )) 3 ()11++ΛΓ13() Λ 23 ()11++ΛΓ13( Λ)23 PDEG= 3 2 = P* cr 2 f cc 1 + Λλ 1 + Λλ where P* is the buckling load of the column as given in eq.(6.16). If both the faces and core are isotropic, then 1+ Λ P = P* f ,cr 1+ Λλ

6.10 LOCAL BUCKLING OR WRINKLING PHENOMENA

Finally, the condition ∂Π/∂Λ = 0 (or ∂Pf/∂Λ = 0) yields that a solution can only be found for the load ratio λ = 1 and where Λ remains undetermined [3]. In essence, the conclusion is that * the critical load in the bi-axial case equals that of the uni-axial case, that is, Pf,cr = P as in eq.(6.16). This means that the local buckling instability in one direction is unaffected by any load in the transverse direction and that the smallest load P*, in either x- or y-direction, should be taken as the critical load. In practice, one should instead use the buckling stress stated in eq.(6.14) calculated in both x- and y-directions and use the smaller of the two as the design load.

6.6 Wrinkling under Multi-Axial Load The results from section 6.2 to 6.4 are only valid for beams or plates under one-dimensional loading and as seen in the previous section there are indications that even the bi-axial loading case for a plate can be treated by simply comparing the lowest wrinkling load in any of the two main directions. For multi-axial loading, i.e., including all edge forces Nx, Ny and Nxy, and generally anisotropic sandwich plates, no solution or solution procedure is known. In [6] some recommendations is given which may help the designer to pursue this task and this is stated as follows;

1. Calculate the principal stresses σ1 and σ2, either by means of analytical or numerical methods. 2. If only one of the principal stresses is in compression, then neglect the tension stress and treat the problem as one-dimensional, i.e., use formulae from sections 6.2 to 6.4. 3. If both principal stresses are in compression then use the interaction formula

3 σ 1 ⎛ σ 2 ⎞ + ⎜ ⎟ ≤ 1 with σ1 <σ2 (6.26) σ 1cr⎝ σ 2 cr ⎠

to determine the wrinkling strength. The critical stresses σ1cr and σ2cr are the one- dimensional wrinkling stresses calculated in the directions of the principal stresses.

One can argue over the validity of this kind of approach but is does in anyway provide some means of determining the wrinkling strength for a more general case. A scheme like the above is also very well suited for implementation in finite element codes or used as a constraint in optimisation programs. The most serious criticism against this scheme is that for highly anisotropic faces the directions for the principal stresses do not coincide with directions for the principal strains. Furthermore, the anisotropy also implies that the wrinkling strengths, here denoted σ1cr and σ2cr, may vary strongly with the in-plane direction. However, for isotropic and moderately anisotropic faces, the approach should be more accurate.

The findings brought by the analysis of the bi-axial case was that wrinkling in one direction is unaffected by any loading in the transverse direction. A simplified model to treat wrinkling of orthotropic sandwich panels subjected to multi-axial loading was proposed in [7]. The basis for this model is shown in Fig.6.6.

6.11 AN INTRODUCTION TO SANDWICH STRUCTURES

Nxy

1 x ϕ α y 2' Nx

Nxy

Figure 6.6 Definition of problem statement for the simplified wrinkling analysis under multi-axial loading. Solid lines slanted and angle ϕ from the x-axis represents the strip and the dashed lines, slanted and angle α from the x-axis represents the local axis of orthotropy of the laminate. Herein, we assume to study a small strip of the laminate at any angular position in the plate and consider its own inherent resistant to wrinkling and the compressive load acting along the strip. Based on this one can postulate that in a general case wrinkling will occur in the direction ϕ at which the applied load Pf(ϕ) first reaches the critical value Pf,cr(ϕ). The applied load may be found in any co-ordinate system by simple transformation of the applied loads

Nx, Ny and Nxy, and Pf(ϕ) is then simply taken as the compressive load in the studied direction

ϕ that act on one of the face sheets. The critical load Pf,cr(ϕ) is simply the one-dimensional wrinkling load as function of the rotation angle and is according to eq.(6.16) equal to

3 EczGcx'z (ϕ) 3 Pf ,cr (ϕ) = 2D fx' (ϕ) 2 (6.27) 2 1−ν c

Hence, both the applied load and the wrinkling load are angular functions and by scaling the applied load until it first reaches the critical load will give both the wrinkling load and direction. The critical load in eq.(6.27) thus follows the classical wrinkling load as derived in eq.(6.16) but is evaluated for an infinite number of angular strips in the laminate. The local bending stiffness Dfx’ can be obtained by means of transformation of the D11 term for the laminate and similarly for the core shear modulus, providing the core is orthotropic. In some sense, each strip is by this theory given its own inherent wrinkling load.

t The applied load on the sandwich is given by some applied load vector (Nx, Ny, Nxy) . This load vector can be transformed to any angular direction by means of

2 2 2 2 Pϕ = N x c + N y s − N xy 2sc or σ ϕ = σ x c + σ y s −τ xy 2sc (6.28)

The hypothesis forming the wrinkling criterion is that wrinkling occurs when the applied load in any direction Pϕ equals the critical uni-axial wrinkling load Pϕ,cr. Thus, by multiplying the applied load by an arbitrary load factor λ, we get that wrinkling occurs when

λPϕ ≥ Pcr,ϕ for any angle ϕ (6.29)

6.12 LOCAL BUCKLING OR WRINKLING PHENOMENA

The load factor λ is thus given the minimum value P λ = min cr,ϕ (6.30) ϕ Pϕ

An example of this is given in Fig.6.7 for a case of a sandwich panel with a unidirectional carbon fibre laminate on a foam core. The laminate is oriented 15 degrees from the loading.

Figure 6.7 Results from the analytical approach calculations on a sandwich plate with fibre angle α =

15°. The diagram shows Pϕ,cr, Pϕ. versus ϕ . An illustration of α is also included in the figure.

In some sense, one can now scale Pϕ until this line for the first time meets the line representing Pϕ,cr. This gives the critical buckling load, but also the direction for the wrinkling, which not necessarily appears in the direction perpendicular to the load. In this particular case, the wrinkling will occur at an angle of –19 degrees from the x-axis, or rather a strip oriented in this direction will first fail in wrinkling so that the wrinkling will occur perpendicular to this direction. This is shown in Fig.6.8.

6.13 AN INTRODUCTION TO SANDWICH STRUCTURES

Figure 6.8 Results from the analytical approach calculations on a sandwich plate with fibre angle α =

15°. Pϕ,cr and λPϕ versus ϕ for the calculated example. Included in the figure ϕ,cr and α are also illustrated. This procedure has been verified properly and has shown to give some fairly good but non- conservative estimates [7]. A photograph of a test of panel with a 15-degree off-axis unidirectional laminate subject to uni-axial compression is shown in Fig.6.9.

Figure 6.9 Wrinkling failure of 15-degree off-axis composite sandwich panel. Note the wrinkling failure not being perpendicular to the load direction [7] The model has also been tested for cases of bi-axial loading giving fair results and predicting trends accurately.

6.14 LOCAL BUCKLING OR WRINKLING PHENOMENA

6.7 Intercellular Buckling Honeycomb cores are highly anisotropic with very high modulus perpendicular to the faces and very low modulus in the plane of the panel, and corrugated cores have very high shear stiffness in one direction and almost negligible stiffness in the other. The effects of orthotropic cores have been investigated by Norris et al. [8] but the results are only applicable to honeycomb cores with a cell size much less than the wavelength of the wrinkles. In practice this may not be the case. But if this restriction is satisfied, the formulae of sections 6.2 to 6.4 can be used. However, when honeycomb or corrugated cores are used, a large part of the face is unsupported and buckling may occur in this region locally. This is called face dimpling or intercellular buckling.

A study of the buckling load for square cell honeycombs [9] leads to a critical stress that can be written in the form

2 ⎛ t ⎞ ⎜ f ⎟ (6.31) σ f ,cr = constant E⎜ ⎟ ⎝ a ⎠ where a is length of the side of the cell, as shown in Fig.6.10..

Figure 6.10 Schematic of honeycomb core. An empirical formula proposed in [10] for hexagonal honeycombs with constant cell size is

2 2E ⎛ t ⎞ σ = f ⎜ f ⎟ (6.32) f ,cr 2 ⎜ ⎟ 1−ν f ⎝ s ⎠ which was verified by tests using several different sandwiches with different face thicknesses and cell sizes. Ef may also here be substituted with the reduced modulus according to eq.(6.23) if the face stress exceeds the elastic limit.

tf l

Figure 6.11 Sandwich with corrugated core. For corrugated cores, such as the one shown in Fig.6.11, intercellular buckling could occur in the plate inscribed by the corrugation. This can be predicted using the buckling theory for an ordinary homogeneous plate that is

6.15 AN INTRODUCTION TO SANDWICH STRUCTURES

2 kπ 2 E ⎛ t ⎞ σ = f ⎜ f ⎟ (6.33) f ,cr 2 ⎜ ⎟ 12(1−ν f ) ⎝ l ⎠ where k is the buckling coefficient that depends on size of the plate, i.e., l/b (b – width) and the restraint, i.e., the edge conditions and the number wavelengths in the buckling mode. This can be studied in e.g. [11].

In a bi-axial case also the intracellular buckling stress can be estimated by the same type of interaction formula as in the wrinkling case, however, slightly modified to [6] σ σ 1 +≤2 1 (6.34) σ1crσ2 cr providing both principal stresses are compressive. Otherwise only a one-dimensional case should be treated.

6.8 Imperfection Induced Wrinkling A theory for the treatment of initial imperfections on the wrinkling stress was presented in [12]. It constitutes a fairly simple and straightforward approach on how to treat imperfections. As a matter of fact, the reason for the usual knockdown on the theoretical wrinkling stress formulae are commonly attributed to imperfection sensitivity. The uses the results from Allen [2] described in section 6.4. We may approach this by assuming a small local imperfection of some shape and amplitude. However, it is more convenient to assume that the imperfection has the same shape as the natural wrinkling pattern with an initial amplitude W0 and length l. This is perhaps not strict in a practical sense, but it would be a conservative estimate. Furthermore, this is the shape the panel “wishes” to assume leading to a minimum energy state. The initial shape of the sandwich face sheet is therefore described as

⎛πx ⎞ w0 =W0 sin⎜ ⎟ (6.35) ⎝ l ⎠

If we apply an in-plane compressive load to a strut or plate with an initial imperfection of this kind, we can use a linear approach to find the new total deformation wt as w w = 0 (6.36) t P 1− Pcr which is also valid for a strut or plate on an elastic foundation [2] and constitutes a first order adaptation to non-linear kinematics. The difference (or increase) in the wrinkling wave wm is the difference between the total wrinkling wave after loading wt and the initial wave w0

wm = wt − w0 (6.37)

The change in wavelength is thereby neglected, implying that the derived formulae for the failure load will give a somewhat non-conservative result. By using the relations described in section 6.4 the change of wrinkling waveform can be expressed as

6.16 LOCAL BUCKLING OR WRINKLING PHENOMENA

w P P ⎛πx ⎞ w = 0 − w = w =W sin⎜ ⎟ (6.38) m P 0 P − P 0 0 P − P l 1− cr cr ⎝ ⎠ Pcr

Thus, the equilibrium path for the imperfect sandwich is known and can be plotted in the P/Pcr-Wm space for different imperfection amplitudes (see examples depicted in Fig.6.12). Here Wm is the effective wrinkling amplitude, being the difference between the total wrinkling amplitude Wt and the initial imperfection amplitude W0 (compare with eq.(6.37)).

Figure 6.12 Equilibrium paths in the P/Pcr -Wm. space for different imperfection amplitudes. The next step is to consider points on this load-deformation path where the different constituent materials may fail. As previously pointed out, the local loss of stability is often considered as a failure criteria which is dependent on the material combination and the design variables. Usually wrinkling and face compression failure are investigated as two different failure modes and the lowest of the two is taken as the estimated failure load to be used as a design constraint for face sheet compression. The compressively loaded imperfect sandwich can fail in two distinct ways; face sheet failure or core/adhesive joint failure. In the case of a perfectly straight panel, the face sheet compression stress is simply when the far field compressive stress reaches the laminate compressive strength, since one assume no local face sheet bending. In the case of an initially imperfect panel, the deformation causes an additional curvature of the face sheet, giving a locally varying bending moment and bending stresses over the thickness of the face sheet. These would vary from tensile to compressive along the face sheet and over the thickness. One would not expect a core failure in a perfectly straight panel, but for one with initial imperfections, local tensile and compressive stresses, normal to the face sheet, will build up as a result of the increasing out-of-plane deformation and may cause core failure. The most critical areas in a compressed imperfect sandwich are where the core and face stresses are highest. It is quite easy to understand that the maximum stress in the core occurs at the face-core interface, at the location where the wrinkling wave reaches its

6.17 AN INTRODUCTION TO SANDWICH STRUCTURES peak or bottom, see (A) and (B) in Fig.6.13. The compressive stress in the face sheet is highest at the inside of the face sheet in point (A) or outside in point (B).

Figure 6.13 Critical areas in the compression loaded imperfect sandwich. The preferred case is often to have face failure occur as the critical failure mode, at a load level close to the pure compression failure load of the face sheet. Thus the expensive face sheet is used to its fullest and gives the most “value for the money spent”. The second failure mode where the core is the critical constituent is undesirable and can occur at a load levels much lower than the face sheet compression failure load if the combination of core and face material is not thought through properly. In theory it is possible to get core compression failure but in practice it is much more common to get core tension failure, which is quite abrupt and the face sheet virtually ruptures away from the core.

Face compression failure

The face sheet material is assumed to be able to sustain a certain compression strain εˆ f . Hooke’s law then gives the relation (P is the applied compressive load per unit width)

c P ε f = (6.29) t f E f and maximum load is calculated from

ˆ c P = t f E f εˆf (6.40)

c where εˆ f is the strain at which the face sheet fails in compression.

Imperfection introduced face bending failure The strain in the face sheet that arises from bending (wrinkling) varies both through the thickness of the face sheet and along the length of the face sheet. It can be described according to ordinary beam bending theory as

2 2 b M d w π π x ε f = − z = z 2 = z(wt − w0 ) 2 sin( ) (6.41) D f d x l l

The maximum local face strain in the bent configuration is obviously at the boundary of the face sheet ( z = ±t f / 2 ), and where the curvature of the face sheet has a maximum ( x = ±l / 2 ), see Fig.6.13. Hence the maximum local strain in the face sheet is

π 2t εˆb = f (w − w ) (6.42) f 2l 2 t 0

If the face sheet is not isotropic and, e.g. one of the outer plies consists of a high strain material, a different local coordinate z must be used. If for example a [90/0]s face sheet is

6.18 LOCAL BUCKLING OR WRINKLING PHENOMENA

studied z = ± tf/4 should be used in order to predict compressive failure in the 0-degree layer. This is in fact the “first ply failure” criterion that can be implemented fully in the described theory. In order to not complicate the following text and examples more then necessary, the first ply failure technique is not further mentioned but the reader should be aware of its existence. By using eqs.(6.38), eq.(6.40) can be rewritten as

2 b π t f P ε f = 2 w0 (6.43) 2l (Pcr − P)

Hence the total maximum face strain is the sum of the strain from the bent and straight configurations (eq.(6.39) and (6.41) respectively).

2 π t f P P ε f = 2 w0 + (6.44) 2l (Pcr − P) t f E f

Imperfection introduced core wrinkling failure The face-core interface normal stress is, according to Allen [2], given in eq.(6.18). The maximum interface normal stress is therefore equal to

a a 2π Ec σ z max = − wm = (wt − w0 ) with a = l l (3 − vc )(1+ vc ) so that the maximum strain in the core is

σ z a a P ε c = = (wt − w0 ) = W0 (6.45) Ec l Ec l Ec Pcr − P

Which gives the critical load with respect to core failure P Pˆ = cr (6.46) c aW 1+ 0 l Ecεˆc

And the critical effective imperfection amplitude l E εˆ Wˆ = c c (6.47) m a

The critical effective wrinkling amplitude is constant and unaffected by the initial imperfection amplitude and a function of the material properties and thicknesses alone.

Failure prediction procedure Using eq.(6.46) and (6.47) core failure can be directly evaluated while for the face sheet a different, iterative approach has to be used. Using eq.(6.38) we increase the applied in-plane load P to calculate the added deformation Wm for a given initial imperfection amplitude W0. At each load step, we check the face strain according to eq.(6.44) being superimposed by a far field compressive stress from the applied in-plane load and a local maximum bending stress resulting from the initial imperfection w0. Failure will then occur at the lower of the two loads

6.19 AN INTRODUCTION TO SANDWICH STRUCTURES predicted for the core and face respectively. This method obviously also provides means to determine which type of failure is most likely to occur.

Comparison with tests The results of the experiments compared to a traditional approach of comparing with the standard wrinkling formula, eq.(6.29) and with the face compression failure stress along with the present method is shown in Fig.6.14. The failure load is given as load per unit width of the face sheet. From the experimental results, the load carried by the core has been subtracted since this (though rather small contribution) is not included in the theory. The classical wrinkling load is given by dashed lines and so also the horizontal line corresponding to face compression failure load. From this classical approach, one can see that as the core modulus increases there is a failure mode shift from wrinkling to face sheet compression failure. The present theory is presented by two shaded areas; one for each face sheet configuration. The upper boundary corresponds to an initial imperfection of 0.01 mm and the lower boundary to an initial imperfection of 0.25 mm. In reality, the imperfection amplitude should lie within these boundaries.

Figure 6.14 Developed theory compared to test results. [0/90]s and [90/0]s carbon fibre vinylester face sheet on Divinycell H-grade core material. Black triangles pointing up shows test results for [0/90]s specimens and white triangles pointing down show test results for [90/0]s specimens. Dashed lines are traditional theory (wrinkling and compression failure) and solid lines are the developed theory. The upper solid line for each material configuration corresponds to an initial imperfection amplitude of 0.01 mm and the lower line in each pair to 0.25 mm. As can be seen from Fig.6.14, almost all test results lie within the assumed initial imperfection amplitude regime. This regime may appear relatively wide but it seems that the upper boundary (W0 = 0.01 mm) is closer to the test results. One can further notice that the traditional approach of estimating the wrinkling load or the face sheet compression failure load is, in almost all cases, non-conservative.

6.20 LOCAL BUCKLING OR WRINKLING PHENOMENA

The conclusions from this approach are multi-fold. However, the most important ones are as follows; As the support stiffness increases, the failure mode shifts from wrinkling to face sheet compression failure. This is, in itself, not that surprising. As we include initial imperfections, though simplified in this approach, the wrinkling load prediction decreases. Actually, the above approach does not inherently treat wrinkling, but material failure in a post-buckling regime similar to wrinkling. In any case, the theory is more conservative than wrinkling and face sheet compression predictions, and more so as the imperfection amplitude increases. What can be seen from Fig.6.14 shows rather clearly what the theory implies. When the wrinkling stress is considerably lower than the face sheet compression strength, the imperfection theory and the classical wrinkling theories are rather close, typically for low stiffness cores. At the other end, for high core stiffness, the theory approaches the load prediction given by face sheet compression failure. However, the intermediate regime is perhaps the most interesting. It would be obvious to design a sandwich plate for which the wrinkling stress is almost the same as the face sheet compressive strength, aiming at some sort of optimum material usage – optimum design. However, the theory also shows that in this particular regime, the panel has the highest imperfection sensitivity. We can perhaps equally well say the following; the knockdown factor required for classical wrinkling analysis is rather small when the wrinkling load is considerably smaller than the face sheet compression strength. The same knockdown factor appears to have a maximum when the wrinkling stress and face sheet compression strength are equal. Thus, what appeared to be an optimum design point may still be optimum but will not be a very robust point in the design space.

6.9 Summary of Buckling Phenomena In summary, the core acts as a support for the faces preventing them from buckling independently of each other, which would occur at very low stress level since the faces are usually thin. However, depending mainly on the properties of the core and of the face, the core will eventually be unable to sustain this support that will result in face wrinkling. This will, however, occur at a very high stress level. This phenomenon is an important feature of sandwich structures. One can argue that the very high compressive stresses in the faces are supported by much lower stresses in the core acting perpendicular to the face stresses but on a much larger area.

To illustrate all the buckling phenomena of sandwich beams, a schematic graph is shown in Fig.6.15 illustrating the buckling load of a sandwich beam in compression as function of beam length.

6.21 AN INTRODUCTION TO SANDWICH STRUCTURES

Pcr

Euler Buckling Thick Faces

Wrinkling

Sandwich Beam

Weak Core L

Figure 6.15 Critical buckling load for a sandwich beam as function of beam length. The equations schematically illustrated are; Euler buckling eq.(5.15b), the sandwich beam eq.(5.15b), thick faces eq.(5.42a), weak core eq.(5.42b) and the face wrinkling eq.(6.14). By studying the graph of Fig.6.15 several important facts can be noted: The use of the ordinary Euler load yields high non-conservative errors. The thickness of the faces only has an effect for non-slender columns, i.e., short or shear weak columns. If the core is weak the buckling load is very low except for very short columns. General or overall buckling depends greatly on the length of the column whereas the wrinkling load is invariable with geometry.

References [1] Hoff N.J. and Mautner S.E., “Buckling of Sandwich Type Panels”, Journal of the Aeronautical Sciences, Vol. 12, No 3, July 1945, pp 285-297.

[2] Allen H.G., Analysis and Design of Structural Sandwich Panels, Pergamon Press, Oxford, 1969.

[3] Plantema F.J., Sandwich Construction, John Wiley & Sons, New York, 1966.

[4] Grenestedt J.G. and Olsson K.-A., "Wrinkling of Sandwich with Layered Core or Non-Symmetric Skins", Proc. Third International Conference on Sandwich Construction, Southampton, 1995, EMAS Ltd, UK.

[5] Norris C.B., Ericksen W.S., March H.W., Smith C.B., and Boller K.H., “Wrinkling of the Facings of Sandwich Constructions Subjected to Edgewise Compression”, U.S. Forest Product Laboratory Report 1810, 1949.

[6] Sullins R.T., Smith G.W. and Spier E.E., "Manual for Structural Stability Analysis of Sandwich Plates and Shells", NASA-Contractor Report No. 1467, 1969.

[7] Fagerberg L. and Zenkert D., “Effects of anisotropy and multi-axial loading on the wrinkling of sandwich panels”, Journal of Sandwich Structures and Materials, Vol. 7, pp. 195-220, May 2005.

6.22 LOCAL BUCKLING OR WRINKLING PHENOMENA

[8] Norris C.B., Boller K.H. and Voss A.W., “Wrinkling of the Facings of Sandwich Constructions Subjected to Edgewise Compression – Sandwich Having Honeycomb Cores”, U.S. Forest Product Laboratory Report 1810A, 1953.

[9] Weikel R.C. and Kobayashi A.S., “In the Local Elastic Stability of Honeycomb Face Plate Subjected to Uniaxial Compression”, J. Aero/Space Sci., Vol. 26, No 10, Oct. 1959, pp 672-674.

[10] Norris C.B., "Short-Column Compressive Strength of Sandwich Construction as Affected by Size of Cells of Honeycomb Core Materials", U.S. Forest Service Research Note FPL-026, Forest Product Laboratory, Wisconsin, 1964. Also in Norris C.B. and Kommers W.J., U.S. Forest Product Laboratory Report 1817, 1950.

[11] Timoshenko S.P. and Gere J.M., Theory of Elastic Stability, 2nd Edition, McGraw- Hill, New York, 1961.

[12] Fagerberg L. and Zenkert D., ”Imperfection Induced Wrinkling Material Failure in Sandwich Panels”, Journal of Sandwich Structures and Materials, Vol. 7, pp. 177- 194, May 2005.

6.23 CHAPTER 7

FAILURE MODES AND DESIGN CRITERIA

Sandwich panels can fail in several ways, each one of these failure modes giving one constraint on the load bearing capacity of the sandwich. Depending on the geometry of the sandwich and the loading, different failure modes become critical and hence set the limits for the performance of the structure. The most important failure modes have already been treated in the analysis but will be highlighted again due to the importance of recognising them. Other failure modes that appear are of a more practical significance. The most common failure modes are schematically illustrated in Fig. 7.1.

(a) (b) (c) (d) (e) (f) (g) (h)

Figure 7.1 Failure modes in sandwich beams. (a) Face yielding/fracture, (b) core shear failure, (c and d) face wrinkling, (e) general buckling, (f) shear crimping, (g) face dimpling and (h) local indentation. There are quite a number of failure criteria for composite laminate face materials varying from quite simple to extremely complex criteria including allowables for several different modes of failure. Such will not be discussed here, but only the criteria most commonly used by engineers in design, sizing and analysis.

7.1 AN INTRODUCTION TO SANDWICH STRUCTURES

7.1 Formulae for Failure Loads (a) Yielding or fracture of the face in tension or compression Depending on the materials used and on the chosen fracture criterion one will consider the face or the core to have failed either if yielding occurs or if the component has actually fractured. Hence, for every material component there will be a maximum allowed stress, whether this stress is a yield or a fracture stress. The criterion for failure is then when the maximum stress in the component reaches this allowable stress. The maximum principal stress in the faces of a panel is according to Mohr's circle of stress [1]

12/ σσ+ ⎡ σσ− 2 ⎤ fx fy⎛ fx fy ⎞ 2 (7.1) σ f = + ⎢⎜ ⎟ + τ fxy ⎥ 22⎣⎢⎝ ⎠ ⎦⎥ since in the face sheet we have that σz = τxz = τyz = 0. In fact, the direct stresses in the faces are usually orders of magnitude higher than the shear stresses in the core and faces. If one uses metallic face sheets the onset of plastic deformations (yielding) may have to be predicted using some yield criterion, e.g. von Mises or Tresca. For a sandwich beam subjected to bending the failure criterion can be reduced to

MzEx fx MzEyfy σσfx =≥$ fx and in the y-direction σσfy =≥$ fy (7.2) Dx Dy which will remain the same for the faces even if a transverse force is applied to the beam. A similar criterion could be stated for the core, but such a failure criterion is very seldom used since most core materials have a higher yield and fracture strain than the faces, implying that tensile or compressive failure (or yielding) will occur in the faces long before anything happens to the core. Note also that the allowables may well differ between tensile and compressive modes, so that, e.g., even if the compressive face has a nominally lower stress level, it may be the first to fail if the compressive strength is lower than the tensile. Hence, the criterion must be used twice, once for each face. If the load is in-plane tension or compression, the criterion is simply

σ fx≥ σ$ fx and σ fy≥ σ$ fy (7.3)

(b) Core shear failure As pointed out in section 3, the core material is mainly subjected to shear and carries almost the entire transverse force. However, the direct stresses in the core could be of the same order of magnitude as the shear stresses. The maximum transverse shear stress in the core is

1/2 2 1/2 ⎡ 2 ⎤ ⎡⎛ σ cx ⎞ 2 ⎤ ⎛ σ cy ⎞ 2 τ = ⎜ ⎟ + ττ and = ⎢ + τ ⎥ (7.4) cxz ⎢⎝ ⎠ cxz⎥ cyz ⎜ ⎟ cyz ⎣ 22⎦ ⎣⎢⎝ ⎠ ⎦⎥ which is used as the fracture criterion. This allowable could now also be either a yield or a fracture stress. Assuming a beam with a weak core, Ec << Ef, then σcx = σcy = 0 and the maximum shear stress can be written as

7.2 FAILURE MODES

Tx Ty ττττcxz =≥$$cxz and cyz =≥cyz (7.5) d d

And, in fact, in many practical cases the direct stress σc is much lower than the shear stress reducing eq.(7.4) to approximately equal τc. This shear stress produces a positive direct stress, equal to τc, at a 45 degree angle from the x-direction which causes cracks inclined 45 degrees. Such cracks are typical of shear failure and are also usually called shear cracks.

(c and d) Face wrinkling Face wrinkling can in practical cases occur in a sandwich either when subjected to an in-plane compressive buckling or in the compressive face during bending, or in a combination of those. The criterion that was derived in section 6 is used as a failure criterion stating that wrinkling will occur in a face when the compressive stress in that face reaches the wrinkling stress suggested in eq.(6.11)

33 σ fx= 05.. EEG fx cx cx and σ fy= 05 EEG fy cy cy (7.6)

The more refined formulae derived in eqs.(6.6-6.10) or for the multi-axial cases in sections 6.5 and 6.6 can be used instead. Experimental work [2] has concluded that the above formula (eq.(6.7)) gives good and conservative predictions. The actual failure can occur in two ways: (i) a wrinkle that becomes unstable causes an indentation in the core if the compressive strength of the core is lower than the tensile strength of the core and the adhesive joint or (ii), the wrinkle causes a tensile fracture if the tensile strength of the core or the adhesive joint is lower than the compressive strength of the core. This formulation is rather vague but is usually correct. Whichever case applies does not really affect the actual wrinkling stress, but in fact a poor adhesive joint will undoubtedly reduce the wrinkling stress of the sandwich.

(e) General buckling Although buckling itself sometimes does not damage a structure, it must still be avoided since a structure which has buckled may have lost its capability of fulfilling its purpose. The actual buckling load may also be the ultimate load bearing capacity of the sandwich since in its buckled shape it may not sustain any more load. Therefore, it is often stated that buckling must not occur and the buckling load hence becomes an allowable. In section 5, the critical buckling load for a beam was derived as

2π4DD π2D f + SL()β 4 ()βL 2 P = (7.7) cr π2D 1+ SL()β 2 for a general case with thick faces, where β depends on the boundary conditions. This expression can be reduced to and approximated by 111 π2 D =+ where Pb = 2 (7.8) PPPcr b s ()βL where β depends on the edge conditions of the beam and Ps is the shear buckling load equal to the shear stiffness S. The same type of equation can be used for plate buckling cases but with

7.3 AN INTRODUCTION TO SANDWICH STRUCTURES

different expressions for Pb and Ps (see section 10). In practice, a buckled sandwich usually retains its original state if unloaded, providing that the faces have not yielded, and hence the buckling has not damaged the structure. If, however, the buckling load equals the maximum load that can be applied to the structure, this means that in a dead load situation the structure has actually failed. If the deformation is controlled, on the other hand, the load will drop after buckling, and if the loading continues the deformation will increase until failure. This final failure can occur in several ways; (i) the face on the compressive side fails in compression, (ii) the face on the compressive side fails by wrinkling (see case c and d), or (iii) the structure fails by core shear fracture. The latter case occurs since as the deformation w increases so does the transverse force, Tx = S dws /dx. Eventually the transverse force has grown in some point to a level that will cause core shear fracture as stated above. This kind of failure mode appears just like the shear crimping in Fig.7.1g.

(f) Shear crimping The shear crimping failure is actually the same as the limit of the general buckling mode considering thin faces, i.e., when the critical load equals Ps = S [3]. The failure itself looks like that illustrated in Fig.7.1g and is shear instability failure. The critical face stress is hence S σ f = (7.9) 2t f in either x- or y-direction or for either the upper or lower face. As mentioned above, a failure of this kind is more likely to happen as a result of large out-of-plane deformations in a post- buckled state when the transverse forces build up due to the deformation. The failure will then appear where this transverse force has a maximum, that is, at the edges for a simply supported column, and at L/4 for a clamped column.

(g) Face dimpling As mentioned in section 6, another instability phenomenon that may occur in sandwich structures with honeycomb or corrugated cores is dimpling or intercellular buckling. For a square cell honeycomb this buckling stress equals [4]

2 ⎛ t f ⎞ σ = 25. E ⎜ ⎟ for ν = 0.3 (7.10) ff⎝ a ⎠ f where a is the length of the side of the cell. For hexagonal honeycombs, which are the most common ones in practice, the intercellular buckling may be estimated by [5]

2 2Etf ⎛ f ⎞ σ f = 2 ⎜ ⎟ (7.11) 1 − ν f ⎝ s ⎠ where s is the radius of the circle that can be inscribed in the honeycomb cell (see Fig.6.4).

For a beam νf is set to zero. For corrugated cores intercellular buckling occurs when [6]

2 2 kEπ f ⎛ tf ⎞ σ f = 2 ⎜ ⎟ (7.12) 12() 1 − ν f ⎝ l ⎠

7.4 FAILURE MODES where k is the buckling coefficient which depends on the size of the plate inscribed by the corrugation and l is the length of this plate in the loading direction. It has been suggested [3] that Ef can replaced by the reduced modulus of elasticity Er, according to eq.(6.12) if the stress exceeds the elastic limit of the face. The critical stress is then found by an iterative procedure between the buckling formula used and the stress-strain relation of the face. It has also been suggested [3] that orthotropic faces can be approximated by using an average elastic modulus Ef = EfxEfy.

(h) Core indentation Indentation of the core occurs at concentrated loads, such as fittings, corners, or joints. Practically they can be avoided by applying the load over a sufficiently large area. This area can be roughly estimated from [1] P A = (7.13) σ$ cz

^ where σcz is the compressive strength of the core material. What actually happens when point loads are applied is that the face will act as a plate on an elastic support. The face will bend independently of the opposite face and if the deformation and thus the elastic stress applied to the core exceeds the compressive strength of the core, the core will fail. This will be discussed in section 12. In practice, there are many ways to enhance the local strength of a sandwich to avoid indentation.

(i) Vibration In some cases there are constraints on the minimum allowed natural frequency. In moving structures there are often imposed movements within a given frequency range and it is then preferable to avoid having a natural frequency of vibration for the structural member lying within that range since if the imposed movement should coincide with the natural frequency the amplitude of deflection would increase to unwanted levels. The constraint would simply be

min ω ≥ ω$ (7.14)

7.2 Failure Mode Maps For a general load case one can write the maximum bending moment and the maximum transverse load as a function of the applied load as

Mmax== kMT PL and T max k P (7.15) where P is the applied load (which could also be a pressure q). The constants kM and kT depend on the loading configuration, e.g., for a cantilever beam loaded with a point load at its free edge, kM = 1 and kT = 1, whereas for a simply supported beam subjected to a uniform pressure (P = qL) kM = 1/8 and kT = 1/2. For plates similar expressions can be defined. Each of the above failure mode formulae contain three sets of variables [1]: those relating to the ^ ^ loading configuration (kM, kT and L), to the material properties (Ef, σf, Ec, Gc, τc) and to the design (tf, tc and maybe even ρc). It is now a question of determining how the failure mode depends on the design for a given combination of materials. The design at which two failure

7.5 AN INTRODUCTION TO SANDWICH STRUCTURES modes occur simultaneously is found by equating the formulae for these two modes. Thereby one can establish the transition between two failure modes.

Following the work in [1] we must first define the principal failure modes. Face yield and core shear are independent modes and of great importance to a structure. If we leave out the case of in-plane compressive loads causing buckling or shear crimping and look only at beams and panels subjected to transverse loads then the third failure mode of importance would be face wrinkling. Depending on the core material type we might have to consider wrinkling or face dimpling. As done in [1], let us use the former. Since tensile/compressive yield in the core is seldom of any practical importance that may be omitted as well. The variables used are, therefore, face and core thicknesses, and also the core density. One may use core density as a variable providing the core properties can be related to its density. One such is

n n m ECcEc==ρρτρ , GCcGc , and $ cc = Cτ (7.16) and other properties like the core tensile and compressive strength can be written in a similar manner. The constants relating the properties to the density can be extracted from tests but generally a curve fit to data from the material supplier should be satisfactory. We can then rewrite the equations for the three studied failure modes, face yield, core shear and face wrinkling to give the critical load as function of the sought variables tf, d and ρc as

σ$ fftd Face yield: P = (7.17a) kLM

Cdρ m Core shear: P = τ c (7.17b) kT

tdf 3 2n Wrinkling: P = ECCfEGcρ (7.17c) 2kLM The transition between the failure modes is now easily found by equating the loads given above, two by two, giving three transition lines in the failure mode map. Performing these exercises one arrives at the following relations between the core density and the ratio (tf / L)

Transition between face yield and core shear

1 σ$ f kT ⎛ t f ⎞ logρc = log ⎜ ⎟ (7.18a) m kCM τ ⎝ L ⎠

Transition between wrinkling and core shear

1 k ⎛ t f ⎞ logρ = log T 3 ECC ⎜ ⎟ (7.18b) c 2n 2kC f EG⎝ L ⎠ m − M τ 3 Transition between wrinkling and face yield

7.6 FAILURE MODES

3 2σ$ f logρc = log (7.18c) 3 2nECCf EG

To illustrate this take an example of a beam in three-point bending having aluminium alloy faces with an ultimate strength of 150 MPa. Thus, kM = 1/4 and kT = 1/2. Assuming a linear relation between the core properties and the core density (n = m = 1), and that CE = 1, CG =

0.40 and Cτ = 0.015, gives the failure mode map graphically represented in Fig.7.2. ρ log c 1000

Face Yield

100 1

m Core Shear

3 Face Wrinkling 3m-2n log( t /L ) 10 f 0.00001 0.0001 0.001 0.01 0.1

Figure 7.2 Failure mode map for a sandwich beam in three-point bending with aluminium faces and a core for which the properties vary linearly with core density.

7.3 Design criteria The design of a sandwich element is often an integrated process of sizing and materials selection in order to get not only a feasible design but also in some way an optimum design with respect to an objective, such as weight, strength, or stiffness. With the introduction of fibrous composites, the choice of face materials has become almost infinite in terms of mechanical properties. Core materials, especially foams, are now also available in wide ranges of densities and properties. All material systems have some advantages and some disadvantages implying that the choice of materials is determined by the objectives of the specific application and cannot be stated in general terms. These objectives are often practical, e.g., chemical or heat resistance, surface wear resistance, thermal insulation, or related to the manufacturing process. Therefore, materials are often, in the practical case, already defined by the service or manufacturing requirements of the structure. However, some material related properties can still be considered as variables even if the material itself is defined, e.g., the density of a specific core material. However, most material properties, as well as thicknesses, do not vary continuously but in discrete steps, for example, the number of plies of a composite laminate, the available thicknesses of sheet metal or of core materials. Anyway, it is convenient to have methods for design which can give an indication of a suitable starting point for the design process giving approximate thicknesses and maybe even materials selection.

The above formulae constitute the design constraints on a structure but often there are other constraints that appear in an application that are not directly of a mechanical nature but rather

7.7 AN INTRODUCTION TO SANDWICH STRUCTURES of a practical one. Depending on the application they may of course differ but some common constraints are

(i) minimum core thickness and specific core materials for thermal insulation purposes. (ii) minimum face thickness, and a combination of face and core materials for given impact resistance. (iii) specific face material and face thickness for surface wear resistance. (iv) specific face material for surface finish. (v) specific face material for environmental resistance. (vi) maximum total thickness for a volume requirement.

Furthermore, there are always some other requirements on the materials depending on the manufacturing or assembly method, working environment, waste or recycling, to mention some.

Methods can also be described for optimum design of sandwich constructions. In this context it means finding sizes and materials to give a minimum weight, maximum strength or stiffness or minimum cost with respect to one or several constraints. The constraints could be stiffness or strengths in different failure modes or a combination thereof. Simple methods of that kind will also serve as a good starting point for the design process. Since there are generally many different constraints on a structure an optimum may be hard to find with respect to all constraints without using general mathematical programming techniques. These are often both complex and costly. Only considering the most important constraints and using a simple optimisation technique could prove useful in the design process. The methodologies outlined in this section may therefore be called “intelligent sizing” rather than optimisation.

The most common and essential constraint of a structure is, apart from strength requirements, its stiffness. This constraint is often stated as a maximum deflection at some location of the structure. The compliance, the inverse of the stiffness, can, assuming thin faces and weak core, in general be written as Δ L33L 2L L L42L C ==kbsb +k =k 2 +ks or for a plate Ck=+bsk (7.19) P D S Etdff Gdc D S where Δ is the deflection (on which there is a constraint), L is some characteristic length (e.g., length of a beam or length of a side of a panel) and kb and ks are constants relating to the geometry and loading. For example, for a cantilever beam, kb = 1/3 and ks = 1. Strength and stability constraints are usually those given by the different failure modes described in section 7. Since the main feature of using sandwich construction is for weight saving purposes the objective is usually one that will minimise the total weight of the structure; this can be written

W = ρctc + 2ρftf ≈ ρcd + 2tfρf (7.20)

The variables in the design process apart from sizes, i.e., tf and tc, may be also materials. A variable that can be easily introduced is the core density, by letting the core properties vary with its density. This variation is supposed to be continuous and written as [7]

7.8 FAILURE MODES

n n m ECcEc==ρρτρ , GCcGc , and $ cc = Cτ (7.21) and other properties like the core tensile and compressive strength can be written in a similar manner. The constants relating the properties to the density can be extracted from tests or data from the material supplier. In reality, the choice of core material density is not continuous but discrete. However, this assumption is a great simplification in the analysis and the results should rather be interpreted as a good first approximation to serve as a starting point in the design process.

7.4 Determination of Thicknesses (i) Core thickness Consider a beam or a panel subjected to a transverse load and find the core thickness required to carry this load. Assume the materials are known and that the face thickness is already given. The structural member must fulfil all constraints placed on it. The ultimate face stress is not only given by the yield or fracture strength of the materials, but may instead be limited by the wrinkling strength or, if a honeycomb core is used, dimpling strength of the face in compression. Hence, from eqs.(7.2),(7.6) and (7.11)

2 ⎪⎧ 2Etf ⎛ f ⎞ ⎪⎫ M 3 max min⎨σ$ ff,, 05. EEGcc ⎜ ⎟ ⎬ ≤=σ$ f (7.22) 1 2 ⎝ s ⎠ td ⎩⎪ − ν f ⎭⎪ f

^ where σf is the lower of the tensile and compressive yield/fracture strengths. The maximum shear stress in the core must be lower than the ultimate shear strength. Thus,

T max ≤ τ$ c (7.23) d and the stiffness must have a certain value according to eq.(7.19). Hence, in this case there are three constraints to be considered when sizing the sandwich. If the face thickness is known then these equations give three values of the thickness d, and the largest of these gives the design value. For example, consider a simply supported beam with a uniformly distributed load, q, where the maximum deflection must be less than Δ. Now, in this case, the maximum 2 bending moment is qL /8, the maximum transverse force equals qL/2, C = Δ/qL, kb = 5/384, and ks = 1/8. The requirement for the minimum allowed thickness d has then to be taken from the maximum of [8]

qL2 from eq.(7.22): d ≥ 8t ffσ$

qL from eq.(7.23): d ≥ 2τ$ c

qL2 ⎡ 20ΔG 2 ⎤ from eq.(7.19): d ≥++⎢11 c ⎥ 16G Δ 3qE t c ⎣⎢ ff⎦⎥

7.9 AN INTRODUCTION TO SANDWICH STRUCTURES

If the core thickness was known beforehand, the face thickness can be calculated in a similar manner. For a sandwich panel, the sizes can be found in the same way; it is all a matter of calculating the deflection as function of the geometry and using formulae valid for sandwich plates for the strength constraints, e.g., wrinkling formulae for biaxial loading.

If a sandwich column subjected to in-plane compressive end loads is considered then the constraints will be buckling constraints. The face strength given in eq.(7.22) can still be used ^ but with σf taken as the compressive strength only. However, this strength should now be compared with the applied stress P/2tf and hence this constraint only influences the choice of face thickness. The overall buckling constraint can be taken from eq.(7.8). This equation will yield a condition for a minimum thickness d as [8]

P ⎡ 8β 22LG 2⎤ d ≥++⎢11 c ⎥ 2G π 2 EtP c ⎣⎢ ff ⎦⎥ taking n = 1 as the lowest buckling load and β is the edge-constraint factor defined in section 5. Once again, the same procedure can be used for plate buckling problems.

A way of improving a design is to recognise that a design that allows for simultaneous failure of the face and the core has a potential of being closer to the optimum design, that is, to find a design that stresses all components to their limits at maximum load. For example, to design for simultaneous face yield and core shear fracture, or face wrinkling and core shear failure. It is merely a matter of recognising the important failure modes that may appear and finding the sizes and even materials that allow for simultaneous failure in these modes. This is done in the same way as when constructing the failure mode maps given in section 7.

7.5 Single Parameter Optimum This section describes the simplest possible type of optimum design criteria, e.g., finding minimum weight of the sandwich using only one constraint [9-10]. The constraints used here will be a given flexural rigidity, flexural strength or dimpling strength. As will be seen in the following section, just using one constraint is far from adequate but since the analysis and results are so simple they may serve as a first choice in a design process.

(i) Flexural rigidity To keep the derivation simple assume thin faces and a weak core. Thus Etd2 D ≈ ff 2 and the total weight, substituting the above, is

4Dρ f Wt=+ρρρcc2 f t f ≈+ c d 2 Edf where ρf and ρc are the density of the face and the core, respectively. The optimum is now found by letting

7.10 FAILURE MODES

∂W 8Dρ f ρ f =−ρc 3 =⇒=04 dtf ∂d Edf ρc

Recognising that the weight of the core Wc ≈ ρcd and of the faces Wf = 2ρftf gives that

Wc = 2Wf (7.24) Thus, for a minimum weight sandwich of a specified flexural rigidity, the weight of the core should be twice that of the faces combined. One must remember that the flexural rigidity is only one of two stiffness parameters governing the stiffness of a sandwich. Performing the same analysis but assuming dissimilar faces yields the same conclusion [9] and that the weight of the two faces should be equal.

(ii) Flexural strength Performing the same reasoning as above but now for a specified flexural strength, that is a given bending moment capacity M*, gives (see eq.(3.9))

* 2σ$ f D M ≈≈σ$ fftd Edf

^ where the critical face stress σf is defined in eq.(7.22). Substituting into the weight equation and differentiation gives

* 2M ρ f Wt≈+ρcc σ$ f d

* ∂W 2M ρ f ρ f =−ρc 2 =⇒=02 dtf ⇒ WWf =c (7.25) ∂d σ$ f d ρc

Thus, for a minimum weight sandwich of specified flexural rigidity the weight of the core should be equal to that of the faces combined. This, however, only considers the bending moment capacity of the sandwich and not the load bearing of transverse forces. Once again, assuming dissimilar faces [9] gives the same result.

(iii) Face dimpling 2 The limiting face stress is for the dimpling case proportional to tf or

2 σ ff= kt

Hence, the bending moment will then be

* 2σ f D 3 M ≈≈ktf d Edf

Elimination of d, insertion into the weight equation and minimisation gives

** M ρ c ∂W 3M ρ c W ≈+3422ρ fft , =−=⇒ρ f 0 kt f ∂t f kt f

7.11 AN INTRODUCTION TO SANDWICH STRUCTURES

* 3 4 3M ρc 3ktf dρc 2t ffρ t f == ⇒=⇒= d WWf 3 c (7.27) 2kρ f 2kρ f 3ρc showing that the optimum face thickness to avoid face dimpling is to choose a core that weighs one-third of the combined weight of the faces.

7.6 Minimum Weight for Given Stiffness The above analysis only considered one constraint at the time but also only the flexural rigidity instead of the stiffness. The constraint is more often on the deflection of the structure than on the flexural rigidity. This section will describe methods to find the optimum face and core thicknesses for a given stiffness, providing the materials are predetermined. It will also be assumed that the core properties can be varied by choosing different densities.

(i) Core properties predetermined

If the core material and its density is predetermined then ρc cannot be used as a variable in the optimisation which is equivalent to letting n = 0 in eq.(7.21). Performing the analysis as above, that is, solving tf as function of d in eq.(7.19) gives

−1 2 2 2kLb ⎡Cd kds ⎤ t f =−⎢ ⎥ E f ⎣ L Gc ⎦ and substituting into the weight equation gives the total weight as function of d alone

−1 2 2 4ρ f kLb ⎡Cd kds ⎤ W =−⎢ ⎥ + ρ c d (7.28) E f ⎣ L Gc ⎦

Differentiating this equation yields a fourth degree equation in d to which zeros must be found. The solution is somewhat difficult but can be found by means of plotting eq.(7.28) and finding the sought minimum graphically. To illustrate this, take an example of a simply supported beam in three-point bending with a maximum mid-point deflection L/100. Thus, kb 3 = 1/48 and ks = 1/4. Take L = 500 mm, faces made of aluminium alloy (ρf = 2700 kg/m ) and the core having a density of 100 kg/m3 and a shear modulus of 40 MPa. Assume unit thickness and a point load of 10 N acting in the middle of the beam. The weight of the beam per unit thickness (eq.(7.28) times length L) can then be plotted versus thickness d as illustrated in Fig.7.3. W 3.5

2.5

2 d (mm) 0 1020304050

Figure 7.3 Beam weight (g/mm thickness) as function of thickness d.

7.12 FAILURE MODES

It is now seen that there is an optimum (not necessarily global) for d ≈ 29 mm at a total weight of 2.06 g/mm thickness. Back substituting gives tf = 0.2256 mm.

The problem of optimum sizes for a column subjected to compressive loads is solved in the same manner since the constraint is a given buckling load according to eq.(7.7 or 7.8). The overall buckling depends on the two stiffness parameters D and S just as the compliance C. A solution may therefore be found in the same manner as outlined above for the bending case. This problem has been solved for a plate buckling case by Kuenzi [9].

(ii) Core properties varying with density In this case there are three variables, two thicknesses and the core density. By using the definition of the core material properties in eq.(7.21) and inserting these in the compliance equation and writing the equations in the same form as in [11] one arrives at the following relation for the core density

1 ⎛ kE PLt d ⎞ n ρ = ⎜ sf f ⎟ (7.29) c ⎜ 23⎟ ⎝ CG ΔEtdff− 2 PLk b⎠

The total surface weight can now be written as function of tf and d as

1 n 11n+ 1 ⎛ kE PLt d ⎞ − Wtt=+2ρρ⎜ sf f ⎟ ≈+22tAtdBtdPLn n 23 − n (7.30) ff c⎜ 23⎟ ff f ()f ⎝ CG ΔEtdff − 2 PLkb ⎠ where

1 ⎛ kEPL⎞ n ΔE A = ⎜ sf ⎟ and B = f ⎝ kCbG⎠ kb

In order to find an optimum, the partial derivatives ∂W/∂tf and ∂W/∂d should be set to zero, giving the two equations for the unknowns tf and d.

Taking n = 1 in eq.(7.21), that is, linearly varying core properties with density, yields no optimum solution. This is easily seen by the fact that for n = 1 there are no zeros to the equation ∂W/∂d = 0, but the weight decreases asymptotically to a finite value when the core thickness is increased while the face thickness and core density simultaneously are decreased. This is, however, in reality no optimum. Letting the face thickness be predetermined leads to the same result as above. Hence, for these cases the optimum design is by choosing as thick and as light a core as possible in conjunction with other constraints on the structure. This leads automatically to a given minimum face thickness. By instead predetermining the core thickness d, which often is stipulated in practical cases due to requirements on the space or thermal insulation, it is possible to find an optimum face thickness and for that a corresponding core density. Let us illustrate this with the same example as used above. Now 3 let the core density be a variable with n = 1 and CG = 0.40 MPa/kg/m (typical for cross- linked PVC foam). Hence, a 100 kg/m3 core will have a shear modulus of 40 MPa. Assume the thickness d is predetermined to equal 20 mm. The total weight can now be written as

7.13 AN INTRODUCTION TO SANDWICH STRUCTURES

function of tf alone. Either one can differentiate the weight equation and find its zeros (leading to a fifth degree equation in tf), or simply plot it and graphically determine the minimum. Doing the latter yields the graph shown in Fig.7.4. W 3.5

2.5

2 tf (mm) 0.3 0.4 0.5 0.6 0.7 0.8

Figure 7.4 Beam weight (g/mm thickness) as function of face thickness tf. The optimum face thickness is found to be 0.58 mm, giving an optimum core density of 90 kg/m3 and a total weight of 2.47 g/mm thickness. There may not be a 90 kg density quality of the core material chosen so the results should be interpreted as a guide-line and an 80 or a 100 kg/m3 density core should be used rather than a 40 or a 200 kg/m3 core.

Following the derivation by Gibson [11] assuming n = 2, which is the theoretical value for foams [7], an optimum can be found from the conditions ∂W/∂tf = ∂W/∂d = 0. After some algebraic exercise one arrives at the optimum thicknesses [11]

1 2 ⎡⎛ ρ k ⎞ PC ⎤ 5 6kPL3 dL= 434. ⎢⎜ fk⎟ G ⎥ and t = b (7.31) ⎜ ⎟ 2 f 2 ⎢ E k ΔΔ⎥ Ed ⎣⎝ f ⎠ s ⎦ f

Substitution into eq.(7.29) gives the optimum core density. Test with different material and thickness combinations performed in [11] showed that designs using these sizing rules give structures with optimum stiffness to weight ratios.

7.7 Minimum Weight for Given Strength If one now instead is aiming at finding a minimum weight design for a given strength it is convenient to use the failure mode maps described in section 7. One approach is to use each of the failure modes as constraints and optimise the structure with respect to this constraint alone, using the methodology used above for the stiffness problem. One may then compare the results for the different “optima” and on the basis thereof choose a minimum weight design. The chosen design should then be plotted on the failure mode map to ensure that it will fail in the given mode. Another approach, used in [12], is to observe that in a minimum weight design, both the faces and core should fail at the same load so that in a true sense all constituents are stressed to their ultimate limit. If considering only the three dominant failure modes, face yield/fracture, face wrinkling and core shear, an optimum sizing technique could be as follows. Assume that the design assumes either simultaneous face yield and core shear or simultaneous face wrinkling and core shear.

7.14 FAILURE MODES

(i) Simultaneous face yield – core shear Assume we want a design for simultaneous face yield and core shear failure which has a minimum weight. From eq.(7.17a) we can express the face thickness governing face yield as

kLM t f = (7.32) σ$ f Pd and the core density according to eqs.(7.15) and (7.21) as

1 m ⎛ kPT ⎞ ρ c = ⎜ ⎟ (7.33) ⎝ Cdτ ⎠

Substituting into the weight equation, differentiating with respect to the thickness d and finding the zeros gives the optimum thickness

m 1 ⎡ − ⎤ 21m− 2ρ PLk ⎛ kP⎞ m m d = ⎢ f M ⎜ T ⎟ ⎥ (7.34) ⎢ ⎥ σ$ f ⎝ Cτ ⎠ m −1 ⎣⎢ ⎦⎥ and the optimum face thickness and core density are found by substituting d into eqs.(7.32) and (7.33).

(ii) Simultaneous face wrinkling – core shear Eq.(7.17c) gives the face thickness for which wrinkling occurs at a given load as

1 2PLk − t = M ECCρ 2n 3 (7.35) f d ()f EGc and along with the core density for which core shear failure occurs in eq.(7.33) one can write the weight as a function of d alone. Differentiating with respect to d and finding the zeros to this equation leads to

1 ⎛ 2n 1 ⎞ 2n 1 ⎜ + ⎟ ⎛ ⎞ ⎝ 3mm⎠ 3 ⎜ −+2 ⎟ ⎛ kP⎞ 33m − ()ECCf EG d ⎝ 3mm⎠ = ⎜ T ⎟ (7.36) ⎝ Cτ ⎠ 324mn− ρ fMPk L

The analysis derived above easily can be modified for use on panels by simply changing the loading parameters kM and kT. Other load cases may be considered in similar analyses. Flügge [13] considered overall buckling and face wrinkling and derived minimum weight design thickness tf and ρf. The number of combinations of different failure modes and other constraints are vast and will not be given here. The methodology outlined here could be used for any other constraints given on the structure.

7.15 AN INTRODUCTION TO SANDWICH STRUCTURES

References [1] Triantafillou T.C. and Gibson L.J., “Failure Mode Maps for Foam Core Sandwich Beams”, Materials Science and Engineering, Vol. 95, 1987, pp 37-53.

[2] Hoff N.J. and Mautner S.E., “Buckling of Sandwich Type Panels”, Journal of the Aeronautical Sciences, Vol 12, No 3, July 1945, pp 285-297.

[3] MIL-HDBK-23A, Structural Sandwich Composites, Dec. 1968, Government Printing Office, Washington D.C., USA.

[4] Weikel R.C. and Kobayashi A.S., “In the Local Elastic Stability of Honeycomb Face Plate Subjected to Uniaxial Compression”, J. Aero/Space Sci., Vol 26, No 10, Oct. 1959, pp 672-674.

[5] Norris C.B., “Short Column Strength of Sandwich Constructions as Affected by the Size of Cells of Honeycomb Core Materials”, U.S. Forest Service Research Note FPL- 026, Forest Product Laboratory, Wisconsin, 1964. Also in Norris C.B. and Kommers W.J., U.S. Forest Product Laboratory Report 1817, 1950.

[6] Plantema F.J., Sandwich Constructions, John Wiley & Sons, New York, 1966.

[7] Gibson L.J. and Ashby M.F., Cellular Solids – Structure and Properties, Pergamon press, Oxford, 1988.

[8] Allen H.G., Analysis and Design of Structural Sandwich Panels, Pergamon Press, Oxford, 1969.

[9] Kuenzi E.W., “Minimum Weight Structural Sandwich”, U.S. Forest Service Research Note FPL-086, 1965, U.S. Forest Product Laboratory, Madison, WI.

[10] Theulen J.C.M. and Peijs A.A.J.M., “Optimization of the Bending Stiffness and Strength of Composite Sandwich Panels”, Composite Structures, Vol. 17, 1991, pp 87- 92.

[11] Gibson L.J., “Optimization of Stiffness in Sandwich Beams with Rigid Foam Cores”, Materials Science and Engineering, Vol. 67, 1984, pp 125-135.

[12] Triantafillou T.C. and Gibson L.J., “Minimum Weight Design of Foam Core Sandwich Panels for a Given Strength”, Materials Science and Engineering, Vol. 95, 1987, pp 55-62.

[13] Flügge W., “The Optimum Problem of the Sandwich Plate”, Journal of Applied Mechanics, Vol. 19, No 1, March 1952, pp 104-108.

7.16 CHAPTER 8

SANDWICH PLATES – FUNDAMENTAL EQUATIONS

The following section gives an overview of the governing equations for the bending, buckling and free vibration of sandwich plates, and is merely a brief summary of the basic small- deformation plate bending analysis by Timoshenko and Woinowsky-Krieger [1] which is extended to account for transverse shear deformation following the work by Libove and Batdorf [2]. Partial deflections are then introduced for the plate problem in a manner similar to, approximate but simpler than, that given by Plantema [3]. The analysis assumes that the transverse normal stiffness of the core is infinite thus keeping the distance between the centroids of the faces, d, constant, also called antiplane core. The theory is developed for orthotropic plates with the x- and y-axes being the principal axes of orthotropy and for which the properties are constant throughout the plate. This means that the properties of the plate are fully described by the seven constants, the flexural rigidities Dx and Dy, the twisting stiffness

Dxy, the Poisson's ratios νyx and νxy, and the shear stiffnesses Sx and Sy.

The coordinate system and the positive directions for loads and bending moments are defined as in Fig.8.1. The transverse forces are denoted Tx and Ty (in some textbooks Q) instead of the more logical Nxz and Nyz, to clearly separate them from in-plane forces. The deformations in x, y and z-directions, respectively, are as usual denoted u, v and w. It is also assumed that the shear strain is constant over the cross-section (thin-face approximation) so that the in-plane deformations for the classical Reissner/Mindlin kinematics can be used. These are

uu=+00 zxxθ , vv= + z yyθ and ww= 0 (8.1) where subscript 0 refers to the middle plane or neutral axis. θ is a positive rotation of the cross-section, as illustrated in Fig.8.1. Since the plate is assumed to be orthotropic the position of the neutral axes may differ in the x- and y-directions. Thus, zx is the out-of-plane coordinate measure from the neutral axis in the x-direction, and zy from the axis in the y- direction (see Fig.8.4). θx and θy are defined as the cross-section rotations in the x- and y- directions, respectively. They will be defined in terms of partial deflection later. The strain- displacement relation assumes strains to be much smaller than unity so that [1]

∂u ∂θ ∂v ∂θ y ε =+= ε z x , ε =+= ε z xxx∂x 0 ∂x yx∂y y0 y ∂y

8.1 AN INTRODUCTION TO SANDWICH STRUCTURES

∂u ∂v ∂θ ∂θ y and γ =+=γ +z x +z (8.2) xy∂y ∂x yx0 x ∂y y ∂x where εx0, εy0 and γxy0 are the neutral plane strains. y,v

θx x,u M θ y x Myx Mx Mxy σ τxy y q(x,y) dy Nxy Nx Mxy dx h(x,y) z,w τyx Mx σx Tx τyz My Nyx τ N zy τxz y Ty τzx Myx

σz Rx

−θx

x z u Figure 8.1 Sign convention used in the plate analysis. However, proceeding in this manner will make the derivation of the governing equations both difficult and rather illogical. There are two ways to overcome this; one is to use the method by Libove and Batdorf [1] which is pursued below, or to use the approach via classical lamination theory extended to account for transverse shear deformations, which is done later in this chapter.

8.1 Governing Equations In the same manner as for the beam analysis given previously, we can write the bending moments and transverse forces as functions of the displacement field w. Commence with pure bending and define curvatures in the same manner as for the beam. The curvature κ (inverse of the radius of curvature) and the new quantity κxy for the twist [1] can be written ∂θ ∂θ ∂θ ∂θ κ ==x , κ y , and κ =+x y (8.3) x ∂xyy ∂ xy ∂ yx∂ which are now written in terms of cross-section rotations, rather than as derivatives of displacements. A straightforward way of deriving the plate equations is to assume that only one load is allowed to act on the plate at a time. The response to, e.g., the bending moment Mx would be a primary curvature ∂2w/∂2x2 in the middle surface of the element and also a secondary curvature ∂2w/∂2y2 which is the Poisson effect. The following can then be defined:

8.2 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

∂2w ∂2w ∂2w MD=−, and =−ν xx∂x 2 ∂y2 xy∂x 2

∂2w ∂2w ∂2w MD=−, and =−ν yy∂y2 ∂x 2 yx∂y2

∂2w MD=− , where D is the torsional stiffness xy xy ∂∂xy xy where the Poisson's ratios now hence are defined as the ratio between the primary and secondary curvatures as (in the case of only applied bending moments) ∂22wy/ ∂ ∂22wx/ ∂ ν =− and ν =− (8.4) xy∂∂22wx/ yx ∂∂22wy/

(rule-of-thumb to remember indices: first index corresponds to loading and the second to the strain, i.e., νxy is found by applying a curvature in the x-direction and measuring the responding curvature in the y-direction.)

The transverse shear response to the shear loads, if only the transverse forces Tx and Ty were allowed to act on the plate, will (as defined in chapter 4) be ∂T ∂2w ∂T ∂2w x ==S and y S ∂x x ∂x 2 ∂x y ∂y2 where S is the shear stiffness, defined in eq.(4.3), which may differ between the x- and the y- direction. The total curvatures can now be obtained by adding the different contributions. The

Rx curvature, for example, gets its contributions from Mx, My, and from Tx as

2 ∂ w Mx My 1 ∂Tx 2 =− +νyx + (8.5a) ∂x Dx DSyx∂x and in the same manner

2 ∂ w My Mx 1 ∂Ty 2 =− +νxy + (8.5b) ∂y Dy DSxy∂y

The third curvature is given by a geometrical study and is [2] ∂2w M 1 ∂T 1 ∂T =−xy +x + y (8.5c) ∂∂xy DSxy2 x ∂yS2 y ∂x

8.3 AN INTRODUCTION TO SANDWICH STRUCTURES

My Myx Mxy ∂M Mx x M x + dx ∂M yx ∂x M yx + dy ∂M ∂y M + xy dx xy ∂x ∂M M + y dy y ∂y Ty Ny T x Nyx Nxy dx ∂N Nx ∂N x dy yx N x + dx N yx + dy ∂x ∂y ∂T T + x dx x ∂x ∂N xy ∂N y N + dx ∂Ty xy N y + dy ∂x ∂y Ty + dy ∂y

Figure 8.2 Forces and bending moments acting on a differential element. Next, derive the equations of equilibrium by studying Fig.8.2, and assuming an increment change in all forces and bending moments over the differential element. From equilibrium of forces in x- and y-directions, according to Fig.8.2, we have ∂N ∂N ∂N ∂N x +=yx 0 , and yxy+=0 (8.6a,b) ∂x ∂y ∂y ∂x

Vertical equilibrium is found by projecting all forces onto the z-axis. As seen in Fig.8.3, the projection of the normal force Nx on the z-axis gives

∂w ⎛ ∂N ⎞ ⎛ ∂w ∂ 2 w ⎞ ∂ 2 w ∂N ∂w −++N dy⎜ N x dx⎟ dy⎜ + dx⎟ =+ N dxdy x dxdy xx∂x ⎝ ∂x ⎠ ⎝ ∂x ∂x 2 ⎠ x ∂x 2 ∂x ∂x

and the projection of the shear force Nyx is similarly

Nyx

Nx ∂w dx ∂x dy ∂w ∂w ∂x ∂x ∂N ∂w ∂ 2w N + x dx + dy dx x ∂x ∂x ∂x∂y ∂N yx N yx + dy ∂y

Figure 8.3 Force projections.

The projection of the shear force Nxy is then

∂w ⎛ ∂N ⎞ ⎛ ∂w ∂ 2w ⎞ −++N dx⎜ N yx dy⎟ dx⎜ + dy⎟ yx∂x ⎝ yx ∂y ⎠ ⎝ ∂x ∂∂xy ⎠

8.4 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

22 ∂w ⎛ ∂N yx ⎞ ⎛ ∂w ∂ w ⎞ ∂ w ∂N yx ∂w = −++N dx⎜ N dy⎟ dx⎜ + dy⎟ =+ N dxdy dxdy yx∂x ⎝ yx ∂y ⎠ ⎝ ∂x ∂∂xy ⎠ yx ∂∂xy ∂y ∂x

Summing all contributions for the in-plane forces and we arrive at ∂2w ∂N ∂w ∂2w ∂N ∂w N dxdy ++x dxdy N dxdy +yx dxdy + x ∂x 2 ∂x ∂x yx ∂∂xy ∂y ∂x

∂2w ∂N ∂w ∂2w ∂N ∂w N dxdy ++y dxdy N dxdy +yx dxdy = y ∂y2 ∂y ∂y xy ∂∂xy ∂y ∂x

∂2w ∂2w ∂22w ∂ w N ++N N +N x∂x 2 y∂y2 xy∂∂xy yx ∂∂xy through the use of eqs.(8.6a,b). The final expression for the force equilibrium in z-direction is then ∂T ∂T ∂2w ∂2w ∂22w ∂ w x +++y qN +N +N +N =0 (8.6c) ∂x ∂y x∂x 2 y∂y2 xy∂∂xy yx ∂∂xy

From equilibrium of bending moments about the y-, x- and z-axes it follows, observing that

Myx = Mxy since τxy = τyx [1] ∂M ∂M T −−x yx =0 (8.7a) x ∂x ∂y

∂M ∂M T −−yxy =0 (8.7b) y ∂y ∂x

Nxy = Nyx (8.7c) By assuming that the normal forces N are constant throughout the plate and that they do not change as the plate bends, eqs.(8.6a,b) are fulfilled and the analysis is restricted to small deformations. It is now convenient to invert eq.(8.5) to get the bending moments. After some rearrangement of these equations one arrives at

D ⎡ ∂ ⎛ ∂w T ⎞ ∂ ⎛ ∂w T ⎞⎤ M =− x ⎢ ⎜ − x ⎟ +−ν ⎜ y ⎟⎥ (8.8a) x 1 − νν ∂x ∂x Syyx ∂ ⎜ ∂y S ⎟ xy yx ⎣⎢ ⎝ x ⎠ ⎝ y ⎠⎦⎥

D ⎡ ∂ ⎛ ∂w T ⎞ ∂ ⎛ ∂w T ⎞⎤ M =− y ⎢ ⎜ − y ⎟ +−ν ⎜ x ⎟⎥ (8.8b) y 1− νν ∂y ⎜ ∂y Sx⎟ xy ∂ ∂x S xy yx ⎣⎢ ⎝ y ⎠ ⎝ x ⎠⎦⎥

D ⎡ ∂ ⎛ ∂w T ⎞ ∂ ⎛ ∂w T ⎞⎤ M =−xy⎢ ⎜ − y ⎟ +−⎜ x ⎟⎥ (8.8c) xy 2 ∂x ⎜ ∂y Sy⎟ ∂ ∂x S ⎣⎢ ⎝ y ⎠ ⎝ x ⎠⎦⎥

8.5 AN INTRODUCTION TO SANDWICH STRUCTURES

Now, eqs.(8.6c), (8.7a,b) and (8.8) constitute six fundamental equations to determine the deflection field, forces and bending moments. It might be more convenient to rewrite these to obtain equations in one single variable, e.g. for the deflection field only, as a function of the applied loads. This leads to fairly complicated formulae which are readily simplified when, for example, studying plates with isotropic characteristics. Thus, by rearranging eq.(8.6c) using eqs.(8.7a,b) one arrives at the equation

∂2 M ∂22M ∂ M ∂T ∂T x ++=+=−2 xy y x y q* (8.9) ∂x 2 ∂∂xy ∂y2 ∂x ∂y

∂2w ∂22w ∂ w with qqN* =+ +2N +N xxyy∂x 2 ∂∂xy ∂y2

It will also prove useful for the solution of these equations to eliminate the bending moments from eqs.(8.7a,b) by using eq.(8.8), since many solutions contain assumed shapes not only of the deflection w but also of the transverse forces T. In doing so, the assumed fields may be solved as functions of the applied loading. The result of this substitution is

D ∂ 3w ⎛ ν D ⎞ ∂ 3w D ∂ 2 T x + ⎜ yx x + D ⎟ − x x 3 ⎜ xy ⎟ 2 2 11− ννxy yx ∂x ⎝ − ννxy yx ⎠ ∂∂xy Sxxyyx() 1− ν ν ∂x (8.10a) ⎛ ν D D ⎞ ∂ 2 T D ∂ 2 T −⎜ yx x + xy ⎟ yxy−+x T ⎜ ⎟ 2 x ⎝ Syxyyx()122− νν Sy ⎠ ∂∂xy Sx ∂y

D ∂ 3w ⎛ ν D ⎞ ∂ 3w D ∂ 2 T y + ⎜ xy y + D ⎟ − y y 3 ⎜ xy ⎟ 2 2 11− ννxy yx ∂y ⎝ − ννxy yx ⎠ ∂∂xy Syxyyx() 1− νν ∂y (8.10b) ⎛ ν D D ⎞ ∂ 2 T D ∂ 2 T −⎜ xy y + xy ⎟ x −+=xy y T 0 ⎜ ⎟ 2 y ⎝ Sxxyyx()122− νν Sx ⎠ ∂∂xy Sy ∂x

Eqs.(8.10) together with eq.(8.9) now constitutes a set of three differential equations in the three variables w, Tx and Ty which in the general case must be solved. By suitable substitution these equations may be separated into three equations in terms of w, Tx, and Ty alone. The result is [2]

[D]w = −[M]q (8.11)

[D]Tx = −[N]q

[D]Ty = −[P]q where the terms in square brackets are differential operators defined by [2]

8.6 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

1 DD ∂ 6 ⎛ DD DD− ν DD ⎞ ∂ 6 D =++xxy ⎜ xxy xy yxxxy⎟ [] 6 ⎜ ⎟ 42 22Sxy ∂ ⎝ Sx Sxyy ⎠ ∂∂

⎛ DD DD− ν DD )⎞ ∂ 6 1 DD ∂ 6 ++⎜ yxy xy xyyxy⎟ + yxy ⎜ ⎟ 24 6 ⎝ 2Sy Sxyx ⎠ ∂∂ 2 Syx ∂

∂ 4 ∂ 4 ∂ 4 ∂ 4 −−D 21DDD() −νν ++ ν ν −D −D (8.12a) xxyxyyxyxxxyyyy∂x 4 []∂∂xy22 ∂y 4 ∂y 4

1 DD ∂ 4 ⎛ D DDD− ν ⎞ ∂ 4 1 DD ∂ 4 ++xxy N ⎜ x y yxxxy⎟ N +−yxy N 4 ⎜ ⎟ 22 4 2 SSxy ∂ x ⎝ SSxy ⎠ ∂∂ x y 2 SSxy ∂ y

⎛ 1 D ()1 − νν D ⎞ ∂ 2 ⎛ 1 D ()1 − νν D ⎞ ∂ 2 ⎜ xy xy yx + x ⎟ − ⎜ xy xy yx + y ⎟ NN+−()1 νν ⎜ ⎟ 2 ⎜ ⎟ 2 xy yx ⎝ 2 Sy Sxx ⎠ ∂ ⎝ 2 Sx Syy ⎠ ∂

∂2 ∂22∂ with NN=+2N +N xxyy∂x 2 ∂∂xy ∂y2

1 DD ∂ 4 ⎛ DD− ν DD ⎞ ∂ 4 1 DD ∂ 4 M =+xxy ⎜ xy yxxxy⎟ + yxy [] 4 ⎜ ⎟ 22 4 2 SSxy ∂ x ⎝ SSxy ⎠ ∂∂ x y 2 SSxy ∂ y (8.12b) ⎛ 1 D ()1 − νν D ⎞ ∂ 2 ⎛ 1 D ()1 − νν D ⎞ ∂ 2 −⎜ xy xy yx + x ⎟ − ⎜ xy xy yx + y ⎟ +−()1 νν ⎜ ⎟ 2 ⎜ ⎟ 2 xy yx ⎝ 2 Sy Sxx ⎠ ∂ ⎝ 2 Sx Syy ⎠ ∂

1 DD ∂ 5 ⎛ DD− ν DD ⎞ ∂ 5 1 DD ∂ 5 N =+xxy ⎜ xy yxxxy⎟ + yxy [] 5 ⎜ ⎟ 32 4 2 Sxy ∂ ⎝ Sxyy ⎠ ∂∂ 2 Sxyy ∂∂ (8.12c) ∂ 3 ∂ 3 −−−+D DD()1 νν ν xxyxyyxyxx∂x 3 ()∂∂xy2

5 5 5 1 DDxxy ∂ ⎛ DDxy− ν xyyxy DD ⎞ ∂ 1 DDyxy∂ []P =+4 ⎜ ⎟ 23+ 5 2 Sxyx ∂∂ ⎝ Sxyx ⎠ ∂∂ 2 Syx ∂ (8.12d) ∂ 3 ∂ 3 −−−+D DD()1 νν ν yxyxyyxxyy∂y 3 ()∂∂xy2

In essence one can see that the equations of equilibrium in eqs.(8.6) and (8.7) constitute 5 independent equations (not the last of eq.(8.7)) which means we must seek solution for 5 field variables. These are commonly the in-plane deformations u and v which are given by eqs.(8.6a,b), the out-of-plane deformation w, and the transverse forces Tx and Ty which are coupled through eqs.(8.6c), and (9,7a,b). The transverse forces are used in stead of the cross- section rotations more commonly used in Reissner/Mindlin plate theory.

8.7 AN INTRODUCTION TO SANDWICH STRUCTURES

It is interesting to notice at this stage some implications of the derived theory. From the kinematic assumptions of eqs.(8.1) and (8.2) we can observe that the bending moments also can be written as h/2 h/2 EEε + ν ε Mzdz==σ xx yxxyzdx xxz∫∫ z −−h/2 h/2 1 − ννxy yx h/2 Ez ⎡ ∂θ ⎛ ∂θ ⎞⎤ = xz ε ++z x νε +z y dx ∫ ⎢ xx00yx⎜ y y ⎟⎥ −h/2 1 − ννxy yx ⎣ ∂x ⎝ ∂y ⎠⎦

Dx ⎡∂θ x ∂θ y ⎤ = ⎢ + ν yx ⎥ 1 − ννxy yx ⎣ ∂xy∂ ⎦ providing zx and zy are defined from the same axis. A similar situation is obtained for the other bending moments. There is hence a discrepancy appearing in the theory by defining rotations about the neutral planes if they are not shared in the different directions, i.e. the plate does not bend about the same plane in x- and y-directions. In the more general type sandwich theory, discussed later, the geometric middle plane is instead used as the basis for the coordinate system. This, however, results in that the so called B- or coupling matrix will be non-zero of non-symmetric sandwich panels, which in turn results in a very complicated set of governing equations. However, assuming zx and zy are defined from the same plane, or close to it, we see that the rotation can be obtained by comparing the above with eq.(8.8) and it is seen that

Tx ∂w Ty ∂w θ x =− and θ y =− (8.13) Sx ∂x Sy ∂y or with another interpretation ∂w ∂w ∂w ∂w θ = γ − and θ = γ − so that γ = +θ and γ = +θ x xz ∂x y yz ∂y xz ∂x x yz ∂y y

8.2 Partial Deflections Partial deflections similar to those used for the beam can also be introduced for the plate. They are the deflections due to bending and shear and are defined by assuming only one mode of deformation at a time, as done above. Here, a slightly different definition is made than by

Plantema [3], where the deflections are defined as w=wbx+wsx=wby+wsy. Here we instead assume that we can separate the displacement fields due to bending wb and that to transverse shear ws and then simply superimpose as in the beam case. We also introduce a specific relation between the transverse forces and the shear part of the deformation as

∂T ∂ 2 w ∂T ∂ 2 w w = w + w , x = S s and y = S s (8.14) b s ∂x x ∂x 2 ∂y y ∂y 2

As seen in eq.(8.13), two independent field variables Tx and Ty have now become linked to single variable ws, and is in essence the simplification. It also means, as will be seen, that the bending moments will be independent of the transverse force field and vice versa. This

8.8 SANDWICH PLATES – FUNDAMENTAL EQUATIONS assumption will only be correct for isotropic plates but will be a very good approximation for plates that have equal rigidities in the both x- and y-directions, or rather for cross-sections sharing the same neutral axis. However, most sandwiches used in practical applications are quite close to that condition and the error made can sometimes be justified by the simplifications made by that assumption. In doing so, we can see that the rotations defined in eq.(8.1) get a more physical interpretation. Bending causes the cross-section to rotate, whereas shearing is a sliding and thus does not add to any rotation. In doing this we have reduced the number of independent field variables to four: u, v, wb and ws. Strictly speaking we now have more governing equations that variables, but this may be overcome by using eq.(8.9) which reduces eqs.(8.6c), and (8.7a,b) to two independent equations. There is also the more tricky problem of defining the boundary conditions for wb and ws separately, which can cause some difficulties as seen in section 4. On the other hand, the very complex governing equations of eqs.(8.11) can now be transformed into much simpler equations to be solved for the bending and shear part separately.

Thus, assuming that zx=zy=z are measured from the geometrical middle plane of the plate, we get the following approximate set of simplified equations. The cross-section rotations can now be written, following eq.(8.13) ∂w ∂w θ =−b and θ =− b x ∂x y ∂y and from this it yields that (letting u0 and v0 = 0) ∂w ∂w ∂2w ∂2w ∂2w uz=−bb, vz =− → ε =−z b , ε =−z b , and γ =−2z b ∂x ∂y x ∂x 2 y ∂y2 xy ∂∂xy and it is here that the conceptual error occurs by using partial deflections in the way they are introduced here. We must assume that bending takes place about the same plane in all directions. This error should, however, not be too large as long as the flexural rigidities and shear stiffnesses are still calculated with respect to the true neutral axis. Equations (8.8) or (8.13) can now be transformed to

2 2 2 2 Dwx ⎡∂ b ∂ wb ⎤ Dy ⎡∂ wb ∂ wb ⎤ M x =− ⎢ 2 + ν yx 2 ⎥ , M y =− ⎢ 2 + ν xy 2 ⎥ (8.15) 11− ννxy yx ⎣ ∂x ∂ννy ⎦ − xy yx ⎣ ∂y ∂x ⎦

∂22w ∂T ∂ w ∂T ∂2w MD=−bx, =S s and y =S s xy xy ∂∂xy ∂x x ∂x2 ∂y y ∂y2

It is now seen that the partial deflection wb represents the classical plate bending deformation and since the shear deflection does not rotate the cross-section, all bending moments will depend solely on wb. The relation between ws and wb is found by substituting the above equations into eq.(8.7a,b), resulting in

2 4 4 4 ∂ ws Dwx ⎡∂ b ∂ wb ⎤ ∂ wb Sx 2 =− ⎢ 4 + ν yx 22⎥ − Dxy 22 (8.16a) ∂ννx 1 − xy yx ⎣ ∂x ∂∂xy⎦ ∂∂xy

8.9 AN INTRODUCTION TO SANDWICH STRUCTURES

2 4 4 3 ∂ ws Dy ⎡∂ wb ∂ wb ⎤ ∂ wb Sy 2 =− ⎢ 4 + ν xy 22⎥ − Dxy 22 (8.16b) ∂ννy 1 − xy yx ⎣ ∂y ∂∂xy⎦ ∂∂xy which is exactly the same as eq.(8.9) and can be derived therefrom directly. By inserting into the equilibrium equation (8.6c) we get ∂2w ∂2w S s +=−S s q* (8.17a) x ∂x 2 y ∂y2 or rewritten in terms of wb

4 ⎡ ⎤ 4 4 Dwx ∂ b ννyxDD x+ xy y ∂ wb Dy ∂ wb * 4 + ⎢ + 2Dxy ⎥ 22+ 4 = q (8.17b) 11− ννxy yx ∂x ⎣⎢ − ννxy yx ⎦⎥ ∂∂xy 1 − ννxy yx ∂y which is the differential equation in pure bending of an ordinary orthotropic plate [4]. One may now, on the basis that the concept of partial deflection can be accepted, assume eq.(8.17a) to valid for the case when the bending stiffnesses are infinite (pure shear), i.e., for * wb = 0. Then all components w in q takes the value of ws so that eq.(8.17a) is truly an equation in ws only and can be solved for that. In the same manner, eq.(8.17b) can be solved for wb by setting all deflection components on the right hand side equal to wb. The two solutions to eqs.(8.17) are then independent and can after being calculated be superimposed.

8.3 Equation of Motion So far only static terms have been included but we can also include the effect of inertia, which this time has two components, the ordinary transverse inertia and the rotary inertia. In order to assess the motion of sandwich plates the governing equations must be derived assuming inertia forces and hence time is introduced as a variable. The following derivation follows the work by Mindlin [5] except that here a more general cross-section is considered. Mindlin [5] used a form of partial deflections which is an agreement with what is used here, but with another notation. Take another look at Fig.8.2 and once again derive the equation of equilibrium. First define the two inertia forces:

(i) Vertical inertia The body force acting on the element when subjected to an acceleration ∂2w/∂t2

∂ 2 w ∂ 2 w −=−=ρ dz ρ ** , thus ρρdz (8.18) ∫∫∂t 2 ∂t 2 where ρ* is the mass per unit area. This mass may of course be a function of x- and y- coordinates if the plate has a varying thickness. Hence, an acceleration in the positive w- direction (downwards) creates a body force with the opposite direction (upwards).

(ii) Rotary inertia The cross-section rotates when bending with the rotations

Tx ∂w Ty ∂w θ x =− and θ y =− Sx ∂x Sy ∂y

8.10 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

(remember that Tx creates is a sliding which does not add to the rotation), and thus the in- plane deformations, u = zxθx and v = zyθy. If subjected to an acceleration the bending moment inertia in the x-direction will be

∂ 2u ∂θ2 ∂ 2 ⎛ T ∂w⎞ ∂ 3w ∂ 3w −=−=−−ρz dzρ z 2 xxdz ρzdz2 ==b ρzdz2 R b (8.18a) ∫∫xx2 2 2 ⎜ ⎟ ∫∫x 2 xx2 ∂t ∂t ∂t ⎝ Sx ∂x ⎠ ∂∂xt ∂∂xt and similarly in the y-direction

∂ 2 v ∂θ2 ∂ 2 ⎛ T ∂w⎞ ∂ 3w ∂ 3w −=−=−−ρz dzρ z 2 yydz ⎜ ⎟ ρzdz2 ==b ρzdz2 R b (8.18b) ∫∫yy2 2 2 ⎜ ⎟ ∫y 2 ∫yy2 ∂t ∂t ∂t ⎝ Sy ∂y ⎠ ∂∂yt ∂∂yt

defined positive in the same directions as Mx and My, and R is the rotary inertia. vertical equilibrium: ∂T ∂T ∂2w x ++−y q**ρ =0 (8.20) ∂x ∂y ∂t 2

Note that in a general case, however seldom in practice, ρ* may be a function ρ*(x,y) if the plate thickness varies. equilibrium of bending moments about y- and x-axes

2 ∂Mx ∂Mxy ∂ ⎛ ∂w Tx ⎞ Tx −− −Rx 2 ⎜ −⎟ = 0 (8.21) ∂x ∂y ∂t ⎝ ∂x Sx ⎠

∂M ∂M ∂ 2 ⎛ ∂w T ⎞ T −−yxy −R ⎜ −y ⎟ = 0 y y 2 ⎜ ⎟ ∂y ∂y ∂t ⎝ ∂y Sy ⎠

The above is a more general form of eqs.(8.7) but the expressions for the bending moments in eq.(8.8), however, remain the same since the bending moments only depend on the deflected shape. The relation between the bending deflection wb and the shear deflection ws of eqs.(8.16) will take a slightly different form

2 4 4 4 4 ∂ ws Dwx ⎡∂ b ∂ wb ⎤ ∂ wb ∂ wb Sx 2 =− ⎢ 4 + ν yx 22⎥ −+Dxy 22 Rx 22 (8.22a) ∂ννx 1 − xy yx ⎣ ∂x ∂∂xy⎦ ∂∂xy ∂∂xt

∂ 2 w D ⎡∂ 4 w ∂ 4 w ⎤ ∂ 4 w ∂ 4 w s y b b b R b Sy 2 =− ⎢ 4 + ν xy 22⎥ −+Dxy 22 y 22 (8.22b) ∂ννy 1 − xy yx ⎣ ∂y ∂∂xy⎦ ∂∂xy ∂∂yt

By substituting eq.(8.22) into eq.(8.21) and then using eq.(8.15) for the bending moments one soon arrives at

8.11 AN INTRODUCTION TO SANDWICH STRUCTURES

4 ⎡ ⎤ 4 4 Dwx ∂ b ννyxDD x+ xy y ∂ wb Dy ∂ wb 4 + ⎢ + 2Dxy ⎥ 22+ 4 11− ννxy yx ∂x ⎣⎢ − ννxy yx ⎦⎥ ∂∂xy 1 − ννxy yx ∂y (8.23a) 2 2 2 2 **∂ w ∂ ⎡ ∂ wb ∂ wb ⎤ =−q ρ 2 +2 ⎢RRx 2 +y 2 ⎥ ∂tt∂ ⎣ ∂x ∂y ⎦

From this, the governing equation can be derived by eliminating w or wb from eq.(8.23) by using eq.(8.22). When rewritten we have that ∂2w ∂2w ∂2w S s +=−+S s q**ρ (8.23b) x ∂x 2 y ∂y2 ∂t 2

8.4 Governing Buckling Equation In the previous sections the in-plane forces N are all included in the equilibrium and governing equations. In ordinary linear bending or free vibration analysis these are all zero and hence excluded from the analysis. In large deformation theory, on the other hand, they must be included since in-plane forces may develop due to a transverse deflection. However, in many cases there are large in-plane forces acting on the plate elements already in its initial state and these may cause the plate to buckle. In such cases, the transverse force q is usually zero and no motion is accounted for. The buckling equation is derived considering some incremental change in the equilibrium of the plate. Prior to buckling the state of equilibrium is given by eq.(8.9) (properties with index 0). After buckling all properties have changed by an increment so that (changes indicated with index 1)

w = w0 + w1

Tx = Tx0 + Tx1 Ty = Ty0 + Ty1

Nx = Nx0 + Nx1 Ny = Ny0 + Ny1 Nxy = Nxy0 + Nxy1

Mx = Mx0 + Mx1 My = My0 + My1 Mxy = Mxy0 + Mxy1 and the equilibrium (post-buckling equation) is now again given by eq.(8.9) but with a new set of variables inserted.

2 2 2 2 ∂ ()MMxx01+ ∂ (MMxy01+ xy) ∂ (MM y01+ y ) ∂ (ww01+ ) + 2 + ++()NNxx01 ∂x 2 ∂∂xy ∂y 2 ∂x 2 (8.24) 2 2 ∂ ()ww01+ ∂ ()ww01+ ++20NN ++NN = ()xy01 xy∂∂xy () y01 y ∂y 2

The change in the two sets of equilibrium is now given by subtracting eq.(8.9) from eq.(8.24). 2 2 In process of doing this one may realise that terms Nx1 ∂ w1/∂x are much smaller Nx0 2 2 2 ∂ w1/∂ x , etc. since the former load represents a small change. Next, if the deflections prior to buckling are zero or constant (as in axial compression for example) then all derivatives of w0 are zero. For such types of problems the governing buckling equation will be given by

8.12 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

∂2 M ∂2 M ∂2 M ∂2w ∂2w ∂2w x1 +++++220xy1 y1 qN 1 N 1 +N 1 = (8.25) ∂x 2 ∂∂xy ∂y2 xxyy0 ∂x 2 0 ∂∂xy 0 ∂y2

By next taking w1 as the field variable w one sees that this is equation is the same as the equilibrium equation given in (8.9), except that the in-plane loads are pre-set from the initial state. Thus, by only considering small deformation problems, for which the internal in-plane loads caused by any bending or vibrations are zero, one can interpret the Nx, Ny, and Nxy terms in all previous equations as pre-set in-plane loads. Hence, all equations given in sections 8.1 to 8.3, and in the following ones, the in-plane loads are interpreted as applied loads so that q* consists of applied loads.

8.5 Isotropic Sandwich Plates For isotropic sandwich plates we have D D = D = D, ν = ν = ν, S = S = S, and D = (8.26) x y xy yx x y xy 1+ ν and as mentioned earlier, the concept of partial deflections as introduced here, will now be exact. By using this, eqs.(8.15) reduces to equations identical to those for an ordinary plate

[1], and the expression relating wb to ws in eq.(8.22), reduces to the simple form D ΔΔw =− 2w (8.27) sbS()1− ν2 and the governing equation in (8.17) to D Δ2wq= * (8.28) 1− ν2 b where Δ is the Laplace operator. The physical significance of ws becomes clear if eq.(8.27) is substituted into (8.28), which yields

* −=SwΔ s q (8.29)

The bending stiffness D has vanished from this equation. If the normal forces N are zero, then the solution to this equation would be independent of D. Hence, ws can be interpreted as the deflection of a plate having finite shear stiffness and infinite bending stiffness. As pointed out by Plantema [3], in general, eq.(8.27) cannot yield edge values ws = 0, since Δwb will vary along the edge. Thus, ws and wb will not vanish separately on the boundary but only their sum will equal zero. Only when Δwb is constant (or zero) along the entire circumference of the plate, can ws and wb be chosen to vanish separately. By combining eqs.(8.27) and (8.28) we can replace the partial deflections and arrive at an equation of the total deflection as

D ⎡ DΔ ⎤ 2 w 1 q* (8.30) 2 Δ =−⎢ 2 ⎥ 1 − νν⎣ S()1− ⎦

This equation derived are found in the now classical papers by Reissner [6,7] and Mindlin [5]. Now simplify the equation of motion by assuming an isotropic plate. Eq.(8.23) becomes when

Rx = Ry =R

8.13 AN INTRODUCTION TO SANDWICH STRUCTURES

D ∂ 2 w ∂ 2 ΔΔ2 wq=−**ρ +R w (8.31) 1 − ν 2 bb∂t 2 ∂t 2 By rewriting

∂ 2 w ∂ 2 ∂ 2 R ∂ 2 ∂ 2 w SwΔΔΔ=− q** +ρ ⇒ R wR =w +()q ** −ρ sb∂t 2 ∂t 2 ∂t 2 St∂ 2 ∂t 2

2 D 2 D 2 D 2 D 2 DΔ ⎡ **∂ w⎤ 2 ΔΔΔΔwbs= 2 w − 2 w = 2 w + 2 ⎢q − ρ 2 ⎥ 1111− νννν− − − S() 1− ν⎣ ∂t ⎦ and inserting into eq.(8.31) and rearranging one arrives at

2 2 2 D 2 ⎛ DΔ R ∂ ⎞ ⎡ **∂ w⎤ ∂ 2 Δ w −−⎜1 2 + 2 ⎟ ⎢q − ρ 2 ⎥ −=R 2 Δw 0 (8.32a) 1 − νν⎝ S()1 − St∂ ⎠ ⎣ ∂t ⎦ ∂t or again rewritten to the form

⎛ DΔ ∂ 2 ⎞⎛ ρ * ∂ 2 ⎞ ∂ 2 w ⎛ DΔ R ∂ 2 ⎞ ⎜ − R ⎟⎜Δ − ⎟w +=−ρ **⎜1 + ⎟q (8.32b) ⎝1 − ν 2 ∂tSt2 ⎠⎝ ∂ 2 ⎠ ∂νt 22⎝ S()1 − St∂ 2 ⎠ which is the well-known Mindlin [5] plate equation. Mindlin considered a homogeneous plate and the notation he uses is as follows: S = G'h, R = ρh3/12, D/(1–ν2) = D, ρ* = ρh, where h is the plate thickness. Eq.(8.32) is also the two-dimensional form of the so called Timoshenko [8] beam equation. The rotary inertia may be omitted by letting R = 0, and the transverse shear may be omitted by letting S approach infinity. If both are omitted, then it reduces to the ordinary plate equation D ∂2w Δ2w +=ρ**q (8.33) 1− ν2 ∂t 2

8.6 Isotropic Sandwich Plates with Thick Faces Hoff [9] used variational principles to derive the governing differential equations for sandwich plates with thick faces, that is, the strain energy stored in the faces due to bending about their individual neutral axes were included. The rather lengthy derivation is omitted here. However, by adopting the same approach as when deriving the differential equation for sandwich beams with thick faces the derivation becomes simple and straightforward. First assume an isotropic plate. The total bending moment consist of two parts, one originating from bending about the neutral axis of the plate as a whole, and secondly, one of bending of the faces about their individual neutral axes (see the section on sandwich beams). Thus

2 2 2 2 DD0 + cb⎡∂ w ∂ w b⎤ 2D f ⎡∂ w ∂ w⎤ M x =− 2 ⎢ 2 + ν 22⎥ − ⎢ 2 + ν 2 ⎥ 1 − ν ⎣ ∂x ∂νy ⎦ 1 − ⎣ ∂x ∂y ⎦

2 2 2 2 Dw⎡∂ ∂ w⎤ DD0 + cs⎡∂ w ∂ w s⎤ =− 2 ⎢ 2 + ν 2 ⎥ + 2 ⎢ 2 + ν 2 ⎥ 11− ν ⎣ ∂x ∂νy ⎦ − ⎣ ∂x ∂y ⎦

8.14 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

with D = D0 +2Df + Dc, and similarly

2 2 2 2 Dw⎡∂ ∂ w⎤ DD0 + cs⎡∂ w ∂ w s⎤ M y =− 2 ⎢ 2 + ν 2 ⎥ + 2 ⎢ 2 + ν 2 ⎥ 11− ν ⎣ ∂y ∂νx ⎦ − ⎣ ∂y ∂x ⎦

DD+ ∂2w 2D ∂22w Dw∂ DD+ ∂2 w M =− 0 cb− f =− + 0 cs xy 1+ ν ∂∂xy 111+ ν∂∂xy + ν∂∂xy + ν ∂∂xy

Together with the governing eq.(8.9) we then get D DD+ DD+ 2D ΔΔΔΔ2w − 0 c 2w = c 2w + f 2wq= * 11− ν2 − ν2 s 1− ν2 b 1− ν2 by using eq.(8.24) this can be rewritten to contain w only. However, first notice that eq.(8.24) was derived for thin faces and therefore D only includes the terms D0 + Dc in eq.(8.24). For simplification assume Dc = 0 (if included, only substitute D0 with D0 + Dc). We then get

2 2D f 3 DS 2 ⎡ S()1 − ν ⎤ * 2 ΔΔΔw −=−w ⎢ ⎥q (8.34) 1 − ν D0 ⎣ D0 ⎦

By assuming thin faces, Df = 0, this equation reduces to eq.(8.27). However, even though the faces may be rather thick, the ratio D/D0 will always be very close to unity. 8.7 Cross-section Properties So far nothing has been mentioned about the cross-sectional properties apart from their definition and that they differ in different directions for the orthotropic plate. It should be mentioned again that even if partial deflections are used assuming bending to occur about the same neutral plane in all directions, the cross-section properties should still account for different positions of the neutral axes.

Firstly, Dx is the relation between the bending moment Mx and the corresponding curvature ∂2w/∂x2 when applied to a thin strip of the plate and is defined in same manner as for a beam

EtEtd2 DzEdz=≈2 xx11 2 2 (8.35a) xxxx∫ Etxx11+ Et 2 2

t E 1 x1 E y1

E xc tc E z yc e y e y x z x

t 2 EE x2 y2

Figure 8.4 Cross-section and position of the neutral axis in x- and y-directions.

8.15 AN INTRODUCTION TO SANDWICH STRUCTURES for thin faces and weak core. 1 and 2 refer to the upper and lower faces, respectively, as defined in Fig.8.4. Since in a general case the position of the neutral axes in x- and y- directions may differ (if the faces have different orthotropy), zx is the out-of-plane coordinate in the xz-plane and zy the out-of-plane coordinate in the yz-plane. Similarly EtEtd2 DzEdz=≈2 yy11 2 2 (8.35b) yyyy∫ Etyy11+ Et 2 2

The position of the neutral axes can be found in the same manner as outlined in section 3. It can soon be shown that the neutral planes will have the same transverse position, ex = ey when E E x2 = y2 (8.36) Ex1 Ey1 and as long as this relation is approximately satisfied the concept of partial deflections can be used. The relation between the Poisson's ratios and the flexural rigidities can be derived as follows. Take the differential element in Fig.8.1 with sides dx and dy and apply the reciprocity theorem.

(i) apply bending moment Mxdy giving curvatures

2 2 ∂ w Mx ∂ w Mx 2 =− , and 2 =νxy ∂x Dx ∂y Dx

(ii) apply bending moment Mydx giving curvatures

2 2 ∂ w My ∂ w My 2 =− , and 2 =νyx ∂y Dy ∂x Dy

The work done by moment Mxdy on curvature νxyMy /Dy must now equal the work done by moment Mydx on curvature νyxMx/Dx. Thus

My Mx −=−Mxyxννdxdy Myxy dxdy Dy Dx from which it follows that ν ν yx = xy (8.37) DDy x which is similar to the in-plane relation for orthotropic materials, νxy /Ex = νyx /Ey [4]. An exact expression for the torsional stiffness Dxy is rather more difficult to assess, but through use of eqs.(8.2), (8.13) and (8.14) we can, under the same assumption as for the partial deflections, write

∂ 2 w 2GtGtd2 Mzdz==−τ b 22zG22 dz ⇒=≈ D zG dz xy11 xy 2 2 (8.38) xy∫∫ xy ∫ xy xy xy ∂∂xy Gtxy11+ Gt xy 2 2 assuming thin faces and weak core [10]. The shear stiffnesses Sx and Sy are computed in the same manner as for the beam, i.e., using eq.(4.4) or approximately

8.16 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

2 2 Gdcx Gdcy Sx == and Sy (8.39) tc tc

The cross-sectional properties ρ* and are easily calculated for a sandwich as

* ρ =++ρ11tttρcc ρ 2 2 (8.40)

* where ρi is the density of the material component. Hence, ρ can also be interpreted as the cross-section surface weight. The way to compute is exactly the same as calculating the flexural rigidity D but with ρ substituted for E. Hence, from eq.(3.21)

3 3 3 2 ρ t ρ t ρ t ⎛ ttcf+ ⎞ R =+++11 cc 22 ρρtd() −+ e2 t⎜ − ete⎟ + ρ2 (8.41) x 12 12 1211 xcc⎝ 2 xx⎠ 22 and similarly for Ry. Note that the contribution from the core (terms 2 and 5) in this expression may be significant since the ratio of ρf /ρc is usually lower than Ef /Ec. One important observation is that all units must be consistent, e.g., SI-units. A good suggestion is to use kilograms, metres, seconds and Newtons only, giving that all terms in eq.(8.23) have the dimension kg/ms2 (or N/m2).

8.8 Energy Relations

An expression for the strain energy produced by moments Mx, My, and Mxy and the shear forces Tx and Ty is obtained by considering the work done by these when distorting an element of size dx-dy as shown in Fig.8.2. The work of the moment Mxdy equals Mxdy/2 times the positive rotation of the element. This rotation is made up of two parts, the rotation caused by moment Mx and the Poisson effect of the moment My. Note that even though the term ∂Tx/∂x contributes to the curvature of the middle surface, it represents a sliding rather than a rate of change of rotation and therefore it makes no contribution to the rotation of one face with respect to the other. Thus, the strain energy will be

1 ⎛ ∂ 2 w ⎞ 1 ⎛ M M ⎞ Mdy− b dx=− M dy⎜ x ν y ⎟dx x ⎜ 2 ⎟ x ⎜ yx ⎟ 2 ⎝ ∂x ⎠ 2 ⎝ Dx Dy ⎠

and similarly for the work of the bending moment My, and the twisting moments Mxy

1 ⎛ M M ⎞ 1 ⎛ ∂ 2 w ⎞ M 2 ⎜ y x ⎟ b xy Mdxy ⎜ − ν xy ⎟dy and 2×− Mxy dy⎜ ⎟dx = dxdy 2 ⎝ Dy Dx ⎠ 2 ⎝ ∂∂xy⎠ Dxy

The work done by the shear force Tx equals Txdy/2 times the displacement over the element, which is ∂ws/∂x dx. Thus, 1 T 2 1 T 2 x dxdy and y dxdy 2 Sx 2 Sy

Integration of the energy over the entire plate gives the total energy

8.17 AN INTRODUCTION TO SANDWICH STRUCTURES

⎡ 2 2 2 2 2 ⎤ 1 M ⎛ν ν ⎞ M M T T U = ⎢ x − ⎜ xy + yx ⎟M M + y + 2 xy + x + x ⎥dxdy (8.42a) se 2 ∫∫⎢ D ⎜ D D ⎟ x y D D S S ⎥ ⎣ x ⎝ x y ⎠ y xy x x ⎦

The strain energy can now be expressed in terms of the distortions by means of eq.(8.8) and the above relation is then transformed into

2 2 ⎡ D ⎧ ⎛ T ⎞⎫ D ⎧ ⎛ T ⎞⎫ 1 x ∂ ∂w x y ⎪ ∂ ⎜ ∂w y ⎟⎪ U se = ⎢ ⎨ ⎜ − ⎟⎬ + ⎨ − ⎬ 2 ∫∫ 1−ν ν ∂x ⎜ ∂x S ⎟ 1−ν ν ∂y ⎜ ∂y S ⎟ ⎣⎢ xy yx ⎩ ⎝ x ⎠⎭ xy yx ⎩⎪ ⎝ y ⎠⎭⎪ ⎧ ⎫ ⎛ν yx Dx +ν xy Dy ⎞⎧ ∂ ⎛ ∂w T ⎞⎫⎪ ∂ ⎛ ∂w Ty ⎞⎪ + ⎜ ⎟⎨ ⎜ − x ⎟⎬⎨ ⎜ − ⎟⎬ (8.42b) ⎜ 1−ν ν ⎟ ∂x ⎜ ∂x S ⎟ ∂y ⎜ ∂y S ⎟ ⎝ xy yx ⎠⎩ ⎝ x ⎠⎭⎩⎪ ⎝ y ⎠⎭⎪ 2 2 ⎧ ⎫ 2 ⎤ Dxy ⎪ ∂ ⎛ ∂w Ty ⎞ ∂ ⎛ ∂w T ⎞⎪ T Ty + ⎜ − ⎟ + ⎜ − x ⎟ + x + ⎥dxdy ⎨ ⎜ ⎟ ⎜ ⎟⎬ 2 ⎪∂x ∂y S y ∂y ∂x S x ⎪ S x S y ⎥ ⎩ ⎝ ⎠ ⎝ ⎠⎭ ⎦

Finally, one can approximately write the energy in terms of partial deflections

2 2 2 2 ⎡ ⎛ ⎞ ν D +ν D ⎛ ⎞⎛ ⎞ 1 Dx ⎜ ∂ wb ⎟ yx x xy y ⎜ ∂ wb ⎟⎜ ∂ wb ⎟ U = ⎢ + + se 2 ∫∫ 1−ν ν ⎜ 2 ⎟ 1−ν ν ⎜ 2 ⎟⎜ 2 ⎟ ⎣⎢ xy yx ⎝ ∂x ⎠ xy yx ⎝ ∂x ⎠⎝ ∂y ⎠ (8.42c) 2 2 ⎛ 2 ⎞ ⎛ 2 ⎞ 2 2 ⎤ Dy ∂ w ∂ w ⎛ ∂w ⎞ ⎛ ∂w ⎞ + ⎜ b ⎟ + 2D ⎜ b ⎟ + S ⎜ s ⎟ + S ⎜ s ⎟ ⎥dxdy 1−ν ν ⎜ 2 ⎟ xy ⎜ ∂x∂y ⎟ x ∂x y ⎜ ∂y ⎟ xy yx ⎝ ∂y ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦⎥

The potential energy of the external forces of the plate is independent of the plate's internal structure and depends only on the displacement of the middle surface. Therefore, the potential energy expression for the general plate is the same as that for a homogeneous isotropic plate, that is, the potential energy of the forces Nx, Ny and Nxy. If the potential energy of the transverse load q is added to that the resulting expression will be

2 2 1 ⎡ ⎛ ∂w ⎞ ⎛ ∂w ⎞⎛ ∂w ⎞ ⎛ ∂w ⎞ ⎤ U = ⎢− 2qw + N ⎜ ⎟ + 2N ⎜ ⎟⎜ ⎟ + N ⎜ ⎟ ⎥dxdy (8.43) p 2 ∫∫ x ∂x xy ∂x ⎜ ∂y ⎟ y ⎜ ∂y ⎟ ⎣⎢ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎦⎥

The kinetic energy per unit area of the plate can be written as

22 2 ρ ⎡⎛ ∂u⎞ ⎛ ∂v⎞ ⎛ ∂w⎞ ⎤ ∫ ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎥dz 2 ⎣⎝ ∂t ⎠ ⎝ ∂t ⎠ ⎝ ∂t ⎠ ⎦

By first integrating this over the thickness of the plate and then over the entire plate we get

⎡ 2 ⎛ 2 2 2 2 ⎞⎤ 1 * ⎛ ∂w ⎞ ⎜ ⎛ ∂ wb ⎞ ⎛ ∂ wb ⎞ ⎟ U k = ⎢ρ ⎜ ⎟ + Rx ⎜ ⎟ + Ry ⎜ ⎟ ⎥dxdy (8.44) 2 ∫∫⎢ ⎝ ∂t ⎠ ⎜ ⎜ ∂x∂t ⎟ ⎜ ∂y∂t ⎟ ⎟⎥ ⎣ ⎝ ⎝ ⎠ ⎝ ⎠ ⎠⎦

The potential of the entire system is then, since the strain energy plus the potential energy of the applied loads must equal the kinetic energy. We can then define an energy potential as

8.18 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

Π = Use + Up + Uk (8.45) which later can be used to solve plate problems by the so called Ritz' method.

8.9 Stresses and Strains The stresses and strains in the plate can be found once the deflection field and the transverse force and bending moment distributions are computed using the above. However, for a generally orthotropic cross-section the neutral axes in the x- and y-directions will be in different positions giving rise to some difficulty when calculating the stresses and strains. This adds severe complexity since the definition of the middle plane is no longer well defined. By simply neglecting this fact in all cases except in calculating the flexural rigidities as above, only the stress calculation becomes rather dubious. The fact that different neutral axes make the definition of the middle plane difficult has not been treated in the open literature. In practical applications one seldom finds plates that have a very high degree of orthotropy so even if the neutral axes in the different principal directions do indeed differ, this difference should be fairly small. The problem arising is really that the secondary stresses appearing due to the Poisson effect relate to a different neutral axis. However, these stresses are always much lower than the primary stresses and the error made by neglecting this difference is supposedly small. Without getting into too much detail of the stress analysis we can derive the following:

Direct stresses By using the generalised Hooke's law [4] and the bending moment relation in eq.(8.15) we get the approximate relation based on the partial deflection assumption

MzExx x Myyz E y σσx == and y (8.46a) Dx Dy where E must be taken as the modulus of the component at the specific z-coordinate. This expression should be fairly accurate providing the plate is moderately orthotropic. For sandwiches with thin dissimilar faces and a weak core we have pure membrane face stresses equalling

MxxE 122E xtd MxxE 121E xtd σσfx1 ≈− , fx2 ≈ (8.46b) DEtxx()11+ Et x 2 2 DEtxx ()11+ Et x 2 2

MyyE 122E ytd MyyE 121E ytd σσfy1 ≈− , fy2 ≈ , and σc ≈ 0 DEtyy()11+ Et y 2 2 DEtyy ()11+ Et y 2 2 and for sandwiches with thin (not necessarily equal) faces and a weak core we get [11]

Mx Mx My My σσσσfx1 ≈− , fx2 ≈ , fy1 ≈− , fy2 ≈ , and σc ≈0 (8.46c) td1 td2 td1 td2 The sign of the stresses will differ between the two faces so that for a positive bending moment the stress in lower face (positive z-coordinate) will exhibit a tensile (positive) stress whereas the upper face will be in compression. The strains are according to the generalised Hooke's law

8.19 AN INTRODUCTION TO SANDWICH STRUCTURES

σ x σ y σ y σ x ε x =−νyx , ε y =−νxy (8.46d) EEx y EEy x

The stresses and strains appearing due to in-plane loads are calculated in the same manner, and appear as

N x ν yx N y N x ν yx N y ε x0 = − = − (8.46e) Ex1t1 + Exctc + Ex2t2 E y1t1 + E yctc + E y2t2 Ax Ay

ν xy N x N y N y ν xy N x ε y0 = − + = − Ex1t1 + Exctc + Ex2t2 E y1t1 + E yctc + E y2t2 Ay Ax

N xy N xy γ xy0 = = Gxy1t1 + Gxyctc + Gxy2t2 Axy where the Poisson ratios are for the entire assembly, i.e., for the plate and not for each layer. The corresponding stresses are calculated using Hooke's law for each layer.

N x N x σ f 1x = Ex1 , σ f 2x = Ex2 (8.46f) Ax Ax

N y N y σ f 1y = E y1 , σ f 2 y = E y2 Ay Ay

N xy N xy τ xy1 = Gxy1 and τ xy2 = Gxy2 Axy Axy

The strains and stresses due to bending and in-plane loads can then be superimposed in linear plate analysis.

In-plane shear stress:

2Mxyz γτγxy == and xyG xy xy (8.47a) Dxy where Gxy should be taken as the shear modulus of the material at coordinate z. For sandwiches with thin faces and a weak core we have

Mxy Mxy γττxy ≈≈±≈ and thus fxy , and cxy 0 (8.47b) Gtdfxy f tdf

Core shear stress: The exact stresses are assessed in the same way as for the beam and can be computed using eqs.(3.13) and (3.14) in the x- and y-directions, respectively. The thin face, weak core approximation leads to

8.20 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

Tx Ty τcxz τcyz ττγcxz === , cyz , cxz , and γ cyz = (8.48) d dGcx Gcy

8.10 Thermal Stresses and Deformations A temperature change in a material induces strains proportional to the temperature change. Since a sandwich in a general case is anisotropic, these strains may cause bending and warping as an effect. The thermoelastic stress-strain relation may be written

εi=+∑ ST ijσ jδ ijα i Δ or inverted to σεδαiijjijj=−∑ CT( Δ ) , i = 1...6 (8.49) j j

where αι are the coefficients of thermal expansion, δij the Kronecker delta function (which equals 1 if i = j but is otherwise zero), Sij the material compliance matrix, and Cij the material stiffness matrix. Note that the coefficients of thermal expansion only affects extensional strains, not the shear strains. Thus, if the axes of symmetry of the plate coincides with the axes of orthotropy, thermal strains will not cause the plate to twist or induce any twisting moments.

In the case of a sandwich, both the faces and the core may in general be treated as orthotropic and faces are usually thin, implying than only in-plane characteristics are of interest, thus reducing the face stiffness matrices to only 4 independent constants (see chapter 2.2.5). If strains are prevented from being developed, e.g., a plate is fixed within a stiff frame, stresses develop according to eq.(8.49). If, on the other hand, the structure is free to move, the thermal strains will induce some thermal deformation field.

The above characteristics do have some important effects on sandwich plates: Consider a sandwich plate with dissimilar faces (see Fig.3.7 or 8.4), also having dissimilar coefficients of thermal expansion, α1x, α1y, α2x, and α2y, and antiplane core (Ec = 0). Further assume that the temperature and temperature change are constant over the face, but not necessarily the same in both faces. If transverse deformation is free to take place, the plate will curve if the temperature is changed on one side of the plate (typically refrigerated tanks, truck-bodies, train-cars or building panels), or even if the temperature is changed over the entire body. The latter case is the most general so assume a temperature change ΔT1 one the side of face 1 and

ΔT2 on the side of face 2.

(i) Deformation constrained In an orthotropic plate, for which the deformations are constrained, stresses will develop according to eq.(8.49). For face 1 the stresses will according to generalised Hooke's law be (plane stress assumed)

()T ETx Δ 1 ()T Ey ΔT1 σ 1x =− []ανα111xyxy+ and σ 1y =− []ανα111yxyx+ (8.50) 1 − νν11xy yx 1 − νν11xy yx and similarly for face 2. These stresses act as force-couples equivalent of a bending moment and an in-plane load. The latter is simply the face stresses times the face thickness on which they act, i.e.

8.21 AN INTRODUCTION TO SANDWICH STRUCTURES

()T ()T ()T ()T ()T ()T NttNttx =+σ11x σ 22x and y =σ 11y + σ 22y (8.51)

Since the face strains in the general case are different (α1x ≠ α2x, α1y ≠ α2y and/or ΔT1 ≠ ΔT2) the asymmetry implies the presence of bending moments which derive to

()T DTx Δ 1 DTx Δ 2 M x = []ανα111xyxy+− []ανα222xyxy+ (8.52a) ()11− νν11xy yx d (− νν22xy yx )d

()T Dy ΔT1 Dy ΔT2 M y = []ανα111yxyx+− []ανα222yxyx+ ()11− νν11xy yx d (− νν22xy yx )d

or if ν1xy = ν2xy = νxy and ν1yx = ν2yx = νyx it simplifies to

D M ()T =− x αανααΔΔTT−+ ΔΔ TT − x []22xxyxyy 11() 22 11 ()1 − ννxy yx d

()T Dy My =− []ααναα22yyxyxxΔΔTT−+ 11() 22 ΔΔ TT − 11 (8.52b) ()1 − ννxy yx d

The expressions simplify considerably if an isotropic plate (all α are equal) is considered and the bending moments can then be written

()T ()T Dα MMx ==−y []ΔΔTT21− (8.53) ()1 − ν d

Thus, if the deformation due to thermal strains is prevented, forces and bending moments are introduced into the structure which has to be accounted for.

(ii) Deformation unconstrained Another case is that when a plate or structure is free to deform as the thermal strains develop. Contrary to the constrained case the plate will now exhibit in-plane deformations and curvatures in stead of in-plane loads and bending moments. Since only direct strains act in the faces there will be no global transverse shear presence and the curvatures may be defined as in chapter 8.1 but without respect to in-plane shear. The thermal strains are then

()T ()T ()T ()T ()T ()T ε x =+ε x00zzxxκ and ε y =ε y + yyκ (8.54)

The in-plane deformation at the neutral axis depends on the average face strain and is given by

()T 11 εαx02211=−+≈[]xxxxΔΔTd() e α Te [] α22221111xx ΔΔTE t+ α xx TE t d Et11xx+ Et 2 2

()T 11 εαy02211=−+≈[]yyyyΔΔTd() e α Te [ α22221111yy ΔΔTE t+ α yy TE t](8.55) d Et11yy+ Et 2 2

8.22 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

where ex is the distance from the neutral axis in the x-direction to face 2 and ey the distance in the y-direction (see chapter 3). The strains in the faces may also be written as function of the curvatures as

()T ()T ε x10=+−=−Δε x ()deκ xxα 11 T

()T ()T ε x20=+ε x eTκ xx= −Δα 22 and by subtracting the first with the second one arrives at

()T 1 ()T 1 κααx =−()22xxΔΔTT 11 and similarly κααy =−22yyΔΔTT 11 (8.56) d d ( ) For an isotropic sandwich (all α, E and ν are equal) these expressions reduce to

()T ()T Eα ()T ()T α εεx00==y []tT11ΔΔ+==− tT 2 2 and κκx 0y 0() ΔΔTT 2 1 (8.57) tt12+ d and now it is clearly seen that if the both faces exhibit the same change in temperature, the isotropic plate only extends in its plane and no curvature will develop.

8.11 General Sandwich Theory for Anisotropic Plates In some textbooks the theory of bending, buckling and free vibration of sandwich plates is derived using an extension of the classical lamination theory. In a sense this perhaps the most logical and straightforward way to do it, but may on the other hand become rather more complex and less a physical approach. Start by recapitulating some basic relations [4] and let the material be built-up of any anisotropic material described by a plane strain stiffness matrix Q in a classical way as (see chapter 2.2.5)

t t σ = Qε, with σ = [σx, σy, τxy] and ε = [εx, εy, γxy] (8.58)

⎡Q Q Q ⎤ ⎡ E1 ν 12E 2 0 ⎤ 11 12 16 1 Q = ⎢Q Q Q ⎥ with Q = ⎢ν E E 0 ⎥ ⎢ 12 22 26 ⎥ l ⎢ 21 1 2 ⎥ 1 − νν12 21 ⎣⎢Q16 Q26 Q66 ⎦⎥ ⎣⎢ 00G12()1 − νν 12 21 ⎦⎥ where Q is given by in the global x, y, z coordinate system given by the plate and is simply a stiffness transformation of the local orthotropic stiffness matrix Ql. Assume that the sandwich is built-up of an arbitrary number of laminar plies in each face sheet and by an orthotropic core material. The local lamina stiffness matrices are assumed known (as they usually are), as are their position in the sandwich assembly and their orientation. If so, the Q-matrix can be obtained for each layer. In the ordinary way we can now derive local A, B and D-stiffness matrices for each face and the core [4]. These are found by integration as

N t f 1 /2 1 1 ABD,, ==−−− Q(,1 zzdz ,2 ) Q⎜⎛ z z , ( z2 z23 ), ( z z3 )⎟⎞ []f 1 ∫ ff11 fkkkkkkk11∑ −−1 −1 −t f 1 /2 ⎝ ⎠ k=1 2 3

8.23 AN INTRODUCTION TO SANDWICH STRUCTURES

3 tc /2 ⎛ t ⎞ ABD,, ==QQ(,10zz ,2 ) dz t , , c (8.59) []c ∫ c cc⎜ ⎟ −tc /2 ⎝ 12⎠

M t f 2 /2 1 1 A,,BD==−−− Q (,1 zzdz ,2 ) Q⎜⎛ z z ,( z2 z23 ),( z z3 )⎟⎞ []f 2 ∫ ff22 fkkkkkkk21∑ −−1 −1 −t f 2 /2 ⎝ ⎠ k=1 2 3 where the middle axis is taken as the middle of the core. Next assume that the faces are thin so that the bending about their individual neutral axes may be disregarded. This assumption makes Bf1 = Df1 = Bf2 = Df2 = 0 and thus the faces are described by their extensional stiffnesses

Af1 and Af1 only. The core, on the other hand may still be treated using both its A and D- matrices.

Now, the sandwich plate stiffnesses may be derived using the above notation. For a sandwich with thin faces and a weak core the stiffness of the sandwich as a whole now becomes

d d 2 A = A + A + A , B = ()A − A and D = ()A + A (8.60) f1 c f2 2 f 2 f 1 4 f 1 f 2 The coupling term B in this case does not arise due to asymmetry in the faces since they were considered thin but as a result of different thickness and/or different stiffness of the two face sheets. This in turn depends on the fact that the middle axis generally does not coincide with the neutral plane of the plate. For an orthotropic plate (B = 0 and D16 = D26 = 0) the bending stiffness terms in D are connected to the former representation as

Dx Dy νyxDD x νxy y D11 = , D22 = , D12 = = , 2D66 = Dxy 1− ννxy yx 1− ννxy yx 11− ννxy yx − ννxy yx whereas the in-plane stiffness matrix components in A are more obvious. In order to account for the shear properties of the core we must also define the relation between out-of-plane shear stresses and strains. This is done in a similar manner as for in-plane strains and curvatures, by defining the relation between transverse forces and out-of-plane shear strains (see chapter 2 or [13])

⎛T1 ⎞ ⎡S11 S12 ⎤⎛γ 13 ⎞ ⎜ ⎟ = ⎢ ⎥⎜ ⎟ or Τ = Sγ (8.61) ⎝T2 ⎠ ⎣S21 S 22 ⎦⎝γ 23 ⎠ where the matrix components Sij denote the anisotropic shear stiffnesses [13], in this case the direct relations between τxz and γxz, τxz and γyz, and τyz and γyz. However, most core materials are at least orthotropic (balsa, honeycombs) or even isotropic (most foams) and in such cases the components in eq.(8.61) reduce to

S11 = tcGcxz ≈ Sx, S12 = S21 = 0, and S22 = tcGcyz ≈ Sy which is the same as prior derivations assuming thin faces. For orthotropic cores, the transverse shear properties can be described in the materials own local co-ordinate and then transformed to the global panel fixed co-ordinate system by transformation, as described in chapter 2, eq.(2.25).

8.24 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

Next assume the classical Reissner/Mindlin kinematic field displacement of eq.(8.1) but let it refer to the geometrical middle-plane (middle of the core) of the plate rather than any neutral plane. In this way we avoid the problem of finding the neutral plane and having to refer to that specific position, and also the problem that the neutral planes in x- and y-axes do not coincide. On the other hand, as in classical lamination theory, we will encounter (and rightfully use) coupling terms between tension-shear, tension-bending, and bending-twisting. Further kinematic relations are those for strains and curvatures [14,15]

t ε = ε0 + zκ with ε0 = [εx0, εy0, γxy0] ,

t t ε = [εx, εy, γxy] , and κ = [κx, κy, κxy] (8.62)

∂θ ∂θ ∂θ ∂θ ∂w ∂w κ ==x ,, κ y κ =+x y , γ =+θ and γ =+θ (8.63) x ∂xyy ∂ xy ∂ yx∂ xz∂x x yz∂y y where ε is the mid-plane strain vector, and κ the curvature vector. The strain energy of the plate may be written in terms of stiffness matrix components. To simplify the equations make the following very reasonable assumptions: the out-of-plane stress σzz is negligible and the face shear stresses are small. These two stress components are then assumed not to add to the total strain energy of the system. This leads to an expression

1 tc / 2+t f 1 W = σ tεdz 2 ∫−tc / 2−t f 2 t / 2+t −t / 2 t / 2 1 c f 1 t 1 c t 1 c t ≈ σ f 1ε f 1dz + σ f 2ε f 2 dz + ()σ cε c +τ xzγ xz +τ yzγ yz dz 2 ∫tc / 2 2 ∫−tc / 2−t f 2 2 ∫−tc / 2

1 ⎛ t d t ⎞ t ⎛ d ⎞ 1 ⎛ t d t ⎞ t ⎛ d ⎞ = ⎜ε 0 + κ ⎟A f 1 ⎜ε 0 + κ ⎟ + ⎜ε 0 − κ ⎟A f 2 ⎜ε 0 − κ ⎟ (8.64) 2 ⎝ 2 ⎠ ⎝ 2 ⎠ 2 ⎝ 2 ⎠ ⎝ 2 ⎠ −t / 2 1 c t t 1 1 + ()ε 0c + zκ Qc ()ε 0 + κ dz + S11γ xzγ xz + S 22γ yzγ yz + S12γ xzγ yz 2 ∫tc / 2 2 2 1 1 1 = ε t Aε + ε t Bκ + κ t Dκ + γ t Sγ 2 0 0 0 2 2 The relations between normal forces and bending moments to mid-plane strains and curvatures are then given from eqs.(8.63) and (8.64) as ∂W N = = Aε + Bκ with N = [N , N , N ]t (8.65) ∂ε 0 x y xy

∂W M = = Bε + Dκ with M = [M , M , M ]t ∂κ 0 x y xy

∂W ⎛ ∂w ⎞ ⎛ ∂w ⎞ Tx = = S11 ⎜ +θ x ⎟ + S12 ⎜ + θ y ⎟ ∂γ xz ⎝ ∂x ⎠ ⎝ ∂y ⎠

8.25 AN INTRODUCTION TO SANDWICH STRUCTURES

∂W ⎛ ∂w ⎞ ⎛ ∂w ⎞ Ty = = S 22 ⎜ +θ y ⎟ + S12 ⎜ +θ x ⎟ ∂γ yz ⎝ ∂y ⎠ ⎝ ∂x ⎠

Let’s now return to the equilibrium equations of eqs.(8.19) and (8.20). First we must realise that inertia effects will take a different from when transverse is included. Vertical and in- plane inertia is first discussed followed by the more intricate rotary or rotatory inertia

Vertical inertia The body force acting on the element when subjected to an acceleration ∂2w/∂t2 is

∂ 2 w ∂ 2 w −=−ρ dz ρ * , with ρρ* = dz (8.66) ∫ ∂t 2 ∂t 2 ∫ where ρ* is the surface mass (mass per unit length per unit width) of the beam which reduces to ρt for a homogeneous cross-section. This mass may of course be a function of x if the beam has a varying cross-section. Hence, an acceleration in positive w-direction (downwards) creates a body force with opposite direction (upwards).

In-plane inertia Similarly to the above, an in-plane acceleration creates inertia forces which may be written as

∂ 2u ∂ 2u ∂θ2 ∂ 2u ∂θ2 −=−−ρ dz ρ 0 dzρ z xxdz =−−ρ * 0 H and (8.67) ∫∫∫∂t 2 ∂t 2 ∂t 2 ∂t 2 ∂t 2

∂ 2 v ∂ 2 v ∂θ2 ∂ 2 v ∂θ2 −=−−ρ dz ρ 0 dzρ z yydz =−−ρ * 0 H ∫∫∫∂t 2 ∂t 2 ∂t 2 ∂t 2 ∂t 2

where Hzdz= ∫ ρ (8.68)

(ii) Rotary inertia (or rotatory inertia) The cross-section rotates when bending, and this rotation equals θx or θy. The x-displacement of a particle somewhere in the cross-section is given by the kinematic assumption in eq.(8.1). An angular acceleration creates an inertia bending moment in the opposite direction with magnitude

∂ 2u ∂ 2u ∂ 2θ ∂ 2u ∂ 2θ − ρz dz = − ρz 0 dz − ρz 2 x dz = −H 0 − R x (8.69) ∫ ∂t 2 ∫ ∂t 2 ∫ ∂t 2 ∂t 2 ∂t 2

where R = ∫ ρz 2 dz (8.70) R is the rotary inertia which reduces to ρh3/12 for a homogeneous cross-section. Note that using the definition of coordinate system of classical lamination theory with the geometrical middle plane as reference, the property R becomes direction independent. The equations take the same form in the y-direction. Now, the equations of equilibrium take the following form

8.26 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

∂T ∂T ∂ 2 w ∂ 22w ∂ w ∂ 2 w Vertical: x ++y N +20N +N +−q ρ * = (8.71a) ∂x ∂y xxyy∂x 2 ∂∂xy ∂y 2 ∂t 2

∂N N ∂ 2u ∂θ2 In-plane x +−xy ρ * ox −H =0 (8.71b) ∂∂x y ∂t 2 ∂t 2

∂N N ∂ 2 v ∂θ2 xy+− y ρ * o −H y =0 ∂∂x y ∂t 2 ∂t 2

∂M ∂M ∂ 2u ∂ 2θ Rotation T − x − xy + H o + R x = 0 (8.72c) x ∂x ∂y ∂t 2 ∂t 2

∂M ∂M ∂ 2v ∂ 2θ T − xy − y + H o + R y = 0 y ∂x ∂y ∂t 2 ∂t 2

The parameters ρ*, H and R are similar to the A, B and D-matrices in terms of definition. For a symmetric laminate, H = 0 but also that a laminate made of the same material for which ρ is equal for each layer H is also zero. By substituting the constitutive relations of eqs.(8.65) into the above equilibrium equations and exercising some algebra, we arrive at a new set of equations instead of (8.28) and (8.29) which now will appear as (see also [13])

∂ 2u ∂ 2u ∂ 2u ∂ 2 v ∂ 2 v ∂ 2 v A 0 ++++++2A 0 A 0 A 0 ()AA 0 A 0 11 ∂x 2 16 ∂∂xy 66 ∂y 2 16 ∂x 2 12 66 ∂∂xy 26 ∂y 2 ∂θ2 ∂θ2 ∂θ2 ∂θ2 ++B xxx2B +B +B y + (8.73a) 11 ∂x 2 16 ∂∂xy 66 ∂y 2 16 ∂x 2 ∂θ2 ∂θ2 ∂ 2u ∂θ2 ()BB++−−=yyB ρ * oxH 0 12 66 ∂∂xy 26 ∂y 2 ∂t 2 ∂t 2

∂ 2u ∂ 2u ∂ 2u ∂ 2 v ∂ 2 v ∂ 2 v A 0 ++()AA 0 +A 0 +A 0 +2A 0 +A 0 16 ∂x 2 12 66 ∂∂xy 26 ∂y 2 66 ∂x 2 16 ∂∂xy 22 ∂y 2 ∂θ2 ∂θ2 ∂θ2 ∂θ2 ++++B xxx()BB B +B y + (8.73b) 16 ∂x 2 12 66 ∂∂xy 26 ∂y 2 66 ∂x 2 ∂θ2 ∂θ2 ∂ 2 v ∂θ2 20B y +−−=B xoρ * H y 26 ∂∂xy 22 ∂y 2 ∂t 2 ∂t 2

8.27 AN INTRODUCTION TO SANDWICH STRUCTURES

∂ 2u ∂ 2u ∂ 2u ∂ 2v ∂ 2v ∂ 2v B 0 + 2B 0 + B 0 + B 0 + (B + B ) 0 + B 0 11 ∂x 2 12 ∂x∂y 66 ∂y 2 16 ∂x 2 12 66 ∂x∂y 26 ∂y 2 ∂ 2θ ∂ 2θ ∂ 2θ ∂ 2θ ∂ 2θ ∂ 2θ + D x + 2D x + D x + D y + (D + D ) y + D x (8.73c) 11 ∂x 2 16 ∂x∂y 66 ∂y 2 16 ∂x 2 12 66 ∂x∂y 26 ∂y 2 ⎛ ∂w ⎞ ⎛ ∂w ⎞ ∂ 2u ∂ 2θ − S θ + − S ⎜θ + ⎟ − H o − R x = 0 11 ⎜ x ⎟ 12 ⎜ y ⎟ 2 2 ⎝ ∂x ⎠ ⎝ ∂y ⎠ ∂t ∂t

∂ 2u ∂ 2u ∂ 2u ∂ 2v ∂ 2v ∂ 2v B 0 + (B + B ) 0 + B 0 + B 0 + 2B 0 + B 0 16 ∂x 2 12 66 ∂x∂y 26 ∂y 2 66 ∂x 2 26 ∂x∂y 22 ∂y 2 ∂ 2θ ∂ 2θ ∂ 2θ ∂ 2θ ∂ 2θ ∂ 2θ + D x + (D + D ) x + D x + D y + 2D y + D x (8.73d) 16 ∂x 2 12 66 ∂x∂y 26 ∂y 2 66 ∂x 2 26 ∂x∂y 26 ∂y 2

2 2 ⎛ ∂w ⎞ ⎛ ∂w ⎞ ∂ v ∂ θ y − S θ + − S ⎜θ + ⎟ − H o − R = 0 12 ⎜ x ⎟ 22 ⎜ y ⎟ 2 2 ⎝ ∂x ⎠ ⎝ ∂y ⎠ ∂t ∂t

⎛ ∂θ ∂ 2 w ⎞ ⎛ ∂θ ∂θ ∂ 2 w ⎞ ⎛ ∂θ ∂ 2 w ⎞ S ⎜ x + ⎟ + S ⎜ x + y + 2 ⎟ + S ⎜ y + ⎟ 11 ⎜ ∂x ∂x 2 ⎟ 12 ⎜ ∂y ∂x ∂x∂y ⎟ 22 ⎜ ∂y ∂y 2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ (8.73e) ∂ 2 w ∂ 2 w ∂ 2 w ∂ 2 w + N + 2N + N + q − ρ * = 0 x ∂x 2 xy ∂x∂y y ∂y 2 ∂t 2

These equations are all coupled in the five field variables

u0, v0, w, θx, and θy If the coupling stiffness matrix B is zero some simplifications may be done. If the faces are dissimilar then B is in fact non-zero, but by calculating all stiffness matrix components about the true neutral axis, rather than from the geometrical middle plane, B may become very small or zero. This is done by recalculating all properties in eqs.(8.59) and (8.60) about the neutral axis, in a similar manner as described in section 8.7. If so, the five coupled differential equations take the form (S12 is assumed to zero)

∂ 2u ∂ 2u ∂ 2u ∂ 2 v ∂ 2 v ∂ 2 v A 0 + 20A 0 + A 0 + A 0 ++()AA 0 + A 0 = (8.74a) 11 ∂x 2 16 ∂∂xy 66 ∂y 2 16 ∂x 2 12 66 ∂∂xy 26 ∂y 2

∂ 2u ∂ 2u ∂ 2u ∂ 2v ∂ 2v ∂ 2v A 0 + (A + A ) 0 + A 0 + A 0 + 2A 0 + A 0 = 0 (8.74b) 16 ∂x 2 12 66 ∂x∂y 26 ∂y 2 66 ∂x 2 26 ∂x∂y 22 ∂y 2

∂ 2θ ∂ 2θ ∂ 2θ ∂ 2θ ∂ 2θ D x + 2D x + D x + D y + ()D + D y 11 ∂x 2 16 ∂x∂y 66 ∂y 2 16 ∂x 2 12 66 ∂x∂y (8.74c) 2 2 ∂ θ y ⎛ ∂w ⎞ ∂ θ x + D26 − S x ⎜ +θ x ⎟ − Rx = 0 ∂y 2 ⎝ ∂x ⎠ ∂t 2

8.28 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

∂ 2θ ∂ 2θ ∂ 2θ ∂ 2θ ∂ 2θ D x + ()D + D x + D x + D y + 2D y 16 ∂x 2 12 66 ∂x∂y 26 ∂y 2 66 ∂x 2 26 ∂x∂y (8.74d) ∂ 2θ ⎛ ∂w ⎞ ∂ 2θ + D y − S ⎜ +θ ⎟ − R x = 0 22 2 y ⎜ y ⎟ y 2 ∂y ⎝ ∂y ⎠ ∂t

⎛ ∂ 2 w ∂θ ⎞ ⎛ ∂ 2 w ∂θ ⎞ ∂ 2 w S ⎜ + x ⎟ + S ⎜ + y ⎟ + q* − ρ * = 0 (8.74e) x ⎜ 2 ⎟ y ⎜ 2 ⎟ 2 ⎝ ∂x ∂x ⎠ ⎝ ∂y ∂y ⎠ ∂t

For this case it is seen that in-plane motion, eqs.(8.74a-b), becomes uncoupled to the out-of- plane deformation in eqs.(8.74c-e). The latter three equations of (8.74) constitutes three coupled differential equations for the deformation of the plate in the three field variables w, θx and θy.

At this stage we can once again make the simplification of introducing partial deflections. As before, this can be done as in eqs.(8.13) and (8.14) by introducing

∂w ∂w w = w + w , θ = b and θ = b b s x ∂x y ∂y

The governing equation (8.74e) is further simplified to

∂T ∂T ∂ 2 w ∂ 2 w x + y + q* = S s + S s + q* = 0 (8.75) ∂x ∂y x ∂x 2 y ∂y 2 i.e., an equation very similar (except for the shear stiffness) to that derived in eq.(8.17a). An equivalent equation for the bending part may be derived from the relation between bending moments in eq.(8.65). For B = 0 the differential equation then takes the form of an ordinary anisotropic plate [4]

∂ 4 w ∂ 4 w ∂ 4 w ∂ 4 w ∂ 4 w D b + 4D b + 2(D + 2D ) b + 4D b + D b = q* (8.76) 11 ∂x 4 16 ∂x3∂y 12 66 ∂x 2∂y 2 26 ∂x∂y 3 22 vy 4

Similar equations for the in-plane deformations may also be obtained [4] [13], and also these will be coupled with the out-of-plane deformation variables wb and ws. Thus, by introducing partial deflections, the plate problem is solved as two separate problems: one of pure bending, eq.(8.76) and one for pure shear, eq.(8.75).

8.12 Boundary Conditions The boundary conditions for the most common types of edge supports in practice are: complete freedom, simple support and clamped. There are others, like elastic boundary support, which are treated in [2] and [12]. For simplicity, study boundary conditions in a Cartesian coordinate system for a rectangular panel, as schematically illustrated in Fig.8.5.

(i) Free edge The boundary condition for a free unloaded edge can be expressed as

8.29 AN INTRODUCTION TO SANDWICH STRUCTURES

x = 0,a Mx = Mxy = Tx = 0 (8.77a)

y = 0,b My = Mxy = Ty = 0

x b q(x,y)

Figure 8.5 Rectangular sandwich panel These are, however, three boundary conditions and only two may be specified at each edge to solve the governing equation. According to classical Kirchhoff theory the two latter conditions in eq.(8.77a) may be rewritten so that the resultant vertical force is set to zero at the edge. From eq.(8.7) we can specify that, for example an edge parallel to the y-axis at x =

0,a, we have that Mx = 0 and that Tx − ∂Mxy/∂y = 0. Thus, eq.(8.74a) is equivalent to ∂M x = 0,a M = 0 and T −=xy 0 (8.77b) x x ∂y

∂M y = 0,b M = 0 and T −=xy 0 y y ∂x

If the edge carries load, the mid-plane forces Nx, Ny and Nxy will not be zero and the boundary condition for the transverse load becomes ∂w ∂w x = 0,a TN++N =0 (8.77c) xx∂x xy∂y

∂w ∂w y = 0,b TN++N =0 yy∂y xy∂x

The boundary conditions must usually be written in terms of deflection, or partial derivatives of the deflection. This can be done by using e.g. eq.(8.8) in terms of deflections and transverse forces, or by inserting the definition of rotations in eq.(8.13), in terms rotations only. The same thing goes for the twisting moment Mxy. Transverse forces are generally primary unknowns and should thus be specified directly in the boundary conditions (see eqs.(8.10) and (8.11)).

(ii) Simply supported edge The principal boundary conditions for a simply supported edge are that deflections and bending moments Mx or My are zero. Now, there are two different conditions that may be imposed as the third boundary condition, as illustrated in Fig.8.6. One may set the twisting moment Mxy to zero thus permitting the edge to shear, e.g. on an edge parallel to the x-axis then γxz ≠ 0. This is a rather unrealistic condition since in most practical cases there will be an edge stiffener, or some symmetry constraint preventing such shearing deformation. If so, the

8.30 SANDWICH PLATES – FUNDAMENTAL EQUATIONS boundary condition should rather be allowing the existence of a twisting moment but restricting the shear of the edge. The former condition, that the twisting moment along the edges are zero (shearing permitted) is called the soft boundary condition, and the latter, preventing shear deformations, is called the hard boundary condition. These are then written as

Mxy = 0 γxz = 0

x γxz Mxy z "soft" "hard"

Figure 8.6 Definition of boundary conditions for sandwich plates (shown for y = 0 or b). Soft boundary: x = 0,a w = Mx = Mxy = 0 (8.78a)

y = 0,b w = My = Myx = 0

In this case, since both ∂My/∂y and ∂Mxy/∂x are zero along a free edge parallel to the y-axis the resultant force (the vertical force acting on the support) equals the transverse force Ty. For the same reason no resulting forces at the corners appear in this case.

Hard boundary: x = 0,a w = Mx = γyz = 0 (8.78b)

y = 0,b w = My = γxz = 0 where γxz also can be written as ∂u/∂z and γyz as ∂v/∂z, since ∂w/∂x = ∂w/∂y = 0 on the boundaries. In this case, on the other hand, the resultant vertical forces are [1] ∂M ∂M VT=− xy for x = 0,a and VT=− xy for y = 0,b xx∂y yy∂x which at the corners add up to a resultant point force equal to 2Mxy, as in ordinary Kirchhoff theory.

In order to specify the boundary conditions in terms of the primary unknowns. A bending moment Mx for example can be written using partial deflections according to eq.(8.15) as

2 2 Dx ⎡∂ wb ∂ wb ⎤ M x = − ⎢ 2 +ν yx 2 ⎥ = 0 1−ν xyν yx ⎣ ∂x ∂y ⎦

∂ 2 w ∂θ → b = 0 or x = 0 along the boundaries x = 0,a ∂x 2 ∂x

8.31 AN INTRODUCTION TO SANDWICH STRUCTURES

2 2 since there ∂ wb/∂y must be zero (no curvature in the y-direction along those boundaries). From eq.(8.16) defining the relation between wb and ws indicates that

2 4 4 4 ∂ ws Dwx ⎡∂ b ∂ wb ⎤ ∂ wb Sx 2 =− ⎢ 4 + ν yx 22⎥ − Dxy 22 at x = 0,a ∂ννx 1 − xy yx ⎣ ∂x ∂∂xy⎦ ∂∂xy

(iii) Clamped edge The principal boundary conditions characterising a clamped edge are zero displacements of the middle surface and zero rotation of the cross-section at the boundary. The latter requirement that the cross-section remains parallel to the z-axis is satisfied by letting ∂w/∂x =

Tx /Sx (note that this is no rotation of the cross-section but a sliding.). This assumes thin faces since the local rotation Tx /Sx of the faces can not take place unless the faces are membranes. If now Sx approaches infinity then this condition reduces to the ordinary boundary condition ∂w/∂x = 0. Just as in the simply supported case there are two ways to describe the third boundary condition. Thus,

Soft boundary:

∂w Tx ∂wb x = 0,a w = 0, − = 0 or = θ x = 0 and Mxy = 0 (8.79a) ∂x S x ∂x

∂w Ty ∂wb y = 0,b w = 0, − = 0 or = θ y = 0 and Mxy = 0 ∂y S y ∂y

Hard boundary:

∂w Tx ∂wb x = 0,a w = 0, − = 0 or = θ x = 0 and γyz = 0 (8.79b) ∂x S x ∂x

∂w Ty ∂wb y = 0,b w = 0, − = 0 or = θ y = 0 and γxz = 0 ∂y S y ∂y

However, if the faces are considered thick, then indeed the rotation of the entire cross-section is prevented, leading to the first of the above condition being changed to ∂w w = 0 and = 0 (8.79c) ∂x which once again is the same as for the ordinary homogeneous plate. The equivalent resultant vertical forces at the edges and in the corners appear in the same way as in the simply supported case for both soft and hard boundary conditions.

References [1] Timoshenko S.P. and Woinowsky-Krieger S., Theory of Plates and Shells, Second edition, McGraw-Hill, London, 1970.

8.32 SANDWICH PLATES – FUNDAMENTAL EQUATIONS

[2] Libove C. and Batdorf S.B., “A General Small-Deflection Theory for Flat Sandwich Plates”, NACA TN 1526, 1948, also in NACA report 899.

[3] Plantema F.J., Sandwich Construction, John Wiley & Sons, New York, 1966.

[4] Jones R.M., Mechanics of Composite Materials, McGraw-Hill, New York, 1975.

[5] Mindlin R.D., “The Influence of Rotary Inertia and Shear on Flexural Motions of Isotropic, Elastic Plates”, Journal of Applied Mechanics, Transactions of the ASME, Vol. 18, 1951, pp 31-38.

[6] Reissner E., “The Effect of Transverse Shear Deformation on the Bending of Elastic Plates”, Journal of Applied Mechanics, Trans. ASME, Vol. 12, 1945, pp A69-A77.

[7] Reissner E., “Finite Deflections of Sandwich Plates”, Journal of the Aeronautical Sciences, Vol. 15, No 7, July 1948, pp 435-440.

[8] Timoshenko S.P., Vibration Problems in Engineering, Second Edition, D. Van Nostrand Company Inc., New York, N.Y., 1937, p 337.

[9] Hoff N.J., “Bending and Buckling of Rectangular Sandwich Plates”, NACA TN 2225, 1950.

[10] Allen H.G., Analysis and Design of Structural Sandwich Panels, Pergamon Press, Oxford, 1969.

[11] March H.W., “Effects of Shear Deformation in the Core of a Flat Rectangular Sandwich Panel – 1. Buckling under Compressive End Load, 2. Deflection under Uniform Transverse Load”, U.S. Forest Products Laboratory Report 1583, 1948.

[12] Reissner E., “On Bending of Elastic Plates”, Quarterly Applied Mathematics, Vol. 5, No 1, 1947, pp 55-68.

[13] Whitney J.M., Structural Analysis of Laminated Anisotropic Plates, Technomic, Lancaster, PA, USA, 1987.

[14] Grenestedt J.L., “Lamination Parameters for Reissner/Mindlin and Sandwich Type Plate Theories”, AIAA Technical Note, Vol 32, No 11, pp.2328-2331, 1994.

[15] Fung Y.C., Foundations of Solid Mechanics, Prentice-Hall Inc., New Jersey, 1965.

8.33 CHAPTER 9

SOLUTIONS TO PLATE PROBLEMS

In this section solutions to some commonly used plate problems will be presented. In some cases the entire solution is derived whereas in others only the solution is presented. First, only isotropic plates are considered but then orthotropic plates are studied also. Primarily the deflection field is calculated, and especially the maximum deflection of the plate, but in some cases also maximum bending moments and transverse forces are given.

A set of approximate solutions to bending and buckling problems are also given. These are in no way intended to be correct but rather to serve as simple closed form equations for estimations of deflections or buckling loads. For more complicated problems where closed form solutions are very complex or not available, engineers will in many cases have access some finite element code whereby very accurate calculations can be done. Most commercial FE-codes of today also implement special sandwich elements making modelling and post- processing work even faster. In such cases, simple closed form solutions may serve as a complement and quick reference check to the finite element solution. The derivation and solution of the approximate formulae also give some valuable insight into how orthotropy and panel aspect ratio influence the behaviour of the panel in general.

All solutions given herein are derived for small deflections and strains although theories accounting for finite deflections exists, e.g. [1]. It has been concluded from analytical, numerical and experimental work that for deflections up to the magnitude of the plate thickness, linear analysis is sufficiently accurate [2,3]. No solutions for finite or large deflection cases are given since they tend to become rather complicated and is still best solved using the finite element method.

9.1 Rotationally-Symmetric Plates In this section solutions to rotational symmetric plate problems will be derived. The solutions

assume isotropic sandwich plates with thin faces, tf << tc, and weak cores, Ec << Ef so that the concept of partial deflections can be used. The circular sandwich plate subjected to a rotationally symmetric transverse load will be considered. It is particularly simple under the above assumptions since the deformation also will be rotationally symmetric. In such cases, the solution will be independent of the of the tangential coordinate ϕ and of course also of

coordinate z. If the deflection is expressed in terms of wb and ws, then they will both be

9.1 AN INTRODUCTION TO SANDWICH STRUCTURES

constant along the circumference of the plate and the problem will become one-dimensional as a beam. It will also imply that shear forces, moments and deflections are identical to those

of an ordinary circular plate, except for the additional term ws.

Consider a circular sandwich plate as illustrated in Fig.9.1. It has radius R and is subjected to

a uniform transverse load q on an eccentric circular area between radii ρ1 and ρ2.

R q r r ρ 1 ρ z 2 ρ 1 ρ 2 R

Figure 9.1 Circular sandwich plate subjected to a uniform load q between ρ1 and ρ2. Assume the plate to be simply supported along the outer radius R. The supporting transverse load equals qπ()ρ − ρ 2 Tr()= 12 r 2πR

Note also that the transverse shear deformation ws is independent of the boundary condition at r = R. Now, the transverse shear load Tr and the corresponding shear deformation according to eq.(4.2) can be directly written as function of r (note that due to the choice of coordinate system the deflection will measured from r = 0 with positive displacement downward). There are three distinct regions to be analysed

()a (a) r ≤ ρ1: Tr = 0, and thus wrs ()= ws (ρ1 ) (9.1)

2 2 2 qrqπ()− 1 qr( −+2 rq11 q ) (b) ρ12≤≤r ρ : T =− =− (9.2) r 2πr 2r

r T qr()()2 − ρρρρ2 qr− q 2 r wr()b ()==−r dr 1 + 11−+ 1lnw (ρ ) s ∫ S 42S S S ρ s 2 ρ1 1

q()ρ − ρ 2 (c) r ≥ ρ : T =− 21 (9.3) 2 r 2r

r T q()ρρ− 2 r wr()c ()==−r dr 12ln+ w (ρ ) s ∫ S 2S ρ s 2 ρ 2 2

Now some examples can be solved using the derived relations (9.1-9.3).

9.2 SOLUTIONS TO PLATE PROBLEMS

(i) Uniform load over the entire plate, ρ1 = 0, ρ2 = R q

The deflection of a simply supported circular sandwich plate subjected to a uniform pressure over the entire surface is according to [4] and eq.(9.2). Note that the shear deformation value

must now be changed to ws(r) + ws(R) since the displacement of the boundary r = R is set to zero.

qR()()22−− r 1 νν 2 5 + qR ()22− r wr()=+= w () r w () r ⎡Rr22− ⎤ + bs 64D ⎣⎢ 14+ ν ⎦⎥ S

and similarly for a clamped plate, since the shear deformation is independent of the boundary condition.

qR()()22−− r 1 ν 2 qR ()22− r wr()=+= w () r w () r Rr22−+ bs 64D [] 4S

This can now be verified by using the relation between wb and ws in eq.(8.24). One may use any of the above bending displacement fields, simply supported or clamped. Using the latter yields (recall the Laplace operator in polar coordinates)

∂ 2 w 1 ∂w q(1 −ν 2 ) Δw = b + b = []16r 2 − 8R 2 b ∂r 2 r ∂r 64D

q Thus w =−16rRC22 − 8 + s 64S []

2 using the boundary condition ws(R) = 0, gives the constant C = qR /8S. This yields the shear deformation q q w =−16rRR222 − 8 − 8 =Rr 22 − s 64S []4S [] which is the same result as derived above.

(ii) Uniform load over the central part of the plate, ρ1 = 0 q

ρ 2

22 2 2 qr()ρρ2 − q 2 R qρ2 R r ≤ ρ2 : ws = + ln , and r ≥ ρ2 : ws = ln 42S S ρ 2 2S r

and the corresponding deflections due to bending can be found in [4] or [5].

9.3 AN INTRODUCTION TO SANDWICH STRUCTURES

(iii) Uniform load but over central part of the plate, ρ2 = R q

ρ 1

2 2 2 qR()()− ρ1 qRρ111− ρ qρ R r ≤ ρ1: ws = + − ln 42S S S ρ1

22 2 qR()()− r qRrρ11− qρ R r ≥ ρ1: ws = + − ln 42S S S r (the bending deformations can be found in [4] or [5])

(iv) Annular line load, ρ1 = ρ2 = ρ P

P R P R r ≤ ρ: w = ln , and r ≥ ρ: w = ln s 2πρS s 2πS r

(the bending deformations can be found in [4] or [5])

(v) Point load in the middle of the plate, ρ1 = ρ2 = 0 P

The case of a point load in the centre of the plate is given by eq.(9.3) by letting ρ1 = 0 and ρ2 2 approach zero. This gives the shear force Tr = −qπρ1/2πr = −P/2πr, so that Tr approaches infinity at r = 0. From eq.(9.3)

r T r P P P P ⎛ R ⎞ w (r) = r dr = − dr = − ln r + ln R = ln s ∫ ∫ ⎜ ⎟ R S R 2πrS 2πS 2πS 2πS ⎝ r ⎠

This yields that ws = ∞ at the centre of the plate, and therefore, the solution contains a singularity at r = 0. This singularity can be removed by taking into account the bending stiffness of the faces and assuming the shear stiffness of the faces to be infinite. In practice, point loads appear over a finite area of the plate. The bending deformation is finite, however, and equals [4,5]

2 P(1 −ν ) ⎡3 +ν 2 2 2 r ⎤ P()1− ν 3 + ν 2 wb = ⎢ (R − r ) − 2r ln ⎥ , with wb ()0 = R 16πD ⎣1 +ν R⎦ 16πD 1+ ν

for a simply supported plate, and

9.4 SOLUTIONS TO PLATE PROBLEMS

2 2 P(1 −ν ) ⎡ 2 2 ⎛ r ⎞⎤ P()1− ν 2 wb = ⎢R − r ⎜1 + 2 ln ⎟⎥ , with wb ()0 = R 16πD ⎣ ⎝ R ⎠⎦ 16πD for a clamped plate.

9.2 Bending of a Rectangular, Simply Supported, Isotropic Sandwich Plate Consider a rectangular simply supported sandwich plate with sides a and b as shown in Fig.9.2.

Figure 9.2 Rectangular simply supported plate. The boundary conditions for a simply supported plate are [6]

w = 0, Mx = 0 at x = 0 and x = a (9.4)

w = 0, My = 0 at y = 0 and y = b The deflection can be represented by a double Fourier sine series [7] ∞ ∞ mxπ nyπ ww= ∑∑ mn sin sin (9.5) n=1 m=1 a b which satisfies the boundary conditions. This becomes clearer if the partial deflections are used; on a boundary x = 0,a

2 2 D ⎡∂ wb ∂ wb ⎤ M x = − 2 ⎢ 2 +ν 2 ⎥ = 0 1−ν ⎣ ∂x ∂y ⎦

2 2 from eq.(8.15). If wb is assumed to be zero along all boundaries, then ∂ wb/∂y must also be 2 2 zero along x = 0,a. Thus, the boundary condition Mx = 0 can be replaced with ∂ wb/∂x = 0. Since, w = wb + ws, the assumption in eq.(9.5) must satisfy this boundary condition, and similarly for edges with y = 0,b.

Similar deflection assumptions can be made for the in-plane deformations u and v [7], based on that bending stresses and therefore the corresponding strains must be zero along the simply supported edges. To make the analysis more general we can also assume the following form for the in-plane deformation and transverse forces

∞ ∞ mπx nπy ∞ ∞ mπx nπy u = ∑∑umn cos sin , and v = ∑∑vmn sincos (9.6a) nm=11= a b n1=1 m= a b

9.5 AN INTRODUCTION TO SANDWICH STRUCTURES

∞ ∞ mxπ nyπ ∞ ∞ mxππny TTx= ∑∑ xmn cos sin , and TTyymn= ∑∑ sin cos (9.6b) n=1 m=1 a b n1= m1= a b and now it becomes more evident that the boundary conditions are satisfied by inserting eq.(9.6) as well as (9.5) in eq.(8.8). However, this field assumption does not imply that Mxy is zero along the plate edges, i.e. the assumption implies a hard boundary condition (see section 8.9) in which ∂v/∂z = 0 for x = 0,a and ∂u/∂z = 0 for y = 0,b. To simplify the analysis express the load in a similar way. A transverse load can in a general form be written as ∞ ∞ mxπ nyπ qq= ∑∑ mn sin sin (9.7) n=1 m=1 a b where qmn are the Fourier coefficients that are determined by 4 a b mxππny q = qxy(,)sin sin dxdy (9.8) mn ∫ ∫ ab 0 0 a b for a given loading q(x,y). For a uniformly distributed load q(x,y) = q the coefficients will equal 16q q = for odd values of n and m, otherwise q = 0 (9.9) mn mnπ2 mn In the case of a concentrated load Q (dimension N) acting in the centre of the plate the coefficients are obtained by a limiting process. Let p be the intensity of a uniformly distributed load over a small square with sides c parallel to those of the plate. Set the point load Q = pc2. Then

ac+ bc+ 442 2 Q mxππny Q q ==sin sin dxdy (9.10a) mn ∫ ∫ 2 ab ac− bc− c a b ab 2 2 for odd values of n and m, otherwise qmn = 0. If the point load is situated at some other location than the centre, say at (x,y) = (x',y') then [4] 4Q mπx′′nπy q = sin sin (9.10b) mn ab a b (summation over all m and n). We can use these expressions in the differential equations derived in section 9 and start with the simplest possible plate problem.

(i) Uniformly distributed load, thin faces.

We can now either use the Mindlin equation (8.27) which assumes Df = 0 and calculate the total displacement field w or use eqs.(8.24-26) to calculate the partial deflections separately. By substituting into eq.(8.27) it yields that

4224 22 Dm⎡⎛ ππππ⎞ ⎛ m ⎞ ⎛ n ⎞ ⎛ n ⎞ ⎤ ⎡ D ⎛⎛ mππ⎞ ⎛ n ⎞ ⎞⎤ ⎜ ⎟ + 21⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ w =+ ⎜⎜ ⎟ + ⎜ ⎟ ⎟ q 2 ⎢⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎥ mn⎢ 2 ⎝ ⎠ ⎝ ⎠ ⎥ mn 1 − ν ⎣ a a b b ⎦ ⎣ S()1 − ν ⎝ a b ⎠⎦

9.6 SOLUTIONS TO PLATE PROBLEMS giving

22 D ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ 1 + 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ q ()1 − ν 2 S()1 − ν ⎝ a ⎠ ⎝ b ⎠ w = mn ⎣ ⎦ (9.11) mn 222 D ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎣⎝ a ⎠ ⎝ b ⎠ ⎦

Hence, the deflection can be written as

22 D ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ 1 + 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ∞ ∞ q ()1− ν 2 S()1− ν ⎝ a ⎠ ⎝ b ⎠ mxππny w = mn ⎣ ⎦ sin sin (9.12) ∑∑ 222 n=1 m=1 D ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ a b ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎣⎝ a ⎠ ⎝ b ⎠ ⎦

It is now seen that this expression can be divided up into the partial deflections as

2 mxπ nyπ mxπ nyπ ∞ ∞ qmn ()sinsin1 − ν ∞ ∞ qmn sin sin w = a b , and w = a b b ∑∑ 222 s ∑∑ 22 n=1 m=1 ⎡ mππn ⎤ n=1 m=1 ⎡ mππn ⎤ ⎛ ⎞ ⎛ ⎞ S ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ D⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎢⎝ ⎠ ⎝ ⎠ ⎥ ⎣⎝ a ⎠ ⎝ b ⎠ ⎦ ⎣ a b ⎦

It is now easily shown that these expressions satisfy eq.(8.24). In fact, the same equation can be derived using the differential equation for wb in eq.(8.25) and then eq.(8.24) to derive the shear deformation ws. It is also seen that the solution satisfies the hard boundary condition since γxz = ∂ws/∂x = 0 for y = 0,b and γyz = ∂ws/∂y = 0 for x = 0,a. By using the coefficient qmn for a uniformly distributed load, substituting into eq.(9.12) and rearranging one arrives at

22 D ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ 1 + 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ 16q() 1 − ν 2 ∞ ∞ S()1 − ν ⎝ a ⎠ ⎝ b ⎠ mxππny w = ⎣ ⎦ sin sin (9.13a) 2 ∑ ∑ 222 π D n=135.,, .m=135.,, . ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ a b mn⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎣⎝ a ⎠ ⎝ b ⎠ ⎦

2 or rewritten using the shear factor φ = D0/b S

2 2 ⎡⎛ mb⎞ 2 ⎤ 1 + πφ⎢⎜ ⎟ + n ⎥ 16qb 42() 1 − ν ∞ ∞ ⎣⎝ a ⎠ ⎦ mxππny w = sin sin (9.13b) π 6 D ∑ ∑ 2 2 a b n=13,5,..m=13,5,.. ⎡⎛ mb⎞ 2 ⎤ mn⎢⎜ ⎟ + n ⎥ ⎣⎝ a ⎠ ⎦ and once again the partial deflections can be extracted from this relation as above. The bending moments and transverse forces are now calculated by using the general equations (8.8) as outlined in [8] (which will be performed further on in this section) or more simply by using partial deflections. Thus, all the necessary equations are available and it is now merely a matter of differentiating eq.(9.13) and performing the summation. The series converge rather quickly for the deflections and bending moments but require more terms to provide

9.7 AN INTRODUCTION TO SANDWICH STRUCTURES accurate numbers for the transverse forces. The maximum deflection and bending moments appear in the middle of the plate at (x,y) = (a/2,b/2) and the maximum transverse forces Tx at

(0,b/2) and (a,b/2) and Ty at (a/2,0) and (a/2,b), i.e., at the middle of the sides. By summation using terms of m and n up to 27, the data in Table 9.1 is obtained.

Dwb Sws Mx My Tx Ty 42 2 2 a/b qb ()1 − ν qb qb qb2 qb qb 1.0 0.00406 0.0737 0.0479 0.0479 0.338 0.338 1.2 0.00564 0.0868 0.0501 0.0626 0.353 0.380 1.4 0.00705 0.0967 0.0506 0.0753 0.361 0.411 1.6 0.00830 0.1042 0.0493 0.0862 0.365 0.435 1.8 0.00931 0.1098 0.0479 0.0948 0.368 0.452 2.0 0.01013 0.1139 0.0464 0.1017 0.370 0.465 3.0 0.01223 0.1227 0.0404 0.1189 0.372 0.493 4.0 0.01282 0.1245 0.0384 0.1235 0.372 0.498 5.0 0.01297 0.1249 0.0375 0.1246 0.372 0.500 ∞ 0.01302 0.1250 0.0375 0.1250 0.372 0.500

Table 9.1 Maximum bending and shear deformations, bending moments and transverse forces for a simply supported rectangular sandwich plate with flexural rigidity D and shear stiffness S. For a/b ratios less than unity all properties are easily found by interchanging a and b so that

b 4 b 2 b 2 wba(/)= wab (/)⎜⎛ ⎟⎞ , wba(/)= wab (/)⎜⎛ ⎟⎞ , Mba(/)= Mab (/)⎜⎛ ⎟⎞ bb⎝ a⎠ ss⎝ a⎠ xy⎝ a⎠

b 2 b b Mba(/)= Mab (/)⎜⎛ ⎟⎞ , Tba(/)= Tab (/)⎜⎛ ⎟⎞ , Tba(/)= Tab (/)⎜⎛ ⎟⎞ yx⎝ a⎠ xy⎝ a⎠ yx⎝ a⎠

(ii) Uniformly distributed load, thick faces. The same approach as above can be used when accounting for thick faces. Take the deflection shape assumed in eq.(9.5), and use the in the governing differential equation accounting for thick faces, eq.(8.31). After some differentiating and inserting in (8.31) one arrives at [7]

∞ ∞ q ⎛ D ⎡ mππ22n ⎤⎞ mx ππny w =+mn ⎜1 0 ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ ⎟ sin sin (9.14) ∑∑ 2 ⎢⎝ ⎠ ⎝ ⎠ ⎥ n=1 m=1 Kmn ⎝ S()1 − ν ⎣ a b ⎦⎠ a b where the denominator equals

223 222 2DDf 0 ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ Dm⎡⎛ ππ⎞ ⎛ n ⎞ ⎤ Kmn = 22⎢⎜ ⎟ + ⎜ ⎟ ⎥ + 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ (9.15) S()11− ν ⎣⎝ a ⎠ ⎝ b ⎠ ⎦ − ν ⎣⎝ a ⎠ ⎝ b ⎠ ⎦

It is seen that this expression equals eq.(9.12) if Df = 0. We can again extract the partial deflection out of this expression as

q D ⎡ mππ22n ⎤ q w ==mn , w 0 ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ mn bmn smn 2 ⎢⎝ ⎠ ⎝ ⎠ ⎥ Kmn S()1 − ν ⎣ a b ⎦ Kmn

9.8 SOLUTIONS TO PLATE PROBLEMS

These expressions are not strictly correct since ws depends on the ratios 2Df /S and D/D0 which appear in the denominator, but is convenient to use when explaining the phenomena occurring. Also, one can no longer express the deflection and the forces and bending moments in non-dimensional terms in such a simple way as for the thin face case. This is due to the fact that the denominator Kmn this time includes terms of plate lengths in different powers, i.e., a4 and a6. It is worth noting though, that when the shear stiffness S is large, the plate deflection equals that computed using the thin face theory, but using D = 2Df + D0. One can also see that when S approaches zero, i.e., no shear rigidity, the term wb approaches zero whereas the term ws defined above equals the deformation of the two faces bending independently of each other (that is, using the thin face theory, but now with D = 2Df). Hence, the asymptotic solution yields physically correct answers in terms of the total deflection. This behaviour can also be explained by the following reasoning; if the shear stiffness is very low then ws >> wb and thus the total deformation would equal ws providing the faces are considered thin. However, if the faces have a bending stiffness of their own they cannot undergo this shear deformation but will stiffen the plate. At the limit, S approaches zero, the shear deformation cannot equal infinity but is restricted by the bending of the two faces, which now bend independently of each other. At the other limit, D0 is very large and Df = 0, then ws equals that given in Table 9.1. Hence, the term wb in this case contains the deflections due to bending about the neutral axes of the sandwich as a whole, just as in the thin face case, whereas ws is the shear deflection with the restriction from the influence of Df imposed on it.

Inserting the value qmn given in eq.(9.9) gives

∞ ∞ 16q ⎡ D ⎡ mππ22n ⎤⎤ mx ππny w =+1 0 ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ sin sin (9.16) ∑ ∑ 2 ⎢ 2 ⎢⎝ ⎠ ⎝ ⎠ ⎥⎥ n=135.,, .m=135.,, .mnπν K mn ⎣ S()1 − ⎣ a b ⎦⎦ a b

In order to clearly show these effects Fig.9.3 has been prepared so that the ratio w/w* is 2 plotted versus Df /D0 = (tf/d) /6 (see chapter 3) for different values of the shear factor φ = D/b2S. Now, w is the deflection computed using the present theory and w* is the deflection computed using the thin face theory given in Table 9.1.

As seen in Fig.9.3, the effect of the bending stiffness of the faces is very small, since φ is usually in the order of 0.1 or less and Df /D0 is usually in the order of 0.01 or less for sandwiches used in practical applications. The approximation that Df can be neglected in the calculation of the total flexural rigidity derived in section 2, stated that 2Df /D0 < 0.01 if d/tf > 5.77. Hence, this approximation could once again be used since, as seen in Fig.9.3, the influence of thick faces is negligible for Df /D0 < 0.01 and φ smaller than 0.1. One can therefore conclude that thick faces must only be considered when the plate is weak in shear, i.e., φ is small, otherwise the deflection can just as well be calculated using the thin face theory but accounting for the thickness of the faces when calculating the flexural rigidity D. The way to use Fig.9.3 is to compute the deflection using the thin face theory in Table 9.1 and then account for the thickness of the faces by multiplying with the factor w/w* from Fig.9.3 for a given φ and D f /D0 ratio.

9.9 AN INTRODUCTION TO SANDWICH STRUCTURES

1 π 2φ =0.01 0.9 0.1 0.8 0.7 1 0.6 w/w* 0.5 0.4 0.3 0.2 10 0.1 0 100 0.0001 0.001D/D 0.01 0.1 f 0

Figure 9.3 Effect of thick faces on the deformation of a square simply supported sandwich plate subjected to a uniform load. The lines are for π2φ = 0, 0.01, 0.1, 1, 10 and 100. The bending moments and transverse forces are more difficult to assess in this problem and to do a proper analysis one must include the field assumptions for the transverse forces given in eq.(9.6), use these together with the assumed field for w in eq.(9.5) and insert in the equations connecting the transverse forces to the deflection shape derived in section 9, eq.(8.9) [9]. However, the very complex and quite lengthy operation and will be omitted here. The relations are given in a general form for an orthotropic plate in [9]. It is also seen that eq.(9.16) satisfies the expression eq.(8.24) for the partial deflections. Hence, it is likely that the distribution of the moments and transverse forces are the same as in the thin face case, at least for reasonably small shear factors φ. However, additional stresses may occur in the faces due to the local bending of the faces taking place.

(iii) Point load, thick faces

Using the same methodology as above, and taking the load qmn as given in eq.(9.10) one rapidly arrives at a solution for the deflections similar to that given above. The deflection field will be similar to that of the uniformly distributed load in eq.(9.16), or written out mπx' nπy' ∞∞4Q sin sin ⎡ ⎡ 2 2 ⎤⎤ a b D0 ⎛ mπ ⎞ ⎛ nπ ⎞ mπx nπy w = ⎢1 + ⎢⎜ ⎟ + ⎜ ⎟ ⎥⎥ sin sin (9.17) ∑∑ abK S(1 −ν 2 ) a b a b nm. mn ⎣⎢ ⎣⎢⎝ ⎠ ⎝ ⎠ ⎦⎥⎦⎥

If the point load is applied at the centre, the expression for the load given in eq.(9.10a) is used instead of the one in eq.(9.10b). In fact, a solution can be derived even assuming thin faces by the method described for the uniformly loaded plate in the beginning of this section. However, when Df/D0 becomes small the term representing the shear deformation will not converge and a solution is not possible. In order to assess the forces and moments at the load point a three-dimensional elasticity approach must used or some other higher order theory. In fact, a concentrated load will in practice be applied to only one face, whereas the other will be loaded over a much larger area since the load must be transferred through the core. Hence, the analysis must account for the transverse stress σz in the core.

9.10 SOLUTIONS TO PLATE PROBLEMS

By letting S approach zero it is seen that the second term in w and the first term in Kmn dominate the expression for the deflection. Thus, ignoring other terms in eq.(9.17) it can be rewritten as

mπx' nπy' 2Q(1−ν 2 )sin sin ∞∞ mπx nπy w = a b sin sin ∑∑ 2 2 2 a b nm ⎡⎛ mπ ⎞ ⎛ nπ ⎞ ⎤ abD f ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎣⎢⎝ a ⎠ ⎝ b ⎠ ⎦⎥ which is the solution for the two faces (D = 2Df) bending independently of each other [4]. If, on the other hand, S approaches infinity then the solution will be same as above but with D

(=D0 + 2Df) substituted for 2Df. 9.3 Rectangular, Simply Supported, Orthotropic Sandwich Plate In this case one must use the full solution in order to assess all variables. In short, if one uses the field assumptions made in eqs.(9.5) and (9.6) in eq.(8.6c) then the relation between the unknown coefficients wmn, Txmn, and Tymn can be derived, or inserted in the separated solution of eq.(8.11) all coefficients can be solved. The bending moments are then derived from eq.(8.8). This is quite a lengthy and complex derivation and is omitted here but the results are given below [9].

∞ ∞ Wq mxππny w =−∑∑ mn mn sin sin (9.18) n=1 m=1 Zmn a b

∞ ∞ Xqmn mn mxπ nyπ Tx = ∑∑ cos sin , and n=1 m=1 Zmn a b

∞ ∞ Yqmn mn mxππny Ty =−∑∑ sin cos n=1 m=1 Zmn a b where the coefficients are found from eq.(8.11) as

⎧ 4 2 2 1 ⎡1 ⎪⎛ mπ ⎞ Dx ⎛ mπ ⎞ ⎛ nπ ⎞ ν xy D y +ν yx Dx Wmn = − ⎢ Dxy ⎨⎜ ⎟ − ⎜ ⎟ ⎜ ⎟ S x S y ⎣2 ⎩⎪⎝ a ⎠ 1 −ν xyν yx ⎝ a ⎠ ⎝ b ⎠ 1 −ν xyν yx 4 2 2 2 ⎛ nπ ⎞ D y ⎪⎫ ⎛ mπ ⎞ ⎛ nπ ⎞ Dx D y ⎤ ⎡⎛ mπ ⎞ D + ⎜ ⎟ ⎬ + ⎜ ⎟ ⎜ ⎟ ⎥ − ⎢⎜ ⎟ x (9.19) ⎝ b ⎠ 1 −ν xyν yx ⎭⎪ ⎝ a ⎠ ⎝ b ⎠ 1 −ν xyν yx ⎦⎥ ⎣⎢⎝ a ⎠ S x (1 −ν xyν yx ) 2 2 2 ⎛ nπ ⎞ D y ⎤ 1 ⎡ 1 ⎛ mπ ⎞ 1 ⎛ nπ ⎞ ⎤ + ⎜ ⎟ ⎥ − Dxy ⎢ ⎜ ⎟ + ⎜ ⎟ ⎥ −1 ⎝ b ⎠ S y (1 −ν xyν yx )⎦⎥ 2 ⎣⎢S y ⎝ a ⎠ S x ⎝ b ⎠ ⎦⎥

9.11 AN INTRODUCTION TO SANDWICH STRUCTURES

1 mπ 532DD mππn ⎡ DD D()νν D+ D ⎤ XS= ⎜⎛ ⎟⎞ xxy + ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ xy − xy xy y yx x mn y ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎢ ⎥ 21a − ννxy yx a b ⎣⎢ 1− ννxy yx 21()− ννxy yx ⎦⎥

422DD ⎡ ⎛ D ⎞⎤ 1 ⎛ mππ⎞⎛ n ⎞ yxy ⎛ mππ⎞ ⎛ m ⎞ Dx ⎛ nπ ⎞ ν yx x + ⎜ ⎟⎜ ⎟ + S ⎜ ⎟ ⎢⎜ ⎟ + ⎜ ⎟ ⎜ D + ⎟⎥ 21⎝ a ⎠⎝ b ⎠ − νν y ⎝ a ⎠ ⎝ a ⎠ 1− νν ⎝ b ⎠ ⎜ xy 1− νν ⎟ xy yx ⎣⎢ xy yx ⎝ xy yx ⎠⎦⎥ 1 nπ 523DD mππn ⎡ DD D()νν D+ D ⎤ YS=− ⎜⎛ ⎟⎞ yxy − ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ xy − xy xy y yx x mn x ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎢ ⎥ 21b − ννxy yx a b ⎣⎢ 1− ννxy yx 21()− ννxy yx ⎦⎥

422⎡ ⎤ 1 ⎛ mππ⎞ ⎛ n ⎞ DDxxy ⎛ nππ⎞ ⎛ n ⎞ Dy ⎛ mπ ⎞ ⎛ ν xyD y ⎞ − ⎜ ⎟ ⎜ ⎟ − S ⎜ ⎟ ⎢⎜ ⎟ + ⎜ ⎟ ⎜ D + ⎟⎥ 21⎝ a ⎠ ⎝ b ⎠ − νν x ⎝ b ⎠ ⎝ b ⎠ 1− νν ⎝ a ⎠ ⎜ xy 1− νν ⎟ xy yx ⎣⎢ xy yx ⎝ xy yx ⎠⎦⎥

mπ nπ Z = ⎜⎛ ⎟⎞ X − ⎜⎛ ⎟⎞Y mn⎝ a ⎠ mn⎝ b ⎠ mn

By using these relations it is possible to write the bending moments as

∞ ∞ −D ⎡ ⎧ mπ 22nππ⎫ m X nπ Y ⎤ M = x W ⎜⎛ ⎟⎞ + ν ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ mn − ν ⎜⎛ ⎟⎞ mn x ∑∑ ⎢ mn⎨⎝ ⎠ yx ⎝ ⎠ ⎬ ⎝ ⎠ yx ⎝ ⎠ ⎥ n=1 m=1 1 − ννxy yx ⎣⎢ ⎩ a b ⎭ a Sx b Sy ⎦⎥ (9.20) q mxππny × mn sin sin Zmn a b

∞ ∞ −D ⎡ ⎧ nπ 22mππ⎫ n Y mπ X ⎤ M = y W ⎜⎛ ⎟⎞ + ν ⎜⎛ ⎟⎞ − ⎜⎛ ⎟⎞ mn + ν ⎜⎛ ⎟⎞ mn y ∑∑ ⎢ mn⎨⎝ ⎠ xy ⎝ ⎠ ⎬ ⎝ ⎠ xy ⎝ ⎠ ⎥ n=1 m=1 1 − ννxy yx ⎢ b a b Sy a Sx ⎥ ⎣ ⎩ ⎭ ⎦ q mxππny × mn sin sin Zmn a b

The load factor qmn can be taken either for a uniformly distributed load as in eq.(9.9) or as a point load as in eq.(9.10). In the following, only a uniform load is considered. These rather complex equations can now be coded and solved numerically. A full solution in the form of graphs is not possible due to the number of variables. Anyway, below there are some results derived using the above equations, prepared in order to show the trends of the different parameters involved. As in Table 9.1 only the maximum values are given, that is the deformation and bending moments at the centre of the plate and the maximum transverse forces appearing mid-way between the corners.

In the following graphs the maximum deflection, bending moments and transverse forces are plotted in terms of the factor β defined as

4 qb 2 2 w = β1 , Mqbx = β2 , Mqby = β3 , Tqbx = β4 , and Tqby = β5 (9.21) Dy

The Poisson's ratios νxy and νyx are taken as νxy = 0.25 Dx/Dy, and νyx = 0.25 Dy/Dx so that the product νxyνyx is kept constant. This also means that the isotropic case is for Dx = Dy = 1.25

9.12 SOLUTIONS TO PLATE PROBLEMS

Dxy. The set of graphs are also derived using m = n = 27, which gives fairly good accuracy for the deflection but less good for the bending moments and transverse forces, as long as a/b is close to unity.

The first set of graphs show the influence of the torsional stiffness Dxy, and the ratio of the flexural stiffness Dx/Dy on the response of a square orthotropic plate subjected to a uniformly distributed load q. The values of Dxy = 0.1Dy and 1.2Dy could be seen as extreme values and most practical values should lie within that regime. The curves are for the shear factor 2 equalling zero (pure bending) and π φy = 1.0 which implies are fairly large amount of shear.

As seen in Fig.9.4, a high torsional stiffness stiffens the plate even though the flexural rigidities are kept constant. Different values of the torsional stiffness Dxy than D(1+ν) can be obtained using fibrous composite faces. It is also seen that increasing the rigidity in one direction not only decreases the deformation but also changes the bending moments and transverse forces. In fact, increasing the rigidity in one direction influences the moments and forces in the same manner as decreasing the plate span in that direction, e.g., increasing Dx increases Mx and Tx whereas My and Ty are decreased relatively, which is what happens if the span in the x-direction, a, is decreased. It is also seen that though changing the shear factor θ has a considerable influence on the deflection, it has little effect on the maximum values of the transverse forces and the bending moments.

To illustrate what effect orthotropy might have on the bending of a square plate let us take an example: A square 1000 by 1000 mm sandwich plate with 1 mm thick equal faces and a 10 mm core. Assume the faces are made of a fibre composite laminate having fibres only in the global x- and y-directions so that Ex = Ey = 25 GPa, νxy = νyx = 0.25 and Gxy = 1 GPa. Depending on the fibre/matrix system and fibre volume fraction of the laminate the ratio

Ex/Gxy could vary between 5 and 50. Now, calculate the mid-point deflection and maximum bending moments and transverse forces using either the orthotropic solution or by approximating the plate as an isotropic plate having D = DxDy. The results are given in Table 9.2. Next, rotate the laminate 45 degrees so that all fibres now will be placed in the ±45 ° direction. By a stiffness transformation using classical lamination theory [10] the global stiffness for the laminate now becomes Ex = Ey = 17.67 GPa and Gxy = 10 GPa. Once again the response of the plate can be calculated as above and the results are given in Table 9.2.

0°/90° ±45°

9.13 AN INTRODUCTION TO SANDWICH STRUCTURES

Dx, Dy Dxy w Mx, My

0/90 isotropic: Ex=Ey=25 GPa 1512500 1210000 2.69 46.1

0/90 orthotropic: Ex=Ey=25GPa, Gxy=1 GPa 1512500 121000 4.08 71.6

±45 isotropic: Ex=Ey=17.7 GPa 1068833 855067 3.80 46.1

±45 orthotropic: Ex=Ey=17.7 GPa, Gxy=10 GPa 1068833 1210000 3.28 39.4 Table 9.2 Comparison between isotropic approximation and correct orthotropic calculation for a sandwich plate with 0/90 and ±45 degree laminate faces. There are two important facts to be noticed from these results; firstly that the isotropic approximation yields poor results for the mid-point deflection and the maximum occurring bending moment. Some small changes occur for the maximum transverse forces but are omitted in the table since these changes are probably due to lack of numerical accuracy. This can also be seen in Fig.9.4. Secondly and perhaps more interesting is the fact the although the same face is used in the two above configurations the plate response differs. The difference is not as anticipated since one might assume that the laminate should be built up with the fibres in the direction of the maximum stresses, i.e., global x- and y-directions (0/90 degrees), to yield maximum stiffness, rather than in ±45 degrees to the global coordinate system. It is seen that by simply rotating the orthotropic faces 45 degrees the mid-point deflection decreases from 4.08 to 3.28 mm and the maximum bending moment decreases from 71.6 to 39.4 Nmm/mm width.

A second set of graphs that follows in Fig.9.5 are for the same assumptions as given above, but now the properties are plotted for different panel aspect ratios, b/a. In order to do so the ratio Dxy/Dy has been set to 0.8 (which yields the isotropic case for Dx = Dy). Maximum 2 bending moments and transverse forces (β2 - β5) are only plotted for π φy = 0 (pure bending) 2 and π φy = 1.0. An increased shear factor of course strongly influences the deflection as the 2 2 transverse shear part increases, but the relative change between the cases π φy = 0 and π φy =

1.0 is very much the same irrespective of the panel aspect ratio a/b. Mx and Tx are as seen almost unaffected by both the change in a/b or the shear factor φ, whereas for My and Ty the relative change for different φ-values is about the same for different panel aspect ratios.

In the third set of graphs of Fig.9.6, the faces are considered isotropic with flexural rigidity D but now the core is orthotropic. The graphs show the deformation, bending moments and 2 transverse forces as a function of the shear parameter φy in the y-direction (φy = D/b Sy) for different values of orthotropy, i.e., Sx/Sy. Some foam cores are slightly orthotropic with a Sx/Sy ratio close to unity whereas most honeycomb cores have a Sx/Sy ratio in the order of 0.4 - 0.7.

In this case, the deflection increases linearly with φ irrespective of shear stiffness ratio Sx/Sy. As also seen, the shear stiffness orthotropy influences the internal force fields. This illustrates the one weakness with the concept of partial deflections under which assumption the bending moments only depend on the bending of the plate and the transverse forces only on the shearing of the plate and hence shear stiffness orthotropy cannot influence the bending moment fields. The same thing can be seen in Fig.9.4 where a bending orthotropy obviously influences the transverse force fields even for the case of pure bending.

It is similarly seen in the graphs of Fig.9.6, that the shear orthotropy influence is similar to that when only the faces were orthotropic; increasing the stiffness in the x-direction has the

9.14 SOLUTIONS TO PLATE PROBLEMS same effect as decreasing the length of span in the x-direction, a. It is also seen that the average shear stiffness used gives quite a good prediction of the deformation even for quite highly orthotropic cores, with a divergence from the true value of less than 15 percent. The same results as the above can be found in [11] for Sx/Sy = 1 and 0.4.

To be comprehensive similar sets of graphs should be produced for every value of the panel aspect ratio b/a which in that case must include inverse values of Sx/Sy to give the full picture. This cannot be done within the scope of this text but two such sets, for b/a = 1.4 and 2.0, are prepared in Fig.9.7 in order to show the trends of each parameter. Once again, the central plate deflection is very much affected by the orthotropy of the core in this case. It is also seen that the maximum bending moments and transverse forces are affected in quite a similar way as for the square plate. It is interesting to note that Tx seems to stay almost unaffected by both the shear factor or the panel aspect ratio and only takes a strong influence from the shear stiffness ratio Sx/Sy.

As mentioned earlier, in order to give all the information similar set of graphs must be drawn for each panel aspect ratio. In reality, however, the case may very well be that both the faces and the core are orthotropic and, in such cases, the only way of achieving good values for the deflection, bending moments and transverse forces is actually to perform the summation outlined above. This can be done by coding the formulae in a PC-computer or equivalent and since the series converge rather rapidly this requires very little computing time. The graphs above have been prepared using odd values of n and m up to 27 in eq.(9.18).

To summarise the behaviour of orthotropic sandwich plates Fig.9.8 shows what happens if some of the stiffnesses are changed compared with an isotropic plate. The effect of increasing one stiffness has the same effect as changing the panel aspect ratio. y x

Increasing Dy or Sy Increasing Dx or Sx

Figure 9.8 The effect of increasing one stiffness relative to the others in an orthotropic sandwich plate.

9.15 AN INTRODUCTION TO SANDWICH STRUCTURES

β 1 0.04

0.03

0.1 0.02 0.8

0.1 1.2 0.01 0.8 1.2 0 0.1 1 10 D /D x y

β β 0.16 2 3 0.1 0.8 0.1 1.2 0.08 0.1 0.8 0.8 0.1 1.2 0.8 1.2 1.2 0 0.1 1 10 0.1 1 10 D/D D/D xy xy

β β 0.5 4 5 0.8 0.1 1.2 0.1 0.4 0.8 1.2 1.2 0.3 0.8 0.1 0.2 1.2 0.8 0.1 0.1 0.1 1 10 0.1 1 10 D/D D/D xy xy

Figure 9.4 Normalised maximum deflection, bending moments, and transverse forces for a square, simply supported plate with orthotropic faces and isotropic core subjected to uniform pressure. Solid 2 2 lines are for π φy = 1.0 and dashed lines for π φy = 0 (pure bending). The lines are for Dxy /Dy = 0.1, 0.8 (isotropic case) and 1.2.

9.16 SOLUTIONS TO PLATE PROBLEMS

β 1 0.05

0.04

0.03 1.0 1.4 1.8 0.02 3.0 1.0 0.01 1.4 1.8 0 3.0 0.1 1 10 D/D x y

β β 0.2 2 1.8 3 3.0 1.4 3.0 1.8 3.0 0.1 1.0 1.4 1.8 1.0 1.4 1.0 0 0.1 1 10 0.1 1 10 D/D D/D xy xy

β4 β5 0.6 1.4 1.8 0.5 3.0 1.0 3.0 0.4 1.8 3.0 1.8 1.4 0.3 1.0 1.4 0.2 1.0 0.1 0.1 1 10 0.1 1 10 D/D D/D xy xy

Figure 9.5 Normalised maximum deflection, bending moments, and transverse forces for orthotropic, simply supported rectangular plates with Dxy/Dy = 0.8 and isotropic core subjected to uniform pressure. 2 2 Solid lines are for π φy = 1.0 and dashed lines for π φy = 0 (pure bending). The lines are for a/b = 1.0, 1.4, 1.8 and 3.0.

9.17 AN INTRODUCTION TO SANDWICH STRUCTURES

β 1 0.12 0.4 0.10 0.7 0.08 1.0 1.43 0.06

2.5 0.04

0.02

0 0246810 2 πφy

β2 β3 0.08 2.5 0.4 0.06 1.43 1.0 0.7 0.04 0.7 1.0 1.43 0.02 0.4 2.5 0 0246810 0246810 2 2 πφy πφy

β4 β 0.5 5

2.5 0.4 0.4 1.43 0.7 1.0 1.0 0.3 0.7 1.43 0.4 2.5 0.2 0246810 0246810 2 2 πφy πφy

Figure 9.6 Normalised maximum deflection, bending moments, and transverse forces for square simply supported plates with isotropic faces and an orthotropic core subjected to uniform pressure.

The lines are for Sx /Sy = 0.4, 0.7, 1.0, 1.43, and 2.5.

9.18 SOLUTIONS TO PLATE PROBLEMS

β 1 0.14 0.4 0.12 1.0 0.10

0.08 2.5 0.06

0.04

0.02

0 0246810 2φ π y

β2 β3 0.14 0.08 2.5 0.4 0.10 1.0 1.0 0.04 0.4 0.06 2.5

0 0.02 0246810 0246810 2 2 πφy πφy

β β 0.5 4 5 2.5 0.4 1.0 0.4 1.0 2.5 0.3

0.4 0.2 02468100246810 2 2 πφy πφy

Figure 9.7 Normalised maximum deflection, bending moments, and transverse forces for rectangular simply supported plates with isotropic faces and an orthotropic core subjected to uniform pressure.

Solid lines are for a/b = 1.4 and dashed lines for a/b = 2.0. The lines are for Sx/Sy = 0.4, 1.0, and 2.5.

9.19 AN INTRODUCTION TO SANDWICH STRUCTURES

As seen from eqs.(9.19), the solution to this problem is highly complicated even though it only involved a rectangular orthotropic sandwich plate with relatively simple boundary conditions. In order to simplify the analysis one can obtain an approximate solution by assuming the theory of partial deflections and solving the bending and shear problems independently through eqs.(8.17). Since the boundary conditions for the simply supported plate are the same for both pure bending and pure shear one can assume the deflection field

∞ ∞ mxππny ∞ ∞ mxππny wWbbmn= ∑∑ sin sin and wWssmn= ∑∑ sin sin n=1 m=1 a b n=1 m=1 a b which obviously satisfies all the boundary conditions

∂wb ∂wb wb = ws = 0 on all boundaries, ≠ 0 , ≠ 0 ∂x xa=0, ∂y yb=0,

∂ 2 w ∂ 2 w T ∂w T ∂w b = 0 , b = 0 , x =≠s 0 and y =≠s 0 ∂x 2 ∂y 2 S ∂x S ∂y xa=0, yb=0, x xa=0, y yb=0, and as before the assumption leads to hard simple support. Inserting the assumptions into eqs.(8.17) leads to the expressions 16q wsmn = 22 and 2 ⎡ ⎛ mππ⎞ ⎛ n ⎞ ⎤ mnπ ⎢ Sxy⎜ ⎟ + S ⎜ ⎟ ⎥ ⎣ ⎝ a ⎠ ⎝ b ⎠ ⎦

16qmn/ π 2 wbmn = D mπ 4224⎛ννDD+ ⎞ mππn D nπ x ⎜⎛ ⎟⎞ + ⎜ yx x xy y + 2D ⎟⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ + y ⎜⎛ ⎟⎞ ⎝ ⎠ ⎜ xy ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 11− ννxy yx a ⎝ − ννxy yx ⎠ a b 1 − ννxy yx b which will give reasonable accuracy when summed up in eq.(9.22). However, note that using this approximate solution method, bending moments and transverse forces will become uncoupled which means that bending moment are independent of orthotropy in the shear stiffness and transverse forces independent of orthotropy in the bending stiffness. As shown in Figs.9.4-7 illustrating the exact solution, this is not correct. Note though that the method is solution is greatly simplified and the deformation calculated from this approximate form should become at least reasonably correct.

9.20 SOLUTIONS TO PLATE PROBLEMS

9.4 Solution by Energy Method - Ritz' Method The Ritz' method is a most versatile and convenient tool for obtaining approximate solutions to various beam and plate problems, and is equally applicable to bending, buckling and free vibration. The method is based on the use of the theorem for the minimum of the potential energy which reads: The total potential energy of a system has a stationary value for all small displacements when the system is in equilibrium and that the stationary value is a minimum. Thus, derivatives to the energy equations of eq.(8.45) with respect to unknown field assumptions are sought. Usually the field assumptions contain the unknown deflections u, v and w, but in sandwich problems it might sometimes be more convenient to use the transverse forces as unknowns rather than the in-plane deflections. The problem is then stated as [12]

U(w, Tx, Ty) = stationary value or U(u, v, w) = stationary value (9.22) The field assumption may take the form

M1 N1 wAwxy= ∑∑ mn mn (,) (9.23) m=1 n=1

M2 N2 M3 N3 TBTxyx= ∑∑ mn xmn (,) and TCTxyy= ∑∑ mn ymn (,) m=1 n=1 m=1 n=1 where Amn, Bmn and Cmn are unknown coefficients and wmn, Txmn and Tymn are chosen functions which are usually taken so that they may be separable in the form Xm⋅Yn. The field assumption functions wmn, Txmn and Tymn must satisfy the boundary conditions and be differentiable at least to the same order as the corresponding differential equation (order 4 for ordinary beams and plates, and order 6 when accounting for thick faces). Inserting these field assumptions into the energy equations leads to one single equation in the unknown coefficients Amn, Bmn and Cmn. The condition (9.22) then takes the form

∂U ⎧mM= 1,2,.., 1 = 0 ⎨ (9.24) ∂Amn ⎩ nN= 1,2,.., 1

∂U ⎧mM= 1,2,.., 2 = 0 ⎨ ∂Bmn ⎩ nN= 1,2,.., 2

∂U ⎧mM= 1,2,.., 3 = 0 ⎨ ∂Cmn ⎩ nN= 1,2,.., 3

This leads to (M1+M2+M3)×(N1+N2+N3) linear equations to be solved. For buckling and free vibration problems the set of equations is solved as an eigenvalue problem so that buckling loads and eigen-frequencies are given by eigenvalues to the matrix system, i.e. so that the determinant of the coefficients of Amn, Bmn and Cmn vanishes.

9.21 AN INTRODUCTION TO SANDWICH STRUCTURES

9.5 Approximate Solutions for Bending of Orthotropic Sandwich Plates As mentioned in section 9.2, the solution for the simply supported plate given in eq.(9.13) converges very rapidly for the deflection w. The same is true for the solution of the orthotropic plate in section 9.3. In fact it converges so rapidly that the first term in itself provides a very good approximation when the panel aspect ratio is close to unity. This fact can be used when deriving bending and buckling formulae for sandwich plates. This section will provide closed form solutions for bending problems of orthotropic sandwich plates with various edge conditions derived using energy principles, the so called Ritz' method, based on approximate deflection assumptions. The bending and buckling of orthotropic sandwich plates have been investigated in several reports from the U.S. Forest Laboratory, e.g., [13,15- 19]. There is a certain beauty about the approach used in all these reports as to the method of solution used. Although the derived solutions are not exact they are simple and yields results which are accurate enough for engineering purposes in terms of maximum deflection. Bending moments and transverse forces can not, however, be accurately extracted from these expressions which of course is a serious drawback. The formulae should therefore rather be view upon as a simple set of approximate deflection estimation equations suitable for quick reference calculations which may prove useful in cases where the stiffness is of primary interest.

The approach used here is based on the approximate theory for partial deflections introduced in section 9 but is physically the same as the approximate approach used by March [13] which leads to the same results.

9.5.1 Simply supported plate A good approximation is to use the first term only in the series of eq.(9.13). Since convergence is very fast for the deflection component the first term yields very accurate approximation. The bending moments depend on the second derivative of eq.(9.13) and the transverse forces on the third derivative. For every differentiation an m or an n is multiplied to the term in the series, that is ∂ k w ij ∝ mwk ∂x k ij making terms higher up in the series more significant. A first approach for a simple estimate of the plate behaviour would be by assuming partial deflections. First, let Sx = Sy = ∞, and thus ws = 0 and w = wb. By inserting the deflection assumption of eq.(9.5) into the governing equation (8.16b) we soon arrive at the deflection mxπ nyπ ()sinsin1− νν 16qb 4 ∞ ∞ xy yx w = a b (9.25a) b 6 ∑ ∑ 42 π n=135.,, .m=135.,, . ⎡ ⎛ mb⎞ ⎛ mnb⎞ 4 ⎤ mn⎢ Dx⎜ ⎟ ++−21[]DD xννν yx() xy yx xy⎜ ⎟ + Dn y ⎥ ⎣ ⎝ a ⎠ ⎝ a ⎠ ⎦ which is exactly the deformation field for a shear stiff orthotropic plate [10]. Next assume Dx

= Dy = Dxy = ∞, so that wb = 0 and w = ws. Inserting into eq.(8.16a) yields

9.22 SOLUTIONS TO PLATE PROBLEMS

mxπ nyπ sin sin 16qb 2 ∞ ∞ w = a b (9.25b) s 4 ∑ ∑ 2 π n=135.,, .m=135.,, . ⎡ ⎛ mb⎞ 2 ⎤ mn⎢ Sxy⎜ ⎟ + Sn ⎥ ⎣ ⎝ a ⎠ ⎦

Even for fairly high panel aspect ratios the mid-point deflection calculated using eq.(9.25) yields results in quite good agreement with the exact solution given in section 9.3 in terms of maximum deflections, whereas the maximum bending moments and transverse forces calculated from the partial deflections in eq.(8.14) agree less well. The reason for the concept of partial deflections not being exact in this case is that the relation between wb and ws in eq.(8.21) is not satisfied and the deflected shape due to bending is allowed to be different from the shape due to shear. The main benefit is however that the solution is much simpler and that it gives fairly good approximate results. By taking only the first term (n=m=1) in the above series and adding them we get for the mid-point deflection

4 2 16qb () 1 − ννxy yx 16qb w = 42+ 2 (9.26a) 6 ⎡ ⎛ b⎞ ⎛ b⎞ ⎤ 4 ⎡ ⎛ b⎞ ⎤ πνννπ⎢Dx⎜ ⎟ ++−21[]DD x yx() xy yx xy⎜ ⎟ + D y⎥ ⎢S x⎜ ⎟ + S y ⎥ ⎣ ⎝ a⎠ ⎝ a⎠ ⎦ ⎣ ⎝ a⎠ ⎦ which for an isotropic plate reduces to

16qb 42() 1 − ν 16qb 2 w = + (9.26b) 2 2 2 ⎡ b ⎤ 4 ⎡ b ⎤ 6 ⎛ ⎞ π S ⎜⎛ ⎟⎞ +1 π D⎢⎜ ⎟ +1⎥ ⎢⎝ ⎠ ⎥ ⎣⎝ a⎠ ⎦ ⎣ a ⎦

Note that this is exactly the same as using the first term (m=n=1) only in the Fourier-series for the isotropic plate given in eq.(9.13). March [13] used the approximate deflection field πx πy ww= sin sin (9.27) a b _ _ _ That means, setting w(x,y)=sin πx/a sin πy/b, Amn= w = wb+ws and M1=N1=1 in eq.(9.24). If one then applies the Ritz method using the simplified equation for the total strain energy in _ _ eq.(8.42c), performing the integration yields an equation in the unknowns wb and wb. Minimisation of the total energy with respect to these unknowns yields the same equation as in (9.23). For a complete exact solution one must use field assumptions equivalent to those in eqs.(9.5) and (9.6) use the exact energy equation of eq.(8.42b) and perform the analysis using several terms in the summation indices m and n. By using the first term only in the exact solution of eq.(9.13), the errors obtained for the deflection, bending moments and transverse forces of Table 9.1 are as in Table 9.3.

9.23 AN INTRODUCTION TO SANDWICH STRUCTURES

a/b Δwb Δws ΔMx ΔMy ΔTx ΔTy 1.0 1.025 1.114 1.115 1.115 0.763 0.763 1.2 1.027 1.117 1.136 1.104 0.719 0.801 1.4 1.035 1.125 1.153 1.103 0.676 0.831 1.6 1.037 1.134 1.190 1.104 0.636 0.853 1.8 1.044 1.143 1.219 1.106 0.595 0.872 2.0 1.052 1.154 1.246 1.111 0.558 0.888 3.0 1.102 1.205 1.354 1.156 0.416 0.942 4.0 1.149 1.242 1.373 1.200 0.326 0.976 5.0 1.186 1.265 1.377 1.234 0.267 0.992 ∞ 1.278 1.314 1.314 1.314 0 1.00

Table 9.3 Errors made by using only the first term in the series of eq.(9.13) for an isotropic sandwich plate. Δ is the ratio of the approximate value to the exact value given in Table 9.1. ν = 0.3. As seen in Table 9.3, the deflection values agree very well by using only the first term, whereas the transverse forces yield very large errors. Equation (9.26b) of course yields the same errors summarised in Table 9.3. The errors between eq.(9.26a) and the exact solution of section 9.3 are surprisingly small even for quite large values of b/a and Dx/Dy, and even in extreme cases the difference is less than 30%, with the approximate value always higher than the exact value. It is also worth noticing that the bending deflection wb is much more accurate than the that due to shear, ws. By comparing the exact solution to the approximate one it is found that ws is less than 10% higher than the exact value when approximately

1 ⎛ ⎞ 4 b Dx 033. < ⎜ ⎟ < 3 (9.28a) a ⎝ Dy ⎠

It also seems that ws is less than 10% higher than the exact value if

b ⎛ Dxy ⎞ 012. < ⎜ ⎟ < 8 (9.28b) a ⎝ Dy ⎠

providing Dx = Dy. For the shear deflection on the other hand there is a more systematic error in the solution so that ws of eq.(9.26) is 11.5% higher than the exact solution even for an isotropic core and a square plate. However, the error is between 11.5 and 20% if

1 ⎛ ⎞ 2 b Sx 033. < ⎜ ⎟ < 3 (9.28c) a ⎝ Sy ⎠

These are all approximate limits and variations of several parameters can of course yield higher errors. Still, for engineering purposes errors in the order of such magnitudes may be acceptable, especially if known beforehand, and also known to be conservative. In practice, the degree of orthotropy is commonly quite low, as well as the panel aspect ratio, thus increasing the possible use of the approximate formulae. A way of estimating the deflection for cases which lie outside the range of eq.(9.28), e.g., for large or small b/a ratios is the following: for a large b/a ratio one can argue that the line at y = b/2 is more or less unaffected

9.24 SOLUTIONS TO PLATE PROBLEMS by the boundary conditions at y = 0,b, and one can thus approximate the deflection in the middle of the plate by πx πx www=+( )sin =w sin (9.29) bs a a By inserting this into the energy equation of (8.42c) we get

24 22 wDbxπ wSsxπ 2qa() w b+ w s U = 3 +− 41a ()− ννxy yx 4a π

Minimisation of the total energy in eq.(8.45) in then performed as ∂U ∂U U ===0 ∂wbs∂w which yields that

1 1 41qa 4 ()− νν 4qa 2 b ⎛ D ⎞ 4 b ⎛ S ⎞ 2 w = xy yx + , when ⎜ x ⎟ >> 11 and/or ⎜ x ⎟ >> (9.30a) 5 3 ⎜ ⎟ ⎜ ⎟ ππDxxS a ⎝ Dy ⎠ a ⎝ Sy ⎠

From beam theory it is seen that this expression is very close to that of a simply supported beam with length a, bending stiffness Dx and shear stiffness Sx, subjected to a uniform pressure q. In the same manner πy ww= sin b

1 1 41qb 4 ()− νν 4qb 2 b ⎛ D ⎞ 4 b ⎛ S ⎞ 2 w = xy yx + , when ⎜ x ⎟ << 11 and/or ⎜ x ⎟ << (9.30b) 5 3 ⎜ ⎟ ⎜ ⎟ ππDyyS a ⎝ Dy ⎠ a ⎝ Sy ⎠

9.5.2 Clamped plate

Figure 9.9 Rectangular sandwich plate with all edges clamped. The boundary conditions for this case are ∂w ∂w w = 0, b = 0 for x = 0,a, and w = 0, b = 0 for y = 0,b ∂x ∂y

9.25 AN INTRODUCTION TO SANDWICH STRUCTURES

The assumed deflection field for the simply supported plate given in eq.(9.5) cannot be used in this case since is does not satisfy the clamped boundary conditions at the plate edges. A function that will satisfy the boundary conditions but only approximately describe the shape of the plate is πx πy πx πy www=+( )sin22 sin =w sin 22 sin (9.31) bs a b a b This assumed deflected shape will satisfy the boundary condition ∂w/∂x or ∂w/∂y along the edges as would be the case for a thick face sandwich, but not the condition ∂wb/∂x or ∂wb/∂y in the thin face case. If so, one could perhaps instead use the function πx πy πx πy ww=+sin22 sinw sin sin bsa b a b The result of doing so implies that the shear part of the solution would remain the same as in the simply supported case, i.e. as given in eq.(9.26). As will be seen using the function of eq.(9.31) gives a result for the shear part very similar to that of the simply supported case and since the approach outlined is approximate the choice makes very little difference. Using the assumed shape of eq.(9.31), one cannot proceed as in the simply supported case since the load q can not be easily written in the same form as the deflected shape. Using Ritz' method implies differentiation and inserting into the equation for the total strain energy, eq.(8.39c). Doing this thoroughly we first have the derivatives

∂ 2 w 2w π 2 πππx x y bb=−⎜⎛cos222 sin⎟⎞ sin ∂x 2 a 2 ⎝ a a ⎠ b

∂ 2 w 2w π 2 ππx y πy bb=−sin22⎜⎛ cos sin 2⎟⎞ ∂y 2 b 2 a ⎝ b b ⎠

∂22w 4w π πx πx πy πy bb= sin cos sin cos ∂∂xy ab a a b b

∂w 2w π πx πx πy ss= sin cos sin2 ∂x a a a b

∂w 2w π πx πy πy ss= sin2 sin cos ∂y b a b b

Inserting into the equation for the total strain energy eq.(8.42c) we get the integrals

2 a b Dw⎛ ∂ 2 ⎞ 3Dw24π b x bxbdxdy ==.... , ∫ ∫ ⎜ 2 ⎟ 3 0 0 1 − ννxy yx ⎝ ∂x ⎠ 4(1− ννxy yx )a

2 a b D ⎛ ∂ 2 w ⎞ 3Dw24π a y b dxdy ==.... yb , ∫ ∫ ⎜ 2 ⎟ 3 0 0 1 − ννxy yx ⎝ ∂y ⎠ 4(1− ννxy yx )b

9.26 SOLUTIONS TO PLATE PROBLEMS

a b 2ν D ⎛ ∂ 2 w ⎞⎛ ∂ 2 w ⎞ νπDw24 yx x bbdxdy ==.... yx x b , since ν D = ν D ∫ ∫ ⎜ 2 ⎟⎜ 2 ⎟ yx x xy y 0 0 121− ννxy yx ⎝ ∂x ⎠⎝ ∂y ⎠ ()− ννxy yx ab

2 a b ⎛ ∂ 2 w ⎞ Dw24π 2D b dxdy ==....xy b , ∫ ∫ xy ⎜ ⎟ 0 0 ⎝ ∂∂xy⎠ 2ab

2 a b ⎛ ∂w ⎞ 3Sw22π b S ⎜ sxs⎟ dxdy ==.... , and finally ∫ ∫ x ⎝ ⎠ 0 0 ∂x 16a

2 a b ⎛ ∂w ⎞ 3Sw22π a S s dxdy ==....ys ∫ ∫ y ⎜ ⎟ 0 0 ⎝ ∂y ⎠ 16b

The potential of the applied load q is then according to eq.(8.40)

a b qab qwdxdy ==.... (ww + ) ∫ ∫ bs 0 0 4

The total energy is then

24⎡ ⎤ wDbbxπ 3 3Day ν yxD x Dxy U = ⎢ 33+ + + ⎥ 2 ⎣⎢41a ()()()− ννxy yx 41b − ννxy yx 21ab − ννxy yx 2ab⎦⎥

22 ws π ⎡3Sbx 3Say ⎤ qab() wbs+ w ++⎢ ⎥ − 2 ⎣ 16a 16b ⎦ 4 _ _ Minimisation of the total energy in eq.(8.42) with respect to wb and ws yields

4 ∂U wDbπ ⎡3 3Dayν yxD xD xy()1 − ν xyν yx ⎤ qab b x 0 = ⎢ 33++ + ⎥ −= ∂wb ()1 − ννxy yx ⎣ 4a 42b ab 24ab ⎦

∂U 2 ⎡3Sbx 3Say ⎤ qab =+wsπ ⎢ ⎥ −=0 ∂ws ⎣ 16a 16b ⎦ 4 which means that the maximum deflection appearing in the middle of the panel

4 2 qb ()1 − ννxy yx 4qb w = 42+ 2 (9.32a) 4 ⎡ ⎛ b⎞ 2 ⎛ b⎞ ⎤ 2 ⎡ ⎛ b⎞ ⎤ 3πνννπ⎢Dx⎜ ⎟ ++−[]DD x yx()1 xy yx xy⎜ ⎟ + D y⎥ 3 ⎢S x⎜ ⎟ + S y ⎥ ⎣ ⎝ a⎠ 3 ⎝ a⎠ ⎦ ⎣ ⎝ a⎠ ⎦ where once again the first and second terms represent bending and shear deflection, respectively, and where the maximum deflection appear in the middle of the plate, i.e., for (x,y) = (a/2,b/2). The same procedure could be used for the simply supported plate, but results in the same relations as using partial deflections. For an isotropic plate eq.(9.32a) reduces to

9.27 AN INTRODUCTION TO SANDWICH STRUCTURES

qb 42()1− ν 4qb 2 w = 42+ 2 (9.32b) 4 ⎡ ⎛ b⎞ ⎛ b⎞ ⎤ 2 ⎡⎛ b⎞ ⎤ ππD⎢323⎜ ⎟ + ⎜ ⎟ + ⎥ 31S⎢⎜ ⎟ + ⎥ ⎣ ⎝ a⎠ ⎝ a⎠ ⎦ ⎣⎝ a⎠ ⎦

The first term is quite close to that given by Timoshenko [4] for an ordinary isotropic plate, and the second one is still quite close to that for the simply supported plate (eq.(9.26)). The shear deflection should be invariable with the boundary condition under the given assumptions and therefore it should remain exactly the same as for the simply supported plate. Another way of assessing the deflection of a clamped plate subjected to uniform pressure loading would then be to simply sum the deflections from the two extreme cases pure bending and pure shear. Since the shear part in this case will be invariable with the boundary condition, the shear deflection ws will equal that of the simply supported plate given in Table 9.1. The bending part is found from, e.g., [4] or [5] and is given in Table 9.4.

It is mentioned in [13] that the use of the approximate deflected shape function in eq.(9.31) provides acceptable accuracy if

1 ⎛ ⎞ 4 b Dx 07..< ⎜ ⎟ < 14 (9.33) a ⎝ Dy ⎠

Dwb My Mx My Mx 2 (,)(/,)xy= a20 2 (,)(/,/)xy= a22 b a/b qb 42()1− ν qb2 (,)(,/)xy= 02 b qb qb2 (,)(/,/)xy= a22 b qb

1.0 0.0013 -0.0513 -0.0513 0.0231 0.0231 1.2 0.0017 -0.0639 -0.0554 0.0299 0.0228 1.4 0.0021 -0.0726 -0.0568 0.0349 0.0212 1.6 0.0023 -0.0780 -0.0571 0.0381 0.0193 1.8 0.0024 -0.0812 -0.0571 0.0401 0.0174 2.0 0.0025 -0.0829 -0.0571 0.0412 0.0158 ∞ 0.0026 -0.0833 -0.0571 0.0417 0.0125

Table 9.4 Maximum bending deformation and moments of a clamped rectangular isotropic sandwich plate with flexural rigidity D. Numbers from ref. [4]. ν = 0.3. Since eqs.(9.32) only yield accurate results for panel aspect ratios close to unity, March [13] suggested the use of πx ww= sin2 (9.34) a as the assumed deflected shape for large b/a ratios, or rather when the relation in eq.(9.33) is very large. The corresponding deflection is then calculated as

1 1 qa 4 ()1 − νν qa 2 b ⎛ D ⎞ 4 b ⎛ S ⎞ 2 w = xy yx + , when ⎜ x ⎟ >> 11 and/or ⎜ x ⎟ >> (9.35a) 4 2 ⎜ ⎟ ⎜ ⎟ 4ππDxxS a ⎝ Dy ⎠ a ⎝ Sy ⎠ and similarly

9.28 SOLUTIONS TO PLATE PROBLEMS

πy ww= sin2 b

1 1 qb 4 ()1 − νν qb 2 b ⎛ D ⎞ 4 b ⎛ S ⎞ 2 w = xy yx + , when ⎜ x ⎟ << 11 and / or ⎜ x ⎟ << (9.35b) 4 2 ⎜ ⎟ ⎜ ⎟ 4ππDyyS a ⎝ Dy ⎠ a ⎝ Sy ⎠ which is very close to the result of a beam with both edges clamped subjected to uniform pressure. Even though these approximate formulae can be used for the central deflection of the clamped plate, other things change more drastically when the edges are clamped. If one assumes thin faces then the boundary condition stated in section 8.9 allows for a slope ∂w/∂x

= Tx/Sx at the boundaries x = 0,a. The distribution of transverse forces remains the same as in the simply supported case, but the bending moment distribution depends on the shear stiffness [8], similar to what happens to a hyperstatic sandwich beam (see examples 3.7.4 and 3.7.5). In fact, the presence of a finite shear stiffness has a relaxing effect on the maximum bending moment appearing in the middle of the sides. Hence, using the maximum bending moment of a clamped shear stiff plate as given in Table 9.4 yields conservative results. The maximum bending moment appearing in the middle of the sides of a square sandwich plate was calculated by Lockwood-Taylor [14] for different shear factors φ. In fact, the maximum bending moment decreases from 0.0531qa2 for the ordinary plate with φ = 0 (infinite shear stiffness) to 0.0347qa2 for a pure shear plate (φ = ∞). However, this calculation assumes faces with no thickness, i.e., membranes and the deflection assumption is for a soft clamped support. In practice the faces have a finite thickness implying that there actually is a local bending stiffness of the faces about their own neutral axes leading to the boundary condition ∂w/∂x = 0 at x = 0 and x = a. This in consequence means that even though the global bending moments decrease so that the membrane stresses in the faces are reduced by the presence of shear deflection, local bending moments act on the faces in the vicinity of the clamped edge. These local bending moments give rise to additional face stresses to be superimposed on the state of membrane stress. This is analogous to the example of the cantilever in section 4.1. As a result of this, the use of globally reduced maximum bending moments in the design of a clamped plate is not only a complex procedure but may also be misleading due to the local stresses appearing. It may therefore, for engineering purposes, be more adequate to use the maximum bending moments of the ordinary plate as given in Table 9.4 as these are an upper bound, and disregard the local bending moments.

9.29 AN INTRODUCTION TO SANDWICH STRUCTURES

9.5.2 Two sides clamped, the other two simply supported First study a plate that has its sides along the x-axis clamped and the two other simply supported, that is, a plate with the boundary conditions

Figure 9.10 Rectangular sandwich plate with two edges clamped and two simply supported.

∂w w = 0, M = 0 for x = 0,a, and w = 0, b = 0 for y = 0,b x ∂y

The assumed deflected shape in this case is πx πy πx πy www=+( )sin sin22 =w sin sin (9.36) bs a b a b which is approximate but satisfies the boundary conditions. Perform the same analysis as for the clamped plate above, that is, differentiation of eq.(9.36), inserting in the equation for the strain energy and finally minimising with respect to the unknown amplitudes. This gives after some algebra

16qb 4 () 1 − νν 16qb 2 xy yx (9.37a) w = 42+ 2 5 ⎡ ⎛ b⎞ ⎛ b⎞ ⎤ 3 ⎡ ⎛ b⎞ ⎤ πνννπ⎢381Dx⎜ ⎟ ++−[]DD x yx() xy yx xy⎜ ⎟ + 16D y⎥ ⎢34S x⎜ ⎟ + S y ⎥ ⎣ ⎝ a⎠ ⎝ a⎠ ⎦ ⎣ ⎝ a⎠ ⎦ which for an isotropic plate reduces to

16qb 42() 1 − ν 16qb 2 w = + (9.37b) 42 2 5 ⎡ ⎛ b⎞ ⎛ b⎞ ⎤ 3 ⎡ ⎛ b⎞ ⎤ ππD⎢3816⎜ ⎟ + ⎜ ⎟ + ⎥ S⎢34⎜ ⎟ + ⎥ ⎣ ⎝ a⎠ ⎝ a⎠ ⎦ ⎣ ⎝ a⎠ ⎦

Now, these expressions will only give acceptable accuracy in approximately the same range as stipulated for the clamped plate. Thus, as above, when the relation in eq.(9.33) is very large or very small we can approximate the plate deflection by πx ww= sin a

1 1 41qa 4 ()− νν 4qa 2 b ⎛ D ⎞ 4 b ⎛ S ⎞ 2 w = xy yx + , when ⎜ x ⎟ >> 11 and/or ⎜ x ⎟ >> (9.38a) 5 3 ⎜ ⎟ ⎜ ⎟ ππDxxS a ⎝ Dy ⎠ a ⎝ Sy ⎠

9.30 SOLUTIONS TO PLATE PROBLEMS

πy and ww= sin2 b

1 1 qb 4 ()1 − νν qb 2 b ⎛ D ⎞ 4 b ⎛ S ⎞ 2 w = xy yx + , when ⎜ x ⎟ << 11 and / or ⎜ x ⎟ << (9.38b) 4 2 ⎜ ⎟ ⎜ ⎟ 4ππDyyS a ⎝ Dy ⎠ a ⎝ Sy ⎠

If we now shift the boundary conditions so that we instead have

Figure 9.11 Rectangular sandwich plate with two edges clamped and two simply supported. The boundary conditions are now ∂w w = 0, b = 0 for x = 0,a, and w = 0, M = 0 for y = 0,b ∂x y The assumed deflected shape in this case can be chosen as πx πy πx πy www=+( )sin22 sin =w sin sin (9.39) bs a b a b Following the same procedure as above one arrives at

4 2 16qb () 1 − ννxy yx 16qb w = 42+ 2 (9.40a) 5 ⎡ ⎛ b⎞ ⎛ b⎞ ⎤ 3 ⎡ ⎛ b⎞ ⎤ πνννπ⎢16Dx⎜ ⎟ ++− 8[]DD x yx() 1 xy yx xy⎜ ⎟ + 3D y⎥ ⎢43S x⎜ ⎟ + S y ⎥ ⎣ ⎝ a⎠ ⎝ a⎠ ⎦ ⎣ ⎝ a⎠ ⎦ which for an isotropic plate reduces to

16qb 42() 1 − ν 16qb 2 w = 42+ 2 (9.40b) 5 ⎡ ⎛ b⎞ ⎛ b⎞ ⎤ 3 ⎡ ⎛ b⎞ ⎤ ππD⎢16⎜ ⎟ + 8⎜ ⎟ + 3⎥ S⎢43⎜ ⎟ + ⎥ ⎣ ⎝ a⎠ ⎝ a⎠ ⎦ ⎣ ⎝ a⎠ ⎦ which would be valid within the range given in eq.(9.33). Outside this range, we have πx ww= sin2 a

1 1 qa 4 ()1 − νν qa 2 b ⎛ D ⎞ 4 b ⎛ S ⎞ 2 w = xy yx + , when ⎜ x ⎟ >> 11 and/or ⎜ x ⎟ >> (9.41a) 4 2 ⎜ ⎟ ⎜ ⎟ 4ππDxxS a ⎝ Dy ⎠ a ⎝ Sy ⎠

9.31 AN INTRODUCTION TO SANDWICH STRUCTURES

πy and ww= sin b

1 1 41qb 4 ()− νν 4qb 2 b ⎛ D ⎞ 4 b ⎛ S ⎞ 2 w = xy yx + , when ⎜ x ⎟ << 11 and/or ⎜ x ⎟ << (9.41b) 5 3 ⎜ ⎟ ⎜ ⎟ ππDyyS a ⎝ Dy ⎠ a ⎝ Sy ⎠

9.6 Buckling of a Simply Supported, Isotropic Sandwich Plate This section covering buckling problems will be briefer than the previous one on the bending of sandwich plates due to transverse load. The fact is that transverse loads are usually of more importance in the design and calculation of sandwich plates for applications outside the aerospace community. Still, in-plane compressive loads may appear and buckling constraints may be imposed even on elements in building, transportation and marine applications. However, there is much published work on the buckling of sandwich plates in which buckling coefficients for almost any possible case can be found. The theoretical background has already been given in section 9 but in contrast to the bending problem, now the transverse load q is zero but the in-plane normal forces Nx and Ny equal –Px and –Py, respectively, in the governing equations given in eqs.(8.16), (8.22), (8.27), (8.30), and (8.31). Once again, start with the simplest buckling problem.

(i) Thin faces Consider a rectangular sandwich plate with sides a and b, subjected to in-plane compressive forces Px and Py according to Fig.9.12.

a Px

Figure 9.12 Sandwich plate subjected to biaxial compression. In this case it is appropriate to use the governing equation in the form given in eq.(8.27) which will now appear as

D ⎛ DΔ ⎞⎛ ∂ 2 w ∂ 2 w⎞ Δ2 w =−⎜1 ⎟⎜−−P P ⎟ (9.42) 1 − νν2 ⎝ S()1 − 2 ⎠⎝ xy∂x 2 ∂y 2 ⎠

In the same manner as for the sandwich beam an expression for the assumed displacement can be chosen as (see the bending problem in section 9.2) ∞ ∞ mxπ nyπ ww= ∑∑ mn sin sin (9.43) n=1 m=1 a b

9.32 SOLUTIONS TO PLATE PROBLEMS which satisfies the boundary condition for the simply supported edges at x = 0,a and y = 0,b just as in the bending case in section 9.2. Inserting eq.(9.43) into the governing equation yields

222 ∞ ∞ ⎪⎧ Dm⎡⎛ ππ⎞ ⎛ n ⎞ ⎤ ⎨ 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ∑∑ 1 − ν ⎝ a ⎠ ⎝ b ⎠ n=1 m=1 ⎩⎪ ⎣ ⎦ (9.44) ⎡ mππ22n ⎤⎡ D ⎧ mππ22n ⎫⎤⎪⎫ mx ππny − P ⎜⎛ ⎟⎞ + P ⎜⎛ ⎟⎞ 1 + ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ w sin sin = 0 ⎢ xy⎝ ⎠ ⎝ ⎠ ⎥⎢ 2 ⎨⎝ ⎠ ⎝ ⎠ ⎬⎥⎬ mn ⎣ a b ⎦⎣ S()1 − ν ⎩ a b ⎭⎦⎭⎪ a b

This must valid for any x and y and hence the term within the bracket must be zero for every combination of m and n. This now constitutes the governing stability equation for which the minimum with respect to the wave form m and n is sought for a given set of Px and Py. Now, m and n are the number of half waves in the buckled form in the x- and y-direction, respectively. To further simplify and take an example assume that the compression is uniaxial with Px = P and Py = 0. It is known that the minimum value of eq.(9.44) always corresponds to n=1 for this case [8]. After insertion and simplification of eq.(9.44) one can write the buckling coefficient K as

−1 2 2 (1 −ν 2 )b 2 ⎛ mb a ⎞ ⎛ ⎡⎛ mb ⎞ ⎤⎞ K = P = ⎜ + ⎟ ⎜1 + π 2φ ⎢⎜ ⎟ + 1⎥⎟ (9.45) 2 ⎜ ⎟ π D ⎝ a mb ⎠ ⎝ ⎣⎢⎝ a ⎠ ⎦⎥⎠ where φ = D/(1–ν2)b2S is the shear factor, as defined in the beam case. Note, however, that the normalisation now is done with respect to b. The same expressions are found in [8] and [11] but written with a slightly different notation. The actual buckling load is given by eq.(9.45) for the value of m giving a minimum load P. The above equation can easily be plotted and although it appears in several of the references given in this section it is shown in Fig.9.13. 5 π 2φ m=1 m=2 m=3 4 0 0.05 0.1 3 K 0.2 0.3 2 0.4

1 1

0 0 0.5 1 1.5 2 2.5 3 a/b

Figure 9.13 Buckling coefficients for a simply supported, isotropic sandwich plate under uniaxial compression. The lines are for π2φ = 0, 0.05, 0.1, 0.2, 0.3, 0.4, 0.6, and 1. As seen in Fig.9.13, the critical buckling load has a finite value even for a/b equal to zero. Recall the buckling of a sandwich column where the buckling load approaches a finite value,

9.33 AN INTRODUCTION TO SANDWICH STRUCTURES the shear buckling load equal to the shear stiffness, when the column length approached zero. 2 Similarly here, as seen in the graph above, the buckling coefficient approaches 1/π φ, or Px approaches S when the panel aspect ratio a/b is small. Hence, in analogy with the bending case, the plate buckling problem can be approximated by a beam with length a when the panel aspect ratio is small. The number of waves, m, in the buckling mode depends on the panel aspect ratio as well as the ratio of the shear factor φ. For the case of pure bending (S equals infinity), we have that

()1 − ν 22b mb a 2 K = P =+⎜⎛ ⎟⎞ π 2 D b ⎝ a mb⎠ which is the buckling coefficient for an ordinary isotropic plate. At the other limit, D equals infinity, we have

22 ()11− ν 22b ⎛ ⎛ a ⎞ ⎞ ⎛ ⎛ a ⎞ ⎞ K = P =+⎜11⎜ ⎟ ⎟, or PS=+⎜ ⎜ ⎟ ⎟ ππφ22D ss⎝ ⎝ mb⎠ ⎠ ⎝ ⎝ mb⎠ ⎠

(ii) Thick faces In order to account for the bending stiffness of the faces about their individual neutral axes in the buckling of a sandwich column in section 5, or in the bending of plates as in section 9.2, one may proceed as above but instead using the differential equation proposed by Hoff [6] as in eq.(8.31). The deflection assumption of eq.(9.43) may once again be used as it satisfies the boundary conditions as well as the governing equation. Performing the same algebraic exercise as above one arrives at (note that now one must distinguish between D and D0)

3 2 2D ⎡ mππ22n ⎤ ()DDS+ 2 ⎡ m ππ22n ⎤ ff⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ + 0 ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ 2 ⎢⎝ ⎠ ⎝ ⎠ ⎥ ⎢⎝ ⎠ ⎝ ⎠ ⎥ 1 − ν ⎣ a b ⎦ D0 ⎣ a b ⎦ (9.46) ⎡ mππνππ22n ⎤⎡S()1 − 2 m 22n ⎤ − P ⎜⎛ ⎟⎞ + P ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ = 0 ⎢ xy⎝ ⎠ ⎝ ⎠ ⎥⎢ ⎝ ⎠ ⎝ ⎠ ⎥ ⎣ a b ⎦⎣ D0 a b ⎦

Once again assuming uniaxial compression, i.e., Px = P, Py = 0 and n = 1, the expression for the critical load becomes

−1 22 2 ⎡ 2 ⎤ ()1 − ν b mb a 2D f ⎛ ⎡ mb ⎤⎞ ⎛ ⎞ ⎢ 2 ⎛ ⎞ ⎥ K = 2 P =+⎜ ⎟ ++⎜11πφ⎢⎜ ⎟ + ⎥⎟ (9.47) π D ⎝ a mb⎠ D ⎝ a ⎠ 0 ⎣⎢ ⎝ ⎣ ⎦⎠ ⎦⎥

2 2 where φ = D0/(1–ν )b S. This is the expression derived by Hoff [6] and as can be seen it is very similar to the expression for thin faces but with the term 2Df /D added. Hence, if the faces are thin (Df = 0) this expression converges to that derived above. In perfect analogy with the beam case, the incorporation of the flexural rigidity of the faces leads to an infinite buckling load as the side length a approaches zero. By plotting eq.(9.47) in the same manner as for the thin face case above, this is clearly seen; when the side length a is large, the buckling load is similar to that of the thin face solution but with D = D0 + 2Df, and as the side length a approaches zero the buckling equals that of the two faces buckling independently of each other, and hence approaches infinity. The buckling coefficients for some values of the

9.34 SOLUTIONS TO PLATE PROBLEMS

shear parameter φ have been plotted in Fig.9.14 for both Df = 0 and for 2Df /D0 = 0.1, where the latter can be seen as a limiting case for practical purposes. 5 π 2φ 4 0.05 3 0.2 K 0.4 2

1 1.0

0 0 0.5 1 1.5 2 2.5 3 a/b

Figure 9.14 Buckling coefficients for a simply supported sandwich plate under uniaxial compression. 2 2 Dashed lines are for Df = 0 (π φ = 0.05, 0.2, 0.4, 1.0) and full lines for 2 Df / D0=0.1 (π φ = 0.05, 0.2, 0.4, 1.0).

9.7 Buckling of a Simply Supported, Orthotropic Sandwich Plate with Thin Faces The solution to the buckling load for an orthotropic simply-supported sandwich plate can be derived using the same approach as in the lateral bending case described in section 9.3. Following once again the work by Robinsson [9] and assuming the following deflection field W mπx nyπ w =− mn sin sin (9.48) Zmn a b

Xmn mπx nπy Ymn mπx nπy Tx = cos sin , and Ty = sin cos Zmn a b Zmn a b where the coefficients are the same as given in eqs.(9.19) apart from the last equation which now reads [9]

mπππ2 m n ZN=− ⎜⎛ ⎟⎞ W +−X Y (9.49) mn x⎝ a ⎠ mna mnb mn when uniaxial compression Nx is acting on the plate. The critical buckling load is found by letting eq.(9.48) approach infinity, or equivalently, making Zmn equal zero. By using the following notation

Dx Dy φ x = 22 and φ y = (9.50) aSxxyyx()11−νν bSyxyyx ()−νν and by identifying that the minimum buckling load corresponds to n = 1, i.e., one wave only in the direction perpendicular to the load, the equations becomes [9]

9.35 AN INTRODUCTION TO SANDWICH STRUCTURES

2 2 ⎡ 42 2 42 bP()1 − ννxy yx x ⎛ mb⎞ πφyyS ⎛ νxyD y ⎞⎪⎧ DDxxy⎛ mb⎞ DDxxyxy− ν ⎛ mb⎞ Dxy ⎪⎫ ⎜ ⎟ ⎢ ⎜1 − ⎟⎨ ⎜ ⎟ + ⎜ ⎟ + ⎬ π 2 D ⎝ a ⎠ S D 22D2 ⎝ a ⎠ D ⎝ a ⎠ D y ⎣⎢ x ⎝ x ⎠⎩⎪ y y y ⎭⎪

⎧ 2 ⎫ ⎤ mb ⎛ DSxy Dxy ⎛ D ⎞⎞ ⎛ Dxy ⎛ D ⎞ Sy ⎞ 2 ⎪⎛ ⎞ ⎜ x 22⎟ ⎜ x ⎟⎪ +πφy ⎨⎜ ⎟ +−⎜ νxy ⎟ ++1 ⎜ − νxy ⎟ ⎬ + 1⎥ ⎝ a ⎠ ⎜ DS 2D ⎜ D ⎟⎟ ⎜ 2D ⎜ D ⎟ S ⎟ ⎥ ⎩⎪ ⎝ yx x ⎝ y ⎠⎠ ⎝ x ⎝ y ⎠ x ⎠⎭⎪ ⎦ ⎛ D νν2 ⎞ ⎡ DD 422DD− D ⎤ S 2 yxy xxy⎛ mb⎞ xxyxy⎛ mb⎞ xy ⎡⎛ mb⎞ y ⎤ =−πφ⎜1 ⎟ ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + y ⎢ 2 ⎝ ⎠ ⎝ ⎠ ⎥⎢⎝ ⎠ ⎥ ⎝ Dx ⎠ ⎣⎢ 22Dy a Dy a Dy ⎦⎥⎣ a Sx ⎦ (9.51) 4 ⎛ D D ν 2 ⎞ 2 Dx ⎛ mb⎞ xy xy xy ⎛ mb⎞ + ⎜ ⎟ +−21⎜ +ν ⎟⎜ ⎟ + ⎝ ⎠ ⎜ xy ⎟⎝ ⎠ Dy a ⎝ Dy Dx ⎠ a

By plotting the in-plane compressive load Px versus a/b for various integral values of m using this rather complex equation, the same kind of buckling curves can be produced as in the isotropic case. It is easily shown that the limiting value of the buckling load Px when the ratio a/b approaches zero equals Sx, which is not surprising since this can be approximated by a beam of length a subjected to an in-plane compressive load Px. It is convenient to introduce the buckling coefficient

2 b ()1− νxyν yx K = 2 Px (9.52) π Dx

2 giving the value of K for a/b = 0 equal to 1/π φx, as defined in eq.(9.50). To illustrate the influence of transverse properties on the buckling of a simply-supported orthotropic sandwich plate subjected to uniaxial compression Px, eq.(9.51) has been plotted in Fig.9.15. These graphs are prepared by keeping the properties in the x-direction constant whilst changing Dxy,

Dy and Sy, respectively. Due to the number of variables involved, the graphs are all prepared 2 for one value of the shear factor in the x-direction, π φx = 0.1. In the same manner, graphs can easily be drawn for other values of φx. In order to keep 1–νxyνyx constant, the Poisson's ratios have been chosen in the same way as in section 9.3.

As can be seen in the graphs of Fig.9.15, the critical buckling load decreases with decreasing torsional stiffness Dxy; it also decreases if the flexural rigidity and/or the shear stiffness in the perpendicular direction Dy and Sy, respectively, decreases. This is analogous to the plate bending case in section 9.3. In fact, the critical buckling load depends on the same parameters as the deflection due to transverse loading, i.e., plate geometry and stiffness. To get a comprehensive picture of the buckling of orthotropic sandwich plates, the same set of graphs as above must be produced for every value of the shear factor φx, and in the general case both the faces and the core could be orthotropic, yielding an almost infinite number of graphs. However, eq.(9.51) may look complex but it is quite easily solved once the geometry and properties of a plate are known.

9.36 SOLUTIONS TO PLATE PROBLEMS

6 D/D xy y 5 1.2 4 0.8 1.2 K 0.8 3 0.1 0.1 2

1 0 0.5 1 1.5 2 2.5 3 a/b

Figure 9.15a Buckling coefficients K for uniaxial compression K versus a/b for different Dxy. All curves 2 2 are for Dx = Dy, νxy = νyx = 0.25 and isotropic core. Solid lines are for π φ x = π φ y = 0.1 and dashed lines for π2φ = 0.

10 D/D 8 xy

6 0.5 K 4 1.0

2 2.0

0 0 0.5 1 1.5 2 2.5 3 a/b

Figure 9.15b Buckling coefficients K for uniaxial compression K versus a/b for different Dx. All curves 2 are for Dxx = 0.8Dx, νxy = 0.25 Dx/Dy and isotropic core. Solid lines are for π θy = 0.1 and dashed lines 2 for π θy = 0. 4.5 S/S xy 4

3.5 2.5 1.43 K 1.0 3 0.7 0.4 2.5

2 0 0.5 1 1.5 2 2.5 3 a/b

Figure 9.15c Buckling coefficients K for uniaxial compression versus a/b for different Sx./Sy. All curves 2 are for isotropic faces with Dxx = 0.8Dx = 0.8Dy νxy = νyx = 0.25. Solid lines are for π φ y = 0.1 and 2 dashed lines for π φ y = 0.

9.37 AN INTRODUCTION TO SANDWICH STRUCTURES

9.8 Approximate Buckling Formulae for Orthotropic Sandwich Plates with Various Edge Conditions. The methods of solution to be used here have already been partly mentioned in section 9.4. Ericksen and March [15] and March [13] used energy relations to derive buckling loads for sandwich plates having various edge conditions. These relations are valid for orthotropic sandwich plates with unequal thick faces and are hence of great generality. Using assumed sinusoidal deflection shapes and computing the total energy of the plate, including the energy stored due to local bending of the faces, results in a set of formulae given in its most general form in [15]. These formulae are used in [16] and [17] where a very large set of buckling coefficient curves are given for various combinations of orthotropy. Before reporting on the formulae from the above references, which results in a quite complex set of equations, first study the cases of orthotropic sandwich plates with thin faces. To do this assume the concept of partial deflections to be applicable, that is [8] 111 =+, with ww =+bs w (9.53) PPPbs by the same reasoning as in the beam case. The buckling load due to bending Pb can usually be found in a handbook, whereas Ps is calculated using eq.(8.16a) together with, e.g., the deflection assumption eq.(9.43). By instead using energy principles, as in the bending case previously and as done in refs.[13] and [15], together with the partial deflections, one can derive closed form solutions for several cases of uniaxial buckling. This yields the same equations as reported in [13] since the derivation follows the same approximate approach.

9.8.1 Simply supported plate

As above we have that for uniaxial buckling (Px = P and Py = 0) that the minimum buckling load is given by n = 1, that is, only one wave length in the buckling mode in the y-direction. Thus, a feasible deflection assumption is mπx πy ww=+= w( w + w )sin sin (9.54) bs bs a b where m is the number of buckling waves in the x-direction. Inserting into the equation for the total energy in eq.(8.42c) yields

24⎡ 4 22⎤ wDmbbxπ Day ν xyDm x Dmxy U = ⎢ 33+ + + ⎥ 241⎣⎢ a ()()()− ννxy yx 41b − ννxy yx 21ab − ννxy yx 2ab ⎦⎥

22 2 2 22 ws ππ⎡Smbx Say ⎤ Pw()bs+ w m b ++⎢ ⎥ − 24⎣ a 4b ⎦ 8a

There are now two ways of determining the buckling load P; the most precise way is stating that the buckling load is found when the total potential energy U has a minimum, that is ∂U ∂U ==0 ∂wbs∂w

9.38 SOLUTIONS TO PLATE PROBLEMS

_ _ which gives two simultaneous equations in the unknowns wb and ws which when found automatically give the value of P. A second, and algebraically much simpler approach is to use eq.(9.53) assuming partial deflections, which can be shown to yield the same result. The buckling load in pure bending is then found by

⎡ ∂U ⎤ ⎢ ⎥ = 0 which gives that ∂wb ⎣ ⎦ ws =0

2 1 b ()1 − ννxy yx = 22 Pb 2 ⎡ ⎛ mb⎞ ⎛ a ⎞ ⎤ πννν⎢Dxy⎜ ⎟ + D ⎜ ⎟ ++−21[] yxxxyxyyxDD()⎥ ⎣ ⎝ a ⎠ ⎝ mb⎠ ⎦ and the buckling load in pure shear is found as

⎡ ∂U ⎤ 11 ⎢ ⎥ = 0 , which gives that = 2 ⎣∂ws ⎦ Ps ⎛ a ⎞ ws =0 SS+ ⎜ ⎟ xy⎝ mb⎠

Following eq.(9.53) we arrive after some algebra at

⎡ 22⎤ D 2 D ()1 − νν Sx ⎛ a ⎞ ⎡⎛ mb⎞ y ⎛ a ⎞ ⎛ xy xy yx ⎞⎤ 12+ ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ++⎜ν ⎟ ⎢ ⎝ ⎠ ⎥⎢⎝ ⎠ ⎝ ⎠ yx ⎥ ⎣⎢ Sy mb ⎦⎥⎣ a Dx mb ⎝ Dx ⎠⎦ K = (9.55a) ⎡ 2 ⎤ 22D 2 D ()1 − νν Sx ⎛ a ⎞ 2 ⎛ a⎞ ⎡⎛ mb⎞ y ⎛ a ⎞ ⎛ xy xy yx ⎞⎤ 12+ ⎜ ⎟ + πφ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ++⎜ ν ⎟ ⎢ ⎝ ⎠ ⎥ x ⎝ ⎠ ⎢⎝ ⎠ ⎝ ⎠ yx ⎥ ⎣⎢ Sy mb ⎦⎥ b ⎣ a Dx mb ⎝ Dx ⎠⎦ where K is as defined in eq.(9.52) and φx as in (9.50). For an isotropic plate, using φ = φy, this equation reduces to

22 2 ⎡ ⎛ a ⎞ ⎤⎡⎛ mb⎞ ⎛ a ⎞ ⎤ ⎢12+ ⎜ ⎟ ⎥⎢⎜ ⎟ + ⎜ ⎟ + ⎥ ⎣ ⎝ mb⎠ ⎦⎣⎝ a ⎠ ⎝ mb⎠ ⎦ K = (9.55b) 2 22 ⎡ ⎛ a ⎞ ⎤ 2 ⎡⎛ mb⎞ ⎛ a ⎞ ⎤ ⎢12+ ⎜ ⎟ ⎥ + πφ⎢⎜ ⎟ + ⎜ ⎟ + ⎥ ⎣ ⎝ mb⎠ ⎦ ⎣⎝ a ⎠ ⎝ mb⎠ ⎦ which can be rewritten as exactly the same equation as derived in eq.(9.45).

9.39 AN INTRODUCTION TO SANDWICH STRUCTURES

9.8.2 Loaded edges simply supported, the other edges clamped

a Px

Figure 9.16 Rectangular orthotropic sandwich plate with loaded edges simply supported and the others clamped. Thus, a feasible deflection assumption in this case that satisfies the boundary conditions is mπx πy ww=+= w( w + w )sin sin2 (9.56) bs bs a b Again one can discuss the feasibility of this function since it is only correct for the thick face case and does not imply that ∂ws/∂y is non-zero along the edges. The assumed shape should then be mπx πy mπx πy ww=+sin sin2 w sin sin bsa b a b

This again implies that the shear part of the calculation, i.e. Ps, becomes the same as in the simply supported case presented previously. Performing the same analysis as above but now integrating the strain energy and the potential energy of the applied load Px using the deflection in eq.(9.56) field one soon arrives at

2 1 31b ()− ννxy yx = 22 , Pb 2 ⎡ ⎛ mb⎞ ⎛ a ⎞ ⎤ πννν⎢3168Dxyyxxxyxyyx⎜ ⎟ + D ⎜ ⎟ ++−[]DD() 1⎥ ⎣ ⎝ a ⎠ ⎝ mb⎠ ⎦

11 and = 2 P 4Sy a s S + ⎜⎛ ⎟⎞ x 3 ⎝ mb⎠ or rewritten as

22 2 ⎡ 4Sy ⎛ a ⎞ ⎤⎡⎛ mb⎞ 16Dy ⎛ a ⎞ 8 ⎛ Dxy()1 − νν xy yx ⎞⎤ 1 + ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ++⎜ν ⎟ ⎢ ⎝ ⎠ ⎥⎢⎝ ⎠ ⎝ ⎠ yx ⎥ ⎣ 3Sx mb ⎦⎣ a 3Dx mb 3 ⎝ Dx ⎠⎦ K = (9.57a) 4S 2 2216D 2 D ()1 − νν ⎡ y ⎛ a ⎞ ⎤ 2 ⎛ a⎞ ⎡⎛ mb⎞ y ⎛ a ⎞ 8 ⎛ xy xy yx ⎞⎤ 1 + ⎜ ⎟ + πφ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ++⎜ ν ⎟ ⎢ ⎝ ⎠ ⎥ x ⎝ ⎠ ⎢⎝ ⎠ ⎝ ⎠ yx ⎥ ⎣ 3Sx mb ⎦ b ⎣ a 3Dx mb 3 ⎝ Dx ⎠⎦ which for an isotropic plate reduces to (with φ = φy)

9.40 SOLUTIONS TO PLATE PROBLEMS

22 2 ⎡ 4 ⎛ a ⎞ ⎤⎡⎛ mb⎞ 16 ⎛ a ⎞ 8⎤ ⎢1 + ⎜ ⎟ ⎥⎢⎜ ⎟ + ⎜ ⎟ + ⎥ ⎣ 3 ⎝ mb⎠ ⎦⎣⎝ a ⎠ 3 ⎝ mb⎠ 3⎦ K = (9.57b) 2 22 ⎡ 4 ⎛ a ⎞ ⎤ 2 ⎡⎛ mb⎞ 16 ⎛ a ⎞ 8⎤ ⎢1 + ⎜ ⎟ ⎥ + πφ⎢⎜ ⎟ + ⎜ ⎟ + ⎥ ⎣ 3 ⎝ mb⎠ ⎦ ⎣⎝ a ⎠ 3 ⎝ mb⎠ 3⎦

Eq.(9.57b) has the graphical shape shown in Fig.9.17. 20

15 K

10 π 2φ 0 0.05 5 0.1 0.2 0.4 0 0 0.5 1 1.5 2 2.5 3 a/b

Figure 9.17 Buckling coefficients K for an isotropic sandwich plate with thin faces subjected to uniaxial compression, having loaded edges simply supported and the others clamped. The lines are for π2φ = 0, 0.05, 0.1, 0.2 and 0.4.

9.8.3 Loaded edges simply supported, the other edges clamped

a Px

Figure 9.18 Rectangular orthotropic sandwich plate with loaded edges clamped, the others simply supported. A feasible deflection assumption in this case that satisfies the boundary conditions is πx mπx πy ww=+= w( w + w )sin sin sin (9.58) bs bs a a b where once again m is the number of waves in the buckling mode in the x-direction. This deflection assumption assumes that the deflection is zero but the slope of the deflection is non-zero along the nodal lines [13], e.g., for m = 2 we have that w = 0 along the line a/2 while the slope does not vanish along this line. Let us perform the calculation for this example thoroughly. First, we have the derivatives

9.41 AN INTRODUCTION TO SANDWICH STRUCTURES

∂w π πx mxπ πx mxπ πy =+()cossinsincossinww⎜⎛ ⎟⎞⎜⎛ + m ⎟⎞ ∂x bs⎝ a ⎠⎝ a a a a ⎠ b

∂w π πx mxπ πy =+()sinsincosww⎜⎛ ⎟⎞ ∂y bs⎝ b ⎠ a a b

∂ 2 w πππ2 x mx πππx mx y =+()ww⎜⎛ ⎟⎞ ⎜⎛21m coscos()sinsinsin−+m 2 ⎟⎞ ∂x 2 bs⎝ a⎠ ⎝ a a a a ⎠ b

∂ 2 w ππ2 x mx ππy =−()sinsinsinww + ⎜⎛ ⎟⎞ ∂y 2 bs⎝ b ⎠ a a b

∂ 22w ⎛ πππ⎞⎛ x mx πππx mx⎞ y =+(ww )⎜ ⎟⎜ cos sin+ m sin cos⎟ cos ∂∂xy bs⎝ ab ⎠⎝ a a a a ⎠ b

In this case the integration of the strain energy in eq.(8.42c) is no longer trivial, but for m = 1 it is easily given since all the energy components can be derived from the bending case with the same boundary conditions.

For m = 1: By integration of the partial derivatives we soon get that

2 a b Dw⎛ ∂ 2 ⎞ Dw24π b x bxbdxdy ==.... , ∫ ∫ ⎜ 2 ⎟ 3 0 0 11− ννxy yx ⎝ ∂x ⎠ ()− ννxy yx a

2 a b D ⎛ ∂ 2 w ⎞ 3Dw24π a y b dxdy ==.... yb , ∫ ∫ ⎜ 2 ⎟ 3 0 0 1 − ννxy yx ⎝ ∂y ⎠ 16() 1 − ννxy yx b

a b 2ν D ⎛ ∂ 2 w ⎞⎛ ∂ 2 w ⎞ νπDw24 yx x bbdxdy ==.... yx x b , ∫ ∫ ⎜ 2 ⎟⎜ 2 ⎟ 0 0 121− ννxy yx ⎝ ∂x ⎠⎝ ∂y ⎠ ()− ννxy yx ab

2 a b ⎛ ∂ 2 w ⎞ Dw24π 2D b dxdy ==....xy b , ∫ ∫ xy ⎜ ⎟ 0 0 ⎝ ∂∂xy⎠ 2ab

2 a b ⎛ ∂w ⎞ Sw22π b S ⎜ sxs⎟ dxdy ==.... , and finally ∫ ∫ x ⎝ ⎠ 0 0 ∂x 4a

2 a b ⎛ ∂w ⎞ 3Sw22π a S s dxdy ==....ys ∫ ∫ y ⎜ ⎟ 0 0 ⎝ ∂y ⎠ 16b

The potential of the applied load Px is then according to eq.(8.43)

9.42 SOLUTIONS TO PLATE PROBLEMS

a b 2 1 ⎛ ∂w⎞ Pw()+ w22π b − P⎜ ⎟ dxdy ==−.... bs ∫ ∫ ⎝ ⎠ 280 0 ∂x a

The total energy is then

24 wDbπ ⎡ 3Dayν yxD xD xy()1 − ν xyν yx ⎤ U b x = ⎢ 33++ + ⎥ 21()− ννxy yx ⎣ a 16b 2ab 2ab ⎦

22 2 2 ws ππ⎡Sbx 3Say ⎤ Pw()bs+ w b ++⎢ ⎥ − 24⎣ a 16b ⎦ 8a

The partial buckling load for the pure bending case is found by

⎡ ∂U ⎤ 1 41b 2 ()− νν = 0 → xy yx ⎢ ⎥ = 22 ∂wb Pb ⎡ b a ⎤ ⎣ ⎦ ws =0 2 ⎛ ⎞ ⎛ ⎞ πννν⎢16Dxy⎜ ⎟ + 3D ⎜ ⎟ ++− 8[] yxxxyxyyxDD() 1 ⎥ ⎣ ⎝ a⎠ ⎝ b⎠ ⎦

Similarly for the pure shear case we get

⎡ ∂U ⎤ 11 = 0 → = ⎢ ⎥ 3S 2 ⎣∂ws ⎦ Ps y ⎛ a⎞ wb =0 S + ⎜ ⎟ x 4 ⎝ b⎠

Combining eq.(9.52) and eq.(9.53) one can write the buckling coefficient

⎡ 3S a 2 ⎤⎡16 b 2 D a 2 8 ⎛ D (1−ν ν ) ⎞⎤ y ⎛ ⎞ ⎛ ⎞ y ⎛ ⎞ ⎜ xy xy yx ⎟ ⎢1+ ⎜ ⎟ ⎥⎢ ⎜ ⎟ + ⎜ ⎟ + ⎜ν yx + ⎟⎥ ⎢ 4S x ⎝ b ⎠ ⎥⎢ 3 ⎝ a ⎠ Dx ⎝ b ⎠ 3 ⎝ Dx ⎠⎥ K = ⎣ ⎦⎣ ⎦ (9.59a) ⎡ S a 2 ⎤ a 2 ⎡16 b 2 D a 2 8 ⎛ D (1−ν ν ) ⎞⎤ y ⎛ ⎞ 2 ⎛ ⎞ ⎛ ⎞ y ⎛ ⎞ ⎜ xy xy yx ⎟ ⎢4 + 3 ⎜ ⎟ ⎥ + π φx ⎜ ⎟ ⎢ ⎜ ⎟ + ⎜ ⎟ + ⎜ν yx + ⎟⎥ ⎣⎢ S x ⎝ b ⎠ ⎦⎥ ⎝ b ⎠ ⎣⎢ 3 ⎝ a ⎠ Dx ⎝ b ⎠ 3 ⎝ Dx ⎠⎦⎥ which for an isotropic plate reduces to (with φ = φy)

2 2 2 ⎡ 3 ⎛ a ⎞ ⎤⎡16 ⎛ b ⎞ ⎛ a ⎞ 8⎤ ⎢1+ ⎜ ⎟ ⎥⎢ ⎜ ⎟ + ⎜ ⎟ + ⎥ ⎢ 4 ⎝ b ⎠ ⎥⎢ 3 ⎝ a ⎠ ⎝ b ⎠ 3⎥ K = ⎣ ⎦⎣ ⎦ (9.59b) 2 2 2 ⎡ ⎛ a ⎞ ⎤ ⎡16 ⎛ b ⎞ ⎛ a ⎞ 8⎤ ⎢4 + 3⎜ ⎟ ⎥ + π 2φ⎢ ⎜ ⎟ + ⎜ ⎟ + ⎥ ⎣⎢ ⎝ b ⎠ ⎦⎥ ⎣⎢ 3 ⎝ a ⎠ ⎝ b ⎠ 3⎦⎥

For m > 1: Secondly we must consider the case where m might take another value than unity. The integral of eq.(8.42c) will now take a slightly different form. Hence,

2 a b Dw⎛ ∂ 2 ⎞ Dw24π b x bxbdxdy ==.... mm42++61 , ∫ ∫ ⎜ 2 ⎟ 3 () 0 0 181− ννxy yx ⎝ ∂x ⎠ ()− ννxy yx a

9.43 AN INTRODUCTION TO SANDWICH STRUCTURES

2 a b D ⎛ ∂ 2 w ⎞ Dw24π a y b dxdy ==.... yb , ∫ ∫ ⎜ 2 ⎟ 3 0 0 181− ννxy yx ⎝ ∂y ⎠ ()− ννxy yx b

a b 2ν D ⎛ ∂ 2 w ⎞⎛ ∂ 2 w ⎞ νπDw24 yx x bbdxdy ==.... yx x b m 2 +1 , ∫ ∫ ⎜ 2 ⎟⎜ 2 ⎟ () 0 0 14(1− ννxy yx ⎝ ∂x ⎠⎝ ∂y ⎠ − ννxy yx )ab

2 a b ⎛ ∂ 2 w ⎞ Dw24π 2D b dxdy ==....xy b m 2 +1 , ∫ ∫ xy ⎜ ⎟ () 0 0 ⎝ ∂∂xy⎠ 4ab

2 a b ⎛ ∂w ⎞ Sw22π b S ⎜ sxs⎟ dxdy ==.... m 2 +1 , and finally ∫ ∫ x ⎝ ⎠ () 0 0 ∂x 4a

2 a b ⎛ ∂w ⎞ Sw22π a S s dxdy ==....ys ∫ ∫ y ⎜ ⎟ 0 0 ⎝ ∂y ⎠ 8b

The potential of the applied load Px is then according to eq.(8.43)

a b 2 1 ⎛ ∂w⎞ Pw()+ w22π b − P⎜ ⎟ dxdy ==−.... bs m 2 + 1 ∫ ∫ ⎝ ⎠ () 280 0 ∂x a

The total energy is then

24 4 2 22 wDbmmb π ⎡ x ()++61Dayννν yxDm x()(+1 Dm xy11−+ xy yx )()⎤ U = ⎢ 33++ + ⎥ 21()− ννxy yx ⎣ 88a b 4ab 4ab ⎦

22 2 22 wSbmsxππ⎡ ()+ 1 Say ⎤ Pw (bs+ w ) b 2 + ⎢ + ⎥ − ()m +1 2 ⎣ 88a b ⎦ 16a

As before we now seek a stationary value of the total energy and assuming once again we can superimpose the cases of pure bending and pure shear we get

2 ⎡ ∂U ⎤ 1 bAmxyyx()1 − νν ⎢ ⎥ = 0 → = 22 , ∂wb Pb ⎡ b a ⎤ ⎣ ⎦ ws =0 2 ⎛ ⎞ ⎛ ⎞ πννν⎢DBxm⎜ ⎟ + D y⎜ ⎟ ++−21ADD m[] yxx xy() xyyx⎥ ⎣ ⎝ a⎠ ⎝ b⎠ ⎦ and

⎡ ∂U ⎤ 1 Am ⎢ ⎥ = 0 → = 2 ⎣∂ws ⎦ Ps ⎛ a⎞ wb =0 SA+ S⎜ ⎟ xm y⎝ b⎠

2 4 2 where Am = (1+m ), and Bm = (m +6m +1). Adding the partial deflections we get

9.44 SOLUTIONS TO PLATE PROBLEMS

22 2 1 ⎡ Sy ⎛ a⎞ ⎤⎡ ⎛ b⎞ Dy ⎛ a⎞ ⎛ Dxy()1 − νν xy yx ⎞⎤ A + ⎜ ⎟ B ⎜ ⎟ + ⎜ ⎟ ++2A ⎜ν ⎟ ⎢ m ⎝ ⎠ ⎥⎢ m ⎝ ⎠ ⎝ ⎠ myx ⎥ Am ⎣ Sx b ⎦⎣ a Dx b ⎝ Dx ⎠⎦ K = (9.60a) ⎡ 2 ⎤ 22D 2 D ()1 − νν Sx ⎛ a⎞ 2 ⎛ a⎞ ⎡ ⎛ b⎞ y ⎛ a⎞ ⎛ xy xy yx ⎞⎤ A + ⎜ ⎟ + πφ⎜ ⎟ B ⎜ ⎟ + ⎜ ⎟ ++2A ⎜ ν ⎟ ⎢ m ⎝ ⎠ ⎥ xm⎝ ⎠ ⎢ ⎝ ⎠ ⎝ ⎠ myx ⎥ ⎣⎢ Sy b ⎦⎥ b ⎣ a Dx b ⎝ Dx ⎠⎦ which for an isotropic plate reduces to (with θ = θy)

222 1 ⎡ ⎛ a⎞ ⎤⎡ ⎛ b⎞ ⎛ a⎞ ⎤ A + ⎜ ⎟ B ⎜ ⎟ + ⎜ ⎟ + 2A ⎢ mm⎝ ⎠ ⎥⎢ ⎝ ⎠ ⎝ ⎠ m⎥ Am ⎣ b ⎦⎣ a b ⎦ K = (9.60b) 2 22 ⎡ ⎛ a⎞ ⎤ 2 ⎡ ⎛ b⎞ ⎛ a⎞ ⎤ ⎢Amm+ ⎜ ⎟ ⎥ + πφ⎢B ⎜ ⎟ + ⎜ ⎟ + 2A m⎥ ⎣ ⎝ b⎠ ⎦ ⎣ ⎝ a⎠ ⎝ b⎠ ⎦

Eq.(9.60b) has the graphical shape shown in Fig.9.19. 20

15 K 10 π 2φ

5 0 0.05 0.1 0.2 0 0.4 0 0.5 1 1.5 2 2.5 3 a/b

Figure 9.19 Buckling coefficients K for an isotropic sandwich plate with thin faces subjected to uniaxial compression, having loaded edges simply supported and the others clamped. The lines are for π2φ = 0, 0.05, 0.1, 0.2 and 0.4.

9.8.4 All edges clamped

P a x

Figure 9.29 Rectangular orthotropic sandwich plate with all edges clamped. A feasible deflection assumption in this case that satisfies the boundary conditions is πx mπx πy ww=+= w( w + w )sin sin sin2 (9.61) bs bs a a b

9.45 AN INTRODUCTION TO SANDWICH STRUCTURES

By the same reasoning as given above, once again we must derive the equations first assuming m = 1 and then for values of m greater than unity.

For m = 1:

2 1 b ()1 − ννxy yx = 22 , Pb 2 ⎡ ⎛ b⎞ ⎛ a⎞ 8 ⎤ πννν⎢44Dxy⎜ ⎟ + D ⎜ ⎟ ++−[] yxxxyxyyxDD()1 ⎥ ⎣ ⎝ a⎠ ⎝ b⎠ 3 ⎦

11 and = P a 2 s SS+ ⎜⎛ ⎟⎞ xy⎝ b⎠

Note that the shear buckling load given here turns out to be the same as in the simply supported case which is not surprising since the shear deflection under the assumption of partial deflections is invariable with the boundary condition. Adding the above according to eq.(9.53) we get

22 2 ⎡ Sy ⎛ a⎞ ⎤⎡ ⎛ b⎞ 4Dy ⎛ a⎞ 8 ⎛ Dxy()1 − νν xy yx ⎞⎤ 14+ ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ++⎜ν ⎟ ⎢ ⎝ ⎠ ⎥⎢ ⎝ ⎠ ⎝ ⎠ yx ⎥ ⎣ Sx b ⎦⎣ a Dx b 3 ⎝ Dx ⎠⎦ K = (9.62a) ⎡ 2 ⎤ 224D 2 D ()1 − νν Sx ⎛ a⎞ 2 ⎛ a⎞ ⎡ ⎛ b⎞ y ⎛ a⎞ 8 ⎛ xy xy yx ⎞⎤ 14+ ⎜ ⎟ + πφ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ++⎜ ν ⎟ ⎢ ⎝ ⎠ ⎥ x ⎝ ⎠ ⎢ ⎝ ⎠ ⎝ ⎠ yx ⎥ ⎣⎢ Sy b ⎦⎥ b ⎣ a Dx b 3 ⎝ Dx ⎠⎦ which for an isotropic plate reduces to (with φ = φy)

22 2 ⎡ ⎛ a⎞ ⎤⎡ ⎛ b⎞ ⎛ a⎞ 8⎤ ⎢144+ ⎜ ⎟ ⎥⎢ ⎜ ⎟ + ⎜ ⎟ + ⎥ ⎣ ⎝ b⎠ ⎦⎣ ⎝ a⎠ ⎝ b⎠ 3⎦ K = (9.62b) 2 22 ⎡ ⎛ a⎞ ⎤ 2 ⎡ ⎛ b⎞ ⎛ a⎞ 8⎤ ⎢144+ ⎜ ⎟ ⎥ + πφ⎢ ⎜ ⎟ + ⎜ ⎟ + ⎥ ⎣ ⎝ b⎠ ⎦ ⎣ ⎝ a⎠ ⎝ b⎠ 3⎦

For m > 1:

2 1 31bAmxyyx()− νν = 22 , Pb 2 ⎡ ⎛ b⎞ ⎛ a⎞ ⎤ πννν⎢3168DBxm⎜ ⎟ + D y⎜ ⎟ ++−ADD m[] yxx xy() 1 xyyx⎥ ⎣ ⎝ a⎠ ⎝ b⎠ ⎦ 1 A and = m P 4S a 2 s SA + y ⎜⎛ ⎟⎞ xm 3 ⎝ b⎠

2 4 2 where Am = (1+m ), and Bm = (m +6m +1). This gives

9.46 SOLUTIONS TO PLATE PROBLEMS

4S 2216D 2 D ()1 − νν 1 ⎡ y ⎛ a⎞ ⎤⎡ ⎛ b⎞ y ⎛ a⎞ 8Am ⎛ xy xy yx ⎞⎤ A + ⎜ ⎟ B ⎜ ⎟ + ⎜ ⎟ ++⎜ν ⎟ ⎢ m ⎝ ⎠ ⎥⎢ m ⎝ ⎠ ⎝ ⎠ yx ⎥ Am ⎣ 3Sx b ⎦⎣ a 3Dx b 3 ⎝ Dx ⎠⎦ K = ⎡ 2 ⎤ 2216D 2 D ()1 − νν 4Sx ⎛ a⎞ 2 ⎛ a⎞ ⎡ ⎛ b⎞ y ⎛ a⎞ 8Am ⎛ xy xy yx ⎞⎤ A + ⎜ ⎟ + πφ⎜ ⎟ B ⎜ ⎟ + ⎜ ⎟ ++⎜ ν ⎟ ⎢ m ⎝ ⎠ ⎥ xm⎝ ⎠ ⎢ ⎝ ⎠ ⎝ ⎠ yx ⎥ ⎣⎢ 3Sy b ⎦⎥ b ⎣ a 3Dx b 3 ⎝ Dx ⎠⎦ (9.63a) which for an isotropic plate reduces to (with φ = φy)

22 2 14⎡ ⎛ a⎞ ⎤⎡ ⎛ b⎞ 16 ⎛ a⎞ 8 ⎤ A + ⎜ ⎟ B ⎜ ⎟ + ⎜ ⎟ + A ⎢ mm⎝ ⎠ ⎥⎢ ⎝ ⎠ ⎝ ⎠ m⎥ Am ⎣ 3 b ⎦⎣ a 3 b 3 ⎦ K = (9.63b) 2 22 ⎡ 4 ⎛ a⎞ ⎤ 2 ⎡ ⎛ b⎞ 16 ⎛ a⎞ 8 ⎤ ⎢Amm+ ⎜ ⎟ ⎥ + πφ⎢B ⎜ ⎟ + ⎜ ⎟ + A m⎥ ⎣ 3 ⎝ b⎠ ⎦ ⎣ ⎝ a⎠ 3 ⎝ b⎠ 3 ⎦

15 K 2 10 π φ 0 0.05 5 0.1 0.2 0.4 0 0 0.5 1 1.5 2 2.5 3 a/b

Figure 9.21 Buckling coefficients K for an isotropic sandwich plate with thin faces having all edges clamped and subjected to uniaxial compression. The lines are for π2φ = 0, 0.05, 0.1, 0.2 and 0.4. The chosen deflection function satisfies the boundary condition but not the governing differential equation which leads to the calculated buckling loads being approximate (except for the simply supported case). Thurston [18] solved the problem of a clamped isotropic sandwich plate exactly by using double trigonometric cosine series, in the same manner as done in section 9.3 in the bending case. By comparison of the clamped plates, the approximate results calculated by the above formulae yield buckling loads that are 5 to 10% too high. Though approximate, this solution has the advantage of being much more general than any other known solution procedure in that it handles sandwich plates with different boundary conditions, having an orthotropic core and orthotropic faces.

In [16] a solution is presented based on the same assumptions as the above but incorporating the effect of thick and dissimilar faces. Even if the derived formulae are complex in the number of parameters used it still constitutes a fairly simple closed form solution easily transformed into computer code. The derived equations have also been plotted for not less than 125 cases which would at least approximately cover almost every plausible practical case. Thus, the approximate buckling formulae derived here, which only assumed thin faces, can be found in [16] although written in slightly different parameters and some of the 125

9.47 AN INTRODUCTION TO SANDWICH STRUCTURES buckling curves given are also plotted assuming thin faces. However, since the buckling load assuming thin faces can be found fairly easily through the closed form solution no other buckling curves than for the special case of an isotropic plate have been given in this text.

9.9 Shear Buckling Kuenzi and Ericksen [20] solved the problem of rectangular isotropic sandwich plates, simply supported or clamped, subjected to shear loads as illustrated in Fig.9.22.

b Pxy a

Figure 9.22 Shear of a simply supported sandwich plate. The problem is solved for a number of cases using Ritz's method where different deflection assumptions are adopted for the various boundary conditions and values of the shear factor φ (φ = D/b2S). The analyses are rather lengthy and are therefore omitted here but Kuenzi and Ericksen [20] suggest some approximate design formulae which are very simple to use. The formulae suggested proved to be slightly conservative. Providing a is larger than b the 2 2 2 formula for simply supported edges reads (with K = Pxy(1-ν )b /π D) K b 2 K = 0 for 01≤≤+πφ2 (9.64a) ⎛ b 2 ⎞ a 2 11+−−πφ2 ⎜ K ⎟ ⎝ 0 a 2 ⎠ and for all edges clamped

K 4 ⎛ b 2 ⎞ K = 0 for 0 ≤≤+πφ2 ⎜1 ⎟ (9.64b) ⎛ 4 ⎛ b 2 ⎞⎞ 3 ⎝ a 2 ⎠ 1 +−+πφ2 ⎜ K ⎜1 ⎟⎟ ⎝ 0 3 ⎝ a 2 ⎠⎠

where K0 is the value of the buckling coefficient for an ordinary plate with φ = 0. These can be written approximately [8] 16 b2 Simply supported: K =+4 (9.65a) 0 3 a2

17 b2 Clamped edges: K =+9 (9.65b) 0 3 a2

For φ larger than the values indicated in eqs.(9.64) then K takes the value 1/π2φ, the same limiting value as in the previous buckling analysis. A closed form solution to buckling of infinitely long plates is given by Plantema [8].

9.48 SOLUTIONS TO PLATE PROBLEMS

9.10 Combined Buckling and Transverse Load The governing equations of section 9 are derived so that in the general case transverse as well as in-plane loading may be considered simultaneously. Doing so may, however, be fairly difficult and only a few solutions of combined loading situations are known, e.g., [9]. However, in the case of an isotropic sandwich plate subjected to transverse loading q and in- plane compressive loads Px and Py the governing equations can be used in the following manner in order to clarify the effects, rather than solving a specific plate problem.

Assume for simplicity reasons the following loading and deflected shape mπx nπy mπx nπy ww= sin sin and qq= sin sin mn a b mn a b

Inserting into the governing equation of eq.(8.27) and writing Nxy = 0, Nx = –Px, and Ny = –Py

= –pyP yields

222 22 22 ⎪⎧ Dm⎡⎛ ππ⎞ ⎛ n ⎞ ⎤ ⎡ ⎛ m π⎞ ⎛ n π⎞ ⎤⎡ D ⎧⎛ mππ⎞ ⎛ n ⎞ ⎫⎤⎪⎫ ⎨ 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ − ⎢Pxy⎜ ⎟ + P ⎜ ⎟ ⎥⎢1 + 2 ⎨⎜ ⎟ + ⎜ ⎟ ⎬⎥⎬w mn 1 ⎝ a ⎠ ⎝ b ⎠ ⎝ a ⎠ ⎝ b ⎠ S()1 − ⎝ a ⎠ ⎝ b ⎠ ⎩⎪ − ν ⎣ ⎦ ⎣ ⎦⎣ ν ⎩ ⎭⎦⎭⎪

qmn ⎡ D ⎧ mππ22n ⎫⎤ =−1 + ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ (9.64) ⎢ 2 ⎨⎝ ⎠ ⎝ ⎠ ⎬⎥ ⎣ S()1 − ν ⎩ a b ⎭⎦

The deflection becomes infinite if the coefficient for wmn is zero, i.e.,

2 Dm⎡ ππ22n ⎤ ⎡ m π22n π⎤⎡ D ⎧ mππ22n ⎫⎤ ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ − P ⎜⎛ ⎟⎞ + p ⎜⎛ ⎟⎞ 1 + ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ = 0 2 ⎢⎝ ⎠ ⎝ ⎠ ⎥ mn⎢⎝ ⎠ y ⎝ ⎠ ⎥⎢ 2 ⎨⎝ ⎠ ⎝ ⎠ ⎬⎥ 1 − ν ⎣ a b ⎦ ⎣ a b ⎦⎣ S()1 − ν ⎩ a b ⎭⎦

Now, this equation is exactly the same as eq.(9.43) and thus constitutes the stability criterion.

The value of P satisfying the stability criterion is denoted Pmn (the buckling load in the (m,n):th mode) and hence equals

222 Dm⎡⎛ ππ⎞ ⎛ n ⎞ ⎤ 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ 1− ν ⎣⎝ a ⎠ ⎝ b ⎠ ⎦ Pmn = ⎡ mππ22n ⎤⎡ D ⎧ mππ22n ⎫⎤ ⎜⎛ ⎟⎞ + p ⎜⎛ ⎟⎞ 1 + ⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ ⎢⎝ ⎠ y ⎝ ⎠ ⎥⎢ 2 ⎨⎝ ⎠ ⎝ ⎠ ⎬⎥ ⎣ a b ⎦⎣ S()1 − ν ⎩ a b ⎭⎦

Then the governing equation (9.56) can be written as

222 22 Dm⎡⎛ ππ⎞ ⎛ n ⎞ ⎤ ⎛ P ⎞ ⎡ D ⎧⎛ mππ⎞ ⎛ n ⎞ ⎫⎤ ⎜ ⎟ + ⎜ ⎟ ⎜11− ⎟w =+ ⎜ ⎟ + ⎜ ⎟ q 2 ⎢⎝ ⎠ ⎝ ⎠ ⎥ mn⎢ 2 ⎨⎝ ⎠ ⎝ ⎠ ⎬⎥ mn 1 − ν ⎣ a b ⎦ ⎝ Pmn ⎠ ⎣ S()1 − ν ⎩ a b ⎭⎦

When P = 0 and if the deflection wmn(P = 0) is denoted w0 we have that

9.49 AN INTRODUCTION TO SANDWICH STRUCTURES

22 D ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ 1 + 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ q ()1 − ν 2 S()1 − ν ⎝ a ⎠ ⎝ b ⎠ w = mn ⎣ ⎦ 0 222 D ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ⎣⎝ a ⎠ ⎝ b ⎠ ⎦ just as in eq.(9.11). Substituting this into the above we get

w0 wmn = (9.65) ⎛ P ⎞ ⎜1 − ⎟ ⎝ Pmn ⎠

Hence, the effect of an in-plane compressive load P can be added into the equations for the deflection of rectangular plates simply by substituting wmn in eqs.(9.12-19) by the expression given above. The complexity introduced is that the value of the buckling load in the (m,n):th mode must be calculated for every step in summation of these equations. A practical way of overcoming this is, if the panel aspect ratio is close to unity and P/Pmn is small [11], to use only the first term in the series of eqs.(9.12-19) in which case only P11 must be known. Using only the first term yields fairly accurate results in terms of deflection and bending moments, whereas the transverse forces are not accurate. Still, the procedure can be seen as a good approximation. March [19] solved the differential equation eq.(8.27) for a simply supported isotropic sandwich plate assuming non-zero transverse load q as well as non-zero compressive in-plane load Px and Py. 9.11 Free Vibration of a Simply Supported Sandwich Plate In this example the natural frequency of undamped free vibration of a simply supported sandwich plate will be derived. The governing equation of motion is given in eq.(8.32a). Assuming free harmonic excitations the deflection can be written as

w(x, y,t) = Φ(x, y)eiωt = Ψ(t) Φ(x, y) (9.66) where Φ(x,y) is the static spatial displacement function and ω the natural frequency. As shown previously, ∞ ∞ mxπ nyπ ΦΦ(xy , )= ∑∑ mn sin sin n=1 m=1 a b can be used for this purpose. In fact, only using the first term in the series expansion yields the same results since Φ itself (the shape and amplitude of the distorted plate) vanishes from the expression. Inserting into eq.(8.32a) yields

2 ∞ ∞ ⎧ 22 * 2 22 ⎪ Dm⎡⎛ ππ⎞ ⎛ n ⎞ ⎤ * 2 Dρω ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ ⎜ ⎟ + ⎜ ⎟ −−ρω ⎜ ⎟ + ⎜ ⎟ ∑∑ ⎨ 2 ⎢⎝ ⎠ ⎝ ⎠ ⎥ 2 ⎢⎝ ⎠ ⎝ ⎠ ⎥ n=1 m=1 ⎪11− ν ⎣ a b ⎦ S()− ν ⎣ a b ⎦ ⎩ * 4 22 Rρω 2 ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤⎫ +−Rω ⎜ ⎟ − ⎜ ⎟ ΦΨ = 0 ⎢⎝ ⎠ ⎝ ⎠ ⎥⎬ S ⎣ a b ⎦⎭⎪

9.50 SOLUTIONS TO PLATE PROBLEMS

A non-trivial solution implies that Φ and Ψ are non-zero and as in the buckling case we get that the term within the bracket must be zero for every combination of m and n. Introducing the quantity α = ρ*ω2(1–ν2)/D we can rewrite the above as

222 22 ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ ⎛ RDα α ⎞ ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ RD α2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ −−α ⎜ ** + 2 ⎟ ⎢⎜ ⎟ + ⎜ ⎟ ⎥ + 2 = 0 ⎣⎝ a ⎠ ⎝ b ⎠ ⎦ ⎝ ρ S()11− ν ⎠ ⎣⎝ a ⎠ ⎝ b ⎠ ⎦ ρνS ()−

For a homogeneous cross-section of isotropic material, the ratio D/S = 1.2(1+ν)h2/6 is in the order of 3 times as large as R/ρ* = h2/12 and hence the effect of shear deformation is more important than the rotary inertia. In a sandwich plate, the ratio D/S is magnitudes larger than for a homogeneous cross-section and thus much more important. By omitting the effect of rotary inertia it is possible to simplify the above expression to

222 22 ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ Dα ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ ⎢⎜ ⎟ + ⎜ ⎟ ⎥ −−α 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ = 0 ⎣⎝ a ⎠ ⎝ b ⎠ ⎦ S()1 − ν ⎣⎝ a ⎠ ⎝ b ⎠ ⎦ or rewritten

D 2 * 42 2 ⎡⎛ mb⎞ 2 ⎤ ρνb ()1 − D ωπ= ⎜ ⎟ + n , with φ = (9.67) mn ⎢⎝ ⎠ ⎥ 2 22 ⎣ a ⎦ 2 ⎡⎛ mb⎞ 2 ⎤ Sb()1−ν 1 + πφ⎢⎜ ⎟ + n ⎥ ⎣⎝ a ⎠ ⎦

This can be seen as the ordinary plate theory expression divided by a correction term (the denominator) that depends on the amount of shear deformation. Once again, the portion of shear deformation is seen to increase as the size of the plate decreases. It is important to notice that the correction also depends on the mode of vibration n, giving a small correction in the first eigen-frequency but larger and larger corrections for higher modes. This is, of course, the same as decreasing the plate size.

If the shear deformation (D/S = 0) and the rotary inertia (R = 0) are omitted, this expression equals that of ordinary plate theory

22 2 ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ Dmb2 ⎡⎛ ⎞ 2 ⎤ D ω mn = ⎢⎜ ⎟ + ⎜ ⎟ ⎥ **2 = π ⎢⎜ ⎟ + n ⎥ 42 (9.68) ⎣⎝ a ⎠ ⎝ b ⎠ ⎦ ρν()11− ⎣⎝ a ⎠ ⎦ ρνb ()−

To exemplify this, use some real numbers. Consider a plate with the following data in SI-units

9 2 6 2 Ef = 20·10 N/m Gc = 40·10 N/m ν = 0.3

tf = 0.002 m tc = 0.05 m a = b = 2 m 3 3 ρf = 1400 kg/m ρc = 100 kg/m

This gives

D = 54.080 Nm2/m width, and S = 2.163.000 N/m width → D/S(1–ν2) = 0.0250

ρ* = 10.6 kg/m2

9.51 AN INTRODUCTION TO SANDWICH STRUCTURES

ρρttd32ρ t 3 R =+ff ff +cc= 1.86 10-6 + 0.0038 + 0.0010 = 0.0048 kg 6212

→ R/ρ* = 0.00045 This shows that the effect of shear deformation is about 55 times larger than the effect of rotary inertia. In calculating the rotary inertia it is seen that the term representing the core cannot be neglected. Hence, neglecting the rotary inertia term seems justified in this case. Now, calculating the natural frequency for the first five modes gives the following results when firstly neglecting the effect of transverse shear and secondly including transverse shear, but in either case omitting the rotary inertia. The results is given in Table 9.5 (note that ωmn =

ωnm ).

As seen in this example, the effect of transverse shear has little effect in the lowest eigen- frequency mode, but the effect increases drastically as the higher modes are considered. This is once again length dependent, just as in the bending or buckling case. If the length in the above example is decreased to 1 m the first eigen-frequency neglecting shear deformation -1 -1 equals 1478 s and if the shear deformation is included ω11 = 1190 s , which is a much larger difference. The plate considered above is quite slender, that is, the amount of shear deformation is low, the ratio wb/ws = 8 (see Table 9.1).

m n ωmn {no shear} ωmn {including shear} s-1 s-1 1 1 370 347 1 2 924 798 2 2 1478 1190 1 3 1848 1426 2 3 2402 1751 1 4 3140 2140

Table 9.5 Natural frequencies for a simply supported sandwich plate including and excluding transverse shear deformations. If a load is to be added, e.g., a uniformly distributed load, this is most easily done by including it in the plate weight ρ*. Add for example a uniformly distributed load of 25 kPa (25,000 N/m2 ≈ 2500 kg/m2). This creates a total static deformation of plate equals to * approximately 30 mm (wb = 27.3 mm and ws = 3.4 mm). Now, replace ρ by 2500 and we get -1 -1 for the first mode: ω11 {no shear} = 24.1 s , and ω11 {including shear} = 22.6 s .

In the case of thick faces the same approach and assumptions can be used but now using the governing equation in eq.(8.31). This results in an expression for the natural frequencies as

2DDf 0 D 2 **42+ 42 2 ⎡⎛ mb⎞ 2 ⎤ Sbρνρν()11− b ()− ωπ= ⎜ ⎟ + n (9.69) mn ⎢⎝ ⎠ ⎥ 2 ⎣ a ⎦ 2 ⎡⎛ mb⎞ 2 ⎤ 1 + πφ⎢⎜ ⎟ + n ⎥ ⎣⎝ a ⎠ ⎦

9.52 SOLUTIONS TO PLATE PROBLEMS

It is seen that for thin faces, Df = 0, the expression equals that in eq.(9.67) and that for a high shear stiffness, S → ∞, it equals that in eq.(9.68). The main difference is, however, that for a very low shear stiffness, the natural frequencies will equal that of the two faces vibrating independently of each other, i.e.

2 2 ⎡⎛ mb⎞ 2 ⎤ 2D f π ⎢⎜ ⎟ + n ⎥ ⎝ a ⎠ ω = ⎣ ⎦ (9.70) mn ρν*b 22()1 −

9.12 Conclusions As has been seen in all the examples given in this chapter, the behaviour of sandwich plates 2 depends strongly on the shear factor φ (=D/b S) and in some cases also upon the ratio Df /D. It is possible to state some simple rules for the use of different theories.

(i) φ < 0.01: Shear deflections are negligible. Deflections and stresses may be predicted using ordinary beam theory (S = ∞)

(ii) 0.01 < φ < 0.1: Shear and bending deflection in the same order of magnitude. Deflections and stresses may be predicted using principles for partial deflections.

(iii) 0.1 < φ < 10 (depending on Df/D): Shear deflections dominant. Deflections and stresses must be calculated using sandwich theory including effect of thick faces.

(iv) φ > 10 (depending on Df /D): Core to weak to connect the faces which act as two plates (or beams) acting independently of each other.

References [1] Reissner E., “Finite Deflections of Sandwich Plates”, Journal of the Aeronautical Sciences, Vol. 15, No. 7, July, 1948, pp 435-440.

[2] Meyer-Piening H.-R., “Remarks on Higher Order Sandwich Stress and Deflection Analyses”, Sandwich Construction 1 (Eds.: K.-A. Olsson and R.P. Reichard), First International Conference on Sandwich Construction, Royal Institute of Technology, Stockholm, Sweden, 19-21 June, 1989, pp 107-127.

[3] Bau-Madsen N.K., Svendsen K.-H. and Kildegaard A., “Large Deflections of Sandwich Plates - An Experimental Study”, Composite Structures, Vol. 23, 1993, pp 47-52.

[4] Timoshenko S.P. and Woinowsky-Kreiger S., Theory of Plates and Shells, Second edition, McGraw-Hill, London, 1970.

[5] Roark R.J. and Young W.C., Formulas for Stress and Strain, 5th edition, McGraw- Hill, Singapore, 1976.

9.53 AN INTRODUCTION TO SANDWICH STRUCTURES

[6] Hoff N.J., “Bending and Buckling of Rectangular Sandwich Plates”, NACA TN 2225, 1950.

[7] Yen K.T., Gunturkun S. and Pohle F.V., “Deflections of a Simply Supported Rectangular Sandwich Plate Subjected to Transverse Loads”, NACA TN 2581, 1951.

[8] Plantema F.J., Sandwich Construction, John Wiley & Sons, New York, 1966.

[9] Robinsson J.R., “The Buckling and Bending of Orthotropic Sandwich Plates with all Edges Simply Supported”, Aero. Quart., Vol. 6, No 2, 1955, pp 125-148.

[10] Jones R.M., Mechanics of Composite Materials, Scripta Book Company, Washington D.C., 1975.

[11] Allen H.G., Analysis and Design of Structural Sandwich Panels, Pergamon Press, Oxford, 1969.

[12] Whitney J.M., “Structural Analysis of Laminated Anisotropic Plates”, Technomic Publishing Company, Lancaster PA, USA, 1987.

[13] March H.W., “Effects of Shear Deformation in the Core of a Flat Rectangular Sandwich Panel – 1. Buckling under Compressive End Load, 2. Deflection under Uniform Transverse Load”, U.S. Forest Products Laboratory Report 1583, 1948.

[14] Lockwood-Taylor J., “Strength of Sandwich Panels”, Proceedings to the VII International Congress of Applied Mechanics, Vol. 1, 1948, pp 187-199.

[15] Ericksen W.S. and March H.W., “Compressive Buckling of Sandwich Panels Having Dissimilar Facings of Unequal Thickness”, US Forest Product Laboratory Report 1583-B, Nov 1950, revised Nov. 1958.

[16] Norris C.B., “Compressive Buckling Curves for Flat Sandwich Panels with Isotropic Facings and Isotropic or Orthotropic Cores”, U.S. Forest Product Laboratory Report 1854, revised Jan. 1958.

[17] Kuenzi E.W., Norris C.B. and Jenkinson P.M., “Buckling Coefficients for Simply Supported and Clamped Flat, Rectangular Sandwich Panels under Edgewise Compresson”, U.S. Forest Service Research Note FPL-070, Dec. 1964.

[18] Thurston G.A., “Bending and Buckling of Clamped Sandwich Plates”, Journal of the Aeronautical Sciences, Vol. 24, No 6, 1957, pp 407-412.

[19] March H.W., “Behavior of a Rectangular Sandwich Panel under a Uniform Lateral Load and Compressive End Loads”, U.S. Forest Products Laboratory Report 1834, 1952.

9.54 SOLUTIONS TO PLATE PROBLEMS

[20] Kuenzi E.W. and Ericksen W.S., “Shear Stability of Flat Panels of Sandwich Construction”, U.S. Forest Products Laboratory Report 1560, Revised 1962.

Exercises and design examples Example 9.1 The fundamental equation for free, undamped vibration of an isotropic sandwich plate reads ρ* D ∂2 ∂2w DwΔΔ2 −+=w ρ* 0 S ∂t 2 ∂t 2 where ρ* the surface density (kg/m2) and Δ the Laplace-operator. The solution can be written separating the spatial and time parts as w(x,y,t) = Φ(x,y) Ψ(t). The plate is rectangular with side lengths a = 800 mm and b = 1200 mm, and has a symmetric cross-section with properties

3 tf = 1 mm Ef = 200 GPa Gf = 77 GPa νf = 0.3 ρf = 7800 kg/m 3 tc = 25 mm Ec = 195 MPa Gc = 75 MPa νc = 0.3 ρc = 200 kg/m

(a) Compute the three lowest eigen-frequencies. (b) Describe how the transverse shear deformation affects eigen-frequencies and how that changes for different vibration modes.

Ans. See eq.(9.69) D = 67,000 Nm2/m, S = 2,028,000 N/m, and ρ* = 20.6 kg/m2 gives the following results -1 -1 -1 -1 ω11 = 975s , ω12 = 1585s , ω21 = 2176s and ω22 = 2571s

Example 9.2

A rectangular isotropic sandwich plate according to the figure is subjected to a uniaxial compressive load Px per unit length. K x

a Px a/b y 012 3

⎛ ()1 − ν 22b ⎞ (a) Derive an expression for the buckling coefficient K. ⎜ K = P ⎟ ⎝ π 2 D x ⎠ (b) The curve K - a/b for infinite shear stiffness is presented in the figure above. Sketch how the curve looks for different values of the shear factor θ = D/(1-ν2)b2S, and point out the important details in them.

Ans. See eq.(9.38)

9.55 AN INTRODUCTION TO SANDWICH STRUCTURES

Example 9.3 A panel in the bottom of a motor yacht with dimensions a × b (a = 1 m, b = 0.6 m) is to be designed according to Det Norske Veritas rules for "High Speed Craft". According to these rules, the surface inscribed between longitudinal and transverse stiffeners is considered as a simply supported panel (which of course may be a matter of discussion but such is the rule). The rules also prescribe safety factors and criteria as

σf < 0.3 σf,cr in the faces τc < 0.35 τc,cr in the core wmax < 0.01 b in the middle of the panel The panel is subjected to slamming, a pressure from the water appearing as the boat "falls" into a wave. The design slamming-pressure is 100 kPa uniformly distributed. The faces of the panel are made of a E- glass/polyester laminate which in the bottom of the boat must be 3 mm thick to support impact. Choose (by calculations) a suitable core material from the table below and find its thickness.

Faces: Ef = 12 GPa σf,cr = 150 MPa νf = 0.3 Core: Core/density Ec (MPa) Gc (MPa) τcr (MPa) H60 55 22 0.6 H100 95 38 1.2 H130 125 47 1.6 H200 195 75 3.0 Hint: Use Table 9.1 for deformations, forces and bending moments.

Ans. Assume thin faces and weak core. T 26. 1 N / mm (i) τ ≈=max max dd Mmax 1034. 4 N / mm (ii) σmax ≈= tdf d qb42()1− ν qb 25438.. 2 3751 2 (iii) wmax = 0. 00830 +=+=0. 1042 2 6 mm D Sd Gdc Gives (design numbers in bold face) H60 H100 H130 H200 (i) d (mm) 124.3 62.1 46.6 24.9 weight (kg) 4.5 3.7 3.6 3.0 (ii) d (mm) 23 23 23 23 weight (kg) 0.83 1.4 1.8 2.8 (iii) d (mm) 47 39 37 34 weight (kg) 1.7 2.3 2.9 4.1 i.e., choose a 43.6 mm thick H130 core.

9.56 SOLUTIONS TO PLATE PROBLEMS

Example 9.4 On top of a building for the drying of wood products there a heat fans and dehumidifiers. This rather expensive equipment must be protected against the environment by a sandwich shell construction according to the figure below. side view "from above" roof

4 7.5 2 wall 3 10 On the horizontal roof the equipment is placed and is not part the shell structure. The incline roof and the walls shall be in a sandwich design with a PUR core with a thickness of 170 mm due to the required thermal insulation. The core has a shear modulus of Gc = 4 MPa, and a shear strength of τc,cr = 0.3 MPa. The faces are of a hand lay-up glass(CSM)/vinylester laminate which can be approximated as quasi-isotropic with E = 5 GPa, νf = 0.3 and strength σf,cr = 90 MPa. For buildings of this type we have that the maximum allowed deflection is 4% of the shortest plate span. The incline roof part is built up of three panels with dimension 5 × 2.5 m which are supported vertically by the side walls and two long supporting beams (dashed lines). The designing load case is a combination of snow, wind and "negative" pressure inside the building. These add up to a total of approximately 4 kPa (kN/m2) uniform pressure perpendicular to the panels. The vertical walls (with length 7.5 m) are subjected to a compressive load equal to 6.5 kN/m. All faces should for simplicity be of equal thickness on both sides and in all panels. Compute the minimum required face thickness and carefully specify all made approximations.

Ans. (i) Incline roof part - simply supported plate with a/b = 2

Assume d ≈ tc, and the ν for the panel is the same as for the faces. qb42()12− ν qb2 w ≈ 0. 01013 2 +≤0. 1139 100 mm → tf > 0.21 mm EtdffGd c 3 σlocal buckling≈ 05. EEG f c c = [assume νc = 0.3 so that Ec = 10.4 MPa] = 30 MPa qb2 0. 465qb σmax =≤0. 1017 30 MPa → tf > 0.5 mm and τmax = = 0.027 MPa tdf d (ii) wall Assume that the panel aspect ratio is large (7.5/2 = 3.75) so that the wall can be approximated by a beam of length 2 m. 2 22 111L 121L ()−ν f 11 =+=+=2 22+≤ < → tf > 0.04 mm PPcr Euler PScπ D Gd π Etdff Gdc 65.N/m Answer: The face should be a least 0.5 mm thick.

9.57 AN INTRODUCTION TO SANDWICH STRUCTURES

Example 9.5 The bottom panel in a coal carrying truck, schematically illustrated below, is supported by two longitudinal beams. The panel has width B = 2500 mm, length L = 8000 mm and the two support beams are located with a distance of 1250 mm from each other. The truck may carry up to 20,000 kg of coal which may be treated as a uniformly distributed load.

L The faces shall be of aluminium for cleaning and wear purposes and must be 2.5 mm thick on both sides. Local reinforcements are placed where the support beams are placed to avoid damage due to locally concentrated loads. A Diviniycell H80 core is used which has the following properties 3 ρc = 80 kg/m Ec = 80 MPa Gc = 30 MPa τcr = 1.0 MPa For the faces we have

Ef = 70 GPa Gf = 27 GPa σcr = 150 MPa Calculate the minimum allowed core thickness subject that the safety factor for fracture is 5 and that the maximum allowed deformation is 10 mm. Clearly state all approximations and give reasons for why they may be used.

Ans. The bottom plate may due to symmetry be divided into four equal parts. As these plate parts have a very large aspect ration they may be treated as beams., Thus, it is sufficient (with good accuracy) to study a cantilever beam with length 625 mm subjected to a uniform pressure q = 0.01 MPa. The maximum bending moment then becomes Mmax = 1953 Nmm/mm width and the maximum transverse force Tmax = 6.25 N/mm width. ηT (a) Core shear: τ =≤max 10. MPa → d = 31 mm cr d ηMmax (b) Face stress: σcr =≤150 MPa → d = 26 mm tdf ql42ql ql 4 ql2 2180 65. 1 (c) Deformation: σcr =+≈2 + =2 + ≤10 mm → d = 19 mm 824D S Etdff 2Gd c d d 3 (d) Local buckling: σ lb = 0.5⋅ E f EcGc = 275 MPa Ok! Other more sophisticated methods may also be fine!

Example 9.6 A simply supported sandwich panel according to the figure is to be designed. The panel shall have face of aluminium and a core of 50 kg/m3 PMI cellular plastics. The panel is subjected to a uniformly distributed load q=150 kPa and can according to the specification have a maximum deflection on 10 mm. Find the face and core thickness so that the panel weight is minimised. Then find maximum stresses in the faces and the core and find if the design is feasible. x a = 1200 mm and b = 700 mm 3 Ef = 70 000 MPa, νf = 0.3, ρf = 2700 kg/m E = 50 MPa, G = 30 MPa, ρ = 50 kg/m3 b c c c σf,cr = 400 MPa (tension and compression) τ = 1 MPa a c,cr

2 Ans. a/b = 1.71 ≈ 1.8, thin faces → D ≈ Eftfd /2, S ≈ Gcd, tc ≈ d −1 qb42()1−ν qb 2 58199⎡ 10 1793⎤ w = 0. 00931 +=0. 1098 10mm → t f =−2 ⎢ ⎥ D S d ⎣ qd⎦

9.58 SOLUTIONS TO PLATE PROBLEMS

116398ρ The total weight W = 2ρ t + ρ d = f + ρ d , minimise with respect to d or use some trail and error. f f c dd67− 1793 c

Doing the latter one arrives at d ≈ 74 mm → tf = 0,25 mm 2 MdEmax f 0, 0948qb The maximum face stress:σ f ,max == = 376 MPa 2D tdf T 0, 452qb The maximum core shear stress: τ ==max = 0.64 MPa c,max d d 13/ Local buckling stress: σ flocal, = 05, ()EEG f c c = 235 MPa Solution not feasible!

Example 9.7 Your first assignment on the new job as an engineering consultant you are to do a pre-study of a concept of building air-cargo pallets in a sandwich design. These pallets have the dimension 3175 by 2235 mm according to the figure and a supported by six frames which run in parallel inside the air-plane. The pallet design used today consists of a solid 4.2 mm thick aluminium plate which weigh approximately 80 kg. Your client believes this can be reduced. The value of one kg of reduced weight is approximately 17 000 SEK per year.

The total load on the pallet is 6800 kg uniformly distributed. On top of this the pallet should be able to sustain an extra acceleration of 1.5g, i.e. the total load times 2.5g. The maximum deformation may not exceed 50 mm. The loading is such that you may approximately the system by studying the panels inscribed between the supports (with dimension 665 by 2235 mm). Furthermore, these rectangular panels may be regarded as simply supported. Other physically reasonable approximation may of course also be done. In order to comply with handling quality the face sheet should be made of 1mm thick aluminium.

You are now to choose a suitable core material and its thickness so your clients will be convinced of the superiority by using a sandwich pallet.

2135

3175

Supports

60 1470 805 CL Airplane 140

Face: Ef = 70 GPa σf,cr = 150 MPa νf = 0.3 tf = 1 mm Core: Core/density Ec (MPa) Gc (MPa) τcr (MPa) WF51* 75 24 0.8 WF71* 105 42 1.3 WF110* 180 70 2.4 WF200* 350 150 5.0 * Rohacell WF-quality (the number indicates the density of the material).

Ans. Assume thin faces and a weak core. The boundary conditions on the plate (simply supported along the long edges and free along the short) and that a/b = 3.4 indicate that one may consider this a case of cylindrical bending, i.e. a beam with length 665 mm and width 2235 mm. Under these circumstances we get: The load is 6800 kg * 9.81m/s2 / 2135mm / 3175mm * 2.5 = 24.6 kPa

9.59 AN INTRODUCTION TO SANDWICH STRUCTURES

T qL 818. N/mm (i) τ ≈==max max d 2dd 3 (ii) σlokalbuckling= 05. EEG f c c > 150 MPa, not active constraint 2 Mmax qL 1360N / mm (iii) σmax ≈== tdff8td d 2 42 2 Etdff 51qL ()− ν qL 1629 1360 (iv) D ≈ , S ≈ Gcd → wmax = +=2 + =50 mm 2 384D 8Sd Gdc

The weight becomes: W = (2ρftf + ρctc)*3.175*2.135. This gives (design values in bold face) WF51 WF71 WF110 WF200

(i) tc (mm) 10.2 6.3 3.4 1.63 weight (kg) 40.1 39.6 39.1 38.8

(iii) tc (mm) 9.1 9.1 9.1 9.1 weight (kg) 39.8 41.0 43.4 48.9

(iv) tc (mm) 6.3 6.0 5.9 5.8 weight (kg) 38.8 39.5 41.0 44.5 hence, choose a 10 mm thick WF51 core.

Example 9.8 A simply supported sandwich panel according to the figure is to be designed. The panel will have faces of aluminium and a PVC core. The panel is subjected to a uniformly distributed pressure q = 100 kPa. Calculate the core thickness and the core density. Three different failure modes are to be considered; face fracture, core shear failure and local buckling (wrinkling). Both faces may be assumed to have the same thickness, tf = 2 mm. After choosing a core and its thickness, calculate the face and core stresses. Give all made approximations.

x a = 1200 mm and b = 600 mm 3 Ef = 70 000 MPa, νf = 0.3, ρf = 2700 kg/m σf,cr = 250 MPa (tension and compression) b The core properties may approximately be described with a 3 2 Ec = CEρc where CE = 1 Nm /kgmm and Gc = Ec/2.5 τ = C ρ where C = 0,015 Nm3/kgmm2 y c,cr τ c τ

Ans. Three different failure modes must be considered, two for the faces and one for the core M 0, 1017qb2 (i) Face failure (tension or compression): σ ≈=max ≤250MPa f max tdf 2d

3 3 22 (ii) Local buckling: σρf, lokal≈=05,,/,E f E c G c 05 E f C E c 25 ≥ 250 MPa T 0, 465qb (iii) Core shear failure: τρ≈=max ≤0, 015 ccmax d d Assume that (i) and (iii) occur simultaneously 3 (i) gives that d ≥ 7,3 mm, and (iii) then yields that ρc = 254 kg/m , and (ii) that σf,lokal = 609 MPa Assume then that (ii) and (iii) occur simultaneously 3 From (ii) ρc = 66,8 kg/m , which gives from (iii) that d = 27,8 mm, and (i) that σf = 66 MPa

3 3 Hence, ρc < 66,8 kg/m gives face tensile/compressive failure and/or core shear failure, and for ρc > 66,8 kg/m one gets local buckling and/or core shear failure.

9.60 SOLUTIONS TO PLATE PROBLEMS

Example 9.9 You should make a preliminary design for a floor to a helicopter which is to be built in a sandwich design. The face material is of sheet aluminium and must from wear and impact constraints be 0.8 mm thick on both sides of the sandwich. The largest floor panel is rectangular with dimensions 1500 mm by 825 mm and is self- supporting, i.e., there are no external support frames except around the edges of the floor. The floor plate may also be assumed to be simply supported around all edges. There is a stiffness constraint implying that the maximum deformation of the floor must be less than 25 mm (so that people don’t get the notion that it is about to break or feels too ”soft”).

You must now choose an appropriate core material and its thickness for the floor so that the weight is minimised. You can choose between different densities of a aluminium honeycomb core material which properties are given in the table below. For simplicity, you are allowed to treat the honeycomb as isotropic and the data given in the table are conservative, i.e., gives the properties for the ”poorest” direction

Face: Ef = 70 GPa σf,cr = 300 MPa νf = 0.3 Core: Density Ec (MPa) Gc (MPa) τcr (MPa) 50 510 151 0.89 72 1030 210 1.5 98 1650 280 2.2 130 2410 370 3.1

The design load 1500 kg/m2 uniformly distributed pressure times 4.5 g acceleration. There must be a safety factor again failure of 1.5.

Ans. Assume thin faces and a weak core. The boundary conditions are simply supported and a/b = 1.8. The design load is 1500*9.91*4.5=67 kPa plus whatever comes from the safety factor. T qb 37.5 N/mm (i) τ ≈ max = 1.5× 0.452 = max d d d 3 (ii) σ lokalbuckling = 0.5 E f EcGc ≥ 876 MPa, not active constraint 2 M max qb 8106N/mm (iii) σ max ≈ = 1.5× 0.0948 = t f d t f d d E t d 2 (iv) D ≈ f f , S ≈ G d → 2 c qb 4 (1−ν 2 ) qb 2 9391 5007 w = 0.00931 + 0.1098 = + = 25 mm max 2 D S d Gc d Since the weight of the faces is the same in all case it is only necessary to compare the weight of the core, or even simpler, d*ρc. 50 72 98 130 (i) d (mm) 42 25 17 12.5 weight (kg) 2.10 (iii) d (mm) 27 27 27 27 weight (kg) 1.95 2.65 3.51 (iv) d (mm) 22 20 20 19 weight (kg) hence, choose a 27 mm thick core with density 72 kg/m2.

Footnote: The actual design of the helicopter in question has 0.8mm aluminium faces on the upper side and 0.6 mm on the lower side and a 30 mm aluminium honeycomb core of 72 kg/m3.

9.61 AN INTRODUCTION TO SANDWICH STRUCTURES

Example 9.10 A hatch on an off-shore platform is made in a sandwich construction. The hatch covers a cell with expensive equipment and above heavy equipment is being moved. A worst case scenario is when this equipment is dropped on top of the hatch. The load on the hatch in this case is 200 kN and it may then not deform more than a maximum of 200 mm. The hatch is 4 by 4 metres and is supported along its edges. In order to obtain the necessary impact strength the face sheets (same on both sides) must be at least 10 mm thick and these are made of glass/phenolic composite laminates in a quasi-isotropic lay-up (can be assumed as isotropic in the plane). Its E-modulus is 10 GPa and the Poisson ratio is 0.3. The core is an isotropic PVC foam with 200 kg/m3 density with an E-modulus of 200 MPa and a Poisson ratio of 0.3. Hence, the problem is that of a simply supported square sandwich panel subjected to a point load in the middle. Calculate the minimum allowed core thickness by estimating the deflection of the panel approximately using Ritz method (energy method).

The strain energy of an isotropic sandwich panel is given by ⎡ 2 2 2 ⎤ D ⎛ ∂ 2 w ⎞ ⎛ ∂ 2 w ⎞ ⎛ ∂ 2 w ⎞⎛ ∂ 2 w ⎞ ⎛ ∂ 2 w ⎞ U = ⎢⎜ b ⎟ + ⎜ b ⎟ + 2ν ⎜ b ⎟⎜ b ⎟ + 2(1−ν )⎜ b ⎟ ⎥dxdy + 1 2 ∫∫ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎜ 2 ⎟⎜ 2 ⎟ ⎜ ⎟ 2(1−ν ) ⎢⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂x ⎠⎝ ∂y ⎠ ⎝ ∂x∂y ⎠ ⎥ ⎣ ⎦ 2 2 S ⎡⎛ ∂w ⎞ ⎛ ∂w ⎞ ⎤ + ⎢⎜ s ⎟ + ⎜ s ⎟ ⎥dxdy 2 ∫∫⎢ ∂x ⎜ ∂y ⎟ ⎥ ⎣⎝ ⎠ ⎝ ⎠ ⎦ Hint: Make a simple deflection assumption w(x,y) which satisfies the boundary conditions, differentiate and insert into the energy equation and integrate!

More hints: the integrals a πx a a πx a ∫ sin 2 dx = , ∫ cos2 dx = 0 a 2 0 a 2

Ans. Assume the deflection assumption πx πy w(x, y) = ()W + W sin sin b s a b which satisfies the kinematic boundary conditions and the natural boundary conditions that bending moments are zero along the edges. Differentiation and inserting into the energy equation gives 2 2 2 2 2 D ⎡⎛ π ⎞ ⎛ π ⎞ ⎤ ab S ⎡⎛ π ⎞ ⎛ π ⎞ ⎤ ab U = ⎢ + ⎥ W 2 + ⎢ + ⎥ W 2 − P W + W 2 ⎜ ⎟ ⎜ ⎟ b ⎜ ⎟ ⎜ ⎟ s ()b s 2(1−ν ) ⎣⎢⎝ a ⎠ ⎝ a ⎠ ⎦⎥ 4 2 ⎣⎢⎝ a ⎠ ⎝ a ⎠ ⎦⎥ 4

Minimisation with respect to Wb gives 2 2 2 dU DW ⎡⎛ π ⎞ ⎛ π ⎞ ⎤ ab = b ⎢ + ⎥ − P = 0 2 ⎜ ⎟ ⎜ ⎟ dWb (1−ν ) ⎣⎢⎝ a ⎠ ⎝ a ⎠ ⎦⎥ 4 and with respect to Ws 2 2 dU ⎡⎛ π ⎞ ⎛ π ⎞ ⎤ ab = SWs ⎢⎜ ⎟ + ⎜ ⎟ ⎥ − P = 0 dWs ⎣⎢⎝ a ⎠ ⎝ a ⎠ ⎦⎥ 4 which leads to 4P(1−ν 2 ) 4P(1−ν 2 ) Pa 2 (1−ν 2 ) W = = = since b = a b 2 4 4 2 2 Dπ ⎡⎛ π ⎞ ⎛ π ⎞ ⎤ 2 ⎛ π ⎞ abD⎢⎜ ⎟ + ⎜ ⎟ ⎥ 4a D⎜ ⎟ a a ⎝ a ⎠ ⎣⎢⎝ ⎠ ⎝ ⎠ ⎦⎥ 4P 4P 2P and W = = = since b = a s 2 2 2 2 ⎡⎛ π ⎞ ⎛ π ⎞ ⎤ 2 ⎛ π ⎞ Sπ abS⎢⎜ ⎟ + ⎜ ⎟ ⎥ 2a S⎜ ⎟ a a a ⎣⎢⎝ ⎠ ⎝ ⎠ ⎦⎥ ⎝ ⎠ The maximum deflection is now

9.62 SOLUTIONS TO PLATE PROBLEMS

Pa 2 (1−ν 2 ) 2P 2Pa 2 (1−ν 2 ) 2P w = W + W = + = + < 200 mm max b s 4 2 4 2 2 Dπ Sπ π E f t f d Gc dπ This leads to a second order equation in d which using the numbers in the problem to d > 56 mm, i.e., tc > 46 mm

Example 9.11 A sandwich panel for a satellite application according to the figure shall be designed. The panel is simply supported and has aluminium face sheet and a 50 kg/m3 aluminium honeycomb core. The panel is subjected to a uniform pressure q = 50 kPa and may according to specifications only deform 2 mm at any point of the panel. Calculate the face and core thicknesses for a minimum weight design. The face sheet are assumed to have the same thickness. Use a safety factor of 1-5 for both the core and the face sheets. Clearly specify all made approximations. x a = 700 mm and b = 400 mm 3 Ef = 70 000 MPa, νf = 0.3, ρf = 2700 kg/m E = 500 MPa, G = 100 MPa, ρ = 50 kg/m3 b c c c σf,cr = 250 MPa τ = 0.5 MPa a c,cr

Ans.

E f t f d 2 a/b = 1.75 ≈ 1.8, thin faces → D ≈ , S ≈ Gcd, tc ≈ d 2 Maximum face sheet stress: 2 2 M max dE f 0.0948qb 0.0948qb 2 σ f ,max = = < 250/1.5 → tfd > 1.5 = 4.55 mm 2D t f d 250 Maximum core shear stress: T 0.452qb τ = max = < 0.5/1.5 MPa → d > 1.356qb = 27.1 mm c,max d d Wrinkling stress: 1/ 3 σ f ,local = 0.5()E f EcGc = 760 MPa - lower than allowable compressive strength!! Stiffness: −1 qb 4 (1−ν 2 ) qb 2 ⎡ 6198 175.7 ⎤ 6198 ⎡ 2 175.7 ⎤ wmax = 0.00931 + 0.1098 = ⎢ + ⎥q = 2 mm → t f = ⎢ − ⎥ D S 2 d 2 q d ⎣⎢t f d ⎦⎥ d ⎣ ⎦ The total weight is: 12393ρ f W = 2ρftf + ρcd = + ρc d 2d 2 / q −175.7d

Minmise w.r.t. d! One then finds that d ≈ 34 mm → tf ≈ 0.154 mm

The solution is feasible since it gives that d > 27.1 mm for maximum core shear stress and that the face sheet 2 2 stress condition tfd = 5.2 mm > 4.55 mm

9.63 AN INTRODUCTION TO SANDWICH STRUCTURES

Example 9.12 The deck in a sailing yacht is load carrying and contributes very much to the overall stiffness of the boat. When the boat is subjected to loads from the rig and waves, large compressive forces build up in the deck. The deck is supported along the side of the boat and rests on a number stringers transverse to the hull. The deck can thus be studied as a number of near rectangular panels according to the figure below. The width of the boat is 4m (b) and the distance between the supports (stingers) is 0.8m (a). The face sheet must 1mm thick to give the required impact strength. These are made of a carbon/epoxy composite in a quasi-isotropic lay-up. The core shall be a Nomex honeycomb. You shall now design the deck by choosing a core type and its thickness so that you will get a panel with the highest load carrying capability (Px) per unit weight. Give explicitly all made approximations in your analysis.

Data P Face sheet x Ef = 50 GPa, νf = 0.25 σˆ f = 700 MPa 3 b ρf = 1500 kg/m safety factor for material failure: 2.0 tf = 1 mm

Core

Density Ec (MPa) Gc (MPa) τcr (MPa) 29 80 17 0.4 48 180 35 0.9 80 330 55 1.5 144 600 80 2.5

Help: Buckling of sandwich beam is given by 1 1 1 = + Pcr Pb Ps

Ans. The deck panel is designed to carry in-plane compressive loads and should thus not buckle or suffer from face sheet compressive failure. Look into all possible compressive failure modes and design the panel. First, study those independent of the choice of core material.

(i) Compressive face sheet failure

The safety factor should be 2.0, and thus the allowable face sheet stress is σˆ f = 350 MPa. The allowable in- plane load is then 2*350MPa*1mm = 700 N/mm.

Local buckling or wrinkling The wrinkling stress is given by 3 3 σ w = 0.5 E f EcGc or Pw = 2 *t f * 0.5 E f EcGc Now, take the lowest of the compressive strength and the wrinkling stress and choose a core and its thickness to avoid global buckling.

(iii) Global buckling (Euler type buckling) If the panel is approximated as a beam with simply supported edges, the buckling load can be written as 1 1 1 a 2 1 2a 2 1 = + = + = + 2 2 2 Pcr Pb Ps π D S π E f t f d Gc d

9.64 SOLUTIONS TO PLATE PROBLEMS

Panel weight The total panel weight is given by 2 W = 2tfρf + tcρc = 3 kg/m + tcρc

Core 29 48 80 144 P (i) 700 700 700 700 P (ii) 487 756 1097 1629 d (iii) 52 54 49 47 W 4.51 5.59 6.92 9.77 P/W 108 125 101 72 i.e., a panel with a 53 mm thick core of density 48 kg/m3 will satisfy all constraints and provide the panel having the largest load carrying capacity per unit weight.

Example 9.13 You are to design a floor to an aeroplane. It can, as a first approximation, be analysed as a simply supported sandwich panel, but since the aspect ratio is high (a >> b) it is possible to further approximate the deformation by using a beam with length a. The load case is a uniformly distributed pressure q. The strength of the panel will be treated later and is given by the type of inserts and load introductions to be used. You are now find the optimal combination of face and core thicknesses for a given stiffness with the given materials chosen for the pre-study. Assume the faces to be much thinner than the core, and the core to be weak. q

Data: a = 500 mm

Ef = 52 000 MPa (carbon fibre composite laminate) 3 ρf = 1 700 kg/m Gc = 40 MPa (Nomex honeycomb) 3 ρc = 48 kg/m Stiffness: Maximu allowed deformation ≤ a/100 = 5 mm Load: q = 200 kPa (or translated, 0,2 N/mm bredd)

Ans. The deformation of the beam can be calculated as 4 ⎡ 4 3 ⎤ qL ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ q 2 w(x) = wb ()x + ws (x) = ⎢⎜ ⎟ − 2⎜ ⎟ + ⎜ ⎟⎥ + ()ax − x 24D ⎣⎢⎝ a ⎠ ⎝ a ⎠ ⎝ a ⎠⎦⎥ 2S qa 4 ⎡ 1 1 1 ⎤ qa 2 ⎛ 1 1 ⎞ 5qa 4 qa 2 Thus, wˆ = ⎢ − 2 + ⎥ + ⎜ − ⎟ = + 24D ⎣16 8 2⎦ 2S ⎝ 2 4 ⎠ 384D 8S 2 2 E f t f d Gc d with D = and S = ≈ Gc d 2 tc −1 a 5qa 4 qa 2 5qa 4 ⎛ qa 2 ⎞ One can then write wˆ = ≤ + or t ≥ ⎜ wˆ − ⎟ 2 f 2 ⎜ ⎟ 100 192E f t f d 8Gc d 192E f d ⎝ 8Gc d ⎠ 4 −1 5qa ρ ⎛ qa 2 ⎞ The total weight W = ρ t + 2ρ t ≈ ρ d + 2t ρ = f ⎜ wˆ − ⎟ + ρ d c c f f c f f 2 ⎜ ⎟ c 96E f d ⎝ 8Gc d ⎠ Minimise this expression! For example, draw W as function of d and find the minimum approximately by visual inspection of the curve. Any approach is fine! After some fiddling I came to the result

d ≈ 75 mm which gives that tf = 0.38 mm.

9.65 AN INTRODUCTION TO SANDWICH STRUCTURES

Example 9.14 A roof to an industrial building made in a sandwich shall be designed. The roof is designed to carry a (snow) load corresponding to a uniform pressure of 5 kPa (5000 N/m2). The roof is made in modules with dimension 4 by 2 metres and these are mounted onto a frame so that each panel can be analysed as simply supported around its edges. The face sheets are prefabricated quasi-isotropic laminates with a thickness of 2 mm and bonded onto the core. The roof may only deform 2% of the short panel span, i.e. a maximum of 40 mm and the safety margin against failure must be 5. Design the roof so that it becomes as cheap as possible.

The materials to choose from are Face: Ef = 10 GPa σf,cr = 110 MPa, νf = 0.3 Kärna/Core: Density Ec Gc τc,cr (SEK/kg) (kg/m3) (MPa) (MPa) (MPa) PS28 28 10 4 0.2 20 PS36 36 18 7 0.35 25 PS45 45 30 12 0.6 30

Ans. 2 a/b = 2,thin faces → D ≈ Eftfd /2, S ≈ Gcd, tc ≈ d qb 4 (1−ν 2 ) qb 2 2qb 4 (1−ν 2 ) qb 2 (i) Stiffness: w = 0.01013 + 0.1139 = 0.01013 + 0.1139 ≤ 60 mm 2 D S E f t f d Gc d 1/ 3 (ii) Wrinkling: σ f ,local = 0,5()E f EcGc > 22 MPa for all materials. Not active constraint! T 0.465qb 4.65 (iii) Core shear failure: τ = y = = ≤ τˆ / 5 c d d d cr 2 M max dE f 0.1017qb 1017 (iv) Laminat fracture: σ f ,max = = = ≤ σˆ f / 5 = 22 MPa 2D t f d d The cost of the face sheet material is the same for all configuration and the cost of the core can be found as 2 V = tc (volume of core per m ) 2 W = V ρc (weight of core per m ) 2 C = W p = tc ρc p ≈ d ρc p (cost of core per m )

Thus, PS28 PS36 PS45 (i) 51 47 45 cost (SEK/m2) 28.6 42.3 60.8 (iii) 116 66 39 cost (SEK/m2) 65.0 59.4 52.7 (iv) 46 46 46 cost (SEK/m2) 25.8 41.4 62.1

Thus, choose a PS36 core with thickness 64 mm (since tc = d − tf)

9.66 CHAPTER 10

SINGLE CURVED SANDWICH SHELLS

As pointed out by Plantema [1], surprisingly little has been written on the theory and application of bending of sandwich shells. It seems that most effort has been spent on buckling of cylindrical shells. In this section, equations are developed for an orthotropic sandwich plate with a constant cylindrical curvature. The derivation follows the work by Stein and Mayers [2,3] and is restricted to plates of constant thickness with orthotropic material characteristics and a single cylindrical curvature with a radius of curvature much greater than the thickness of the plate. The derived equations can, of course, be extended to be valid for a plate with constant double curvature. There exists a more refined but similar theory that accounts for thick faces as described by Fulton [4] based on the same approach as outlined here, but in most practical cases the thin face approach will suffice. Solutions to some simple cases of uniaxial buckling of isotropic sandwich cylinders and buckling due to external pressure [1] are given herein. In [1] one may find much further information on the topic, such as buckling of orthotropic shells, buckling due to torsion, buckling of circular cylindrical shells, and buckling of spherical shells.

The case of a cylindrical sandwich shell subjected to bending moments is then briefly discussed based on some recent findings of Smidt [5,6] which gives some advice on how to design cylindrical corners.

10.1 Fundamental Equations First consider a small element of the plate with geometry as shown in Fig.10.1. The plate has a constant curvature κy ≠ 0 while the in other direction κx is zero. Starting with the equilibrium equations we have that the force equilibrium in x- and y-directions yields the same results as for a flat plate, that is, the same as in eqs.(8.6a,b). Vertical equilibrium on the other hand yields

∂T ∂T ∂ 2 w ⎛ ∂ 2 w 1 ⎞ ∂ 2 w x +++y qN +N ⎜ +⎟ +=2N 0 (10.1) xy2 ⎜ 2 ⎟ xy ∂x ∂y ∂x ⎝ ∂yRy ⎠ ∂∂xy

Equilibrium of bending moments once again yields the same equations as for the flat plate, i.e., eqs.(8.7), which in turn means that bending moments can be written as a function of the

10.1 AN INTRODUCTION TO SANDWICH STRUCTURES

deflection field w and transverse forces Tx and Ty as in eqs.(8.8). Thus, only one equilibrium equation changes due to the single curvature. M M y M yx M xy x ∂M ∂M M x dx yx x+ ∂ M + dy x yx ∂y ∂Mxy Mxy+ dx ∂My ∂x M + dy y ∂y x R T y y z T N y x N yx N xy y N x ∂N N x dx x+ ∂x

∂Tx ∂Nyx T + dx N + dy x ∂x yx ∂y ∂N ∂N ∂Ty xy y T + dy N + dx N + dy y ∂ xy ∂x y ∂y y

Figure 10.1 Forces and bending moments acting on a small curved sandwich element. Next consider the relation between the middle-surface stresses and strains. The middle- surface strains for a cylindrical section in the y-direction can be found from the geometrical relation illustrated in Fig.10.2.

y,v w

z,w

dθ

Figure 10.2 Deformation in the transverse direction of small curved element. Hence, ∂u ∂v w ∂u ∂v ε xy==−, ε , and γ xy =+ (10.2) ∂x ∂y Ry ∂y ∂x providing Ry is much larger than the thickness of the plate. This gives that

A A ⎡∂u ⎛ ∂v w ⎞⎤ N = x []ε +ν ′ ε = x ⎢ +ν ′ ⎜ − ⎟⎥ (10.3) x 1 −ν ′ ν ′ x yx y 1 −ν ′ ν ′ ∂x yx ⎜ ∂y R ⎟ xy yx xy yx ⎣⎢ ⎝ y ⎠⎦⎥

10.2 SINGLE CURVED SANDWICH SHELLS

Ay ⎡∂v w ∂u ⎤ N = ⎢ − +ν ′ ⎥ y ′ ′ xy 1 −ν xyν yx ⎣⎢∂y R y ∂x ⎦⎥

⎛ ∂u ∂v ⎞ N xy = Axy ⎜ + ⎟ ⎝ ∂y ∂x ⎠ where A are the extensional stiffnesses of the plate, i.e.,

Ax = Ex1t1 + Ecxtc + Ex2t2, Ay = Ey1t1 + Ecytc + Ey2t2, and Axy = Gxy1t1 + Gcxytc + Gxy2t2 The next step is to start the reduction of the equilibrium and constitutive equations to a set of three independent differential equations as for the plane case (see eqs.(8.6c) and (8.9)). In fact, since eqs.(8.7) and (8.8) remain unchanged for the curved plate, eqs.(8.9) are still valid which together with the slightly different equilibrium equation of (10.1) now constitute the new set of governing equations. These equations could then be used to solve bending problems of different kinds.

10.2 Governing Buckling Equation

When buckling is considered however, the term Nyκy in eq.(10.1) gives rise to some problems: these can be overcome by the following reasoning [2]. Vertical equilibrium prior to buckling (properties with index 0) can be written

2 2 2 ∂Tx0 ∂Ty0 ∂ w0 ∂ w0 ∂ w0 Ny0 +++qNxy0 2 +N 0 2 +20N xy0 += (10.1a) ∂x ∂y ∂x ∂y ∂∂xy Ry whereas after buckling all properties have changed by an increment so that

w = w0 + w1

Tx = Tx0 + Tx1 Ty = Ty0 + Ty1

Nx = Nx0 + Nx1 Ny = Ny0 + Ny1 Nxy = Nxy0 + Nxy1

Mx = Mx0 + Mx1 My = My0 + My1 Mxy = Mxy0 + Mxy1 The vertical equilibrium equation now takes the same form as eq.(10.1) but with the properties interpreted as sums of the value prior to buckling plus the change after buckling (properties with index 1). Subtracting eq.(10.1a) from eq.(10.1), i.e., the change in the equilibrium, gives the equation which applies to buckling problems as

2 2 2 ⎡ 2 ⎤ ∂Tx1 ∂Ty1 ∂ w1 ∂ ()ww01+ ∂ w1 1 ∂ ()ww01+ ++N xx0 2 +N 1 2 +++N yy0 2 N 1 ⎢ 2 ⎥ ∂x ∂y ∂x ∂x ∂y ⎢ Ry ∂y ⎥ ⎣ ⎦ (10.4) ∂ 2 w ∂ 2 ()ww+ ++22N 1 N 01= 0 xy0 ∂∂xy xy1 ∂∂xy

2 2 2 2 One may discard the terms Nx1 ∂ w1/∂x for the term Nx0 ∂ w1/∂x , etc. since product of load changes and deflection changes are small compared with similar products containing pre- buckling loads. Next, if the deflections prior to buckling are zero or constant (as in axial

10.3 AN INTRODUCTION TO SANDWICH STRUCTURES

compression for example) then all derivatives of w0 are zero. For such types of problems eq.(10.4) simplifies to

2 2 2 ∂Tx1 ∂Ty1 ∂ w1 ∂ w1 ∂ w1 Ny1 ++Nxy0 2 +N 0 2 +20N xy0 += (10.5) ∂x ∂y ∂x ∂y ∂∂xy Ry in which Nx0, Ny0 and Nxy0 can be seen as the applied edge loads whereas Ny1 on the other hand is a load caused by the deformations u, v, and w as seen in eq.(10.3) and thus only appears after the plate has buckled. All of a sudden one of the governing equations, namely eq.(10.5) which must used instead of eq.(10.1) for buckling problems, contains derivatives of u and v which must be eliminated. This can be done by using the equilibrium equation of eqs.(8.6a,b) in combination with the constitutive relation in eq.(10.3). By combining these and taking w0 =

0 and thus writing w1 = w we get

Eh ⎡∂ 2u ⎛ ∂ 22v 1 ∂w⎞⎤ ⎛ ∂ u ∂ 2 v ⎞ x ⎢ + ν ′ ⎜ − ⎟⎥ ++Gh⎜ ⎟ = 0 1 − νν ∂x 2 yx ⎜ ∂∂xy R ∂x ⎟ xy ⎝ ∂y 2 ∂∂xy⎠ ′′xy yx ⎣⎢ ⎝ y ⎠⎦⎥ or rewritten

2 222 ∂ u ∂ v ν ′yx ∂w Gxy()1− ν ′ xyν ′ yx ⎛ ∂ u ∂ v ⎞ 2 + ν ′yx − + ⎜ 2 + ⎟ = 0 (10.6) ∂x ∂∂xy Ry ∂x Ex ⎝ ∂y ∂∂xy⎠

2 222 ∂ v 1 ∂w ∂ u Gxy()1− ν ′ xyν ′ yx ⎛ ∂ v ∂ u ⎞ 2 −+ν ′xy + ⎜ 2 + ⎟ = 0 ∂yRy ∂y ∂∂xy Ey ⎝ ∂x ∂∂xy⎠

Now, get back to eq.(10.5) and insert the relations in eqs.(8.9) for the transverse forces by differentiation of eq.(8.9a) with respect to x and eq.(8.9b) with respect to y, and using the expression for Ny1 in eq.(10.3). We then get

D ∂ 4 w ⎛ ν D ⎞ ∂ 4 w D ∂ 4 w x + 2⎜ yx x + D ⎟ + y 4 ⎜ xy ⎟ 22 4 1− ννxy yx ∂x ⎝11− ννxy yx ⎠ ∂∂xy − ννxy yx ∂y

1 ⎡ DT∂ 3 ⎛ ν D ⎞ ∂ 3T ⎤ − ⎢ x x ++⎜ D xy y ⎟ x ⎥ S ()11− νν ∂x 3 ⎜ xy − νν ⎟ ∂∂xy2 x ⎣⎢ xy yx ⎝ xy yx ⎠ ⎦⎥ (10.7) 1 ⎡ DT∂ 3 ⎛ ν DT⎞ ∂ 3 ⎤ − ⎢ y y ++⎜ D yx x ⎟ y ⎥ S ()11− νν ∂y3 ⎜ xy − νν ⎟ ∂∂xy2 y ⎣⎢ xy yx ⎝ xy yx ⎠ ⎦⎥

⎡ ⎤ 2 2 2 Ehy ∂v w ∂u ∂ w1 ∂ w1 ∂ w1 − ⎢ −+ν ′xy⎥ −−−N x0 2 N y0 2 20N xy0 = Ryxyyxy()1− νν′′⎣⎢∂y R ∂x ⎦⎥ ∂x ∂y ∂∂xy

The next step is to find relations for the in-plane deflection u and v as function of w. This is most easily done by differentiation of the first of eqs.(10.6) with respect to x twice giving a relation for the derivative ∂v/∂x3∂y, and then taking the same equation and differentiating with respect to y twice to find a relation for ∂4v/∂x∂y3. These relations are then inserted into the

10.4 SINGLE CURVED SANDWICH SHELLS second of eq.(10.6) after it has been differentiated with respect to x and y. This procedure gives

4 4 4 3 3 Gxy ∂ u ⎡ νν′xyG xy ′yxG xy ⎤ ∂ u Gxy ∂ u 1 ⎡ν ′yxG xy ∂ w Gxy ∂ w ⎤ 4 +−⎢1 − ⎥ 22+=4 ⎢ 3 − 2 ⎥ Ey ∂x ⎣⎢ Ex Ey ⎦⎥ ∂∂xy Exy∂yR⎣⎢ Ey ∂x Ex ∂∂xy ⎦⎥

4 4 4 3 3 Gxy ∂ v ⎡ νν′xyG xy ′yxG xy ⎤ ∂ v Gxy ∂ v 1 ⎡⎛ ν ′yxG xy ⎞ ∂ w Gxy ∂ w⎤ +−⎢1 − ⎥ +=−⎢⎜1 ⎟ + ⎥ E ∂x 4 E E ∂∂xy22 E ∂yR4 ⎜ E ⎟ ∂∂xy2 E ∂y3 y ⎣⎢ x y ⎦⎥ xy⎣⎢⎝ y ⎠ x ⎦⎥

By introducing the differential operator

4 4 4 Gxy ∂ ⎡ νν′xyG xy ′yxG xy ⎤ ∂ Gxy ∂ LE =+−4 ⎢1− ⎥ 22+ 4 (10.8) Exy ∂ ⎣⎢ Ex Exyy ⎦⎥ ∂∂ Eyx ∂ these equations can be written as

3 3 ν′yxG xy ∂ w Gxy ∂ w RLuyE = 3 − 2 Ey ∂x Ex ∂∂xy

⎛ ν ′ G ⎞ ∂ 3 w G ∂ 3w RLv=−⎜1 yx xy ⎟ + xy (10.9) yE ⎜ ⎟ 2 3 ⎝ Ey ⎠ ∂∂xy Ex ∂y

One may now, at least symbolically, write the unknown derivatives as

∂u ⎛ν ′ G ∂ 4 w G ∂ 4 w ⎞ = L−1 ⎜ yx xy − xy ⎟ E ⎜ 4 22⎟ ∂x ⎝ REyy∂x REyx∂∂xy⎠

∂v ⎡ 1 ⎛ ν ′ G ⎞ ∂ 4 w G ∂ 4 w⎤ =−L−1 ⎢ ⎜1yx xy ⎟ + xy ⎥ ∂y E R ⎜ E ⎟ ∂∂xy22 RE ∂y 4 ⎣⎢ y ⎝ y ⎠ yx ⎦⎥

-1 -1 -1 where LE is defined by LE (LEw) = LE(LE w) = w. Substituting the above expressions into eq.(10.7) yields

4 2 2 2 Ghxy −1 ∂ w ⎛ ∂ w1 ∂ w1 ∂ w1 ⎞ LwD +−++2 LEx4 ⎜ N 0 2 N y0 2 2N xy0 ⎟ Ry ∂x ⎝ ∂x ∂y ∂∂xy⎠

1 ⎡ DT∂ 3 ⎛ ν D ⎞ ∂ 3T ⎤ − ⎢ x x ++⎜ D xy y ⎟ x ⎥ (10.10) S ()11− νν ∂x 3 ⎜ xy − νν ⎟ ∂∂xy2 x ⎣⎢ xy yx ⎝ xy yx ⎠ ⎦⎥

1 ⎡ DT∂ 3 ⎛ ν DT⎞ ∂ 3 ⎤ − ⎢ y y ++⎜ D yx x ⎟ y ⎥ = 0 S ()11− νν ∂y3 ⎜ xy − νν ⎟ ∂∂xy2 y ⎣⎢ xy yx ⎝ xy yx ⎠ ⎥⎦ where the operator LD is defined as

10.5 AN INTRODUCTION TO SANDWICH STRUCTURES

D ∂ 4 w ⎛ ν D ⎞ ∂ 4 w D ∂ 4 w L = x + 2⎜ yx x + D ⎟ + y D 4 ⎜ xy ⎟ 22 4 1 − ννxy yx ∂x ⎝11− ννxy yx ⎠ ∂∂xy − ννxy yx ∂y

For the special case of an isotropic plate one first discovers that the operator LE equals 2 2 2 Δ /2(1+ν), and secondly that LD equals DΔ /(1−ν ). The extensional moduli Eih are now equal in both directions and equal to 2Eftf + Ectc ≈ 2Eftf and consequently Gxyh ≈ Eftf /1+ν. Since in the isotropic case we have that ν' = ν, eq.(10.10) reduces to

2 −2 4 2 2 2 DwΔ 2EtffΔ ∂ w ⎛ ∂ w1 ∂ w1 ∂ w1 ⎞ 2 +−++2 4 ⎜ N xy0 2 N 0 2 2N xy0 ⎟ 1 − ν Ry ∂x ⎝ ∂x ∂y ∂∂xy⎠

DΔ ⎛ ∂T ∂Ty ⎞ − ⎜ x + ⎟ = 0 S()1 − ν 2 ⎝ ∂x ∂y ⎠ but as we can use the expression in eq.(10.5) it is possible to omit the terms containing transverse forces and the equation simplifies further to

DwΔ2 2EtΔ−2 ∂ 4 w DΔ ⎛ N ⎞ ++ff ⎜ y ⎟ 2 2 42⎜ ⎟ 1 − ν Ry ∂νx S()1 − ⎝ Ry ⎠

⎛ DΔ ⎞⎛ ∂ 2 w ∂ 2 w ∂ 2 w ⎞ −−⎜1 ⎟⎜ N 1 ++N 1 20N 1 ⎟ = ⎝ S()1 − ν 2 ⎠⎝ xy0 ∂x 2 0 ∂y 2 xy0 ∂∂xy⎠

However, as before, the term Ny/Ry may be rewritten as

4 4 Ny Gxyh −−1 ∂ w 2E fft 2 ∂ w ==2 LE 42Δ 4 Ry Ry ∂x Ry ∂x and thus we have that D Δ2 w 1 − ν 2 (10.11) 2 2 2 4 ⎛ D ⎞⎛ ∂ w ∂ w ∂ w 2Etff ∂ w⎞ −−1 ΔΔ⎜ N 1 ++N 1 2N 1 −−2 ⎟ = 0 ⎜ 2 ⎟⎜ xy0 2 0 2 xy0 2 4 ⎟ ⎝ S()1− ν ⎠⎝ ∂x ∂y ∂∂xy Ry ∂x ⎠

As seen, eq.(10.11) simplifies to the flat plate case of eq.(8.27) when the radius of curvature approaches infinity.

10.3 Buckling of a Simply Supported Isotropic Plate Subjected to Uniaxial Load Consider a curved plate with length a and width b where the width is along the curved part of the plate, as illustrated in Fig.10.3. The plate is subjected to a uniform uniaxial load Px per unit length along the side b.

10.6 SINGLE CURVED SANDWICH SHELLS

Figure 10.3 Sandwich panel with radius of curvature Ry under uniaxial compression. As for the flat plate case the function mπx nπy ww= sin sin (10.12) a b satisfies the boundary conditions. Differentiation and substitution into the governing equation of eq.(10.11) with Nx = −Px, Ny = Nxy = 0 yields

2 4 D ⎛ D ⎞⎛ ∂ w 2Etff ∂ w⎞ ΔΔ2 w +−1 ⎜ P 1 + Δ−2 ⎟ = 0 or rewritten as 2 ⎜ 2 ⎟⎜ x 22 4 ⎟ 1 − νν⎝ S()1 − ⎠⎝ ∂x Ry ∂x ⎠

2 4 D 42⎛ D 3 ⎞ ∂ w1 ⎛ D ⎞ 2Etff∂ w 2 ΔΔw +−⎜ 2 Δ⎟ Px 222+−⎜1 Δ⎟ 4 = 0 11− νν⎝ S()− ⎠ ∂νx ⎝ S ()1 − ⎠ Ry ∂x yields

2 4 D mπ ⎛ DΩΩ⎞ 2Etff mπ ⎛ D ⎞ ΩΩ4 − P ⎜⎛ ⎟⎞ 2 1 + + ⎜⎛ ⎟⎞ 1 + = 0 2 x ⎝ ⎠ ⎜ 22⎟ ⎝ ⎠ ⎜ 2 ⎟ 1 − ν a ⎝ S()1− ν ⎠ Ry a ⎝ S ()1− ν ⎠

mππ22n with Ω=⎜⎛ ⎟⎞ + ⎜⎛ ⎟⎞ ⎝ a ⎠ ⎝ b ⎠

Simplification of this expression soon leads to that one can write the buckling coefficient as

2 2 2 ⎡mb na⎤ 2Etff⎛ mb⎞ + ⎜ ⎟ ()1 − ν 22b ⎢ a mb ⎥ R2 ⎝ a ⎠ PK== ⎣ ⎦ + y 2 x 222 222 π D D ⎡⎛ mππ⎞ ⎛ n ⎞ ⎤ Dm⎡⎛ ππ⎞ ⎛ n ⎞ ⎤ 1 + 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ 2 ⎢⎜ ⎟ + ⎜ ⎟ ⎥ ()1 − ν S ⎣⎝ a ⎠ ⎝ b ⎠ ⎦ 1 − ν ⎣⎝ a ⎠ ⎝ b ⎠ ⎦ or rewritten using the shear factor (see section 10) and introducing the property Zy defined as

42 D 2 21Etbff ()− ν φ y = 22 and Z y = 2 (10.13) Sb()1 − ν RDy makes the buckling coefficient take the following form

10.7 AN INTRODUCTION TO SANDWICH STRUCTURES

2 ⎡ na 2 ⎤ 2 2 ⎛ ⎞ Z y ⎛ a ⎞ ⎢1 + ⎜ ⎟ ⎥ ⎜ ⎟ ⎣ ⎝ mb⎠ ⎦ π 4 ⎝ mb⎠ K = + (10.14) 2 2 2 2 ⎛ a ⎞ 2 ⎡ ⎛ na ⎞ ⎤ ⎡ na ⎤ ⎜ ⎟ ++πφ 1 ⎜ ⎟ ⎛ ⎞ ⎝ ⎠ y ⎢ ⎝ ⎠ ⎥ ⎢1 + ⎜ ⎟ ⎥ mb ⎣ mb ⎦ ⎣ ⎝ mb⎠ ⎦ The remaining problem is to minimise this equation with respect to m and n for a given set of a, b, φy, and Zy. Now concentrate on two special cases: In the special case of an infinitely long curved plate (a,m → ∞), the minima of eq.(10.14) can be simplified to [3]

2 2 2 4 Z yy1 − πφ Z y 41− πφy K ≈ + when < , (10.15a) 2 2 π 4 4 π 2 2 2 ()1 + πφy ()1 + πφy

2 ⎛ ⎞ 2 2 ZZ/ π φ 41− πφy Z 1− πφy K ≈ y ⎜2 − yy ⎟ when <≤y (10.15b) 22⎜ ⎟ 2 2 2 2 1 1 π πφy − πφy ⎝ − πφy ⎠ ()1+ πφy

2 1 1 − πφy K = 2 when Z y ≥ (10.15c) πφy φ y

Eqs.(10.15) are not exact but are quite accurate for the indicated curvature parameter ranges. In Fig.10.4 buckling coefficients for an infinitely long curved panel are plotted versus the shear factor φy.

For the special case of a cylinder for which b = 2πR, the buckling equation is preferably rewritten in terms of a instead of b. This results in

2 2 ⎡ ⎛ na ⎞ ⎤ Z 2 1 ⎢1 + ⎜ ⎟ ⎥ x 22 ⎝ ⎠ 42 ()1 − ν a ⎣ mb ⎦ π m PKx == + (10.16) 2 2 2 2 Dπ 1 2 ⎡ ⎛ na ⎞ ⎤ ⎡ na ⎤ ++πφ 1 ⎜ ⎟ ⎛ ⎞ 2 x ⎢ ⎝ ⎠ ⎥ ⎢1 + ⎜ ⎟ ⎥ m ⎣ mb ⎦ ⎣ ⎝ mb⎠ ⎦

42 D 2 21taEff()− ν where φ x = 22 and Z x = 2 (10.17) Sa()1−ν RDy

10.8 SINGLE CURVED SANDWICH SHELLS

10,000.0

2 π φy Px

b 0.0 1,000.0 0.001 ∞

100.0 0.01

10.0 0.1

0.2

0.5

1.0 1.0

0.1 1 10 100 1000 10000

Figure 10.4 Compressive buckling coefficients for simply supported infinitely long curved isotropic sandwich plates. When applied to a cylinder the value of n must be set to 0 or even integers greater than or equal to 4. The minimisation is performed analytically with respect to m and n and the resulting equations are [3]:

2 1 Z x Z x 1 K = 2 + 4 when 22≤ , (10.18a) 1+ πφx π ππφ1 + x

Z x 1 Za 1 K =−2 ()2Z xxφ when 222≤≤ , (10.18b) π 1 + πφx π πφx

1 1 K = 2 when Z x ≥ 2 (10.18c) πφx πφx

Buckling coefficients for simply supported sandwich cylinders are plotted versus the shear factor φx in Fig.10.5.

10.9 AN INTRODUCTION TO SANDWICH STRUCTURES

10000 P 2 x π φ x

0 1000

Px 100 0.01

10 0.1

0.2

0.5

1 1.0

0.1 1 10 100 1000 10000

Z x

Figure 10.5 Compressive buckling coefficients for simply supported sandwich cylinders. Buckling tests conducted on sandwich cylinders subjected to uniaxial compressive loading, intended to verify the theoretical results, are reported in [7], and reasonable agreement was found. It is also argued in [1] that sandwich cylinders are much less affected by initial imperfections than flat plates. Extension of the buckling theory for large-deflection and post- buckling analysis revealed that the small-deflection result as given here is valid.

10.4 Buckling due to External Lateral Pressure This sub-section will only briefly cover some special cases of buckling of infinitely long cylinders subjected to a lateral pressure as indicated in Fig.10.6. In doing so one might assume that all generators of the cylinder will remain straight so that all derivatives with respect to x are zero. q

N f1

R N R f2 R 2 1

Figure 10.6 Sandwich cylinder part subjected to uniform external pressure.

10.10 SINGLE CURVED SANDWICH SHELLS

First we need to account for the curvature change due to a lateral deflection w. It is realised that the curvature changes from 1/R to 1/(R − w) which equals 11 w w −= ≈ Rw− R RR()− w R2

The correct load distortion is, assuming partial deflection, then written as

2 2 ⎡∂ w ∂ ws w ⎤ MDyy=−⎢ 2 −22 + ⎥ ⎣ ∂y ∂y R ⎦

Using this equation along with the relation between transverse forces and bending moment Ty

= ∂My/∂y, the equilibrium equation of eq.(10.1) now takes the form

2 2 ∂Ty ⎛ 1 ∂ w⎞ ∂Ty ∂ w ++qN⎜ + ⎟ =−qR =0 ∂y y ⎝ R ∂y 2 ⎠ ∂y ∂y 2

since the normal force Ny equals −qR (which is easily seen by studying the equilibrium of half the cylinder). This gives the two unknown equations ∂2w ∂2w S s −=qR 0 (10.19) y ∂y2 ∂y2

3 3 ∂ws ⎡∂ w ∂ ws 1 ∂w⎤ Sy =−Dy ⎢ 3 −32 + ⎥ ∂y ⎣ ∂y ∂yR∂y ⎦ These can be solved using the simple assumption ny ny ny wA= sin , wB= sin , and/or wC= sin R b R s R By using the first and third of these and inserting into eq.(10.19) one soon arrives at

2 ()nD−1 y q = (10.20) ⎛ nD2 ⎞ R3 ⎜1 + y ⎟ ⎜ 2 ⎟ ⎝ SRy ⎠

The solution takes a form very similar to that of straight beams and flat plates, the solution to a shear stiff section divided by a term containing 1 plus some kind of non-dimensional correction term which in turn depends on the ratio of shear deformation to the bending deformation. The lowest admissible value for n, which is 2, gives the buckling load for an 3 ordinary cylinder (with no shear deformation) which is q = 3Dy/R [8]. 10.5 Local Buckling of Curved Sandwich Panels Tests performed on curved sandwich panels in uniaxial compression [5] as in Fig.10.3 seem to verify that curved panels have a local buckling load equal to that of a similar flat plate with the same materials and geometry (face and core thickness). Therefore, for curved plates, Hoff's buckling formulae as given in eq.(6.11) may well be used for this case.

10.11 AN INTRODUCTION TO SANDWICH STRUCTURES

The test series performed in [5] also seems to indicate that the local buckling stress in the circumferential direction (y-direction in Fig.10.1) is of the same order as for the flat plate case, i.e., Hoff's formula of eq.(6.11) may be used also in this direction. This has not been experimentally verified yet and hence the statement is so far only based on indications.

10.6 Bending of Single Curved Sandwich Beams a Practical Approach In this section some practical aspects on the bending of curved sandwich beams, or panels in cylindrical bending, will be discussed. The issue addressed is to find appropriate fracture criteria for curved beams in the sense that a curved beam part may exhibit a different failure mode from surrounding flat parts. First, study the radial stress appearing in a curved beam subjected to a constant bending moment Mx. The beam, or corner, has a straight part and a curved part according to Fig.10.7.

R t1 R2 R1

M M

Figure 10.7 Sandwich corner construction; 90 degree corner with straight connections. Since the structure is subjected to a bending moment only the straight beam parts may carry a bending moment equal to

Mtdxff= σ$ (10.21) where the face strength is either a tensile, compressive or local buckling stress, whichever is the lowest in the particular case. This bending moment capacity must now in some way be compared with the bending moment capacity of the curved part. Assume that a curved beam part has the same local buckling load as a straight beam with the same materials. There are test results indicating that [5], and if so, the beam parts have in that sense equal strength. One may also assume that the curved beam part has the same bending moment capacity as the straight part, that is, as in eq.(10.21) in terms of face stresses. This is at least approximately true if the curvature R is large compared to the thickness of the sandwich. Hence, the straight and the curved beam parts have equal bending moment capacity providing only face fracture or local buckling is considered.

In a curved beam, as opposed to its straight counterpart, radial stresses appear in the core that may cause fracture. Estimating these can be done by adopting the following simplified approach. Study the curved beam part of Fig.10.8.

10.12 SINGLE CURVED SANDWICH SHELLS

σz

z σ σ 2 2

Figure 10.8 Radial stresses in a curved sandwich beam. Find the force equilibrium in the z-direction along the x-axis (along the beam) at a position z. For a differential element of length dx we have that dx = Rdθ and the curvature at this position is R + z. The vertical load created by σz is then σz (R + z)dθ. Since the face sheet normal stress must be constant in the entire structure, the lower face sheet stress in the curved part of the beam is the same as in the straight part. However, in this curved part, this stress component will have a vertical component as shown in Fig.10.8. This will equal σ2t2 times the curvature, i.e. 1/R2, times dx. This may be written as

Rd2 θ σθz ()Rzd+−σ 22t =0 R2 which leads to the relation

Mx Mx σ z ()z = since σ 2 = for an opening moment (10.22) dR()+ z tdf

The radial core stress at the two interfaces is thus

MxxM σz (/)d 2 = = (10.22) dR(/)+ d 2 dR1

MxxM σz (/)−=d 2 = dR(/)− d 2 dR2 If the bending moment is reversed, then the radial stress will be negative but of the same magnitude as indicated by eq.(10.22). These simple formulae based on such simplified assumptions yield surprisingly accurate estimates of the radial stress component in the core [6], as compared with FE-analyses.

Now introduce a so called bending efficiency factor ηb [6] which is defined by the ratio of the bending moment capacity of a straight beam to the bending moment capacity of a curved beam, as

Mc σ$ c dR σ$ c R η b == = (10.23) M x σ$ fftd σ$ fft where σˆ c is the core strength in the radial direction. Note that depending on the direction of the applied bending moment the core stress may be negative, and if so σˆ c should be taken as the core compressive strength rather than its tensile strength. This is important since many

10.13 AN INTRODUCTION TO SANDWICH STRUCTURES

core materials exhibit very different strengths in tension than in compression. The factor ηb is set to unity if the curved beam part fails by face tension, compression, or local buckling prior to core tensile/compressive fracture. This new fracture definition may be graphically represented by plotting ηb versus the ratio R/tf for different ratios of σˆ f/σˆ c as shown in Fig.10.9.

The ratio R/tf may in practical cases be in the order of 10 up to infinity whereas the ratios σˆ f

/σˆ c for practical cases are in the order of 0.1 (high density honeycomb or foam and SMC faces) to 0.005 (low density foam and unidirectional composite laminates). 1.0

0.9

0.8

0.7

0.6

0.05 η b 0.5 0.025 0.01 0.4 0.005 0.0025 0.3 0.001 0.0005

0.2 0.00025

0.1 0.0001

0.0

10 100 1000 10000 R tf

Figure 10.9 Bending efficiency factor for curved beams with different strength ratios σˆ f/σˆ c. An important aspect of the above reasoning is the following. Consider a curved section as in Fig.10.7. If the beam fails by face fracture under the assumptions given there are equal chances of failure occurring in the straight or the curved beam part. ηb then equals unity and the only way of increasing the strength is to improve the strength and/or the thickness of the faces or the thickness of the core in both the curved and the straight beam parts. The same thing goes for face local buckling and the way to increase the strength is to improve the face and/or the core moduli everywhere in the beam. An interesting aspect of the above comes to light when the beam fails by core fracture in the curved part, i.e. for ηb < 1. It is now realised that increasing the strength and/or the thickness of the faces has no effect on the strength of

10.14 SINGLE CURVED SANDWICH SHELLS the beam, except increasing the strength of the straight part of the beam, which is not critical. The only efficient design is to use a higher strength core in the curved part. This core should be chosen so that ηb is as close to unity as possible. By choosing a too high strength core the failure mode is transformed to face fracture and since the core strength is usually in some way proportional to the core density one should try to keep ηb as close to unity as possible for a most efficient design. Tests performed in [5] also indicate that a design having ηb < 1 benefits considerably from choosing a higher density core in the curved part of the beam, whereas increasing the face thickness only adds substantial weight but no strength.

References [1] Plantema F.J., Sandwich Construction, John Wiley & Sons, New York, 1966.

[2] Stein M. and Mayers J., “A Small-Deflection Theory for Curved Sandwich Plates”, NACA Report 1008, 1951.

[3] Stein M. and Mayers J., “Compressive Buckling of Simply Supported Curved Plates and Cylinders of Sandwich Construction”, NACA TN 2601, 1952.

[4] Fulton, “Effect of Face-Sheet Stiffness on Buckling of Curved Plates and Cylindrical Shells of Sandwich Construction in Axial Compression”, NASA TN D-2783, 1965.

[5] Smidt, S., “Testing of Curved Sandwich Panels and Comparison with Calculations Based on the Finite Element Method”, Proc. Second International Conference on Sandwich Construction, Eds.: D. Weissman.Berman and K.-A. Olsson, Gainesville, 1992, EMAS, Ltd., UK, pp 670-680.

[6] Smidt S., “Curved Sandwich Beams and Panels: Theoretical and Experimental Studies”,Licentiate thesis, Department of Lightweight Structures, Royal Institute of Technology, Stockholm, Sweden, Report 93-10, 1993.

[7] Gerard G., “Compressive and Torsional Instability of Sandwich Cylinders”, Symposium on Structural Sandwich Constructions, ASTM STP 118, pp 56-69, 1951.

[8] Timoshenko S.P., Theory of Elastic Stability, McGraw-Hill, Tokyo, 1961.

10.15 CHAPTER 11

LOCALISED LOADS by Ole Thybo Thomsen

It is a well known fact that sandwich structures are notoriously sensitive to failure by the application of highly localised external lateral loads such as point loads, line loads or concentrated surface loads of high intensity. This pronounced sensitivity towards the application of localised external loads is due to the associated inducing of significant local deflections of the loaded face into the core material of the sandwich panel, thus causing high local stress concentrations. Actually, the stress state corresponding to the ideal load transfer mechanism of sandwich panels, where the core material is loaded in a stress state of pure shear and the faces are loaded in a membrane state of stress, can be completely destroyed, and instead a complex multiaxial state of stress is obtained in the near vicinity of the area of external load application. The result of this may be a premature failure.

The local bending problem is illustrated in Fig.11.1 showing the simplest possible case, i.e., a symmetrically supported and symmetrically loaded sandwich beam in three point bending. The deformed sandwich beam configuration, shown in Fig. 11.1, consists of two contributing parts:

(i) Overall bending of the sandwich beam due to the overall bending and shearing. (ii) Local bending of the loaded face about its own neutral axis.

Figure 11.1 Undeformed and deformed sandwich beam in 3-point bending.

11.1 AN INTRODUCTION TO SANDWICH STRUCTURES

11.1 Elastic Foundation Analogy Relatively few references, [1-12], have treated the problems associated with local bending of the faces of sandwich panels subjected to concentrated loads, and, among those, not one gives an explicit description of the onset and development of irreversible failure. Instead, they restrict themselves to the development of structural analyses based on the somewhat unrealistic assumption of linear elastic behaviour of the constituent materials. Thus, the results obtained do not reflect the actual sequence of events leading to failure of the sandwich structure, but they do have the potential of giving valuable information about the fundamental mechanics of the local bending problem as well as the parameters controlling the onset of failure.

Some of the references cited have been based on the use of an elastic foundation model for the loaded face, i.e., to consider the relative deflection of the loaded face of a sandwich panel with respect to the unloaded face as being governed by an appropriate elastic foundation model. The simplest possible elastic foundation model is the well known Winkler foundation model, which corresponds to the modelling of the core material (supporting the loaded face) as continuously distributed linear tension-compression springs (in principle equivalent to a liquid base).

The Winkler foundation model, however, suffers a serious drawback, as it does not account for the possible existence of shear interactions between the loaded face and the supporting medium (the core). This feature of the Winkler foundation model suggests that it becomes inadequate for deflection modes with short wave-lengths where shear deformations become important.

The analysis presented herein [8-10] introduces a two-parameter elastic foundation model, which includes the shear interaction effects. This additional feature, compared to the Winkler model, allows for the calculation of the interface shear stresses (interface between the loaded face and the core), which may build up adjacent to the area of external load application. For some cases these interface shear stresses are attributed significant importance (together with the interface peel stresses) in the development of failure of sandwich panels subjected to localised loading.

In this section, the elastic foundation approach and its implications will be described. Special attention will be focused on the physics of the local bending problem, while the mathematical details will given in an abbreviated form. The presentation of the fundamental theory will be succeeded by a section in which the application of the suggested approach, as well as a brief parametric study, is presented.

11.1.1 Classical Winkler foundation model Considering the problem of a sandwich beam subjected to arbitrary localised lateral loading (or a sandwich plate in cylindrical bending) the idea behind the application of the elastic foundation approach can be formulated in simple terms as follows. It seems reasonable to assume that the deflection of the loaded face with respect to the unloaded face of the sandwich beam can be regarded as governed by an appropriate elastic foundation model (given that the constituent materials behave linear elastically). The simplest possible elastic

11.2 LOCALISED LOADS foundation model is the well known Winkler foundation model, which assumes that the supporting medium can be modelled as continuously distributed linear tension/compression springs. The Winkler foundation model and its application is extensively treated in refs.[13,14].

p (x) z p (x) x face beam x,u z,w supporting medium

Figure 11.2 Loaded face of sandwich beam supported by the core material. Fig.11.2 illustrates the problem considered with its constituent parts: the loaded face of a sandwich beam and the supporting medium (i.e., the core material). The face considered can be subjected to arbitrary distributions of external transverse normal and shear loads (pz(x) and px(x)) along the upper surface. According to the Winkler foundation model (see ref. [13] for further details) the elastic response of the supporting medium is expressed in the form

qzz=−K w()x (11.1) where qz(x) is the interfacial transverse normal stress component measured per unit length of the sandwich beam (see Fig.11.2), w(x) is the lateral deflection of the loaded face, Kz is the foundation modulus and x is the longitudinal coordinate. It should be noticed that qz(x) represents the foundation transverse normal stress component per unit length of the sandwich beam, i.e., it can be obtained by multiplying the transverse normal stress component σz at the interface by the width of the beam. In the forthcoming derivations qz will be referred to as the interlaminar stress distribution function.

The relationship defined by eq.(11.1) is developed under the assumption that the thickness tf (see Fig.11.1 and 11.2) of the loaded face is small compared to the wave-length of the applied load. Furthermore, it is assumed that the wave-length of the elastic response is sufficiently small so that the local bending effects induced in the loaded face can be assumed not to influence the stress state of the lower (not loaded) face. Thus, it is assumed implicitly that the decay of the transverse normal stress component σz (induced by local bending) through the thickness of the core material is complete, meaning that the behaviour of the lower face is assumed to obey the classical theory of sandwich beams.

The foundation modulus Kz, which represents the stiffness of the supporting medium, is suggested to be expressed by the following empirically obtained expression (obtained by comparison with the 2-D elasticity solution to the problem of an infinitely long beam attached to an elastic medium which extends to infinity on one side of the beam) [13]:

Ec KEzc= 028. 3 (11.2) Df

11.3 AN INTRODUCTION TO SANDWICH STRUCTURES

where Ec is the elastic modulus of the core material, and Df is the flexural rigidity of the loaded face. If the depth of the elastic foundation (core thickness) is small, i.e., of the same order of magnitude as the thickness tf of the face, the following relation is suggested instead of eq.(11.2) [1,8,13]

Ec Kz = (11.3) tc The loaded face is modelled by application of the ordinary theory of bending of beams. Thus, it is assumed that normals to the undeformed neutral axis of the loaded face remain straight, normal and rigidly stiff during deformation so that transverse normal and shearing strains may be neglected in deriving the beam kinematic relations (εz = γxz = 0). p dx z

p dx MT x x N+dN N M+dM z T+dT q dx z

Figure 11.3 Notation and sign convention for equilibrium element of the loaded face of the considered sandwich beam. Referring to Fig.11.3 for notation and sign convention the three equilibrium equations for the elastically supported face can be written as dN F =⇒0 =−p ∑ xxdx

dT F =⇒0 =−− (qp ) (11.4) ∑ zzzdx

dM t M =⇒0 =+T f p ∑ dx 2 x According to the classical theory of bending of beams, and by application of the equilibrium eqs.(11.4) along with the force-displacement relationship defined by eq.(11.1), the following is obtained

2 4 2 dw M dw 11dM ⎛ t f dpx ⎞ 2 =− ⇒ 4 =−2 =−⎜ pqzz − + ⎟ ⇒ dx Df dx Dffdx D ⎝ 2 dx ⎠

4 dw Kz 1 ⎛ t f dpx ⎞ 4 +=−+w ⎜ pz ⎟ (11.5) dx DffD ⎝ 2 dx ⎠ The latter of eqs. (11.5) can be written in an alternative form

11.4 LOCALISED LOADS

4 dw 4 1 ⎛ t f dpx ⎞ 4 +=−+4κ w ⎜ pz ⎟ (11.6) dx D f ⎝ 2 dx ⎠ where

K κ 4 = z (11.7) 4Df

The governing eq.(11.6) is an ordinary non-homogeneous fourth order differential equation with constant coefficients. κ, given by eq.(11.7), includes the flexural rigidity of the face, Df, as well as the elasticity of the core, Kz, and is an important factor influencing the shape of the elastic line (neutral axis of the face beam). For this reason κ is known as the characteristic of the system, and, since its dimension is length-1, the term 1/κ is frequently referred to as the characteristic length.

The general solution can divided into two parts, the solution to the homogeneous part of eq.(11.6) and the particular solution to the non-homogeneous differential equation (11.6)

w()x =+whp ()x w ()x (11.8)

The derivation of the homogeneous solution part wh(x) is straightforward and is therefore omitted, and it can be shown that wh(x) can be written in the form w (x )=+ C sinh(κx )sin(κx )C sinh(κx )cos(κx ) h 12 (11.9) ++ CxxCxx34cosh(κκ )sin( ) cosh( κ )cos( κ )

Cj (j = 1,2,3,4) are constants which have to be determined from the boundary conditions, i.e., from the support and external loading conditions imposed on the loaded face of the considered sandwich beam. The determination of the particular solution part of the complete solution given by eq.(11.8) is accomplished by assuming a suitable particular solution function by “qualified guessing”. The derivation of the particular solution wp(x) is of course strongly influenced by the explicit form of the external shear and transverse normal surface loads px(x) and pz(x), as these load distribution functions determine the non-homogeneous terms on the right-hand side of the governing differential equation (11.6).

Considering for a moment the homogeneous solution part given by eq.(11.9), it is recognised that the lateral deflection of the loaded face wh(x), and thereby the interfacial stress distribution function qz(x) (see eq.(11.1)), display a wavy harmonic behaviour (due to the presence of the trigonometric functions sin(κx) and cos(κx)) as well as an exponential dependency of the x-coordinate (due to the presence of the hyperbolic functions sinh(κx) and cosh(κx)). Recognising the wavy harmonic nature of the elastic response implies that the deflection of the loaded face into the core can be attributed some characteristic wave-length λ which can be defined by

2π 4Df λ =⇒ λπ=2 4 (11.10) κ Kz

11.5 AN INTRODUCTION TO SANDWICH STRUCTURES

Inserting eq.(11.2) along with the expression for the flexural rigidity Df into eq.(11.10), the wave-length λ of the elastic response can be estimated in the form

E f 3 λ ≈ 533. t f (11.11) Ec

In order to illustrate the solution of a specific problem, the case of a simply supported sandwich beam in 3-point bending shown in Fig.11.1 is considered, i.e., it is assumed that the sandwich beam is subjected to a point load of magnitude P (in reality a line load since P is assumed to be distributed uniformly over the width of the beam).

Due to the symmetry of the problem, only one half of the loaded face need be considered. Furthermore, the complete solution only consists of the homogenous solution part, as no surface loads are applied. Thus, the only thing to be done is to determine the four constants Cj defined by (11.9). Referring to Figs. 11.1 and 11.2 the four boundary conditions for the loaded face can be written as

dw Pdw⎡ 3 ⎤ P ⎡ ⎤ = 0, T =− ⇒ =− ⎢ ⎥ []x=0 ⎢ 3 ⎥ ⎣ dx ⎦ x=0 22⎣ dx ⎦ x=0 Df

⎡dw2 ⎤ ⎡dw3 ⎤ M 00 , T 00 (11.12) []xL= =⇒⎢ 2 ⎥ = []xL= =⇒⎢ 3 ⎥ = ⎣ dx ⎦ xL= ⎣ dx ⎦ xL=

Solving this system of equations with respect to the unknown constants Cj yield P sinh22 (κLL )+ sin (κ ) C1 = 3 (11.13) 8κ Df cosh(κκLL )sinh( )+ cos( κκ LL )sin( )

P P cosh22 (κLL )+ cos (κ ) C2 ==−3 , CC32, C4 =− 3 8κ Df 8κ Df cosh(κκLL )sinh( )+ cos( κκ LL )sin( )

The outline of the solution corresponding to the classical Winkler foundation hypothesis given in the above will not be explored any further. However, the nature of the obtained solution, its application for engineering design purposes as well as the bounds of applicability will be discussed later.

11.1.2 Two-parameter elastic foundation model The Winkler foundation model which was discussed in the preceding section is the simplest possible foundation model, and it may be expected to give reasonable results as long as the restrictive assumptions outlined in the preceding chapter are reasonably fulfilled. However, this model suffers a serious drawback, as it does not account for the possible existence of shearing interactions between the loaded face and the supporting medium (core). This feature suggests, that the Winkler foundation model becomes inadequate for deformations of short wave-length in which the shearing deformations of the core material becomes important.

11.6 LOCALISED LOADS

The approach to the local bending problem presented in this chapter is essentially similar to the approach of the preceding chapter, i.e., that the deflection of the loaded face relative to the unloaded face can be modelled by application of an elastic foundation formulation. However, in order to account for the possible existence of shearing interactions between the loaded face and the supporting medium (core), which was described earlier, a two-parameter elastic foundation model is introduced. The problem considered, i.e., the loaded face of a sandwich beam (subjected to arbitrary surface loads pz(x), px(x)) and the supporting medium, is illustrated in Fig.11.2. As before, it is assumed that the faces as well as the core material are linear elastic, even though this assumption generally has to be considered somewhat unrealistic (especially for FRP-faces and polymeric foam-core materials).

The elastic response of the supporting medium is suggested to be expressed in the following form, relating the deflections of the loaded face to the interfacial stress components measured per unit length of the beam [8-10] t qx()==− Kux (,f ) and qx () Kwx () (11.14) xx2 zz where qx(x), qz(x) are the foundation shear and transverse normal stress resultants (referred to as the interlaminar stress distribution functions); Kx, Kz are the shearing and transverse foundation moduli of the supporting medium; u(x,tf/2) is the longitudinal displacement of the lower fibre of the loaded face, and w(x) is the lateral displacement of the loaded face.

The transverse foundation modulus Kz is suggested to be expressed by eq.(11.2) (or eq.(11.3)) presented in the preceding chapter.The second foundation modulus Kx, which is referred to as the shearing foundation modulus of the elastic foundation, is suggested to be related to Kz through the following expression [8-10] (assuming isotropic or nearly isotropic core material)

Kz Kx = (11.15) 21()+ νc where νc is the Poisson's ratio of the core material.

As was the case in the preceding section, the loaded face is modelled using the ordinary Bernoulli-Euler beam theory. It should be mentioned that adoption of a Timoshenko beam theory including the effects of the transverse shear deformations could be achieved without any significant difficulty.

The displacement components u, w of an arbitrary point in the loaded face may be expressed in terms of the neutral axis quantities u0, w0 dw uu=− z and ww = (11.16) 00dx By inserting the first of eqs.(11.16) into the first of eqs.(11.14) the following is obtained

⎛ t f dw⎞ qx()=− K ⎜ u ⎟ (11.17) xx⎝ 0 2 dx ⎠

11.7 AN INTRODUCTION TO SANDWICH STRUCTURES

from which it is seen that the interfacial shear stress distribution function qx(x) is proportional to the angular rotation dw/dx of the face-beam neutral axis.

pdxz pdx M T x x N+dN N q dx M+dM z x T+dT q dx z

Figure 11.4 Notation and sign convention for equilibrium element of the loaded face of a sandwich beam (two-parameter foundation model). Referring to Fig.11.4 for notation and sign convention, the three equilibrium equations for the elastically supported face beam can be written in the form dN F =⇒0 =−qp ∑ xxxdx

dT F = 0 ⇒ = −()q − p (11.18) ∑ z dx z z

dM t M = 0 ⇒ = T + f ()q + p ∑ dx 2 x x By application of the ordinary theory of bending of beams, and by use of the equilibrium eqs.(11.18), the following relations are obtained

2 3 dw M d w 11dM ⎛ t f ⎞ 2 =− ⇒ 3 =− =−⎜T +()qpxx + ⎟ ⇒ dx Df dx Dffdx D ⎝ 2 ⎠

4 dw 1 ⎛ dT t f ⎛ dqxxdp ⎞⎞ 1 ⎛ t f ⎛ dqxxdp ⎞⎞ =− +⎜ + ⎟ =−pq − +⎜ + ⎟ (11.19) 4 ⎜ ⎝ ⎠⎟ ⎜ zz ⎝ ⎠⎟ dx Df ⎝ dx 2 dx dx⎠ Df ⎝ 2 dx dx ⎠

From the last of eqs.(11.19) it is seen that if the interfacial normal and shear stress distribution functions qz(x), qx(x) are known in advance, then the governing differential equation for w(x) is formulated explicitly, and the solution can be obtained by direct integration. This integration procedure will give four integration constants which can be determined from the boundary conditions of the sandwich beam problem considered. The situation is, however, that qz(x), qx(x) are unknown, and the objective of the preceding derivations is to formulate and solve the differential equations determining these functions. First, the longitudinal normal strain εx at the interface between the loaded face and the core material is expressed as follows

t d ⎛ txt⎞ σ (, /2 ) 16⎛ M⎞ ffff⎜ ⎟ ε x (,x )= ⎜ux (, )⎟ ==+⎜ N ⎟ (11.20) 22dx ⎝ ⎠ EEtffff⎝ t ⎠

11.8 LOCALISED LOADS

From the first of eqs.(11.14) the following is derived

dq() x d ⎛ t f ⎞ x = K ⎜ux(, )⎟ (11.21) dx x dx ⎝ 2 ⎠

By combination of eqs.(11.20) and (11.21) followed by successive differentiation as well as introduction of the equilibrium eqs.(11.18) the following expressions appear

dq K ⎛ 6M⎞ dq2 K ⎛ 6 ⎞ xx=+⎜ N ⎟ ⇒ xx=++⎜42qp T⎟ ⇒ ⎜ ⎟ 2 ⎜ xx ⎟ dx Etff⎝ t f⎠ dx Etff⎝ t f ⎠

dq3 K ⎛ dq dp 6 dT ⎞ K ⎛ dq dp 6 ⎞ xx= ⎜42xx++⎟ =++−x ⎜42xxpq⎟ (11.22) 3 ⎜ ⎟ ⎜ ()zz⎟ dx Etff⎝ dx dx t f dx ⎠ Etff⎝ dx dx t f ⎠

By rearrangement, the latter of eqs.(11.22) can be written in the form

3 dqxxK dq xx62K K x dpx 32−=−−+()qpzz (11.23) dx Etff dx Etff Etff dx

Eq.(11.23) establishes a relationship between the two unknown interfacial stress distribution functions qx(x) and qz(x). A similar equation can be found by combining the latter of the force- displacement eqs.(11.14) with the latter of eqs.(11.19)

4 4 dq dw K ⎛ t f ⎛ dq dp ⎞⎞ q =−K w ⇒ z =−K =z pq − +⎜ xx + ⎟ (11.24) zz 4 z 4 ⎜ zz ⎝ ⎠⎟ dx dx Df ⎝ 2 dx dx ⎠

Rearrangement of eq.(11.24) gives

4 dq K t f K ⎛ dq dp ⎞ K zz+=q z ⎜ xx +⎟ + z p (11.25) 4 z ⎝ ⎠ z dx D f 2 D f dx dx D f

Eq.(11.25) is the second of the sought relations connecting the two unknown interfacial stress distribution functions. Thus, eqs.(11.23) and (11.25) constitutes a set of two coupled ordinary constant-coefficient differential equations expressed in the two unknown functions qx(x), qz(x).

In order to obtain two differential equations expressed solely in the two unknown functions qx(x) and qz(x), eq.(11.23) can be employed to eliminate qz(x) and its derivatives from eq.(11.25). Similarly eq.(11.25) can be employed to eliminate the derivatives of qx(x) from eq.(11.23). From the latter the following can be obtained

Et2 ⎛ dq3 4K dq ⎞ t dp q =−ff⎜ xx − x ⎟ ++⇒f x p z ⎜ 3 ⎟ z 6K x ⎝ dx Etff dx ⎠ 3 dx

dq4 Et2 ⎛ dq7 4K dq5 ⎞ t dp5 dp4 z =−ff⎜ xx − x ⎟ ++f x z (11.26) 4 ⎜ 7 5 ⎟ 5 4 dx 6K x ⎝ dx Etff dx ⎠ 3 dx dx

11.9 AN INTRODUCTION TO SANDWICH STRUCTURES

Similarly eq.(11.25) gives

dq 2D ⎛ dq4 K ⎞ 2 dp x =+f ⎜ zzq ⎟ −−⇒p x ⎜ 4 z ⎟ z dx Ktzf⎝ dx D f ⎠ t f dx

dq3 2D ⎛ dq6 K dq2 ⎞ 2 dp2 dp3 x =+f ⎜ zzz ⎟ −−z x (11.27) 3 ⎜ 6 2 ⎟ 2 3 dx Ktz f ⎝ dx Df dx⎠ t f dx dx

Inserting eqs.(11.26) and (11.27) into eqs.(11.23) and (11.25) yields a set of two ordinary non-homogeneous differential equations of seventh and sixth order respectively, where each of the two differential equations is expressed solely in one of the unknown interfacial stress distribution functions

7 5 3 4 5 dqxxxxdq dq dq 3αα2 dpz 2 ⎛ dpxx α1 dp ⎞ 7 +++=−−ααα2 5 1 3 0 4 ⎜ 5 −⎟ (11.28) dx dx dx dx222 t f dx ⎝ dx dx ⎠

6 4 2 2 3 dq dq dq ⎛ dp α ⎞ α 1t f ⎛ dp α dp ⎞ zzz+++=+ααααq ⎜ z 2 p ⎟ ++⎜ xx2 ⎟ (11.29) dx 6 2 dx 4 1 dx 2 01z ⎝ dx 2 42z ⎠ ⎝ dx 3 4dx ⎠

4Kx Kz α12α ααα2 =−,, 10 = = (11.30) Etff Df 4

The complete solution to the set of two ordinary non-homogeneous constant-coefficient differential equations given by (11.28) and (11.29) can be written in the general form

qx()=+ qx () qx () , qx () =+ qx () qx () (11.31) xx()()h x p zz( )h ( z ) p where subscript h denotes the solutions to the homogeneous parts of eqs. (11.28), (11.29) and subscript p denotes the particular solutions to eqs.(11.28) and (11.29).

The derivation of the homogeneous solution parts is straightforward and is given by Thomsen [8,10] to which the reader is referred for further details. It can be shown that the solution to the homogeneous part of eq.(11.28) can be represented in the form [8,10]

qx( )=+ AA cosh(φφ xA ) + sinh( xA ) + cosh( ξη x )cos( x ) ()x h 01 1 2 1 3 (11.32)

+++AxxAxxAxx4 sinh(ξ )cos( η )5 sinh( ξη )sin( )6 cosh( ξη )sin( ) where Aj (j = 0,...,6) are seven integration constants which have to be determined from the statement of the boundary conditions for the elastically supported loaded face. φ1, ξ and η are coefficients expressed in terms of α2, α1 and α0 as defined by eq.(11.30)

α 2 ⎛θ ⎞ ⎛θ ⎞ φ =+−SY,cos,sin ξ = R ⎜ ⎟ η = R ⎜ ⎟ (11.33) 1 322⎝ ⎠ ⎝ ⎠

11.10 LOCALISED LOADS

2 ⎛ SY+ α 2 ⎞ ⎛ 3 ⎞ -1 ⎛ 33()SY− ⎞ R =−3 ⎜ − ⎟ +−⎜ (),SY⎟ θ = tan ⎜ ⎟ ⎝ ⎠ 33⎝ 2 ⎠ ⎝ 2()−−SY −α 2 ⎠

SWDYWDDVW=+33,, = − =32 +

9272α α − α − α3 VW=−3αα2 , =12 0 2 12 54 From the first of eqs.(11.26) (omitting the non-homogeneous terms) it follows that the homogeneous part of the transverse normal stress distribution function, (qz(x))h, can be expressed in terms of the first and third derivatives of (qx(x))h

2 3 Et ⎛ dqx()x ()dq( x () x) ⎞ qx() =−ff⎜ h +α h ⎟ ⇒ ()z h ⎜ 3 2 ⎟ 6K x ⎝ dx dx ⎠

Et2 qx( )=−ff φαφ2 +AxAx sinh φ + cosh φ ()z h {()1 211()() 1 2 () 1 6K x

++Axxxx3 ()ΛΩ sinh(ξη )cos( ) cosh( ξη )sin( )

++Axxxx4 ()ΛΩ cosh()cos()ξη sinh ()sin() ξη

+−Axxxx5 ()ΛΩ cosh(ξη )sin( ) sinh( ξ )cos( η )

+−Axxxx6 ()ΛΩ sinh(ξη )sin( ) cosh ( ξ )cos( η ) } where

32 32 ΛΩ=−ξ 33ξη + α2ξ, =η − ηξ − α2 η (11.35) Thus, the general solutions to the homogeneous parts of the governing differential eqs.(11.28) and (11.29) have been established. What still needs to be done is to establish the particular solution parts (qx(x))p, (qz(x))p, and to determine the integration constants Aj (j = 0,...,6).

The determination of the particular solution parts of the governing eqs.(11.28) and (11.29) is accomplished by assuming suitable particular solution functions by “qualified guessing”, as was the case for the particular solution in the context of the classical Winkler foundation model discussed in the preceding chapter. The derivation of the particular solution functions is of course strongly influenced by the form of the external loads, as these load distribution functions px(x) and pz(x) determine the non-homogeneous terms appearing in the governing differential equations (see ref. [10]).

At this stage it seems appropriate to present some reflections concerning the results obtained, i.e., to give some comments on the solution obtained by application of the Winkler foundation model and the solutions obtained by application of the two-parameter foundation model. Considering at first the governing ordinary differential equations obtained by application of

11.11 AN INTRODUCTION TO SANDWICH STRUCTURES the Winkler hypothesis and the two-parameter foundation model respectively, it is observed that the order is four for the Winkler foundation formulation (see eq.(11.6); notice that qz(x) =

–Kz w(x)) and six for the two-parameter foundation formulation (see eq.(11.29)). This reduced order of the governing differential equation, for the Winkler foundation model as compared to the two-parameter model, of course facilitates the solution to the problem, i.e., the determination of qz(x).

Comparing the solutions given by eqs.(11.9) and (11.34) for the Winkler and two-parameter foundation models, respectively, it is seen that, although the expressions appear to be quite complicated (especially for the two-parameter model), the component parts of the solutions consist of simple products of hyperbolic and trigonometric functions. Thus, it is seen that the principal form of the two expressions for qz(x) is similar for the two foundation formulations. The most dominant difference between the two foundation models is that the two-parameter model includes the existence of an interfacial shear stress contribution qx(x). The governing differential equation for qx(x) is of order seven (see eq.(11.28)), and it is observed that the mathematical form of the solution (eq.(11.32)) is very similar to the solution for qz(x), i.e., it is composed of simple products of hyperbolic and trigonometric functions. From an application point of view, the principal difference between the Winkler and two-parameter foundation models is the number of integration constants which have to be determined from the boundary conditions for the loaded face of a specific problem (Winkler foundation Cj (j = 1,2,3,4); two- parameter foundation Aj (j = 0,...,6)).

11.1.3 Specification of boundary conditions

The determination of the seven integration constants Aj (j = 0,...,6) appearing in the solution obtained for the two-parameter foundation model is to be accomplished from the formulation of seven natural boundary conditions for the loaded face of a specific problem. Actually, it turns out that the formulation of these seven natural boundary conditions is not as straightforward and self-explanatory as is usually the case when classical beam theory is involved, and in order to elucidate this point the formulation of the necessary boundary conditions will be sketched for a special type of problem. The type of problem considered is the class of symmetrically supported as well as symmetrically loaded sandwich beams. Fig.11.5 shows the loaded face of such a sandwich beam, where only one half of the face need be considered due to the symmetry of the problem. The half length and the thickness of the loaded face are denoted by L and tf respectively (notation from Fig.11.1 is retained).

The upper surface of the loaded face can be subjected to shearing and transverse normal load distribution functions px(x), pz(x) which can be attributed arbitrary forms as long as the property of symmetry about x = 0 is retained. This means that px(x) must be an odd function of the x-coordinate (i.e., px(x) = –px(–x)), and pz(x) must be an even function (i.e., pz(x) = pz(– x)).

11.12 LOCALISED LOADS

p (x) z p (x) x

x,u z,w free end L

axis of symmetry

Figure 11.5 Symmetrically loaded face of symmetrically supported sandwich beam. Referring to Fig.11.5, six of the necessary boundary conditions for the loaded face can be expressed as follows

dw ⎡dw2 ⎤ ⎡dw3 ⎤ ⎡ ⎤ ==00, q , = 00, = ⎢ ⎥ []x x=0 ⎢ 2 ⎥ ⎢ 3 ⎥ ⎣ dx ⎦ x=0 ⎣ dx ⎦ xL==⎣ dx ⎦ xL

L L L L qdxpdxqdxpdx==, (11.36) ∫∫∫∫x x z z 00 00

The first and the second of eqs.(11.36) represent the conditions of symmetry about the centre of the beam span (i.e., about x = 0); the third and the fourth of eqs.(11.36) represent the conditions at the free end of the loaded face (M = T = 0 at x = L); and the last two of eqs.(11.36) represent the conditions of overall horizontal and vertical equilibrium, respectively, for the loaded face subjected to surface loads px(x) and pz(x).

The last of the seven boundary conditions is derived by combination of the second of eqs.(11.22) and the second of eqs.(11.19). In the derivation of eqs.(11.22) an intermediate result, relating qx(x) and its second derivative to the transverse shear resultant T(x) as well as the external shear load function px(x), was obtained. This relationship can be expressed in the following form involving the coefficient α2 defined by the first of eqs.(11.30)

dq2 α ⎛ 6 ⎞ 2t ⎛ dq2 ⎞ t x =++2 ⎜42qp T⎟ ⇒ T =−f x α q − f p (11.37) 2 ⎜ xx ⎟ ⎜ 2 2 x ⎟ x dx 4 ⎝ t f ⎠ 33α 2 ⎝ dx ⎠

Another intermediate result relating the third derivative of qz(x) to qx(x), T(x) and px(x) can be obtained from the second of eqs.(11.19) (notice that qz(x) = –Kz w(x))

3 3 dw 1 ⎛ t f ⎞ dqzzK ⎛ t f ⎞ 3 =−⎜T +()qpxx + ⎟ ⇒ 3 =++⎜T ()qpxx⎟ (11.38) dx Df ⎝ 2 ⎠ dx Df ⎝ 2 ⎠

By inserting T(x) expressed by the latter of eqs.(11.37) into eq.(11.38) an expression is derived, relating qx(x) and its second derivative to the third derivative of qz(x) and to the external shear load distribution function qx(x)

3 2 dqz 2t f α1 dqx t f t f 3 −+=2 αα11qx px (11.39) dx 366α2 dx

11.13 AN INTRODUCTION TO SANDWICH STRUCTURES

The last boundary condition (seventh), supplementing the already established boundary conditions for the class of problems considered (given by eqs.(11.36)), is then specified by requiring fulfilment of eq.(11.39) at some position, say x = 0 (centre of beam span)

3 2 ⎡dqz 2t f α 1 dqx t f ⎤ ⎢ 3 −+2 α 1qx ⎥ = 0 (11.40) dx 36dx ⎣ α 2 ⎦ x=0

This completes the formulation of the full system of boundary conditions necessary to determine the seven integration constants Aj (j = 0,...,6) for the considered class of problems.

It should be noted here that the determination of the seven integration constants Aj (j = 0,...,6), even though straightforward in principle, in practice is too complicated to perform by solely analytical means. This means that, even though a closed form solution has been achieved, the constants appearing in the expressions have to be determined by use of some appropriate numerical algorithm for solving systems of linear algebraic equations.

11.1.4 Superposition with classical sandwich beam theory The two preceding sections have been concentrating on the description of the behaviour of the loaded face of a sandwich beam subjected to arbitrary surface shearing and transverse normal loads. The content of these chapters has been focused on the derivation and solution of the governing equations for the Winkler and two-parameter foundation models respectively, i.e., quite a lot of attention has been placed on the mathematics involved (especially the case for the two-parameter model). The mathematical side of the problem is of course important (and necessary as well) in order to quantify the mechanical behaviour, but it is not very important for the reader to go into the mathematical description, as well as the mathematical differences between the two foundation models, in great detail. What is really important here, however, is for the reader to recognise the physics of the suggested approach, i.e., to grasp the idea of considering the mechanical behaviour of the loaded face of a sandwich beam as being describable in terms of some appropriate elastic foundation model. It is also important to notice that the suggested two-parameter elastic foundation model can be considered as the simplest possible extension of the classical Winkler foundation model:

Winkler model: corresponds to considering the supporting medium as continuously distributed linear tension/compression springs.

Two-parameter model: corresponds to considering the supporting medium as continuously distributed linear tension/compression and shear springs.

Thus, it is recognised that the principal difference between the two models is that the two- parameter foundation model accounts for the possibility of transferring shear stresses between the loaded face and the supporting material (core).

In order to obtain a complete solution to a specific problem of a sandwich beam subjected to concentrated loads, it is necessary to investigate the overall bending and shearing effects, i.e., to establish an overall solution supplementing the local bending solution presented in the preceding chapters. The simplest way to establish an overall solution to a specific sandwich beam problem is to apply the classical sandwich beam, plate and shell theory as given earlier.

11.14 LOCALISED LOADS

The overall solution could also be a finite element solution obtained by use of a simple sandwich beam or plate element.

Assuming that an overall solution has been established (by use of classical sandwich theory or

FEM), the approximate total lateral displacement wtotal of the loaded face can be obtained by superposition of the overall solution and the approximate local solution derived earlier

wtotal()x =+w overall ()x w local ()x (11.41)

The local lateral displacement wlocal is defined by eqs.(11.8) and (11.9) (Winkler foundation model), or by the latter of eqs.(11.14) together with eqs.(11.31) and (11.34) (two-parameter model). The local displacement component wlocal only defines the displacement of the “elastic line” of the loaded face, i.e., the superposition of the overall and local displacement components suggested by eq.(11.41) will only describe the local bending effects in the loaded face itself and at the interface between the loaded face and the core of a sandwich beam. It is not possible to decide, in explicit terms, the decay through the thickness of the core material of the lateral displacements induced by local bending of the loaded face.

The stress field induced by overall bending and shearing together with local bending of the loaded face is another subject of major interest. The longitudinal normal stress distribution in the loaded face can be written as

σ f,, total(,)x z =+σ f overall (,)x z σ f , local (,)x z (11.42) where σf,overall(x,z) represents the overall solution part, and σf,local(x,z) is expressed in terms of the elastic response function qz(x) defined by eqs.(11.1), (11.8) and (11.9) for the Winkler foundation model, or by eqs.(11.31) and (11.34) for the two-parameter foundation model

2 Mlocal z Ezz dqz t fft σ f, local (,)xz =⇒E f σ f, local (,)xz =≥≥−2 , z (11.43) Df Kz dx 22

The lateral coordinate z referred to in eq.(11.43) corresponds to the local face-beam coordinates shown in Fig.11.2. The transverse normal stress component σz is not defined in the context of classical sandwich beam theory, but from the condition of continuity across the interface between the loaded face and the core, it is clear that the transverse normal stress component at the interface (denoted by σint) is given by qz. Thus, the interfacial transverse normal stress component can be written

σint,total ()x = qz ()x (11.44)

Similarly the total shear stress τint,total(x) at the interface between the loaded face and the core material can be expressed as

τint,total()x =+τc, overall ()x τ int,total ()x (11.45) where τc,overll(x) represents the overall solution part, and τint,total(x) is defined exclusively by the elastic shearing response function qx(x) (only defined for the two-parameter foundation model) given by eqs.(11.31) and (11.32)

(11.46) τint,local()x = q x ()x

11.15 AN INTRODUCTION TO SANDWICH STRUCTURES

As mentioned earlier for the lateral displacements, the superposition of the overall and local stress field components only describes the state of stress in the loaded face and in the interface between the loaded face and the core material. This last statement indicates that it is implicitly assumed that the behaviour of the lower face can be described completely in terms of the classical sandwich beam theory (analytical or FEM solution), i.e., the decay of the local bending effects through the thickness of the core material is assumed to be complete. Whether this last statement is true, that is, whether the classical solution obtained for the lower face is realistic may be questioned. An experimental investigation presented in ref. [12] has shown that local bending may occur in the lower face as well. However, it was shown that the results obtained (using the two-parameter foundation model) for the loaded (upper) face of a sandwich beam were very good [12]. From a practical engineering point of view it is not a serious drawback that the local bending behaviour of the lower face is not included in the analysis, as the loaded face and the interface between the loaded face and the core (where good results are obtained) are, by far, the most severely loaded parts of a sandwich beam subjected to localised loading.

11.1.5 Range of applicability – Winkler vs. two-parameter foundation model The introduction of the elastic foundation analogy in the context of local bending of sandwich panels subjected to localised loads, has been carried out in order to present a simple method of obtaining detailed information about the displacement and stress fields induced locally (loaded face and interface between loaded face and core material) as a result of localised lateral loads. The basic idea behind the application of an elastic foundation analogy is, as described earlier, to approximate the supporting medium by continuously distributed linear tension/compression and shear springs (only included in the two-parameter foundation model). The quality of this approximation, however, is strongly dependent on the quality of the basic assumptions of the elastic foundation models employed.

The basic assumptions of the Winkler and two-parameter elastic foundation models are given by eq.(11.1) and eq.(11.14), respectively, which state that the elastic response of the supporting medium, at a given position specified by the longitudinal coordinate x (see Fig.11.2), is directly proportional to the displacements of the lower boundary of the loaded face. This implies that the foundation moduli Kz and Kx (Kx only defined for the two-parameter foundation model) are assumed to be constants, which can be related to the elastic coefficients of the core material as well as the elastic coefficients and the geometrical characteristics of the loaded face. Suggestions for such expressions are given by eqs.(11.2), (11.3) for the transverse modulus Kz (in accordance with ref. [13]), and by eq.(11.15) for the foundation shear modulus Kx. Whether assumptions of the type described are generally justified for typical core materials used for structural sandwich structures is a matter of discussion, and the most important questions posed are:

(i) Is the assumption of “constant-value” foundation moduli Kz, Kx justified ? (ii) Is the assumption of linear elastic material properties realistic ? (iii) Can the elastic foundation analogy be used to model all types of core materials typically in use for structural sandwich constructions ?

11.16 LOCALISED LOADS

(iv) What is the difference between the Winkler and two-parameter foundation models from a practical engineering point of view, i.e., when is it recommended to use one model instead of the other ?

Concerning the first of the questions posed, the answer demands some additional considerations. The assumption of constant value elastic foundation moduli Kx, Kz, is obviously not generally justified, but can, at the most, be legitimate if some, as yet unspecified, further restrictions are imposed on the class of problems for which the proposed elastic foundation analogy can be used successfully.

It can be shown [1,8] that the key to the specification of the necessary restrictions, on the applicability of elastic foundation models is the wave-length, λ, of the deflections of the elastically supported loaded face. It turns out that the elastic foundation models become inadequate for deformations of short wave-length. This is caused by the circumstance that the shearing deformations of the supporting medium becomes increasingly important as the wave- length of the deflections decreases. Thus, it is recognised that the simplifications introduced by the application of an elastic foundation model, instead of a continuum model (general theory of elasticity problem) of the core material, are not justified for deformations of short wave-length. A natural question in this context is whether it is possible to give an explicit quantification of the concept of deformations of short wave-length. Unfortunately the answer to this question turns out to be negative if the term quantification means definition of very precise bounds on the applicability of the foundation models. This is caused by the fact that the wave-length of the deflections is related to the characteristic material and geometrical parameters of the problem in a very complex manner. For practical sandwich panels, however, the bounds imposed by the vaguely formulated concept of deformations of short wave-length are not likely to be active, since the typical face thicknesses (0.5 mm < tf < 10.0 mm ), and the typical modular ratios (25 < Ef /Ec < 1500), will usually ensure sufficiently large deflection wave-lengths to ensure the justification of the simple elastic foundation approach instead of using a very complex two or three dimensional continuum formulation.

The second question posed was whether it is reasonable to assume linear elastic properties of the constituent materials of typical structural sandwich panels. Without going into details about this topic, it can be said that the constituent materials of typical structural sandwich structures may well behave far more non-linearly than is usually expected, and this is especially so for sandwich panels with polymer based fibre-reinforced faces and polymeric foam core materials. The non-linear behaviour observable for such sandwich panel combinations typically includes viscous, plastic as well as hygrothermal effects. However, the service conditions, under which structural sandwich panels are employed, are usually arranged to be well within the safe domain specified by the proportional limit of the constituent materials. Thus, it is expected that the suggested approach, based on linear elastic assumptions, is capable of giving a fairly good estimate of the magnitude of the stress concentrations induced by local bending effects.

The third question posed was whether the elastic foundation approach can be used to model all types of core materials in use for structural sandwich panels, and in order give a proper answer to this question it is necessary to include some additional comments. There are three

11.17 AN INTRODUCTION TO SANDWICH STRUCTURES major types of core materials used for structural sandwich panels: polymeric foam core materials, honeycomb core materials and balsa cores. The first and third of the listed core material types can be considered as being homogeneous materials (at least from the macroscopic point of view which is usually relevant in the context of engineering applications), since the cellular structure of these materials is usually at least one order of magnitude smaller than the characteristic dimensions of the sandwich panel considered (face thickness, core thickness, width, span etc.). Thus, it seems reasonable to model the foam and balsa core materials respectively by use of a continuum or an elastic foundation formulation as suggested. With respect to the honeycomb type cores, it is a bit more complicated to present an answer to the question posed. Honeycomb cores are discrete by nature, i.e., they do not support the faces of a sandwich panel continuously but rather in a discrete manner along the edges of the honeycomb cells. This means that continuum formulations or continuous spring support formulations may be incapable of modelling the core material realistically. Whether this is the case is determined from the cell size of the honeycomb cell structure: if the characteristic dimensions (width and edge length of cells) of the cells are small compared to the thickness of the faces, continuum or continuous spring support formulations will be capable of giving a fair estimate of the mechanical behaviour from an engineering point of view. If, on the other hand, the cell size of the honeycomb structure is comparable to, or even larger than, the face thicknesses, it is not likely that good results can be obtained, as the faces will tend to act like plates within the boundaries of each cell in the honeycomb structure.

The above mentioned considerations deal with the justification of elastic foundation models in general as opposed to elastic continuum models, with a discussion about the assumption of linear elastic material properties and with a discussion about the capability of continuum or elastic foundation formulations to describe the mechanical behaviour of various types of core materials. No distinction between the very simple Winkler foundation model and the more complicated two-parameter model has been introduced in the discussion.

It is clear that none of the foundation models will give satisfactory answers for deformations of short wave-length, as it can be shown (refs.[1,8]), that it is not possible to select constant values of the foundation moduli Kz and Kx which are appropriate for displacements of any wave-length. The reason for this is, as described earlier in this section, the fact that the shearing deformations of the core material become very influential for deformations of short wave-length. This last statement, however, makes it possible to distinguish between the “quality” of the Winkler and two-parameter foundation models, because the latter actually does take into account the possible shearing interaction (although in very simple form) between the loaded face and the core material of a sandwich beam subjected to strongly localised lateral loads. Thus, it is recognised that the two-parameter foundation model is superior to the Winkler foundation model, as it predicts the existence of the interfacial shear stress distribution function qx(x), although the formulation of the shearing interaction effects is not sophisticated enough to handle problems characterised by displacements of very short wave-length. The application of the two-parameter foundation model instead of the simpler, from a mathematical point of view, Winkler foundation model can be recommended for structural sandwich panels characterised by the following approximate relationship

11.18 LOCALISED LOADS

Ef λ ≈<533. t f 3 50- 60 mm (11.47) Ec where λ is the wave-length of the elastic deformations (elastic line) as defined by eq.(11.11) for the simple Winkler foundation model. The guidelines defined by the inequality (11.47) should be considered as only a rough estimate of the recommended bounds of applicability of the two foundation models under consideration. The values of λ suggested by the inequality (11.47) as an appropriate application separation between the Winkler and two-parameter foundation models, however, will usually ensure that the peak value of the locally induced interface shear stress component τint(x) is negligible compared to the locally induced interface peel stress component σint(x). 11.2 Discussion: Application, Results and Parametric Effects 11.2.1 Application of the method for solving engineering design problems In order to illustrate the applicability of the theory presented and discussed in the preceding chapters an approach is suggested for using the local bending analysis results as a sort of adjustment to calculations based on classical sandwich beam theory. In order to convince the reader that such an approach can be justified from a physical point of view, it is necessary to focus the attention on the structure of the local bending solutions obtained by either of the two elastic foundation formulations.

As mentioned earlier, the component parts of the derived solutions (see eq.(11.9) and eqs.(11.32) and (11.34)) consist of products of simple trigonometric and hyperbolic functions, implying that the solutions exhibit a wavy harmonic nature as well as an exponential (increase or decay) dependency of the longitudinal coordinate. The exponential terms appearing in the solutions ensure a very steep decay of the solutions with increasing longitudinal coordinate. Thus, it is observed that the very nature of the elastic foundation solutions ensures that the local bending effects can be looked upon as local disturbances of the ideal stress and deformation states predicted from classical sandwich theory.

The following steps are suggested for the approximate solution procedure:

(i) The sandwich beam (or plate) considered is analysed by use of classical sandwich theory or by a rough finite element model using sandwich plate or beam elements. Thus, an overall solution is obtained.

(ii) The areas of the considered sandwich beam (or plate), where localised loads are introduced, are analysed by use of either of the elastic foundation formulations, following the general recommendation given by eq.(11.47). The superposition of the local bending solution and the overall solution can be carried out if the following conditions are satisfied: (a) The local bending effects induced by each of the localised loads applied upon the sandwich beam (or plate) are not allowed to interfere with the bending effects induced by the other localised loads or to interfere with boundaries (if any) of the considered sandwich beam. The meaning of this is that the local bending solutions should exhibit a sufficiently steep decay, so that the local bending solutions or edge effects will not

11.19 AN INTRODUCTION TO SANDWICH STRUCTURES

influence each other. If this condition is not fulfilled the suggested simple approach (superposition) will not be valid. To determine the extension of the zone, in which the local bending effects are of importance, the so-called decay length x*, known from the theory of bending of circular cylindrical shells, is defined by use of eqs.(11.10) and (11.11)

∗∗λ 4Df E f 3 3 x ==π ⇒ xt ≈266. f (11.48) 2 Kz Ec

If the distance from the point of introduction of some localised load to other local disturbances (external loads or boundaries) exceeds the decay zone distance x*, then the superposition suggested can be carried out without any problems, as the local bending effects (disturbances) do not influence each other significantly.

(b) If the results obtained by the suggested superposition are to be capable of giving a fairly accurate description of the state of stress present in the lower face of the considered sandwich beam (or plate), the thickness of the core material (see Fig.11.1) should be larger than the decay zone distance x*. If this condition is fulfilled, no significant local bending effects are transmitted from the loaded face to the unloaded face (through the core material), and the overall solution obtained by classical sandwich theory or a simple FEM-solution will give adequately accurate results. If the condition is not fulfilled, however, the results obtained for the lower face are not very good, but as the results for the loaded face and the core-material adjacent to the loaded face are very good for this case as well (ref. [12]), this is no serious problem from a practical point of view, since the loaded face is the most severely loaded part of the structure.

11.2.2 Example The actual solution of a problem is illustrated by considering the simple case of a unit width sandwich beam in 3-point bending (point load). The geometrical notations for the problem are adopted from Fig.11.1 and the geometry, material data and the point load are as follows:

GEOMETRY: L = 500.0 mm tf = 4.0 mm tc = 50.0 mm.

FACES: Ef = 15.0 GPa (corresponding to 50 vol. % E-glass/epoxy). 3 CORE: Ec = 0.1 GPa νc = 0.35; (PVC-foam, ρc = 100.0 kg/m ). POINT LOAD: P = 100.0 N/mm (force per unit width)

Following the suggested two-step procedure the first task is to analyse the sandwich beam considered using classical sandwich beam theory. When this is accomplished the local bending analysis can be carried out, and in order to check whether the suggested method of superposition is valid, the decay zone distance x* is evaluated using the approximate expression given by eq.(11.48). x* is found to be x* ≈ 57 mm , and, as x∗ ≈ 57 mm < L = 500 mm , it is concluded that the method will give sufficiently accurate results.

For the present example, the local bending solution could be obtained by use of the simple Winkler foundation model because the wave-length of the elastic response is λ ≈ 113 mm (see the guide lines in eq.(11.47)), but the two-parameter foundation model is used anyway.

11.20 LOCALISED LOADS

Figure 11.6 Lateral deflection of sandwich beam in 3-point bending, where only one half of the beam span is shown due to the symmetry of the problem.

Fig.11.6 shows the lateral deflection woverall obtained from classical sandwich beam theory, and the lateral deflection of the loaded face wtotal obtained by superposition of the overall and local solution parts. It is observed that the midpoint (x = 0) deflection of the sandwich beam is about 35 mm , and it is clear that the local bending effects really are very localised, as the two curves shown are identical except very near the point of load application (x = 0).

Fig.11.7 shows the distribution of longitudinal normal stresses σf in the upper and lower fibre of the loaded face, respectively. From Fig.11.7 it is observed that the local bending effects are of significant importance near the point of external load application (x = 0), as severe stress concentrations are induced in this area. The curve paths in the right hand side of Fig.11.7, which are straight lines, correspond to the classical sandwich beam solution, and by prolonging these straight lines to their intersection with the ordinate axis, the peak stresses obtained from classical sandwich theory are found to be: upper boundary: σf,peak ≈ −120 MPa; lower boundary: σf,peak ≈ −105 MPa. Due to the local bending of the loaded face, however, the real stress state present is very much different from the results obtained by classical sandwich theory. At the upper boundary of the loaded face a compressive state of stress is present, and the peak value encountered (x = 0) is about −290 MPa. Thus, it is recognised that the peak stresses found in the upper fibre are about 2.4 times larger than predicted by ordinary sandwich beam theory. At the lower boundary of the loaded face a tensile state of stress is present in the regions very near the centre of the beam span, whereas the stress state is compressive some distance away from x = 0; say for x > 5-10 mm . The change of sign of the longitudinal normal stress component at the lower boundary of the loaded face can be attributed to the significant local bending contribution. The peak tensile stress in the lower fibre is σf ≈ 62 MPa.

11.21 AN INTRODUCTION TO SANDWICH STRUCTURES

Figure 11.7 Distribution of longitudinal normal stresses σf in the upper and lower fibre of the loaded face.

Fig.11.8 shows the distribution of transverse normal and shear stresses, σint and τint, at the interface between the loaded face and the core material of the sandwich beam considered. The overall tendency observed is that severe stress concentrations (see eq.(11.44)) are present at the centre of the beam span (x = 0) and in the zones very near the centre of the beam span.

The wavy harmonic pattern as well as the characteristic decay of σint with increasing value of x is clearly observed, and the decay zone distance x* determined earlier to x*≈ 57 mm corresponds to the positive (tensile) peak σint-value (σint ≈ 0.12 MPa), which is only about 4% * of the overall peak value encountered at x = 0 (σint ≈ –2.75 MPa). For x > x the σint-values fades out and tends to zero. Thus, it is recognised that the presence of transverse normal stresses is truly a local phenomenon.

Fig.11.8 also shows the longitudinal distribution of the total interfacial shear stress τint (see eq.(11.45)), and as expected (due to the symmetry of the considered problem) no shear stresses are present at the centre of the beam span (x = 0). The overall tendency is that the shear stress component τint builds up and attains its maximum, τint,max ≈ 1.0 MPa, adjacent to the centre of the beam span after which the τint - value approaches a constant value, τint ≈ 0.9 MPa, which equals the constant shear stress predicted to be present in the sandwich beam by classical sandwich beam theory. For the example considered it is seen from Fig.11.8 that the peak interfacial shear stresses induced by local bending are not very significant in magnitude compared to the shear stresses predicted by the classical sandwich beam theory. This was expected in advance as the guide lines, specified by the inequality eq.(11.47), indicated that the Winkler foundation model would give sufficiently accurate results, i.e., the shearing interaction between the loaded face and the core-material is not important in the present example.

11.22 LOCALISED LOADS

Figure 11.8 Longitudinal distribution of σint and τint at the interface between the loaded face and the core material.

11.2.3 Parametric effects The local bending effects induced in sandwich panels by localised loading are influenced especially by two parameters: the modular ratio Ef /Ec, and the face-thickness tf. Other parameters, such as the geometrical parameters L and tc in the earlier example given of a sandwich beam in three point bending also influence the stress distribution, but they only exert influence on the overall bending and shearing of the specific sandwich panel considered, i.e., their influence is included in the description supplied by classical sandwich theory.

In order to illustrate the influence of the modular ratio Ef /Ec and the face-thickness tf on the interfacial stress components, some results obtained from a brief parametric study, based on the two-parameter foundation model, will be presented. The problem considered, once again, is a sandwich beam in three point bending (see Fig.11.1), but this time only the loaded face elastically supported by the core material is considered, as only the local bending effects are of interest. The base line geometry and the material parameters, from which all variations are made, are assumed to be as quoted in the preceding section.

Starting with the effect of altering the modular ratio Ef /Ec, the distribution of the interfacial transverse normal stress component σint is shown in Fig.11.9 for three different values of Ef /Ec

(Ec is kept fixed Ec = 0.1 GPa). The thickness of the face is fixed tf = 4.0 mm. It is observed that the lower the value of the modular ratio Ef /Ec, the higher the peak value of σint (at x = 0). Furthermore, it is observed that the wave-length of the elastic response increases significantly as Ef /Ec is increased, and this observation is in close agreement with the relation eq.(11.47) which gives an estimate of the wave-length of the elastic line (λ) expressed in terms of Ef /Ec and tf.

11.23 AN INTRODUCTION TO SANDWICH STRUCTURES

Figure 11.9 σint vs. x for three different values of Ef /Ec (Ec = 0.1 GPa, tf = 4.0 mm ).

Fig.11.10 shows the longitudinal distribution of the interfacial shear stress component τint (interface between loaded face and core material) for the same three different modular ratios used in Fig.11.9. It should be emphasised that the shear stress component τint shown in Fig.11.10, only represents the shear stress contribution induced by local bending of the loaded face, i.e., the contribution given by eq.(11.46). From Fig.11.10 it is observed that the peak value of τint decreases dramatically as Ef /Ec is increased. Furthermore, it is seen that the wave- length of the shearing response increases with increasing Ef /Ec-values.

Figure 11.10 τint (interface shear stress induced by local bending) vs. x for three different values of Ef/Ec (Ec = 0.1 GPa, tf = 4.0 mm ).

11.24 LOCALISED LOADS

Figure 11.11 σint vs. x for three different values of tf (Ef /Ec = 150.0 and Ec = 0.1 GPa).

Having discussed the effects of altering the modular ratio Ef /Ec, the next issue is to evaluate the influence of altering the thickness tf of loaded face. Fig.11.11 shows the longitudinal distribution of the interface transverse normal stresses σint for three different thicknesses of the loaded face (tf = 1.0, 4.0 and 16.0 mm ), and Fig.11.12 shows the distribution of interface shear stresses τint for the same three tf-values. For the calculations leading to the results shown in Figs. 11.11 and 11.12 the modular ratio has been set Ef /Ec = 150.0 (Ec = 0.1 GPa).

The overall tendencies shown by Figs. 11.11-12 are similar to those shown by Figs. 11.9-10.

Thus, it is recognised that the peak values of σint and τint, respectively, decrease significantly as the thickness tf of the loaded face is increased, i.e., as the flexural rigidity of the face is increased. Furthermore, it is observed that the wave-length of the elastic response functions increases with increasing tf-values, just as predicted by eq.(11.47).

11.25 AN INTRODUCTION TO SANDWICH STRUCTURES

Figure 11.12 τint (local bending contribution) vs. x for three different values of tf (modular ratio fixed Ef /Ec = 150.0, Ec = 0.1 GPa). The plots shown in Figs. 11.9-12 illustrate that the local bending effects are strongly influenced by the modular ratio Ef /Ec as well as the thickness tf of the loaded face. The results shown, which represent the interfacial stress distributions, indicate that the local bending phenomenon really is very local, and also reveal that it is possible to extract the most important results by plotting the peak values of the interface stresses σint and τint, as well as the maximum bending stress σf in the loaded face, against the modular ratio Ef /Ec and thickness tf.

With the aid of such graphs (design-charts), representing locally induced peak stresses as functions of the modular ratio and the face-thickness, it will be possible to adjust engineering design calculations, based on the ordinary theory of structural sandwich panels (analytical or FEM calculations), in the sense that very good estimates of the stress concentrations induced by local bending effects can be accomplished. Thus, it is clear that by using such design- charts it is possible to design structural sandwich panels against local bending (indentation problems) with very little computational effort involved.

11.26 LOCALISED LOADS

11.3 Concluding Remarks A method of approximate analysis of the local stress and displacement fields in the near vicinity of strongly localised external loads applied to sandwich beams (or plates in cylindrical bending) has been presented. The formulation presented is based on the assumption that the deflection of the loaded face against the other (not loaded) face can be modelled properly by use of some elastic foundation model. Furthermore, it is assumed that the faces as well as the core material of the sandwich beams considered behave linearly elastically.

The principle of the solution procedure is very simple: the local stress and displacement fields derived by application of an elastic foundation model are superimposed on the overall stress and displacement fields obtained by use of, for instance, classical sandwich beam theory, and an approximate complete solution, which includes overall bending and shearing effects as well as local bending effects, is obtained.

Two elastic foundation models have been discussed: the classical Winkler foundation model, and the two-parameter foundation model (described in ref. [8-10,12]). The use of the classical Winkler foundation model corresponds to the modelling of the supporting medium as continuously distributed linear tension/compression-springs (equivalent to a liquid base), i.e., the shearing interaction between the loaded face and the core material is not included in the model. To include the possibility of such shearing interaction, the two-parameter foundation model has been introduced, and the use of this model corresponds to the modelling of the supporting medium as continuously distributed tension/compression and shear springs.

The applicability of the two models has been discussed, and the application of the two- parameter model instead of the simpler (from a mathematical point of view) Winkler foundation model is recommended for elastic responses with wave-lengths below 50-60 mm (see eq.(11.47)). For elastic wave-lengths exceeding 50-60 mm, it turns out that the two models give similar results, meaning that the shearing interaction between the loaded face and the core material becomes negligible.

As mentioned it is possible for specified types of external loading and support conditions to construct graphical design-charts, representing locally induced peak stresses as functions of the modular ratio Ef /Ec and the face-thickness tf. Such design-charts can be used to estimate the severity of the stress concentrations induced by local bending, and among the type of questions which can be answered are: (i) Considering a sandwich with given face and core materials with given strengths, given face thickness and core thickness, what magnitude of external load could be applied ? (ii) Considering a sandwich with given face material with given strength, given core material with given strength, what face thickness is required to prevent failure of the loaded face or the core ? (iii) Considering a sandwich with given face and core materials with given strengths, given face and core thicknesses, given total external load, over how large an area should the external load be distributed to prevent the maximum allowable stresses (core and face) from being exceeded ?

11.27 AN INTRODUCTION TO SANDWICH STRUCTURES

The construction and application of graphical design-charts is introduced and discussed for three specific load cases in the chapter “Localised Loads” contained in “Handbook of Sandwich Constructions” [15].

As mentioned, the approximate solution procedure presented is only strictly valid for sandwich beams and sandwich plates in cylindrical bending. Plate analysis in general, however, cannot be accomplished directly with the solutions presented so far, but general sandwich plate solutions (rectangular sandwich plates with orthotropic faces, circular sandwich plates with isotropic faces), based on an extended version of the two-parameter elastic foundation approach, will be presented in the near future.

Acknowledgement The author wishes to thank the Danish Technical Research Council (STVF), under the “Programme of Research on Computer Aided Engineering Design”, for the financial support received during the period of the work presented in this chapter.

References [1] Allen, H.G., Analysis and Design of Structural Sandwich Panels, Pergamon Press, Oxford, 1969.

[2] Plantema, F.J., Sandwich Construction, John Wiley & Sons, New York, 1966.

[3] Stamm, K. and Witte, H., Sandwichkonstruktionen (in German), Springer-Verlag, Wien, Austria, 1974.

[4] Weissman-Berman, D., Petrie, G.L. and Wang, M.-H., “Flexural response of foam- cored sandwich panels”, The Society of Naval Architects and Marine Engineers (SNAME), November 1988.

[5] Meyer-Piening, H.-R., “Remarks on higher order sandwich stress and deflection analyses”, Proc. First International Conference on Sandwich Construction, Eds.: K.- A. Olsson and R.P. Reichard, Stockholm, 1989, EMAS, Ltd., UK, pp 107-127.

[6] Frostig, Y., Baruch, M., Vilnay, O. and Sheinman, I., “Bending of nonsymmetric sandwich beams with transversely flexible core”, ASCE Journal of Engineering Mechanics, Vol. 117, No. 9, 1991, pp 1931-1952.

[7] Ericsson, A. and Sankar, A.V., “Contact stiffness of sandwich plates and applications to impact problems”, Proc. Second International Conference on Sandwich Construction, Eds.: D. Weissman.Berman and K.-A. Olsson, Gainesville, 1992, EMAS, Ltd., UK, pp 139-159.

11.28 LOCALISED LOADS

[8] Thomsen, O.T., “Flexural response of sandwich panels subjected to concentrated loads”, Special Report No. 7, Institute of Mechanical Engineering,, Aalborg University, Denmark, May 1991.

[9] Thomsen, O.T., “Analysis of local bending effects in sandwich panels subjected to concentrated loads”, Proc. Second International Conference on Sandwich Construction, Eds.: D. Weissman.Berman and K.-A. Olsson, Gainesville, 1992, EMAS, Ltd., UK, pp 417-440..

[10] Thomsen, O.T., “Further remarks on local bending analysis using a two-parameter elastic foundation model”, Report No. 40, Institute of Mechanical Engineering, Aalborg University, Denmark, March 1992.

[11] Allison, I.M., “Localised loading of sandwich beams”, Ninth International Conference on Experimental Mechanics, Technical University of Denmark, Copenhagen, Denmark, 20-24 August, 1990, pp 1604-1612.

[12] Thomsen, O.T., “Photoelastic investigation of local bending effects in sandwich beams”, Report No. 41, Institute of Mechanical Engineering, Aalborg University, Denmark, April 1992.

[13] Hétenyi, M., “Beams on elastic foundations”, The University of Michigan Press, Ann Arbor, Michigan, 1946.

[14] Kerr, A.D., “Elastic and viscoelastic foundation models”, Journal of Applied Mechanics, Sept. 1964, pp 491-498.

[15] Handbook of Sandwich Constructions: localised loads (Edited by Dr. Dan Zenkert); Nordic Fund for Industrial Development, to be published in 1993.

11.29 CHAPTER 12

SANDWICH AND FEM

This section contains some practical aspects of analysis of sandwich constructions using the Finite Element Method - FEM. The theory or implementation of special sandwich elements is not emphasised as that is a specialist topic in a text book on FEM. The section discusses some special considerations due to the nature of sandwich constructions. There is a quite thorough description of how the fundamental theory of sandwich beams can be used to derived special sandwich beam elements and also what kinds of problems that can occur in such elements. There is also a more brief description of sandwich plate elements, whereas the theory of sandwich shell finite elements are only discussed. A thorough knowledge of FEM is required as well as a fundamental understanding of the mechanical behaviour of sandwich structures before anyone can accurately and efficiently use FE programs for sandwich analysis and design.

12.1 General Remarks on FEM The finite element method is the most general and versatile engineering tool for analysis and design of structures [1]. Commercially available FE programs such as NASTRAN, ANSYS, ABAQUS, NISA etc. can treat very complex problems. The key to a successful use of FEM is, however, that all of the following types of errors are kept under control:

(i) Modelling error. The word “modelling” refers to the mathematical modelling of the real physical problem. Hence, the user must have a basic understanding of the physics of the problem to be solved. Adequate constitutive models for the materials have to be employed (linear elastic; elasto-plastic; visco-elastic; visco-plastic; rate dependent etc.), and an appropriate level of analysis must be selected (small or large displacements; small or large strains; static or dynamic analysis etc.). Apart from engineering experience, skill and intuition of the user, a good way of certifying the FE-data is to compare computer results with analytical ones.

(ii) Discretisation errors. Due to the approximate nature of FEM, errors will emerge in the results due to the use of a finite number of elements. The fineness of the FE mesh is crucial to the outcome of the analysis. In areas of high stress gradients, smaller elements are required than in areas of little variation. Since a too fine mesh might be too costly to run through the computer, the user is faced with an optimisation problem on how to choose the FE mesh. Methods of automatic re-meshing by the FE program during computations are being developed but are still not generally available in commercial codes (“adaptive” FE analysis).

12.1 AN INTRODUCTION TO SANDWICH STRUCTURES

Therefore, the experience and skill of the user and numerical and/or analytical verifications will be required to ensure reliable results. This means that the same problem should be analysed by two or more different meshes with successively finer elements, and that comparisons with closed-form analytical solutions should be carried out if possible.

Another error source is the solution of non-linear problems as a set of linear ones, often with iterative corrections of the equilibrium equations [1].

(iii) Computer errors. When the computer solves the sets of equations that result from the FE discretisation, numerical handling (truncation) errors may emerge. A good way of checking such errors is to investigate if the entire structure and parts of the structure are kept in equilibrium under applied loads and calculated support forces and/or section forces. If so, the solution is probably satisfactory from a numerical point of view. If not, higher computer accuracy or remodelling is required [1].

12.2 Special Considerations for Sandwich Structures Apart from the general comments on FE analysis given in section 12.1, special care has to be taken when analysing sandwich constructions due to the inhomogeneous and often anisotropic build-up of a sandwich. In general, the following items will have to be kept in mind when analysing sandwich constructions by finite elements:

(i) Core shear. The shear deformation in the core has to be considered in most cases, be it static or dynamic; beam or shell; linear or non-linear analysis. General finite elements do not normally include shear deformation (since it is usually negligible for metallic structures), and the user has to select elements that do so for adequate sandwich analysis and design. Specific problems may occur for cores with very low shear modulus (e.g. low density PUR cores). If the FE formulation of the beam, plate or shell element is incorrect (which it might be also for commercial codes), the shear stiffness of the faces might dominate over the shear stiffness of the core even if the faces are very thin. It is advised to check this by comparing FE-results of a simple beam or plate problem with analytical solutions or experiments or full 3D FE analyses.

(ii) Anisotropy. The anisotropy of composite face materials has to be accounted for (even in the case of a 0°/90° fabric due to other properties in the 45° directions). Honeycomb cores have different shear moduli in different directions and this has also to be included.

(iii) Local effects. Due to the low stiffness and strength in the thickness direction of the sandwich, local load introductions, corners and joints have to be checked by 2D or 3D analyses to a far larger extent than is the case for metal structures. By the same reason curved panels with small radii of curvature (less than 10 times the sandwich thickness) will have to be analysed in 2D or 3D to account for the transverse normal stresses not included in shell elements.

(iv) Boundary conditions. As described in section 9, boundary conditions for sandwich plates may be different from ordinary structures. For example, a simple support could be soft or hard. This will be discussed more thoroughly later on.

12.2 SANDWICH AND FEM

12.3 Beam Analysis For a structural component to be defined as a beam, the width of its cross-section should be considerably smaller than the height. Sandwich beams are not frequently used as structural elements. However, beam theory is quite useful since it can be applied to sandwich panels subjected to cylindrical bending. In this case, the apparent Young's modulus E'=E/(1−ν2) should be used for isotropic faces and a corresponding modification be made for anisotropic faces. Hence, the current description of sandwich beam elements is limited to plane beams.

The two most common beam elements are the two-node and the three-node elements, Fig.12.3. Of these, the three-node element is the most efficient one [3]. Before the three-node elements are assembled to a beam structure, the interior node is usually condensed. To account for shear deformations, the beam rotation degrees of freedom must be independent of the slope w' of the beam. The in-plane degree of freedom u is essential in the case of axial loading and/or unequal faces of the sandwich.

For beams with constant cross-section and concentrated loads only, these beam elements will give exact results according to the sandwich beam theory if the element mesh is constructed so that loads and supports are situated at element interconnections only. For beams with uniformly distributed loads, two to three three-node elements or three to four two-node elements per span will usually give accurate results (errors within a few percent) [3].

θ1 θ2

u1 u2

ww1 (a) 2

θ θ1 θ2 3

u1 u2 u3

w1 w2 w3 (b)

Figure 12.3 Sandwich beam elements. (a) two-node element and (b) three-node element As a sample problem, a case analysed in [3] is given below. A sandwich beam, clamped at one end and simply supported at the other, is loaded by a point load. Material and geometric data are given in Fig.12.4.

P P = 500 N L = 1.0 m E = 124 GPa E t f f f1 G = 22 GPa E t c c c b = 10 mm E t f f2 tc = 24.71 mm t = 0.2875 mm L/2 L/2 b f1 tf2 = 0.5750 mm

Figure 12.4 Sandwich beam with concentrated load

12.3 AN INTRODUCTION TO SANDWICH STRUCTURES

Shear strains in the faces and normal strains in the core are neglected. Analytical solutions for midpoint deflection and support force at the roller support are: ()()25PRL− 3 PRL− P()524+ θ w = + with R = 48D 2S 28()+ 24θ as given in section 3.12. The numerical values obtained using two three-node beam elements were identical to the analytical solution: w = 55.436 mm, and R = 163.202 N

12.4 Sandwich Beam Finite Element We shall herein take a more closer look at how finite elements can be constructed for sandwich beams using the very fundamental equations of sandwich beam theory. Consider a beam illustrated in Fig.12.8, and consider this as one beam finite element. This time we shall derive a finite element for which the rotation θ is independent of the slope w. This is done so that transverse shear deformation may be accounted for, so called sandwich beam theory, or Timoshenko beam theory. The rotation θ is here defined as positive rotation about the y-axis and has hence not the same sign as ∂w/∂x. However, this is common notation in most textbooks on the subject.

For a two-node element the element force vector is defined by and related to cross-section transverse forces and bending moments as

⎛ T ⎞ ⎛ −T(0) ⎞ ⎜ 1 ⎟ ⎜ ⎟ ⎜ M 1 ⎟ ⎜− M (0)⎟ F e = = (12.1) ⎜ T ⎟ ⎜ T (L) ⎟ ⎜ 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ M 2 ⎠ ⎝ M (L) ⎠

The corresponding element displacement vector is then

t d = ()w1 θ1 w2 θ 2 (12.2)

In the following, we are seeking a method to rewrite the sandwich beam theory into an algebraic equation system of the type F = K d (12.3)

12.4 SANDWICH AND FEM

q(x) q (x) m M(L) M(0) M(x) T(x) M(x+dx) N (0) x Nx(L) Nx(x+dx)

Nx(x)

T(0) x, u T(L) T(x+dx)

z, w

M1, θ1 M2, θ2 ξ w1

−θ1 -1 +1 w2

T1, w1 T2, w2

x −θ2

x,u −θ

z,w γ −θ u=zθ w

Figure 12.8 Beam with definitions and beam element

12.4.1 Derivation of stiffness matrix by beam calculations If we now approach the problem by the stiffness method commonly used in structural t mechanics, first apply the displacements u = (w1, 0, 0, 0) as illustrated in Fig.12.9.

T2 T1

M1 M2 1 2 w1

Figure 12.9 Beam with unit w1 displacement Then apply sandwich beam theory to the find the loads necessary to create such a deformation. First, the boundary conditions are given by the deflection of the nodes, i.e.

dwb w(0) = wb(0) + ws(0) = 1, θ = − (0) = 0 , 1 dx

12.5 AN INTRODUCTION TO SANDWICH STRUCTURES

dwb w(h) = wb(h) + ws(h) = 0, and θ = − (h) = 0 2 dx since by using this coordinate system, the rotation θ equals −dwb/dx (see chapter 4). We can now solve this by integrating eqs.(4.18) and (4.19) separately in order to get the bending and shear deformations, respectively. Taking the beam part deflection first and then the shear part, these equations are integrate in general terms to

dw4 x 3 x 2 D b = q → Dw=++++ qdx4 A A Ax A dx 4 b ∫∫ ∫∫ 1 622 34

dw2 S s =−q → Sw=− qdx2 − A x − A dx 2 s ∫∫ 12 where the A’s are integration constants. The total deflection can thus be written as

wx()=+= wbs () x w () x 1 ⎧ x 3 x 2 ⎫ 1 qdx4 ++++ A A Ax A −++qdx2 A x A ⎨∫∫∫∫ 1 2 34⎬ {} ∫∫ 12 D ⎩ 62 ⎭ S However, the rigid body translations given by A A A 42+=5 D S D may be replaced by a new integration constant. In deriving the stiffness matrix for the sandwich beam element, we only consider nodal loads T and M and may therefore set q = 0. By applying the boundary conditions to the first unit deflection u = (1, 0, 0, 0)t the following is obtained

A5 = D

A3 = 0

3 2 1 ⎧ h h ⎫ Ah1 ⎨A1 +++A2 Ah3 A5 ⎬ −=0 D ⎩ 62 ⎭ S

h 2 A ++=Ah A 0 1 2 23 Solving for A gives 12D 12D 6D A1 = = 3 , A2 =− 2 , A3 = 0, A5 = D 3 ⎛ 12D⎞ h 112+ φ h 112+ φ h ⎜1 + ⎟ () () ⎝ hS2 ⎠

The forces and bending moments are then obtained as

12.6 SANDWICH AND FEM

⎛ dws ⎞ 12D TT1 =−()0 =− S⎜ ⎟ ==A1 3 ⎝ dx ⎠ x=0 h ()112+ φ

⎛ d 2 w ⎞ 6D M = −M (0) = D⎜ b ⎟ = A = − 1 ⎜ dx 2 ⎟ 2 h 2 1 + 12φ ⎝ ⎠ x=0 ()

⎛ dws ⎞ 12D TThS21==() ⎜ ⎟ =−A =− 3 ⎝ dx ⎠ xh= h ()112+ φ

⎛ d 2 w ⎞ 6D M = M (h) = −D⎜ b ⎟ = −A h − A = − 2 ⎜ dx 2 ⎟ 1 2 h 2 1 + 12φ ⎝ ⎠ x=h ()

This now gives the first row in the stiffness matrix.

(ii) Next apply n = (0, 1, 0, 0) t and solving for A gives 6D 6D(4 + 12φ) A = − , A = , A3 = −D, A5 = 0 1 h 2 ()1+12φ 2 h()1 + 12φ

Continuing like this for the other two unit displacements gives the entire stiffness matrix. The result of this is

⎡ 12 − 6h −12 − 6h ⎤ ⎢ 2 2 ⎥ D − 6h h (4 + 12φ) 6h h (2 −12φ) K e = ⎢ ⎥ (12.4) h 3 (1 + 12φ) ⎢−12 6h 12 6h ⎥ ⎢ 2 2 ⎥ ⎣− 6h h (2 −12φ) 6h h (4 + 12φ)⎦ where φ again is the shear factor, defined as D/kGAh2 or D/Sh2. However, to be able to use these elements, proper definitions for the corresponding mass and initial stress matrices must be obtained using the same derivation, and these are in none of the references given in an explicit form. This is for example the two-node shear deformable beam element used by ANSYS (BEAM3), but with the in-plane displacement u also included.

12.4.2 Governing equations The above method has its restrictions due to its derivation. It is usually much more powerful to derive the finite element equation system by using the governing differential equations. This type of derivation will automatically yield all boundary conditions and all the necessary stiffness and mass matrices required for bending, buckling and free vibration analyses.

The kinematics of the shear deformable (sandwich) beam are du dθ u = u + zθ → ε(z) = = ε + z (12.5) 0 dx 0 dx

12.7 AN INTRODUCTION TO SANDWICH STRUCTURES

⎛ dθ ⎞ dθ and thus M = σzdz = E ε + z zdz = D (12.6) ∫ ∫ ⎜ 0 ⎟ z z ⎝ dx ⎠ dx where the bending moment is defined so that a positive moment yields positive normal stress in positive z-direction. The rotation θ is the same as −∂wb/∂x as defined in Chapter 4. The shear strain (in the core) γ is then according to Fig.12.8

∂w T ∂w dw γ = + θ = = s , θ = − b (12.7) ∂x S ∂x dx

This shear strain is hence equivalent to ∂ws/∂x as defined in Chapter 4. From previous theory we also have that

⎛ ∂w ⎞ ⎛ ∂w ⎞ T = kGAγ = kGA⎜ + θ ⎟ = S⎜ + θ ⎟ (12.8) ⎝ ∂x ⎠ ⎝ ∂x ⎠ where k is the so called shear correction factor, which equals 1 for a thin face sandwich and 5/6 for a rectangular homogeneous cross-section. The product kGA is more commonly referred to as the shear stiffness, usually denoted S. Note that this parameter has dimension [force/width] and since the analysis herein assumes a unit width of the beam, the cross-section area A should equal h (since the width is unity).

The equilibrium equations take the usual form

∂T ∂ 2 w ∂ 2 w Vertical: + q + N − ρ * = 0 (12.9) ∂x x ∂x 2 ∂t 2

∂ 2θ ∂M Moment : − T − R + + q = 0 (12.10) ∂t 2 ∂x m where R is the rotary inertia. By inserting eq.(12.6) and (12.8) into eq.(12.9) we get

⎛ ∂ 2 w ∂θ ⎞ ∂ 2 w ∂ 2 w S⎜ + ⎟ + q + N − ρ * = 0 (12.11) ⎜ 2 ⎟ x 2 2 ⎝ ∂x ∂x ⎠ ∂x ∂t and eq.(12.10) takes the form

⎛ ∂w ⎞ ∂ 2θ ∂ 2θ − S⎜ + θ ⎟ − R + D + qm = 0 (12.12) ⎝ ∂x ⎠ ∂t 2 ∂x 2

These two (partial) differential equations now constitute the foundation for the construction of the shear deformable beam element. As seen, only second derivatives of w and θ appear in this strong form of the equations and one can thus now see that only C0 continuity will be required for the variables. However, there is a coupling between the slope and deflection. Eqs.(11.11) and (11.12) now constitutes a bending moment and transverse force equilibrium equation, respectively.

12.8 SANDWICH AND FEM

12.4.3 Weak form of DE using virtual work method The virtual work done by these forces is then simply the equation times a virtual rotation and translation, respectively. If we denote a virtual displacement in the z-direction with wv and a virtual rotation (with the same sign as θ) as θv, add these to the respective equation and integrate over the element length, we obtain

h ⎛ ∂ 2 w ∂θ ⎞ h h ∂ 2 w h ∂ 2 w S⎜ + ⎟w dx + qw dx + N w dx − ρ * w dx = 0 ∫ ⎜ 2 ⎟ v ∫ v ∫ x 2 v ∫ 2 v 0 ⎝ ∂x ∂x ⎠ 0 0 ∂x 0 ∂t

h ⎛ ∂w ⎞ h ∂ 2θ h ∂ 2θ h − S⎜ + θ ⎟θ dx + D θ dx − R θ dx + q θ dx = 0 ∫ v ∫ 2 v ∫ 2 v ∫ m v 0 ⎝ ∂x ⎠ 0 ∂x 0 ∂t 0

Note here that the integration is only done over the length of the element and not over the entire beam. Really, the integration should be done over the entire length of the beam, but the summation of element contributions are done in the element matrix assembly and is therefore omitted herein.

We now see that both equations have obtained the same dimension and may hence be added together. However, before doing so, integrate the equations by parts (as in common Galerkin method). In the first, we only need to integrate the term including the bending stiffness D, whereas the other terms may be left as given.

h ⎡ ⎛ ∂w ⎞ ⎤ h ⎛ ∂w ⎞ ∂w h S +θ w − S +θ v dx + qw dx ⎢ ⎜ ⎟ v ⎥ ∫ ⎜ ⎟ ∫ v ⎣ ⎝ ∂x ⎠ ⎦ 0 0 ⎝ ∂x ⎠ ∂x 0 h h h 2 ⎡ in ∂w ⎤ ∂w ∂wv * ∂ w + N wv − N x dx − ρ wv dx = ⎣⎢ ∂x ⎦⎥ ∫ ∂x ∂x ∫ ∂t 2 0 0 0 h h h ⎛ ∂w ⎞ ∂w w T − S +θ v dx + qw dx + []v 0 ∫ ⎜ ⎟ ∫ v 0 ⎝ ∂x ⎠ ∂x 0 h ⎡ ∂w⎤ h ∂w ∂w ∂ 2 w h w N − v N dx − ρ *w dx = 0 ⎢ v x ⎥ ∫ x 2 ∫ v ⎣ ∂x ⎦ 0 0 ∂x ∂x ∂t 0

The second equation takes the form

h h ⎛ ∂w ⎞ ⎡ ∂θ ⎤ h ∂θ ∂θ ∂ 2θ h h − S +θ θ dx + D θ − D v dx − Rθ dx + q θ dx = ⎜ ⎟ v ⎢ v ⎥ 2 v m v ∫ ⎝ ∂x ⎠ ⎣ ∂x ⎦ ∫ ∂x ∂x ∂t ∫ ∫ 0 0 0 0 0 h h 2 h h ⎛ ∂w ⎞ h ∂θ ∂θ ∂ θ − S +θ θ dx + θ M − v D dx − Rθ dx + q θ dx = 0 ∫ ⎜ ⎟ v []v 0 ∫ 2 ∫ v ∫ m v 0 ⎝ ∂x ⎠ 0 ∂x ∂x ∂t 0 0

Since both these equations have the dimension virtual work they may be added to constitute the total virtual work of the beam element. This then becomes

12.9 AN INTRODUCTION TO SANDWICH STRUCTURES

h ∂θ ∂θ h ⎛ ∂w ⎞ h ⎛ ∂w ⎞ ∂w h ∂w ∂w − D v dx − S +θ θ dx − S +θ v dx − v N dx − ∫ ∫ ⎜ ⎟ v ∫ ⎜ ⎟ ∫ x 0 ∂x ∂x 0 ⎝ ∂x ⎠ 0 ⎝ ∂x ⎠ ∂x 0 ∂x ∂x ∂ 2θ h ∂ 2 w h h h Rθ dx − ρ *w dx + q θ dx + qw dx + 2 ∫ v 2 ∫ v ∫ m v ∫ v ∂t 0 ∂t 0 0 0 h ⎡ ∂w⎤ h h + w N + θ M + w T = 0 ⎢ v x ⎥ [][]v 0 v 0 ⎣ ∂x ⎦ 0 or slightly rewritten

h ∂θ ∂θ h ⎛ ∂w ⎞ ⎛ ∂w ⎞ h ∂w ∂w v D dx + v +θ S +θ dx + v N dx + ∫ ∫ ⎜ v ⎟ ⎜ ⎟ ∫ x 0 ∂x ∂x 0 ⎝ ∂x ⎠ ⎝ ∂x ⎠ 0 ∂x ∂x ∂ 2θ h ∂ 2 w h (12.13) Rθ dx + ρ *w dx = 2 ∫ v 2 ∫ v ∂t 0 ∂t 0 h h h ⎡ ∂w⎤ h h q θ dx + qw dx + w N + θ M + w T ∫ m v ∫ v ⎢ v x ⎥ [][]v 0 v 0 0 0 ⎣ ∂x ⎦ 0 where M(h) and M(0) are the nodal concentrated bending moments (boundary values), T(h) and T(0) the boundary transverse forces, and N(h) and N(0) are the boundary in-plane forces. This is now the weak form of the governing differential equation for a so called Timoshenko beam.

12.4.4 Displacement finite element formulation If we now proceed with the finite element discretisation, we approximate the deflection field and the virtual deflections with

w = Nwd, θ = Nθd (12.14) and

wv = Nw dv and θv = Nθ dv (12.15) where

t t d = (w1, θ1, w2, θ2) and dv = (wv1, θv1, wv2, θv2) or for elements with more than two nodes, the displacement vector may contain more entries, i.e., w3, θ3, etc. Thus, the shape functions may then be written (two-node case)

Nw = (Nw1, 0, Nw2, 0) and Nθ = (0, Nθ1, 0, Nθ2) and their respective derivatives are defined as

∂w ∂θ ∂w ∂θ = B d , = B d , v = B d , and v = B d (12.16) ∂x w ∂x θ ∂x w v ∂x θ v where B is the derivative of N with respect to x. By inserting these into the weak form of the differential equation one obtains

12.10 SANDWICH AND FEM

h h h d t B t DB ddx + d t B + N t S B + N ddx + d t B t N B ddx + ∫ v θ θ ∫ v ()()w θ w θ ∫ v w x w 0 0 0 h ∂ 2d h ∂ 2d h h d t N t RN dx + d t N t ρ * N dx = d t N t q dx + d t N t qdx + ∫ v θ θ 2 ∫ v w w 2 ∫ v θ m ∫ v w 0 ∂t 0 ∂t 0 0 t t h t t h t t h [][][]d v N w N x Bwd 0 + d v Nθ M 0 + d v N wT 0

The virtual deflection vector v may now be taken as an arbitrary deflection field so that the above equation may be simplified to

h h h B t DB dx ⋅ d + B + N t S B + N dx ⋅ d + B t N B dx ⋅ d + ∫ θ θ ∫ ()()w θ w θ ∫ w x w 0 0 0 h ∂ 2d h ∂ 2d h h N t RN dx ⋅ + N t ρ * N dx ⋅ = N t q dx + N t qdx + (12.17) ∫ θ θ 2 ∫ w w 2 ∫ θ m ∫ w 0 ∂t 0 ∂t 0 0 t h t h t h [][][]N w N x Bwd 0 + Nθ M 0 + N wT 0

This equation is now in a form that can be used for finite element calculations and is in the form

∂ 2d Kd + K d + M = F (12.18) σ ∂t 2 where the stiffness matrix K is

h h K = B t DB dx + B + N t S B + N dx (12.19) ∫ θ θ ∫ ()()w θ w θ 0 0 the initial stress matrix is

h K = B t N B dx (12.20) σ ∫ w x w 0 the mass matrix is

h h M = N t RN dx + N t ρ * N dx (12.21) ∫ θ θ ∫ w w 0 0 in which the first term represents the rotary inertia and the second term translational inertia. Finally, the load vector is

h h h h h F = N t qdx + N t q dx + N t M + N t T + N t N B w (12.22) ∫ w ∫ θ m [][][θ 0 w 0 w x w ]0 0 0

For eigenvalue buckling analysis, the last term in the above equation, containing the in-plane load Nx is of no consequence for the analysis and may therefore be omitted.

12.11 AN INTRODUCTION TO SANDWICH STRUCTURES

12.4.5 Two-node shear deformable beam element The simplest possible element for the shear deformable beam can now be constructed using a very simple shape function, i.e., a linear interpolation, as for a rod element. We can now interpolation functions derived in the x-coordinate system or an isoparametric formulation in the ξ-coordinate. The result remain the same! Now, a simple and appropriate shape function is

2 ⎛1 − ξξ⎞ ⎛1 + ⎞ ⎛1 −+ ξξ1 ⎞⎛ x1 ⎞ xx= ⎜ ⎟ 12+ ⎜ ⎟ xNx==∑ ii ⎜ , ⎟⎜ ⎟ = Nx (12.23) ⎝ 2 ⎠ ⎝ 2 ⎠ i=1 ⎝ 2 2 ⎠⎝ x2 ⎠

In the x-coordinate these would have appeared as x x N =−1 and N = 1 h 2 h These can then be used to obtain

⎛1 −+ξξ1 ⎞ t w = Nww = ⎜ ,, 0 , 0⎟()ww ,θθ , , (12.24) ⎝ 2 2 ⎠ 11 2 2

⎛ 1 −+ξξ1 ⎞ t θ = Nθw = ⎜0,,,,,, 0 ⎟()wwθθ (12.25) ⎝ 2 2 ⎠ 11 2 2

Since we need derivatives with respect x in all equations, first note that dx 1 1 hdξ 2 =−xx + = → = dξ 2 1222 dx h

The derivatives of w and θ with respect to x can then hence be written

∂w ∂w ∂ξ 2 ∂N w 2 2 ⎛ 1 1 ⎞ t = = w = Bw w = ⎜− , 0, , 0⎟()w1 ,θ1 , w2 ,θ 2 ∂x ∂ξ ∂x h ∂ξ h h ⎝ 2 2 ⎠

∂θ ∂θ ∂ξ 2 ∂Nθ 2 2 ⎛ 1 1 ⎞ t = = w = Bθ w = ⎜0, − , 0, ⎟()w1 ,θ1 , w2 ,θ 2 ∂x ∂ξ ∂x h ∂ξ h h ⎝ 2 2 ⎠

The stiffness matrix then takes the form following eq.(12.19) and converting the vectors N and B from global co-ordinates x to local co-ordinates ξ

h h K e = B t DB dx + B + N t S B + N dx ∫ θ θ ∫ ()()w θ w θ 0 0 t 1 2 2 h 1 ⎛ 2 ⎞ ⎛ 2 ⎞ h = B t D B dξ + B + N S B + N dξ (12.26) ∫ θ θ ∫ ⎜ w θ ⎟ ⎜ w θ ⎟ −1 h h 2 -1 ⎝ h ⎠ ⎝ h ⎠ 2 t 1 2 1 ⎛ 2 ⎞ ⎛ 2 ⎞ h = B t DB dξ + B + N S B + N dξ ∫ θ θ ∫ ⎜ w θ ⎟ ⎜ w θ ⎟ −1 h -1 ⎝ h ⎠ ⎝ h ⎠ 2 which may be integrated to take the form

12.12 SANDWICH AND FEM

⎡ S S S S ⎤ − − − ⎢ h 2 h 2 ⎥ ⎢ S Sh D S Sh D ⎥ ⎢− + − ⎥ K e = ⎢ 2 3 h 2 6 h ⎥ (12.27) S S S S ⎢− ⎥ ⎢ h 2 h 2 ⎥ ⎢ S Sh D S Sh D⎥ ⎢− − + ⎥ ⎣ 2 6 h 2 3 h ⎦

This looks fine so far! However, as mentioned in the textbook by Bathe [7], this element formulation suffers from a serious deficiency for cases when the shear stiffness is large, i.e., for pure bending, and this is illustrated and discussed below.

12.4.6 Shear Locking Study a cantilever beam of one two-node shear deformable beam, as derived above and illustrated Fig.12.10.

θ1 ,M1 θ ,M ξ 2 2

T11 ,w T22 ,w

Figure 12.10 Two-node element as a cantilever beam subjected to edge bending moment The shape functions for w and θ gives that 1 + ξ 1 + ξ w(ξ ) = w and θ (ξ ) = θ 2 2 2 2 and hence, the shearing is

∂w ∂w ∂ξ 2 ∂w 2 w w 1 + ξ γ (ξ ) = + θ = + θ = + θ = 2 + θ = 2 + θ ∂x ∂ξ ∂x h ∂ξ h 2 h 2 2

For an applied bending moment, the shear strain should be zero, i.e.,

w 1 + ξ 0 = 2 + θ h 2 2 However, for this expression to be zero all along the beam (for any value of ξ) we must have that both w2 and θ2 are zero. Hence, a zero shear strain can only be obtained when there are no deformations. This means that for beams with high shear stiffness, the element becomes too stiff, and this phenomenon is often referred to as shear locking. This behaviour is encountered for two-node beam elements based on a pure displacement formulation, and such elements should therefore not be used. There are two methods to overcome this problem; one

12.13 AN INTRODUCTION TO SANDWICH STRUCTURES is to use a three- or four-node displacement based element or to use a mixed-formulation for the two-node beam element. The first of these possible improvements is discussed next.

12.4.7 Three-node Shear Deformable Beam Although the two-node element derived by a pure displacement formulation has a serious deficiency, the three-node beam is free from the so called shear locking problem. As mentioned in e.g. Bathe [7] and Cook [1], this is in any case valid as long as the middle node is located mid-way between the end nodes. Although the three-node element has considerable improved accuracy compared to the two-node displacement based element, it is still not exact. This will be seen in examples that follow. This incorrectness is however not that severe and is of the same kind as for any other quadratic element type. The derivation of the three-node shear deformable element is now quite simple, since all equations used above can be used again. θ ,M θ ,M 0 1 1 2 2 θ3 ,M3 -1ξ +1

1 23 x T11 ,w T22 ,w T33 ,w

Figure 12.11 Definition of three-node beam element For a three-node beam element of length h, defined in Fig.12.11, we again use an isoparametric description which may appear as

3 1 2 1 xNx===−−+−++∑ ii Nx ξξ(11 ) x1 ( ξ ) x23 ξξ (1 ) x (12.28) i=1 2 2 where the shape functions are 1 1 N =−ξξ(),1 − N =−()1 ξ 2 and N =+ξξ(1 ) (12.29) 1 2 2 3 2 In the x-coordinate these would have appeared as

23x 2 x 44x 2 x 2x 2 x N =−+1, N =− + and N =− 1 h 2 h 2 h 2 h 3 h 2 h These can then be used to obtain

⎛ 1 2 1 ⎞ t w = Nwd = ⎜−−ξξ(1010 ),, − ξ ,, ξξ (10 + ), ⎟()ww11 , θ , 2 , θ 2 , w 33 , θ (12.30) ⎝ 2 2 ⎠

⎛ 1 2 1 ⎞ t θ = Nθd = ⎜0,(),,,,(),,,,, −−ξξ1010 − ξ ξξ1 +⎟()ww11 θ 2 θ 2 w 33 θ (12.31) ⎝ 2 2 ⎠

Since we need derivatives with respect x in all equations, first note that

12.14 SANDWICH AND FEM

dx 1 1 1 h =−⎜⎛ −ξξ⎟⎞ xx−++2 ⎜⎛ ξ⎟⎞ x=−+−+=()( xxxxx ξ2 ) dξ ⎝ 2 ⎠ 12⎝ 2 ⎠ 32 311232

dξ 2 →= dx h The derivatives of w and θ with respect to x can then hence be written

∂w ∂w ∂ξ 2 ∂N 2 = = w d = B d = ∂x ∂ξ ∂x h ∂ξ h w

2 ⎛ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎞ t ⎜− ⎜ − ξ ⎟, 0, − 2ξ, 0 ⎜ + ξ ⎟, 0⎟()w1 ,θ1 , w2 ,θ 2 , w3 ,θ3 h ⎝ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎠

∂θ ∂θ ∂ξ 2 ∂N 2 = = θ d = B d = ∂x ∂ξ ∂x h ∂ξ h θ

2 ⎛ ⎛ 1 ⎞ ⎛ 1 ⎞⎞ t ⎜0, − ⎜ − ξ ⎟, 0, − 2ξ, 0 ⎜ + ξ ⎟⎟()w1 ,θ1 , w2 ,θ 2 , w3 ,θ3 h ⎝ ⎝ 2 ⎠ ⎝ 2 ⎠⎠

The stiffness matrix then takes the same form as for the two-node element, i.e.,

h h K e = B t DB dx + B + N t S B + N dx ∫ θ θ ∫ ()()w θ w θ 0 0 t 1 2 2 h 1 ⎛ 2 ⎞ ⎛ 2 ⎞ h = B t D B dξ + B + N S B + N dξ (12.32) ∫ θ θ ∫ ⎜ w θ ⎟ ⎜ w θ ⎟ −1 h h 2 -1 ⎝ h ⎠ ⎝ h ⎠ 2 t 1 2 1 ⎛ 2 ⎞ ⎛ 2 ⎞ h = B t DB dξ + B + N S B + N dξ ∫ θ θ ∫ ⎜ w θ ⎟ ⎜ w θ ⎟ −1 h -1 ⎝ h ⎠ ⎝ h ⎠ 2 which may be integrated to take the form

⎡0 0 0 0 0 0 ⎤ ⎢ ⎥ ⎢0 7 0 −8 0 1 ⎥

e D ⎢0 0 0 0 0 0 ⎥ K D = ⎢ ⎥ and (12.33a) 3h ⎢0 −8 0 16 0 −8⎥ ⎢0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎣⎢0 1 0 −8 0 7 ⎦⎥

⎡ 70 −15h − 80 − 20h 10 5h ⎤ ⎢ 2 2 2 ⎥ ⎢−15h 4h 20h 2h − 5h − h ⎥

e S ⎢ − 80 20h 160 0 − 80 − 20h⎥ K S = ⎢ 2 2 2 ⎥ (12.33b) 30h ⎢− 20h 2h 0 16h 20h 2h ⎥ ⎢ 10 − 5h − 80 20h 70 15h ⎥ ⎢ 2 2 2 ⎥ ⎣⎢ 5h − h − 20h 2h 15h 4h ⎦⎥

12.15 AN INTRODUCTION TO SANDWICH STRUCTURES where

e e e K = K D + K S (12.34)

The load vector for a uniform transverse pressure q(x) = q, and a uniform distributed bending moment qm(x) = qm is then calculated as

h h 1 h 1 h F e = N t qdx + N t q dx = N t q dξ + N t q dξ (12.35) ∫ w ∫ θ m ∫ w ∫ θ m 0 0 −1 2 −1 2 which results in 1 F e = ()qh q h 4qh 4q h qh q h t (12.36) 6 m m m In terms of input for FE-models we cannot specify any pressure distribution function and the only feasible option is to model this distribution by specifying discrete pressure intensities at the nodes and use the shape functions for interpolation. If so, a linear element can handle a linearly varying pressure over the element, a quadratic element a quadratic distribution, etc. In doing this we can approximate the pressure over the three-node quadratic element by

t q(ξ ) = N w1q1 + N w2 q2 + N w3q3 = N w (q1 ,0,q2 ,0,q3 0,) = N wq and similarly

t qm (ξ) = Nθ1qm1 + Nθ 2 qm2 + Nθ 3qm3 = N θ (0,qm1 ,0,qm2 ,0,qm3 ) = Nθ qm

The resulting vector of consistent nodal loads is then

h h 1 h h F e = N t qdx + N t q dx = N t N q dξ + N t N q dx ∫ w ∫ θ m ∫ w w ∫ θ θ m 0 0 −1 2 0

For example, for a hydrostatic pressure of the type

t ⎛ q ⎞ q = ⎜0 0 0 q 0⎟ ⎝ 2 ⎠ the vector of consistent nodal loads becomes 1 F e = ()0 0 2qh 0 qh 0 t 6 The initial stress matrix used in eigenvalue buckling analysis is obtained through

h 1 2 2 h K e = B t N B dx = B t N B dξ (12.37) σ ∫ w x w ∫ w x w 0 −1 h h 2 which will appear as

12.16 SANDWICH AND FEM

⎡ 7 0 − 8 0 1 0⎤ ⎢ ⎥ ⎢ 0 0 0 0 0 0⎥

e N x ⎢− 8 0 16 0 − 8 0⎥ Kσ = ⎢ ⎥ (12.38) 3h ⎢ 0 0 0 0 0 0⎥ ⎢ 1 0 − 8 0 7 0⎥ ⎢ ⎥ ⎣⎢ 0 0 0 0 0 0⎦⎥ The mass matrix is

h h 1 h 1 h M e = N t RN dx + N t ρ * N dx = N t RN dξ + N t ρ * N dξ (12.39) ∫ θ θ ∫ w w ∫ θ θ ∫ w w 0 0 −1 2 −1 2 which take the form

⎡ 4ρ * 0 2ρ * 0 − ρ * 0 ⎤ ⎢ ⎥ ⎢ 0 4R 0 2R 0 − R⎥ h ⎢ 2ρ * 0 16ρ * 0 2ρ * 0 ⎥ M e = ⎢ ⎥ (12.40) 30 ⎢ 0 2R 0 16R 0 2R ⎥ ⎢− ρ * 0 2ρ * 0 4ρ * 0 ⎥ ⎢ ⎥ ⎣⎢ 0 − R 0 2R 0 4R ⎦⎥

12.4.8 Stiffness matrix assembly Before solving a real problem one must assemble the structural stiffness matrices. This is done by adding element stiffness matrices and load vector entries for each element to describe the structure to be analysed. For a beam for example, this is done by connecting beam element via each node that connects elements to each other. Each degree-of-freedom will then receive “stiffness” and “mass” from each element connected to the node. Schematically this looks like what is shown in Fig.12.12.

u1 u2 u3 u4 u5

e e u1 u2

e e u2 u3

e e u3 u4

ue ue 4 5

Figure 12.12 Assembly of rod elements What really is done here is that the integration of virtual work equations should be performed over the entire length of the beam. However, we can perform this integration piece-wise, i.e.,

L h ∫ Kdx = ∑∫ K e dx 0 elem 0

12.17 AN INTRODUCTION TO SANDWICH STRUCTURES and similarly for the load vector. But since the components in the stiffness matrix and load vector are assigned to one node pair, the summation will actually become a matrix assembly. The element stiffness matrices are now schematically assembled as

⎡xx000⎤ ⎡00000⎤ ⎡00000⎤ ⎡00000⎤ ⎡xx000⎤ ⎢xx000⎥ ⎢000 xx ⎥ ⎢00000⎥ ⎢00000⎥ ⎢xxx00⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢00000⎥ + ⎢000xx⎥ + ⎢00 xx 0⎥ + ⎢00000⎥ = ⎢0 xxx 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢00000⎥ ⎢00000⎥ ⎢00xx 0⎥ ⎢000 xx⎥ ⎢00xxx⎥ ⎣⎢00000⎦⎥ ⎣⎢00000⎦⎥ ⎣⎢00000⎦⎥ ⎣⎢000xx⎦⎥ ⎣⎢000xx⎦⎥

For the structural stiffness matrix, the structural initial stiffness matrix and structural mass matrix. For the load vector, this is done schematically as

⎛ x⎞ ⎛0⎞ ⎛0⎞ ⎛0⎞ ⎛ x⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ xx⎟ ⎜ ⎟ ⎜0⎟ ⎜0⎟ ⎜ x⎟ ⎜0⎟ + ⎜ xx⎟ + ⎜ ⎟ + ⎜0⎟ = ⎜ x⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜0⎟ ⎜0⎟ ⎜ xx⎟ ⎜ ⎟ ⎜ x⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎝0⎠ ⎝ x⎠ ⎝ x⎠

12.4.9 Boundary conditions In the bending case, the equation system to be solved appear as Kd = F The properties of the stiffness matrix are (the same for the initial stress matrix and the mass matrix): (i) Contains mainly entries on the diagonal.

(ii) All entries on the diagonal are positive.

(iii) det(K) = 0 if no boundary conditions are applied. The last property implies that the solution of the equation system of eq.(12.4) is singular, that is, have an infinite number of solutions. To avoid this, we must apply some appropriate boundary conditions. To obtain a solution we must use

d = K −1F However, this cannot at this stage be done since det(K) = 0. What must now be done is to apply appropriate boundary conditions. This is done by reducing the equation system. If, for example, we are studying a cantilever beam, then the first two (or the last two) degrees-of- freedom should be set to zero, corresponding to the first translations degree-of-freedom and the first rotation one (w1 = θ1 = 0). Schematically this will look like

12.18 SANDWICH AND FEM

⎡y y 0 0 0⎤⎛ u = 0 ⎞ ⎛ z ⎞ ⎜ 1 ⎟ ⎜ ⎟ ⎢y y x 0 0⎥ u = 0 z ⎢ ⎥⎜ 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢0 x x x 0⎥ u3 = x (12.41) ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢0 0 x x x⎥⎜ u4 ⎟ ⎜ x⎟ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎣0 0 0 x x⎦⎝ u5 ⎠ ⎝ x⎠

The boundary conditions are then applied by removing the rows and columns corresponding to the degrees-of-freedom with active boundary conditions. In this case this means removing rows 1 and 2 and columns 1 and 2. This gives us the reduced equations system, schematically appearing as

⎡x x 0⎤⎛u ⎞ ⎛ x⎞ ⎜ 3 ⎟ ⎜ ⎟ ⎢x x x⎥ u = x or K d = F (12.42) ⎢ ⎥⎜ 4 ⎟ ⎜ ⎟ ff f f ⎜ ⎟ ⎜ ⎟ ⎣⎢0 x x⎦⎥⎝u5 ⎠ ⎝ x⎠

The new reduced stiffness matrix Kr has a finite value of its determinant and may therefore be inverted according to

−1 ⎛u ⎞ ⎡x x 0⎤ ⎛ x⎞ ⎜ 3 ⎟ ⎜ ⎟ d = K −1F or u = ⎢x x x⎥ x (12.43) f ff f ⎜ 4 ⎟ ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝u5 ⎠ ⎣⎢0 x x⎦⎥ ⎝ x⎠

Kff becomes non-singular providing all rigid body translations and rigid body rotations are prevented for the structure. All degrees-of-freedom are obtained by expanding the displacement vector to its initial size by adding the boundary condition values, e.g.

⎛ 0 ⎞ ⎜ ⎟ ⎜ 0 ⎟ d = ⎜u ⎟ ⎜ 3 ⎟ ⎜u4 ⎟ ⎜ ⎟ ⎝u5 ⎠

There is also a possibility to apply prescribed displacements to the structure. This is in essence the same as prescribing zero nodal displacement. In general we can rewrite the equation system of eq.(12.18) into a set of prescribed displacement and another set of "free" displacements. This is done by partitioning the matrix. If we denote the free displacement by a vector wf and the prescribed displacement with wp. Similarly we can partition the load vector into one part of prescribed loads Fp and one part of free loads Ff, the former corresponding to loads appearing at boundary conditions (where the displacements are prescribed) and the latter the loads applied to the structure. The system Kd = F can then be partitioned into

12.19 AN INTRODUCTION TO SANDWICH STRUCTURES

⎡K K ⎤⎛d ⎞ ⎛ F ⎞ pp pf ⎜ p ⎟ ⎜ p ⎟ ⎢ ⎥⎜ ⎟ = ⎜ ⎟ (12.44) ⎣K fp K ff ⎦⎝d f ⎠ ⎝ F f ⎠ which basically implies restructuring the matrix system so that the displacement and load vectors are rearranged in known (prescribed) and unknown entries. For eq.(12.41) this will symbolically look like

⎡y y⎤ ⎡0 0 0⎤ t ⎛0⎞ ⎛ z ⎞ K pp = ⎢ ⎥ , K pf = ⎢ ⎥ = K fp , w p = ⎜ ⎟ and Fp = ⎜ ⎟ ⎣y y⎦ ⎣x 0 0⎦ ⎝0⎠ ⎝ z ⎠ We can also see that due to symmetry of K we must have that

t K pf = K fp

The lower part of equation (44) can be written on the form

Kff df = Ff − Kfp dp (12.45) If the prescribed displacements are all zero, then

Kff df = Ff

Once the unknowns uf we can calculate the remaining forces through

Fp = Kpp dp + Kpf df = Kpf df (12.46) if the prescribed displacements are zero. The term Fp on right hand side of eq.(12.44) may also contain prescribed loads, i.e., there may be entries in the load vector where displacements are prescribed. One such case is for example a cantilever beam with an applied uniform pressure (see Fig.5). In eq.(12.44) the right hand term for Fp should then be replaced with Fp + FR, where the latter term are the prescribed load vector entries at the nodes which have prescribed displacements. This will have no effect on the solution of the equation system, as shown in eq.(12.45). However, the reaction force of eq.(12.46) must now be written as

FR = Kpp dp + Kpf df − Fp

Usually, the vector Fp is zero, as in the case of a cantilever subjected to an edge point load.

If there are m degrees-of-freedom in the system and we have n number of constraints, the size of each matrix and vector must be

K [m×m] d [m×1] F [m×1]

Kff [m−n× m−n] df [m−n ×1] Ff [m−n ×1] Kpf [n×m−n]

Kpp [n×n] dp [n×1] Fp [n×1] Kfp [m−n ×n] One of the advantages about this finite element implementation is the ease in which boundary conditions can be applied. We only need to specify kinematic constraints to the structure, i.e., constraints on w or θ, in separate nodes. This can be done as

12.20 SANDWICH AND FEM

Free edge Simple support Clamped support No constraints w = 0 w = θ = 0

12.4.10 Solution of equation systems There are now three different case to be solved; bending, buckling and free vibration and these will imply different schemes for the solution of the sought after properties. This beam equation is in the form (eq.(12.18))

∂ 2d Kd + K d + M = F σ ∂t 2 and this can be solved for various sub-problems resulting in linear equation systems. For the combined equation to be solved one would need to use non-linear approaches. For example, the combination of transverse loads (a non-zero F) and in-plane loading (non-zero Kσ) the equations system becomes non-linear since Kσ depends on the in-plane load applied. Analytically, this problem can be solved for some problems, using so called Berry functions.

Bending For the bending case, there is a non-zero load vector F, but no in-plane load or dynamic terms to be included. Thus, for this case we take Kσ and M to be zero. The equation to be solved is a linear equation system of the type Kd = F (12.47) where appropriate boundary conditions has been applied to all vector entries (index f omitted here). Resulting forces and bending moments at each node may then also be calculated by t taking the nodal displacement vector for each element, wi = (wi, θi, wi+1, θi+1) and multiplying it with the element stiffness matrix so that the element force vector is obtained. Thus,

e Fi = K i d i (12.48)

Buckling In the case of buckling of a sandwich beam, we assume that the transverse loading is zero, i.e., F = 0. By also omitting dynamic terms eq.(12.18) takes the form

Kd + K σ d = F (12.49)

Assume that Kσ is known for a given loading situation, hence the load N = −P give rise to a certain initial stress matrix. If λ is an arbitrary scalar multiplier, then λKσ represents the same initial stress matrix for another load intensity. If the system is linear, then neither Kσ nor the ordinary stiffness matrix K are functions of the initial displacement vector d (small deformations). Assume now that the initial displacement field is perturbed by some virtual displacement d1 while the applied loads remain constant, what must λ be so that both d and (d+d1) are equilibrium configurations? In other words, what is the value of λ for which the structure no longer remains in equilibrium for a small virtual displacement away from the initial state of equilibrium? In mathematical terms, this is written as

(K + λKσ)d = (K + λKσ)(d+d1) = F

12.21 AN INTRODUCTION TO SANDWICH STRUCTURES

By subtracting the first term from the second gives the eigenvalue problem or stability condition as

(K + λKσ)d1 = 0 (12.50) If the in-plane loading (stresses) are compressive, real valued eigenvalues λ can usually be found to this equation. The number of eigenvalues are theoretically infinite but by using this formulation one can calculate as many eigenvalues as there are degrees of freedom in the system. The lowest value of λ represents the actual buckling load. The eigenvalues are found by determining

det(K + λKσ) = 0 (12.51) although this is done in a more or less complicated numerical fashions in the computer codes. Once λ has been determined the eigenvector belonging to that particular value of λ is calculated. The displacements in d1 corresponding to an eigenvalue identify the shape of buckling, but not its magnitude. Hence, each eigenvalue (load) has a given buckling shape (mode of buckling) associated with it. Compare this with e.g. Euler buckling which a factor m associated with each buckling load, and recall that this value m describes the shape of buckling since

mEI2 mπx P = assuming wx()= sin cr L2 L Free vibration In the case of vibration of a sandwich beam, we assume that the in-plane load N also is zero so that Kσ is zero. Then eq.(12.18) takes the form

∂ 2d Kd + M = F (12.52) ∂t 2 The matrix M is the mass matrix of structure, or sometimes called the consistent mass matrix, since it discretises the mass by using the shape functions (isoparametric formulation). It is also possible to include structural damping into this equation by the use of a damping matrix C, and the FE-discretisation will appear as

∂d ∂ 2d Kd + C + M = F ∂t ∂t 2 The nature of the damping matrix will not be discussed further in this document since that in itself is subject for a course on its own.

A simpler formulation of the consistent mass matrix is the lumped mass matrix, which is a diagonal mass matrix. This means that parts of the mass of the elements surrounding a node is placed as a local mass point in that particular node and the all the mass in the elements are placed as local lumps of mass in the nodes. The physical meaning of replacing the mass of the element into ”lumps” of mass in each node implies that there is only translational inertia for each node, i.e., there is no rotary inertia.

12.22 SANDWICH AND FEM

The problem at hand is commonly to find the natural frequencies of a structure. That might be a very important design criterion not only for structural safety, but it may also appear as constraints on the design. To find the natural frequencies of vibration it is usually sufficient to assume that the damping is zero, i.e., C = 0, and also that the external forces are zero. Free, undamped vibration is harmonic and if we assume harmonic motion for all dof’s we can write that

∂ 2d d(t) = deiωt ⇒ = −ω 2deiωt (12.53) ∂t 2 where w are the nodal degrees of freedom and ω the circular frequency (radians per second). The cyclic frequency f = ω/2π. Eqs.(45) and (46) yields that

(K − ω2M)d = 0 (12.54) which is exactly the same type of equation as for the buckling case except that λ is substituted 2 for ω and Kσ for M. If the problem is linear, then neither K nor M is a function of ω, then the problem is a linear eigenvalue problem. To each eigenvalue ω2 there is a corresponding eigenvector di which describes the mode of vibration associated with that particular eigenfrequency.

If K is real, symmetric and non-singular, the number of eigenfrequencies of the system is equal to the number of unrestrained dof’s of the system. The eigenvalues will also be real- valued and positive since both K and M are positive definite matrices.

12.4.11 Post-processing Once a solution has been obtained one often needs to calculate properties such as strain, stress, or study particular buckling of vibration modes. The result from a finite element calculation is as seen solely a vector of nodal displacements (displacements and rotations) however being a solution to static bending, or an eigenvector to a buckling or free vibration mode. However, after the solution is obtained one can “post-process” the solution vector to calculate the properties required.

Bending From the theory of the sandwich beam that has been implemented one can calculate the element bending moment distribution from

dθ dθ dξ 2D dθ 2D M (ξ) = D = D = = B d = dx dξ dx h dξ h θ

2D ⎡ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎤ ⎢− ⎜ −ξ ⎟θ1 − 2ξθ 2 + ⎜ + ξ ⎟θ 3 ⎥ h ⎣ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ which implies that the bending moment varies linearly over the element. Similarly, one can find the transverse force distribution through

12.23 AN INTRODUCTION TO SANDWICH STRUCTURES

⎛ dw ⎞ ⎛ 2 ⎞ T(ξ ) = S⎜ +θ ⎟ = S⎜ Bw + Nθ ⎟d = ⎝ dx ⎠ ⎝ h ⎠

2S ⎡ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎤ ⎡ 1 2 1 ⎤ ⎢− ⎜ −ξ ⎟w1 − 2ξw2 + ⎜ + ξ ⎟w3 ⎥ + S⎢− ξ()1−ξ θ1 + (1−ξ )θ 2 + ξ ()1+ ξ θ 3 ⎥ h ⎣ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ ⎣ 2 2 ⎦

The transverse force is then obviously a linear function in displacements wi and a quadratic function in θi. By this scheme one calculates the bending moment and transverse force twice in nodes which join two elements together and since the result depends on the local deflection vector for an individual element only, they may differ slightly going from one element to the next. This means that M and T in such nodes will obtain two slightly different values. If the difference is only small then one can take either value or simply use the average value.

The reason for the mis-match in bending moment and transverse force over element boundaries stem from the problem itself. As indicated, a shear deformable beam element requires only C0 continuity. Thus, w and θ are continuous over element boundaries but not their derivatives. This is seen by e.g. looking at the rotation at the right edge of one element, which is

dθ ⎡ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎤ 1 3 = ⎢− ⎜ − ξ ⎟θ1 − 2ξθ 2 + ⎜ + ξ ⎟θ 3 ⎥ = θ1 − 2θ 2 + θ 3 dx ξ =+1 ⎣ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ ξ =+1 2 2 and comparing it with the rotation at the left edge of the next adjacent element, which is

dθ ⎡ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎤ 3 1 = ⎢− ⎜ − ξ ⎟θ 3 − 2ξθ 4 + ⎜ + ξ ⎟θ 5 ⎥ = − θ 3 + 2θ 4 − θ 5 dx ξ =−1 ⎣ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ ξ =−1 2 2 and it is seen that these derivatives will differ unless θ is constant in both elements.

Another point of some interest is that transverse forces and bending moments are obtained by differentiation of the displacements. Since the displacement vector calculated is approximate (since the interpolation of displacements through shape functions is approximate) the derivation of such results further deteriorates the accuracy. Thus, bending moments and transverse forces are obtained will less accuracy than displacements and the only way to improve the accuracy is to use more and smaller elements.

Once bending moments and transverse forces are calculated, the strains and stresses are given through sandwich theory, i.e. Mz T ε = and γ = D S Buckling and free vibration The results from a buckling or free vibration analysis is given by a number of eigenvalues to the equation systems in eqs.(44) and (47). A general eigenvalue analysis calculates as many eigenvalues as there a free degree-of-freedom (the size of the reduced structural equation system). To each eigenvalue one can then obtain the corresponding eigenvector. This vector contains the shape of the buckling or free vibration mode, not the actual displacement values

12.24 SANDWICH AND FEM but the mutual displacements and rotations between the nodes. These eigenvectors are suitable for graphical representation of the mode shapes by simply plotting for example the displacement entires of the vector versus nodal coordinates.

12.4.12 Example of FEM-Calculation for a Cantilever Sandwich Beam As shown previously, the two-node pure displacement element suffered from so called shear locking. This three-node element will behave much better although it will be overly stiff in the pure bending case, i.e., when S is large. To exemplify this, take the example from Fig.2 where one element is used to calculate the displacement of a cantilever beam. Now use a three-node beam element and calculate the edge displacement, i.e. for node 3 when subjected to an edge bending moment M and an edge transverse force T.

This is done by removing the first and second rows and columns of the stiffness matrices in eqs.(33a) and (33b) and adding them together and then applying a force vector of the type

F = ()0 0 0 0 0 M t or F = (0 0 0 0 T 0)t

By reducing the matrix system using the boundary conditions w(ξ = −1) = θ(ξ = −1) = 0 we get

⎡ 16S 8S 2S ⎤ ⎢ 0 − − ⎥ 0 3 3h 3 w ⎛ ⎞ ⎢ 16D 8Sh 2S 8D Sh⎥⎛ 2 ⎞ ⎜ ⎟ ⎢ 0 + − + ⎥⎜ ⎟ ⎜ 0 ⎟ ⎜θ 2 ⎟ = ⎢ 3h 15 3 3h 15 ⎥ or K d = F ⎜ ⎟ 8S 2S 7S S ⎜ ⎟ ff f f 0 ⎢ ⎥ w3 ⎜ ⎟ ⎢− ⎥⎜ ⎟ ⎜ M ⎟ 3h 3 3h 2 ⎜θ ⎟ ⎝ ⎠ ⎢ 2S 8D Sh S 7D 2Sh ⎥⎝ 3 ⎠ ⎢− − + + ⎥ ⎣ 3 3h 15 2 3h 15 ⎦

In the first case we get a displacement vector

2 ⎛ Mh 2 Mh Mh 2 Mh ⎞ ⎜ ⎟ d f = ⎜− , ,− , ⎟ ⎝ 8D 2D 2D D ⎠ where the edge displacement and rotation thus equals

1 Mh 2 Mh w = − and θ = 3 2 D 3 D which is exactly the same as the analytical result. The submatrix Kpf consists of the first two rows and columns 3 to 6 of the element stiffness matrix K and is found to be (see eq.(12.33))

⎡ 8S 2S S S ⎤ ⎢− − ⎥ K = 3h 3 3h 6 pf ⎢ 2S Sh 8D S Sh D ⎥ ⎢ − − − + ⎥ ⎣ 3 15 3h 6 30 3h⎦

The reaction force vector is then

t Fp = Kpf df = (0, −M)

12.25 AN INTRODUCTION TO SANDWICH STRUCTURES

This element will thus behave correctly for this case.

In the second case, for an edge point load T, we get an expression of the type

1 240φ + 84 + hT φ w = 3 4 S(1+ 60φ) which for the cases of pure bending (φ = 0) and pure shear (φ large)

Th 3 Th w = and w = , respectively 3 4D 3 S This should be compared with analytical solution, which is given by

Th 3 Th w(h) = + 3D S The FE solution is thus seen to differ in the case of pure bending to give an overly stiff structure. However, by using a few more elements the error becomes very small even for cases of pure bending. As mentioned by Bathe [7], the pure displacement element must have 4 nodes or more per element to be exact. However, as shown below, the just derived three- node element is shown to give very good results if only more than one or two elements are used.

It is interesting to elaborate about the reasons for this discrepancy in the accuracy of the solutions using one element compared to the exact solution. The problem appears mainly for the case of pure bending, i.e., when the shear stiffness S is large. It is known that a finite element construction for a non shear deformable element (based on engineering beam theory – Euler-Bernoulli beam theory) requires C1 continuity, i.e., that displacements and their derivatives are continuous over element boundaries. This was not a requirement when deriving the present element. In order to prescribe continuity in both displacements and their derivatives at each end-point of the element (end nodes) a third degree polynomial of the kind

2 3 w(x) = c1 + c2x + c3x + c4x must be used. The governing differential equation to the beam problem when transverse shear deformations are omitted appear as

d 4 w D = q dx 4 which when integrated has a solution of the kind

qx 4 Dw = + Ax 3 + Bx 2 + Cx + D 24 For the beam element all loads are transformed into consistent nodal load so that q = 0. If a transverse force exists, then A ≠ 0 and the solution is a third degree polynomial (which is used as shape functions for an Euler-Bernoulli type beam element). This explains why the case of

12.26 SANDWICH AND FEM the cantilever with an edge point does not give the exact result in the pure bending case. It becomes overly stiff since the used shape function is a second degree polynomial for a problem which require a higher order polynomial to obtain an exact solution. In the case of cantilever with an edge bending moment the problem is slightly different. In such cases, the transverse force is zero and thus A in the above equation is zero. The exact solution is then given by a second order polynomial, which is also perfectly represented by the element shape functions. The solution for that problem will thus be exact even when using only one element.

12.4.13 Example of FEM-Calculation for a Simply Supported Sandwich Beam To exemplify the performance of the three-node shear deformable element, let’s take an example. A simply sandwich beam is studied according to Fig.12.13. q

Figure 12.13 Simply supported sandwich beam If we now calculate the edge displacement for a uniform applied pressure, q, the first two eigenfrequencies and the first two buckling modes, these can be compared with closed form analytical expression.

The mid-point displacement is given by

5qL4 qL2 w(L / 2) = + (12.55) 384D 8S where D is the bending stiffness [Nm2/m] and S the shear stiffness [N/m]. The first two free vibration eigen-frequencies are given by

D D ωπ= m 22 = m 22π (12.56) m 2 * 422 * 4 ⎡ D ⎛ mπ ⎞ ⎤ ρπφLm()1 + ρ L ⎢1 + ⎜ ⎟ ⎥ ⎣ S ⎝ L ⎠ ⎦ where ρ* is the mass per unit length of the beam, and φ is the shear factor, defined by φ = D/L2S. m is the vibration mode index (integer) for which vibration modes higher than 1 are used to calculate the eigen-frequencies. In this analytical expression, the effect of rotary inertia is neglected, i.e., R = 0.

The buckling loads are given by the expression

n 2π 2 D / L2 P = (12.57) cr,n 1+ n 2π 2 D / SL2 where n is buckling mode index (integer) which gives the successive buckling loads, i.e., n = 1 for the lowest buckling mode, n = 2 for the second, etc.

Take a 1 m long simply sandwich beam with faces of aluminium and a core of PMI foam. The materials and thickness are given below (all calculated for a unit width beam).

12.27 AN INTRODUCTION TO SANDWICH STRUCTURES

Face thickness: tf = 2 mm = 0.002 m 9 2 Face modulus: Ef = 70 GPa = 70·10 N/m 3 Face density: ρf = 2700 kg/m Core thickness: tc = 50 mm = 0.05 m 6 2 Core shear modulus: Gc = 30 MPa = 30·10 N/m 3 Core density: ρc = 75 kg/m Beam length: L = 1000 mm = 1 m Face neutral axis distance: d = tc + tf = 0.052 m Applied uniform pressure: q = 10 kPa = 10000 N/m2

The bending stiffness is then approximately equal to

Etd2 D ≈=ff 189 280 Nm2/m width 2 The beam cross-section shear stiffness is

G d 2 S ≈ c = 1 622 400 N/m width tc

The beam mass per unit length equals

* ρ = 2tfρf + tcρc = 14.55 kg/m and the rotary inertia is

ρρttd32ρ t 3 ρ td2 Rzdz==++≈=ρ 2 ff ff cc ff 0.0073 kgm/m ∫ 62122 The shear factor is hence for this beam equal to D φ = = 0.1167 LS2 We can now make some analytical calculation of beam bending, buckling and free vibration, and compare the results with those obtained by the use of the three-node shear deformable beam finite element. Let’s also perform the same calculations for the two extreme case; pure bending (S large) and pure shear (D large) to examine the performance of the element. The results are summarised in Table 1.

The general comments about the results obtained by the three-node shear deformable elements are as follows:

• The displacement gives very accurate results even for a small number of elements. By using only four elements mixed and pure shear cases compares almost exactly with the analytical results. In fact, the shear case should be exact using only element. In the bending case, the finite element solution is overly stiff, due to shear locking, as discussed above. However, by using 4 or more elements, the error becomes negligibly small.

12.28 SANDWICH AND FEM

No. of elements 1 2 4 8 16 64 analytical w(L/2) [mm] 1.321 1.454 1.458 1.458 1.458 1.458 1.458

Pcr(n=1) [N/mm] 946.4 876.7 868.9 868.3 868.3 868.3 868.3 Pcr(n=2) [N/mm] - 1377 1337 1333 1333 1333 1333 -1 ω1 [s ] 806.1 773.7 767.5 767.1 767.1 767.1 767.5 -1 ω2 [s ] 14907 1945 1911 1902 1901 1901 1902 Pure bending w(L/2) [mm] 0.550 0.550 0.654 0.679 0.686 0.688 0.688

Pcr(n=1) [N/mm] 22713 22713 1966 1892 1874 1868 1868 Pcr(n=2) [N/mm] - 9085 9085 7864 7569 7478 7472 -1 ω1 [s ] 1246 1246 1152 1130 1124 1123 1126 -1 7 ω2 [s ] 1.4 10 4948 4948 4575 4488 4461 4503 Pure shear w(L/2) [mm] 0.770 0.770 0.770 0.770 0.770 0.770 0.770 6 6 6 6 6 6 6 Pcr(n=1) [N/mm] 1.6 10 1.6 10 1.6 10 1.6 10 1.6 10 1.6 10 1.6 10 6 6 6 6 6 6 Pcr(n=2) [N/mm] - 1.6 10 1.6 10 1.6 10 1.6 10 1.6 10 1.6 10 -1 ω1 [s ] 1056 1053 1049 1049 1049 1049 1049 -1 ω2 [s ] 14908 2112 2105 2098 2098 2098 2098 Table 1 Comparison between analytical calculations and FEM for a simply supported beam • The buckling loads also compare well with analytical results, although a more than eight elements are required to give good results for the pure bending case. It should be noted here, however, that higher buckling modes will require more elements to give good results, since these are accompanied with more complicated deformation shapes.

• The first two eigen-frequencies give good results in the two cases mixed and pure shear by using only four elements. Again, the extreme case pure bending requires more element to yield accurate results.

Finally, let us compare the free vibration results for the case when the rotary inertia is omitted. This is done by letting R = 0, in the mass matrix of the element. For sixteen elements we get the following results when doing so:

-1 -1 ω1 [s ] ω2 [s ] Mixed case R = 0.0073 767.1 1901 R = 0 767.5 1902 Pure bending (S large) R = 0.0073 1125 4487 R = 0 1127 4532 Pure shear (D large) R = 0.0073 1049 2098 R = 0 1049 2098 Table 2 Comparison of FEM results, using 16 three-node shear deformable elements, when including or omitting the rotary inertia. As seen in Table 2, the effect of the rotary inertia is relatively small, at least for the first vibration modes.

12.29 AN INTRODUCTION TO SANDWICH STRUCTURES

12.4.14 More Examples First take an example of bending of sandwich beams with various edge supports. In all cases the beam is subjected to a uniform pressure and all other data are also the same as for cantilever beam example above. The effect of the support loads for a hyperstatic beam case, a beam with one edge clamped and other simply supported (C-S) is given in Chapter 4, and thus only the mid-point deflection will be calculated in this example. Three examples are considered; both edges simply supported (S-S), one edge clamped and other simply supported (C-S) and one with both edges clamped (C-C).

No. of elements 2 4 8 16 64 Analytical w(L/2) [mm] C-free 9.667 9.685 9.686 9.686 9.686 9.686 S-S 1.454 1.458 1.458 1.458 1.458 1.458 C-S 1.147 1.152 1.153 1.153 1.153 1.153 C-C 0.903 0.908 0.908 0.908 0.908 0.908 Table 3 Bending of sandwich beams using three-node finite element compared to analytical calculations Next, let’s do the same exercise but calculating the first buckling mode and its corresponding critical load for the same configurations as in the bending case.

No. of elements 2 4 8 16 64 Analytical Pcr(n=1) [N/mm] C-free 363.4 362.7 362.6 362.6 362.6 362.6 S-S 876.7 868.9 868.3 868.3 868.3 868.3 C-S 1090 1074 1072 1072 1072 1072 C-C 1384 1337 1333 1333 1333 1333 Table 4 Buckling of sandwich beams using three-node finite element compared to analytical calculations Finally, the same exercise again but calculating the first free vibration mode and its corresponding eigen frequency for the same configurations as in the bending case.

No. of elements 2 4 8 16 64 Analytical -1 ω1 [s ] C-free 322.0 321.3 321.3 321.3 321.3 321 S-S 773.7 767.5 767.1 767.1 767.1 767.5 C-S 871.9 862.8 862.1 862.1 862.1 862 C-C 985.9 972.7 971.8 971.8 971.8 972 Table 5 Free vibration of sandwich beams using three-node finite element compared to analytical calculations

12.4.15 Mixed Formulation for Two-Node Beam In mixed finite element formulations, not only displacements and rotations are used as variables, but also strains or stresses. This method will not be discussed herein since it belongs in a course on general finite element procedures. However, by using a mixed formulation, two-node shear deformable elements can be derived that does not have the shear locking problem. In the textbook by Bathe [7] a two-node beam element is derived using a mixed formulation for which the shear strain degree-of-freedom is condensed (removed) and

12.30 SANDWICH AND FEM a two-node element is formulated. The result is very close to that of the two-node pure displacement element derived above, and the stiffness matrix obtained is

⎡ S S S S ⎤ − − − ⎢ h 2 h 2 ⎥ ⎢ S Sh D S Sh D ⎥ ⎢− + − ⎥ e 2 4 h 2 4 h K = ⎢ ⎥ (12.58) ⎢ S S S S ⎥ − ⎢ h 2 h 2 ⎥ ⎢ S Sh D S Sh D⎥ ⎢− − + ⎥ ⎣⎢ 2 4 h 2 4 h ⎦⎥ where the underlined denominators (all 4’s) denote the difference between this element and the pure displacement element in eq.(12.27). It is reported by Bathe [7] that this element will have a much better predictive capability.

A two-node element derived from an assumed stress field is derived by Severn [8], and also referred to by Cook [1]. The results of this exercise is a stiffness matrix exactly the same as is if a common displacement method based unit deflections of a Timoshenko beam, see eq.(12.4).

12.5 Plate analysis Analogously to the beam elements above, the four-node and the eight-node plate elements are most common. Fig.12.5 shows the eight-node element [4]. The four-node element has nodes only at its corners. Here also, the (quadratic) eight-node element is superior. Both types of elements may take general quadrilateral shapes as is normal for isoparametric elements (the eight-node element can also have curved boundaries). The comment on the in-plane degrees of freedom in the previous section is equally valid for the plate elements. z, w top face core 1 2 x, u 3 bottom face

4 5 t t 6 7 8 b h θy t b θx y, v a

Figure 12.5 Eight-node sandwich plate element For uniformly loaded panels, meshes of two to three elements in each direction are usually sufficient, as in the sample problem in Fig.12.6. A square simply supported square sandwich plate with uniform load is considered [4], with data as

12.31 AN INTRODUCTION TO SANDWICH STRUCTURES

x tf = 0.2875 mm 1 by 1~mesh 2 by 2~mesh Ef = 142 GPa NE = 1 NE = 4 νf = 0.3 NN = 8 NN = 21 NV = 40 tc = 24.71 mm NA = 10/5 NV = 105 NA = 37/28 Ec = 0 4 by 4~mesh 6 by 6~mesh Gc = 22 and 220 GPa NE = 16 NE = 36 q = 0.05 MPa NN = 65 NN = 133 a = b = 1 m NV = 325 NV = 665 NA = 145/128 NA = 325/300

Figure 12.6 Element subdivisions. (NE= number of elements, NN = number of nodes, NV = number of variables, NA = number of active variables) One quarter of the plate (due to symmetry conditions) is analysed by four different meshes as shown in Fig.12.6. Here, NA indicates the number of active variables in the analyses. Due to the non-existence of in-plane loads, and a symmetric sandwich, the in-plane variables u are inactive (set to zero).

Results using eight-node plate elements for two different shear moduli of the core are given in Table 12.1 for the four different meshes.

Mesh Gc = 22 MPa Gc = 220 MPa normal stress in shear stress in core face (MPa) (kPa) 1x1 20.356 14.365 347 675 2x2 21.243 15.156 333 673 4x4 21.262 15.165 333 679 6x6 21.263 15.166 333 681 exact 21.3 15.2 333 684

Table 12.1 Midpoint deflection and maximum stresses for simply supported plate

12.6 Shear Deformable Plate and Shell Elements In this section we shall have a brief look at how a sandwich plate element is formulated. It is very similar to the derivation of the beam element. We shall further take a look at the most common type of plate or shell element, the isoparametric element. As for the beam, we start with the governing equations for the sandwich plate and construct an expression for the virtual work of the plate. We shall however restrict the analysis to plates that has well-defined neutral plane so that there is no coupling between bending moments and in-plane displacement and vice versa. In a common composites language this is usually formulated as a plate with zero bending-stretching action, i.e., with a zero B-matrix.

12.6.1 Governing equations For a generally anisotropic shear deformable plate, the equilibrium equations corresponding to eqs.(9) and (10) for the beam are (see chapter 8)

12.32 SANDWICH AND FEM

∂T ∂T ∂2w ∂2w ∂22w ∂ w x +++y qN +N +N +N =0 (12.59a) ∂x ∂y x∂x 2 y∂y2 xy∂∂xy yx ∂∂xy

∂M ∂M x + yx − T = 0 (12.59b) ∂x ∂y x

∂M ∂M y + xy − T = 0 (12.59c) ∂y ∂x y

Observe that the two last equations are written with a negative sign so that they constitute a bending moment in the same direction as the rotations, see Fig.12.7). These three equations now constitute the equilibrium equations for translation in the z-direction (w), rotation about the y-axis (θx) and rotation about the x-axis (θy), respectively. Now, if we multiply each and one of these equations with a virtual displacement and virtual rotations, denoted wv, θxv and θyv, we get the virtual work done by the internal forces on a differential element dx-dy. Integrating these over the entire plate gives us the virtual work for the entire plate. Thus,

⎛ ∂T ∂T ∂ 2 w ∂ 2 w ∂ 2 w ∂ 2 w ∂ 2 w ⎞ ⎜ x + y + q + N + N + N + N − ρh ⎟w = 0 ⎜ x 2 y 2 xy yx 2 ⎟ v ⎝ ∂x ∂y ∂x ∂y ∂x∂y ∂x∂y ∂t ⎠

⎛ ∂M ∂M ⎞ ⎜ x yx ⎟ ⎜ + − Tx ⎟θ xv = 0 ⎝ ∂x ∂y ⎠

⎛ ∂M ∂M ⎞ ⎜ y xy ⎟ ⎜ + − Ty ⎟θ yv = 0 ⎝ ∂y ∂x ⎠

These equations now form the virtual work done by the internal and applied loads per unit area. We can thus sum up the three contributions (they all form parts of the total virtual work and all have dimension load/length). To get the total work we merely integrate the sum over the entire plate, which also gives the weak form of the governing equation. This then becomes

⎛ ∂T ∂T ∂ 2 w ∂ 2 w ∂ 2 w ∂ 2 w ∂ 2 w ⎞ ⎜ x + y + q + N + N + N + N − ρh ⎟w dxdy + ∫∫⎜ x 2 y 2 xy yx 2 ⎟ v Ω ⎝ ∂x ∂y ∂x ∂y ∂x∂y ∂x∂y ∂t ⎠ (12.60) ⎛ ∂M ∂M ⎞ ⎛ ∂M ∂M ⎞ ⎜ x + yx − T ⎟θ dxdy + ⎜ y + xy − T ⎟θ dxdy = 0 ∫∫⎜ x ⎟ xv ∫∫⎜ y ⎟ xv Ω ⎝ ∂x ∂y ⎠ Ω ⎝ ∂y ∂x ⎠

We now need to integrate this by parts, but let’s do this term by term. Take the first term

⎛ ∂T ∂T ⎞ ⎛ ∂w ∂w ⎞ ⎜ x + y ⎟w dxdy = T dy + T dx w − ⎜T v + T v ⎟dxdy (12.61) ∫∫⎜ ⎟ v ∫ ()x y v ∫∫⎜ x y ⎟ Ω ⎝ ∂x ∂y ⎠ S Ω ⎝ ∂x ∂y ⎠

If we then proceed to the second integral, we integrate this by parts to become

12.33 AN INTRODUCTION TO SANDWICH STRUCTURES

⎛ ∂M ∂M ⎞ ⎜ x + yx − T ⎟θ dxdy ∫∫⎜ x ⎟ xv Ω ⎝ ∂x ∂y ⎠ (12.62) ⎛ ∂θ ∂θ ⎞ = M dy + M dxθ − ⎜ M xv + M xv + T θ ⎟dxdy ∫ ()x yx xv ∫∫⎜ x yx x xv ⎟ S Ω ⎝ ∂x ∂y ⎠ and similarly for the third integral

⎛ ∂M ∂M ⎞ ⎜ y + xy − T ⎟θ dxdy ∫∫⎜ y ⎟ yv Ω ⎝ ∂y ∂x ⎠ (12.63) ⎛ ∂θ ∂θ ⎞ = M dy + M dxθ − ⎜ M yv + M yv + T θ ⎟dxdy ∫ ()xy y yv ∫∫⎜ xy y y yv ⎟ S Ω ⎝ ∂x ∂y ⎠

The remaining part of the first integral does not need to be altered much, apart from the terms including the in-plane loads. These become

⎛ ∂ 2 w ∂ 2 w ∂ 2 w ∂ 2 w ⎞ ⎜ N + N + N + N ⎟w dxdy = ∫∫⎜ x 2 y 2 xy yx ⎟ v Ω ⎝ ∂x ∂y ∂x∂y ∂x∂y ⎠ ⎛ ∂w ∂w ∂w ∂w ⎞ ⎜ N dy + N dx + N dy + N dx⎟w (12.64) ∫⎜ x y xy yx ⎟ v S ⎝ ∂x ∂y ∂x ∂y ⎠ ⎛ ∂w ∂w ∂w ∂w ∂w ∂w ∂w ∂w ⎞ − ⎜ N v + N v + N v + N v ⎟dxdy ∫∫⎜ x y xy yx ⎟ Ω ⎝ ∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x ⎠

It may be convenient to make a little note about this term already at this stage. The boundary integral in the expression forms boundary loads. If the displacements of the boundaries are prevented, i.e., for simply supported or clamped plates, this term will be zero, since wv must zero just as w on such boundaries. If the edges are free, the integral may be non-zero, but only if the applied loads are non-conservative, i.e., if e.g., Nx dw/dx is non-zero. For conservative loads, the integral is zero. This follows the same reasoning as for the cantilever column where a distinction is made between Euler type I buckling and Beck’s column. Anyhow, we shall assume that this integral is zero from now on.

We can now write the complete expression as (after some rearrangement of terms)

⎛ ∂w ∂w ⎞ ⎛ ∂θ ∂θ ⎞ − ⎜T v + T v ⎟dxdy − ⎜ M xv + M xv + T θ ⎟dxdy ∫∫⎜ x y ⎟ ∫∫⎜ x yx x xv ⎟ Ω ⎝ ∂x ∂y ⎠ Ω ⎝ ∂x ∂y ⎠ ⎛ ∂θ ∂θ ⎞ − ⎜ M yv + M yv + T θ ⎟dxdy ∫∫⎜ xy ∂x y ∂y y yv ⎟ Ω ⎝ ⎠ (12.65) ⎛ ∂w ∂w ∂w ∂w ∂w ∂w ∂w ∂w ⎞ ⎛ ∂ 2 w ⎞ − ⎜ N v + N v + N v + N v ⎟dxdy − ⎜ ρh ⎟w dxdy ∫∫⎜ x y xy yx ⎟ ∫∫⎜ 2 ⎟ v Ω ⎝ ∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x ⎠ Ω ⎝ ∂t ⎠ = − T dy + T dx w − M dy + M dxθ − M dy + M dxθ − qw dxdy ∫ ()x y v ∫ ()x yx xv ∫ ()xy y yv ∫∫ v S S S Ω or rewritten slightly

12.34 SANDWICH AND FEM

⎛ ⎛ ∂w ⎞ ⎛ ∂w ⎞⎞ − ⎜T v θ + T ⎜ v θ ⎟⎟dxdy ∫∫⎜ x ⎜ xv ⎟ y ⎜ yv ⎟⎟ Ω ⎝ ⎝ ∂x ⎠ ⎝ ∂y ⎠⎠ ⎛ ∂θ ∂θ ∂θ ∂θ ⎞ − ⎜ M xv + M xv + M yv + M yv ⎟dxdy ∫∫⎜ x ∂x yx ∂y xy ∂x y ∂y ⎟ Ω ⎝ ⎠ (12.66) ⎛ ∂w ∂w ∂w ∂w ∂w ∂w ∂w ∂w ⎞ ⎛ ∂ 2 w ⎞ − ⎜ N v + N v + N v + N v ⎟dxdy − ⎜ ρh ⎟w dxdy ∫∫⎜ x y xy yx ⎟ ∫∫⎜ 2 ⎟ v Ω ⎝ ∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x ⎠ Ω ⎝ ∂t ⎠ = − T dy + T dx w − M dy + M dxθ − M dy + M dxθ − qw dxdy ∫ ()x y v ∫ ()x yx xv ∫ ()xy y yv ∫∫ v S S S Ω

The kinematic relations are obtained through Fig.12.7

θy,θyv

θx,θxv Mxy Nx dy q(x,y)

y z,w,wv dx Nx M N x My yx Tx

Myx Ty

Ny Rx

τxy σy w

τyx σx −θx τyz x τzy τzx τxz z u σ z

Figure 12.7 Sign convention used in the plate analysis.

u(x,y) = zθx, v(x,y) = zθy, w(x,y) = w (12.67) so that

∂u ∂θ ∂v ∂θ y ∂u ∂v ⎛ ∂θ ∂θ y ⎞ z x , ε = = z and z⎜ x ⎟ (12.68) ε x = = y γ xy = + = ⎜ + ⎟ ∂x ∂x ∂y ∂y ∂y ∂x ⎝ ∂y ∂x ⎠

From this, we can also identify the curvatures as (see eq.(8.3))

12.35 AN INTRODUCTION TO SANDWICH STRUCTURES

∂θ ∂θ ∂θ ∂θ κ ==x , κ y , and κ =+x y x ∂xyy ∂ xy ∂ yx∂ or in vector form as

⎛ ∂ ⎞ ⎜ 0 ⎟ t ⎜ ∂x ⎟ ⎛ ∂θ ∂θ ∂θ ∂θ ⎞ ⎜ ∂ ⎟⎛θ x ⎞ κ = ()κ κ κ t = ⎜ x y x + y ⎟ = 0 ⎜ ⎟ (12.69) x y xy ⎜ ⎟ ⎜ ⎟⎜θ ⎟ ⎝ ∂x ∂y ∂y ∂x ⎠ ⎜ ∂y ⎟⎝ y ⎠ ⎜ ∂ ∂ ⎟ ⎜ ⎟ ⎝ ∂y ∂x ⎠ so that eq.(12.67) takes the form

∂θ ∂θ ⎛ ∂θ ∂θ ⎞ z x z , ε = z y = zκ and ⎜ x y ⎟ (12.70) ε x = = κ x y y γ xy = z⎜ + ⎟ = zκ xy ∂x ∂y ⎝ ∂y ∂x ⎠

Note here the definition of rotations θ are positive rotations. The shear strains may now be written as ∂w ∂w γ = + θ and γ = + θ (12.71) xz ∂x x yz ∂y y and the transverse forces T are written as (from eq.(8.58))

⎛ ∂w ⎞ +θ ⎛T ⎞ ⎡S S ⎤⎛γ ⎞ ⎡S S ⎤⎜ x ⎟ ⎜ 1 ⎟ 11 12 ⎜ 13 ⎟ 11 12 ⎜ ∂x ⎟ ⎜ ⎟ = ⎢ ⎥⎜ ⎟ = ⎢ ⎥ ∂w or Τ = Sγ (12.72) ⎝T2 ⎠ ⎣S 21 S 22 ⎦⎝γ 23 ⎠ ⎣S 21 S22 ⎦⎜ ⎟ ⎜ +θ y ⎟ ⎝ ∂y ⎠

The bending moments are functions of plate rotations only and may be written

⎛ ∂θ ⎞ ⎜ x ⎟ ⎛ M ⎞ ⎡D D D ⎤⎜ ∂x ⎟ ⎜ x ⎟ 11 12 16 ⎢ ⎥⎜ ∂θ y ⎟ M = ⎜ M ⎟ = D D D ⎜ ⎟ = Dκ (12.73) y ⎢ 12 22 26 ⎥ ∂y ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎝ M xy ⎠ ⎣D16 D26 D66 ⎦ ⎜ ∂θ x ∂θ y ⎟ ⎜ + ⎟ ⎝ ∂y ∂x ⎠

For an orhtotropic plate with the notation given in chapter 8, these terms relate as

Dx Dy νyxDD x νxy y D11 = , D22 = , D12 = = , 2D66 = Dxy, D16 = D26 = 0 1− ννxy yx 1− ννxy yx 11− ννxy yx − ννxy yx

S11 = Sx, S12 = S21 = 0, and S22 = Sy (12.74) By using these kinematic expressions, the weak form of the equation becomes

12.36 SANDWICH AND FEM

− γ t Sγ dxdy − κ t Dκ dxdy ∫∫()v ∫∫ (v ) Ω Ω ⎛ ∂w ∂w ∂w ∂w ∂w ∂w ∂w ∂w ⎞ ⎛ ∂ 2 w ⎞ − ⎜ N v + N v + N v + N v ⎟dxdy − ⎜ ρh ⎟w dxdy (12.75) ∫∫⎜ x y xy yx ⎟ ∫∫⎜ 2 ⎟ v Ω ⎝ ∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x ⎠ Ω ⎝ ∂t ⎠ = − T dy + T dx w − M dy + M dxθ − M dy + M dxθ − qw dxdy ∫ ()x y v ∫ ()x yx xv ∫ ()xy y yv ∫∫ v S S S Ω

This equation has many similarities with eq.(12.13), especially if we include the expressions for the bending moments in eq.(12.73) and the shear definitions according to eq.(12.72).

12.6.2 Loads on an arbitrary plane On the right hand side we still have a few terms to study. First, look at the term

T dy + T dx w ∫ ()x y v S

The second term in this expression can be rewritten. If we have an arbitrary boundary of the plate we can deduce the following from Fig.12.8.

Tx Ty dx

x dy ds

s n Tn

Figure 12.8 Transverse force on an arbitrary surface From Fig.12.8 we can see the that the transverse force in the normal direction can be written as

t / 2 T = τ dz = T cosϕ + T sinϕ n ∫ nz x y −t / 2

We can also see that on the boundary we can write

dx = ds sinϕ and dy = ds cosϕ so that

Txdy + Tydx = (Tx cosϕ + Ty sinϕ)ds = Tnds The first term in the weak form of the equation then takes the form

T dy + T dx w = T w ds ∫ ()x y v ∫ n v S S

12.37 AN INTRODUCTION TO SANDWICH STRUCTURES

We can now similarly rewrite the boundary integrals

M dy + M dxθ + M dy + M dxθ ∫ ()x yx xv ∫ ()xy y yv S S in the last two obtained expressions by realising that

Mxdy + Myxdx = Mx ds cosϕ + Myx ds sinϕ

Mxydy + Mydx = Mxy ds cosϕ + My ds sinϕ We can also see that on the boundary we can transform the coordinate systems by

x = n cosϕ − s sinϕ and y = n sinϕ + s cosϕ

n = x cosϕ + y sinϕ and s = −x sinϕ + y cosϕ The rotation transform in the same manner as shear angles, thus

θx = θn cosϕ − θs sinϕ and θy = θn sinϕ + θs cosϕ We can then write

M dy + M dx θ + M dy + M dx θ ∫ []()()x yx xv xy y yv S = θ M cosϕ + M sinϕ +θ M cosϕ + M sinϕ ds ∫[]xv ()()x yx yv xy y S = θ M cos 2 ϕ + 2M cosϕ sinϕ + M sin 2 ϕ ds ∫[]nv ()x xy y S + θ − M cosϕ sinϕ + M cosϕ sinϕ + M cos 2 ϕ − sin 2 ϕ ds ∫[]sv ()x y xy () S = θ M +θ M ds ∫[]nv n sv ns S

The two boundary load terms are next to be evaluated further. These read

T w ds − θ M +θ M ds ∫ n v ∫[]nv n sv ns S S but

⎡ ⎛ ∂w ⎞⎤ ⎡ ∂w ⎤ θ M ds = M v − γ ds = M v − M γ ds ∫[]sv ns ∫ ⎢ ns ⎜ zsv ⎟⎥ ∫ ⎢ ns ns zsv ⎥ S S ⎣ ⎝ ∂s ⎠⎦ S ⎣ ∂s ⎦

⎡ ∂ ⎤ ⎡ ∂M ⎤ ⎡ ∂M ⎤ = w M ds − w ns ds − M γ ds = − w ns ds − M γ ds ∫ ⎢ ()v ns ⎥ ∫ ⎢ v ⎥ ∫[]ns zsv ∫ ⎢ v ⎥ ∫[]ns zsv S ⎣∂s ⎦ S ⎣ ∂s ⎦ S S ⎣ ∂s ⎦ S so that

12.38 SANDWICH AND FEM

⎡ ⎛ ∂M ns ⎞⎤ − θ nv M n ds + ⎢wv ⎜Tn + ⎟⎥ds − []M nsγ zsv ds ∫ ∫ ∂s ∫ S S ⎣ ⎝ ⎠⎦ S = − θ M ds + w V ds − M γ ds ∫ nv n ∫[]v n ∫[]ns zsv S S S where Vn is the support load, according to ordinary Kirchhoff plate theory. This expression actually provide the boundary conditions. We do not need to evaluate these all that much further, but we can have a quick discussion of them already at this point. At a free or simply supported edge, the bending moment Mn must be zero. For a simple or clamped edge we can stipulate two versions of the boundary condition; a soft support is given by zero twisting moment on the edge, i.e., with Mns = 0. For this case, the support load Vn = Tn. Another choice is to stipulate the shear to zero, called hard support, given by γzs = 0. This will get us back to the same type of boundary condition as in Kirchhoff plate theory.

The governing weak form of the equation then takes the final form

⎛ ∂ 2 w ⎞ γ t Sγ dxdy + κ t Dκ dxdy + ⎜ ρh ⎟w dxdy ∫∫()v ∫∫ (v ) ∫∫⎜ 2 ⎟ v Ω Ω Ω ⎝ ∂t ⎠ ⎛ ∂w ∂w ∂w ∂w ∂w ∂w ∂w ∂w ⎞ + ⎜ N v + N v + N v + N v ⎟dxdy (12.76) ∫∫⎜ x y xy yx ⎟ Ω ⎝ ∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x ⎠ = θ M ds − w V ds + M γ ds + qw dxdy ∫ nv n ∫[]v n ∫[]ns zsv ∫∫ v S S S Ω

We can already now identify the terms in eq.(12.76) providing the terms in the final matrix equation for the FE-analysis. The first two terms

γ t Sγ dxdy + κ t Dκ dxdy (12.77) ∫∫()v ∫∫ (v ) Ω Ω will provide the stiffness matrix. As seen, we can also here, given the assumptions made of a zero material coupling, divide this up in two parts; one for bending and one for transverse shear, just as in the beam case. The second term

⎛ ∂ 2 w ⎞ ⎜ ρh ⎟w dxdy ∫∫⎜ 2 ⎟ v Ω ⎝ ∂t ⎠ will provide the mass matrix for the element. The third term

⎛ ∂w ∂w ∂w ∂w ∂w ∂w ∂w ∂w ⎞ + ⎜ N v + N v + N v + N v ⎟dxdy (12.78) ∫∫⎜ x y xy yx ⎟ Ω ⎝ ∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x ⎠ gives us the geometric stiffness matrix, used for eigenvalue buckling analysis. The last term

θ M ds − w V ds + M γ ds + qw dxdy (12.79) ∫ nv n ∫[]v n ∫[]ns zsv ∫∫ v S S S Ω will give the boundary load vector and the vector of consistent nodal loads.

12.39 AN INTRODUCTION TO SANDWICH STRUCTURES

It may also be worth to make a notice on the derived equations. The first part of the weak governing equation in eq.(12.77) includes terms of the type

γ t Sγ dxdy ∫∫()v Ω where γ includes first derivatives of the displacement (w) and rotations (θ). For this term, only C0-continuity is obviously required. The second term

κ t Dκ dxdy ∫∫()v Ω includes term of curvature. In Kirchhoff plate theory, these would be second derivatives of the displacement (w), and would require C1-continuity. However, now the curvatures are defined as first derivatives of the rotations, uncoupled to the displacements, and thus only requires C0-continuity. Thus, a shear deformable plate element has a lower continuity requirement than an ordinary plate.

12.6.3 Element construction We can now proceed to form the element. We look, for simplicity, to a plane plate element with 3 degrees-of-freedom per node. It has 1 translations (w) and 2 rotations (θx, θy), see Fig.12.9.

θy η

θx w

Figure 12.9 Degrees-of-freedom for plane sheardeformable plate element. We take that the nodal degrees-of-freedom are

t d = ()w θ x θ y (12.80)

For element, the vector of nodal displacements, and rotations, will be

t d = ()w1 θ x1 θ y1 ...... wn θ xn θ yn (12.81) where n is the number of nodes in the element. The FE discretisation is done by means of shape functions, functions that interpolate the displacements and rotations using discrete nodal values. This is done by

n w(ξ,η) = ∑ N i (ξ,η)wi = Nw (12.82a) i=1

12.40 SANDWICH AND FEM

n θ x (ξ,η) = ∑ N i (ξ,η)θ xi = Nθx (12.82b) i=1

n θ y (ξ,η) = ∑ N i (ξ,η)θ yi = Nθ y (12.82c) i=1 where

t t t w = ()w1 ... wn , θ x = ()θ x1 ... θ xn and θ y = (θ y1 ... θ yn )

We can then write

⎛ w ⎞ ⎜ 1 ⎟ ⎜θ x1 ⎟ ⎜θ ⎟ ⎜ y1 ⎟ ⎛ w ⎞ ⎡N1 0 0 .. .. N n 0 0 ⎤ ⎜ ⎟ ⎢ ⎥⎜ .. ⎟ ⎜θ ⎟ = 0 N 0 .. .. 0 N 0 = Nd (12.83) x ⎢ 1 n ⎥⎜ .. ⎟ ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎝θ y ⎠ ⎣ 0 0 N1 .. .. 0 0 N n ⎦ ⎜ wn ⎟ ⎜ ⎟ ⎜θ xn ⎟ ⎜ ⎟ ⎝θ yn ⎠

We can now insert the material law into the relations. Eq.(12.72), including the shear stiffness, gives us that

⎛ ∂w ⎞ ⎡ ∂ ⎤⎛ w ⎞ ⎜ +θ x ⎟ ⎢ 1 0⎥⎜ ⎟ γ = ⎜ ∂x ⎟ = ∂x θ ∂w ⎢ ∂ ⎥⎜ x ⎟ ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ +θ y ⎟ 0 1 θ ⎝ ∂y ⎠ ⎣⎢∂y ⎦⎥⎝ y ⎠ ⎛ w ⎞ ⎜ 1 ⎟ ⎜θ x1 ⎟ ⎜θ ⎟ (12.84) ⎡∂N ∂N ⎤ y1 1 N 0 .. .. n N 0 ⎜ ⎟ ⎢ 1 n ⎥⎜ .. ⎟ = ∂x ∂x = B d ⎢∂N ∂N ⎥⎜ ⎟ s ⎢ 1 0 N .. .. n 0 N ⎥⎜ .. ⎟ ⎢ ∂y 1 ∂y n ⎥ ⎣ ⎦⎜ wn ⎟ ⎜ ⎟ ⎜θ xn ⎟ ⎜ ⎟ ⎝θ yn ⎠

Eq.(12.73), including the bending moments – curvature relations may now be written

12.41 AN INTRODUCTION TO SANDWICH STRUCTURES

⎛ w ⎞ ⎜ 1 ⎟ θ ⎡ ⎤ ⎡ ⎤⎜ x1 ⎟ ∂ ∂N1 ∂N n ⎜ ⎟ ⎢0 0 ⎥ ⎢0 0 .. .. 0 0 ⎥ θ y1 ⎢ ∂x ⎥⎛ w ⎞ ⎢ ∂x ∂x ⎥⎜ ⎟ ∂ ⎜ ⎟ ∂N ∂N ⎜ .. ⎟ κ = ⎢0 0 ⎥⎜θ ⎟ = ⎢0 0 1 .. .. 0 0 n ⎥ = B d (12.85) ⎢ ⎥ x ⎢ ⎥⎜ ⎟ b ∂y ⎜ ⎟ ∂y ∂y ⎜ .. ⎟ ⎢ ⎥ θ y ⎢ ⎥ ∂ ∂ ⎝ ⎠ ∂N1 ∂N1 ∂N n ∂N n ⎜ ⎟ ⎢0 ⎥ ⎢ ⎥ wn 0 .. .. 0 ⎜ ⎟ ⎣⎢ ∂y ∂x ⎦⎥ ⎣⎢ ∂y ∂x ∂y ∂x ⎦⎥ ⎜θ xn ⎟ ⎜ ⎟ ⎝θ yn ⎠

We can now insert these relations into the weak form of the governing equations (eq.(12.77)), leading to

γ t Sγ dxdy + κ t Dκ dxdy ∫∫()v ∫∫ (v ) Ω Ω = d t B t SB ddxdy + d t B t DB ddxdy ∫∫ v s s ∫∫ v b b Ω Ω

= d t B t SB dxdy ⋅ d + d t B t DB dxdy ⋅ d v ∫∫ s s v ∫∫ b b Ω Ω ⎛ ⎞ = d t ⎜ B t SB dxdy + B t DB dxdy⎟ ⋅ d = d t K + K ⋅ d v ⎜ ∫∫ s s ∫∫ b b ⎟ v ()s b ⎝ Ω Ω ⎠

The stiffness matrix can thus be decoupled, just as for the beam element, so that we can write the bending stiffness matrix as

K = B t DB dxdy (12.86) b ∫∫ b b Ω and the shear stiffness matrix as

K = B t SB dxdy (12.87) s ∫∫ s s Ω

The size of the matrix Bb is 3×3n, n being the number of nodes and D has the size 3×3, which means that Kb has the size 3n×3n. The size of the matrix Bs is 2×3n and S has the size 2×2, which means that Ks also has the size 3n×3n.

The problem remaining is to map the element geometry, which not necessarily has a square nice shape in the global xy-coordinate system. We can construct a displacement formulated element using a so called isoparametric formulation. Isoparametric implies that we use the same functions to map the geometry as we use for interpolation of displacements within the elements. To do this, we map the element in its global xy-coordinate system to an a so called parent element in a local ξη-coordinate system, as shown in Fig.12.10. The geometry of the parent element is square, with a intrinsic coordinate system ξ−η ranging from –1 to +1. This is the space in which the stiffness matrix is calculated.

12.42 SANDWICH AND FEM

y (x1,y1) (-1,1) η (1,1) (x ,y ) 1 2 1 2 2 2 x ξ

3 4 MAP 3 (-1,-1) (1,-1) (x3,y3) 4

Parent element Actual element (x4,y4)

Figure 12.10 Mapping of isoparametric element (example of 4-node element) The mapping is performed by means of the shape functions, as

n n x(ξ,η) = ∑ N i (ξ,η)xi = Nx and y(ξ,η) = ∑ N i (ξ,η)yi = Ny i=1 i=1

The derivatives are then readily found as

∂x ∂N ∂N ∂N ∂N ∂N = x = 1 x + 2 x + 3 x + 4 x ∂ξ ∂ξ ∂ξ 1 ∂ξ 2 ∂ξ 3 ∂ξ 4

∂x ∂N ∂N ∂N ∂N ∂N = x = 1 x + 2 x + 3 x + 4 x ∂η ∂η ∂η 1 ∂η 2 ∂η 3 ∂η 4

∂y ∂N ∂N ∂N ∂N ∂N = y = 1 y + 2 y + 3 y + 4 y ∂ξ ∂ξ ∂ξ 1 ∂ξ 2 ∂ξ 3 ∂ξ 4

∂y ∂N ∂N ∂N ∂N ∂N = y = 1 y + 2 y + 3 y + 4 y ∂η ∂η ∂η 1 ∂η 2 ∂η 3 ∂η 4 which then form the terms in the Jacobian. We then use the same shape functions N to interpolate the displacement and the rotations

n w(ξ,η) = ∑ N i (ξ,η)wi = Nw i=1

n θ x (ξ,η) = ∑ N i (ξ,η)θ xi = Nθx i=1

n θ y (ξ,η) = ∑ N i (ξ,η)θ yi = Nθ y i=1

12.43 AN INTRODUCTION TO SANDWICH STRUCTURES

where N is a vector of shape functions and w is a vector of nodal displacements, θx and θy are vectors of nodal rotations, as defined previously.

However, we wish to perform the integration in the element local co-ordinate system, and not in the global co-ordinate system. The co-ordinate transformation is performed using the shape functions, which also describe the mapping of the element. By using

⎛ ∂ ⎞ ⎛ ∂ ⎞ ⎡ ∂y ∂y ⎤⎛ ∂ ⎞ ⎜ ⎟ ⎜ ⎟ − ⎜ ⎟ ∂ξ 1 ⎢ ∂η ∂ξ ⎥ ∂ξ ⎜ ∂x ⎟ = J −1 ⎜ ⎟ = ⎢ ⎥⎜ ⎟ (12.88) ⎜ ∂ ⎟ ⎜ ∂ ⎟ J ⎢ ∂x ∂x ⎥⎜ ∂ ⎟ ⎜ ⎟ ⎜ ⎟ − ⎜ ⎟ ⎝ ∂y ⎠ ⎝ ∂η ⎠ ⎣⎢ ∂η ∂ξ ⎦⎥⎝ ∂η ⎠ we can write

∂θ x 1 ⎛ ∂y ∂θ x ∂y ∂θ x ⎞ 1 ⎛ ∂y ∂N i ∂y ∂N i ⎞ = ⎜ − ⎟ = ⎜ − ⎟θ x ∂x J ⎝ ∂η ∂ξ ∂ξ ∂η ⎠ J ⎝ ∂η ∂ξ ∂ξ ∂η ⎠

∂θ x 1 ⎛ ∂x ∂θ x ∂x ∂θ x ⎞ 1 ⎛ ∂x ∂N i ∂x ∂N i ⎞ = ⎜− + ⎟ = ⎜− + ⎟θx ∂y J ⎝ ∂η ∂ξ ∂ξ ∂η ⎠ J ⎝ ∂η ∂ξ ∂ξ ∂η ⎠

∂θ 1 ⎛ ∂y ∂θ ∂y ∂θ ⎞ 1 ⎛ ∂y ∂N ∂y ∂N ⎞ y ⎜ y y ⎟ ⎜ i i ⎟ = ⎜ − ⎟ = ⎜ − ⎟θ y ∂x J ⎝ ∂η ∂ξ ∂ξ ∂η ⎠ J ⎝ ∂η ∂ξ ∂ξ ∂η ⎠

∂θ 1 ⎛ ∂x ∂θ ∂x ∂θ ⎞ 1 ⎛ ∂x ∂N ∂x ∂N ⎞ y ⎜ y y ⎟ ⎜ i i ⎟ = ⎜− + ⎟ = ⎜− + ⎟θ y ∂y J ⎝ ∂η ∂ξ ∂ξ ∂η ⎠ J ⎝ ∂η ∂ξ ∂ξ ∂η ⎠ so that

⎡ ∂ ⎤ ⎢0 0 ⎥ ⎢ ∂x ⎥⎛ w ⎞ ⎡0 ai 0 ⎤⎛ w ⎞ ∂ ⎜ ⎟ ⎢ ⎥⎜ ⎟ κ = ⎢0 0 ⎥⎜θ ⎟ = 0 0 b ⎜θ ⎟ = B d (12.89) ⎢ ∂y ⎥ x ⎢ i ⎥ x b ⎜ ⎟ ⎢0 b a ⎥⎜ ⎟ ⎢ ∂ ∂ ⎥⎝θ y ⎠ ⎣ i i ⎦⎝θ y ⎠ ⎢0 ⎥ ⎣⎢ ∂y ∂x ⎦⎥ where

1 ⎛ ∂y ∂N i ∂y ∂N i ⎞ 1 ⎛ ∂x ∂N i ∂x ∂N i ⎞ ai = ⎜ − ⎟ and bi = ⎜− + ⎟ J ⎝ ∂η ∂ξ ∂ξ ∂η ⎠ J ⎝ ∂η ∂ξ ∂ξ ∂η ⎠ We then get that

K = B t DB dxdy = B t DB J dξdη (12.90) ∫ b b ∫ b b A A where the matrix Bb now is given by

12.44 SANDWICH AND FEM

⎡0 a1 0 .... 0 an 0 ⎤ B = ⎢0 0 b .... 0 0 b ⎥ b ⎢ 1 n ⎥ ⎣⎢0 b1 a1 .... 0 ai an ⎦⎥

Furthermore,

⎛ w ⎞ ⎜ 1 ⎟ ⎜θ x1 ⎟ ⎜θ ⎟ ⎡ ∂ ⎤⎛ w ⎞ ⎜ y1 ⎟ ⎢ 1 0⎥⎜ ⎟ ∂x ⎡a1 N1 0 .. .. an N n 0 ⎤⎜ .. ⎟ γ = ⎢ ⎥⎜θ x ⎟ = ⎜ ⎟ = Bs d (12.91) ∂ ⎢b 0 N .. .. b 0 N ⎥ .. ⎢ 0 1⎥⎜θ ⎟ ⎣ 1 1 n n ⎦⎜ ⎟ ⎢∂y ⎥⎝ y ⎠ ⎣ ⎦ ⎜ wn ⎟ ⎜ ⎟ ⎜θ xn ⎟ ⎜ ⎟ ⎝θ yn ⎠

What have we actually done with this? Well, we have assumed that the displacements w are uncoupled from the rotations θ. The continuity requirement, C0-continuity, implies that we should be able to use linear interpolation functions and construct a four-node plate element. These rotations would then vary linearly over the elements. Such an element can thus display a constant curvature since

∂θ ∂w κ = x , etc. and γ = +θ , etc. x ∂x xz ∂x x and a shear strain that varies linearly. However, such an element would suffer from shear locking, similar to what was discussed for the linear beam element previously. By using quadratic interpolation functions, creating an 8- or 9-node element, some of this shear locking would vanish and we would get an element that would perform reasonably well. The rotations would then vary quadratically giving an element that has bilinear variation in curvatures and some quadratic form for the shear strains. Actually, to remove shear locking problems, the element would require cubic interpolations, giving an element with 16 nodes.

There are ways to overcome a lot of the shear locking problems. Bathe [7] shows that using a so called mixed interpolation, a family of plate bending elements can be constructed, called MITC-elements (mixed interpolation of tensorial components). In this formulation, the displacement w, the rotation θ and the transverse shear strains γ are included. For more information on this, refer to Bathe [7].

For a shell element, the plate equations must be generalised to account for curvature of the t element and usually five degrees-of-freedom per node are used, i.e., ui = (u, v, w, θx, θy) . In some elements, even the sixth degree-of-freedom θz, is included. For a curved shell, displacement u and v correspond to in-plane deformations (in the local plane of the shell) and w to the displacement transverse to the plane of the shell, and the same for the rotations, i.e., θz, is the rotation in the transverse direction.

12.45 AN INTRODUCTION TO SANDWICH STRUCTURES

θz

θy u θy u

θx v θx v w w

Figure 12.11 Schematic examples of shell elements with 5 or 6 degrees-of-freedom per node

12.7 Boundary Conditions In section 9 different boundary conditions for Reissner/Mindlin plates were discussed and it was concluded that what seems to be a simple boundary condition can appear in different ways, e.g. hard or soft simple support or hard or soft clamped edge. Study any edge of a sandwich, let it be a beam, a plate or a shell, as depicted in Fig.12.12. Denote the normal to the surface by n and the tangential direction by t.

Mns = 0 γzs = 0

s γzs ≠ 0 Mns ≠ 0 z

”soft” ”hard”

Figure 12.12 Definition of boundary conditions for sandwich plates. In the general case (large deflections, buckling, etc.) the boundary conditions are: (i) Hard simple support

Mn = σn(n,s,z) = 0, w(n,s,z) = 0 and ∂vs(n,s,z)/∂s = 0 (ii) Soft simple support

Mn = σn(n,s,z) = 0, w(n,s,z) = 0 and Mnt = τzs(n,s,z) = 0 (iii) Hard clamped edge

un(n,s,z) = 0, w(n,s,z) = 0 and ∂vs(n,s,z)/∂s = 0 (iv) Soft clamped edge

un(n,s,z) = 0, w(n,s,z) = 0 and Mns = τnt(n,s,z) = 0 There are also other possibilities like the supersoft support [2] which states that the average transverse displacement w is zero and that the shear stress τzs is constant for the edge. The problem with the above boundary conditions is to model the prevented rotations in the hard boundary conditions while still allowing for tangential displacements, i.e. to keep ∂vs(n,s,z)/∂s

= 0 while vs ≠ 0. A soft boundary is much more easily created as the condition of a shear

12.46 SANDWICH AND FEM stress free edge implies no restrictions on the structure. Similar problems occur when modelling a clamped support in a case of buckling loading, e.g., where ∂un(n,s,z)/∂n = 0 while un ≠ 0.

When using special sandwich beam and shells elements (see Figs.12.3 and 12.5) these different boundary conditions are quite easily given since the elements contain rotational degrees of freedom, e.g. a clamped edge is given by preventing the rotation due to bending by setting θn=0. Hence, for beams and shells Hard simple support: Soft simple support:

θt(n,s,z) = 0 w(n,s,z) = 0 w(n,s,z) = 0 Since on the boundary we have that ∂w = γ −θ = 0 from eq.(8.13) and hence θs = 0 for a hard simple support. ∂z zs s Hard clamped edge: Soft clamped edge:

θn(n,s,z) = 0, θs(n,s,z) = 0 θn(n,s,z) = 0 w(n,s,z) = 0 w(n,s,z) = 0 In FE-analysis using shell elements, the application of hard or soft boundary conditions is rather straight-forward. An example of a simply supported edge with soft and hard boundaries is shown in Fig.12.13.

θx=0 θx=0 θx=0

w=0 w=0 w=0 w=0 w=0 w=0

”Soft simple support” ”Hard simple support”

Figure 12.13 Examples of soft and hard simple support For a sandwich structure built-up by rods and membranes (2D sandwich beam) or by membranes and solids (3D sandwich plate or shell) the boundary conditions are the same but become a little more complicated. For instance, at a simple support or clamped edge it is not strictly correct to assume that w = 0 on the entire boundary. This would mean that there is locally no deformation in the thickness direction of the plate or shell at the boundaries. Instead one can place the restriction on, e.g., the lower face only, if that simulates the real case in a more accurate manner. However, the stress perpendicular to the faces are usually of minor importance unless local effects, such as point loads, inserts and joints, are studied.

12.47 AN INTRODUCTION TO SANDWICH STRUCTURES

In linear small deformation analysis (not buckling) the in-plane deformation of the middle plane un is automatically set to zero which means that the hard boundary conditions become easy to obtain. The nodal restrictions are then Hard simple support: Soft simple support:

vs(n,s,z) = 0 all nodes w(n,s,z) = 0 all nodes/along a line w(n,s,z) = 0 all nodes/along one line Hard clamped edge: Soft clamped edge:

un(n,s,z) = 0, vs(n,s,z) = 0 all nodes un(n,s,z) = 0 all nodes w(n,s,z) = 0 all nodes/along one line w(n,s,z) = 0 all nodes/along one line For example, for a simply supported rectangular plate as shown in Fig.9.2 the hard boundary conditions are stated as v = w = 0 for x = 0,a u = w = 0 for y = 0,b whereas a soft boundary conditions states that only w = 0 for the nodes on the edges.

12.8 Spurious zero energy modes When the Jacobian is constant and the order of integration is such that one obtains the exact result it is denoted as full integration. A four-node linear 2D membrane requires 2×2 integration points and an eight-node 2D element requires 3×3 integration points for full integration. However, when the elements are distorted, the Jacobian matrix is not constant. Thus, for best results, elements should be as little distorted as possible. Due to this, and the fact that many plate and shell elements use a too low order of approximation (non- conforming) yields that an exact integration scheme often leads to an overestimation of the stiffness. A full integration scheme is in fact also costly. Therefore, in many cases on may choose to use a lower order integration scheme, reduced integration. This leads to less computation time and "softens" the elements and can lead to better results. The reason for the overly stiff structure is that the order of the shape functions may be too low to accurately represent the true displacement field. This provides a constraint that prevents the structure from deforming the way it wants too. As in any energy minimisation procedure, a too low order displacement field assumption leads to a too high stiffness. Now, although there are advantages for using reduced integration, additional problems appear. This problem is referred to as spurious zero energy modes

Zero energy modes arise when a pattern of nodal degrees-of-freedom produces a strain field that is zero at all integration points. The numerical integration scheme appears in general as

n p t K = ∑α i Bi C i Bi i=1

12.48 SANDWICH AND FEM

where Bi and Ci means that B and C are evaluated at Gauss point i. The term αi depends on the Gauss weights and the determinant of the Jacobian. These are both positive so that αi > 0. The elastic strain energy can be written

n p n p 1 t 1 t t 1 t U (d) = d Kd = ∑α i d Bi C i Bi d = ∑α iε i C iε i 2 2 i=1 2 i=1 where we have used that ε = Bd. By definition, a zero-energy mode exists when U = 0 for a displacement vector d ≠ 0. The constitutive relation C is a positive definite matrix. It then follows from the above that the strain energy U can only be zero if εi is zero for i = 1, … np. This gives us a system of equations of the type

⎡ B1 ⎤ ⎢ B ⎥ ⎢ 2 ⎥

⎢ B3 ⎥d = 0 ⎢ ⎥ ⎢ ... ⎥ ⎢B ⎥ ⎣ n p ⎦

We now look for a non-trivial solution to this equation system so that d ≠ 0. If we take, for example 2D elasticity, the dimension of Bi is 3×2n, n being the number of nodes. The displacement vector d has dimension 2n×1. The dimension of the resulting vector is then 3np×1. For the equation system we then have 2n unknowns and 3np equations. Then, if 2n > 3np there are more unknowns that equations and we have at least 2n − 3np non-trivial solutions of d. Therefore, there are 2n − 3np zero energy modes. Considering that 3 of them represent rigid body motion, there are 2n − 3np − 3 spurious zero energy modes.

We can also see it the following way. The maximum rank of

t K = α i Bi C i Bi is the same as the maximum rank of C. In the 2D case, the rank of C is 3. The maximum rank of the summation is then rank(C) × np. The number of equations is n (number of nodes) × dof (the number of degrees-of-freedom per node). The number of spurious zero energy modes is then (at least)

(n × dof) − rank(C) × np − 3 We can also write n × dof as the order of the equation system. For one element the order is the size of the element stiffness matrix.

Let's again have a look at 2D elasticity. A four-node membrane element is integrated using

n p t K = ∑α i Bi C i Bi i=1

Since B in general is a linear polynomial, the stiffness matrix is then obtained by integrating a second order polynomial. To do this using Gauss integration requires 2×2=4 Gauss points. However, if we choose to use 1×1 Gauss integration, we get that the number of spurious zero

12.49 AN INTRODUCTION TO SANDWICH STRUCTURES

energy modes are (at least) (n × dof) − rank(C) × np − 3 = 4 × 2 − 3 × 1 − 3 = 2, i.e., two such modes. These are depicted in Fig.12.14

η ξ

Figure 12.14 Spurious zero energy modes for four-node membrane element using one integration point However, by applying some additional boundary conditions one can prevent these modes. It should be pointed out one must in all cases ensure that these modes cannot appear by applying additional conditions. In the 8-node quadratic 2D membrane element, the order of the polynomial degree to evaluate K is four and thus 3×3 integration is required. By a reduced integration scheme (2×2) the number of spurious zero energy modes is (at least) (n × dof) − rank(C) × np − 3 = 8 × 2 − 3 × 4 − 3 = 1.

η ξ

Figure 12.15 Spurious zero energy modes for four-node membrane element using one integration point We have the same situation with 3D elements. A linear solid element with eight nodes integrated with one Gauss point exhibits (at least) (n × dof) − rank(C) × np − 3 = 8 × 3 − 6 × 1 − 6 = 12, and a quadratic 20-node brick with reduced integration results in (at least) (n × dof) − rank(C) × np − 3 = 20 × 3 − 6 × 8 − 6 = 6 zero energy modes.

Going back to the Reissner/Mindlin plate element we can make the following observations. The stiffness matrix is given by (in decoupled form)

K = B t D B dxdy = K + K = B t DB dxdy + B t SB dxdy ∫∫ b tot b b s ∫∫ b b ∫∫ s s Ω Ω Ω

The rank of the total bending and shear stiffness matrix Dtot is 5, or less. With one integration point the rank of K is thus 5, or less. With a 2×2 integration scheme the highest possible rank of K is 20. Furthermore, there are 3 possible rigid body displacements. Now, the element stiffness matrix for a 4-node plate has a size of 12, i.e., the order is 12. Integrating K using one Gauss point implies that the number of possible eigenvalues to a matrix of order 12 is exactly 12. Of these 12, there are 3 rigid body motions. The rank is 5 (or less) meaning that there are 4 (or more) zero energy modes.

12.50 SANDWICH AND FEM

We can use different integration schemes in the decoupled form. These are given in the table below.

Integration rule Integration rule Rank Zero energy Shear Kb [rank 3] Ks [rank 2] modes constraints (lower bound) 1×1 1×1 5 12-5-3=4 2 2×2 1×1 4×3+1×2=14 12-14-3<0 2 12 dof (actually 2) 2×2 2×2 12+8=20 24-20-3=1 8 27+8=35 0 8 3×3 2×2 24 dof 3×3 3×3 27+18=45 0 18 2×2 2×2 12+8=20 27-20-3=4 8 27+8=35 27-34-3<0 8 3×3 2×2 27 dof (actually 1) 3×3 3×3 27+18=45 0 18

Shear constraints are applied for thin plates, which reduce the number of spurious zero energy modes.

12.9 Effects of reduced integration We can discuss the effect of reduced integration in the context of shear locking effects as a complement to previous discussions. Lets use the shear deformable beam element as a reference. A two-node displacement formulated element suffers from shear locking. The strain energy (excluding energy from applied loads) reads

2 2 2 1 h ⎛ ∂θ ⎞ 1 h ⎛ ∂w ⎞ 1 h ⎛ ∂θ ⎞ 1 h U = ∫ D⎜ ⎟ dx + ∫ S⎜ +θ ⎟ dx = ∫ D⎜ ⎟ dx + ∫ Sγ 2 dx 2 0 ⎝ ∂x ⎠ 2 0 ⎝ ∂x ⎠ 2 0 ⎝ ∂x ⎠ 2 0

From this, we obtain the stiffness matrix K = Kb + Ks, where Kb represents the bending part (the first term) and Ks the shear part (the second term), so that

(Kb + Ks) d = F

The displacement vector d for a thin beam is governed by Kb because transverse shear effects are small, or negligible. However, we cannot simply disregard Ks since it couples rotations θ to displacements w. But, as the beam becomes more slender, Ks grows in relation to Kb. The term D is proportional to the thickness cube and the shear stiffness S to the thickness (for a homogeneous beam). The shear term will thus dominate. This in turn enforces that γ = ∂w/∂x − θ = 0. A linear element implies that a two point integration scheme is required for exact integration (second order polynomial). Now, every additional element bring with it two additional degrees-of-freedom (one θ and one w). Integrated by a two-point rule, it also brings two additional shear constraints γ = 0. This implies that the mesh locks since any addition of more elements in fact does not add to the total number of degrees-of-freedom. However, if the element is integrated by a one-point rule (reduced integration), there is only one additional shear constraint, Ks becomes singular and the elements works better. Furthermore, since ∂θ/∂x is constant, Kb is integrated exactly by a one-point rule.

12.51 AN INTRODUCTION TO SANDWICH STRUCTURES

For a plate or shell element, the same arguments apply. Using reduced integration for Ks makes Ks singular and shear locking effects are reduced. However, it introduces possible spurious zero energy modes.

12.10 Point loads in plate formulations Study a circular plate subjected to a point load in the middle, as depicted in Fig.12.16.

Figure 12.16 Circular plate with radius R subjected to point load The exact solution can be obtained analytically by solving the Kirchhoff plate equation in cylindrical co-ordinates. It reads

P(1−ν ) ⎡3 +ν r ⎤ P(1−ν ) 3 +ν w = (R 2 − r 2 ) − 2r 2 ln with w(0) = R 2 16πD ⎣⎢1+ν R⎦⎥ 16πD 1+ν

for a simply supported plate, and

2 2 P(1−ν ) ⎡ 2 2 ⎛ r ⎞⎤ P(1−ν ) 2 w = ⎢R − r ⎜1+ 2ln ⎟⎥ with w(0) = R 16πD ⎣ ⎝ R ⎠⎦ 16πD for a clamped plate. The bending moment can be obtained through

D ⎡∂ 2 w ν ∂w⎤ M r (r) = − 2 ⎢ 2 + ⎥ (1−ν ) ⎣ ∂r r ∂r ⎦ using (clamped edge case)

∂w P(1−ν 2 ) ⎡ ⎛ r ⎞ ⎤ P(1−ν 2 ) r = ⎢− 2r⎜1+ 2ln ⎟ − 2r⎥ = − r ln ∂r 16πD ⎣ ⎝ R ⎠ ⎦ 4πD R

∂ 2 w P(1−ν 2 ) ⎡ r ⎤ = − 2 + ln 2 ⎢ ⎥ ∂r 4πD ⎣ R⎦

P ⎡ r ⎤ ν P r P ⎡ r r ⎤ M r (r) = 2 + ln + r ln = 2 + ln +ν ln 4π ⎣⎢ R⎦⎥ r 4π R 4π ⎣⎢ R R⎦⎥

This means that the bending moment

1 M ∝ ln → ∞ as r approaches zero r r

12.52 SANDWICH AND FEM

Thus, the displacements are finite, but the stresses (proportional to the bending moments) are singular!!!

In Reissner-Mindlin theory the situation is somewhat different. Let's just look at the shear deformation part of the total deformation. In pure bending (S is large) the displacements are finite. In the rotational symmetric problem at hand we can solve for the shear deformation as follows; Study a circular plate subjected to a point load P in the middle, as shown in Fig.12.17. P

Figure 12.17 Circular plate subject to uniform pressure q over radius a The transverse shear force is P T = − r 2πr The shear deformation is then

r T r P P P P ⎛ R ⎞ w (r) = r dr = − dr = − ln r + ln R = ln s ∫ ∫ ⎜ ⎟ R S R 2πrS 2πS 2πS 2πS ⎝ r ⎠

This means that ws = ∞ at the centre of the plate, and therefore, the solution contains a singularity at r = 0. The energy is thus not finite under a point load, just as in 3D. This means that for thin plate and shell elements (Kirchhoff type), point loads can be applied, whereas for thick plate and shell elements (Mindlin type), they cannot as the solution will not converge.

12.11 Shell Elements Since plate elements are often obtained by degenerating shell elements [5,6], most of what is mentioned above for plates is also valid here. The specific issue that has to be dealt with for shells is the curvature(s) of the middle surface of the shell. If the smallest radius of curvature is in the order of magnitude of the thickness of the shell (say less than ten times), then local effects such as normal stresses in the core have to be considered in the analysis.

12.12 Alternative Modelling of Sandwich Structures A sandwich can be modelled by assembling standard finite elements to form the components of the sandwich, the faces and core. This is perhaps the most rigorous approach but is also a very costly one. A sandwich beam, for example, can be modelled by membrane elements for the core and rod elements for the faces. The membrane and the rod should have the same type of implementation so that compatibility is obtained for the common edges of the elements, e.g., if an eight-node isoparametric membrane is used for the core, three-node isoparametric rods should be used for the faces. If so, all nodes have only translation degrees of freedom, that is, u, v and w, and since the shape functions are common to both element types complete compatibility is obtained. Using rods to model the faces implies that the bending stiffness of the faces is omitted, i.e. Df = 0. Another discrepancy occurs in this type of modelling, as schematically shown in Fig.12.18; the distance between the centroids of the faces is placed at

12.53 AN INTRODUCTION TO SANDWICH STRUCTURES

the distance tc rather then d since the centreline of the rod is modelled along and on the face of the membrane.

d tc

Figure 12.18 Schematic of a sandwich beam assembly of membrane and rod finite elements.

Thus, the extensional stiffness of the face (Eftf1) is all placed at the position tc/2 from the middle axis. If the faces are relatively thick an error corresponding to that is made. If the core instead is modelled with thickness d, the bending stiffness D0 will be correct but a minor error 3 is made in Dc which then instead equals Ecd /12.

If the bending stiffness of the faces must be accounted for, the rods must be substituted with beam elements (a beam has extensional and bending stiffness, a rod only extensional). However, as a beam is formulated with rotational degrees of freedom as well as translation, an incompatibility arises. The nodal displacements are shared and they are the same for the membrane and the beams, but the lines between the nodes may have different shapes. This should normally not influence the result very much but should be noticed. If the computational efforts allows, the faces could instead be modelled with membrane elements, greatly increasing the number of degrees of freedom, but removing the incompatibility. There are usually some modelling problems occurring when doing so, since membrane elements should be kept within certain side length aspect ratios, normally 1:10, in order to give good accuracy. For thin face sandwiches this becomes an almost impossible task.

For plate or shell problems the same reasoning as above holds, but now the core is modelled with solid (brick) elements and the faces with membranes. Once again, the solids and the membranes must have the same formulation. If the face bending stiffness must be accounted for, shells can be used to model the faces but then the incompatibility occurs again as with the beams discussed above. Here also the faces may be modelled with solids as well, but here even greater problems arise since most solids should have a side length aspect ratio in the order of 1:4 to give good accuracy.

12.54 SANDWICH AND FEM

In order to avoid large problems with many elements and degrees of freedom, and further avoiding complex and difficult geometrical modelling due to element side length aspect ratio, many FE-codes have now implemented special sandwich beam, plate and shell elements. In such elements the assembly shown in Fig.12.18 is reduced to a three-node beam which accounts for transverse shear deformation, and instead of using solid elements for plates eight-node shells are used.

References [1] Cook R.D., Malkus R.S., and Plesha M.E., Concepts and Applications of Finite Element Analysis, J Wiley & Sons, New York 1989.

[2] Szabó B. and Babuška I, Finite Element Analysis, John Wiley & Sons, Inc. New York, 1991.

[3] Wennerström H. and Bäcklund J., Static, Free Vibration and Buckling Analysis of Sandwich Beams, Department of Lightweight Structures, Royal Institute of Technology, Report 86-3, Stockholm 1986.

[4] Wennerström H. and Bäcklund J., Plane Finite Elements for Static Analysis of Stiffened Sandwich Constructions, Department of Lightweight Structures, Royal Institute of Technology, Report 83-6, Stockholm 1983.

[5] Wennerström H. and Bäcklund J., Cylindrical Finite Elements for Static Analysis of Stiffened Sandwich Constructions, Department of Lightweight Structures, Royal Institute of Technology, Report 84-3, Stockholm 1984.

[6] Wennerström H. and Bäcklund J., General Curved Finite Elements for Static Analysis of Sandwich Shell Structures, Department of Lightweight Structures, Royal Institute of Technology, Report 85-1, Stockholm 1985.

[7] Bathe K.-J. Finite Element Procedures, Prentice-Hall, New Jersey, 1996.

[8] Severn R.T., ”Inclusion of Shear Deflection in the Stiffness Matrix for a Beam Element”, Journal of Strain Analysis, Vol 5, No 4, pp 239-241, 1970.

12.55 CHAPTER 13

JOINTS AND LOAD INTRODUCTIONS by Jørgen Kepler and Arne Kildegaard

This chapter will present some practical aspects of the problems of joining sandwich elements and how to take care of large point loads. There is limited theoretical or analytical analysis involved, rather solutions to practical problems as they have been solved for real structures. These are of course based on some fundamental issues concerning the sandwich concept.

Joints and load introductions are present in every structure, not always because the designer or the demands on structural behaviour so specifies, but rather because there is no possibility to manufacture, assemble, or repair the structure unless it consists of several parts which need to be joined. Although sandwich structures, especially those using composite materials, often have the advantage of allowing integrated manufacturing thus producing fewer parts than using conventional materials and structural concepts, components must be joined and loads introduced. The joint or load introduction may in itself be quite a complex structure in its own, consisting of several parts that must be joined. 13.1 Inserts Loads and reactions on a structural panel may in general be applied to the edges or somewhere on the surface of the panel. Usually, the edges are supported and the load is applied to the surface. Whatever the case, any localised load directly applied to a sandwich panel will cause deformations rather unlike those in a similar steel or aluminium panel.

An insert is a local change in stiffness and strength of the sandwich panel, the purpose of which is to distribute a localised load in an appropriate manner to the sandwich panel.

Sandwich plates are basically unsuited for carrying localised loads, for the simple reason that the core does not have the stiffness necessary to distribute the forces effectively. Therefore, a sandwich structure should ideally be constructed so that inserts are avoided. As this would clearly be impractical, it is important to choose an appropriate type of insert. This chapter will in particular treat circular inserts. Other types of inserts falling within the definition above will be mentioned in passing.

13.1 AN INTRODUCTION TO SANDWICH STRUCTURES

Consider first the case of a concentrated load on the surface of a panel. At any closed section surrounding the load, the conditions of equilibrium must be satisfied (assuming a static situation). The closed path is denoted Π, and the concentrated load is in this case a transverse force Q, see Fig.13.1. This means that by integrating the transverse force components T along Π, the resultant must be −Q. Π Q M r r T

Figure 13.1 Equilibrium of a section defined by a closed (circular) path around a concentrated load. In this case, the concentrated load is a force Q [N], T [N/m] is the transverse reaction at any given point on the path Π. Since the length of Π increases by r, the reaction T must decrease by r. In the cases of a concentrated force and a concentrated bending moment on a plate, the dependency is 1 Force F: T ≈ (total shear component T proportional to 1/r) r

1 Moment M: T ≈ (total shear component T proportional to 1/r2 ) r 2

The 1/r or 1/r2 dependency indicates that the affected zone is small, and a calculation model needs not generally include the whole panel.

The above considerations are true for thin plates. However, a sandwich plate is not necessarily thin compared to the length or width. Thus, a significant amount of bending moment M can be transferred through the face sheets as in-plane tension/compression, particularly for small r (until the radius is sufficiently large compared to the thickness). This is shown in Fig.13.2. τ

M σ

Figure 13.2 Primary bending moment-transferring reactions in sandwich panel. In-plane stresses σ and τ are transferred primarily by the face-sheets, whereas T, at small radii, is composed of significant contributions from both face sheets and core. The region considered for calculation purposes must be estimated with this in mind. For a sandwich plate, T itself will also be distributed between face-plates and core. The distribution will, as hinted above, depend upon the parameters of the actual situation (geometry, materials and loads).

13.2 JOINTS AND LOAD INTRODUCTIONS

13.1.1 The elements involved Apart from the basic structural element, the sandwich panel, two more elements are of interest: the insert and sometimes a fixing agent (adhesive or potting compound which holds the insert in place). Inserts and fixing agents are briefly described below.

Fig.13.3 shows some different types of inserts, divided into 4 categories. 1234

Figure 13.3 Insert types, shown in cross-section and as seen from above. The categories are: 1: Self- tapping screws and rivets, affecting one face sheet. 2: Through-the-thickness inserts (cylindrical, skin tie, diaphragm), affecting both face sheets and core. 3: Partial inserts, directly affecting only one face sheet and the core. 4: Flared cylinder (top hat) bonded onto one face sheet The categories 1 and 4 may not exactly qualify as inserts in the strict sense of the word. They are included because they are also used for applying loads to the panel surface. 1. Self-tapping screws and rivets are “low-grade” inserts, not intended to carry bending moments. They may be used for attaching light equipment to panels at points where these panels are not otherwise subjected to any significant loads. 2. Through-the-thickness inserts can directly transfer shear force to the core and bending moments and in-plane forces to the face sheets. They may be used for applying significant loads to the sandwich panels. 3. The partial insert is directly attached to one face sheet and the core. It is used where through-the-thickness inserts are undesirable (in ship-building, it may be necessary to leave at least one face sheet undamaged). It should, like self-tappng screws and rivets, be avoided for transfer of bending moment and in panels subjected to significant shear force. 4. The adhesively bonded cylinder relies on the adhesive joint for transfer of forces and bending moments. Thus, for transfer of in-plane forces, it is not fail-safe in the same manner as the partial insert. However, it is useful in panels subjected to a high, uniform shear load, since it does not disrupt the core.

This chapter is concerned primarily with cylindrical inserts made from metal or some other stiff and strong material. Since the modulus of elasticity of the insert thus is considerably higher than that of FRP face-sheets and much higher than the core material, calculations may usually be made under the assumption of infinitely rigid inserts, to within a fair degree of precision.

13.3 AN INTRODUCTION TO SANDWICH STRUCTURES

An insert is held in place by a fixing agent which is considered isotropic; it may be a potting compound or adhesive. The stiffness of the fixing agent is usually about ten times that of the core and applied in thin layers. Therefore, the fixing agent can most often be assumed infinitely rigid relative to the core-insert interface i.e. it is considered part of the insert. The assumptions at the insert/face-sheet interface vary, depending on the exact conditions.

13.1.2 Face sheet/insert interface In Fig.13.4, an interface element at the edge of an insert is shown. In general, the force resultants Nrϕ , Nr , Tr [N/m] and bending moment resultants Mr , M rϕ [Nm/m] are all present at any such sections of a plate, according to common plate theory. However, the choice of insert will decide which ones are significant.

t Nrϕ M z Nr rϕ Tr Mr ϕ r

Figure 13.4 Differential interface element at face-sheet/insert interface, and reaction bending moments and forces. The height of the element is equal to the thickness of the face sheet, and the width is rdϕ. The interface position is taken according to Fig.13.5 for some different inserts. Note that in cases (b) and (c), the interface is chosen at the edge of the flared part. z z z z r r r r

(a) (b) (c) (d)

Figure 13.5 Interface positions (at the small squares). (a): Simple cylindrical insert. (b): Cylindrical insert with flared top. (c): “Top Hat”. (d): Rivet or self-tapping screw Case (a): The adhesive or potting compound that holds the insert in place is usually much softer and weaker than both insert and face plate. The moments Mr and M rϕ are made up of stress variations across the thickness of the face layer, but since this thickness is small, Mr and M rϕ are also small and this load transfer path can usually be ignored when considering the global stiffnes of the insert design. In this case, the adhesive layer acts as a hinge between the face sheet and the insert in case of a transverse load. However, the bending moment stresses may easily be large enough to break the adhesive itself, causing a circular crack in the face- sheet/insert interface. This will further reduce the transferable reactions to only compressive Nr. If the insert is subjected to an in-plane force or (in the case of through-the-thickness inserts) bending moment-vector, this is often sufficient.

Cases (b) and (c): The function of the flared disc, apart from giving a larger bonded surface and increasing the radius (due to the 1/r-dependence) is to allow the transfer of bending moment and significant transverse force T by the face sheets, thus stiffening the joint and

13.4 JOINTS AND LOAD INTRODUCTIONS minimising any stress concentrations in the core. While the face-sheet bending moment will certainly contribute to the stiffness, it may cause large bending stresses in the face sheets (because the deformation of the face sheets is often largely dictated by the core). The obvious solution to this is either locally increasing the thickness of the face sheets or reducing the stiffness of the adhesive layer underneath the flared disc, so that the bending moment does not become critical.

Case (d): The screw or rivet is often chosen for being convenient and cheap. Since the radius r of the “insert” is very small, and because of the geometric discontinuities arising in the core, a screw or rivet cannot transfer as much load as the larger inserts. In general, Tr is reliable in this case; both screws and rivets will grip the face sheet from underneath. Only the compressive part of Nr should be considered, since neither screws nor rivets are normally fixed by use of adhesive. Bedning moments on single screws should be avoided. Instead, a bending moment load can be distributed to several screws, the distance between these giving a much larger lever arm.

13.1.3 Core/insert interface In the core, an insert is usually held in place by a adhesive or potting compound. Structurally, the adhesive should be considered part of the insert, since it is generally 5-10 times as stiff as the core. With a proper bond, all stress components are present at any point in the interface, as outlined in Fig.13.6.

σz τ τzϕ zr

τ rϕ σ z z r τrz ϕ ϕ r r

Figure 13.6 Interface elements between insert and core, at bottom (for partial insert) and side of insert. The core is the weaker part and in the case of partial inserts in particular, stress concentrations in the core tends to cause cracks. In an isotropic core material, these grow through the core perpendicular to the direction of largest principal stress. Screws and rivets penetrating the face sheet are a special case, since they are not normally connected to the core by adhesive. They may cause serious discontinuities in the core.

13.1.4 Stress concentrations due to inserts An insert is a disruption of the basic structure, and will cause stress concentrations even if no load is applied to the insert itself. This is referred to as the passive insert. Points of particular interest are mentioned hereafter.

Most of the insert types mentioned in this chapter involve drilling holes in the face-sheets. This causes a concentration of the in-plane stresses. The concentration factor will be 3 for isotropic face sheets under compression or tension. The adhesive connection between insert and face sheet is not necessarily reliable, so generally, this factor applies (again, note that the

13.5 AN INTRODUCTION TO SANDWICH STRUCTURES stress concentration will be different for anisotropic face sheets, where the adhesive layer has significant relative stiffness, and for bending).

If the insert has flared ends bonded onto the face sheets, the relevant concentration factor is reduced. The local offset of bending stiffness may cause significant peel stresses between face sheet and insert flare, depending on the actual parameters.

13.1.5 Inserts in panels subjected to shear In shear, the face sheets of a sandwich panel will move in-plane relative to each other, causing large deformations compared to for example a thin steel plate. Depending upon the insert, the relative movement of the face-sheets will be prevented to a certain extent, the total shear stiffness of the panel increased by the insert. In Fig.13.7, some different cases are outlined.

(a) (b) (c) (d)

Figure 13.7 Different passive inserts in shear-subjected panel. Points of severe stress-concentration are marked by circles. (a) Very soft insert or diaphragm, (b) stiff through-thickness insert with flared ends, (c) stiff simple cylindrical through-thickness insert and (d) partial insert A soft insert or diaphragm will remain at right angles to the face sheets, getting a “stretched S”-shape. This causes severe bending in the insert near the face sheets, and thereby large stresses in the insert/face-sheet interface. The stiff through-thickness inserts will not change shape much. This gives rise to anti-symmetric deflection of the face sheets. If the insert is simple cylindrical, the sudden, sharp deflection will cause stress concentrations in the core and the core/face-sheet interface. This effect can be somewhat reduced by the use of flared ends; instead, the face sheets will bend, causing large peel-stresses between flared edge and face sheet. A partial insert should be avoided in places of large shear deformation. The stress concentration in the core will most likely cause core failure or debonding between core and insert.

13.1.6 Summary Based on the circumstances outlined above, the Table 13.1 summarises the abilities of different inserts to carry different loads. It is assumed that forces are applied so that no rotation will occur (that is, at the apparent centre of stiffness).

13.6 JOINTS AND LOAD INTRODUCTIONS

Transverse In-plane Bending Comments force force moment

Use where one face sheet must be kept intact + + ÷ ÷ Avoid bending moments Do not use in panels subjected to much Partial shear Beware of stress concentration in core

Use for heavy loads Note stress-concentrations in core and + + + + + + + in core/face-sheet interface Through-thickness

Use for attaching light equipment Use mainly in thick face-sheets ÷ + ÷ ÷ ÷ Avoid bending moments and transverse Rivet / screw forces Beware of stress-concentration in core, in particular when using long screws.

Use for improved stiffness and pull-out strength + + + + Beware of bending stress in face sheets Flared top and peel-stresses underneath flared edge Table 13.1 Summary of insert performances.

13.2 Insert Calculation Examples Four different inserts are investigated, using finite element models. In each case, a transverse force F is applied to an insert in a sandwich panel. The panel has the following properties:

Face sheets: Aluminium (Ef = 70000 MPa, νf = 0.30), thickness tf = 2 mm

Core: Divinycell H130 PVC foam (Ec = 140 MPa, νc = 0.32), thickness tc = 30 mm At first, the modelling radius r must be estimated. In this case, the important local effects are assumed to be described well within a radius of 100 mm, roughly 3 times the height of the core. Now, the structural boundary conditions must be decided. By having assumed infinitely rigid inserts, we can conveniently choose to “invert” the loading by supporting the model at the insert and applying an equivalent load to the boundary at the radius r. This has two advantages: 1: We don’t have to actually model the insert. Instead we impose suitable constraints at the interface between sandwich panel and insert. 2: The problems of assuming certain support-conditions at a hypothetical radius r can be circumvented. By choosing the radius r, we assume that at this point, the local effects have “died out”, and the transverse force is distributed evenly across the height of the core. In this case, it means distributing shear stress τ = −Q/(2πr tc) on the core at the radius r. Setting Q at

13.7 AN INTRODUCTION TO SANDWICH STRUCTURES

1000 N, τ becomes 0.0531 MPa. In Fig.13.8, the choice of element modelling radius and boundary conditions is outlined.

Q r r

Figure 13.8 Left: Sandwich panel, simply supported at the edges, with insert subjected to transverse force. Right: Cut-out, radius r=100 mm, with equivalent shear stress distributed evenly on core and supported at the insert. Note that the centre support shown in Fig.13.8 is symbolic; the actual support will be determined by the insert shape. Finally, since model and load are in this case axi-symmetric, the FE-model is essentially plane. This reduces model complexity and calculation time significantly.

Initially, however, it is worthwhile to clarify the primary load-carrying components in the insert/sandwich-panel interface for the pullout load. These are shown in Fig.13.9 for three 1 of the four cases considered. (a) (b) (c) Q Q Q

τ σ

Figure 13.9 Primary interface reactions for cases. (a) Self-tapping screw or rivet, (b) partial insert and (c) through-the-thickness insert (simple cylindrical insert) The initial rough estimate of load-carrying capacity may be:

^ Case a: Qmax = πDT , where D: screw/rivet minimum diameter ^ T : allowable face-sheet shear resultant

2 ^ ^ Case b: Qmax = π (r σ + 2rhτ) where r: insert radius h: insert height σ^ : allowable core material tensile stress ^ τ: allowable core material shear stress (roughly ½ σall)

1 The fourth case, the through-the-thickness insert with flared edges, is not shown in figure 9.13. The basic interface reactions are somewhat like case (c) (simple cylindrical through-the-thickness insert).

13.8 JOINTS AND LOAD INTRODUCTIONS

^ Case c: Qmax = 2πrhτ

These are very rough estimates, assuming uniform distribution of stress on the relevant surfaces and disregarding any stress concentrations. The calculated load carrying capacity may therefore have to be reduced significantly.

13.2.1 Self-tapping screw or rivet The self-tapping screw or rivet is schematically depicted in Fig.13.10.

O10 2

R5 30 τ

2 r 100

Figure 13.10 Actual geometry (left) and modelling (right) of sandwich panel with rivet in upper face sheet. The load τ is equivalent to 1000 N transverse to the panel-plane, applied to the insert. Since a rivet or screw is in this case assumed to transfer a transverse force T and nothing else, the equivalent support is as shown (transverse movement at inner edge of upper face sheet restricted). Also, at the centre-line, radial displacements must be 0 (this is not shown).

The shear stress distribution across the thickness of the panel indicates which parts carries what. Two sections have been chosen, one at a radius 5 mm larger than the face-sheet/insert interface and one at radius r = 80 mm. For ease of comparison, the shear stresses τrz have in each case been multiplied by the relevant radius. In this case, the radii are 10 and 80 mm, and the shear distribution is as shown in Fig.13.11. r =10 r =80

4 −τ r −τ r [N/mm] [N/mm]

Figure 13.11 Shear distribution at radii 10 and 80 mm in the case of a rivet/screw.

The maximum face sheet bending stress σr at r = 10 mm is 22 MPa. This rather large value is not surprising, considering the small radius.

Several pullout tests of self-tapping screws in FRP-face sheets have been performed. The main parameters were as shown in Fig.13.12.

13.9 AN INTRODUCTION TO SANDWICH STRUCTURES

Q 1

Os 2

3 O d

Figure 13.12 Parameters for pullout tests of self-tapping screws from FRP face-sheets. The parts are: (1) self-tapping screw, core diameter Øs, (2) face-sheet, GFRP, pre-drilled screw hole diameter Ød and (3) core material The screws had core diameters 2.0, 3.0 and 4.8 mm. The face sheets consisted of 2-6 layers of combi-mat (chopped-strand + woven) 800T2/300, with about 60% glass fibre and a “hand lay-up” quality. One layer is roughly 1.1 mm thick, and the total face sheet thickness is thus about 2-7 mm. It should be noted that in most cases, the pullout caused local interface debonding between face-sheet and core, and local face-sheet delamination. The experimentally obtained failure loads are given in Table 13.2.

Screw core diameter 2.0 3.0 4.8 Øs [mm] Drilling diameter 2.3 3.5 4.8 Ød [mm] Pullout force [kN], 0.78 1.97 2.64 2 layer face-sheet Pullout force [kN], 1.19 2.99 4.89 4 layer face-sheet Pullout force [kN], 1.70 3.61 5.63 6 layer face-sheet Table 13.2 Pullout force for different screw inserts The dependence upon drill diameter was found to be insignificant at small variations. The drill diameter should be a bit larger than the screw core diameter for ease of mounting.

Rivets of various kind are often used for low load carrying inserts. There are several on the market. One such example is schematically illustrated in Fig.13.13.

Figure 13.13 Schematic of rivet for bolt mounting

13.10 JOINTS AND LOAD INTRODUCTIONS

This type of rivet is mounted by drilling a hole through the outer laminate and then mounting the rivet using a special tool. The rivet is threaded to fit standard bolts. These exist is various sizes and variants. An advantage with using this type of single-sided insert is that any equipment that is fastened can be detached, albeit not moved without applying a new rivet.

13.2.2 Partial insert The partial insert is schematically depicted in Fig.13.14. O 20 2

15 30 τ R5

2 r 100

Figure 13.14 Actual geometry (left) and modelling (right) of partial insert, penetrating upper face sheet The hatched area denotes a full restriction of all displacements in the interface. The load τ is equivalent to 1000N transverse to the panel-plane, applied to the insert.

Since the insert does not have a flared top, no significant face sheet bending moment Mr can contribute to load load transfer, and only the force-components T and Nr are present at the face-sheet/insert interface. In this case, the shear distribution is evaluated at radii 15 and 80 mm, see Fig.13.15. r =15 r =80

9 −τ r −τ r

[N/mm] [N/mm]

Figure 13.15 Shear distribution in the case of partial insert.

The maximum bending stress σr in the upper face sheet at r = 15 mm is 12 MPa. The transverse tensile stress σz is shown in Fig.13.16 as a function of the radius, from r = 0 to r = 100 mm. The section chosen lies 17 mm underneath the upper face-sheet/core interface, that is, 2 mm underneath the bottom of the insert.

13.11 AN INTRODUCTION TO SANDWICH STRUCTURES

σz [MPa] 0.8

0.6

0.4

0.2

r [mm]

20 40 60 80 100

Figure 13.16 Transverse tensile stress 2 mm underneath insert bottom. The insert has a radius of 10 mm. From Fig.13.16 it is seen that the mean transverse tensile 2 stress from r = 0 to 10 mm is about 0.7 MPa, which means that the force πr σz,mean = 220 N is transferred at the bottom of the insert. This is about 1/4 of the total 1000 N, and the rest is thus transferred as shear stress acting upon the cylindrical surface. The stress concentration at the insert bottom causes fracture. A test specimen for pullout tests is shown in Fig.13.17, after the insert has been pulled out.

Figure 13.17 Cross-section example of pullout test of partial insert. The fracture sequence is: a) Linear or roughly linear force-displacement Q[kN] 3 relation until fracture in interface at bottom of a 2.5 insert. The resulting conical crack immediately 2 b c propagates most of the way to the upper face 1.5 sheet 1 b) Propagation of conical crack, accompanied by 0.5 acoustic emission. Crack starts at face- 0 d [mm] sheet/insert interface and propagates towards 0123456 bottom crack c) Insert detached from panel, held only by friction.

The core fracture mode is similar to that of a punching problem, where a transversely loaded thick plate is supported by columns. A few pullout tests of different partial inserts were performed for determining the fracture sequence, in each case characterised by the transverse force at point a as described above.

13.12 JOINTS AND LOAD INTRODUCTIONS

The sandwich panel used had the following main characteristics:

Face sheets: Aluminium, thickness 1.5 mm Core: Divinycell H130, thickness 30 mm

Three partial insert shapes were tested, all with an interface area of 1000 mm2, but with varying diameters: Øi = 12, 16, 20 mm. The parameters are summarised in Fig.13.18.

1.5 R3 hi

1.5

Figure 13.18 Sandwich and insert parameters

The insert heights hi were adjusted to reach the desired interface area. The maximum pullout force values were as shown in Table 13.3.

Insert diameter Øi [mm] Max. force [kN] 12 3.28 16 2.93 20 3.40 Table 13.3 Results of pullout tests. Fatigue tests of similar partial inserts show that the mode of initial fracture is the same as that of static fracture. By rounding the core end of the partial insert, as shown above, the stress concentration decreases somewhat compared to a sharp edge, giving a better lifetime expectancy.

Söderlund et al (1998) studied the re-design of the partial insert. They used numerical optimisation methods together with FE-analysis in a combined framework. They used a cylindrical partial insert as the starting point. However, this type is confined underneath the face sheet, as shown in Fig.13.19(a). The insert is thus placed in the core prior to the attachemnet of face sheet. After manufacturing a hole is drilled through the laminate, into the insert, which is consequently threaded and used as an attachment point with a bolt.

13.13 AN INTRODUCTION TO SANDWICH STRUCTURES

Figure 13.19 (a) Cylindrical partial insert and (b) optimised partial insert The optimisation scheme was set-up to maximise the load carrying capacity while maintaining the same weight (same insert volume). The result of this exercise is depicted in Fig.13.19(b) which not surprisingly shows a smoother shaped insert which has much lower stress concentrations. Both the initial and optimised configurations were then manufactured and tested using a simple pull-out test. The optimised insert had twice the load carrying capacity. A drawback of using an insert type resembling the optimised version is a higher manufacturing cost.

13.2.3 Through-thickness insert The through-thickness insert is schematically depicted in Fig.13.20.

O 20 2

30 τ

2 r 10 100

Figure 13.20 Actual geometry (left) and modelling (right) of through-thickness insert. The load τ is equivalent to 1000 N transverse to the panel-plane, applied to the insert. Conditions at the face-sheet/insert interface are similar to those for the partial insert (section 13.2.2), and the shear distribution is again evaluated at radii 15 and 80 mm, see Fig.13.21. r =15 r =80

25 −τ r −τ r [N/mm] [N/mm]

Figure 13.21 Shear distribution in the case of through-the-thickness insert.

At r = 15 mm, the maximum face sheet bending stress σr is 10 MPa.

13.14 JOINTS AND LOAD INTRODUCTIONS

13.2.4 Through-the-thickness insert with flared ends The through-thickness insert with flared ends is schematically depicted in Fig.13.22. O 40 O 20 2

30 τ O 40

2 r 10 100

Figure 13.22 Actual geometry (left) and modelling (right) of through-the-thickness insert with flared edges. The flared edges have diameter 40 mm, i.e. twice the diameter of the cylindrical part of the insert. The load τ is equivalent to 1000 N transverse to the panel plane, applied to the insert. The shear distribution is in this case evaluated at radii 25 and 80 mm, see Fig.13.23. r =25 r =80

34 −τ r −τ r [N/mm] [N/mm]

Figure 13.23 Shear distribution in the case of through-the-thickness insert with flared edges. Note similarity with distribution for simple cylindrical through-the-thickness insert.

At r = 25 mm, the maximum face sheet bending stress σr is 19 MPa. This is nearly as much as the face sheet bending stress at r = 10 mm in the case of a rivet/screw, described in section 13.2.1.

13.15 AN INTRODUCTION TO SANDWICH STRUCTURES

13.3 Joints Large and/or complicated sandwich structures are often manufactured by shaping prefabricated sandwich panels and then joining them. For example, the bulkheads in sandwich ships are fitted to the hull in this fashion. Joints between sandwich panels are practically unavoidable in most structures. This may be considered unfortunate, since it will inevitably compromise the basic principles of sandwich construction. This chapter will deal with methods of stiffening/strengthening the joints, so that the composite action is improved.

The primary joint geometry may be defined in the following manner: Two panels, each defined by a middle surface, joined at the intersection curve of said middle surfaces. Examples of mid-surface intersections are shown in Fig.13.24. 1 y 1 1 2 2 x 2 z

Figure 13.24 Panel mid-surface intersections. The panels may in general, as shown, be plane or curved, and intersect each other at an angle ≠ 0°, 180°. Herein, only plane panels are treated, and cases where the intersection line is long compared to the thickness of the individual sandwich panels. This means that problems of stress/strain analysis are reduced to cases of plane strain.

13.3.1 Basic types Some basic joint-types are of particular importance. These are shown in Fig.13.25.

TL V

Figure 13.25 Basic joint-types: T-joint: Bulkheads, stiffeners, walls, L-joint: Deck-hull, wall-roof, and V-joint: hull-bottom The purpose of the joint is to transfer reactions between the joined panels. The force reactions for the T-, L- and V-joints are shown in Fig.13.26.

13.16 JOINTS AND LOAD INTRODUCTIONS

T M N M T T M M M T N M N T T

Figure 13.26 Force reactions in the joints Basically, sandwich panels should act in the following manner: • The face-sheets act as membranes and transfer in-plane force N directly and bending moment M as force couples. • The core transfers the transverse reaction force T

This is clearly violated in all the joint-types shown in Figs.13.25 and 13.26, the general problem being that in-plane reactions on one panel are out-of-plane reactions on the other. This means that the stiffness/geometry is “discontinuous” relative to the direction of the reactions.

13.3.2 T-joints The T-joints are used wherever one sandwich panel is joined transversely to another (sandwich-) panel. In larger sandwich structures, such as ships, several kilometres of T-joints are not uncommon. Thus, T-joints have been subjected to extensive research in an effort to develop cheap, reliable designs.

Transverse stiffening of ship hulls is basically a matter of transferring force from the hull to a bulkhead, as shown by the component N at the T-joint in Fig.13.26. Since the bulkhead is usually perpendicular to the hull, the large stiffness of the bulkhead face sheets will cause a reaction as two parallel line-loads on the hull inner face-sheet. This is shown in Fig.13.27.

P/2 P/2

Figure 13.27 Reaction between hull panel and bulkhead. The overall deflection, the bending of the hull inner face sheet and the compression of the core is shown to the right. The rather direct transfer of force between bulkhead face sheets and hull core may cause severe bending of the hull inner face sheet and compression of the hull core. Furthermore, when only a thin layer of adhesive bonds the bulkhead to the hull (as is assumed to be the case in Fig.13.27), only a negligible amount of tensile force P may be transferred. In a fast boat moving at high speed, slamming2 occurs, the dynamics are such that tensile force may be

2 Slamming: when the bow of a boat rises on one wave and slams back onto the ridge of the next wave.

13.17 AN INTRODUCTION TO SANDWICH STRUCTURES present and non-negligible. Thus, a T-joint should distribute the loads so that interface stresses between the panels are sufficiently low.

Fig.13.28 shows several components that may be used in combination or individually. The gap-filler (5) serves two purposes: It fills the gap between the hull and the bulkhead, and, importantly, distributes the force between the bulkhead face sheets and the hull panel. The bonded tapered laminate (3) is normally used in combination with a fillet (4) and a gap-filler (5). This distributes the force and enables the joint to transfer bending moment (whether this is desirable or not). For producing a cheap joint, a putty-fillet (made from Crestomer, Divilette or some similar filler) may be used alone. A joint of that type should not be subjected to fatigue loads. Reinforcement of the hull inner face-sheet (6) increases the bending stiffness, thus distributing the load transverse to the hull panel on a larger portion of the core. Similarly, the local high density core (7) is able to withstand a higher local stress arising from the bulk-head.

3 2

4 5

Figure 13.28 Reinforced T-joint. The components are: (1) hull panel (face-sheets and core), (2) bulkhead (face-sheets and core), (3) bonded tapered laminate, (4) fillet (5) gap filler, (6) reinforcement of hull inner face-sheet and (7) local high density hull core.

13.3.3 L-joints The L-joint is used for joining the roof and sides of lightweight containers and trailers3, for joining a sandwich ship’s deck and hull and similar. If the joint interface is primarily subjected to a limited amount of compression, and insignificant bending, the joint needs not to be very elaborate. However, if the bending moment is significant, problems may arise, as illustrated in Fig.13.29.

3 For refrigerated trailers and containers, a sandwich construction with foam core is often used, due to the high thermal insulation of the foam.

13.18 JOINTS AND LOAD INTRODUCTIONS

Figure 13.29 Bending of L-joint. Significant reactions between the face-sheets of the upright and the horizontal panel are shown in cut-out. Deformation is outlined to the right. In a manner quite similar to T- and V-joints, reactions are transferred from a stiff region (in- plane-direction of upright face-sheets) to a soft region (transverse to horizontal panel). Furthermore, since the horizontal panel is cut off flush with the upright, the outermost portion of the horizontal panel core possesses little shear stiffness. This means that the upper face sheet of the horizontal panel cannot contribute significantly to the stiffness. As indicated in Fig.13.29, the combination of circumstances causes compression/extension of the horizontal panel core, and local bending of the lower face sheet. Fig.13.30 shows some ways to reinforce the L-joint. 8

5 4

Figure 13.30 Reinforcement of L-joint. The parts are: (1) upright panel, (2) horizontal panel, (3) fillet, (4) bonded tape laminate, (5) reinforced face-sheet, (6) bulk laminate reinforcement, (7) edge laminate, and (8) high density core. The fillet (3), used in combination with the bonded tapered laminate (4), distributes interface reactions over a larger area. The reinforcement of the horizontal panel (5) inner face-sheet increases the face-sheet bending stiffness, thus distributing the reaction from the upright face- sheets to a larger portion of the horizontal panel core. Using a high density core in the edge of the horizontal panel (8) serves two purposes: It is able to withstand greater stress arising from

13.19 AN INTRODUCTION TO SANDWICH STRUCTURES the reactions from the upright face-sheets, and it has greater shear stiffness to better transfer shear stresses to the upper face-sheet. The bulk laminate (6) may be used to improve shear transfer between the face-sheets of the horizontal panel, similar to using a high density core. It may have to be very thick (and heavy) to be sufficiently stiff. An edge laminate (7), used in combination with a high density core and/or bulk laminate, increases the bending stiffness. It is often used in any case, since it seals the edge neatly.

Some other ways to make an L-joint are shown in Fig.13.31. These may be more demanding in terms of precision and preparation, since the panels may have to be manufactured specially in order to accommodate the edges. (a) (b) (c)

Figure 13.31 Other L-joints. (a) panels bonded into metallic edge profile, (b) laminated “lattice”-edge, and (c) prefabricated panels with metallic edges, welded together The welded-edge joint in Fig. 13.31(c) can only transfer small bending moments.

13.3.4 V-joints The V-joint is used at the bottom of boat-hulls and for assembling other similarly symmetric halves (half shells). It has a superficial similarity to the butt joint, where two core-plates are joined to form a continuous plate. The difference is that in the V-joint, the planes of the plates are joined at some angle to each other. The resulting discontinuity may cause problems, as illustrated in Fig.13.32.

N Ny MM Nx

Ny N

Figure 13.32 V-joint subjected to bending moment. The cut-out shows the components of the force- pair N caused by M. Deflection and resulting bending of face sheets and compression of core is shown to the right. The V-joint is deliberately shown as a two part core with continuous face sheets. Boat-hulls are often made by attaching core materials to an upside down male mould. The core is then coated with several layers of fibre to form the outer face sheet. When the resin is cured, the shell is removed from the framework and turned around so that the inner face sheet may be applied.

13.20 JOINTS AND LOAD INTRODUCTIONS

The components Nx from the left and right parts of the V-joint are equal and opposite, due to symmetry. The components Ny , however, will cause severe transverse compression/extension of the comparatively soft core.

An improved design should aim to increase the strength of the whole joint. Some of the means by which this may be obtained are shown in Fig.13.33.

2 3 4

6 5

Figure 13.33 Reinforcement of V-joint. The parts are: (1) sandwich panel, (2) reinforced inner face sheet,(3) fillet core, (4) fillet face-sheet, (5) higher density core, and (6) reinforced outer face sheet One or more of the methods shown can be used (apart from the fillet parts (3) and (4), which should be used in combination). Reinforcement of the inner face sheet (2) is done by applying a strip of extra laminae. This increases the bending stiffness of the face sheet locally, thereby distributing the vertical force-component more evenly on the core. When reinforcing the outer face sheet (6) in this manner, due consideration should be given to the required shape (i.e. the shape of a boat keel may affect stability). The core in the immediate vicinity of the V-joint (5) may have a higher density, enabling it to accommodate the local stress-concentration. The fillet reinforcement (4) transforms the joint-region to a multi-skin, multi-core sandwich. It may be difficult to manufacture, since the fillet core (3) must be shaped to fit properly in the V.

13.3.5 Localised deflection This is discussed in detail in chapter 11. One can use the analysis method outlined there to conceptually study joints. For calculating the local deflection, a one-parameter elastic foundation model (Winkler foundation model) is convenient. The deflection is in this case given4 by:

Pe −β x yx()=+[] cos(ββxx ) sin( ), x ≥ 0 (13.1) 8Dβ 3 where P [N/mm]: Load per width (line load extending perpendicular to section) D [Nmm]: Face-sheet bending stiffness,

4 The equation can be found by combining elementary cases from “Formulas for Stress and Strain” or similar litterature.

13.21 AN INTRODUCTION TO SANDWICH STRUCTURES

E, t, ν are the modulus of elasticity, the thickness and the Poisson’s ratio for the face sheet, respectively, for an isotropic face-sheet

-1 k β [mm ]: Parameter of relative stiffness, β = 4 4D

k [N/mm3]: Foundation stiffness

The foundation stiffness k may be estimated by Ec /tc, where Ec is the modulus of elasticity of the sandwich panel core, and tc is the thickness of the core. The wavelength of the deflection is given by λ = 2π /β (13.2)

Although the above expression for deflection is based on a panel extending infinitely in the ±x-directions, the decay is so rapid that the calculation is usually valid if there is no free 5 sandwich panel edge within a distance of approximately λ or 3tc, whichever is largest .

The simple case of a single line load may be combined by superposition. Fig.13.34 shows a few cases for the T-joint.

P P M 2a~ tc 2a~ tc 1 2 1 2

y y 1 y 1 x x tot y y x 2 x 2 x y y tot x tot x

Figure 13.34 The local deflection y of the upper face-sheet is the sum of the individual deflections, y1 and y2, at each point x. Left to right: Left: Thin plate transferring a single force to the sandwich panel, essentially this is a single line load. Middle: Sandwich bulkhead transferring force to hull. Bulkhead face sheets 1 and 2 act as parallel line loads with magnitude P/2, placed a distance c from each other. Right: Sandwich bulkhead transferring a bending moment to the hull. The bending moment M acts as two opposite line loads P = M/tc..

Application for V-joint The V-joints may also be considered in this manner. At the concave notch (the inner face- sheet) in particular, the sharp angle in the panel may cause something similar to a line load, as shown in Fig.13.35.

5 The number 3tc may have to be varied, depending on the relation between core shear stiffness and face sheet extensional stiffness. For soft cores, the number should be larger.

13.22 JOINTS AND LOAD INTRODUCTIONS

~t M c M/t c M/tc α

Figure 13.35 Left: Face-sheet forces acting on a cut-out section, width Δ, caused by an external bending moment M on the V-joint. Right: Vector parallelogram of forces. Py is the resulting line load when Δ is very small.

From the geometry, Py is given by Py = −2Msin(α)/tc. Note that the core thickness used for calculating the foundation modulus k should be based upon the thickness in the direction of the force Py . The foundation stiffness for the V-joint is then given by k = Ec cos(α)/tc. This means that the stiffness is less than for a plane panel.

By taking the second derivative of the expression for deflection (eq.(13.1)), the approximate curvature of the face sheet is found. P yx''( )=⋅⋅exx−β x (sin(ββ ) − cos( )) (13.3) 4Dβ

The face sheet bending stress for isotropic face sheets is then

y''E t + t σ = ⋅ c f (13.4) b 1−ν 2 2

13.3.6 Calculation example, T-joint This example considers the case of a sandwich ship hull, subjected to a load q from one side and supported by a number of bulkheads, see Fig.13.36. The thickness of the sandwich panels is assumed to be small, compared to the span l. All loads and geometric entities are expressed per width (perpendicular to the plane of the paper), assuming plane strain. l /2 l

y x /2 lf

Figure 13.36 Left: Long sandwich panel supported by bulkheads and subjected to uniform pressure q. Right: Cut-out of representative section. lf /2 marks the “free length” of the hull panel, where the bending moment is zero. The deflection y(x) of the hull is shown below the cut-out. At the bulkhead, the reactions at relevant sections will be considered, as shown in Fig.13.37. Initially, the shear force and bending moment in the hull panel should be controlled.

13.23 AN INTRODUCTION TO SANDWICH STRUCTURES

t2 t2 tc2

T T

T M t1 tc1 t 1 M T

Figure 13.37 Dimensions and reactions at joint. The reactions are expressed per width of the panel (width: perpendicular to plane of paper). The nominal shear stress in the hull panel at the bulkhead is T ql τ == (13.5) tcc112t assuming that the shear is carried by the core alone. This should be considerably smaller than the allowable core shear stress, since the bulkhead will cause stress concentrations.

The tension/compression stress in the hull face-sheets due to the panel moment are probably about 1-2 orders of magnitude larger than the tension or compression stresses in the bulkhead face-sheets. For a known free length lf , the bending moment in the hull panel at the bulkhead is

l l M = ql ( − f ) (13.6) f 4 8

For uniform load q, lf /2 is approximately 0.21l. Assuming thin face sheets carrying the entire bending moment as a force-pair, this alone causes the compression/extension stress M σ M =± (13.7) ttc11

The force 2T = ql is supported by the bulkhead, where each of the bulkhead face sheets, being much stiffer than the core, acts on the hull upper face sheet with the force P ≅−⏐T⏐, P measured positive upwards. This is then treated by superposing two forces P at a distance tc2, as shown in Fig.13.37.

13.24 JOINTS AND LOAD INTRODUCTIONS

13.3.7 Calculation example, L-joint This example is included to demonstrate the stiffness-increasing effect of modifying an L- joint, by adding three different reinforcements one by one. The calculations are done using the finite-element method. The geometry and loading is as shown in Fig.13.38.

The load τ = 0.1 MPa is applied across the height of the core, at end (a) as shown in Fig.13.38.

The basic panel features are: • Panel face-sheets: thickness tf = 4 mm, GFRP, Ef = 10 000 MPa, ν = 0.30 • Panel core: thickness tc = 40 mm, Divinycell H130, Ec = 140 MPa, ν = 0.32

The reinforcements are: 1. High density core material (1), E = 300 MPa replaces original core at outer 90 mm of horizontal panel 2. Divinycell H130 fillet (2), covered by GFRP bonded tape laminate (thickness and stiffness as panel face sheets), added at inner corner 3. GFRP bonded tape laminate (3) (thickness and stiffness as panel face sheets) added on outer corner 400 1 t 3 f

tf tc (a) tf 2 R

t 400 tf f

Figure 13.38 L-joint, geometry and loading. 1, 2 and 3 represents the different reinforcements. Plane strain is assumed in the FE-model. It should be noted that the geometry shown in Fig.13.38 is simplified. Gap-fillers, thick adhesive-layers and common imperfections should

13.25 AN INTRODUCTION TO SANDWICH STRUCTURES be included if the local stresses and strains are investigated. For the purpose of investigating the stiffness, however, a coarser model may be used. For comparison of the effects of increasing the stiffness, the vertical deflection (y) is taken at the end (a) of the horizontal panel (at the end where the shear stress τ is applied). The deflections for the four different cases is as shown in the diagram below, Fig.13.39.

-y [mm] 14 12

10 Unreinforced 8 Reinf. 1 6 Reinf. 1&2 4 Reinf. 1&2&3 2 0

Figure 13.39 Deflection of L-joint configurations Clearly, the reinforcements increases the stiffness significantly. Whether the joint should actually be stiffened, and how much, is usually a matter of performance and cost.

13.3.8 T-joints - tests and bbservations The T-joint has been subjected to much research. Here we look at some primary results from tests and calculations. Three joint configurations have been considered. These are shown in Fig.13.40.

(a) (b) (c)

Figure 13.40 T-joints - (a) circular fillet, filler between intersecting panels, (b) triangular fillet, filler between intersecting panels and (c) circular fillet, bulkhead face-sheets cut off The face-sheets and the bonded tape laminate (covering the fillet) are CFRP, the core and fillet are made from PVC-foam (Divinycell), and the filler is a polyester compound.

For testing the joints, the most common method is to support a bulkhead panel at the ends and apply a tensile force to the bulkhead. This, and the resulting fracture modes for joints (a), (b) and (c), is shown in Fig.13.41.

(a) (b) (c) F 2 2 2 1 3 1 3 1 3

Figure 13.41 Load case (left) and fracture modes (dashed lines). 1 denotes initiation of cracks.

13.26 JOINTS AND LOAD INTRODUCTIONS

Common to the fractures is that cracks tend to initiate in the same place, as marked by “1” in Fig.13.41. The stress concentration in this place is caused by the stiffness discontinuity and the difficulty of avoiding voids when manufacturing the joint.

Case (a): For the circular fillet, the joint strength clearly depends upon the fillet radius; larger fillet gives greater strength. The stress concentration at the point where the bonded tape laminate meets the hull upper face-sheet will eventually initiate a crack.

Case (b): In the case of a triangular fillet, the flat sides of the bonded tape laminate means that significant transverse (vertical) force can be transferred between the bulkhead and the hull upper face-sheet. This causes a large stress concentration at the point where the bonded tape laminate touches the hull upper face-sheet. Different fillet sizes were tested, with side lengths ranging from ½ to 2 times the thickness of the cores. No significant difference in strength was detected for different triangular fillet sizes. The circular fillet will in general be stronger than the triangular fillet.

Case (c): The main difference from type (a) is that the bulkhead face-sheets and the bonded tape laminate form a continuous sheet. Burchardt concluded, among other things, that the thickness of the bonded tape laminate is not of primary importance; it should merely delay fracture from points “2”. The fillet material carries the actual load.

Three joint-configurations tested by Kildegaard (1993) with a geometry as shown in Fig.13.42. Note that a simple fillet, using a polyester filler, is included.

I II III

1000

Figure 13.42 Overall geometry and joint-configurations I, II and III.

• Panel core and foam fillet: Divinycell H100, Thickness tc = 60 mm • Face-sheets: GFRP, three layers combi-mat, 900 g/m2 (600 g/m2 0°/90° woven mat and 300 g/m2 chopped strand mat) and one layer 450 g/m2 chopped strand mat • Bonded tape laminate: six layers 450 g/m2 chopped strand mat • Face-sheet and bonded tape laminate matrix: Isopolyester

13.27 AN INTRODUCTION TO SANDWICH STRUCTURES

• Filler: Polyester filler

The following table contains the joint-configuration parameters, test number, initial fracture force and final fracture force. The fracture forces are given per joint-length (perpendicular to the plane of the paper) in Table 13.4.

Joint type a Test number Initial fracture force Final fracture force [mm] [N/mm] [N/mm] I 30 Δ30-1 102.0 142.7 I 30 Δ30-2 118.5 149.3 I 60 Δ60-1 127.3 138.7 I 60 Δ60-2 106.7 137.3 I 120 Δ120-1 125.3 144.0 I 120 Δ120-2 108.0 134.7 II 30 R30-1 - 142.7 II 30 R30-2 - 146.7 II 60 R60-1 - 162.0 II 60 R60-2 - 142.7 II 120 R120-1 - 193.3 II 120 R120-2 - 200.0 III 15 F1 134.7 134.7 III 15 F2 112.0 137.3 Table 13.4 Test results from testing of different T-joints. The type-II joints tend to break suddenly, and the initial and the final fractures are coincident. Bibliography Burchardt C., ”Bonded Sandwich T-joint for Maritime Applications”, IME, AAU, ISSN 0905-2305, Special Report No. 32, Ph.D.-dissertation, 1996.

Erichsen C., Nielsen E.S. and Brønnum N., ”Undersøgelse af skrueudtræksstyrke og mekaniske egenskaber for GFRP”, IME, Aalborg University.

Frostig, Y., ”On Stress Concentration in the Bending of Sandwich Beams with Transversely Flexible Core”, Composite Structures, Vol 24, 1993, pp 161-169.

Insert Design Handbook, ESA PSS-03-1202

Kildegaard C., ”Konstruktiv udformning af samlinger i maritime FRP-sandwichkonstruktioner”, IME, AAU, Special Report No 20, 1993, ISBN 87-89206-28-2, Ph.D.- dissertation (in Danish).

Kristensen K., ”Stress Singularities caused by Inserts in Sandwich Structures”, IME, AAU, Report No. 64, December 1994.

13.28 JOINTS AND LOAD INTRODUCTIONS

Mortensen C. and Kristensen K., ”Lastindføring i Sandwichkonstruktioner - En Numerisk og Eksperimentel Analyse af Lokaleffekter”, IME, AAU, ISSN 0905-4219. Report No. 65, 1995 (in Danish).

Rosander M., “T-joints in Sandwich Structures”, Licentiate thesis, Report No. 97-29, Kungliga Tekniska Högskolan, Dept. Aeronautical and Vehicle Engng., 1997,

Shipsha A., Söderlund J. and Zenkert D., ”Shape Optimisation of Internal Metal Doubler for Load Introduction in Sandwich Structures”, in Proceedings of Fourth International Conference on Sandwich Construction, Ed. Karl-Axel Olsson, EMAS, Warley, UK, pp 839- 851, 1998

Thomsen O.T. and Kildegaard A., ”Teknisk note: Indtrykning og lokale bøjningseffekter i lastbærende FRP-sandwichpaneler”, IME, AAU, Report No. 56, 1993 (in Danish).

Thomsen O.T. and Rits W., ”Analysis and Design of Sandwich Plates with Inserts - A Higher-order Sandwich Plate Theory Approach”, IME, AAU, ISSN 0905-4219 Report No 69, June 1996

13.29 CHAPTER 14

1 MANUFACTURING by K.F. Karlsson and B.T. Åström

Techniques to manufacture sandwich components for structural applications are summarised and discussed in terms of processing steps, characteristics of both technique and manufactured component, and application examples. The emphasis is on the commercially most common manufacturing techniques, though less common processing routes are briefly discussed as well. The intentions of this review are two-fold: First, the paper aims to provide a means of comparing available manufacture techniques in terms of feasibility for a specific application; second, the paper aims to provide a baseline for further research and development efforts in the field of sandwich manufacture. A discussion of recent developments and future trends in terms of both materials and processing routes rounds up the paper.

The structural sandwich concept involves the combining of two thin and stiff faces with a thick and relatively weak core. By sandwiching the core between the two faces and integrally bonding them together, a structure of superior bending stiffness and low weight is obtained. Since the core often has exceptional insulative properties, the entire sandwich structure may further be characterised by excellent thermal insulation and also acoustic damping at certain frequencies.

Sandwich structures are used in a wide variety of applications, such as in automobiles, refrigerated transportation containers, pleasure boats and commercial vessels, aircraft, building panels, etc. The face materials in common use include sheet metal and fibre-reinforced polymers, while common core materials are balsa wood, honeycomb, and expanded polymer foam. These materials and material combinations all have their own market share and one or more of advantages such as low cost, high mechanical and thermal properties, thermal and acoustic insulation, fire retardancy, low smoke emission, compliance, ease of machining, ease of forming, etc.

While the sandwich concept is used in an impressive variety of applications, the techniques employed to manufacture components tend to be few and usually involve a large degree of manual labour. This paper describes and discusses the most common manufacturing techniques in some detail, but also briefly cover some of the more unorthodox techniques that have been

1 Based entirely on the paper KF. Karlsson and B. T. Å;ström”Manufacturing and applications of structural sandwich components”, Composites Part A, Vol. 28:2, 1997, pp 97-111 AN INTRODUCTION TO SANDWICH STRUCTURES documented in the literature. By attempting to compile the state of the art of sandwich manufacture, which the authors found was a field where comparatively little work has been documented in the open literature, the authors hope to create a baseline for future research and development efforts into the relatively immature field of sandwich manufacture.

14.1. Face Materials Face materials used in sandwich applications may, for the purpose of this paper, be divided into fibre-reinforced polymer composites and other materials. Polymer composites are unique as face material in that they under certain conditions may be laminated directly onto the core. Alternatively and regardless of material type, the face may be manufactured in one step and then bonded to the core in another step at a later point in time.

In terms of fibre-reinforced polymers, all constitutive materials commonly used in composites applications are used also as sandwich face materials. Reinforcements thus include all kinds of glass, carbon, and aramid; similarly, virtually all types of thermoset and thermoplastic resins are used. When performance so dictates, preimpregnated unidirectional or woven reinforcement (prepreg) is used to obtain high-performance faces and/or to gain some manufacturing advantage. Prepregs are usually composed of glass- or carbon-fibre-reinforced epoxy (EP). When composite faces are laminated directly onto the core in marine and transportation applications, glass reinforcement in the form of random mats (CSM), fabrics, and combinations thereof dominate, while resins used in such applications normally are unsaturated polyesters (UP) and sometimes vinylesters (VE). The norm when laminating faces directly onto the core in aerospace applications is to use epoxy prepregs reinforced with carbon fibres. However, also more advanced marine and transportation applications tend to employ various kinds of prepregs to an increasing degree. Naturally, all types of composite faces may be premanufactured in any conventional composites-manufacturing process, meaning that no material combination is excluded from being used as face material.

In addition to the previously mentioned material forms, faces may readily be premanufactured using moulding compounds in compression moulding. These moulding compounds include sheet-moulding compound (SMC), bulk-moulding compound (BMC) and glass-mat–reinforced thermoplastic (GMT). SMC and BMC are thermoset-based compounds, normally UP, whereas GMT is thermoplastic-based, normally polypropylene (PP). In most cases, all these moulding compounds are reinforced with discontinuous and randomly oriented glass fibres.

Non-composite face materials are always manufactured in one step and bonded to the core in another. Common examples include wood veneer, sheet metal, and unreinforced polymers, although the latter rarely results in a structurally capable sandwich component. The most common face material in this category by far is sheet metal, which offers good properties at reasonable cost but with a weight penalty; applications include refrigerated transportation containers and construction elements.

14.2 Core Materials The earliest material used as core in sandwich components was balsa wood, which still is used in some applications although alternative core materials tend to replace balsa in an increasing number of applications.

14.2 MANUFACTURING

The most common core materials used in all applications except aerospace are expanded polymer foams, which are often thermoset to achieve reasonably high temperature tolerance, though thermoplastic foams are used as well. Almost any polymer may be expanded, but the most common ones in sandwich applications are polyurethanes (PUR), polystyrenes (PS), polyvinylchlorides (PVC), polymethacrylimides (PMI), polyetherimides (PEI), and polyphenolics (PF). PUR may be in-situ foamed between the faces and thus does not need to be prefoamed into blocks. In-situ foaming thus may eliminate the need to form or machine complex core geometries.

Although some high-performance cores, such as PMI and PEI, are used in aerospace applications, honeycomb cores clearly dominate over alternative materials. Any of several materials may be used to manufacture a honeycomb core: sheet metals; fibre-reinforced polymers; unreinforced polymers; and papers. The most common honeycomb cores are based on aluminium and aramid-fibre paper dipped in phenolic resin, the latter having the trade name Nomex.

14.3 Wet Lay-Up Wet lay-up is one of the oldest but still one of the most commonly used methods to manufacture sandwich components with composite faces. The method is very flexible yet labour-intensive and thus best suited for short production series of especially large structures. The Procedure section commences with descriptions of wet lay-up of single-skin laminates (e.g. [1,2]) followed by a description of how these methodologies also may be used to produce sandwich components (e.g. [1,3-6]).

14.3.1 Procedure Wet lay-up of laminates may be performed either by hand lay-up or spray-up. A schematic of the hand lay-up process is shown in Fig.14.1. The process uses a one-sided mould, male or female, which is treated with a mould-release agent. Normally, a neat resin layer, a gel coat, is deposited directly onto the mould and is allowed to gel before lamination starts. The gel-coat resin usually is of high-quality and has good environmental resistance, thus allowing the use of a lower- quality, cheaper resin within the actual laminate. The gel-coat also produces a smooth, cosmetically appealing surface that hides the reinforcement structure, which otherwise may be visible on the composite surface. Dry reinforcement Resin Roller Gel coat

Female mould (tool)

Figure 14.1 Schematic of wet hand lay-up.

14.3 AN INTRODUCTION TO SANDWICH STRUCTURES

An appropriate amount of resin is applied and distributed on top of the mould whereupon the dry reinforcement, typically in mat and fabric form, is placed on top, see Fig.14.1. The resin is worked into the reinforcement (upwards, through the reinforcement) with a hand-held roller, which also compacts the laminate and works out voids. After one reinforcement layer has been satisfactorily impregnated and compacted, the impregnation step is repeated until the desired number of reinforcement layers have been applied or the desired laminate thickness has been reached. The lamination is often finished with a top-coat that is similar to the gel-coat in function and composition.

Spraying of a mixture of resin and discontinuous fibres is an alternative to hand lay-up that reduces the manual work required for impregnation and lay-up. A special gun, which may be hand-held or mounted on a robot, is employed to deposit the fibre-resin mixture onto a one-sided mould, see Fig.14.2. The chopped roving may be combined with fabrics or mats on the mould if the fibre content in the sprayed mixture intentionally is kept low. Resin and accelerator Chopper Roving Resin and curing agent

Female mould

Figure 14.2 Schematic of wet spray-up. Although much of the manual work of wet lay-up is reduced by employing spray-up, manual rolling still may be required and may be important in the final compaction of the laminate to remove voids and improve impregnation. The use of discontinuous and randomly oriented fibres, as opposed to a certain degree of continuous and directed fibres in hand lay-up, results in inferior mechanical properties.

The lamination procedure described above may also be used to manufacture faces of sandwich components. In most cases, the faces are laminated directly onto the core, which then acts as the mould. With this procedure, the core is often primed to improve adhesion to the faces. The primer, usually pure resin of the same type as that used in the faces, is applied to the core and is allowed to gel or even cure before lamination commences. Alternatively, if traditional moulds (such as in Figs.14.1 and 14.2) are used, the core may be placed on top of the laminated but not cross-linked laminate or it may be adhesively bonded onto the fully cross-linked laminate as further discussed in the Adhesive Bonding section below. If good surface finish is required, fully closed moulds may be used; a gel coat is then applied onto each mould half and faces laminated on top, whereupon the mould is closed with the core sandwiched in between.

Pleasure boats and larger ships built according to the sandwich concept are manufactured in a fashion somewhat different from that described above (e.g. [1,3-6]). Core planks are nailed onto a male wooden frame using finishing nails, thus crudely forming the hull, see Fig.14.3. For plane

14.4 MANUFACTURING and low-curvature surfaces large, flat core blocks can be used, while sharply curved structures can be formed by using smaller blocks or by grooving the core in one or several directions to allow for bending and shaping. Alternatively, cores bonded to a light-weight fabric may be periodically sawed almost all the way through to the fabric in two perpendicular directions to form small cubes, thus allowing a significant degree of draping. In all these cases, the gaps between the core sections are spackled using a putty that is typically based on the same resin as that used as laminate matrix. When the putty is fully cross-linked and the nails are removed, the structure is carefully sanded and sanding dust removed to provide a smooth surface for subsequent lamination. The core structure is then used as a male mould and the aforementioned lamination takes place directly onto the core. Since the cross-linked resin act as an adhesive, the bonding between the face and the core is sufficient if the lamination is correctly performed. Priming of the core as described above is common. When the lamination is completed and the laminate fully cross-linked, the hull is rotated and the wooden frame removed. The lamination procedure is then repeated on the inside of the hull. Since no exterior mould is used with this manufacturing method, at least the external laminate must be spackled, sanded and polished if a smooth surface with good dimensional tolerances is required. Top coat and/or paint layers also may be applied to protect the laminate as well as for purely cosmetical reasons.

Figure 14.3 Core planks are nailed to a wooden frame, thus creating a male mould onto which the outer face is laminated. Reprinted from [1] with permission from Prof. K.-A. Olsson. Loading points may be integrated into the core prior to application of faces through introduction of high-strength (i.e. high-density) core or metal inserts. Hull members such as bulk heads, deck sections, etc. are normally prefabricated separately and are then laminated onto the hull.

Regardless of whether a laminate has been hand laid-up or sprayed-up, as well as whether it is part of a sandwich structure or not, it must be consolidated through cross-linking of the resin. Successful cross-linking normally requires precise control of laminate temperature and pressure as functions of time. The simplest cross-linking requirements possible clearly are ambient conditions, i.e. room temperature and no externally applied pressure, which with some resins result in reasonably well-consolidated laminates. Resins that start to cross-link at room temperature heat up as the exothermal cross-linking process progresses, thus further fuelling the reaction, which normally is sufficient to ensure proper cross-linking. However, many resins require an increased temperature for cross-linking to start and often also require that a specific temperature-time relationship is followed throughout cross-linking. Although specific temperature requirements may be sufficient to properly cross-link the resin, the quality of the

14.5 AN INTRODUCTION TO SANDWICH STRUCTURES laminate is unlikely to end up very good unless externally applied pressure is used to compact the laminate as it sets, i.e. until the cross-linking is more or less complete. Further, vacuum is often drawn out of the laminate to minimise voids.

The laminate temperature during cross-linking may be controlled through heating of the mould and/or the surrounding atmosphere. In the simplest case, heating of the surrounding air may be achieved by blowing hot air under a tarpaulin “tent“ covering the laminate. External pressure may be applied using a closed mould, a gas-filled bladder, weights, a hydraulic press etc. To draw vacuum, a flexible rubber membrane called a vacuum bag is placed on top of or around the laminate and the edges of the membrane are sealed to prevent leakage, see Fig.14.4. Since the vacuum bag also applies a certain degree of pressure onto the laminate, this may for many applications be sufficient to properly compact the laminate. However, the most refined way to exactly control the cross-linking conditions is to use an autoclave, which is a pressure vessel capable of exactly controlling the temperature and pressure of the internal atmosphere. In addition to controlling pressure of the internal atmosphere, i.e. outside a vacuum bag, the autoclave may be used to draw vacuum under the vacuum bag. In terms of temperature, the autoclave can control both the temperature of the internal atmosphere and of the mould, if required. Nevertheless, it is rare that an autoclave is used to cross-link wet laid-up laminates.

Gel coat Vacuum bag Seal Laminate Core

Vacuum Vacuum

Mould

Figure 14.4 Schematic of vacuum-bag arrangement. The core is placed on top of the laminated but not cross-linked laminate whereupon vacuum is drawn and the laminate cross-linked. The “top“ laminate is then laminated directly onto the core and vacuum preferably drawn to improve compaction. Alternatively, the core and the top laminate may be applied concurrently before drawing vacuum. Rolling on top of the vacuum bag is common in order to work out voids and remove excess resin. Unsaturated polyesters heavily dominate the wet lay-up processes, but vinylesters are also used to some degree in more demanding applications. Although not very common, epoxies formulated for low-temperature cross-linking may be used as well. Just as clearly as polyesters dominate as resin, glass—and specifically E glass—dominates as reinforcing fibre. In hand lay-up, woven fabrics, random mats and stitched combinations thereof are used, whereas spray-up usually employs roving that is chopped to lengths in the range 10-40 mm. Both balsa and a variety of expanded foam cores are used; by far the most common is cross-linked PVC.

14.3.2 Characteristics The wet lay-up methods: • require small capital investments • typically use resins that cross-link at room temperature with little or no applied pressure and that are tolerant to variations in processing temperature • use simple tooling due to modest cross-linking requirements • are labour intensive • are very cost-effective for short production series and prototype production

14.6 MANUFACTURING

• are suitable for any size structures, notably very large • bring on worker health concerns due to the active chemistry of the resin (especially spray-up)

Sandwich components manufactured through the wet lay-up methods are characterised by: • modest mechanical properties (especially with spray-up) • matrix and void contents in the faces strongly depends on the skill of the workers (high contents with spray-up) • laminate quality depends on the skill of the workers • no well-controlled exterior surface if both laminates are laid-up directly onto the core (but one or two controlled surfaces obtainable with use of rigid, closed moulds)

14.3.3 Applications Due to low capital and high labour costs, wet hand lay-up is used for products manufactured in short or very short series and where the requirements on structural and environmental properties are not excessive, typically meaning low to moderate loads and ambient temperatures. Applications include [1,4,5,8]: • motor and sailing yachts • mine-sweepers and high-speed passenger ships • refrigerated truck and railroad containers • storage tanks

Particularly interesting applications are ships, since they are the largest composite components built. Mine-sweepers, or mine-counter-measure vessels (MCMV), longer than 50 meters, wider than 10 meters and with displacements up to 400 metric tons have been built [7]. Although sometimes built as single-skin structures, they are more commonly built according to the sandwich concept using foam cores. In this application, composite materials and wet hand lay-up offer substantial advantages; the non-magnetic material is highly relevant so as not to risk detonating magnetically sensitive mines; the foam core sandwich structure is damage tolerant to under-water detonations; hand lay-up is highly competitive, since large structures are manufacturable and few look-alike MCMVs are built.

Similarly, surface-effect ships (SES) are capable of transporting up to four hundred passengers at speeds in excess of 50 knots. The SES is a catamaran with curtains between the two keels in bow and stern allowing an air cushion to be maintained between the keels; the air cushion lifts the hull out of the water to significantly reduce drag, thus allowing high speeds to be reached. The largest composite sandwich SES built measures approximately 40 by 15 meters [7]. Fig.14.5 shows the Swedish stealth vessel SMYGE, which is a 30 m long SES prototype craft built according to the sandwich concept using aramid- and glass-fibre-reinforced vinylester/polyester in the faces and PVC foam cores. SMYGE has a displacement of 140 metric tons and permits speeds up to 50 knots in calm waters. In SES applications composites help achieve a low enough weight to suspend an entire ship on an air cushion and hand lay-up is once again highly competitive for size and series-size reasons.

14.7 AN INTRODUCTION TO SANDWICH STRUCTURES

Figure 14.5 The Swedish SES SMYGE is built using stealth technology. While MCMVs and SESs are spectacularly successful case studies of wet hand lay-up of sandwich structures, they are nevertheless relatively rare. In contrast, the same techniques are in wide-spread use in manufacture of yachts and various tanks and containers.

The main differences in applications between hand laid-up and sprayed-up composites are due to the differences in labour costs and mechanical properties. The lower labour cost of spray-up implies that longer series are economically feasible and the inferior mechanical properties achieved mean that commodity-type products are more common. Sprayed-up sandwich components include:

• Small pleasure boats • Storage containers and tanks

14.4 Prepreg Lay-Up Popular as wet lay-up is, it is at best limited to moderately loaded structures due to the materials used, the cross-linking conditions employed, and the manner in which the impregnation is accomplished. Both single-skin laminates and sandwich structures for more advanced structures, e.g. for competition yachts and in aerospace applications, tend to be laid up using prepregs. The use of prepregs ensures that the reinforcement is well impregnated and resins used in prepregs also tend to have better properties than the ones available for wet lay-up. However, prepreg resins typically require well-controlled cross-linking conditions, meaning that the temperature must be increased above room temperature. For consolidation, a vacuum bag is likely required and in advanced applications also an autoclave.

14.4.1 Procedure Manufacturing of sandwich components with faces made from prepregs may be accomplished in two overall manners. Under certain conditions, the laminate may be laid directly onto the core in a fashion similar to that used in wet hand lay-up (e.g. [9-11]). Alternatively, previously manufactured single-skin laminates may be adhesively bonded to the core in a separate process (see further the Adhesive Bonding section below). Following the previous section, also this section commences with a description of how a single-skin laminate is manufactured (e.g. [12]),

14.8 MANUFACTURING whereupon descriptions of how sandwich components with faces made from prepregs may be laid-up directly onto the core.

To manufacture a single-skin laminate from prepregs, a one-sided mould is first treated with a mould-release agent, whereupon prepreg sheets are placed one at a time on top of one another, carefully ensuring that no voids or other contaminants (such as prepreg backing paper) are entrapped and that the sheets are thoroughly tacked to each other. Lay-up of prepregs is most often performed manually but may be automated. Automated tape lay-up combining the cutting, lay-up and compaction processes is becoming an accepted process in the aerospace industry to manufacture unidirectionally reinforced flat or nearly flat components (e.g. wing skins) [13,14]. In high-performance applications, the lay-up is performed in specially conditioned rooms to further minimise contaminants in the laminate.

Since almost all resins used in prepregs require controlled temperature and pressure to varying degrees to achieve intended properties, cross-linking usually occurs under a vacuum bag and with heat applied; in high-performance applications autoclaves are used [15]. To prepare the laid-up prepreg stack for cross-linking, it is covered with a perforated separator, a bleeder ply, a second separator, a barrier, a breather ply, and a vacuum bag (see Fig.14.6). The separator ensures that the part can be released, while the bleeder ply absorbs excess resin squeezed out of the prepreg stack. The barrier prevents the resin from diffusing into the breather, which ensures that the vacuum pressure of the vacuum bag is evenly applied. In several types of applications moulds are complemented with so-called caul plates which may be elastomeric or rigid, normally of metal. Elastomeric caul plates may be added on top of the prepreg stack to improve the surface finish of the part by ensuring more even pressure. Cast or moulded elastic caul plates may also be used to eliminate bridging over concave areas through application of localised pressure. Rigid caul plates are used to allow precise geometrical control at edges, holes, flanges, etc. by not allowing resin bleeding and thus tapering of the part. Bleeder Barrier Vacuum bag Separator Bleeder Part Perforated separator Separator Seal

Mould Vacuum Vacuum

Figure 14.6 Schematic of vacuum-bag assembly. After the vacuum-bag assembly is completed, the consolidation process starts with evacuation of the bag; vacuum may or may not be maintained throughout the moulding operation. Pressure is then applied and the temperature gradually increased to the specified resin cross-linking temperature, which is maintained for a significant amount of time. After cross-linking has been completed, vacuum and pressure are released and the temperature gradually lowered. Note that the processing conditions vary with material system but, in general, a typical processing sequence for an epoxy prepreg system may be: • apply release agent to mould

14.9 AN INTRODUCTION TO SANDWICH STRUCTURES

• arrange vacuum-bag assembly on mould, e.g. as illustrated in Fig.14.6 • place in autoclave and apply vacuum • apply specified pressure and release vacuum

• increase temperature to specified temperature at specified rate • maintain temperature and pressure for specified time • cool at specified rate

It is often possible to lay prepregs directly onto the core and cross-link the face laminate already in place in a process similar to that common in wet lay-up. In aerospace applications, both direct lay-up onto the core [9,11,16] and separate face manufacture with subsequent bonding [9,16-18] are used. Prepreg lay-up directly onto the core is of increasing interest also in the ship-building industry [10] to improve properties and to reduce the worker health hazards associated with wet lay-up. Also with prepreg lay-up, balsa and foam cores may need to be primed and/or a separate adhesive film added to achieve sufficient face-core bonding.

Since the aim of using prepregs instead of wet lay-up usually is to achieve improved composite properties, higher-performance materials are more common. Thus, carbon and aramid fibres are commonplace, although S glass is also used; the fibres are continuous and oriented in weaves or merely aligned. In most cases the matrix is an epoxy, especially for aerospace and other advanced applications. With such high-performance prepregs, it is natural to use high- performance cores with high temperature tolerance, e.g. honeycombs and PMI and PEI foams. Recent developments in prepregs with low-temperature cross-linking polyesters and epoxies, typically glass- or perhaps carbon-reinforced, appear to be attractive as direct replacements of wet lay-up polyesters; in such applications the likely cores are expanded foams, e.g. PUR and PVC, and balsa.

14.4.2 Characteristics The prepreg lay-up method: • requires medium capital investments, high if autoclave is used • uses resins that require increased temperature, vacuum, and often externally applied pressure to cross-link as intended and that are fairly intolerant to variations in processing conditions • is labour intensive • is suitable for short production series • is suitable for structures of any size Sandwich components manufactured through prepreg lay-up are characterised by: • good to excellent mechanical properties • low void contents in the laminate faces • consistent laminate quality • no well-controlled exterior surface if both laminates are laid-up directly onto the core (but one or two controlled surfaces obtainable with use of rigid, closed moulds)

14.10 MANUFACTURING

14.4.3 Applications Prepreg lay-up is most suitable for manufacturing of products in short series where the need for good or exceptionally good properties can motivate the higher costs. Lay-up of prepregs directly onto the core is common in the aerospace industry, where applications include [9,16,18-20]: • vertical and horizontal stabilisers • control surfaces • landing-gear doors • floors • rotor blades and servo flaps As an example of high-performance applications, Fig.14.7 shows the Swedish military aircraft JAS 39 Gripen where the canard wing, the vertical stabiliser and access doors are sandwich elements manufactured from carbon-fibre-reinforced epoxy prepregs and aluminium honeycomb cores.

Figure 14.7 The Swedish military aircraft JAS 39 Gripen. As previously mentioned, lay-up of prepregs directly onto the core is of significant interest in the ship-building industry to improve both properties and work environment when compared to the wet lay-up processes. Nevertheless, this practice is still typically limited to competition yachts and the established wet hand lay-up technique still dominates this industry.

14.5 Adhesive Bonding The two previous sections have concentrated on methods to manufacture sandwich components in a one-step procedure. Although clearly desirable from an economical perspective, it is not always possible or desirable to do so. In such cases the faces and the core may instead be bonded to each other in a separate manufacturing step [9,17,18,21-25], which for example is the case when the faces are not polymeric in origin where there is no alternative to a separate bonding step.

14.11 AN INTRODUCTION TO SANDWICH STRUCTURES

14.5.1 Procedure Conceptually, adhesive bonding of faces and core is quite simple and independent of face and core materials, see Fig.14.8. Adhesive layers are interleaved between the faces and the core and the whole stack is subjected to increased temperature and pressure as required by the adhesive resin, whereupon the sandwich is cooled. For high-performance applications the bonding process likely takes place using a vacuum bag and an autoclave, whereas for less demanding applications it may be sufficient to use a vacuum bag and/or weights or a hydraulic press. Since there should be little or no resin bleeding if the bonding is correctly performed, the vacuum-bagging arrangement is simplified when compared to laminate manufacture. Face sheet

Adhesive Honeycomb

Face sheet Bonded sandwich component

Figure 14.8 Schematic of adhesive bonding of sandwich structures. It is normally necessary to prepare the surfaces to be bonded in order to achieve a good enough bond. Unless already done, foam cores are typically sintered and all loose particles removed; they may also be primed. Laminates intended for bonding are often manufactured with a peel ply, which is removed right before bonding to leave a clean and somewhat rugged surface. The surface still should be abraded to ensure proper adhesion. Metal faces should be abraded and chemically treated to promote adhesion. The processing conditions vary with material system but in general a typical processing sequence for bonding of composite laminates to a honeycomb core using an epoxy adhesive may be: • remove peel plies, abrade surfaces, and wash with solvent • apply adhesive film onto faces and place on core • arrange vacuum-bag assembly on mould • place in autoclave and apply vacuum • apply specified pressure and release vacuum • increase temperature to specified temperature at specified rate • maintain and pressure for a specified period of time • cool at specified rate

Face materials used may be composite laminates manufactured through prepreg lay-up or any other composites manufacturing technique capable of producing the required face geometry, or sheet metal. In advanced applications, the faces tend to be fibre-reinforced epoxies and the core Nomex or aluminium honeycomb or high-performance and high-temperature–tolerant expanded foams, e.g. PMI or PEI. Metal-faced sandwich structures typically have foam cores, such as PUR and PVC. The adhesive is used in film or liquid form depending on application and is usually epoxy or PUR. While thermoset adhesives dominate, thermoplastic (“hot-melt“) adhesives are used as well. Adhesives are discussed in references [21-23,25,26].

14.12 MANUFACTURING

14.5.2 Characteristics Adhesive bonding: • requires small to medium capital investments, high if autoclave is used • typically uses adhesive resins that require increased temperature and externally applied pressure to achieve intended properties • is labour intensive • is suitable for short production series • is suitable for small to medium-sized structures

Sandwich components manufactured through adhesive bonding are characterised by: • good to excellent mechanical properties • well-controlled surfaces (assuming the faces have at least one good surface) • potentially having partially failed bonds in curved sections due to geometric mismatch between preformed faces and core

14.5.3 Applications Adhesive bonding of faces and core is rather common in the aerospace industry and the applications are the same as for direct prepreg lay-up [9,16-20,25]. The most common applications of metal-faces sandwich structures are various shipping containers for refrigerated goods, aircraft cargo containers, and construction elements. An interesting application is the external structure of the Stockholm Globe Arena shown in Fig.14.9 which is made of sandwich panels with aluminium faces bonded to the core material.

Figure 14.9 The outer structure of the Stockholm Globe arena is made of sandwich panels with aluminium faces adhesively bonded to the core material.

14.6 Liquid Moulding Several related liquid-moulding processes are used to manufacture sandwich components. They all have in common that the reinforcement is first placed in the mould whereupon the liquid resin is infused into the reinforcement structure. The liquid-moulding processes used include: • resin transfer moulding (RTM) [27-33]

14.13 AN INTRODUCTION TO SANDWICH STRUCTURES

• structural reaction injection moulding (SRIM) [27-29] • vacuum-injection moulding [27,29,34,35]

Liquid moulding has received much interest in recent years due to its capability of producing geometrically complex structures in an economical fashion, without creating an unhealthy work environment since the processes use closed moulds. Especially RTM is becoming increasingly popular in the automobile industry to manufacture components for vehicles produced in short series.

14.6.1 Procedure The initial procedure description below is generic and covers all three aforementioned liquid- moulding processes, whereupon the distinguishing differences are pointed out.

The reinforcement, in the form of fabrics and mats or preforms, is placed in the mould together with the core, normally by hand. However, not only cores, but also inserts and fasteners are easily integrated into the reinforcement or the core before impregnation. After the mould is closed, the resin is introduced into the mould to impregnate the reinforcement using pressure and/or vacuum. The resin is often heated to lower the viscosity and thus facilitate impregnation. The resin infusion is stopped when the resin front has reached all the ventilation holes in the mould and the resin starts to leak out. The resins used may cross-link at ambient temperature or, alternatively, the mould may be heated. The cross-linking reaction should not begin until the mould is nearly filled, as gelation of the resin will prevent it from impregnating the reinforcement completely, thus creating dry spots and voids.

The processes of RTM, SRIM, and vacuum-injection moulding are distinguished from one another primarily by the type of resins, moulds, and impregnation technique used, see Table 1. In vacuum-injection moulding conventional preformulated resins, similar to those used in the wet lay-up processes, are used. In contrast, SRIM employs highly reactive resins that are mixed right before injection. Although the resins used in RTM also are similar to those used in vacuum- injection moulding, they may either be preformulated or mixed right before injection. In RTM and SRIM rigid, matching moulds are used, whereas vacuum-injection moulding employs a one- sided mould, often a marginally modified version of a wet lay-up mould, covered by a vacuum bag. The mould halves are closed using clamps or a press, depending on part size and injection pressure. In RTM and SRIM the resin is injected into the mould under pressure, in RTM sometimes assisted by drawing vacuum at the ventilation ports, whilst in vacuum-injection moulding the sole force driving the impregnation is vacuum drawn from under the vacuum bag. Fig.14.10 illustrates RTM or SRIM and Fig.14.11 vacuum-injection moulding.

Core Dry reinforcement/ preform Clamping force

Resin in Air out

Mould Clamping force

Figure 14.10 Schematic of RTM or SRIM.

14.14 MANUFACTURING

Resin Core Seal Reinforcement

Resin in Vacuum

Vacuum bag Mould

Figure 14.11 Schematic of vacuum-injection moulding. The differences in resin reactivity between RTM and vacuum-injection moulding on the one hand and SRIM on the other translate into two major differences: with the former the mould fill times range from a few minutes to a few hours; in the latter case enabling filling of large parts before the increasing resin viscosity prohibits further impregnation. In contrast, the fill times in SRIM are usually less than a minute due to the much higher resin reactivity, meaning that smaller parts than with the other two processes are manufacturable. On the other hand, the cross- linking times are in the range of a few minutes to a few hours for RTM and a few hours with vacuum-assisted moulding, whilst in SRIM parts can be demoulded in a matter of a few of minutes. The higher injection rates of SRIM increase the potential problem of so-called fibre washout, which is when the reinforcement is moved by the advancing resin front. The manufacture of large, complex parts with high fibre volume fractions thus is more difficult with SRIM than with RTM.

In SRIM, the two-component resin is mixed right before injection in an impingement-mixing nozzle using dedicated pumps. In RTM injection may be achieved with a dedicated pump or a simple pressure pot, which is a closed vessel containing the resin; in the latter case the resin is forced out of the pot through injection of pressurised air. With dedicated pumps the resin is mixed right before injection also in RTM. In all three processes multiple inlet ports may be necessary for large components. An alternative way to facilitate complete wetout of large components, especially in RTM, is to inject the resin into a channel running around the entire exterior of the part, thus allowing the resin to impregnate the reinforcement from all sides concurrently. Vacuum is drawn or air allowed to escape from the centre of the part in one or more locations, so the impregnation occurs from the perimeter and into the centre of the part.

RTM SRIM Vacuum-Injection Moulding Resin formulation Preformulated or mixed right Mixed right before injection Preformulated before injection Resin reactivity Low High Low Mould type Matching, rigid Matching, rigid Open-faced; vacuum bag Mould material Composite, metal Composite, metal Composite, metal; vacuum bag Impregnation Pressurised injection, may Pressurised injection Vacuum be complemented by vacuum Cycle time Long Short Long Part size Small to large Small to medium-sized Small to large Table 1 Comparison of liquid-moulding processes.

14.15 AN INTRODUCTION TO SANDWICH STRUCTURES

Of the several resin systems that perform adequately in RTM and vacuum-injection moulding, polyesters are most often used, although vinylesters, epoxies, and several others may be used as well. In SRIM polyurethanes clearly dominate usage due to the flexibility in altering the resin reactivity to achieve very rapid cross-linking. Reinforcements may be of virtually any type, although E glass dominates for cost reasons. The reinforcement form is often mats and fabrics, although rovings may well be braided onto a foam core acting as a mandrel. High-volume production of geometrically complex parts generally requires preforming of the reinforcement (e.g. [36,37]). Either chopped reinforcement and a binder are sprayed onto a screen mould thus forming the desired geometry or a reinforcement mat held together with a thermoplastic binder is heated and compression moulded into the desired geometry. Using such reinforcement preforming techniques allows the entire manufacturing process to be highly automated and for example automobile parts can be manufactured in a matter of minutes. However, the use of randomly oriented, chopped fibres lowers the structural performance of the composite; if needed, fabrics therefore may be added locally to reinforce a preform. All kinds of foam cores are used; the most common ones being PVC, PUR, and PMI.

14.6.2 Characteristics The RTM method: • requires relatively small capital investments (for short series and simple parts) • uses resins that cross-link at room-temperature or at elevated temperature • uses simple tooling for moderately long production series and low injection pressures; long series, high injection pressures, and high-class surfaces require metal tooling • may be semi-automated, thus suitable for moderate to long production series • is cost effective for short and medium-sized production series • yields moderate to long cycle times • allows manufacture of large, highly integrated structures • may cause reinforcement movement during injection • causes low emission of volatiles due to closed moulds

Sandwich components manufactured through RTM are characterised by: • good mechanical properties • good control over reinforcement orientation • allowing very high fibre fractions • potentially having resin- and void-rich areas • having well-controlled surfaces that may be of high class

The SRIM method: • requires relatively large capital investments • uses resins that cross-link at elevated temperature • uses simple tooling for moderately long production series and low injection pressures; long series, high injection pressures, and high-class surfaces require metal tooling • may be fully automated, thus suitable for long production series • is cost effective for long production series • yields short cycle times

14.16 MANUFACTURING

• allows manufacture of small to medium-sized, highly integrated structures • may cause reinforcement movement during injection • causes low emission of volatiles due to closed moulds

Sandwich components manufactured through SRIM are characterised by: • good mechanical properties • good control over reinforcement orientation • allowing high fibre fractions • potentially having resin- and void-rich areas • having well-controlled surfaces that may be of high class

Vacuum-injection moulding: • requires small capital investments • uses resins that cross-link at room-temperature or at elevated temperature • uses very simple tooling • is labour-intensive • is cost effective for short production series • yields long cycle times • allows manufacture of large, highly integrated structures • causes low emission of volatiles due to closed moulds

Sandwich components manufactured through vacuum-injection moulding are characterised by: • modest mechanical properties • good control over reinforcement orientation • potentially having resin- and void-rich areas • having one well-controlled surface

14.6.3 Applications RTM and SRIM have become very popular in the automobile industry [35,37-40]; RTM for short to medium-sized production series and SRIM for long series. The major reasons for this interest is that complex parts may be economically produced in one step using low-cost moulds and that the parts may have high-class surfaces. Applications include almost any conceivable, but most often exterior panels of semi-exclusive cars. However, the interest in manufacturing structural parts, such as the entire base plate, is increasing. Vacuum-injection moulding is suitable for large products manufactured in short series, such as in ship building, offshore, exterior vehicle panels, etc. (e.g. [34,35]).

An interesting application where the advantages of RTM have been utilised to a significant degree is in the manufacture of the entire self-supporting body of the Swedish all-terrain military caterpillar vehicle no. 206 shown in Fig.14.12a (e.g. [8,31]). The vehicle is unique in that large composite-faced sandwich structures are manufactured entirely through RTM using E glass and polyester. The panels (shown in Fig.14.12b) are sandwich elements with PVC foam cores in roof, floor and sides and PUR foam cores in doors. The panels are bonded together using an epoxy adhesive to form a self-supporting structure. The composite body weighs approximately

14.17 AN INTRODUCTION TO SANDWICH STRUCTURES

300 kilograms. The size of the floor panel is 1.9 by 3 meters and the height of the composite structure 2 meters.

(a) (b) Figure 14.12 (a) All-terrain vehicle (b) the individual sandwich panels of the self-supporting body are manufactured in one piece using RTM. Similarly, the self-supporting driver’s cab of a Swiss locomotive is manufactured entirely through RTM [32]. The cab is prefabricated in four individual sections—roof, front and two side panels—that are bonded together using an epoxy adhesive. The sections are sandwich panels with glass-fibre-reinforced polyester faces and PUR foam cores. To bond the driver’s cab module to the steel locomotive body a highly elastic rubber-modified PUR adhesive is used. A similar application is the self-supporting front cab of an Italian high-speed train where the panels are vacuum-injection moulded using glass-fibre reinforcement—stacked continuous strand mats and woven fabrics—and a low-viscosity polyester resin system on PUR foam cores [34].

Another interesting RTM case study is the manufacture of composite propeller blades for a variety of regional aircraft and hovercraft, such as Fokker F50, Saab SF340, and Bell LCAC hovercraft (see reference [33]). The basic construction of the blade is shown in Fig.14.13. Preformed layers of carbon fibres and fabric held together with a binder are stacked and inserted into the blade mould where a PUR core is foamed in-situ. The braided glass-fibre skin is draped over the foam-filled preform whereupon metal fittings are added together with leading edge reinforcement layers and an aluminium braid for lightning protection. The complete assembly is then placed in the RTM blade mould, which is heated to reduce the epoxy resin viscosity during vacuum-assisted injection

45° ±45€ glass cloth 0° skin layers

Foam core

0° carbon fibre spar

Membrane and interfacial bond layer

Figure 14.13 Rotor blade construction. Reprinted from [33] with kind permission from Butterworth- Heinemann journals, Elsevier Science Ltd, The Boulevard, Langford Lane, Kidlington 0X5 1GB, UK.

14.18 MANUFACTURING

Other high-performance applications include an 8.5 meter long hollow wind turbine blade with an internal axial stiffening web separating the hollow chambers and an 18 meter long hollow sailing boom with an internal furling (sail-rolling) mechanism [41]. Both the wind turbine blade and the sailing boom are manufactured through RTM using a “smart core“ technique, which is a lost-core method said to offer advantages over conventional RTM [41]. Materials used are carbon fibre reinforcement, low-viscosity epoxy and thermoformed PVC core.

14.7 Continuous Lamination From an economical perspective a continuous manufacturing process naturally is preferable. A suitable way to manufacture continuous sandwich panels is using a double-belt press, as shown in Fig.14.14. With a double-belt press it is possible to both heat and cool the material while at the same time subjecting it to a specified pressure profile, thus making it a potentially useful device to impregnate and/or laminate composites. Face material

Core material

Pressure and Cutting/trimming heating/cooling area area

Figure 14.14 Schematic of continuous lamination using a double-belt press. When using a double-belt press to manufacture sandwich components, the face sheets are likely coiled up in very long lengths. Two rolls of face sheets are first uncoiled and guided in between the belts of the press. The core is then, in any of a number of fashions, inserted between the face sheets, possibly together with adhesive layers. The faces and the core are then bonded to one another through concurrent application of heat and pressure, whereupon the sandwich is cooled under pressure to consolidate it [42,43].

Face materials may be sheet metal, unreinforced polymers, and composite laminates or prepregs. To obtain a truly continuous core it may prove convenient to in-situ foam it between the faces through injection and subsequent expansion of for example PUR [44-46]. An alternative route is to sandwich a thermoplastic polymer film containing a foaming agent between the faces; as soon as the double-belt press melts the polymer film, the foaming agent is free to expand, thus filling the gap between the faces with a foam [47]. It is naturally possible to insert discrete blocks of wood or prefoamed core between the faces, although this procedure brings on concerns of core- core disbonds.

A slightly different approach to continuous sandwich manufacture involves a vertical facility similar in concept to a double-belt press [48]. In this process, reinforcement fabrics are simultaneously impregnated and laminated onto core blocks in a continuous process. The materials used were glass fabrics and mats, polyester resin, and polymer-foam cores.

14.19 AN INTRODUCTION TO SANDWICH STRUCTURES

Since there is little conceptual difference between the continuous processes described above and the most well-known continuous composites-manufacturing technique of them all—pultrusion— it should not be surprising to find that pultrusion of sandwich panels has been tried [49]. Just as in the vertical semi-continuous process described above, such manufacture would involve concurrent manufacture of faces and sandwich in the same process.

All continuous processes have in common that they can only with the greatest technical difficulty produce anything but constant cross-section components. With the exception of pultrusion, even a deviation from flat, constant thickness components requires complex technical solutions.

14.8 Other Processes The manufacturing techniques described above are merely the more common ones used to date. As so often holds true with composites, there is nothing preventing an innovative mind from inventing new or combining/modifying established techniques to enable manufacturing of components. Unconventional processes that have been tried in sandwich manufacturing include compression moulding, filament winding and various in-situ foaming techniques. These processes are briefly discussed in the following.

Compression moulding of structural sandwich components is similar to the compression moulding of single-skin laminates (e.g. [50]). The process is schematically shown in Fig.14.15. When using thermoplastic-based faces, e.g. GMT, the material is heated in an oven to a temperature exceeding the softening point of the matrix and thereafter placed in a cooled mould with the core sandwiched inbetween. The mould closes very fast and the material is forced to conform to the mould before it consolidates, whereupon the component is ejected. Rapid closing is essential to achieve high surface finish. The choice of core material is important to ensure that it has a compression strength adequate to withstand the moulding pressure (between 0.2 and 4 MPa) which is important for dimensional stability and surface finish. Also essential is good bonding of the faces to the core. Thermoformability is advantageous, especially for complex- shaped parts; a thermoplastic foam core then may be appropriate since it easily may be reshaped and compacted. Improved bonding between faces and core and improved dimensional tolerance may be obtained if the thermoformable core is slightly over-dimensioned since the increased pressure caused by core compaction may reduce surface irregularities. The use of thermoplastic cores may further enhance bonding since heating of the face sheets will cause the core surface to melt. Although the compression-moulding process described above applies to thermoplastic materials, it is naturally also possible to use thermoset-based face materials, such as SMC and BMC [9,35,40,51].

14.20 MANUFACTURING

Upper mould half Resin/reinforcement compound Guide pin Core Lower mould half

Compression force

Moulded part

Figure 14.15 Schematic of compression moulding. Similarly, the self-supporting driver’s cab of a Swiss locomotive is manufactured entirely through RTM [32]. The cab is prefabricated in four individual sections—roof, front and two side panels—that are bonded together using an epoxy adhesive. The sections are sandwich panels with glass-fibre-reinforced polyester faces and PUR foam cores. To bond the driver’s cab module to the steel locomotive body a highly elastic rubber-modified PUR adhesive is used. A similar application is the self-supporting front cab of an Italian high-speed train where the panels are vacuum-injection moulded using glass-fibre reinforcement—stacked continuous strand mats and woven fabrics—and a low-viscosity polyester resin system on PUR foam cores [34].

Semi-structural applications of compression-moulded sandwich components found in the automotive industry include compression moulded bumper beams with PP foam core and glass- mat-reinforced PP faces [52,53], a compression moulded bonnet of a passenger-car with SMC faces and a foam core [40], partition walls of glass-mat-reinforced PP faces combined with a styrene-acrylonitrile (SAN) foam core in large trucks [52], trunk-floors of passenger cars [52] and folding rear seat backs of a passenger car manufactured using glass-mat-reinforced PP faces and expanded PP foam core [54]. The major reasons why glass-mat-reinforced thermoplastics are gaining interest in the automotive industry [51,52,54,55] are that they look, handle and are

14.21 AN INTRODUCTION TO SANDWICH STRUCTURES processed in a manner similar to sheet metal and have the advantages of being lightweight and having high chemical and corrosion resistance.

In filament winding of sandwich components the inner face would likely be wound first and the winding stopped to apply a flexible or prefoamed core onto the inner face. The winding would then resume to apply the outer face. A particularly interesting application of filament wound sandwich structures is the self-supporting inner body of a railway passenger-car [56-58]. The body is wound in three sections in the manner described above and windows and doors are cut out, whereupon the modules are assembled inside a metal outer body. The body sections are mounted through the roof and slid in to position whereupon the roof section is put in place as illustrated in Fig.14.16. The interior of the car is either integrated in the wound structure or pre- assembled.

Figure 14.16 Mounting of the wound inner body of a railway passenger-car. Reprinted from [56] with permission from Dr. Kurt Anderegg. A similar technique has been used to manufacture prototypes of self-supporting external structures of complete railway cabs mounted on conventional steel chassis. The outer shell- structure is manufactured through filament winding using PMI cores and the windows and doors are cut out. Thermal insulation, wiring and ventilation ducts are added to the wound structure before it is completely cross-linked. Another example of a railway application is a covered hopper car [59] where the car body is filament wound using E glass-reinforced polyester. Bulkheads and side walls are manufactured using E glass-reinforced polyester faces and balsa wood cores. Hat section stiffeners and wide flange beams are prefabricated using hand lay-up and pultrusion, respectively. For joints a combination of adhesive bonding using an acrylic adhesive system and bolting is employed. A slightly different technique has been employed in manufacturing a box girder [60]. The preformed core, fitted on a pipe, serves as mandrel. The face is wound in two steps: the inner laminate layer is wound directly onto the core and the outer is wound after unidirectional prepreg sheets have been laid up on the flanges to further stiffen the structure.

A technique to fabricate an all-thermoplastic sandwich structure employing a thermoplastic foaming hot-melt adhesive as core material has been developed [61,62]. The foaming hot-melt is sandwiched between the face sheets in a “precompact“ and placed in either a hot platen press or in a shaped mould. The sandwich structure forms as heat is applied and the thermoplastic hot-

14.22 MANUFACTURING melt foams. The fully thermoplastic character of the sandwich system allows finishing treatment, thermoforming, such as edge forming and folding [61-64]. All-thermoplastic sandwich structures are gaining interest partly due to the possibility of producing shaped articles from flat sheets using local heating and ordinary sheet metal forming techniques [65-67].

Methods to manufacture net-shaped all-thermoplastic structures in-situ, i.e. in one step, have also been developed [68,69]. Generally in these processes, the core materials (powdered thermoplastic resins, foaming agents and fillers) are mixed and sandwiched between preformed fabric—either commingled, cowoven or unimpregnated—placed in a closed mould and subjected to heat. The face sheets are impregnated with resin during the process as the core foams. These processes potentially offer advantages over other manufacturing techniques in that core and face sheet prefabrication, bonding and machining are eliminated.

14.9 Outlook The development of structural sandwich components in terms of manufacturing science and constitutive materials primarily does not appear to be performance-driven, since high-performing sandwich components indeed already are manufacturable—albeit at very high cost. Rather, the development appears to be driven by desires to lower component price, improve recyclability, and improve work environment. As obvious from the project descriptions above, today’s prevailing manufacturing techniques—with the exceptions of double–belt–press laminating and to some degree SRIM which both have limited application areas—tend to involve a large degree of manual labour and long cycle times. The labour intensity translates into high marginal costs commonly dwarfing other costs, and most manufacturing techniques thus can only be economically justified in relatively short production series. Further, the long cycle times are unacceptable in many industries, most notably the automotive industry. It is therefore hardly surprising that few structural sandwich components have been manufactured in long series.

Much recent work has been dedicated towards automating techniques to manufacture structural sandwich components of complex geometry with the overall aim of satisfying the requirements of the automobile industry in terms of both processing rate and part cost (e.g. [70]). The ever more important issue of recycling is having an increasing impact on the constituent materials used in sandwich applications. The perhaps most obvious solution to the recyclability issue is to use thermoplastic resins only and to utilise the same resin in both faces and core [52,53,70-72]. Although less straightforward in concept, work is underway to develop methods to recycle thermoset-based mixed-material sandwich components, e.g. the combination of glass/polyester faces and PVC cores so common in the ship-building industry [73]. Work-environment issues in composites manufacturing tend to be synonymous with solvent (mainly styrene) emissions and (epoxy-related) allergies. Approaches to reduce the solvent problem are to use closed-mould processes [29], for example liquid moulding, or to use prepregs; the latter approach has been the subject of much interest as direct replacement for wet-lay up of laminates in for example the ship-building industry. A route to complete elimination of both solvent emissions and allergies is to entirely abandon thermosets for thermoplastics, which is an approach potentially solving all three issues of improved productivity/reduced cost, recyclability and improved work environment simultaneously [47,53,61-64,68-71].

14.23 AN INTRODUCTION TO SANDWICH STRUCTURES

There is little doubt that the manufacturing science of sandwich components will develop further to enable low-cost, long-series manufacture of structurally capable sandwich components. To a not insignificant degree this development will be aided by new material systems and forms, such as for example three-dimensional reinforcing fabrics that integrate faces and core and thus have the potential of significantly improving structural integrity [74,75]. There should be equally little doubt that the issues of recyclability and work environment will be satisfactorily solved. One solution to these problems is the aforementioned use of thermoplastics in favour of thermosets, while other possible solutions ought to be within reach through enhancements in constituent materials and manufacturing techniques.

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14.24 MANUFACTURING

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14.25 AN INTRODUCTION TO SANDWICH STRUCTURES

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14.26 MANUFACTURING

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14.27 AN INTRODUCTION TO SANDWICH STRUCTURES

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[60] Meier, U., Müller, R. and Puck, A., “GFK-biegeträger unter quasistatischer und schwingender beanspruchung“, AVK, 18, Fredenstadt, 1982.

[61] A. Beukers, “A new technology for sandwich plates and structures based on the use of in- situ foamable thermoplastic films“, Advanced Materials: Cost Effectiveness, Quality Control, Health and Environment, A. Kwakernaak and L. v. Arkel, Eds., SAMPE/Elsevier, 393-405, 1991.

[62] A. Beukers, “Cost effective composite plate & shell structures for transports by manufacturing technologies like foaming, thermoforming and pressure forming continuous fibre reinforced thermoplastic sheets“, Advanced Materials and Structures from Research to Application, J. Brandt, H. Hornfeld and M. Neitzel, Eds., SAMPE/Elsevier, 427-443, 1992.

[63] van Dreumel, W. H. M., “Origami-Technology—Creative Manufacturing of Advanced Composite Parts“, Ten Cate Advanced Composites, 1989.

[64] Airsan, promotional brochure, Schreiner Composites, Zoetermeer, The Netherlands.

[65] Rietdijk, B. and Brambach, J. A., “Method of making an article from a thermoplastic sandwich material“, US Patent 5,032,443, 1991.

[66] Brambach, J. A., “Method of making a shaped article from a sandwich construction“, US Patent 5,043,127, 1991.

[67] Offringa, A. R., “Application of advanced thermoplastics“, International Symposium on Advanced Materials for Lightweight Structures, ESTEC, 191-195, 1992.

[68] Saatchi, H., “Innovative, low-cost composite fabrication processes“, Fabricating Composites ‘89, SME, EM89-582, 1989.

[69] Saatchi, H., Murray, P. J., Coleman, R. E., Smith, K. A., “Enhanced foaming of thermoplastics for in-situ fabrication of advanced articles“, US Patent 5,122,316, 1992.

14.28 MANUFACTURING

[70] “Efficient manufacturing of all-thermoplastic composite sandwich components (EMATS)“, Brite-EuRam contract n° BRE2-CT94-0912, commenced 1994.

[71] “Manufacturing of high-performance thermoplastic composite and sandwich components (TpCaSC)“, Nordic Industrial Fund project n° P93214, commenced 1993.

[72] Gellhorn, E. v., “Bauteile aus Sandwichelementen herstellen“, Kunststoffe, Vol 81, No 11, 1009-1013, 1991.

[73] “Återvinning av härdplastkompositer“, Nordic Industrial Fund project n° P93216, commenced 1993.

[74] Verpoest, I., Wevers, M. and De Meester, P., “2.5D and 3D-fabrics for delamination resistant composite laminates and sandwich structures“, SAMPE Journal, 25(3), 51-56, 1989.

[75] van Vuure, A. W, Ivens, J. and Verpoest, I., “Sandwich panels“, International SAMPE Symposium, Vol 40, 1281-1292, 1995.

14.29