Composite Mechanics (Advanced Structured Materials, 184) 3031323890, 9783031323898

Table of contents :
Preface
Contents
Symbols and Abbreviations
Latin Symbols (Capital Letters)
Latin Symbols (Small Letters)
Greek Symbols (Small Letters)
Mathematical Symbols
Indices, Superscripted
Indices, Subscripted
Abbreviations
1 Introduction
1.1 Composite Materials
1.2 Continuum Mechanical Modeling
References
2 Micromechanics
2.1 Reference Numbers to Characterize Composition
2.2 Prediction of Elastic Properties of Laminae
2.2.1 Mechanics of Materials Approach
2.2.2 Elasticity Solutions with Contiguity after Tsai
2.2.3 Halpin–Tsai Relationships
2.3 Comparison with Experimental Results
2.3.1 Extraction of Experimental Values
2.3.2 Comparison Between Theoretical Predictions and Experimental Results
2.3.3 Optimized Representation of Theoretical Predictions
References
3 Macromechanics of a Lamina
3.1 Introduction
3.2 Kinematics
3.2.1 Plane Elasticity Element
3.2.2 Classical Plate Element
3.2.3 Combined Plane Elasticity and Classical Plate Element
3.3 Constitutive Equation
3.3.1 Isotropic Material: Plane Elasticity Element
3.3.2 Isotropic Material: Classical Plate Element
3.3.3 Isotropic Material: Combined Plane Elasticity and Classical Plate Element
3.3.4 Orthotropic Material: Combined Plane Elasticity and Classical Plate Element
3.4 Equilibrium
3.4.1 Plane Elasticity Element
3.4.2 Classical Plate Element
3.4.3 Combined Plane Elasticity and Classical Plate Element
3.5 Partial Differential Equations
3.5.1 Plane Elasticity Element
3.5.2 Classical Plate Element
3.5.3 Combined Plane Elasticity and Classical Plate Element
3.6 Failure Criteria
3.6.1 Maximum Stress Criterion
3.6.2 Maximum Strain Criterion
3.6.3 Tsai–Hill Criterion
3.6.4 Tsai–Wu Criterion
References
4 Macromechanics of a Laminate
4.1 Introduction
4.2 Generalized Stress–Strain Relationship
4.3 Special Cases of Laminates
4.4 Failure Analysis of Laminates
References
5 Example Problems
5.1 Introduction
5.2 Problem 1: Stresses and Strains in a Symmetric Laminate
5.3 Problem 2: Stresses and Strains in an Asymmetric Laminate
5.4 Problem 3: Failure Criteria
5.5 Problem 4: Ply-By-Ply Failure Loads
5.6 Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae
5.7 Problem 6: Failure Envelopes for Unidirectional Laminae
References
6 Supplementary Problems
6.1 Problems
6.1.1 Supplementary Problem 1: Stresses and Strains in a Symmetric Laminate
6.1.2 Supplementary Problem 2: Stresses and Strains in an Asymmetric Laminate
6.1.3 Supplementary Problem 3: Failure Criteria
6.1.4 Supplementary Problem 4: Ply-by-Ply Failure Loads
6.1.5 Supplementary Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae
6.1.6 Supplementary Problem 6: Failure Envelopes for Unidirectional Laminae
6.2 Short Solutions
6.2.1 Supplementary Problem 1: Stresses and Strains in a Symmetric Laminate
6.2.2 Supplementary Problem 2: Stresses and Strains in an Asymmetric Laminate
6.2.3 Supplementary Problem 3: Failure Criteria
6.2.4 Supplementary Problem 4: Ply-by-Ply Failure Loads
6.2.5 Supplementary Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae
6.2.6 Supplementary Problem 6: Failure Envelopes for Unidirectional Laminae
Reference
7 Composite Laminate Analysis Tool—CLAT
7.1 Introduction
7.2 Analysis Options
Reference
Index

Citation preview

Advanced Structured Materials

Andreas Öchsner

Composite Mechanics

Advanced Structured Materials Volume 184

Series Editors Andreas Öchsner, Faculty of Mechanical Engineering, Esslingen University of Applied Sciences, Esslingen, Germany Lucas F. M. da Silva, Department of Mechanical Engineering, Faculty of Engineering, University of Porto, Porto, Portugal Holm Altenbach , Faculty of Mechanical Engineering, Otto von Guericke University Magdeburg, Magdeburg, Sachsen-Anhalt, Germany

Common engineering materials are reaching their limits in many applications, and new developments are required to meet the increasing demands on engineering materials. The performance of materials can be improved by combining different materials to achieve better properties than with a single constituent, or by shaping the material or constituents into a specific structure. The interaction between material and structure can occur at different length scales, such as the micro, meso, or macro scale, and offers potential applications in very different fields. This book series addresses the fundamental relationships between materials and their structure on overall properties (e.g., mechanical, thermal, chemical, electrical, or magnetic properties, etc.). Experimental data and procedures are presented, as well as methods for modeling structures and materials using numerical and analytical approaches. In addition, the series shows how these materials engineering and design processes are implemented and how new technologies can be used to optimize materials and processes. Advanced Structured Materials is indexed in Google Scholar and Scopus.

Andreas Öchsner

Composite Mechanics

Andreas Öchsner Faculty of Mechanical Engineering Esslingen University Applied Sciences Esslingen am Neckar, Baden-Württemberg, Germany

ISSN 1869-8433 ISSN 1869-8441 (electronic) Advanced Structured Materials ISBN 978-3-031-32389-8 ISBN 978-3-031-32390-4 (eBook) https://doi.org/10.1007/978-3-031-32390-4 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

Composite materials, especially fiber-reinforced composites, are gaining increasing importance since they can overcome the limits of many structures based on classical metals. Particularly the combination of a matrix with fibers provides far better properties than the constituents alone. Despite their importance, many engineering degree programs do not treat the mechanical behavior of this class of advanced structured materials in detail, at least on the bachelor degree level. Thus, some engineers are not able to thoroughly apply and introduce these modern engineering materials in their design process. This volume in the Advanced Structured Materials series provides first an introduction to the micromechanics of fiber-reinforced laminae, which deals with the prediction of the macroscopic mechanical lamina properties based on the mechanical properties of the constituents, i.e., fibers and matrix. The focus is on unidirectional lamina which can be described based on orthotropic constitutive equations. Three classical approaches to predict the elastic properties, i.e., the mechanics of materials approach, the elasticity solutions with contiguity after Tsai, and the Halpin–Tsai relationships, are presented. The quality of each prediction is benchmarked based on two different sets of experimental values. This chapter concludes with optimized representations, which were obtained based on the least square approach for the used experimental data sets. The second part of this monograph provides a systematic and thorough introduction to the classical laminate theory based on the theory for plane elasticity elements and classical (shear-rigid) plate elements. The focus is on unidirectional lamina which can be described based on orthotropic constitutive equations and their composition to layered laminates. In addition to the elastic behavior, failure is investigated based on the maximum stress, maximum strain, Tsai-Hill, and the Tsai-Wu criteria. The continuum mechanical approach reviews that partial differential equations lay the foundation to mathematically describe the mechanical behavior of any classical structural member known in engineering mechanics, including composite materials. Based on the three basic equations of continuum mechanics, i.e., the kinematic relationship, the constitutive law, and the equilibrium equation, these partial differential equations describe the physical problem. The here introduced classical v

vi

Preface

laminate theory provides a simplified stress analysis and a subsequent failure analysis, without the solution of the system of coupled differential equations for the unknown displacements in the three coordinate directions. This theory provides the solution of the statically indeterminate system based on a generalized stress–strain relationship under consideration of the constitutive relationship and the definition of the so-called stress resultants. Nevertheless, the fundamental knowledge from the first years of engineering education, i.e., higher mathematics, physics, materials science, applied mechanics, design, and programming skills, might be required to master this topic. The monograph concludes with a short introduction to a calculation program, the so-called Composite Laminate Analysis Tool (CLAT), which allows the application of the classical laminate based on a sophisticated Python script. Esslingen, Germany February 2023

Andreas Öchsner

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Composite Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Continuum Mechanical Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 6 9

2 Micromechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Reference Numbers to Characterize Composition . . . . . . . . . . . . . . . 2.2 Prediction of Elastic Properties of Laminae . . . . . . . . . . . . . . . . . . . . . 2.2.1 Mechanics of Materials Approach . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Elasticity Solutions with Contiguity after Tsai . . . . . . . . . . . . 2.2.3 Halpin–Tsai Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Comparison with Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Extraction of Experimental Values . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Comparison Between Theoretical Predictions and Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3 Optimized Representation of Theoretical Predictions . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13 13 15 15 25 31 34 34

3 Macromechanics of a Lamina . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Plane Elasticity Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Classical Plate Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3 Combined Plane Elasticity and Classical Plate Element . . . . 3.3 Constitutive Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Isotropic Material: Plane Elasticity Element . . . . . . . . . . . . . . 3.3.2 Isotropic Material: Classical Plate Element . . . . . . . . . . . . . . 3.3.3 Isotropic Material: Combined Plane Elasticity and Classical Plate Element . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.4 Orthotropic Material: Combined Plane Elasticity and Classical Plate Element . . . . . . . . . . . . . . . . . . . . . . . . . . .

57 57 58 59 59 61 63 63 64

38 47 55

64 65

vii

viii

Contents

3.4 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Plane Elasticity Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Classical Plate Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.3 Combined Plane Elasticity and Classical Plate Element . . . . 3.5 Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Plane Elasticity Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Classical Plate Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Combined Plane Elasticity and Classical Plate Element . . . . 3.6 Failure Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 Maximum Stress Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 Maximum Strain Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.3 Tsai–Hill Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.4 Tsai–Wu Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

70 70 73 78 79 79 80 81 82 82 82 83 83 84

4 Macromechanics of a Laminate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Generalized Stress–Strain Relationship . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Special Cases of Laminates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Failure Analysis of Laminates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

85 85 85 88 93 94

5 Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Problem 1: Stresses and Strains in a Symmetric Laminate . . . . . . . . 5.3 Problem 2: Stresses and Strains in an Asymmetric Laminate . . . . . . 5.4 Problem 3: Failure Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Problem 4: Ply-By-Ply Failure Loads . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Problem 6: Failure Envelopes for Unidirectional Laminae . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

95 95 95 106 117 132

6 Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Supplementary Problem 1: Stresses and Strains in a Symmetric Laminate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Supplementary Problem 2: Stresses and Strains in an Asymmetric Laminate . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.3 Supplementary Problem 3: Failure Criteria . . . . . . . . . . . . . . . 6.1.4 Supplementary Problem 4: Ply-by-Ply Failure Loads . . . . . . 6.1.5 Supplementary Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae . . . . . . . . . . . . . . . . . . . 6.1.6 Supplementary Problem 6: Failure Envelopes for Unidirectional Laminae . . . . . . . . . . . . . . . . . . . . . . . . . . . .

159 159

145 151 158

159 160 161 161 162 162

Contents

ix

6.2 Short Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Supplementary Problem 1: Stresses and Strains in a Symmetric Laminate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.2 Supplementary Problem 2: Stresses and Strains in an Asymmetric Laminate . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.3 Supplementary Problem 3: Failure Criteria . . . . . . . . . . . . . . . 6.2.4 Supplementary Problem 4: Ply-by-Ply Failure Loads . . . . . . 6.2.5 Supplementary Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae . . . . . . . . . . . . . . . . . . . 6.2.6 Supplementary Problem 6: Failure Envelopes for Unidirectional Laminae . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

163

7 Composite Laminate Analysis Tool—CLAT . . . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Analysis Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

199 199 199 202

163 165 169 184 188 191 197

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

Symbols and Abbreviations

Latin Symbols (Capital Letters) A C E G K N V

Area, cross-sectional area Fiber contiguity Modulus of elasticity Shear modulus Bulk modulus Internal normal force Volume

Latin Symbols (Small Letters) a b k m t

Geometric dimension, numerical coefficient Geometric dimension Fiber misalignment factor Mass Thickness

Greek Symbols (Small Letters) α γ ε η ν ξ

Rotation angle Shear strain Normal strain Coefficient (Halpin–Tsai approximations) Poisson’s ratio Coefficient (Halpin–Tsai approximations)

xi

xii

ρ σ τ φ ψ

Symbols and Abbreviations

Relative density, specific gravity Volumetric mass density Normal stress Shear stress Volume fraction Mass fraction

Mathematical Symbols Indices, Superscripted . . .upper Upper bound . . .lower Lower bound

Indices, Subscripted · · ·1 · · ·2 · · ·12 · · ·f · · ·m · · ·ref

Direction 1 Direction 2 Plane 12 Fiber Matrix Reference substance

Abbreviations CLAT CLT MMA SI

Composite laminate analysis tool Classical laminate theory Mechanics of materials approach International system of units

Chapter 1

Introduction

Abstract The first chapter introduces the major concept of composite material, i.e., the combination of different components, to obtain in total superior properties than a single component for itself. The focus is on fiber-reinforced composites where a single layer has unidirectionally aligned reinforcing fibers. The second part of this chapter introduces to major idea and the continuum mechanical background to model structural members. It is explained that physical problems are described based on differential equations. In the context of structural mechanics, these differential equations are obtained by combining the three basic equations of continuum mechanics, i.e., the kinematics relationship, the constitutive law, and the equilibrium equation. Furthermore, some explanations on the simplifications of the chosen theory, i.e., the classical laminate theory, are provided.

1.1 Composite Materials Many classical engineering materials (metals) reach their limits in advanced technical applications, and scientists permanently seek to improve the properties or to propose new materials. To judge the lightweight potential of a material in the elastic range, one may use the so-called specific modulus, i.e., the ratio between the Young’s modulus and the mass density; see Table 1.1. One can see that a typical value for metals is around 20. A typical concept to design materials with better performance is to compose different constituents to a so-called composite material, where the entire composite reveals better properties (sometimes evaluated as a spectrum of different properties) than any of the single constituents. Out of the various types of composite materials, fiber-reinforced plastics are a common choice for high-performance requirements in lightweight applications. A typical basic layer, the so-called lamina, can be composed of unidirectional fibers which are embedded in a matrix. In a second step, layers of laminae may be stacked under different fiber angles to a so-called laminate, which reveals—depending on the stacking sequence— different types of anisotropy/isotropy. Typical fiber and matrix properties are summarized in Tables 1.2 and 1.3. It can be seen that most of the fibers clearly outperform classical engineering © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, Composite Mechanics, Advanced Structured Materials 184, https://doi.org/10.1007/978-3-031-32390-4_1

1

2

1 Introduction

Table 1.1 Properties of different classical engineering materials: E: Young’s modulus; ν: Poisson’s ratio; e mass density, E/e specific modulus (here calculated based on averaged density in case of density intervals) E,

N mm2

E 109 N mm e, kg

ν, –

e, 10−6

0.293

7.9

24.81

kg mm3

Iron

196,000

Iron-based superalloys

193,000–214,000

7.9–8.3

23.83–26.42

Ferritic steels, low alloy

196,000–207,000 0.27–0.3

7.8–7.85

25.05–26.45

Stainless austenitic steels

190,000–200,000 0.25–0.3

7.5–8.1

24.36–25.64

Mild steel

200,000

7.8–7.85

25.56

Cast irons

170,000–190,000 0.26–0.275

6.9–7.8

23.13–25.85

Aluminium

69,000

0.345

2.7

25.56

Aluminium alloys 69,000–79,000

0.32–0.40

2.6–2.9

25.09–28.73

Titanium

116,000

0.361

4.5

25.78

Titanium alloys

80,000–130,000

4.3–5.1

17.02–27.66

Magnesium

44,700

1.74

25.69

Magnesium alloys

41,000–45,000

1.74–1.88

22.65–24.86

0.27–0.3

0.291

Adapted from Ashby and Jones (2005); Gale and Totemeir (2004)

metals. For example, carbon and boron fibers reveal a specific modulus above 100. Obviously, the mixing of fibers and matrix will decrease again these values to a certain extent, but the superior properties of fiber-reinforced plastics compared to classical metals will remain. It should be noted here in regards to Table 1.2 that some fibers do not show isotropic properties, i.e., the properties in axial and transverse direction differ significantly, for example, in the case of carbon and aramid; see Kaw (1997). The mechanical description of laminae can be divided into the so-called macromechanics and the micromechanics of laminae. The macromechanics of fiber-reinforced laminae covers the mechanical response under a macroscopic external load. In this context, a stress and/or strain analysis is done, and the failure may be predicted; see Altenbach et al. (2018); Öchsner (2021). The micromechanics of laminae predicts the macroscopic mechanical lamina properties based on the mechanical properties of the constituents, i.e., fibers and matrix; see Fig. 1.1, see Buryachenko (2007); Dvorak (2013); Herakovich (2012) for more details. These macroscopic properties (such as E 1 , E 2 , . . .) are commonly provided as a function of some fractions, for example, the fiber volume fraction; see Sect. 2.1.

1.1 Composite Materials

3

Table 1.2 Properties of typical fibers: E: Young’s modulus; ν: Poisson’s ratio; e mass density, E/e specific modulus E 109 N mm e, kg

N mm2

ν, –

e, 10−6

E-glass

74,000

0.25

2.6

28.46

R-glass

86,000

0.2

2.5

34.4

Carbon (high modulus)

390,000

0.35

1.8

216.67

Carbon (high strength)

230,000

0.3

1.75

131.43

Aramid (Kevlar® )

130,000

0.4

1.45

89.66

Boron

400,000

0.2

2.6

153.85

E,

kg mm3

Adapted from Berthelot (1999); Chamis (1970); Gay et al. (2003) Table 1.3 Properties of typical matrices: E: Young’s modulus; ν: Poisson’s ratio; e mass density, E/e specific modulus E,

N mm2

ν, –

e, 10−6

kg mm3

E 109 N mm e, kg

Thermosets Epoxy

4500

0.4

1.2

3.75

Polyester

4000

0.4

1.2

3.33

Polycarbonate

2400

0.35

1.2

2.00

Polypropylene (PP)

1200

0.4

0.9

1.33

Polyamide (PA)

2000

0.35

1.1

1.82

Polyetheretherketone (PEEK)

4000

0.38

1.3

3.08

Thermoplastics

Adapted from Berthelot (1999); Gay et al. (2003); Rae et al. (2007)

Fig. 1.1 Schematic illustration of the micromechanics concept (index ‘f’: fiber; index ‘m’: matrix; index ‘1,2’: lamina)

Composite materials, in general, refer to a wide class of advanced materials which are composed of different materials or phases. Typical particularities include matrices (e.g., polymers, metals, or ceramic materials) which contain a second material in the form of particles or fibers as the reinforcing elements (see Fig. 1.2 for a schematic

4

1 Introduction

(a)

(b)

(c)

(d)

Fig. 1.2 Different types of composite materials (matrix = gray, reinforcement = black): a unidirectional fibers, b woven fibers, c short fibers, and d particles

representation) or materials which are shaped in a particular way such as cellular materials (e.g., metals foams Ashby et al. (2000) or hollow sphere structures Öchsner and Augustin (2009)). In the case of cellular materials, the macroscopic properties are defined by the base material as well as the shape of the cells (voids, cavities, free space, air) and their arrangement (e.g., periodic or stochastic). For the particular group of fiber-reinforced materials, one may distinguish between classical fibers such as glass, carbon and boron Chawla (1987), natural fibers Moudood et al. (2019), and nanofibers (e.g., carbon nanotubes) Yengejeh et al. (2017). The following chapters will focus on an important configuration, i.e., unidirectional fiber-reinforced thin layers. Such single layers or plies, also called a lamina, provide superior mechanical properties compared to short fibers with random arrangement. The common approach is to stack several of such lamina with different orientations with respect to some reference direction to form a so-called laminate; see Fig. 1.3 for an example. Thus, the physical properties are dependent on the matrix material, the fibers, and the number/orientation/sequence (lay-up) of the plies which allows to tailor the macroscopic properties by adjusting the different parameters. This allows a much greater flexibility to adjust properties compared to classical engineering materials. To clearly indicate the lay-up of a laminate, it is common to use the so-called laminate orientation code, i.e., some common rules to describe the different layers (laminae) of a laminate; see MIL-HDBK-17/2F (2002) for further details:

1.1 Composite Materials

5

Fig. 1.3 Unbonded view of single laminae forming a 5-layer laminate

• Square brackets are used to indicate the beginning and the end of the laminate code: [. . .]. • The orientation of each lamina is indicated by the angle between the fiber direction and the global x-axis. • Orientations of successive laminae are separated by a virgule (/). They are listed in order from the first layer up to the last towards the positive z-direction: [. . . / . . . / . . .]. • Adjacent laminae with the same orientation are indicated by adding a subscript to the angle, equal to the number of repetitions. • A subscript of ‘s’ is used if the first half of the lay-up is indicated and the second half is symmetric: [. . . / . . . / . . .]s . For a symmetric lay-up with an odd number of laminae, the layer which is not repeated is indicated by overlining the angle of that lamina: [. . . / . . . / . . .]s . These rules imply that each layer has the same material properties and thickness. Should this be not the case, more information must be provided. Some examples of different laminates and the corresponding laminate codes are provided in Table 1.4. It can be seen that the same laminate may be described by different representations, e.g., due to the symmetry or particular sequences. Some of the textbooks which cover the mechanics of classical fiber-reinforced composite materials are summarized in Table 1.5. The interested reader may choose some of them as supplementary reading material to deepen and widen the knowledge presented in this textbook. Topics which are not covered in this textbook, e.g., manufacturing Hall and Javanbakht (2021) can be found in the provided references at the end of this chapter.

6 Table 1.4 Example laminate orientation codes Case Cross section

(a)

1 Introduction

Laminate code

[0◦ /45◦ / − 45◦ /90◦ /90◦ / − 60◦ /60◦ /0◦ ] = [0◦ / ± 45◦ /902◦ / ∓ 60◦ /0◦ ]

(b)

[0◦ /90◦ /90◦ /0◦ ] = [0◦ /90◦ ]s

(c)

[90◦ /0◦ /90◦ ] = [90◦ /0◦ ]s

1.2 Continuum Mechanical Modeling Typical thin laminates composed of othotropic laminae (plies) as described in the previous section can be modeled under particular assumptions as a simple superposition of a plane elasticity and a classical plate element. These elements are described separately in Chap. 3 and subsequently composed to a combined plane elasticity and classical plate element. Let us highlight here that the derivations in the following chapters follow a common approach from continuum mechanics Altenbach and Öchsner (2020); see Fig. 1.4. A combination of the kinematics equation (i.e., the relation between the strains and deformations) with the constitutive equation (i.e., the relation between the stresses and strains) and the equilibrium equation (i.e., the equilibrium between the internal reactions and the external loads) results in a partial differential equation, or a corresponding system of differential equations. Limited to simple problems and configurations, analytical solutions are possible. These analytical solutions are then exact in the frame of the assumptions made. However, the solution of complex problems requires the application of numerical methods such as the finite element method; see Öchsner and Öchsner (2018); Öchsner (2020) for general details on the method and Table 1.6 for specialized literature on finite element simulation of composite materials. These numerical solutions are, in general, no longer exact since the numerical methods provide only an approximate solution. Thus, the major task of an engineer is then to ensure that the approximate solution is as good as possible. This requires a

1.2 Continuum Mechanical Modeling

7

Table 1.5 Some of the English textbooks which cover the mechanics of classical fiber-reinforced composite materials (no claim to completeness) Year (1st ed.)

Author

Title

Ref.

1969

J.E. Ashton, J.C. Halpin, P.H. Petit

Primer on Composite Materials: Analysis

Ashton et al. (PH)

1969

L.R. Calcote

The Analysis of Laminated Composite Structures

Calcote (1969)

1970

J.E. Ashton, J.M. Whitney

Theory of Laminated Plates

Ashton and Whitney (1970)

1975

R.M. Jones

Mechanics of Composite Materials

Jones (1975)

1975

J.R. Vinson, T.W. Chou

Composite Materials and their Use in Structures

Vinson and Chou (1975)

1979

R.M. Christensen

Mechanics of Composite Materials

Christensen (1979)

1980

B.D. Agarwal, L.J. Broutman, K. Chandrashekhara

Analysis and Performance of Fiber Composites

Agarwal et al. (1980)

1980

S.W. Tsai, H.T. Hahn

Introduction to Composite Materials

Tsai (1980)

1981

D. Hull

An Introduction to Composite Materials

Hull (1981)

1987

K.K. Chawla

Composite Materials: Science and Engineering

Chawla (1987)

1987

J.R. Vinson, R.L. Sierakowski

The Behavior of Structures Composed of Composite Materials

Vinson and Sierakowski (1987)

1987

J.M. Whitney

Structural Analysis of Laminated Anisotropic Plates

Whitney (1980)

1990

V.K.S. Choo

Fundamentals of Composite Materials

Choo (1987)

1994

I.M. Daniel, O. Ishai

Engineering Mechanics of Composite Materials

Daniel and Ishai (1994)

1994

R.F. Gibson

Principles of Composite Material Mechanics

Gibson (1969)

1994

J.N. Reddy

Mechanics of Composite Materials: Selected Works of Nicholas J. Pagano

Reddy (1994)

1997

A.K. Kaw

Mechanics of Composite Materials

Kaw (1997)

1998

C.T. Herakovich

Mechanics of Fibrous Composites

Herakovich (1998)

1999

J.-M. Berthelot

Composite Materials: Mechanical Behavior and Structural Analysis

Berthelot (1999)

2001

V.V. Vasilev, E.V. Morozov

Mechanics and Analysis of Composite Materials

Vasiliev and Morozov (2001)

2002

C. Decolon

Analysis of Composite Structures

Decolon (2002)

2003

L.P. Kollár, G.S. Springer, W. Kissing

Mechanics of Composite Structures

Kollár and Springer (2003)

2004

H. Altenbach, J. Altenbach, W. Kissing

Mechanics of Composite Structural Elements

Altenbach et al. (2004)

2004

M.E. Tuttle

Structural Analysis of Polymeric Composite Materials

Tuttle (2004)

2008

R. de Borst, T. Sadowski

Lecture Notes on Composite Materials: Current Topics and Achievements

de Borst and Sadowski (2008)

2010

C. Kassapoglou

Design and Analysis of Composite Structures with Applications to Aerospace Structures

Kassapoglou (2010)

8

1 Introduction

external loads (forces, moments)

deformations (displacements, rotations)

equilibrium

kinematics

stresses

constitutive equation

measure for loading

strains

measure for deformation

partial differential equation(s)

solution - analytical methods - numerical methods (FDM, FEM, FVM, BEM)

Fig. 1.4 Continuum mechanical modeling of structural members Table 1.6 Some of the early textbooks which cover the finite element simulation of composite materials (no claim to completeness) Author Title Ref. Year (1st ed.) 1992 1998 2000

2008

O.O. Ochoa, J.N. Finite Element Analysis Reddy of Composite Laminates L.T. Tenek, J. Argyris Finite Element Analysis for Composite Structures F.L. Matthews, G.A.O. Finite Element Davies, D. Hitchings, Modelling of Composite C. Soutis Materials and Structures E.J. Barbero Finite element analysis of composite materials

Ochoa and Reddy (1992) Tenek and Argyris (1998) Matthews et al. (2000)

Barbero (1996)

lot of experience and a solid foundation in the basics of applied mechanics, materials science, and mathematics. The Chaps. 3 and 4 present a simplified approach, the so-called classical laminate theory (CLT)Dong et al. (1962); Stravsky (1961), which aims to provide a stress and a subsequent strength/failure analysis without the solution of the system of cou-

References

9

pled differential equations for the unknown displacements in the three coordinate directions. This theory provides the solution of the statically indeterminate system based on a generalized stress-strain relationship under consideration of the constitutive relationship and the definition of the so-called stress resultants or generalized stresses and strains. The common assumptions of the classical laminate theory and the derivations in the following chapters can be summarized as follows (e.g., Daniel and Ishai (1994)): 1. Each lamina is considered quasi-homogeneous and orthotropic (in general, the properties can range from isotropic to anisotropic). 2. Only unidirectional and flat lamina are considered in the following chapters. 3. The laminate consists of perfectly bonded laminae and the bond lines are infinitesimally thin as well as non-shear-deformable. 4. The laminate is thin, i.e., the thickness is small compared to the lateral dimensions, and represents a state of plane stress. 5. Displacements (in thickness and lateral directions) are small compared to the thickness of the laminate. 6. Displacements are continuous throughout the laminate (non-shear-deformable bond lines). 7. In-plane displacements are linear functions of the thickness. 8. Shear strains in planes perpendicular to the middle surface are negligible. 9. Assumptions 7 and 8 imply that a line originally straight and perpendicular to the laminate middle surface remains so after deformation (Kirchhoff hypothesis of classical plate theory). 10. Kinematics and constitutive relations are linear. 11. Normal distances from the middle surface remain constant. Thus, the transverse normal strain is negligible compared to the in-plane normal strains (Kirchhoff hypothesis of classical plate theory).

References Agarwal BD, Broutman LJ, Chandrashekhara K (1980) Analysis and performance of fiber composites. Wiley, New York Altenbach H, Altenbach J, Kissing W (2018) Mechanics of composite structural elements. Springer, Singapore Altenbach H, Altenbach J, Kissing W (2004) Mechanics of composite structural elements. Springer, Berlin Altenbach H, Öchsner A (eds) (2020) Encyclopedia of continuum mechanics. Springer, Berlin Ashby M, Evans AG, Fleck NA, Gibson LJ, Hutchinson JW, Wadley HNG (2000) Metal foams: a design guide. Butterworth-Heinemann, Boston Ashton JE, Halpin JC, Petit PH (1969) Primer on composite materials: analysis. Technomic Publishing Company, Stamford Ashton JE, Whitney JM (1970) Theory of laminated plates. Technomic Publishing, Stamford Ashby MF, Jones DRH (2005) Engineering materials 1: an introduction to properties, applications and design. Elsevier, Amsterdam

10

1 Introduction

Barbero EJ (1996) Finite element analysis of composite materials. CRC Press, Boca Raton Berthelot J-M (1999) Composite materials: mechanical behavior and structural analysis. Springer, Berlin de Borst R, Sadowski T (2008) Lecture notes on composite materials: current topics and achievements. Springer, Dordrecht Buryachenko VA (2007) Micromechanics of heterogeneous materials. Springer, New York Calcote LR (1969) The analysis of laminated composite structures. Van Nostrand Reinhold Company, New York Chamis CC (1970) Characterization and design mechanics for fiber-reinforced metals. NASA Technical Note NASA TN D-5784. National Aeronautics and Space Administration, Washington Chawla KK (1987) Composite materials: science and engineering. Springer, New York Chakrabarty J (2010) Applied plasticity. Springer, New York Choo VKS (1987) Fundamentals of composite materials. Knowen Academic Press, Delaware Christensen RM (1979) Mechanics of composite materials. Wiley, New York Daniel IM, Ishai O (1994) Engineering mechanics of composite materials. Oxford University Press, New York Decolon C (2002) Analysis of composite structures. Hermes Penton Science, London Dong SB, Pister KS, Taylor RL (1962) On the theory of laminated anisotropic shells and plates. J Aerosp Sci 29:696–975 Dvorak GJ (2013) Micromechanics of composite materials. Springer, New York Gale WF, Totemeir TC (eds) (2004) Smithells metals reference book. Elsevier ButterworthHeinemann, Burlington Gay D, Hoa SV, Tsai SW (2003) Composite materials: design and application. CRC Press, Boca Raton Gibson RF (1969) Principles of composite material mechanics. McGraw-Hill, New York Hall W, Javanbakht Z (2021) Design and manufacture of fibre-reinforced composites. Springer, Cham Herakovich CT (1998) Mechanics of fibrous composites. Wiley, New York Herakovich CT (2012) Mechanics of composites: A historical review. Mech Res Commun 41:1–20 Hull D (1981) An introduction to composite materials. Cambridge University Press, Cambridge Jones RM (1975) Mechanics of composite materials. Scripta Book Co., Washington Kassapoglou C (2010) Design and analysis of composite structures with applications to aerospace structures. Wiley, Chichester Kaw AK (1997) Mechanics of composite materials. CRC Press, Boca Raton Kollár LP, Springer GS (2003) Mechanics of composite structures. Cambridge University Press, Cambridge Matthews FL, Davies GAO, Hitchings D, Soutis C (2000) Finite element modelling of composite materials and structures. Woodhead Publishing Limited, Cambridge MIL-HDBK-17/2F (2002) Composite materials handbook -, vol 2. Polymer matrix composites materials properties, Department of Defence, USA Moudood A, Rahman A, Khanlou HM, Hall W, Öchsner A, Francucci G (2019) Environmental effects on the durability and the mechanical performance of flax fiber/bio-epoxy composites. Compos Part B-Eng 171:284–293 Ochoa OO, Reddy JN (1992) Finite element analysis of composite laminates. Kluwer Academic Publishers, Dordrecht Öchsner A, Augustin C (2009) Multifunctional metallic hollow sphere structures: manufacturing, properties and application. Springer, Berlin Öchsner A (2018) A project-based introduction to computational statics. Springer, Cham Öchsner A, Öchsner M (2018) A first introduction to the finite element analysis program MSC Marc/Mentat. Springer, Cham Öchsner A (2020) Computational statics and dynamics: an introduction based on the finite element method. Springer, Singapore Öchsner A (2021) Foundations of classical laminate theory. Springer, Cham

References

11

Rae PJ, Brown EN, Orler EB (2007) The mechanical properties of poly(ether-ether-ketone) (PEEK) with emphasis on the large compressive strain response. Polymer 48:598–615 Reddy JN (1994) Mechanics of composite materials: selected works of Nicholas J. Pagano. Kluwer Academic Publishers, Dordrecht Stravsky Y (1961) Bending and stretching of laminated aeolotropic plates. J Eng Mech Div-ASCE 87:31–56 Tenek LT, Argyris J (1998) Finite element analysis for composite structures. Springer, Dordrecht Tsai SW (1980) Introduction to composite materials. Technomic Publishishing Company, Westport Tuttle ME (2004) Structural analysis of polymeric composite materials. Marcel Dekker, New York Vasiliev VV, Morozov EV (2001) Mechanics and analysis of composite materials. Elsevier, Oxford Vinson JR, Chou TW (1975) Composite materials and their use in structures. Wiley, New York Vinson JR, Sierakowski RL (1987) The behavior of structures composed of composite materials. Kluwer Academic Publisher, Dordrecht Whitney JM (1980) Structural analysis of laminated anisotropic plates. Technomic Publishing Company, Lancaster Yengejeh SI, Kazemi SA, Öchsner A (2017) Carbon nanotubes as reinforcement in composites: a review of the analytical, numerical and experimental approaches. Comp Mater Sci 136:85–101

Chapter 2

Micromechanics

Abstract This chapter introduces three classical theories to predict the macroscopic mechanical lamina properties based on the mechanical properties of the constituents, i.e., fibers and matrix. The focus is on unidirectional lamina which can be described based on orthotropic constitutive equations. In detail, predictions for the modulus of elasticity in and transverse to the fiber direction, the major Poisson’s ratio, as well as the in-plane shear modulus are provided. The mechanics of materials approach, the elasticity solutions with contiguity after Tsai, and the Halpin–Tsai relationships are presented.

2.1 Reference Numbers to Characterize Composition As mentioned in the previous chapter, macroscopic lamina properties are commonly provided as functions of some fractions. This allows to estimate the property based on given volume or weight/mass fractions. The fiber and matrix volume fractions are defined as Vf , V Vm φm = , V φf =

(2.1) (2.2)

where Vf is the volume of the fibers, Vm is the volume of the matrix, and V is the volume of the entire composite. The following relationships hold between the volume fractions and the volumes themselves: φf + φm = 1 and Vf + Vm = V .

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, Composite Mechanics, Advanced Structured Materials 184, https://doi.org/10.1007/978-3-031-32390-4_2

(2.3)

13

14

2 Micromechanics

Fiber and matrix mass fractions are defined as mf , m mm ψm = , m ψf =

(2.4) (2.5)

where m f is the mass of the fibers, m m is the mass of the matrix, and m is the mass of the entire composite. The following relationships hold between the mass fractions and the masses themselves: ψf + ψm = 1 and m f + m m = m.

(2.6)

One may find also in scientific literature the so-called relative density (or sometimes called the specific gravity), which is defined as ρf =

f

,

(2.7)

,

(2.8)

ref

ρm =

m ref

where f = m f /Vf is the mass density of the fibers, m = m m /Vm is the mass density of the matrix, and ref is the mass density of a reference substance1 . The relationship between the volume and mass fractions can be expressed as follows Altenbach et al. (2018); Tsai et al. (1963): ρf ρm

f

φm =

m

1 −1+ ψm

= f m

1 ρf −1+ ρm ψm

,

(2.9)

or f

φf = 1 −

m

1 −1+ ψm

=1− f m

ρf ρm ρf 1 −1+ ρm ψm

.

(2.10)

1 If the reference is not explicitly stated then it is normally assumed to be water (H O) at 2 3.98◦ C, which is the temperature at which water reaches its maximum density: H2 O (3.98◦ C) = 999.95 kg/m3 ≈ 1000 kg/m3 or H2 O (3.98◦ C) = 0.99995 g/cm3 ≈ 1.0 g/cm3 .

2.2 Prediction of Elastic Properties of Laminae

15

The following sections contain many graphical representations of mechanical properties and the common approach, i.e., the properties are plotted over the fiber volume fraction (φf ), is followed.

2.2 Prediction of Elastic Properties of Laminae 2.2.1 Mechanics of Materials Approach The prediction of elastic constants for two-phase materials was already a long time before fiber-reinforced laminae discussed in scientific literature. These early predictions were related, for example, to a metal with finely dispersed particles of an alloying material which has higher mechanical properties than the base material Paul (1960). The following sections will repeat the derivations of these classical approximations, which are known as the mechanics of materials approach (MMA). All these derivations are made under the assumption that each constituent, i.e., the fiber and the matrix, behaves itself as an isotropic material.

2.2.1.1

Modulus of Elasticity in Fiber Direction

To estimate the modulus of elasticity in fiber direction (E 1 ), it is assumed that the strains of the fiber (εf,1 ), matrix (εm,1 ) as well as the entire composite (ε1 ) are the same in the loading direction; see Fig. 2.1 for schematic representations of the constituents in parallel arrangement. This equal strain condition is expressed as: ε1 = εf,1 = εm,1 =

∆a . a

(2.11)

Fig. 2.1 Mechanical models to calculate the elastic modulus in fiber direction: a two-phase representation, b springs-in-parallel representation

16

2 Micromechanics

The stress in each constituent, as well as in the entire composite, can be expressed as: (2.11)

σf,1 = E f εf,1 = E f ε1 ,

(2.12)

(2.11)

σm,1 = E m εm,1 = E m ε1 , σ1 = E 1 ε1 .

(2.13) (2.14)

Considering the internal normal force (N ), the force equilibrium in fiber (or loading) direction yields: N1 = σ1 A = Nf,1 + Nm,1 = σf,1 Af + σm,1 Am ,

(2.15)

which yields under the assumption of a constant thickness in Fig. 2.1a and under consideration of Eqs. (2.12)–(2.13) the following expression: Af Am + E m ε1 A A = E f ε1 φf + E m ε1 φm ,

E 1 ε1 = E f ε1

(2.16) (2.17)

or finally under consideration of Eq. (2.3): E 1 = E f φf + E m (1 − φf ).

(2.18)

This estimation is also known as the rule of mixtures or Voigt rule. A graphical representation of the estimation for the modulus of elasticity in fiber direction is given in Fig. 2.2. A linear transition between the modulus of the matrix and the modulus of the fibers is obtained.

Ef Lamina modulus E1

Fig. 2.2 Modulus of elasticity in fiber direction as a function of the fiber volume fraction (MMA)

MMA

Em 0

0.5 Fiber volume fraction φf

1

2.2 Prediction of Elastic Properties of Laminae

17

Equation (2.18) can be normalized either by the modulus of the matrix or the modulus of the fibers to obtain the following normalized representations of the modulus in fiber direction: Ef E1 = φf + (1 − φf ), Em Em Em E1 = φf + (1 − φf ), Ef Ef

(2.19) (2.20)

which allows a graphical representation of the modulus of elasticity in fiber direction for different ratios of the moduli; see Fig. 2.3.

(a) 20

Normalized lamina modulus

E1 Em

MMA

100 25

10

10

Ef Em

1 0

= 1.0

0.5

1

Fiber volume fraction φf

E1 Ef

(b)

Normalized lamina modulus

Fig. 2.3 Normalized modulus of elasticity in fiber direction as a function of the fiber volume fraction (MMA): a normalized by matrix modulus, b normalized by fiber modulus

MMA

1

Ef Em

= 1.0

0.5 10 100 0.1

25 0

0.5 Fiber volume fraction φf

1

18

2 Micromechanics

Fig. 2.4 Mechanical models to calculate the elastic modulus transverse to the fiber direction: a two-phase representation, b springs-in-series representation

2.2.1.2

Modulus of Elasticity Transverse to Fiber Direction

To estimate the modulus of elasticity transverse to the fiber direction (E 2 ), it is assumed that the stresses of the fiber (σf,2 ), matrix (σm,2 ) as well as the entire composite (σ2 ) are the same in the loading direction; see Fig. 2.4 for schematic representations of the constituents in series arrangement. This equal stress condition is expressed as: σ2 = σf,2 = σm,2 = ε2 E 2 = εf,2 E f = εm,2 E m .

(2.21)

The total deformation in loading direction is the sum of the deformations of both constituents: (2.22) ∆b = ∆bf + ∆bm , or expressed in terms of strains: ∆bf ∆bm ∆b = ε2 = + . b b b

(2.23)

Considering the volume fractions, Eq. (2.23) can be written as: ε2 =

∆bf φf b ∆bm φm b × + × , φm b φf b b b

(2.24)

which yields under consideration of2 φf b = bf and φm b = bm : ε2 =

∆bf ∆bm φf + φm = εf,2 φf + εm,2 φm . bm bf

(2.25)

This assumes for the width a = af = am and for the thickness t = tf = tm . Strictly speaking, this would be the case if Poisson’s ratios of the fiber and the matrix would be the same. 2

2.2 Prediction of Elastic Properties of Laminae

19

Introducing the equal stress condition (2.21) in the last relationship gives: ε2 = or

σ2 σ2 σ2 = φf + (1 − φf ), E2 Ef Em

(2.26)

1 φf 1 − φf = + , Ef Em E2

(2.27)

or finally as: E2 =

Ef Em . E m φf + E f (1 − φf )

(2.28)

This estimation is also known as the inverse rule of mixtures or Reuss rule. A graphical representation of the estimation for the modulus of elasticity transverse to the fiber direction is given in Fig. 2.5. A nonlinear transition between the modulus of the matrix and the modulus of the fibers is obtained. Equation (2.28) can be normalized either by the modulus of the matrix or the modulus of the fibers to obtain the following normalized representations of the modulus transverse to the fiber direction: E2 = Em

1 Em Ef

E2 = Ef φf +

(2.29)

1 Ef (1 Em

− φf )

,

(2.30)

Ef Lamina modulus E2

Fig. 2.5 Modulus of elasticity transverse to fiber direction as a function of the fiber volume fraction (MMA)

,

φf + (1 − φf )

MMA

Em 0

0.5 Fiber volume fraction φf

1

20

2 Micromechanics

(a) 10 E2 Em

MMA 25

Normalized lamina modulus

Fig. 2.6 Normalized modulus of elasticity transverse to fiber direction as a function of the fiber volume fraction (MMA): a normalized by matrix modulus, b normalized by fiber modulus

100 5 10 Ef Em

1 0

= 1.0

0.5

1

Fiber volume fraction φf

Normalized lamina modulus

E2 Ef

(b)

MMA

1

Ef Em

= 1.0

0.5 10 25 100

0.1 0

0.5

1

Fiber volume fraction φf

which allows a graphical representation of the modulus of elasticity transverse to the fiber direction for different ratios of the moduli; see Fig. 2.6. 2.2.1.3

Major Poisson’s Ratio

To estimate the major Poisson’s ratio (ν12 ), it is assumed that the strains of the fiber (εf,1 ), matrix (εm,1 ) as well as the entire composite (ε1 ) are the same in the loading direction; see Fig. 2.7 for schematic representations of the problem. This equal strain condition is expressed as: ε1 = εf,1 = εm,1 .

(2.31)

2.2 Prediction of Elastic Properties of Laminae

21

Fig. 2.7 Mechanical models to calculate the major Poisson’s ratio

The total deformation perpendicular to the loading direction is the sum of the deformations of both constituents: ∆b = ∆bf + ∆bm (< 0 for σ1 > 0).

(2.32)

The strains perpendicular to the loading direction can be stated based on the general definition of normal strain and Poisson’s ratio as: ∆b = −ν12 ε1 , b ∆bf = −νf εf,1 , = bf ∆bm = = −νm εm,1 . bm

ε2 = εf,2 εm,2

(2.33) (2.34) (2.35)

Considering again the classical definition of a normal strain and that φf b = bf and φm b = bm , Eq. (2.32) can be written as: ∆b = εf,2 bf + εm,2 bm = εf,2 φf b + εm,2 φm b,

(2.36) (2.37)

which results after division by b and the consideration of the definitions in Eqs. (2.34)– (2.35) to the following expression: ∆b = ε2 = −νf εf,1 φf − νm εm,1 φm , b ε1

(2.38)

ε1

which gives finally under consideration of ε2 /ε1 = −ν12 : ν12 = νf φf + νm (1 − φf ).

(2.39)

2 Micromechanics

Fig. 2.8 Major (lamina) Poisson’s ratio as a function of the fiber volume fraction (MMA)

Lamina Poisson’s ratio ν12

22

νf

MMA νm 0

0.5

1

Fiber volume fraction φf

A graphical representation of the estimation for the major (lamina) Poisson’s ratio is given in Fig. 2.8. A linear transition between the ratio of the matrix and the ratio of the fibers is obtained. Equation (2.39) can be normalized either by the ratio of the matrix or the ratio of the fibers to obtain the following normalized representations of the Poisson’s ratio: νf ν12 = φf + (1 − φf ), νm νm νm ν12 = φf + (1 − φf ), νf νf

(2.40) (2.41)

which allows a graphical representation of the major Poisson’s ratio for different proportions of the ratios; see Fig. 2.9.

2.2.1.4

In-Plane Shear Modulus

To estimate the in-plane shear modulus (G 12 ), it is assumed for simplicity that the shear stress of the fiber (τf,12 ), matrix (τm,12 ) as well as the entire composite (τ 12 ) are the same over the entire cross section3 ; see Fig. 2.10 for schematic representations of the constituents. The shear stresses, which are based on the assumption to be equal (τf,12 = τm,12 = τ12 ), can be expressed due to Hooke’s law for a pure shear state as

3

This simplification is similar to the Timoshenko beam theory, which assumes a constant shear stress over the beam, see Öchsner (2021).

2.2 Prediction of Elastic Properties of Laminae

23

(a) 1 Normalized Poisson’s ratio

ν12 νm

νf νm

= 1.0

0.75 0.5 0.5

0.1

MMA 0

0.5

1

Fiber volume fraction φf

Normalized Poisson’s ratio

ν12 νf

(b) 3

2 0.5 0.75

1

0

νf νm

= 1.0

MMA 0

0.5

1

Fiber volume fraction φf

Fig. 2.9 Normalized major (lamina) Poisson’s ratio as a function of the fiber volume fraction (MMA): a normalized by matrix ratio, b normalized by fiber ratio

Fig. 2.10 Mechanical models to calculate the in-plane shear modulus: a undeformed geometry and load condition, b deformed shape

24

2 Micromechanics

τ12 = G 12 γ12 ,

(2.42)

τf,12 = G f γf,12 , τm,12 = G m γm,12 ,

(2.43) (2.44)

whereas the shear strains can be approximated as ∆a , b ∆af ≈ , bf ∆am . ≈ bm

γ12 ≈ γf,12 γm,12

(2.45) (2.46) (2.47)

The total horizontal deformation can be expressed as

or after dividing by b as:

∆a = ∆af + ∆am ,

(2.48)

∆af ∆am ∆a = + . b b b

(2.49)

Considering the definitions of the normal strains (see Eqs. (2.45)–(2.47)) and Hooke’s law (see Eqs. (2.42)–(2.44)), as well as that φf b = bf and φm b = bm , Eq. (2.49) can be written as: γ12 =

τ12 bf bm = γf,12 + γm,12 G 12 b b τm,12 τf,12 φf + φm , = Gm Gf

(2.50) (2.51)

or finally as: G 12 =

GfGm . G m φf + G f (1 − φf )

(2.52)

A graphical representation of the estimation for the in-plane shear modulus is given in Fig. 2.11. A nonlinear transition between the modulus of the matrix and the modulus of the fibers is obtained. Equation (2.52) can be normalized either by the modulus of the matrix or the modulus of the fibers to obtain the following normalized representations of the inplane shear modulus:

2.2 Prediction of Elastic Properties of Laminae

Gf Lamina modulus G12

Fig. 2.11 In-plane shear modulus as a function of the fiber volume fraction (MMA)

25

MMA

Gm 0

0.5

1

Fiber volume fraction φf

G 12 = Gm

1

Gm Gf

, φf + (1 − φf )

G 12 = Gf φf +

1 Gf (1 Gm

− φf )

,

(2.53) (2.54)

which allows a graphical representation of the in-plane shear modulus for different ratios of the moduli; see Fig. 2.12.

2.2.2 Elasticity Solutions with Contiguity after Tsai The theory of elasticity solutions with contiguity after Tsai et al. Tsai (1964) considers that the fibers do not remain perfectly straight and parallel during the manufacturing process of the composite. In addition, fibers are in contact with one another and not perfectly surrounded and isolated by the matrix. These geometric deviations are considered by two factor, i.e., the empirical factor k, which accounts for the fiber misalignment, and the factor C, which accounts for the fiber contiguity; see Fig. 2.13.

2.2.2.1

Modulus of Elasticity in Fiber Direction

The prediction of the modulus of elasticity in fiber direction is based on the classical expression (2.18), which is now corrected by the fiber misalignment such that E 1 = k × (E f φf + E m (1 − φf )) ,

(2.55)

26

2 Micromechanics

(a) 10 G12 Gm

MMA 25

Normalized shear modulus

Fig. 2.12 Normalized in-plane shear modulus as a function of the fiber volume fraction (MMA): a normalized by matrix modulus, b normalized by fiber modulus

100 5 10 Gf Gm

1 0

= 1.0

0.5

1

Fiber volume fraction φf

Normalized shear modulus

G12 Gf

(b)

MMA

1

Gf Gm

= 1.0

0.5 10 25 100

0.1 0

0.5

1

Fiber volume fraction φf

where k is the filament (fiber) misalignment factor. Typically, this factor, which considers that the fibers are not perfectly parallel or not straight, is in the range of 0.9 ≤ k ≤ 1; see Jones (1999). The graphical representation of this modulus is given in Fig. 2.14 for different values of the fiber misalignment factor k. It should be noted here that the curve for k = 1 is identical with the representation in Fig. 2.2.

2.2.2.2

Modulus of Elasticity Transverse to Fiber Direction

The derivation considers two different limiting cases which are finally linear superposed. The first contribution to the modulus of elasticity transverse to fiber direction

2.2 Prediction of Elastic Properties of Laminae

(a)

27

(b)

Fig. 2.13 a Perfectly isolated fibers (C = 0) and b contiguous fibers (C = 1). Adapted from Jones (1999)

Ef k=1 Lamina modulus E1

Fig. 2.14 Modulus of elasticity in fiber direction as a function of the fiber volume fraction (Tsai)

0.9

Tsai

Em 0

0.5

1

Fiber volume fraction φf

comes from the assumption that the fibers are isolated and the matrix is contiguous. This means that C = 0 (see Fig. 2.13a) and the corresponding lower bound for E 2 can be obtained as follows: E 2(m) = 2 [1 − ν2 ] ×

K f (2K m + G m ) − G m (K f − K m )φm , (2K m + G m ) + 2(K f − K m )φm

(2.56)

where ν2 = νf − (νf − νm )φm is Poisson’s ratio in the isotropic plane normal to the fibers and this value should not be confused with the major Poisson’s ratio (ν12 ) of the unidirectional lamina. The upper bound for E 2 can be obtained by the assumption C = 1 (see Fig. 2.13b) and is obtained from Eq. (2.56) by interchanging the subscripts ‘m’ and ‘f’ and replacing φm by (1 − φm ):

28

2 Micromechanics

K m (2K f + G f ) − G f (K m − K f )(1 − φm ) (2K f + G f ) + 2(K m − K f )(1 − φm ) K f (2K m + G f ) + G f (K m − K f )φm . = 2 [1 − ν2 ] × (2K m + G f ) − 2(K m − K f )φm

E 2(f) = 2 [1 − ν2 ] ×

(2.57)

Linear superposition of expressions (2.56) and (2.56) gives the final expression for E 2 as (2.58) E 2 = E 2(m) + C(E 2(f) − E 2(m) ) = (1 − C)E 2(m) + C E 2(f) , or in components: E 2 = 2 [1 − νf + (νf − νm )φm ] × (1 − C) +C

K f (2K m + G m ) − G m (K f − K m )φm (2K m + G m ) + 2(K f − K m )φm

K f (2K m + G f ) + G f (K m − K f )φm , (2K m + G f ) − 2(K m − K f )φm

(2.59)

where the elastic moduli can be calculated according to Jones (1999); Tsai (1964) as Ef , 2(1 − νf ) Ef Gf = , 2(1 + νf ) Em Km = , 2(1 − νm ) Em Gm = , 2(1 + νm ) Kf =

(2.60) (2.61) (2.62) (2.63)

and the fiber contiguity is in the range of 0 ≤ C ≤ 1. The graphical representation of the modulus E 2 is given in Fig. 2.15 for different values of the fiber contiguity C.

2.2.2.3

Major Poisson’s Ratio

It is important to take the relationship for the major Poisson’s ratio from Ref. Tsai et al. (1963) since the classical textbook by Jones Jones (1999) contains a few typos4 4

The faulty equation reads: K f νf (2K m + G m )φf + K m νm (2K f + G m )φm K f (2K m + G m ) − G m (K f − K m )φm K m νm (2K f + G f )φm + K f νf (2K m + G f )φf . +C K f (2K m + G m ) + G f (K m − K f )φm

ν12 = (1 − C)

.

2.2 Prediction of Elastic Properties of Laminae

Ef Lamina modulus E2

Fig. 2.15 Modulus of elasticity transverse to fiber direction as a function of the fiber volume fraction (Tsai)

29

C=1 0.5 0 Tsai Em 0

0.5

1

Fiber volume fraction φf

for this relation. In addition, scientific literature contains more faulty versions for ν12 which will be not mentioned here. The derivation considers again two different limiting cases which are finally linear superposed. The first contribution to the major Poisson’s ratio comes from the assumption that the fibers are isolated and the matrix is contiguous. This means that C = 0 (see Fig. 2.13a) and the corresponding bound for ν12 can be obtained as follows: (m) = ν12

K f νf (2K m + G m )(1 − φm ) + K m νm (2K f + G m )φm . K m (2K f + G m ) + G m (K f − K m )(1 − φm )

(2.64)

The other bound for ν12 can be obtained by the assumption C = 1 (see Fig. 2.13b) and is obtained from Eq. (2.64) by interchanging the subscripts ‘m’ and ‘f’ and replacing φm by (1 − φm ): (f) ν12 =

K m νm (2K f + G f )φm + K f νf (2K m + G f )(1 − φm ) . K f (2K m + G f ) + G f (K m − K f )φm

(2.65)

Linear superposition of expressions (2.64) and (2.65) gives the final expression for ν12 as (m) (f) (m) (m) (f) + C(ν12 − ν12 ) = (1 − C)ν12 + Cν12 , (2.66) ν12 = ν12

2 Micromechanics

Fig. 2.16 Major (lamina) Poisson’s ratio as a function of the fiber volume fraction (Tsai)

Lamina Poisson’s ratio ν12

30

νm

0 0.5 C=1

νf

Tsai 0

0.5

1

Fiber volume fraction φf

or in components: K f νf (2K m + G m )φf + K m νm (2K f + G m )φm K m (2K f + G m ) + G m (K f − K m )φf K m νm (2K f + G f )φm + K f νf (2K m + G f )φf +C . K f (2K m + G f ) + G f (K m − K f )φm

ν12 = (1 − C)

(2.67)

The graphical representation of the major (lamina) Poisson’s ratio ν12 is given in Fig. 2.16 for different values of the fiber contiguity C. 2.2.2.4

In-Plane Shear Modulus

The original reference by Tsai Tsai et al. (1963) states only an upper and lower bound for the in-plane shear modulus, i.e.,

G

upper

=

2ν12 1 1 + + E1 E2 E1

−1

,

(2.68)

C=k=1

15(1 − νm ) G lower = G m +

Gf − 1 (1 − φm )G m Gm

7 − 5νm + 2(4 − 5νm ) 1 +

Gf − 1 φm Gm

.

(2.69)

Thus, the following provides the final prediction which is again derived by considering two limiting cases, i.e., C = 0 and C = 1, and the subsequent superposition Tsai (1964):

2.2 Prediction of Elastic Properties of Laminae Fig. 2.17 In-plane shear modulus as a function of the fiber volume fraction (Tsai)

31

Lamina modulus G12

Gf

C=1 0.5 0 Tsai Gm 0

0.5

1

Fiber volume fraction φf

2G f − (G f − G m )φm 2G m + (G f − G m )φm (G f + G m ) − (G f − G m )φm . + C Gf (G f + G m ) + (G f − G m )φm

G 12 = (1 − C)G m

(2.70)

The graphical representation of the in-plane shear modulus G 12 is given in Fig. 2.17 for different values of the fiber contiguity C.

2.2.3 Halpin–Tsai Relationships The so-called Halpin–Tsai approximations Halpin (1969); Jones (1999) are based on a mathematical model by Hermans Hermans (1967), which consists of concentric cylinders surrounded by an unbounded composite. Based on an interpolation procedure, reasonable accurate approximations of the original predictions by Hermans were obtained, which can be applied over a wide range of fiber volume fractions.

2.2.3.1

Modulus of Elasticity in Fiber Direction and Major Poisson’s Ratio

The Halpin–Tsai relationships for the modulus of elasticity in fiber direction and the major Poisson’s ratio are identical to the expressions obtained by the mechanics of materials approach; see Eqs. (2.18) and (2.39) and Figs. 2.2 and 2.8:

32

2.2.3.2

2 Micromechanics

E 1 = E f φf + E m (1 − φf ) ,

(2.71)

ν12 = νf φf + νm (1 − φf ) .

(2.72)

Modulus of Elasticity Transverse to Fiber Direction and In-Plane Shear Modulus

The Halpin–Tsai relationships for the modulus of elasticity transverse to the fiber direction and the in-plane shear modulus have the same structure, i.e., 1 + ξ ηφf , 1 − ηφf 1 + ξ ηφf = Gm × , 1 − ηφf

E2 = Em × G 12

(2.73) (2.74)

where the coefficient η is given by η= η=

Ef Em Ef Em Gf Gm Gf Gm

−1 +ξ

, (for E 2 )

−1 +ξ

. (for G 12 )

(2.75)

(2.76)

The parameter ξ is a measure of the reinforcement and depends on the boundary conditions, i.e., geometry of inclusions and loading conditions. The graphical representations of the moduli E 2 and G 12 are given in Figs. 2.18 and 2.19 for different values of the parameter ξ . There are different predictions and estimates of factor ξ , which in general may vary between ξ = 0 and ξ = ∞; see Halpin (1969). For circular fibers in a square array, the values ξ E2 = 2 for the modulus E 2 and ξG 12 = 1 for the modulus G 12 were obtained to provide reasonable results. In the case of rectangular cross-sectional fibers of aspect ration a/b (width a and thickness b) in a hexagonal array, the following estimates provide good approximations:

2.2 Prediction of Elastic Properties of Laminae

Halpin-Tsai

Ef Lamina modulus E2

Fig. 2.18 Modulus of elasticity transverse to fiber direction as a function of the fiber volume fraction (Halpin–Tsai)

33

ξ = 20 8 2 Em

0

0.5

1

Fiber volume fraction φf

Halpin-Tsai

Gf Lamina modulus G12

Fig. 2.19 In-plane shear modulus as a function of the fiber volume fraction (Halpin–Tsai)

ξ = 50 10 1 Gm

0

0.5

1

Fiber volume fraction φf

a ξ E2 = 2 × , b √ a . log ξG 12 = 3 × log b

(2.77) (2.78)

Other proposal for ξG 12 can be found, for example, in Hewitt and de Malherbe (1970).

34

2 Micromechanics

2.3 Comparison with Experimental Results This section presents two classical experimental data sets for the modulus of elasticity in and transverse to the fiber direction, the major Poisson’s ratio, as well as the inplane shear modulus. These data sets were reported by Tsai and Whitney/Riley. The experimental values are used to benchmark the theoretical predictions based on the mechanics of materials approach, the elasticity solutions with contiguity after Tsai, and the Halpin–Tsai relationships. The section concludes with the presentation of optimized material parameters (i.e., fitting parameters) to predict the corresponding properties as a function of the fiber volume fraction.

2.3.1 Extraction of Experimental Values The following sections report two classical data sets for the elastic lamina properties as function of some fraction. These data sets go back to Tsai et al. (1963); Tsai (1964) and Whitney and Riley Whitney and Riley (1966). The tables in the following subsections report the experimental data as given in the corresponding publications and in addition, the conversion to SI units and other reference numbers. Since some experimental data was reported only in the form of diagrams5 , it is hoped that the collection of numerical values in tables allows an easier further processing of the experimental results.

2.3.1.1

Data Set by Tsai

The data set by Tsai Tsai et al. (1963); Tsai (1964) was obtained from unidirectional glass filament-epoxy resin systems, which were manufactured by the hand lay-up technique. Scotch-ply No. 1009-33 W2 38 (Minnesota Mining and Manufacturing Company, U.S.A.) in the range 0.2 ≤ ψm ≤ 0.35 and E-787-NUF (U.S. Polymeric Company) in the range 0.13 ≤ ψm ≤ 0.20 were used. The properties of the constituents, i.e., fiber and matrix, are summarized in Table 2.1. The values for the moduli E 1 (see Table 2.2) and E 2 (see Table 2.3) were obtained from bending or uniaxial tension tests on 0◦ or 90◦ specimens (machined parallel or transverse to the fibers). The strain measurement was done by strain rosettes or the cross-head movement. The values for the lamina Poisson’s ratio ν12 (see Table 2.4) were obtained from bending or uniaxial tension tests on 0◦ specimens, whereas the strain measurement was done by strain rosettes.

5

These diagrams were digitalized, and a specialized software (Digitizeit) was used to extract the numerical values of data points.

2.3 Comparison with Experimental Results

35

Table 2.1 Fiber and matrix properties of a glass filament-epoxy resin system, Tsai et al. (1963); Tsai (1964) Fiber Matrix E in 106 psi in MPa ν in – ρ in –

10.6 73084.42 0.22 2.6

0.5 3447.38 0.35 1.15

Table 2.2 Experimental results for the lamina modulus E 1 in case of a glass filament-epoxy resin system, Tsai et al. (1963); Tsai (1964) ψm in % Tsai et al. E 1 in 106 psi Tsai φf E 1 in MPa (1963); Tsai (1964) et al. (1963); Tsai (1964) 12.9 13.8 14.9 14.9 14.9 16.0 16.1 16.8 19.9 20.9 21.9 21.9 23.0 23.1 23.2 23.3 28.8

7.53 8.02 7.58 7.14 6.83 6.98 6.60 7.63 6.65 5.88 6.11 6.84 6.56 6.84 6.19 5.87 5.32

0.749 0.735 0.716 0.716 0.716 0.699 0.698 0.687 0.640 0.626 0.612 0.611 0.596 0.596 0.594 0.593 0.522

51,900 55,300 52,200 49,300 47,100 48,200 45,500 52,600 45,900 40,500 42,100 47,200 45,200 47,200 42,700 40,400 36,700

The values for the lamina in-plane shear modulus G 12 (see Table 2.5) were obtained from pure twisting of 0◦ square plates. The experiments were realized by four corner forces which were perpendicular to the plate (two upwards on opposing corners and two downwards on opposing corners).

2.3.1.2

Data Set by Whitney and Riley

The data set by Whitney and Riley Whitney and Riley (1966) was obtained from unidirectional boron filament-epoxy resin systems (Union Carbide ERL 2256-ZZLB

36

2 Micromechanics

Table 2.3 Experimental results for the lamina modulus E 2 in case of a glass filament-epoxy resin system, Tsai et al. (1963); Tsai (1964) ψm in % Tsai et al. E 2 in 106 psi Tsai φf E 2 in MPa (1963); Tsai (1964) et al. (1963); Tsai (1964) 12.5 14.4 15.0 15.1 15.1 15.1 16.0 16.0 16.1 17.2 18.2 20.0 20.1 21.1 22.0 22.1 23.0 23.1

2.87 2.58 2.29 2.77 1.98 3.10 2.12 1.86 2.53 2.55 2.50 1.85 2.13 1.61 2.40 1.81 2.04 2.36

0.755 0.724 0.715 0.713 0.713 0.713 0.700 0.699 0.698 0.680 0.665 0.638 0.638 0.623 0.611 0.609 0.596 0.595

19,800 17,800 15,800 19,100 13,700 21,400 14,600 12,800 17,400 17,600 17,200 12,700 14,700 11,100 16,500 12,500 14,000 16,300

Table 2.4 Experimental results for the lamina Poisson’s ratio ν12 in case of a glass filament-epoxy resin system, Tsai et al. (1963); Tsai (1964) ψm in % Tsai et al. (1963); ν12 Tsai et al. (1963); Tsai φf Tsai (1964) (1964) 13.8 14.7 15.7 16.2 22.6 30.7 33.2

0.247 0.243 0.244 0.251 0.250 0.277 0.284

0.734 0.719 0.704 0.695 0.602 0.500 0.471

2.3 Comparison with Experimental Results

37

Table 2.5 Experimental results for the lamina in-plane shear modulus G 12 in case of a glass filament-epoxy resin system, Tsai et al. (1963); Tsai (1964) ψm in % Tsai et al. G 12 in 106 psi Tsai φf G 12 in MPa (1963); Tsai (1964) et al. (1963); Tsai (1964) 14.1 13.8 14.5 14.0 16.0 17.7 23.2 23.1

1.49 1.24 1.28 1.08 1.22 1.25 1.03 0.945

0.729 0.734 0.723 0.730 0.698 0.672 0.595 0.595

10,300 8580 8860 7410 8410 8600 7110 6520

Table 2.6 Fiber and matrix properties of a boron filament-epoxy resin system, Whitney and Riley (1966) Fiber Matrix E in 106 psi in MPa ν in – G in 106 psi in MPa

60.0 413,685.42 0.20 25.0 172,368.93

0.6 4136.85 0.35 0.22 1516.85

0820), which were manufactured either by the hand lay-up technique or wound on a special machine. The properties of the constituents, i.e., fiber and matrix, are summarized in Table 2.6. The values for the modulus E 1 and for the lamina Poisson’s ratio ν12 (see Table 2.7 for the original data and Table 2.8 for the data in SI units) were obtained from single layer tensile specimens (manufactured by the hand lay-up technique for high fiber volume content and winding for low fiber content). The strain measurement was done by strain gages. The values for the transverse modulus E 2 were obtained from wound specimens of 3 to 8 layers. The values for the lamina in-plane shear modulus G 12 were calculated in the following way: filament wound specimens were cut in such a way that the fiber direction was at 45◦ to the sides. These particular specimens were used to measure the tensile modulus at 45◦ to the fibers. Based on these results, the shear modulus G 12 was calculated.

38

2 Micromechanics

Table 2.7 Experimental results for the lamina properties in case of a boron filament-epoxy resin system, as reported in Whitney and Riley (1966) E 1 in 106 psi E 2 in 106 psi G 12 in 106 psi ν12 in – φf in % 0.20 0.55 0.60 0.65 0.70 0.75

11.7 30.1 35.7 35.5 34.5 –

– – 3.10 3.40 3.88 4.90

– – – – 1.77 2.43

0.154 0.141 0.133 0.130 0.125 –

Table 2.8 Experimental results for the lamina properties in case of a boron filament-epoxy resin system (converted to SI units), Whitney and Riley (1966) E 1 in MPa E 2 in MPa G 12 in MPa ν12 in – φf in % 0.20 0.55 0.60 0.65 0.70 0.75

80,669 207,532 246,143 244,764 237,869 –

– – 21,374 23,442 26,752 33,784

– – – – 12,204 16,754

0.154 0.141 0.133 0.130 0.125 –

2.3.2 Comparison Between Theoretical Predictions and Experimental Results 2.3.2.1

Data Set by Tsai

The approximation of the modulus of elasticity in fiber direction (E 1 ) as a function of the fiber volume fraction is presented in Fig. 2.20 based on the predictions of Eqs. (2.18), (2.55), and (2.71). It can be seen that the simple straight line approximation between E m and E f gives already a quite good representation of the experimental data points by Tsai. The fiber misalignment factor k allows some kind of fine tuning of the prediction; see Fig. 2.20b. The approximation of the modulus of elasticity transverse to the fiber direction (E 2 ) as a function of the fiber volume fraction is presented in Fig. 2.21 based on the predictions of Eqs. (2.28), (2.59), and (2.73). It can be seen that the prediction based on the mechanics of materials approach (see Fig. 2.21a) clearly underestimates the experimental data. The fiber contiguity (C) or the Halpin–Tsai coefficient (ξ ) allows a much better adjustment to the experimental points; see Figs. 2.21b and c. The approximation of the Lamina Poisson’s ratio (ν12 ) as a function of the fiber volume fraction is presented in Fig. 2.22 based on the predictions of Eqs. (2.39), (2.67), and (2.72). It can be seen that the simple straight line approximation between νm and νf gives already a quite good representation of the experimental data points

2.3 Comparison with Experimental Results (a) Ef

Theory: Eq. (2.18) Exp.: Tsai [9, 10]

Lamina modulus E1

Fig. 2.20 Modulus of elasticity in fiber direction as a function of the fiber volume fraction: comparison of experimental values and theoretical predictions (a MMA, b Tsai, c Halpin–Tsai)

39

Em 0

0.5

1

Fiber volume fraction φf (b) Ef

k=1

Lamina modulus E1

Theory: Eq. (2.55) Exp.: Tsai [9, 10]

k = 0.9

Em 0

0.5

1

Fiber volume fraction φf (c) Ef Lamina modulus E1

Theory: Eq. (2.71) Exp.: Tsai [9, 10]

Em 0

0.5

1

Fiber volume fraction φf

by Tsai, possibly with a small tendency to overestimate the values; see Figs. 2.22a and c. The fiber contiguity C allows some kind of fine tuning of the prediction; see Fig. 2.22b. The approximation of the in-plane shear modulus (G 12 ) as a function of the fiber volume fraction is presented in Fig. 2.23 based on the predictions of Eqs. (2.52), (2.70), and (2.74). It can be seen that the prediction based on the mechanics of materials approach (see Fig. 2.23a) clearly underestimates the experimental data.

40

2 Micromechanics (a)

Ef

Theory: Eq. (2.28) Exp.: Tsai [9, 10]

Lamina modulus E2

Fig. 2.21 Modulus of elasticity transverse to fiber direction as a function of the fiber volume fraction: comparison of experimental values and theoretical predictions (a MMA, b Tsai, c Halpin–Tsai)

Em 0

0.5

1

Fiber volume fraction φf Ef

Theory: Eq. (2.59) Exp.: Tsai [9, 10]

Lamina modulus E2

(b)

C=1 0.5 0 Em 0

0.5

1

Fiber volume fraction φf Ef

Theory: Eq. (2.73) Exp.: Tsai [9, 10]

Lamina modulus E2

(c)

ξ=5 2.5 0 Em 0

0.5

1

Fiber volume fraction φf

The fiber contiguity (C) or the Halpin–Tsai coefficient (ξ ) allows a much better adjustment to the experimental points; see Figs. 2.23b and c. 2.3.2.2

Data Set by Whitney and Riley

The approximation of the modulus of elasticity in fiber direction as a function of the fiber volume fraction is presented in Fig. 2.24 based on the predictions of Eqs. (2.18), (2.55), and (2.71). It can be seen that the simple straight line approximation between

2.3 Comparison with Experimental Results

Lamina Poisson’s ratio ν12

(a) Theory: Eq. (2.39) Exp.: Tsai [9, 10]

νm

νf 0

0.5

1

Fiber volume fraction φf

Lamina Poisson’s ratio ν12

(b) Theory: Eq. (2.67) Exp.: Tsai [9, 10]

νm

0 0.5 C=1

νf 0

0.5

1

Fiber volume fraction φf (c) Lamina Poisson’s ratio ν12

Fig. 2.22 Major (lamina) Poisson’s ratio as a function of the fiber volume fraction: comparison of experimental values and theoretical predictions (a MMA, b Tsai, c Halpin–Tsai)

41

Theory: Eq. (2.72) Exp.: Tsai [9, 10]

νm

νf 0

0.5 Fiber volume fraction φf

1

42 (a) Lamina modulus G12

Fig. 2.23 In-plane shear modulus as a function of the fiber volume fraction: comparison of experimental values and theoretical predictions (a MMA, b Tsai, c Halpin–Tsai)

2 Micromechanics

Theory: Eq. (2.52) Exp.: Tsai [9, 10]

Gf

Gm 0

0.5

1

Fiber volume fraction φf

Lamina modulus G12

(b)

Theory: Eq. (2.70) Exp.: Tsai [9, 10]

Gf

C=1 0.5 0

Gm 0

0.5

1

Fiber volume fraction φf

Lamina modulus G12

(c)

Theory: Eq. (2.74) Exp.: Tsai [9, 10]

Gf

ξ=5 2.5 0 Gm 0

0.5

1

Fiber volume fraction φf

E m and E f gives already a quite good representation of the experimental data points by Tsai, possibly with a small tendency to overestimate the values. The fiber misalignment factor k allows some kind of fine tuning for k < 1 of the prediction; see Fig. 2.24b. The approximation of the modulus of elasticity transverse to the fiber direction (E 2 ) as a function of the fiber volume fraction is presented in Fig. 2.25 based on the predictions of Eqs. (2.28), (2.59), and (2.73). It can be seen that the prediction

2.3 Comparison with Experimental Results (a)

Ef

Lamina modulus E1

Theory: Eq. (2.18) Exp.: Whitney [11]

Em

0

0.5

1

Fiber volume fraction φf Ef

k=1

Theory: Eq. (2.55) Exp.: Whitney [11]

Lamina modulus E1

(b)

k = 0.9

Em

0

0.5

1

Fiber volume fraction φf (c)

Ef

Theory: Eq. (2.71) Exp.: Whitney [11]

Lamina modulus E1

Fig. 2.24 Modulus of elasticity in fiber direction as a function of the fiber volume fraction: comparison of experimental values and theoretical predictions (a MMA, b Tsai, c Halpin–Tsai)

43

Em

0

0.5 Fiber volume fraction φf

1

44 (a)

Ef

Theory: Eq. (2.28) Exp.: Whitney [11]

Lamina modulus E2

Fig. 2.25 Modulus of elasticity transverse to fiber direction as a function of the fiber volume fraction: comparison of experimental values and theoretical predictions (a MMA, b Tsai, c Halpin–Tsai)

2 Micromechanics

Em

0

0.5

1

Fiber volume fraction φf Ef

Theory: Eq. (2.59) Exp.: Whitney [11]

Lamina modulus E2

(b)

C=1 0.5 0 Em

0

0.5

1

Fiber volume fraction φf Ef

Theory: Eq. (2.73) Exp.: Whitney [11]

Lamina modulus E2

(c)

ξ=5 Em

2.5 0 0

0.5

1

Fiber volume fraction φf

based on the mechanics of materials approach (see Fig. 2.25a) underestimates the experimental data. The fiber contiguity (C) or the Halpin–Tsai coefficient (ξ ) allows a much better adjustment to the experimental points; see Figs. 2.25b and c. The approximation of the lamina Poisson’s ratio (ν12 ) as a function of the fiber volume fraction and the corresponding comparison with the experimental data is not meaningful in this case since the experimental results for Poisson’s ratio are, as

2.3 Comparison with Experimental Results (a) Lamina Poisson’s ratio ν12

Fig. 2.26 Major (lamina) Poisson’s ratio as a function of the fiber volume fraction: comparison of experimental values and theoretical predictions (a MMA, b Tsai, c Halpin–Tsai)

45

Theory: Eq. (2.39) Exp.: Whitney [11]

νm

νf

0

0.5

1

Fiber volume fraction φf Lamina Poisson’s ratio ν12

(b) Theory: Eq. (2.67) Exp.: Whitney [11]

νm

0 0.5

νf

C=1

0

0.5

1

Fiber volume fraction φf Lamina Poisson’s ratio ν12

(c) Theory: Eq. (2.72) Exp.: Whitney [11]

νm

νf

0

0.5

1

Fiber volume fraction φf

clearly stated in the publication by Whitney and Riley Whitney and Riley (1966), obviously incorrect; see Fig. 2.26. It is not possible that the measured values are considerably less than the ratio of either constituent material, i.e., the fiber or the matrix. The approximation of the in-plane shear modulus (G 12 ) as a function of the fiber volume fraction is presented in Fig. 2.27 based on the predictions of Eqs. (2.52), (2.70), and (2.74). It can be seen that the prediction based on the mechanics of

46 (a)

Theory: Eq. (2.52) Exp.: Whitney [11]

Gf Lamina modulus G12

Fig. 2.27 In-plane shear modulus as a function of the fiber volume fraction: comparison of experimental values and theoretical predictions (a MMA, b Tsai, c Halpin–Tsai)

2 Micromechanics

Gm

0

0.5

1

Fiber volume fraction φf (b) Theory: Eq. (2.70) Exp.: Whitney [11]

Lamina modulus G12

Gf

C=1

0.5

0 Gm

0

0.5

1

Fiber volume fraction φf (c) Theory: Eq. (2.74) Exp.: Whitney [11]

Lamina modulus G12

Gf

ξ=5 Gm

2.5 0

0.5

0 1

Fiber volume fraction φf

materials approach (see Fig. 2.27a) underestimates the experimental data. The fiber contiguity (C) or the Halpin–Tsai coefficient (ξ ) allows a much better adjustment to the experimental points; see Figs. 2.27b and c.

2.3 Comparison with Experimental Results

47

2.3.3 Optimized Representation of Theoretical Predictions The expressions for the effective lamina moduli according to Eqs. (2.55), (2.59), (2.67), (2.70), (2.73), and (2.74) contain a single parameter, i.e., k, C, or ξ . This parameter can also be understood as a fitting parameter to approximate the proposed equations to a given data set from experiments. In a more mathematical sense, we can state that the task is to fit a set of data points (xi ; yi ) i = 1, . . . , N to a linear or general model of the form y(x) = y(x; a), (2.79) where y is one of the moduli mentioned in the above cited equations, x is the fiber volume fraction (x = φf ), and a is one of the fitting parameters, i.e., k, C, or ξ . To obtain the best value of the fitting parameter for a given data set, one may use, for example, the method of least squares, which minimizes the sum of the squares of the errors; see Press et al. (1997); Wolberg (2006): N

[yi − y(xi ; a)]2 .

minimize

(2.80)

i =1

Many of the computer algebra systems such as Maple, Matlab ® or Maxima, contain built-in functions to fit parameters in a regression analysis. This approach avoids a programming of the method of least squares.

2.3.3.1

Elasticity Solutions with Contiguity after Tsai

Considering the prediction of the modulus of elasticity in fiber direction according to Eq. (2.55) and the particular properties of fiber and matrix according to Tables 2.1 and 2.6, the following linear models are obtained: 3481852φf + 172369 (Exp.: Tsai), 50 40954857φf + 413685 (Exp.: Whitney), E 1 (φf ) = k × 100 E 1 (φf ) = k ×

(2.81) (2.82)

where the parameter k (fiber misalignment factor) must be adjusted to the corresponding data sets. A graphical comparison between the experimental data sets and the theoretical predictions based on optimized k values is provided in Fig. 2.28.

48

2 Micromechanics (a)

Ef

Theory: Eq. (2.55) Exp.: Tsai [9, 10]

Lamina modulus E1

Fig. 2.28 Modulus of elasticity in fiber direction as a function of the fiber volume fraction: comparison of experimental values and optimized theoretical predictions according to Tsai (a experiments by Tsai Tsai et al. (1963); Tsai (1964), b experiments by Whitney Whitney and Riley (1966))

k = 0.9491 Em 0

0.5

1

Fiber volume fraction φf Ef

Theory: Eq. (2.55) Exp.: Whitney [11]

Lamina modulus E1

(b)

k = 0.8975 Em

0

0.5

1

Fiber volume fraction φf

Considering the prediction of the modulus of elasticity transverse to the fiber direction according to Eq. (2.59) and the particular properties of fiber and matrix according to Tables 2.1 and 2.6, the following cubic/quadratic rational functions are obtained: E 2 (φf ) = −

(a1 C + a2 ) φf 3 + (a3 C + a4 ) φf 2 + (a5 − a6 C) φf − a7 C − a8 a9 φf 2 − a10 φf + a11

(Exp.: Tsai),

(2.83)

with a1 = 3135709377982879890894860017499640, a2 = 139619248519860028523054394399456, a3 = 12542837511931523159713456729094862, a4 = 1125920976338332530221756813884962,

(2.84) (2.85)

a5 = 1267454655750478428831281190461955, a6 = 15678546889914385078718077629705531,

(2.88) (2.89)

(2.86) (2.87)

2.3 Comparison with Experimental Results

a7 = 18024569304188444205,

49

(2.90)

a8 = 4358345064723417546005715094882275, a9 = 841156006107698637560211379680, a10 = 2080428612705313795337309567440,

(2.91) (2.92) (2.93)

a11 = 1264248520535424960898886893600,

(2.94)

as well as for the experimental data set by Whitney (a1 C + a2 ) φf 3 + (a3 C + a4 ) φf 2 + (a5 − a6 C) φf − a7 C − a8 a9 φf 2 − a10 φf + a11 (Exp.: Whitney), (2.95)

E 2 (φf ) = −

with a1 = 995987161425546119580256822212, a2 = 8932611131537881869017695044, a3 = 3319957204751823706411665142572, a4 = 64327785852811125960671737446, a5 = 60155315591632534584085178430, a6 = 4315944366177355525588707318066, a7 = 14478769212781406, a8 = 220409957302426216647804777672,

(2.96) (2.97) (2.98) (2.99) (2.100) (2.101) (2.102) (2.103)

a9 = 38867012429186834553773763, a10 = 91936380605065472990704827,

(2.104) (2.105)

a11 = 53279658992331499531538644.

(2.106)

Now the parameter C (fiber contiguity) must be adjusted to the corresponding data sets. A graphical comparison between the experimental data sets and the theoretical predictions based on optimized C values is provided in Fig. 2.29. Considering the prediction of the major Poisson’s ratio according to Eq. (2.67) and the particular properties of fiber and matrix according to Tables 2.1 and 2.6, the following quadratic/quadratic rational functions are obtained: (a1 C − a2 ) φf 2 + (a3 − a4 C) φf − a5 C + a6 a 7 φf 2 + a 8 φf + a 9 (Exp.: Tsai),

ν12 (φf ) =

(2.107)

50

2 Micromechanics (a)

Ef

Theory: Eq. (2.59) Exp.: Tsai [9, 10]

Lamina modulus E2

Fig. 2.29 Modulus of elasticity transverse to fiber direction as a function of the fiber volume fraction: comparison of experimental values and optimized theoretical predictions according to Tsai (a experiments by Tsai Tsai et al. (1963); Tsai (1964), b experiments by Whitney Whitney and Riley (1966))

C = 0.2233 Em 0

0.5

1

Fiber volume fraction φf Ef

Theory: Eq. (2.59) Exp.: Whitney [11]

Lamina modulus E2

(b)

C = 0.0744 Em

0

0.5

1

Fiber volume fraction φf

with a1 = 4140605357312146633053948893727337966267, a2 = 2721617016872532974292667772393662202875,

(2.108) (2.109)

a3 = 11128694042822532708468835587366165345325, a4 = 4140605357312119192324335093526641644301, a5 = 15858259201696365057056250, a6 = 2923468752745199442685485991627088408150,

(2.110) (2.111)

a7 = 7555849134581552661327999526075070836700, a8 = 35593863812683115662420081375807122918675,

(2.114) (2.115)

a9 = 8352767864986289673215855054641582682425,

(2.116)

(2.112) (2.113)

as well as for the experimental data set by Whitney (a1 C − a2 ) φf 2 + (a3 − a4 C) φf − a5 C + a6 a 7 φf 2 + a 8 φf + a 9 (Exp.: Whitney),

ν12 (φf ) =

(2.117)

2.3 Comparison with Experimental Results

51

with a1 = 109619182172310036727587028801117122240, a2 = 75864993070667534588181512860800914880,

(2.118) (2.119)

a3 = 255055291702006802021082477677925861840, a4 = 109619182172309554983436606350189932128,

(2.120) (2.121)

a5 = 801666919101774197101518, a6 = 12901505516719430436697459149663650640,

(2.122) (2.123)

a7 = 175325149985446265500881788453100169920, a8 = 748272426421362197074893250976958110604, a9 = 36861444333484151986152971697850138647.

(2.124) (2.125) (2.126)

The parameter C (fiber contiguity) must be again adjusted to the corresponding data sets. A graphical comparison between the experimental data sets and the theoretical predictions based on optimized C values is provided in Fig. 2.30. Obviously, the experimental values by Whitney are not very meaningful since only values below the fiber ratio are reported. Nevertheless, some kind of reasonable fitting is obtained, which fulfills the boundary conditions for φf = 0 and φf = 1. Considering the prediction of the in-plane shear modulus according to Eq. (2.70) and the particular properties of fiber and matrix according to Tables 2.1 and 2.6, the following quadratic/quadratic rational functions are obtained: (a1 C + a2 ) φf 2 + (−a3 C − a4 ) φf − a5 a 6 φf 2 − a 7 φf + a 8 (Exp.: Tsai),

G 12 (φf ) = −

(2.127)

with a1 = 105348535905710634784768513, a2 = 4690703647187015371675802,

(2.128) (2.129)

a3 = 105348535905710634784768513, a4 = 4690703647187015371675802, a5 = 10671742601445530465163180,

(2.130) (2.131) (2.132)

a6 = 3673775402597027743830, a7 = 11675633988407280502770,

(2.133) (2.134)

a8 = 8358145903237511459700,

(2.135)

52

2 Micromechanics (a) Lamina Poisson’s ratio ν12

Fig. 2.30 Major (lamina) Poisson’s ratio as a function of the fiber volume fraction: comparison of experimental values and optimized theoretical predictions according to Tsai (a experiments by Tsai Tsai et al. (1963); Tsai (1964), b experiments by Whitney Whitney and Riley (1966))

Theory: Eq. (2.67) Exp.: Tsai [9, 10]

νm

C = 0.1715

νf 0

0.5

1

Fiber volume fraction φf

Lamina Poisson’s ratio ν12

(b) Theory: Eq. (2.67) Exp.: Whitney [11]

νm

νf

C = 2.1625

0

0.5

1

Fiber volume fraction φf

as well as for the experimental data set by Whitney (a1 C + a2 ) φf 2 + (−a3 C − a4 ) φf − a5 a 6 φf 2 − a 7 φf x + a 8 (Exp.: Whitney),

(2.136)

a1 = 1246811558802726329728, a2 = 11069377165147274960,

(2.137) (2.138)

a3 = 1246811558802726329728, a4 = 11069377165147274960,

(2.139) (2.140)

a5 = 22731897742320419745, a6 = 7297608310081600,

(2.141) (2.142)

a7 = 22151981907792800, a8 = 14986252920407700.

(2.143) (2.144)

G 12 (φf ) = −

with

2.3 Comparison with Experimental Results (a) Theory: Eq. (2.70) Exp.: Tsai [9, 10]

Gf

Lamina modulus G12

Fig. 2.31 In-plane shear modulus as a function of the fiber volume fraction: comparison of experimental values and theoretical predictions according to Tsai (a experiments by Tsai Tsai et al. (1963); Tsai (1964), b experiments by Whitney Whitney and Riley (1966))

53

C = 0.2290 Gm 0

0.5

1

Fiber volume fraction φf (b) Theory: Eq. (2.70) Exp.: Whitney [11]

Lamina modulus G12

Gf

C = 0.0603 Gm

0

0.5

1

Fiber volume fraction φf

The parameter C (fiber contiguity) must be again adjusted to the corresponding data sets. A graphical comparison between the experimental data sets and the theoretical predictions based on optimized C values is provided in Fig. 2.31. 2.3.3.2

Halpin–Tsai Relationships

Considering the prediction of the modulus of elasticity transverse to the fiber direction according to Eq. (2.73) and the particular properties of fiber and matrix according to Tables 2.1 and 2.6, the following linear/linear rational functions are obtained: (600163347388φf + 29711072161) ξ + 629874419549 8618450ξ − 174092600φf + 182711050 (Exp.: Tsai). (2.145)

E 2 (φf ) =

(1129494001203φf + 11409018615) ξ + 1140903019818 2757900ξ − 273032380φf + 275790280 (Exp.: Whitney). (2.146)

E 2 (φf ) =

54

2 Micromechanics (a)

Ef

Theory: Eq. (2.73) Exp.: Tsai [9, 10]

Lamina modulus E2

Fig. 2.32 Modulus of elasticity transverse to fiber direction as a function of the fiber volume fraction: comparison of experimental values and optimized theoretical predictions according to Halpin–Tsai (a experiments by Tsai Tsai et al. (1963); Tsai (1964), b experiments by Whitney Whitney and Riley (1966)).

ξ = 1.3027 Em 0

0.5

1

Fiber volume fraction φf Ef

Theory: Eq. (2.73) Exp.: Whitney [11]

Lamina modulus E2

(b)

ξ = 1.6701 Em

0

0.5

1

Fiber volume fraction φf

where the parameter ξ (Halpin–Tsai coefficient) must be adjusted to the corresponding data sets. A graphical comparison between the experimental data sets and the theoretical predictions based on optimized ξ values is provided in Fig. 2.32. Considering the prediction of the in-plane shear modulus according to Eq. (2.74) and the particular properties of fiber and matrix according to Tables 2.1 and 2.6, the following linear/linear rational functions are obtained: (19798752361245φf + 881550741187) ξ + 20680303102432 690433605ξ − 15506451675φf + 16196885280 (Exp.: Tsai). (2.147) (518313955096φf + 4601667845) ξ + 522915622941 G 12 (φf ) = 3033700ξ − 341704160φf + 344737860 (Exp.: Whitney). (2.148) G 12 (φf ) =

The parameter ξ (Halpin–Tsai coefficient) must be again adjusted to the corresponding data sets. A graphical comparison between the experimental data sets and the theoretical predictions based on optimized ξ values is provided in Fig. 2.33.

References (a) Lamina modulus G12

Fig. 2.33 In-plane shear modulus as a function of the fiber volume fraction: comparison of experimental values and theoretical predictions according to Halpin–Tsai (a experiments by Tsai Tsai et al. (1963); Tsai (1964), b experiments by Whitney Whitney and Riley (1966))

55

Theory: Eq. (2.74) Exp.: Tsai [9, 10]

Gf

ξ = 2.7124 Gm 0

0.5

1

Fiber volume fraction φf

Lamina modulus G12

(b)

Theory: Eq. (2.74) Exp.: Whitney [11]

Gf

ξ = 2.5632 Gm

0

0.5

1

Fiber volume fraction φf

References Altenbach H, Altenbach J, Kissing W (2018) Mechanics of composite structural elements. Springer, Singapore Halpin JC (1969) Effects of environmental factors on composite materials. Air Force Materials Laboratory Technical Report AFML-TR-67-423. Air Force Materials Laboratory, Ohio Hermans JJ (1967) The elastic properties of fiber reinforced materials when the fibers are aligned. P K Akad Wet-Amsterd B70:1–9 Hewitt RL, de Malherbe MC (1970) A review and catalogue of an approximation for the longitudinal shear modulus of continuous fibre composites. J Compos Mater 4:280–282 Jones RM (1999) Mechanics of composite materials. Taylor & Francis, Philadelphia Öchsner A (2021) Classical beam theories of structural mechanics. Springer, Cham Paul B (1960) Prediction of elastic constants of multiphase materials. T Metall Soc AIME 218:36–41 Press WH, Teukolsky SA, Vetterling WT, Flannery BP (1997) Numerical recipes in Fortran 77: the art of scientific computing (vol 1 of Fortran numerical recipes). Cambridge University Press, Cambridge Tsai SW, Springer GS, Schultz AB (1963) The composite behavior of filament-wound materials. In: Fourteenth International Astronautical Congress, Paris, France, Sept 1963, Paper No. 139

56

2 Micromechanics

Tsai SW (1964) Structural behavior of composite materials. Technical report NASA-CR-71, Washington Whitney JM, Riley MB (1966) Elastic properties of fiber reinforced composite materials. AIAA J 4:1537–1542 Wolberg J (2006) Data analysis using the method of least squares. Springer, Berlin

Chapter 3

Macromechanics of a Lamina

Abstract This chapter covers the mechanical modeling of a single layer with unidirectionally aligned reinforcing fibers embedded in a homogeneous matrix, a so-called lamina. It is shown that a lamina can be treated as a combination of a plane elasticity element and a classical plate element. For both classical structural elements and their combination, the continuum mechanical modeling based on the three basic equations, i.e., the kinematics relationship, the constitutive law, and the equilibrium equation, is presented. Combining these three questions results in the describing partial differential equations. The chapter closes with different orthotropic failure criteria, i.e., the maximum stress, the maximum strain, the Tsai–Hill, and the Tsai–Wu criterion.

3.1 Introduction Let us consider in the following a unidirectional lamina as schematically shown in the Fig. 3.1. The 1-axis is aligned to the fibers, the 2-axis is perpendicular to the fibers in the plane, and the 3-axis indicates the thickness direction. Furthermore, the geometrical assumptions hold as outlined in Sect. 1.2, i.e., the lamina is assumed as flat and thin. The general difference between a plane elasticity and a classical plate element in regards to the possible loads is illustrated in Fig. 3.2. A plane elasticity element can be only loaded by in-plane normal or shear forces; see Fig. 3.2a. Contrary to that, the classical plate element allows forces with lines of action in the thickness direction (3) or moments at which the moment vector lies in the 1-2 plane; see Fig. 3.2b. A similar distinction can be done in regards to the deformations: the plane elasticity element has only in-plane translations, whereas the classical plate element reveals translations perpendicular to the 1-2 plane and corresponding rotations. The derivations of the basic equations of continuum mechanics, i.e., the kinematics relationship, the constitutive law, and the equilibrium equation, are presented first separately for each of these members and afterward superimposed to the combined plane elasticity and classical plate element. This approach allows for a better understanding of the continuum mechanical basics and the procedure to combine basic structural elements. Such a combination of structural elements is normally done in the early years of © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, Composite Mechanics, Advanced Structured Materials 184, https://doi.org/10.1007/978-3-031-32390-4_3

57

58

3 Macromechanics of a Lamina

Fig. 3.1 Schematic representation of a unidirectional lamina

Fig. 3.2 General configuration for a a plane elasticity and b a classical plate element

degree programs in mechanical engineering by superimposing the tensile rod with a thin beam to form the so-called generalized beam element which can elongate and bend Öchsner (2021). The configuration shown in Fig. 3.2 is simply the two-dimensional generalization of the rod/beam superposition.

3.2 Kinematics The kinematics or strain–displacement relations extract the strain field contained in a displacement field.

3.2 Kinematics

59

3.2.1 Plane Elasticity Element Using engineering definitions of strain, the following relations can be obtained Chen and Han (1988); Eschenauer et al. (1997): ε1 =

∂u 1 ∂u 2 ∂u 1 ∂u 2 ; ε2 = ; γ12 = 2ε12 = + . ∂1 ∂2 ∂2 ∂1

(3.1)

In matrix notation, these three kinematics relationships can be written as ⎡

⎤ ⎡∂ ⎤ 0 [ ] ε1 ∂1 ⎣ ε2 ⎦ = ⎣ 0 ∂ ⎦ u 1 , ∂2 u2 ∂ ∂ 2ε12 ∂2 ∂1

(3.2)

ε = L1 u1,2 ,

(3.3)

or symbolically as where L1 is the differential operator matrix.

3.2.2 Classical Plate Element Let us first derive a kinematics relation which relates the variation of u 1 across the plate thickness in terms of the displacement u 3 . For this purpose, let us imagine that a plate element is bent around the 2-axis; see Fig. 3.3a. The following definition of the rotational angle ϕ2 is assumed: the angle ϕ2 is positive if the vector of the rotational direction is pointing in positive 2-axis. Looking at the right-angled triangle 0, A, B, , we can state that1 sin(−ϕ2 ) =

B, 0, , ,

0A

=

− u1 , 3

(3.4)

which results for small angles (sin(−ϕ2 ) ≈ −ϕ2 ) in: u 1 = +3ϕ2 .

(3.5)

Looking at the curved center line in Fig. 3.3a, it holds that the slope of the tangent line at 0, equals: du 3 ≈ −ϕ2 . (3.6) tan(−ϕ2 ) = + d1 Note that according to the assumptions of the classical thin plate theory the lengths 0A and 0, A, remain unchanged. 1

60

3 Macromechanics of a Lamina

If Eqs. (3.5) and (3.6) are combined, the following results: u 1 = −3

du 3 . d1

(3.7)

Considering a plate which is bent around the 1-axis (see Fig. 3.3b) and following the same line of reasoning (the angle ϕ1 is assumed positive if the vector of the rotational direction is pointing in positive 1-axis.), similar equations can be derived for u 2 : du 3 , d2 u 2 = −3ϕ1 , ϕ1 ≈

u 2 = −3

Fig. 3.3 Configurations for the derivation of kinematics relations: in a 1-3 plane and b 2-3 plane. Note that the deformation is exaggerated for better illustration

du 3 . d2

(3.8) (3.9) (3.10)

3.2 Kinematics

61

One may find in the scholarly literature other definitions of the rotational angles Blaauwendraad (2010); Reddy (2006); Ventsel and Krauthammer (2001); Wang et al. (2000). The angle ϕ2 is introduced in the 13-plane (see Fig. 3.3a), whereas ϕ1 is introduced in the 23-plane (see Fig. 3.3b). These definitions are closer to the classical definitions of the angles in the scope of finite elements but not conform with the definitions of the stress resultants (see M1n and M2n in Fig. 3.9). Other definitions assume, for example, that the rotational angle ϕ1 (now defined in the 1-3 plane) is positive if it leads to a positive displacement u 1 at the positive 3-side of the neutral axis. The same definition holds for the angle ϕ2 (now defined in the 2-3 plane). Using classical engineering definitions of strain, the following relations can be obtained Blaauwendraad (2010); Timoshenko and Woinowsky-Krieger (1959): ( ) ∂u 3 ∂ 2u3 ∂u 1 (3.7) ∂ = (3.11) −3 = −3 2 = 3κ1 , ε1 = ∂1 ∂1 ∂1 ∂1 ( ) ∂u 2 (3.10) ∂ ∂u 3 ∂ 2u3 ε2 = = (3.12) −3 = −3 2 = 3κ2 , ∂2 ∂2 ∂2 ∂2 γ12 =

∂u 1 ∂u 2 + ∂2 ∂1

(3.7),(3.10)

=

−23

∂ 2u3 = 3κ12 . ∂1∂2

In matrix notation, these three relationships can be written as ⎡ 2 ⎤ ⎡ ⎤ ⎡ ⎤ ∂ κ1 ε1 ∂122 ∂ ⎥u = 3⎣κ ⎦, ⎣ ε2 ⎦ = −3 ⎢ ⎣ ∂22 ⎦ 3 2 2∂ 2 γ12 κ12

(3.13)

(3.14)

∂1∂2

or symbolically as ε = −3L2 u 3 = 3κ.

(3.15)

3.2.3 Combined Plane Elasticity and Classical Plate Element Combining Eqs. (3.2) and (3.14), the following kinematics description for a superposed element is obtained

62

3 Macromechanics of a Lamina

⎡ 2 ⎤ ⎤ ⎡∂ ⎤ ∂ 0 [ ] ε1 ∂122 ∂1 u ⎢ 1 ∂ ⎥u ⎣ ε2 ⎦ = ⎣ 0 ∂ ⎦ − 3 ⎣ ∂2 2 ⎦ 3 ∂2 u 2 ∂ ∂ 2∂ 2 γ12 ∂2 ∂1 ∂1∂2 ⎡∂ ⎡ ⎤ ⎤ [ ] 0 κ1 ∂1 ∂ ⎦ u1 = ⎣ 0 ∂2 + 3 ⎣ κ2 ⎦ , u2 ∂ ∂ κ12 ∂2 ∂1 ⎡

(3.16)

(3.17)

or symbolically as ε = L1 u1,2 − 3L2 u 3 = L1 u1,2 +3κ, ᷅ ᷆᷆ ᷆

(3.18) (3.19)

ε0

where ε 0 collects the middle-surface strains. Alternatively, one may collect the single components of the plane elasticity and the plate contribution in matrix form as follows: ⎤

⎡

∂ ⎢ ⎡ 0 ⎤ ⎢ ∂1 ε1 ⎢ 0 ⎢ 0⎥ ⎢ ⎢ ε2 ⎥ ⎢ ⎢ ⎥ ⎢ ∂ ⎢γ 0 ⎥ ⎢ ⎢ 12 ⎥ ⎢ ⎢ ∂1 ⎢ ⎥=⎢ ⎢ κ1 ⎥ ⎢ ⎢ ⎥ ⎢0 ⎢ ⎥ ⎢ ⎢ κ2 ⎥ ⎢ ⎣ ⎦ ⎢ ⎢0 κ12 ⎢ ⎣ 0

0

0

⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎥⎡ ⎤ ⎥ u1 0 ⎥ ⎥⎣ ⎦ u , ⎥ ∂2 ⎥ 2 0 − 2 ⎥ u3 ∂12 ⎥ ∂ ⎥ ⎥ 0 − 2⎥ ∂22 ⎥ 2∂ ⎦ 0 − ∂1∂2

∂ ∂2 ∂ ∂2

(3.20)

or symbolically as e = L, u.

(3.21)

The column matrix of generalized strains can be also expressed as: e=

[ 0] ε . κ

(3.22)

3.3 Constitutive Equation

63

3.3 Constitutive Equation For isotropic materials, the mechanical properties are the same for all directions and it can be shown that the components of the elasticity matrix are determined by two independent constants. In mechanical engineering, many times the set of elasticity modulus E and Poisson’s ration ν are chosen as the independent constants. Orthotropic materials reveal three principal, mutually orthogonal axes. These axes are also called the material principal axes.

3.3.1 Isotropic Material: Plane Elasticity Element The two-dimensional plane stress case (σ3 = σ23 = σ13 = 0) shown in Fig. 3.4 is commonly used for the analysis of thin, flat plates loaded in the plane of the plate (1-2 plane). It should be noted here that the normal thickness stress is zero (σ3 = 0), whereas the thickness normal strain is present (ε3 /= 0). The plane stress Hooke’s law for a linear-elastic isotropic material based on the Young’s modulus E and Poisson’s ratio ν can be written for constant temperature as ⎤ ⎡ ⎤⎡ ⎤ 1ν 0 ε1 σ1 E ⎣ν 1 0 ⎦ ⎣ ε2 ⎦ , ⎣ σ2 ⎦ = 2 1 − ν σ12 2 ε12 0 0 1−ν 2 ⎡

(3.23)

or in matrix notation as σ = Cε,

Fig. 3.4 Two-dimensional problem: plane stress case

(3.24)

64

3 Macromechanics of a Lamina

where C is the so-called elasticity matrix. It should be noted here that the engineering shear strain γ12 = 2ε12 is used in the formulation of Eq. (3.23). Rearranging the elastic stiffness form given in Eq. (3.23) for the strains gives the elastic compliance form ⎤ ⎡ ⎤⎡ ⎤ ε1 0 σ1 1 1 −ν ⎣ ε2 ⎦ = ⎣−ν 1 ⎦ ⎣ σ2 ⎦ , 0 E 0 0 2(ν + 1) 2 ε12 σ12 ⎡

(3.25)

or in matrix notation as ε = Dσ ,

(3.26)

where D = C −1 is the so-called elastic compliance matrix. The general characteristic of a plane Hooke’s law in the form of Eqs. (3.24) and (3.25) is that two independent material parameters are used. It should be finally noted that the thickness strain ε3 can be obtained based on the two in-plane normal strains ε1 and ε2 as: ε3 = −

( ) ν · ε1 + ε2 . 1−ν

(3.27)

The last equation can be derived from the three-dimensional formulation; see Öchsner (2020a). Equations (3.23) and (3.25) indicate that a plane isotropic material requires two independent material constants2 .

3.3.2 Isotropic Material: Classical Plate Element The classical plate theory assumes a plane stress state and the constitutive equation can be taken from Sect. 3.3.1, i.e., Eqs. (3.23) and (3.25) are still valid.

3.3.3 Isotropic Material: Combined Plane Elasticity and Classical Plate Element Since there is no difference in the constitutive description of a plane elasticity and a classical plate element, a combined element is still described based on the the set of equations (3.23)–(3.24) and (3.25)–(3.26).

2

Three-dimensional isotropy requires as well only two independent elastic constants.

3.3 Constitutive Equation

65

3.3.4 Orthotropic Material: Combined Plane Elasticity and Classical Plate Element Since the plane elasticity element follows the same constitutive description as the classical plate element, it is sufficient to indicate the the relationship for the combined element. Let us start with the compliance form, i.e., ⎤ ν21 1 − 0 ⎥⎡ ⎤ ⎤ ⎢ E ⎡ ⎤⎡ ⎤ ⎡ E2 ⎥ σ1 ⎢ 1 ε1 D11 D12 0 σ1 ⎥ ⎢ ν12 1 ⎥ ⎦ ⎣ ⎦ ⎣ ⎣ ⎣ ε2 ⎦ = ⎢− σ D 0 σ2 ⎦ , D = 0 ⎥ 2 12 22 ⎢ E E ⎥ σ12 ⎢ 1 2 2 ε12 σ 0 0 D 44 12 ⎣ 1 ⎦ 0 0 G 12 ⎡

(3.28)

or in matrix notation as ε = Dσ ,

(3.29)

where D = C −1 is again the so-called elastic compliance matrix. The constant G in Eq. (3.29) is the shear modulus in the 1-2 plane. It should be noted that in Eq. (3.28) the identity ν21 ν12 =− (3.30) − E1 E2 holds and we can conclude that four independent elastic constants are required to describe a plane orthotropic material3 . The stress–strain relationship can be obtained from Eq. (3.28) by inverting the compliance matrix to give the following representation: ⎡ ⎤ ν21 E 1 E1 ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ 0 ⎥⎡ ⎢1 − ν ν 1 − ν ν C11 C12 0 σ1 ε1 12 21 12 21 ⎢ ⎥ ε1 ⎥ ⎣ ε2 ⎦ = ⎣C12 C22 0 ⎦ ⎣ ε2 ⎦ , ⎣ σ2 ⎦ = ⎢ ν12 E 2 E2 ⎢ ⎥ 0 ⎣ ⎦ 2 ε12 σ12 2 ε12 0 0 C44 1 − ν12 ν21 1 − ν12 ν21 0 0 G 12 (3.31) or in matrix notation as σ = Cε, (3.32) where C = D−1 is again the so-called elasticity matrix. Let us consider in the following the rotation of the coordinate system and the corresponding transformations of stresses and strains; see Fig. 3.5. This is important in the case of laminates with different laminae at different orientations (see Chap. 4). 3

Three-dimensional orthotropy requires nine independent elastic constants.

66

3 Macromechanics of a Lamina

Fig. 3.5 Rotated lamina: (1, 2) principal material coordinates and (x, y) arbitrary coordinates. The rotational angle α is from the x-axis to the 1-axis (counterclockwise positive for the sketched coordinate systems)

⎤ ⎡ ⎤⎡ ⎤ cos2 α sin2 α −2 sin α cos α σx σ1 ⎣ σ y ⎦ = ⎣ sin2 α cos2 α 2 sin α cos α ⎦ ⎣ σ2 ⎦ , σx y σ12 sin α cos α − sin α cos α cos2 α − sin2 α ⎡

(3.33)

or in matrix notation as σ x,y = T −1 σ σ 1,2 ,

(3.34)

or in the inverse representation ⎤ ⎡ ⎤⎡ ⎤ cos2 α sin2 α 2 sin α cos α σ1 σx ⎣ σ2 ⎦ = ⎣ sin2 α cos2 α −2 sin α cos α ⎦ ⎣ σ y ⎦ , σ12 σx y − sin α cos α sin α cos α cos2 α − sin2 α ⎡

(3.35)

or in matrix notation as σ 1,2 = T σ σ x,y .

(3.36)

The corresponding transformations for the strain field read: ⎡

⎤ ⎡ ⎤ ⎤⎡ cos2 α sin2 α − sin α cos α εx ε1 ⎣ ε y ⎦ = ⎣ sin2 α cos2 α sin α cos α ⎦ ⎣ ε2 ⎦ , 2εx y 2ε12 2 sin α cos α −2 sin α cos α cos2 α − sin2 α

(3.37)

or in matrix notation as ε x,y = T −1 ε ε 1,2 ,

(3.38)

or in the inverse representation ⎡

⎤ ⎡ ⎤ ⎤⎡ cos2 α sin2 α sin α cos α ε1 εx ⎣ ε2 ⎦ = ⎣ sin2 α cos2 α − sin α cos α ⎦ ⎣ ε y ⎦ , 2ε12 2εx y −2 sin α cos α 2 sin α cos α cos2 α − sin2 α

(3.39)

3.3 Constitutive Equation

67

or in matrix notation as ε 1,2 = T ε ε x,y .

(3.40)

It should be noted here that the relationship between the strain and stress transformation matrices is as follows T ε = RT σ R−1 ,

(3.41)

where the matrix R has the following entries: ⎡

⎤ 1 0 0 R = ⎣0 1 0⎦ . 0 0 2

(3.42)

Based on these transformations of the stresses and strains, the stress–strain relationship in the arbitrary (rotated) x-y coordinate system can be obtained as σ 1,2 = Cε1,2 , T −1 σ σ 1,2

(3.43)

T −1 σ Cε1,2 ,

(3.44)

−1 σ x,y = T −1 σ C T ε T ε ε1,2 , ᷅ ᷆᷆ ᷆

(3.45)

᷅ ᷆᷆ ᷆

=

σ x,y

I −1 = T −1 σ CT ε T ε ε1,2 , ᷅ ᷆᷆ ᷆

(3.46)

εx,y

σ x,y =

T −1 σ CT ε

᷅ ᷆᷆ ᷆

εx,y .

(3.47)

C

Thus, the transformed elasticity matrix C can be obtained from the following triple matrix product: ⎡

⎤⎡ ⎤ cos2 α sin2 α −2 sin α cos α C11 C12 C14 cos2 α 2 sin α cos α ⎦ ⎣C12 C22 C24 ⎦ C = ⎣ sin2 α C14 C24 C44 sin α cos α − sin α cos α cos2 α − sin2 α ⎤ ⎡ 2 2 cos α sin α sin α cos α ⎣ sin2 α cos2 α − sin α cos α ⎦ −2 sin α cos α 2 sin α cos α cos2 α − sin2 α

(3.48)

68

3 Macromechanics of a Lamina

⎡

E1 ν21 E 1 cos2 α sin2 α −2 sin α cos α ⎢ 1 − ν ν 1 − ν12 ν21 12 21 ⎢ ν E E cos2 α 2 sin α cos α ⎦ ⎢ = ⎣ sin2 α 12 2 2 ⎢ sin α cos α − sin α cos α cos2 α − sin2 α ⎣ 1 − ν12 ν21 1 − ν12 ν21 0 0 ⎤ ⎡ 2 2 sin α sin α cos α cos α ⎣ sin2 α cos2 α − sin α cos α ⎦ −2 sin α cos α 2 sin α cos α cos2 α − sin2 α ⎡ ⎤ C 11 C 12 C 14 = ⎣C 12 C 22 C 24 ⎦ , C 14 C 24 C 44 ⎡

⎤

⎤ 0 ⎥ ⎥ ⎥ 0 ⎥ ⎦ G 12 (3.49)

(3.50)

where the single matrix entries C i j are given as follows: C 11 = C11 cos4 α + 2(C12 + 2C44 ) sin2 α cos2 α + C22 sin4 α,

(3.51)

C 12 = (C11 + C22 − 4C44 ) sin α cos +C12 (sin α + cos α),

(3.52)

2

2

4

4

C 22 = C11 sin α + 2(C12 + 2C44 ) sin α cos α + C22 cos α, 4

2

2

4

(3.53)

C 14 = (C11 − C12 − 2C44 ) sin α cos α + (C12 − C22 + 2C44 ) sin α cos α, (3.54) 3

3

C 24 = (C11 − C12 − 2C44 ) sin3 α cos α + (C12 − C22 + 2C44 ) sin α cos3 α, (3.55) C 44 = (C11 + C22 − 2C12 − 2C44 ) sin2 α cos2 α + C44 (sin4 α + cos4 α). (3.56) In a similar way, the strain-stress relationship in the x-y coordinate system can be obtained as ε1,2 = Dσ 1,2 , T −1 ε ε 1,2

(3.57)

T −1 ε Dσ 1,2 ,

(3.58)

−1 ε x,y = T −1 ε D T σ T σ σ 1,2 , ᷅ ᷆᷆ ᷆

(3.59)

᷅ ᷆᷆ ᷆

=

ε x,y

I −1 = T −1 ∈ DT σ T σ σ 1,2 , ᷅ ᷆᷆ ᷆

(3.60)

σ x,y

ε x,y =

T −1 ε DT σ

᷅ ᷆᷆ ᷆

σ x,y .

(3.61)

D

Thus, the transformed elastic compliance matrix D can be obtained from the following triple matrix product:

3.3 Constitutive Equation

69

⎡

⎤⎡ ⎤ cos2 α sin2 α − sin α cos α D11 D12 D14 cos2 α sin α cos α ⎦ ⎣ D12 D22 D24 ⎦ D = ⎣ sin2 α D14 D24 D44 2 sin α cos α −2 sin α cos α cos2 α − sin2 α ⎡ ⎤ 2 2 cos α sin α 2 sin α cos α ⎣ sin2 α cos2 α −2 sin α cos α ⎦ (3.62) − sin α cos α sin α cos α cos2 α − sin2 α ⎡ ⎤ ν21 1 ⎤⎢ E − E 0 ⎥ ⎡ 1 2 ⎥ cos2 α sin2 α − sin α cos α ⎢ ⎢ ν12 1 ⎥ 2 2 ⎢ ⎦ ⎣ sin α cos α sin α cos α = 0 ⎥ − ⎢ ⎥ E1 E2 ⎥ 2 sin α cos α −2 sin α cos α cos2 α − sin2 α ⎢ ⎣ 1 ⎦ 0 0 G 12 ⎤ ⎡ sin2 α 2 sin α cos α cos2 α ⎣ sin2 α cos2 α −2 sin α cos α ⎦ (3.63) − sin α cos α sin α cos α cos2 α − sin2 α ⎡ ⎤ D 11 D 12 D 14 (3.64) = ⎣ D 12 D 22 D 24 ⎦ , D 14 D 24 D 44 where the single matrix entries D i j are given as follows: D 11 = D11 cos4 α + (2D12 + D44 ) sin2 α cos2 α + D22 sin4 α, D 12 = (D11 + D22 − D44

) sin2 α cos2 +D

4 4 12 (sin α + cos α),

D 22 = D11 sin4 α + (2D12 + D44 ) sin2 α cos2 α + D22 cos4 α, D 14 = (2D11 − 2D12 − D44 ) sin α cos3 α − (2D22 − 2D12 − D44 ) sin3 α cos α, D 24 = (2D11 − 2D12 − D44 ) sin3 α cos α − (2D22 − 2D12 − D44 ) sin α cos3 α, D 44 = 2(2D11 + 2D22 − 4D12 − D44 ) sin2 α cos2 α + D44 (sin4 α + cos4 α).

(3.65) (3.66) (3.67) (3.68) (3.69) (3.70)

It can be shown that the elements D i j of the transformed elastic compliance matrix (see Eq. (3.64)) can be related to the apparent engineering constants of the orthotropic lamina in the rotated x-y coordinate system as follows (see Ashton and Whitney (1970); Clyne and Hull (2019); Daniel and Ishai (1994); Jones (1975) for details): ⎤ ⎡ νx y ηx 1 − ⎥ ⎡ ⎤ ⎢ Ex Ex ⎥ ⎢ Ex D 11 D 12 D 14 ⎢ ν ηy ⎥ ⎥ ⎢ xy 1 D = ⎣ D 12 D 22 D 24 ⎦ = ⎢− (3.71) ⎥, ⎢ Ex E y E y ⎥ ⎥ ⎢ D 14 D 24 D 44 ηy 1 ⎦ ⎣ ηx Ex E y G x y

70

3 Macromechanics of a Lamina

where the shear–extension coupling coefficients are generally defined as follows: γx y γx y E x = , (for uniaxial tension with σx ) εx σx γx y γx y E y = . (for uniaxial tension with σ y ) ηy = εy σy

ηx =

(3.72) (3.73)

Furthermore, the Poisson’s ratios (extension–extension coupling coefficients) are defined as: εy , (for uniaxial tension with σx ) εx εx = − . (for uniaxial tension with σ y ) εy

νx y = −

(3.74)

ν yx

(3.75)

3.4 Equilibrium The equilibrium equations relate the external loads (i.e., forces and moments) to the corresponding internal reactions (i.e., stresses).

3.4.1 Plane Elasticity Element Figure 3.6 shows the normal and shear stresses which are acting on a differential volume element in the 1-direction. All forces are drawn in their positive direction at each cut face. A positive cut face is obtained if the outward surface normal is directed in the positive direction of the corresponding coordinate axis. This means that the 1 d1)d2d3 is oriented in right-hand face in Fig. 3.6 is positive and the force (σ1 + ∂σ ∂x the positive 1-direction. In a similar way, the top face is positive, i.e., the outward surface normal is directed in the positive 2-direction, and the shear force4 is oriented in the positive 1-direction. Since the volume element is assumed to be in equilibrium, forces resulting from stresses on the sides of the cuboid and from the body forces f i (i = 1, 2, 3) must be balanced. These body forces are defined as forces per unit volume which can be produced by gravity5 , acceleration, magnetic fields, and so on. 4 In the case of a shear force σ , the first index i indicates that the stress acts on a plane normal to ij the i-axis and the second index j denotes the direction in which the stress acts. 5 If gravity is acting, the body force f results as the product of density times standard gravity: mkg m f = VF = mg V = V g = eg. The units can be checked by consideration of 1 N = 1 s2 .

3.4 Equilibrium

71

Fig. 3.6 Stresses and body forces which act on a plane differential volume element in 1-direction (note that the three directions d1, d2, and d3 are differently sketched to indicate the plane problem)

The static equilibrium of forces in the 1-direction based on the five force components—two normal forces, two shear forces and one body force—indicated in Fig. 3.6 gives ( ) ) ∂σ1 ∂σ21 d2d3 − σ1 d2d3 + d1d3 σ1 + ∂1 ∂2

(

− σ21 d1d3 + f 1 d1d2d3 = 0,

(3.76)

or after simplification and canceling with dV = d1d2d3: ∂σ1 ∂σ21 + + f 1 = 0. ∂1 ∂2

(3.77)

Based on the same approach, a similar equation can be specified in the 2-direction: ∂σ2 ∂σ21 + + f 2 = 0. ∂2 ∂1

(3.78)

These two balance equations can be written in matrix notation as ⎡

⎤ ∂ ∂ ⎡ ⎤ [ ] [ ] 0 ⎢ ⎥ σ1 ∂2⎥ ⎣ σ2 ⎦ + f 1 = 0 , ⎢ ∂1 ⎣ ∂ ∂⎦ f2 0 σ12 0 ∂2 ∂1

(3.79)

or in symbolic notation: LT1 σ + b = 0

(3.80)

where L1 is the differential operator matrix and b is the column matrix of body forces.

72

3 Macromechanics of a Lamina

Fig. 3.7 Stress resultants for a plane elasticity element

The internal reactions expressed as stresses as indicated in Fig. 3.6 can be alternatively stated as forces based on an integration over the corresponding reference area. These integral values are then called the stress resultants or generalized stresses. Based on Fig. 3.7, the stress resultant Nin , i.e., normalized with the corresponding side length of the plate element, for the plane elasticity element can be indicated as follows: (t/2 N1 n = σ1 d3, (3.81) N1 = d2 −t/2

N2n

n N12

N2 = = d1

(t/2 σ2 d3,

(3.82)

−t/2

N12 = = d1

(t/2 σ12 d3,

(3.83)

−t/2

or all three relations combined in matrix notation: ⎡ ⎤ ⎡ n⎤ (t/2 σ1 N1 ⎣ σ2 ⎦ d3. ⎣ N2n ⎦ = n N12 σ12 −t/2

(3.84)

Thus, we can get the following alternative formulation of the balance equations by introducing the stress resultants in Eq. (3.79). The two balance equations can be alternatively written in matrix notation as ⎤ ∂ ∂ ⎡ n⎤ [ ] [ ] 0 ⎥ N1 ⎢ ∂2⎥ ⎣ N n ⎦ + q1 = 0 , ⎢ ∂1 2 ⎣ q2 0 ∂ ∂⎦ n N12 0 ∂2 ∂1 ⎡

(3.85)

3.4 Equilibrium

73

or in symbolic notation: LT1 N n + q1,2 = 0,

(3.86)

where q1 = f 1 d3 and q2 = f 2 d3 are the area-specific loads.

3.4.2 Classical Plate Element Let us first look at the stress distributions through the thickness of a classical plate element d1d2t as shown in Fig. 3.8. Linear distributed normal stresses (σ1 , σ2 ), linear distributed shear stresses (τ21 , τ12 ), and parabolic distributed shear stresses (τ23 , τ13 ) can be identified. These stresses can be expressed by the so-called stress resultants, i.e., bending moments and shear forces as shown in Fig. 3.9. These stress resultants are taken to be positive if they cause a tensile stress (positive) at a point with positive 3-coordinate. These stress resultants are obtained by integrating over the stress distributions. In the case of plates, however, the integration is only performed over the thickness, i.e., the moments and forces are given per unit length (normalized with the corresponding side length of the plate element). The normalized (superscript ‘n’) bending moments are obtained as: M1n

M2n

M1 = = d2 M2 = = d1

(t/2 3σ1 d3,

(3.87)

3σ2 d3.

(3.88)

−t/2

(t/2 −t/2

Fig. 3.8 Stresses acting on a classical plate element

74

3 Macromechanics of a Lamina

Fig. 3.9 Stress resultants acting on a classical plate element: a bending and twisting moments and b shear forces. Positive directions are drawn

The twisting moment per unit length reads:

n M12

=

n M21

M12 M21 = = = d2 d1

(t/2 3τ12 d3,

(3.89)

−t/2

or all three relations combined in matrix notation: ⎡ ⎤ ⎡ n⎤ (t/2 σ1 M1 ⎣ M2n ⎦ = 3 ⎣σ12 ⎦ d3. n M12 σ12 −t/2

(3.90)

3.4 Equilibrium

75

Furthermore, the shear forces per unit length are calculated in the following way: Q1 = = d2

(t/2 τ13 d3,

(3.91)

τ23 d3,

(3.92)

] (t/2 [ ] τ13 Q n1 = d3. Q n2 τ23

(3.93)

Q n1

Q n2

Q2 = = d1

−t/2

(t/2 −t/2

or both relations combined in matrix notation: [

−t/2

It should be noted that a slightly different notation when compared to the beam problems is used here. The bending moment around the 2-axis is now called M1n (which directly corresponds to the causing stress σ1 ) while in the beam notation it was M2 ; see Öchsner (2020a). Nevertheless, the orientation remains the same. The shear force, which was in the case of the beams given as Q 3 is now either Q n1 or Q n2 . Thus, in the case of this plate notation, the index refers rather to the plane (check the surface normal vector) in which the corresponding resultant (vector) is located. The equilibrium condition will be determined in the following for the vertical forces. Assuming that the distributed force is constant (q3 (1, 2) → q3 ) and that forces in the direction of the positive 3-axis are considered positive, the following results: − Q n1 (1)d2 − Q n2 (2)d1 + Q n1 (1 + d1)d2 + Q n2 (2 + d2)d1 + q3 d1d2 = 0. (3.94) Expanding the shear forces at 1 + d1 and 2 + d2 in a Taylor’s series of first order, meaning ∂ Q n1 d1, ∂1 ∂ Q n2 d2, Q n2 (2 + d2) ≈ Q n2 (2) + ∂2 Q n1 (1 + d1) ≈ Q n1 (1) +

(3.95) (3.96)

Eq. (3.94) results in ∂ Q n2 ∂ Q n1 d1d2 + d2d1 + q3 d1d2 = 0, ∂1 ∂2

(3.97)

76

3 Macromechanics of a Lamina

or alternatively after simplification to: ∂ Q n1 ∂ Q n2 + + q3 = 0. ∂1 ∂2

(3.98)

The equilibrium of moments around the reference axis at 1 + d1 (positive if the moment vector is pointing in positive 2-axis) gives: n n (2 + d2)d1 − M21 d1 M1n (1 + d1)d2 − M1n (1)d2 + M21

+ Q n2 (2 + d2)d1 d1 − Q n1 (1)d2d1 + q3 d1d2 d1 = 0. − Q n2 (2)d1 d1 2 2 2

(3.99)

Expanding the stress resultants at 1 + d1 and 2 + d2 into a Taylor’s series of first order, meaning ∂ M1n d1, ∂1 n ∂ M21 n n (2 + d2) = M21 (2) + d2, M21 ∂2 n ∂ Q2 d2, Q n2 (2 + d2) = Q n2 (2) + ∂2 M1n (1 + d1) = M1n (1) +

(3.100) (3.101) (3.102)

Eq. (3.99) results in n ∂ M21 ∂ Q n2 d1 d1 ∂ M1n d1d2 + d2d1 + d2d1 − Q n1 (1)d2d1 + q3 d1d2 = 0. 2 2 ∂1 ∂2 ∂2 (3.103)

Seeing that the terms of third order (d1d2d1) are considered as infinitesimally n n small and because of M21 = M12 , finally the following results: n ∂ M1n ∂ M12 + − Q n1 = 0. ∂1 ∂2

(3.104)

In a similar way, the equilibrium of moments around the reference axis at 2 + d2 finally gives: n ∂ M2n ∂ M12 + − Q n2 = 0. (3.105) ∂2 ∂1

3.4 Equilibrium

77

Thus, the three equilibrium equations can be summarized as follows: ∂ Q n1 ∂ Q n2 + + q3 = 0, ∂1 ∂2 n ∂ M1n ∂ M12 + − Q n1 = 0, ∂1 ∂2 n ∂ M2n ∂ M12 + − Q n2 = 0. ∂2 ∂1

(3.106) (3.107) (3.108)

Let us recall here that the following equilibrium equations for the Euler–Bernoulli beam can be obtained Öchsner (2021): dM2 (1) d2 M2 (1) dQ 3 (1) = Q 3 (1), = = −q3 . d1 d1 d12

(3.109)

Rearranging Eqs. (3.107) and (3.108) for Q n and introducing in Eq. (3.106) finally gives the combined equilibrium equation as: ∂ 2 M1n ∂12

+2

∂ 2 M12 ∂ 2 M2n + + q3 = 0. ∂1∂2 ∂22

(3.110)

The last equation can be written in matrix notation as [

∂ ∂ 2∂ ∂12 ∂22 ∂1∂2 2

2

2

]

⎡

⎤ M1n ⎣ M2n ⎦ + q3 = 0, n M12

(3.111)

or symbolically as LT2 M n + q3 = 0.

(3.112)

Equations (3.107) and (3.108) can be rearranged to obtain a relationship between the moments and shear forces similar to Eq. (3.109)1 : LT1 M n = Qn ,

(3.113)

where the first-order differential operator matrix L1 is given by Eqs. (3.79) and (3.80).

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3 Macromechanics of a Lamina

3.4.3 Combined Plane Elasticity and Classical Plate Element Combining Eqs. (3.85)–(3.86) and (3.111)–(3.112), the following equilibrium description for a superposed element is obtained ⎡

∂ ⎢ ∂1 ⎢ ⎢ ⎢0 ⎢ ⎢ ⎢ ⎢ ⎣ 0

∂ 0 ∂2 ∂ ∂ 0 ∂2 ∂1

⎤

⎡

N1n

⎤

⎥ ⎢ 0 ⎥ ⎢ N2n ⎥ ⎡ ⎤ ⎥ ⎢ ⎥⎢ ⎡ ⎤ q1 ⎥ ⎢Nn ⎥ 0 ⎥ ⎢ ⎥ 12 ⎥ ⎥ ⎢q2 ⎥ ⎣ ⎦ 0 0 ⎥⎢ 0 , = + ⎥ ⎢ ⎥ ⎢ Mn ⎥ ⎣ ⎦ ⎥⎢ 1 ⎥ 0 q3 ⎥⎢ ⎥ 2 2 2 ⎦ ⎢ Mn ⎥ ∂ ∂ 2∂ 2 ⎦ ⎣ 0 0 n 2 2 ∂1∂2 M ∂1 ∂2 12

0

0

(3.114)

or in symbolic notation: LT sn + q = 0,

(3.115)

where sn is the column matrix of the stress resultants per unit length (generalized stresses per unit length). Let us note here that the derivation of the stress resultants Nin and Min as given in Eqs. (3.84) and (3.90) must be revised for the combined case. This comes from the fact that the total strain in the form ε = ε 0 + 3κ (see Eq. (3.19)) must be used in the derivation: ⎡

N1n

⎤

⎢ n⎥ ⎣ N2 ⎦ = n N12

(t/2

(t/2

(3.32)

σ d3 =

−t/2

(t/2

(t/2 Cε 0 d3 +

−t/2

= tCε 0 +

(t/2

Cεd3 =

−t/2

=

(3.19)

( ) C ε0 + 3κ d3

(3.116)

−t/2

t/2

t/2

C3κd3 = Cε 0 [3]−t/2 + Cκ[32 /2]−t/2

−t/2

t2 Cκ. 4

(3.117)

3.5 Partial Differential Equations

79

A corresponding derivation for the internal bending moments gives: ⎡

⎤ (t/2 (t/2 (t/2 M1n ( ) (3.32) (3.19) n ⎣ M2 ⎦ = 3σ d3 = 3Cεd3 = 3C ε 0 + 3κ d3 n M12 −t/2 −t/2 −t/2 (t/2

(t/2

2

t/2

C32 κd3 = Cε 0 [32 /2]−t/2 + Cκ[33 /3]−t/2

0

−t/2

=

t/2

Cε 3d3 +

=

(3.118)

−t/2 3

t t Cε 0 + Cκ. 4 12

(3.119)

Equations (3.117) and (3.119) can be combined to obtain a single matrix representation: ⎡ ⎤ ⎤ ε10 ⎡ n⎤ ⎡ 2 t N1 0⎥ C ⎥⎢ tC ⎢ ε2 ⎥ ᷅ ᷆᷆ ᷆ ⎢ N2n ⎥ ⎢ 4 ⎥ ⎥ ⎢ ⎢ ᷅ ᷆᷆ ᷆ ⎢ n⎥ ⎢ A 0⎥ ⎥ ⎢γ12 ⎢ N12 ⎥ ⎢ B ⎥ ⎥ ⎢ ⎢ n⎥ = ⎢ (3.120) ⎥⎢ ⎥ , ⎢ M ⎥ ⎢ t2 3 κ ⎥ ⎥ ⎢ 1 t 1 ⎢ n⎥ ⎢ ⎥ ⎥ ⎢ C ⎣ M2 ⎦ ⎣ C ⎥ 4 12 ⎦ ⎢ ⎣ κ2 ⎦ n ᷅ ᷆᷆ ᷆ ᷅ ᷆᷆ ᷆ M12 B D κ12 or in symbolic notation:

sn = C ∗ e,

(3.121)

where C ∗ is the generalized elasticity matrix.

3.5 Partial Differential Equations 3.5.1 Plane Elasticity Element Introducing the constitutive equation according to (3.32) in the equilibrium Eq. (3.80) gives: LT1 Cε + b = 0.

(3.122)

Introducing the kinematics relations in the last equation according to (3.3) finally gives the Lamé–Navier equations: LT1 CL1 u1,2 + b = 0

(3.123)

80

3 Macromechanics of a Lamina

Alternatively, the displacements may be substituted and the differential equations are obtained in terms of stresses. This formulation is known as the Beltrami–Michell equations. If the body forces vanish (b = 0), the partial differential equations in terms of stresses are called the Beltrami equations.

3.5.2 Classical Plate Element Let us combine the three equations for the resulting moments according to Eqs. (3.87)– (3.89) in matrix notation as ⎡

⎡ ⎤ ⎤ (t/2 (t/2 M1n σ1 n n 3 ⎣ σ2 ⎦ d3 = 3σ d3. M = ⎣ M2 ⎦ = n M12 τ 12 −t/2 −t/2

(3.124)

Introducing Hooke’s law (3.32) and the kinematics relation (3.15) gives for a constant elasticity matrix C (t/2 M =− n

(t/2 3 CL2 u 3 d3 = −CL2 u 3

32 d3 = −

2

−t/2

−t/2

t3 CL2 u 3 . 12

(3.125)

᷅ ᷆᷆ ᷆ t3 12

Using the kinematics relation in the curvature form (see Eq. (3.15)), it can be stated that t3 Cκ. (3.126) Mn = 12 Introducing the moment–displacement relation (3.125) in the equilibrium equation (3.112) results in the plate bending differential equation in the form: ( LT2

) t3 CL2 u 3 − q3 = 0. 12

(3.127)

Assuming isotropic material behavior and the definitions for L2 and C given in Eqs. (3.111) and (3.23), the following classical form of the plate bending differential equation can be obtained: ( ) ∂ 4u3 ∂ 4u3 ∂ 4u3 Et 3 +2 2 2+ = q3 . 12(1 − ν 2 ) ∂14 ∂1 ∂2 ∂24

(3.128)

3.5 Partial Differential Equations

81

Let us recall here that the following partial differential equation for the Euler– Bernoulli beam can be obtained under the assumption of isotropic material behavior: E I2

d4 u 3 (x) d14

= q3 (x),

(3.129)

where q3 is now defined as a load per unit length.

3.5.3 Combined Plane Elasticity and Classical Plate Element Starting from the equilibrium equation as given in Eq. (3.114) and inserting Eq. (3.120), i.e., the equation which is based on the definition of the stress resultants under considering the constitutive equation and the relation for the sum of the strain components, gives with the kinematics relation Eq. (3.20) the following set of partial differential equations: ⎡

⎡

∂ ⎢ ⎢ ∂1 ⎢ ⎢0 ⎢ ⎢ ⎣ 0

∂ 0 ∂2 ∂ ∂ 0 ∂2 ∂1 ∂2 0 0 ∂12 0

⎤⎡ tC 0 ⎥⎢ ᷅ ᷆᷆ ᷆ ⎥⎢ A ⎥⎢ ⎢ 0 0 ⎥ ⎥⎢ 2 ⎥⎢ ⎢t ∂ 2 2∂ 2 ⎦ ⎣ 4 C ᷅ ᷆᷆ ᷆ ∂22 ∂1∂2 B 0

∂ ⎢ ∂1 ⎢ ⎤⎢ ⎢0 2 t ⎢ C ⎥⎢ ⎢ 4 ⎥ ᷅ ᷆᷆ ᷆ ⎥ ⎢ ∂ ∂1 B ⎥⎢ ⎥⎢ ⎢ t3 ⎥ ⎢ 0 C⎥ ⎦⎢ ⎢ ᷅12 ᷆᷆ ᷆ ⎢ ⎢0 D ⎢ ⎢ ⎣ 0

⎡ ⎤ ⎡ ⎤ q1 0 + ⎣q2 ⎦ = ⎣0⎦ , q3 0

⎤ 0

0

∂ ∂2 ∂ ∂2

0 0

0 − 0 −

∂2 ∂12 ∂2

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎡ ⎤ ⎥ u1 ⎥ ⎥ ⎣u 2 ⎦ ⎥ ⎥ u3 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

∂22 2∂ 2 0 − ∂1∂2

(3.130)

or ⎡

⎡

∂ ⎢ ⎢ ∂1 ⎢ ⎢0 ⎢ ⎢ ⎣ 0

∂ 0 0 ∂2 ∂ ∂ 0 ∂2 ∂1 ∂2 0 0 ∂12

⎤⎡

A11 0 ⎥⎢ A ⎥ ⎢ 12 ⎥ ⎢ A14 ⎢ 0 0 ⎥ ⎥ ⎢ B11 ⎢ ⎥ ∂ 2 2∂ 2 ⎦ ⎣ B12 B14 ∂22 ∂1∂2 0

A12 A22 A24 B12 B22 B24

A14 A24 A24 B14 B24 B44

B11 B12 B44 D11 D12 D14

B12 B22 B24 D12 D22 D24

∂ ⎢ ∂1 ⎢ ⎢ ⎤⎢ 0 B14 ⎢ ⎢ ⎢ B24 ⎥ ⎥⎢ ∂ ⎢ B44 ⎥ ⎥ ⎢ ∂1 ⎢ D14 ⎥ ⎥⎢ 0 ⎦ D24 ⎢ ⎢ D44 ⎢ ⎢0 ⎢ ⎢ ⎣ 0

⎤ 0 ∂ ∂2 ∂ ∂2 0 − 0 −

0 0 0 ∂2 ∂12 ∂2

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

∂22 2∂ 2 0 − ∂1∂2

82

3 Macromechanics of a Lamina

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ u1 q1 0 ⎣u 2 ⎦ + ⎣q2 ⎦ = ⎣0⎦ , u3 q3 0

(3.131)

or in symbolic notation: LT C ∗ L, u + q = 0.

(3.132)

Once the solution of the displacements u in Eq. (3.132) is obtained, for example, based on the finite element method Öchsner (2020a) or the finite difference method Öchsner (2020b), the kinematics relationship (3.20) provides the strains from which the constitutive equation (3.31) allows the calculation of the stresses.

3.6 Failure Criteria The following subsections present some of the common orthotropic failure criteria. The corresponding criteria for isotropic material behavior can be found, for example, in Öchsner (2016).

3.6.1 Maximum Stress Criterion This criterion assumes that there is no failure, i.e., no fiber failure in the 1-direction, no transversal stress failure in the 2-direction, and no in-plane shear failure, as long as the stress components are in the following limits of the corresponding strength values k: k1c < σ1 < k1t , k2c < σ2 < k2t , |σ12 | < k12s ,

(3.133) (3.134) (3.135)

which assumes that there is no interaction between the different stress components. Some limitations of this criterion are discussed in Tsai (1968).

3.6.2 Maximum Strain Criterion The maximum strain criterion is similar to the maximum stress criterion and assumes no interaction between the strain components. No failure occurs as long as the strain components are in the following limits:

3.6 Failure Criteria

83 ε ε k1c < ε1 < k1t , ε k2c

(3.136)

ε k2t ,

< ε2 < ε |γ12 | < k12s ,

(3.137) (3.138)

where the k ε represent the corresponding failure strains. Some limitations of this criterion are discussed in Tsai (1968).

3.6.3 Tsai–Hill Criterion The Tsai–Hill criterion is based on Hill’s three-dimensional yield criterion for orthotropic materials which was applied by Tsai to lamina Hill (1950); Tsai (1968). This criterion can be stated for a plane stress state as follows: σ12 k12

−

σ1 σ2 k12

+

σ22 k22

+

2 σ12 2 k12s

< 1,

(3.139)

where k1 = k1t for σ1 > 0 or k1 = k1c for σ1 < 0. The same convention holds for direction 2. This criterion, which accounts for an interaction of the different stress components, showed a good agreement between the theoretical prediction and experimental values for E-glass-epoxy laminae at different rotational angles Tsai (1968).

3.6.4 Tsai–Wu Criterion The Tsai–Wu criterion can be expressed for orthotropic materials under plane stress states as follows Tsai and Wu (1971): 2 + 2a1,2 σ1 σ2 < 1, a1 σ1 + a2 σ2 + a11 σ12 + a22 σ22 + a12 σ12

(3.140)

where the coefficients a can be related to the classical failure stresses as follows Altenbach et al. (2018); Kaw (2006): 1 1 + , k1t k1c 1 1 + , a2 = k2t k2c 1 , a11 = − k1t k1c 1 , a22 = − k2t k2c a1 =

(3.141) (3.142) (3.143) (3.144)

84

3 Macromechanics of a Lamina

a12 =

1 2 k12s

a1,2 ≈ −

1 2

,

(3.145)

/ 1 1 × . k1t k1c k2t k2c

(3.146)

Excellent agreement between the Tsai–Wu criterion and experimental data for boron-epoxy is reported in Pipes and Cole (1973).

References Altenbach H, Altenbach J, Kissing W (2018) Mechanics of composite structural elements. Springer Nature, Singapore Ashton JE, Whitney JM (1970) Theory of laminated plates. Technomic Publishing, Stamford Blaauwendraad J (2010) Plates and FEM: surprises and pitfalls. Springer, Dordrecht Chen WF, Han DJ (1988) Plasticity for structural engineers. Springer, New York Clyne TW, Hull D (2019) An introduction to composite materials. Cambridge University Press, Cambridge Daniel IM, Ishai O (1994) Engineering mechanics of composite materials. Oxford University Press, New York Eschenauer H, Olhoff N, Schnell W (1997) Applied structural mechanics: Fundamentals of elasticity, load-bearing structures, structural optimization. Springer, Berlin Hill R (1950) The mathematical theory of plasticity. Oxford University Press, London Kaw AK (2006) Mechanics of composite materials. CRC Press, Boca Raton Jones RM (1975) Mechanics of composite materials. Scripta Book Co., Washington Öchsner A (2016) Continuum damage and fracture mechanics. Springer, Singapore Öchsner A (2020a) Computational statics and dynamics: an introduction based on the finite element method. Springer, Singapore Öchsner A (2020b) Structural mechanics with a pen: a guide to solve finite difference problems. Springer, Cham Öchsner A (2021) Classical beam theories of structural mechanics. Springer, Cham Pipes RB, Cole BW (1973) On the off-axis strength test for anisotropic materials. J Compos Mater 7:246–256 Reddy JN (2006) An introduction to the finite element method. McGraw Hill, Singapore Timoshenko S, Woinowsky-Krieger S (1959) Theory of plates and shells. McGraw-Hill Book Company, New York Tsai SW (1968) Strength theories of filamentary structures. In: Schwartz RT, Schwartz HS (eds) Fundamental aspects of fiber reinforced plastic composites. Wiley Interscience, New York Tsai SW, Wu EM (1971) A general theory of strength for anisotropic materials. J Compos Mater 5:58–80 Ventsel E, Krauthammer T (2001) Thin plates and shells: theory, analysis, and applications. Marcel Dekker, New York Wang CM, Reddy JN, Lee KH (2000) Shear deformable beams and plates: relationships with classical solution. Elsevier, Oxford

Chapter 4

Macromechanics of a Laminate

Abstract This chapter covers the stacking of single laminae, generally under different angles, to a so-called laminate. Based on the approach of the classical laminate theory, a simplified stress analysis, a subsequent failure analysis is derived, without the solution of the system of coupled differential equations for the unknown displacements in the three coordinate directions. This theory provides the solution of the statically indeterminate system based on a generalized stress–strain relationship under consideration of the constitutive relationship and the definition of the so-called stress resultants.

4.1 Introduction Let us consider in the following a laminate, which is composed of n layers; see Fig. 4.1. Each layer k is a single lamina with its own orientation expressed in a local or lamina-specific coordinate system (1k , 2k , 3k ). Thus, the global or laminate-specific coordinate system (x, y, z) is used to describe the orientation of the laminate. The height of a layer k (1 ≤ k ≤ n) can be expressed based on the thickness coordinate as (4.1) tk = z k − z k−1 , and the total height of the laminate results as t=

n ∑

tk .

(4.2)

k =1

4.2 Generalized Stress–Strain Relationship Let us focus in the following on the evaluation on the stress resultants as introduced in Eqs. (3.116) and (3.118) for a single lamina. The internal normal forces can be expressed in the laminate-specific coordinate system (x, y, z) as © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, Composite Mechanics, Advanced Structured Materials 184, https://doi.org/10.1007/978-3-031-32390-4_4

85

86

4 Macromechanics of a Laminate

Fig. 4.1 Geometry of a laminate with n layers

⎡

⎤ ∫t/2 n ∫z k n ∫z k N xn ∑ ∑ n ⎣ Ny ⎦ = σ dz = σ k dˆz = C k εk dˆz N xny k = 1 k = 1 z k−1 z k−1 −t/2 n ∫z k ∑

=

k = 1z n ∑

=

⎛

(4.4) ⎞

k−1

⎝

k =1 n ( ∑

=

( ) C k ε0 + zˆ κ dˆz

k =1

∫zk

∫zk

(4.3)

C k zˆ κ dˆz ⎠

(4.5)

) ) 1( 2 2 C k (z k − z k−1 ) ε + C k (z k ) − (z k−1 ) κ , , ,, , , 2 ,, ,

(4.6)

C k ε 0 dˆz + z k−1

z k−1 0

Ak

Bk

and the corresponding derivation for the internal bending moments: ⎤ ∫t/2 n ∫z k n ∫z k Mxn ∑ ∑ ⎣ M yn ⎦ = zσ dz = σ k zˆ dˆz = C k ε k zˆ dˆz Mxny k = 1z k = 1z −t/2 k−1 k−1 ⎡

=

n ∫z k ∑ k = 1z

=

n ∑

⎛

k−1

⎝

k =1

=

n ( ∑ k =1

( ) C k ε0 zˆ + zˆ 2 κ dˆz ∫zk

∫zk C k ε 0 zˆ dˆz +

(4.8) ⎞ C k zˆ 2 κ dˆz ⎠

(4.9)

z k−1

z k−1

(4.7)

) ) ) 0 1( 1( 3 3 2 2 C k (z k ) − (z k−1 ) ε + C k (z k ) − (z k−1 ) κ . (4.10) , 3 , 2 ,, ,, , , Bk

Dk

4.2 Generalized Stress–Strain Relationship

87

Equations (4.3) and (4.7) can be combined in a single matrix form to give tension-shear ⎡ n⎤ coupling ⎡ Nx A A A ⎢ Nn ⎥ ⎢ 11 12 14 ⎢ y⎥ ⎢ A12 A22 A24 ⎢ n ⎥ ⎢ ⎢ Nx y ⎥ ⎢ A14 A24 A44 ⎢ ⎥ ⎢ n⎥ = ⎢ ⎢ B11 B12 B14 ⎢ Mx ⎥ ⎢ ⎢ ⎥ ⎣ B12 B22 B24 ⎢ Mn ⎥ ⎣ y⎦ B14 B24 B44 Mxny bending-tension coupling or more symbolically as

tension-twist, bending-shear coupling ⎤ B11 B12 B14 ⎥ B12 B22 B24 ⎥ ⎥ B44 B24 B44 ⎥ ⎥ D11 D12 D14 ⎥ ⎥ D12 D22 D24 ⎦ D14 D24 D44 bending-twist coupling

⎡

εx0

⎤

⎢ ε0 ⎥ ⎢ y⎥ ⎢ 0⎥ ⎢γ x y ⎥ ⎢ ⎥ ⎢ ⎥, ⎢ κx ⎥ ⎢ ⎥ ⎢ κy ⎥ ⎣ ⎦ κx y

][ ] [ 0] ε AB Nn = , κ Mn B D , ,, , , ,, , , ,, ,

(4.11)

[

C∗

s

(4.12)

e

where s is the column matrix of stress resultants (generalized stresses), C ∗ is the generalized elasticity matrix, and e is the column matrix of generalized strains. The corresponding submatrices are given as follows: A= B= D=

n ∑ k =1 n ∑ k =1 n ∑ k =1

Ak = Bk = Dk =

n ∑

C k (z k − z k−1 ) ,

(4.13)

k =1 n ∑

( ) 1 C k (z k )2 − (z k−1 )2 , 2 k =1

(4.14)

n ( ) 1∑ C k (z k )3 − (z k−1 )3 . 3 k =1

(4.15)

It should be noted here that A is called the extensional submatrix, D the bending submatrix, and B the coupling submatrix. Equation (4.12) can be inverted to obtain the strains and curvatures as function of the generalized stresses as ][ ] [ 0 ] ([ ])−1 [ n ] [ ' A B' AB ε Nn N = , = κ BD Mn Mn (B' )T D' ,, , , , ,, , , ,, , e

(C ∗ )−1

s

(4.16)

88

4 Macromechanics of a Laminate

where the submatrices are given as follows (Altenbach et al. 2018; Chawla 1987): ( )( )−1 ( −1 )T A' = A−1 + −A−1 B D − BA−1 B −A B , ( −1 ) ( )−1 ' −1 B = −A B D − BA B , ( ) −1 D' = D − BA−1 B .

(4.17) (4.18) (4.19)

As can be concluded from Eq. (4.11), there is a coupling between tension and shear (A14 : N xn → γx0y ; A24 : N yn → γx0y ), bending and tension (B11 , B12 : Mxn → εx0 , ε0y ; B12 , B22 : M yn → εx0 , ε0y ), tension and twist (B14 : N xn → κx y ; B24 : N yn → κx y ), bending and shear (B14 : Mxn → γx0y ; B24 : M yn → γx0y ), and bending and twist (D14 : Mxn → κx y ; D24 : M yn → κx y ). Based on the generalized strains obtained from Eq. (4.16), it is now possible to calculate the stresses in each layer k expressed in the x–y coordinate system: ( ) σ kx,y (z) = C k ε0 + zκ ,

(4.20)

where the vertical coordinate z ranges for the kth layer in the following boundaries: z k−1 ≤ z ≤ z k . The stresses may be evaluated at the bottom (z = z k−1 ), middle (z = (z k + z k−1 )/2) or top (z = z k ) of each layer. Based on relation (3.36), we can transform the stress values into the 1–2 coordinate system (see Fig. 3.5): σ k1,2 = T kσ σ kx,y .

(4.21)

The obtained stress values may serve for a subsequent failure analysis of layer k (see Chap. 5).

4.3 Special Cases of Laminates The following section covers special cases of the generalized elasticity matrix C ∗ as provided in Eq. (4.11); see MIL-HDBK-17/2F (2002) and Altenbach et al. (2018) for details. • Symmetric laminates: Symmetric laminates have a symmetric buildup sequence in regard to laminae (plies) orientations, thicknesses, and material properties about the middle surface; see Fig. 4.2 for some examples. The generalized stiffness matrix (4.11) simplifies for symmetric laminates to the following form

4.3 Special Cases of Laminates

89

Fig. 4.2 Examples of symmetric laminates: a regular [45◦ /0◦ ]s and b [45◦ /30◦ /0◦ ]s with different layer thicknesses

⎡ ⎢ ⎢ ⎢ ⎢ C∗ = ⎢ ⎢ ⎢ ⎣

⎤ A11 A12 A14 0 0 0

A12 A22 A24 0 0 0

A14 A24 A44 0 0 0

0 0 0 D11 D12 D14

0 0 0 D12 D22 D24

0 0 0 D14 D24 D44

⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎦

(4.22)

where we have now B = 0. • Symmetric laminates with isotropic layers: Symmetric laminates with isotropic layers contain only isotropic laminae (plies), i.e., laminae where the elastic properties are independent of the direction and characterized by two independent properties (e.g., Young’s modulus and Poisson’s ratio); see Fig. 4.3. The generalized stiffness matrix (4.11) simplifies for symmetric laminates with isotropic layers to the following form: ⎡

A11 A12 ⎢ ⎢ A12 A11 ⎢ ⎢ ⎢ 0 0 ∗ C =⎢ ⎢ ⎢ 0 0 ⎢ ⎢ 0 0 ⎣ 0 0

⎤ 0 0 A44 0 0 0

0 0 0

0 0 0

D11 D12 D12 D11 0

0

Fig. 4.3 Example of a symmetric laminate with isotropic layers

0 0 0

⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ 0 ⎥ ⎥ 0 ⎥ ⎦ D44

(4.23)

90

4 Macromechanics of a Laminate

Fig. 4.4 Examples of symmetric cross-ply laminates: a regular [90◦ /0◦ ]s and b [90◦ /0◦ ]s with different layer thicknesses

• Symmetric cross-ply laminates: Symmetric cross-ply laminates only contain orthotropic laminae (plies) at right angles to one another; see Fig. 4.4 for some examples. Laminae may have pair-wise (i.e., to ensure symmetry) different thicknesses and material properties. The generalized stiffness matrix (4.11) simplifies for symmetric cross-ply laminates to the following form: ⎡ A11 A12 ⎢ A ⎢ 12 A22 ⎢ ⎢ ⎢ 0 0 C∗ = ⎢ ⎢ ⎢ 0 0 ⎢ ⎢ 0 0 ⎣ 0 0

⎤ 0 0 A44 0 0 0

0 0 0

0 0 0

D11 D12 D12 D22 0

0

0 0 0

⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ 0 ⎥ ⎥ 0 ⎥ ⎦ D44

(4.24)

• Symmetric angle-ply laminates: Symmetric angle-ply laminates only contain orthotropic laminae (plies) at which adjacent laminae have opposite orientation signs; see Fig. 4.5 for some examples.

Fig. 4.5 Examples of symmetric angle-ply laminates: a regular [30◦ / − 30◦ /30◦ ]s and b [45◦ /− 45◦ ]s with different layer thicknesses

4.3 Special Cases of Laminates

91

Fig. 4.6 Examples of symmetric balanced laminates: a regular [45◦ / − 45◦ /0◦ ]s and b [45◦ / − 45◦ /0◦ ]s with different layer thicknesses

Laminae may have pair-wise (i.e., to ensure symmetry) different thicknesses and material properties. The generalized stiffness matrix (4.11) simplifies for symmetric angle-ply laminates to the following form: ⎡ ⎢ ⎢ ⎢ ⎢ ∗ C =⎢ ⎢ ⎢ ⎣

⎤ A11 A12 A14 0 0 0

A12 A22 A24 0 0 0

A14 A24 A44 0 0 0

0 0 0 D11 D12 D14

0 0 0 D12 D22 D24

0 0 0 D14 D24 D44

⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎦

(4.25)

It should be noted here that we obtain for the special case [45◦ /− 45◦ ]s (see Fig. 4.5b) the further simplification A22 = A11 . However, [30◦ /− 30◦ ]s yields A22 /= A11 . • Symmetric balanced laminates: Symmetric angle-ply laminates only contain orthotropic laminae (plies) at which all identical laminae at angles other than 0◦ and 90◦ occur only in ± pairs (not necessarily adjacent); see Fig. 4.6 for some examples. These laminates only contain an even number of laminae.1 The generalized stiffness matrix (4.11) simplifies for symmetric balanced laminates to the following form:

1

Thus, symmetric angle-ply laminates do not belong to the group of symmetric balanced laminates.

92

4 Macromechanics of a Laminate

⎡ ⎢ ⎢ ⎢ ⎢ ∗ C =⎢ ⎢ ⎢ ⎣

⎤ A11 A12 0 0 0 0

A12 A22 0 0 0 0

0 0 A44 0 0 0

0 0 0 D11 D12 D14

0 0 0 D12 D22 D24

0 0 0 D14 D24 D44

⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎦

(4.26)

• Symmetric quasi-isotropic laminates: Symmetric quasi-isotropic laminates are balanced and only contain orthotropic laminae (plies) for which a constitutive property of interest, at a given point, displays isotropic behavior in the plane of the laminates; see Fig. 4.7. The generalized stiffness matrix (4.11) simplifies for symmetric quasi-isotropic laminates to the following form: ⎡

A11 A12 ⎢ ⎢ A12 A11 ⎢ ⎢ ⎢ 0 0 ∗ C =⎢ ⎢ ⎢ 0 0 ⎢ ⎣ 0 0 0 0

⎤ 0 0 A44 0 0 0

0 0 0

0 0 0

0 0 0

D11 D12 D14 D12 D22 D24 D14 D24 D44

⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎦

(4.27)

It should be noted here that symmetric quasi-isotropic laminates require that all layers have the same thickness and the same orthotropic material properties. As an example, t1 = t6 but t2 = t3 = t4 = t5 /= t1 would yield A11 /= A22 . In a similar way, Mat1 = Mat6 but Mat2 = Mat3 = Mat4 = Mat5 /= Mat1 would yield A11 /= A22 . Thus, such laminates would be no longer quasi-isotropic.

Fig. 4.7 Example of a quasi-isotropic symmetric laminate: [0◦ /60◦ / − 60◦ ]s

4.4 Failure Analysis of Laminates

93

4.4 Failure Analysis of Laminates Let us summarize here the recommended steps for a calculation according to the classical laminate theory (internal forces Nin and internal moments Min given, see Table 4.1. In case that the generalized strains are given, see Table 4.2). Once the stresses and/or strains in each lamina are known, the failure analysis can be performed as outlined in Sect. 3.6. Table 4.1 Recommended steps for a calculation according to the classical laminate theory. Case: internal forces Nin and internal moments Min given No.

Steps to perform

①

Define for each lamina k the material (E 1,k ; E 2,k ; ν12,k ; G 12,k ) and the geometrical (z k ; z k−1 ; αk ) properties

②

Define the column matrix of generalized stresses: [ ]T s = N xn N yn N xny Mxn M yn Mxny

③

Calculate for each layer k the elasticity matrix C k in the 1–2 lamina system according to Eq. (3.31). Transform each matrix according to Eq. (3.49) to obtain the elasticity matrix C k in the x – y laminate system

④

Calculate the submatrices A, B, and C according to Eqs. (4.13)–(4.15). Assemble the generalized elasticity matrix C ∗ according to Eq. (4.12)

⑤

Calculate the generalized compliance matrix (C ∗ )−1 based on Eqs. (4.17)–(4.19)

⑥

[ ]T Calculate the generalized strains e = ε0 κ according to Eq. (4.16)

⑦

) ( Calculate the stresses in each layer k according to Eq. (4.20), σ kx,y (z) = C k ε0 + zκ , in the x – y laminate ) ( system (z k−1 ≤ z ≤ z k ). The strains in the x – y laminate system are obtained from ε0 + zκ

⑧

Transform the stresses in each layer to the 1–2 lamina system according to Eq. (4.21): σ k1,2 = T kσ σ kx,y . The strains in the 1–2 lamina system are obtained from Eq. (3.33): εk1,2 = (C k )−1 σ k1,2

Table 4.2 Recommended steps for a calculation according to the classical laminate theory. Case: ]T [ generalized strains e = εx0 ε0y γx0y κx κ y κx y given No.

Steps to perform

①

Define for each lamina k the material (E 1,k ; E 2,k ; ν12,k ; G 12,k ) and the geometrical (z k ; z k−1 ; αk ) properties

②

Define the column matrix of generalized strains: [ ]T e = εx0 ε0y γx0y κx κ y κx y

③

Calculate for each layer k the elasticity matrix C k in the 1–2 lamina system according to Eq. (3.31). Transform each matrix according to Eq. (3.49) to obtain the elasticity matrix C k in the x – y laminate system

④ ⑤

Calculate the submatrices A, B, and C according to Eqs. (4.13)–(4.15). Assemble the generalized elasticity matrix C ∗ according to Eq. (4.12) [ ] Calculate the generalized stresses s = N n M n according to Eq. (4.11)

⑥

In case that the stresses and strains in each layer are required, go to steps ⑦ and ⑧ in Table 4.1

94

4 Macromechanics of a Laminate

References Altenbach H, Altenbach J, Kissing W (2018) Mechanics of composite structural elements. Springer, Singapore Chawla KK (1987) Composite materials: science and engineering. Springer, New York MIL-HDBK-17/2F (2002) Composite materials handbook—volume 2. Polymer matrix composites materials properties. Department of Defence, USA

Chapter 5

Example Problems

Abstract This chapter presents the solution of some standard calculation problems based on the classical laminate theory. Each solution is based on a step-by-step approach in order to present all the details of the solution procedure. The presented problems comprise the calculation of stresses and strains in symmetric and asymmetric laminates, the application of failure criteria, and the ply-by-ply failure of laminates. In addition, the characterization of basic properties of unidirectional laminae based on pole diagrams for the elastic properties and failure envelopes is presented.

5.1 Introduction The following sections will cover some standard calculation problems which are solved in great detail. All the presented problems are based on the same AS4 carbon/3501-6 epoxy prepreg1 system. The effective lamina properties can be taken from Table 5.1. Other sets of complete mechanical properties for unidirectional laminae can be found, for example, in Soden et al. (1998) and Kaw (2006). The theoretical prediction of lamina properties, i.e., the so-called micromechanics, is not covered in this chapter within example problems and the interested reader is referred to Sect. 2.2 and the specialized literature, e.g., Herakovich (2012), Buryachenko (2007), and Dvorak (2013).

5.2 Problem 1: Stresses and Strains in a Symmetric Laminate Given is a symmetric 8-layer laminate of thickness t = 8 mm as shown in Fig. 5.1. Each lamina has the same thickness (t/8) and identical material properties, which can be taken from Table 5.1. However, the orientation of each lamina changes from layer to layer as indicated in Fig. 5.1. 1

A prepreg is a ready to mold or cure sheet which may be tow, tape, cloth, or mat impregnated with resin (MIL-HDBK-17/1F 2002). © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, Composite Mechanics, Advanced Structured Materials 184, https://doi.org/10.1007/978-3-031-32390-4_5

95

96

5 Example Problems

Table 5.1 Selected mechanical properties of a unidirectional lamina (fiber: AS4 carbon; matrix: 3501-6 epoxy; fiber volume fraction of the prepreg: 0.6) Value Property Elastic constants Longitudinal modulus, E 1 in MPa Transverse modulus, E 2 in MPa In-plane shear modulus, G 12 in MPa Major Poisson’s ratio, ν12 in – Strength parameters Longitudinal tensile strength, k1t in MPa Longitudinal compressive strength, k1c in MPa Transverse tensile strength, k2t in MPa Transverse compressive strength, k2c in MPa In-plane shear strength, k12s in MPa ε in % Longitudinal tensile failure strain, k1t ε in Longitudinal compressive failure strain, k1c % ε in % Transverse tensile failure strain, k2t ε in % Transverse compressive failure strain, k2c ε In-plane shear failure strain, k12s in %

126000 11000 6600 0.28 1950 − 1480 48 − 200 79 1.38 − 1.175 0.436 − 2.0 2

Adapted from Soden et al. (1998) Fig. 5.1 Symmetric laminate: [± 45◦ /0◦ /90◦ ]s

Consider (a) a pure tensile load case with N xn = 1000 N/mm (all other forces and moments are equal to zero) and (b) a pure bending load case with Mxn = 1000 Nmm/mm (all other moments and forces are equal to zero) to determine the stresses and strains in each lamina in the global x–y coordinate system as well as in the principal material coordinates (1–2 coordinate systems) of each lamina. • Laminate loaded by N xn = 1000 N/mm The solution procedure will follow the recommended steps provided in Table 4.1. ① Define for each lamina k the material (E 1,k ; E 2,k ; ν12,k ; G 12,k ) and the geometrical (z k ; z k−1 ; αk ) properties.

5.2 Problem 1: Stresses and Strains in a Symmetric Laminate

97

Table 5.2 Geometrical properties (z k ; z k−1 ; αk ) of the laminate shown in Fig. 5.1 Lamina k Coordinate z k−1 in Coordinate z k in mm Angle αk (◦ ) mm −4 −3 −2 −1 0 1 2 3

1 2 3 4 5 6 7 8

−3 −2 −1 0 1 2 3 4

+ 45 − 45 0 90 90 0 − 45 + 45

The material properties (E 1,k ; E 2,k ; ν12,k ; G 12,k ) are the same for each lamina and can be taken from Table 5.1. The geometrical properties for each lamina are summarized in Table 5.2. ② Define the column matrix of generalized stresses: ⎤T ⎡ s = N xn N yn N xny Mxn M yn Mxny .

The load matrix for case (a) is given as follows: ⎡ ⎤T s = 1000 0 0 0 0 0 N/mm .

(5.1)

③ Calculate for each layer k the elasticity matrix C k in the 1–2 lamina system according to Eq. (3.31). Transform each matrix according to Eq. (3.49) to obtain the elasticity matrix C k in the x–y laminate system. Since each lamina is composed of the same materials and the thickness is equal to t/8 = 1 mm, Eq. (3.31) yields for each lamina the same matrix: ⎤ E1 ν21 E 1 ⎤ 0 ⎥ ⎡ ⎢1 − ν ν 1 − ν ν 126868.343 3101.226 0 ⎥ 12 21 12 21 ⎢ N ⎥ = ⎣ 3101.226 11075.808 0 ⎦ E2 . Ck = ⎢ ⎥ ⎢ ν12 E 2 mm2 0 ⎦ ⎣1 − ν ν 1 − ν ν 0 0 6600.0 12 21 12 21 0 0 G 12 (5.2) ⎡

98

5 Example Problems

The transformation to the x–y laminate system according to Eq. (3.49) gives the following matrices: ⎡

C [±45◦ ]

C [0◦ ]

C [90◦ ]

⎤ 42636.651 29436.651 ±28948.134 N , = ⎣ 29436.651 42636.651 ±28948.134⎦ 2 ±28948.134 ±28948.134 32935.425 mm ⎡ ⎤ 126868.343 3101.226 0.0 N = ⎣ 3101.226 11075.808 0.0 ⎦ , 2 0.0 0.0 6600.0 mm ⎡ ⎤ 11075.808 3101.226 0.0 N = ⎣ 3101.226 126868.343 0.0 ⎦ . 2 0.0 0.0 6600.0 mm

(5.3)

(5.4)

(5.5)

④ Calculate the submatrices A, B, and C according to Eqs. (4.13)–(4.15). Assemble the generalized elasticity matrix C ∗ according to Eq. (4.12).

The evaluation of the submatrices A, B, and C according to Eqs. (4.13)–(4.15) gives ⎡

⎤ 446434.906 130151.508 0.0 N ⎦ 0.0 A= Ak = ⎣130151.508 446434.906 , 0.0 0.0 158141.699 mm k =1 ⎡ ⎤ 8 0.0 0.0 0.0 ∑ B= Bk = ⎣0.0 0.0 0.0⎦ N , 0.0 0.0 0.0 k =1 ⎡ ⎤ 8 2191204.439 1115508.171 347377.607 ∑ D= Dk = ⎣1115508.171 1728034.297 347377.607 ⎦ Nmm , 347377.607 347377.607 1264789.188 k =1 8 ∑

(5.6)

(5.7)

(5.8)

from which the generalized elasticity matrix (written without units) can be assembled as follows: ⎡

⎤ 446434.906 130151.508 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 ⎢130151.508 446434.906 ⎥ ⎢ ⎥ 0.0 158141.699 0.0 0.0 0.0 0.0 ⎢ ⎥ ∗ C =⎢ ⎥. 0.0 0.0 0.0 2191204.439 1115508.171 347377.607 ⎥ ⎢ ⎣ 0.0 0.0 0.0 1115508.171 1728034.297 347377.607 ⎦ 0.0 0.0 0.0 347377.607 347377.607 1264789.188

(5.9)

5.2 Problem 1: Stresses and Strains in a Symmetric Laminate

99

⑤ Calculate the generalized compliance matrix (C ∗ )−1 based on Eqs. (4.17)– (4.19). The evaluation of the submatrices A' , B' , and C ' according to Eqs. (4.17)–(4.19) gives ⎡

24.480334 −7.136880 A' = ⎣−7.136880 24.480334 0.0 0.0 ⎡ ⎤ 0.0 0.0 0.0 1 B' = ⎣0.0 0.0 0.0⎦ , 0.0 0.0 0.0 N ⎡ 6.856755 −4.284248 D' = ⎣−4.284248 8.801992 −0.706546 −1.240810

⎤ 0.0 10−7 mm 0.0 ⎦ , N 63.234429

(5.10)

(5.11) ⎤ −0.706546 10−7 −1.240810⎦ , 8.441302 Nmm

(5.12)

from which the generalized compliance matrix (written without units) can be assembled as follows: ⎡

(C ∗ )−1

⎤ 0.0 0.0 0.0 0.0 24.480334 −7.136880 0.0 0.0 0.0 0.0 ⎢−7.136880 24.480334 ⎥ ⎢ ⎥ 0.0 0.0 63.234429 0.0 0.0 0.0 ⎢ ⎥ −7 =⎢ ⎥ 10 . 0.0 0.0 0.0 6.856755 −4.284248 −0.706546⎥ ⎢ ⎣ 0.0 0.0 0.0 −4.284248 8.801992 −1.240810⎦ 0.0 0.0 0.0 −0.706546 −1.240810 8.441302

(5.13)

⎡ ⎤T ⑥ Calculate the generalized strains e = ε0 κ according to Eq. (4.16).

The evaluation of Eq. (4.16) gives the generalized strains as follows: ⎤ ⎡ ⎤ εx0 2.448033 0 ⎢ ⎥ ⎢−0.713688⎥ ⎢ ⎥ ⎡ 0 ⎤ ⎢ ε0y ⎥ ⎢γx y ⎥ ⎢ ⎥0 ε 0.0 ∗ −1 ⎥ ⎢ ⎢ ⎥ /00 . = ⎢ ⎥ = (C ) s = ⎢ ⎥ 0.0 κ κx ⎥ ⎢ ⎢ ⎥

  ⎣ κ ⎦ ⎣ ⎦ 0.0 y e 0.0 κx y ⎡

(5.14)

100

5 Example Problems

⑦ Calculate the stresses in each layer k according to Eq. (4.20), σ kx,y (z) = ( ) C k ε0 + zκ , in the x–y laminate system ) z ≤ z k ). The strains in the ( (z k−1 ≤ x–y laminate system are obtained from ε0 + zκ .

Evaluation of Eq. (4.20) for each layer k, i.e., ) ( σ kx,y (z) = C k ε0 + zκ ,

(5.15)

under consideration of the generalized strains given in Eq. (5.14) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 5.2 gives the stresses for each layer in the x–y laminate system. ( In case) that the strains in the x–y laminate system are required, the evaluation of ε 0 + zκ provides these values. A graphical representation of the normal stress (σ x ) and normal strain (ε x ) in loading direction (x) is provided in Fig. 5.2 as a function of the laminate thickness (z-coordinate). It can be seen that a uniaxial load by N xn results in the case of this symmetric laminate in symmetric distributions of the normal stress and strain with respect to z = 0. The strain distribution is constant (see Fig. 5.2b) across the entire thickness of the laminate. The stress distribution (see Fig. 5.2a) reveals a layer-wise constant distribution with the characteristic that the stress is proportional to the stiffness value of each layer in the loading direction. Furthermore, it can be concluded that the ± 45◦ layers show the same behavior in the x-direction. ⑧ Transform the stresses in each layer to the 1–2 lamina system according to Eq. (4.21): σ k1,2 = T kσ σ kx,y . The strains in the 1–2 lamina system are obtained from Eq. (3.33): ε k1,2 = (C k )−1 σ k1,2 . The transformation of the stresses and strains into the 1–2 lamina systems is performed according to Eqs. (4.21) and (3.33). The corresponding numerical values for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Tables 5.3 and 5.4. These values could be used for a subsequent failure analysis in each lamina. • Laminate loaded by Mxn = 1000 Nmm/mm The solution procedure will follow again the recommended steps provided in Table 4.1. However, subtasks ① and ③–⑤ are identical to load case (a) and can be omitted here. Thus, only the steps with different content/results will be presented in the following.

5.2 Problem 1: Stresses and Strains in a Symmetric Laminate (a)

101

z in mm

4 symmetric laminate 3 2 1

−300

−200

−100

100

200

300

1

2

3

σx in MPa

−1 −2 −3 −4

(b)

z in mm

4 3 2 1

−3

−2

−1

εx in

−1 −2 −3 −4

Fig. 5.2 Distribution of selected field quantities over the laminate thickness for the tensile load case: a normal stress σx (z) and b normal strain εx (z)

102

5 Example Problems

Table 5.3 Stresses in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm, symmetric laminate Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

4

5

6

7

8

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

112.706 112.706 112.706 112.706 112.706 112.706 308.365 308.365 308.365 − 82.953 − 82.953 − 82.953 − 82.953 − 82.953 − 82.953 308.365 308.365 308.365 112.706 112.706 112.706 112.706 112.706 112.706

12.294 12.294 12.294 12.294 12.294 12.294 − 0.313 − 0.313 − 0.313 24.901 24.901 24.901 24.901 24.901 24.901 − 0.313 − 0.313 − 0.313 12.294 12.294 12.294 12.294 12.294 12.294

− 20.867 − 20.867 − 20.867 20.867 20.867 20.867 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 20.867 20.867 20.867 − 20.867 − 20.867 − 20.867

② Define the column matrix of generalized stresses: ⎤T ⎡ s = N xn N yn N xny Mxn M yn Mxny .

The load matrix for case (b) is given as follows: ⎡ ⎤T s = 0 0 0 1000 0 0 Nmm/mm .

⎡ ⎤T ⑥ Calculate the generalized strains e = ε0 κ according to Eq. (4.16).

(5.16)

5.2 Problem 1: Stresses and Strains in a Symmetric Laminate

103

Table 5.4 Strains in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm, symmetric laminate Position ε1 in 0/00 ε2 in 0/00 γ12 in 0/00 Lamina No. 1

2

3

4

5

6

7

8

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

0.867173 0.867173 0.867173 0.867173 0.867173 0.867173 2.448033 2.448033 2.448033 − 0.713688 − 0.713688 − 0.713688 − 0.713688 − 0.713688 − 0.713688 2.448033 2.448033 2.448033 0.867173 0.867173 0.867173 0.867173 0.867173 0.867173

0.867173 0.867173 0.867173 0.867173 0.867173 0.867173 − 0.713688 − 0.713688 − 0.713688 2.448033 2.448033 2.448033 2.448033 2.448033 2.448033 − 0.713688 − 0.713688 − 0.713688 0.867173 0.867173 0.867173 0.867173 0.867173 0.867173

− 3.161721 − 3.161721 − 3.161721 3.161721 3.161721 3.161721 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 3.161721 3.161721 3.161721 − 3.161721 − 3.161721 − 3.161721

The evaluation of Eq. (4.16) gives the generalized strains as follows: ⎤ ⎡ ⎤ εx0 0.0 ⎢ 0⎥ ⎢ ⎥ 0.0 ⎢ ⎥ ⎡ 0 ⎤ ⎢ ε0y ⎥ ⎥ ⎢ ⎢ ⎥0 γx y ⎥ ε 0.0 ∗ −1 ⎢ ⎥ /00 . =⎢ ) s = = (C ⎥ ⎢ ⎢ 0.685676 ⎥ κ κx ⎥ ⎢ ⎢ ⎥

  ⎣ κ ⎦ ⎣−0.428425⎦ y e −0.070655 κx y ⎡

(5.17)

104

5 Example Problems

⑦ Calculate the stresses in each layer k according to Eq. (4.20), σ kx,y (z) = ( ) C k ε0 + zκ , in the x–y laminate system ) z ≤ z k ). The strains in the ( (z k−1 ≤ x–y laminate system are obtained from ε0 + zκ .

Evaluation of Eq. (4.20) for each layer k, i.e., ) ( σ kx,y (z) = C k ε0 + zκ ,

(5.18)

under consideration of the generalized strains given in Eq. (5.17) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 5.2 gives the stresses for each layer in the x–y laminate system. ( In case) that the strains in the x–y laminate system are required, the evaluation of ε 0 + zκ provides these values. A graphical representation of the normal stress (σ x ) and normal strain (ε x ) in loading direction (x) is provided in Fig. 5.3 as a function of the laminate thickness (z-coordinate). It can be seen that a pure bending load by Mxn results in the case of this symmetric laminate in a perfect linear distribution of the strain across the entire thickness of the laminate (see Fig. 5.3b). The stress distribution (see Fig. 5.3a) reveals a layer-wise linear distribution with the characteristic that the stress is proportional to the stiffness value of each layer in the loading direction. Again, it can be concluded that the ± 45◦ layers show the same behavior in the x-direction. Furthermore, the stress distribution shows the classical characteristics known from isotropic bending problems, i.e., the neural fiber (σx = 0) is in the middle of the cross section and the classical tension (z > 0) and compression parts (z < 0) are as well obtained. ⑧ Transform the stresses in each layer to the 1–2 lamina system according to Eq. (4.21): σ k1,2 = T kσ σ kx,y . The strains in the 1–2 lamina system are obtained from Eq. (3.33): ε k1,2 = (C k )−1 σ k1,2 . The transformation of the stresses and strains into the 1–2 lamina systems is performed according to Eqs. (4.21) and (3.33). The corresponding numerical values for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Tables 5.5 and 5.6. These values could be used for a subsequent failure analysis in each lamina.

5.2 Problem 1: Stresses and Strains in a Symmetric Laminate

(a)

105

z in mm

4 symmetric laminate 3 2 1

−200

−100

100

σx in MPa

200

−1 −2 −3 −4

(b)

z in mm

4 3 2 1

−3

−2

−1

1

2

3

εx in

−1 −2 −3 −4

Fig. 5.3 Distribution of selected field quantities over the laminate thickness for the bending load case: a normal stress σx (z) and b normal strain εx (z)

106

5 Example Problems

Table 5.5 Stresses in the local 1–2 coordinate systems for an external loading of Mxn = 1000 Nmm/mm, symmetric laminate Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

4

5

6

7

8

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

− 37.035 − 43.208 − 49.380 − 42.179 − 52.724 − 63.269 − 85.662 − 128.493 − 171.323 0.0 26.114 52.227 − 52.227 − 26.114 0.0 171.324 128.493 85.662 63.269 52.724 42.179 49.380 43.208 37.035

− 6.316 − 7.368 − 8.421 − 3.084 − 3.855 − 4.625 2.619 3.928 5.237 0.0 − 3.133 − 6.266 6.266 3.133 0.0 − 5.237 − 3.928 − 2.619 4.625 3.855 3.084 8.421 7.368 6.316

22.059 25.736 29.412 − 14.706 − 18.383 − 22.059 0.466 0.699 0.933 0.0 − 0.233 − 0.466 0.466 0.233 0.0 − 0.933 − 0.699 − 0.466 22.059 18.383 14.706 − 29.412 − 25.736 − 22.059

5.3 Problem 2: Stresses and Strains in an Asymmetric Laminate Given is an asymmetric 8-layer laminate of thickness t = 8 mm as shown in Fig. 5.4. Each lamina has the same thickness (t/8) and identical material properties, which can be taken from Table 5.1. However, the orientation of each lamina changes from layer to layer as indicated in Fig. 5.4. Consider (a) a pure tensile load case with N xn = 1000 N/mm (all other forces and moments are equal to zero) and (b) a pure bending load case with Mxn = 1000 Nmm/mm (all other moments and forces are equal to zero) to determine the stresses and strains in each lamina in the global x–y coordinate system as well as in the principal material coordinates (1–2 coordinate systems) of each lamina.

5.3 Problem 2: Stresses and Strains in an Asymmetric Laminate

107

Table 5.6 Strains in the local 1–2 coordinate systems for an external loading of Mxn = 1000 Nmm/mm, symmetric laminate Position ε1 in 0/00 ε2 in 0/00 γ12 in 0/00 Lamina No. 1

2

3

4

5

6

7

8

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

Fig. 5.4 Asymmetric laminate: [± 45◦ /0◦ /90◦ /0◦ /90◦ / ± 45◦ ]

− 0.279894 − 0.326543 − 0.373192 − 0.327905 − 0.409882 − 0.491858 − 0.685676 − 1.028513 − 1.371351 0.0 0.214212 0.428425 − 0.428425 − 0.214212 0.0 1.371351 1.028513 0.685675 0.491855 0.409882 0.327905 0.373192 0.326543 0.279894

− 0.491858 − 0.573834 − 0.655810 − 0.186596 − 0.233245 − 0.279894 0.428425 0.642637 0.856850 0.0 − 0.342838 − 0.685676 0.685675 0.342838 0.0 − 0.856850 − 0.642637 − 0.428425 0.279894 0.233245 0.186596 0.655810 0.573834 0.491858

3.342301 3.899351 4.456401 − 2.228201 − 2.785251 − 3.342301 0.070655 0.105982 0.141309 0.0 − 0.035327 − 0.070655 0.070655 0.035327 0.0 − 0.141309 − 0.105982 − 0.070655 3.342301 2.785251 2.228201 − 4.456401 − 3.899351 − 3.342301

108

5 Example Problems

Table 5.7 Geometrical properties (z k ; z k−1 ; αk ) of the laminate shown in Fig. 5.4 Lamina k Coordinate z k−1 in Coordinate z k in mm Angle αk (◦ ) mm 1 2 3 4 5 6 7 8

−4 −3 −2 −1 0 1 2 3

−3 −2 −1 0 1 2 3 4

+ 45 − 45 0 90 0 90 + 45 − 45

• Laminate loaded by N xn = 1000 N/mm The solution procedure will follow the recommended steps provided in Table 4.1. ① Define for each lamina k the material (E 1,k ; E 2,k ; ν12,k ; G 12,k ) and the geometrical (z k ; z k−1 ; αk ) properties. The material properties (E 1,k ; E 2,k ; ν12,k ; G 12,k ) are the same for each lamina and can be taken from Table 5.1. The geometrical properties for each lamina are summarized in Table 5.7. ② Define the column matrix of generalized stresses: ⎤T ⎡ s = N xn N yn N xny Mxn M yn Mxny .

The load matrix for case (a) is given as follows: ⎡ ⎤T s = 1000 0 0 0 0 0 N/mm .

(5.19)

③ Calculate for each layer k the elasticity matrix C k in the 1–2 lamina system according to Eq. (3.31). Transform each matrix according to Eq. (3.49) to obtain the elasticity matrix C k in the x–y laminate system. Since each lamina is composed of the same materials and the thickness is equal to t/8 = 1 mm, Eq. (3.31) yields for each lamina the same matrix:

5.3 Problem 2: Stresses and Strains in an Asymmetric Laminate

109

⎤ E1 ν21 E 1 ⎤ 0 ⎥ ⎡ ⎢1 − ν ν 1 − ν ν 126868.343 3101.226 0 12 21 12 21 ⎥ ⎢ N ⎥ = ⎣ 3101.226 11075.808 0 ⎦ E2 Ck = ⎢ . ⎥ ⎢ ν12 E 2 mm2 0 ⎦ ⎣1 − ν ν 1 − ν ν 0 0 6600.0 12 21 12 21 0 0 G 12 (5.20) The transformation to the x–y laminate system according to Eq. (3.49) gives the following matrices: ⎡

⎡

C [± 45◦ ]

C [0◦ ]

C [90◦ ]

⎤ 42636.651 29436.651 ±28948.134 N = ⎣ 29436.651 42636.651 ±28948.134⎦ , 2 ±28948.134 ±28948.134 32935.425 mm ⎡ ⎤ 126868.343 3101.226 0.0 N = ⎣ 3101.226 11075.808 0.0 ⎦ , 2 0.0 0.0 6600.0 mm ⎡ ⎤ 11075.808 3101.226 0.0 N = ⎣ 3101.226 126868.343 0.0 ⎦ . 2 0.0 0.0 6600.0 mm

(5.21)

(5.22)

(5.23)

④ Calculate the submatrices A, B, and C according to Eqs. (4.13)–(4.15). Assemble the generalized elasticity matrix C ∗ according to Eq. (4.12).

The evaluation of the submatrices A, B, and C according to Eqs. (4.13)–(4.15) gives ⎡

⎤ 446434.906 130151.508 0.0 N ⎦ 0.0 A= , Ak = ⎣130151.508 446434.906 0.0 0.0 158141.699 mm k =1 ⎡ ⎤ 8 −115792.536 0.0 −57896.268 ∑ 0.0 115792.536 −57896.268⎦ N , B= Bk = ⎣ −57896.268 −57896.268 0.0 k =1 ⎡ ⎤ 8 1959619.368 1115508.171 0.0 ∑ ⎦ Nmm , 0.0 D= Dk = ⎣1115508.171 1959619.368 0.0 0.0 1264789.188 k =1 8 ∑

(5.24)

(5.25)

(5.26)

from which the generalized elasticity matrix (written without units) can be assembled as follows:

110

5 Example Problems ⎡

446434.906 ⎢ 130151.508 ⎢ 0.0 ⎢ C∗ = ⎢ ⎢−115792.536 ⎣ 0.0 −57896.268

⎤ 130151.508 0.0 −115792.536 0.0 −57896.268 446434.906 0.0 0.0 115792.536 −57896.268 ⎥ ⎥ 0.0 158141.699 −57896.268 −57896.268 0.0 ⎥ ⎥. 0.0 −57896.268 1959619.368 1115508.171 0.0 ⎥ ⎦ 115792.536 −57896.268 11115508.171 1959619.368 0.0 −57896.268 0.0 0.0 0.0 1264789.188

(5.27)

⑤ Calculate the generalized compliance matrix (C ∗ )−1 based on Eqs. (4.17)– (4.19). The evaluation of the submatrices A' , B' , and C ' according to Eqs. (4.17)–(4.19) gives ⎡

⎤ 25.035205 −7.030225 0.448216 10−7 mm , A' = ⎣−7.030225 25.035205 −0.448216⎦ N 0.448216 −0.448216 64.130861 ⎡ ⎤ 1.847077 −0.622790 0.824186 10−7 B' = ⎣−0.622790 −1.847077 0.824186⎦ , N 1.224287 1.224287 0.0 ⎡ ⎤ 7.764843 −4.420746 0.113059 10−7 D' = ⎣−4.420746 7.764843 −0.113059⎦ , 0.113059 −0.113059 7.981910 Nmm

(5.28)

(5.29)

(5.30)

from which the generalized compliance matrix (written without units) can be assembled as follows: ⎡

(C ∗ )−1

25.035205 ⎢−7.030225 ⎢ ⎢ 0.448216 =⎢ ⎢ 1.847077 ⎣−0.622790 0.824186

−7.030225 25.035205 −0.448216 0.622790 −1.847077 0.824186

0.448216 −0.448216 64.130861 1.224287 1.224287 0.0

1.847077 0.622790 1.224287 7.764843 −4.420746 0.113059

⎤ −0.622790 0.824186 −1.847077 0.824186 ⎥ ⎥ 1.224287 −3.874189 10−18 ⎥ −7 ⎥ 10 . −4.420746 0.113059 ⎥ 7.764843 −0.113059 ⎦ −0.113059 7.981911

(5.31)

⎡ ⎤T ⑥ Calculate the generalized strains e = ε0 κ according to Eq. (4.16).

5.3 Problem 2: Stresses and Strains in an Asymmetric Laminate

The evaluation of Eq. (4.16) gives the generalized strains as follows: ⎡ 0⎤ ⎡ ⎤ εx 2.503521 ⎢ 0⎥ ⎢−0.703023 ⎥ ⎢ ⎥ ⎡ 0 ⎤ ⎢ ε0y ⎥ ⎥ ⎢ ⎢ 0.044822⎥ 0 γx y ⎥ ε ∗ −1 ⎢ ⎥ =⎢ ) s = = (C ⎢ κx ⎥ ⎢ 0.184708⎥ /00 . κ ⎥ ⎢ ⎢ ⎥

  ⎣ κ ⎦ ⎣−0.062279 ⎦ y e 0.082419 κx y

111

(5.32)

⑦ Calculate the stresses in each layer k according to Eq. (4.20), σ kx,y (z) = ( ) C k ε0 + zκ , in the x–y laminate system ) z ≤ z k ). The strains in the ( (z k−1 ≤ x–y laminate system are obtained from ε0 + zκ .

Evaluation of Eq. (4.20) for each layer k, i.e., ( ) σ kx,y (z) = C k ε0 + zκ ,

(5.33)

under consideration of the generalized strains given in Eq. (5.32) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 5.7 gives the stresses for each layer in the x–y laminate system. ( In case) that the strains in the x–y laminate system are required, the evaluation of ε 0 + zκ provides these values. A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) is provided in Fig. 5.5 as a function of the laminate thickness (z-coordinate). It can be seen that a uniaxial load by N xn results in the case of this asymmetric laminate in an asymmetric distribution of the normal stress and strain with respect to to z = 0. The strain distribution is linear and continuous (see Fig. 5.5b) across the entire thickness of the laminate. The stress distribution (see Fig. 5.5a) reveals a layer-wise linear distribution with the characteristic that the stress is proportional to the stiffness value of each layer in the loading direction. It can be seen that we have a clear tension–bending coupling, i.e., a superposition of a constant and linear distribution for the stresses and strains. This is also manifested in the B-matrix where the bending–tension entries are now unequal to zero, see Eq. (5.25). ⑧ Transform the stresses in each layer to the 1–2 lamina system according to Eq. (4.21): σ k1,2 = T kσ σ kx,y . The strains in the 1–2 lamina system are obtained from Eq. (3.33): ε k1,2 = (C k )−1 σ k1,2 . The transformation of the stresses and strains into the 1–2 lamina systems is performed according to Eqs. (4.21) and (3.33). The corresponding numerical values for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina

112

5 Example Problems

(a)

z in mm

4 asymm. laminate 3 2 1

−300

−200

−100

100

200

300

1

2

3

σx in MPa

−1 −2 −3 −4

(b)

z in mm

4 3 2 1

−3

−2

−1

εx in

−1 −2 −3 −4

Fig. 5.5 Distribution of selected field quantities over the laminate thickness for the tensile load case: a normal stress σx (z) and b normal strain εx (z)

5.3 Problem 2: Stresses and Strains in an Asymmetric Laminate

113

Table 5.8 Stresses in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm, asymmetric laminate Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

4

5

6

7

8

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

80.610 74.081 67.553 108.520 107.092 105.664 292.197 280.577 268.956 − 81.427 − 77.763 − 74.099 338.678 327.057 315.437 − 96.084 − 92.420 − 88.756 158.948 152.420 145.891 125.654 124.226 122.798

10.967 10.697 10.427 10.549 9.950 9.352 0.094 0.153 0.211 25.548 24.622 23.696 − 0.140 − 0.081 − 0.023 29.254 28.327 27.401 14.202 13.932 13.663 17.727 17.129 16.531

− 16.273 − 15.458 − 14.643 17.903 17.088 16.273 − 0.248 − 0.520 − 0.792 − 0.296 − 0.024 0.248 0.840 0.568 0.296 − 1.384 − 1.112 − 0.840 − 26.054 − 25.238 − 24.423 27.684 26.869 26.054

are summarized in Tables 5.8 and 5.9. These values could be used for a subsequent failure analysis in each lamina. • Laminate loaded by Mxn = 1000 Nmm/mm The solution procedure will follow again the recommended steps provided in Table 4.1. However, subtasks ① and ③–⑤ are identical to load case (a) and can be omitted here. Thus, only the steps with different content/results will be presented in the following. ② Define the column matrix of generalized stresses: ⎤T ⎡ s = N xn N yn N xny Mxn M yn Mxny .

114

5 Example Problems

Table 5.9 Strains in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm, asymmetric laminate Position ε1 in 0/00 ε2 in 0/00 γ12 in 0/00 Lamina No. 1

2

3

4

5

6

7

8

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

0.615389 0.564177 0.512965 0.837828 0.827826 0.817823 2.318813 2.226459 2.134105 − 0.703023 − 0.671883 − 0.640744 2.688228 2.595874 2.503521 − 0.827581 − 0.796441 − 0.765302 1.229931 1.178719 1.127507 0.957858 0.947856 0.937853

0.817823 0.807821 0.797818 0.717813 0.666601 0.615389 − 0.640744 − 0.609604 − 0.578465 2.503521 2.411167 2.318813 − 0.765302 − 0.734162 − 0.703023 2.872936 2.780582 2.688228 0.937853 0.927851 0.917848 1.332354 1.281142 1.229931

− 2.465583 − 2.342090 − 2.218596 2.712570 2.589076 2.465583 − 0.037597 − 0.078806 − 0.120016 − 0.044822 − 0.003612 0.037597 0.127240 0.086031 0.044822 − 0.209659 − 0.168449 − 0.127240 − 3.947503 − 3.824010 − 3.700516 4.194490 4.070996 3.947503

The load matrix for case (b) is given as follows: ⎡ ⎤T s = 0 0 0 1000 0 0 Nmm/mm .

⎡ ⎤T ⑥ Calculate the generalized strains e = ε0 κ according to Eq. (4.16).

(5.34)

5.3 Problem 2: Stresses and Strains in an Asymmetric Laminate

115

The evaluation of Eq. (4.16) gives the generalized strains as follows: ⎤ ⎡ ⎤ εx0 0.184708 ⎢ 0⎥ ⎢ 0.062279 ⎥ ⎢ ⎥ ⎡ 0 ⎤ ⎢ ε0y ⎥ ⎢γx y ⎥ ⎢ 0.122429 ⎥ 0 ε ∗ −1 ⎥ ⎢ ⎢ ⎥ /00 . = ⎢ ⎥ = (C ) s = ⎢ 0.776484 ⎥ κ κx ⎥ ⎢ ⎢ ⎥

  ⎣ κ ⎦ ⎣−0.442075⎦ y e 0.011306 κx y ⎡

(5.35)

⑦ Calculate the stresses in each layer k according to Eq. (4.20), σ kx,y (z) = ( ) C k ε0 + zκ , in the x–y laminate system ) z ≤ z k ). The strains in the ( (z k−1 ≤ x–y laminate system are obtained from ε0 + zκ .

Evaluation of Eq. (4.20) for each layer k, i.e., ) ( σ kx,y (z) = C k ε0 + zκ ,

(5.36)

under consideration of the generalized strains given in Eq. (5.35) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 5.7 gives the stresses for each layer in the x–y laminate system. ( In case) that the strains in the x–y laminate system are required, the evaluation of ε 0 + zκ provides these values. A graphical representation of the normal stress (σ x ) and normal strain (ε x ) in loading direction (x) is provided in Fig. 5.6 as a function of the laminate thickness (z-coordinate). It can be seen that a pure bending load by Mxn results in the case of this asymmetric laminate in a perfect linear distribution of the strain across the entire thickness of the laminate (see Fig. 5.6b). The stress distribution (see Fig. 5.6a) reveals a layer-wise linear distribution with the characteristic that the stress is proportional to the stiffness value of each layer in the loading direction. We have again a tension–bending coupling, i.e., a superposition of a constant and linear distribution for the stresses and strains. This is also manifested in the B-matrix where the bending–tension entries are now unequal to zero, see Eq. (5.25). ⑧ Transform the stresses in each layer to the 1–2 lamina system according to Eq. (4.21): σ k1,2 = T kσ σ kx,y . The strains in the 1–2 lamina system are obtained from Eq. (3.33): ε k1,2 = (C k )−1 σ k1,2 . The transformation of the stresses and strains into the 1–2 lamina systems is performed according to Eqs. (4.20) and (3.33). The corresponding numerical values

116

5 Example Problems z in mm

(a)

4 asymm. laminate 3 2 1

−200

−100

100

σx in MPa

200

−1 −2 −3 −4

z in mm

(b)

4 3 2 1

−3

−2

−1

1

2

3

εx in

−1 −2 −3 −4

Fig. 5.6 Distribution of selected field quantities over the laminate thickness for the bending load case: a normal stress σx (z) and b normal strain εx (z)

5.4 Problem 3: Failure Criteria

117

Table 5.10 Stresses in the local 1–2 coordinate systems for an external loading of Mxn = 1000 Nmm/mm, asymmetric laminate Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

4

5

6

7

8

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

− 43.667 54.882 − 66.098 − 33.590 − 44.106 − 54.622 − 73.514 − 122.084 − 170.654 8.474 35.313 62.151 120.767 72.197 23.627 − 98.880 − 72.042 − 45.203 90.920 79.705 68.489 92.602 82.086 71.570

− 5.714 − 6.876 − 8.039 − 2.592 − 3.800 − 5.008 3.751 4.995 6.239 2.239 − 1.376 − 4.990 − 1.226 0.018 1.263 16.697 13.083 9.468 8.239 7.076 5.913 11.901 10.693 9.486

23.319 27.341 31.362 − 15.277 − 19.298 − 23.319 0.733 0.696 0.659 − 0.808 − 0.771 − 0.733 0.883 0.845 0.808 − 0.957 − 0.920 − 0.883 − 24.935 − 20.914 − 16.893 32.978 28.957 24.935

for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Tables 5.10 and 5.11. These values could be used for a subsequent failure analysis in each lamina.

5.4 Problem 3: Failure Criteria Given is an asymmetric 4-layer laminate of thickness t = 8 mm as shown in Fig. 5.7. Each lamina has the same thickness (t/4) and identical material properties, which can be taken from Table 5.1. However, the orientation of each lamina changes from layer to layer as indicated in Fig. 5.7. Consider (a) a biaxial tensile load case with N xn = 1000 N/mm and N yn = 500 N/mm (all other forces and moments are equal to zero) and (b) a biaxial bending

118

5 Example Problems

Table 5.11 Strains in the local 1–2 coordinate systems for an external loading of Mxn = 1000 Nmm/mm, asymmetric laminate Position ε1 in 0/00 ε2 in 0/00 γ12 in 0/00 Lamina No. 1

2

3

4

5

6

7

8

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

− 0.333866 − 0.420295 − 0.506724 − 0.260825 − 0.341601 − 0.422377 − 0.591777 − 0.980019 − 1.368261 0.062279 0.283316 0.504354 0.961192 0.572950 0.184708 − 0.821870 − 0.600833 − 0.379796 0.703281 0.616852 0.530423 0.708487 0.627711 0.546935

− 0.422377 − 0.503153 − 0.583929 − 0.161008 − 0.247437 − 0.333866 0.504354 0.725391 0.946428 0.184708 − 0.203534 − 0.591777 − 0.379796 − 0.158758 0.062279 1.737676 1.349434 0.961192 0.546935 0.466159 0.385383 0.876139 0.789710 0.703281

3.533248 4.142527 4.751807 − 2.314689 − 2.923969 − 3.533248 0.111123 0.105470 0.099817 − 0.122429 − 0.116776 − 0.111123 0.133735 0.128082 0.122429 − 0.145041 − 0.139388 − 0.133735 − 3.778105 − 3.168826 − 2.559546 4.996664 4.387384 3.778105

Fig. 5.7 Asymmetric laminate: [+45◦ / 0◦ / + 30◦ / − 45◦ ]

load case with Mxn = 1000 Nmm/mm and M yn = 500 Nmm/mm (all other moments and forces are equal to zero) to determine the stresses and strains in each lamina in the global x–y coordinate system as well as in the principal material coordinates (1–2 coordinate systems) of each lamina. Use these stress and strain values to perform a failure analysis based on the maximum stress, maximum strain, Tsai–Hill, and Tsai–Wu failure criteria. • Laminate loaded by N xn = 1000 N/mm and N yn = 500 N/mm The solution procedure will follow the recommended steps provided in Table 4.1.

5.4 Problem 3: Failure Criteria

119

① Define for each lamina k the material (E 1,k ; E 2,k ; ν12,k ; G 12,k ) and the geometrical (z k ; z k−1 ; αk ) properties. The material properties (E 1,k ; E 2,k ; ν12,k ; G 12,k ) are the same for each lamina and can be taken from Table 5.1. The geometrical properties for each lamina are summarized in Table 5.12. ② Define the column matrix of generalized stresses: ⎤T ⎡ s = N xn N yn N xny Mxn M yn Mxny .

The load matrix for case (a) is given as follows: ⎡ ⎤T s = 1000 500 0 0 0 0 N/mm .

(5.37)

③ Calculate for each layer k the elasticity matrix C k in the 1–2 lamina system according to Eq. (3.31). Transform each matrix according to Eq. (3.49) to obtain the elasticity matrix C k in the x–y laminate system. Since each lamina is composed of the same materials and the thickness is equal to t/4 = 2 mm, Eq. (3.31) yields for each lamina the same matrix: ⎤ E1 ν21 E 1 ⎤ 0 ⎥ ⎡126868.343 3101.226 ⎢1 − ν ν 1 − ν ν 0 ⎥ 12 21 12 21 ⎢ N ⎥ = ⎣ 3101.226 11075.808 0 ⎦ E2 Ck = ⎢ . ⎥ ⎢ ν12 E 2 mm2 0 ⎦ ⎣1 − ν ν 1 − ν ν 0 0 6600.0 12 21 12 21 0 0 G 12 (5.38) ⎡

The transformation to the x–y laminate system according to Eq. (3.49) gives the following matrices: ⎡

C [±45◦ ]

C [0◦ ]

42636.651 = ⎣ 29436.651 ±28948.134 ⎡ 126868.343 = ⎣ 3101.226 0.0

⎤ 29436.651 ±28948.134 N 42636.651 ±28948.134⎦ , 2 ±28948.134 32935.425 mm ⎤ 3101.226 0.0 N 11075.808 0.0 ⎦ , 2 0.0 6600.0 mm

(5.39)

(5.40)

120

5 Example Problems

⎡

C [+30◦ ]

⎤ 78168.641 22852.795 36473.393 N = ⎣22852.795 20272.373 13666.246⎦ . 2 36473.393 13666.246 26351.569 mm

(5.41)

④ Calculate the submatrices A, B, and C according to Eqs. (4.13)–(4.15). Assemble the generalized elasticity matrix C ∗ according to Eq. (4.12).

The evaluation of the submatrices A, B, and C according to Eqs. (4.13)–(4.15) gives ⎡

⎤ 580620.572 169654.645 72946.785 N Ak = ⎣169654.645 233242.965 27332.492 ⎦ , A= 72946.785 27332.492 197644.836 mm k =1 ⎡ ⎤ 4 −97399.405 39503.137 −274430.821 ∑ 18393.131 −320045.115⎦ N , B= Bk = ⎣ 39503.137 −274430.821 −320045.115 39503.1370 k =1 ⎡ ⎤ 4 2138533.590 1168179.021 97262.381 ∑ Dk = ⎣1168179.021 1675363.448 36443.323 ⎦ Nmm , D= 97262.381 36443.323 1317460.038 k =1 4 ∑

(5.42)

(5.43)

(5.44)

from which the generalized elasticity matrix (written without units) can be assembled as follows: ⎡

580620.572 ⎢ 169654.645 ⎢ ⎢ 72946.785 C∗ = ⎢ ⎢ −97399.405 ⎣ 39503.137 −274430.821

169654.645 233242.965 27332.492 39503.137 18393.131 −320045.115

72946.785 27332.492 197644.836 −274430.821 −320045.115 39503.137

−97399.405 39503.137 −274430.821 2138533.590 1168179.021 97262.381

39503.137 18393.131 −320045.115 1168179.021 1675363.448 36443.323

⎤ −274430.821 −320045.115⎥ ⎥ 39503.137 ⎥ ⎥. 97262.381 ⎥ 36443.323 ⎦ 1317460.038

(5.45)

⑤ Calculate the generalized compliance matrix (C ∗ )−1 based on Eqs. (4.17)– (4.19). The evaluation of the submatrices A' , B' , and C ' according to Eqs. (4.17)–(4.19) gives ⎡

⎤ 24.132386 −13.453542 −11.136057 10−7 mm , A' = ⎣−13.453542 79.436193 −17.624231⎦ N −11.136057 −17.624231 88.407562

(5.46)

5.4 Problem 3: Failure Criteria

121

⎡

⎤ 2.005649 −3.991813 2.054893 10−7 B' = ⎣−4.496308 −1.164721 17.387305 ⎦ , N 3.266024 15.282979 −9.915766 ⎡ ⎤ 8.025880 −4.943500 −1.228186 10−7 D' = ⎣−4.943500 12.476002 −1.552848⎦ , −1.228186 −1.552848 12.673173 Nmm

(5.47)

(5.48)

from which the generalized compliance matrix (written without units) can be assembled as follows: ⎡

(C ∗ )−1

24.132386 ⎢−13.453542 ⎢ ⎢−11.136057 =⎢ ⎢ 2.005649 ⎣ −3.991813 2.054893

−13.453542 79.436193 −17.624231 −4.496308 −1.164721 17.387305

−11.136057 −17.624231 88.407562 3.266024 15.282979 −9.915766

2.005649 −4.496308 3.266024 8.025880 −4.943500 −1.228186

−3.991813 −1.164721 15.282979 −4.943500 12.476002 −1.552848

⎤ 2.054893 17.387305 ⎥ ⎥ −9.915766⎥ −7 ⎥ 10 . −1.228186⎥ −1.552848⎦ 12.673173

(5.49)

⎡ ⎤T ⑥ Calculate the generalized strains e = ε0 κ according to Eq. (4.16).

The evaluation of Eq. (4.16) gives the generalized strains as follows: ⎤ ⎡ ⎤ εx0 1.740561 ⎢ 2.626455 ⎥ ⎢ 0⎥ ⎢ ⎥ ⎡ 0 ⎤ ⎢ ε0y ⎥ ⎥ ⎢−1.994817⎥ 0 ⎢ γx y ⎥ ε ∗ −1 ⎢ ⎢ ⎥ /00 . = ⎢ ⎥ = (C ) s = ⎢ −0.024251⎥ κ κx ⎥ ⎢ ⎢ ⎥

  ⎣ κ ⎦ ⎣−0.457417⎦ y e 1.074855 κx y ⎡

(5.50)

⑦ Calculate the stresses in each layer k according to Eq. (4.20), σ kx,y (z) = ( ) C k ε0 + zκ , in the x–y laminate system ) z ≤ z k ). The strains in the ( (z k−1 ≤ x–y laminate system are obtained from ε0 + zκ .

Evaluation of Eq. (4.20) for each layer k, i.e., ( ) σ kx,y (z) = C k ε0 + zκ ,

(5.51)

122

5 Example Problems

Table 5.12 Geometrical properties (z k ; z k−1 ; αk ) of the laminate shown in Fig. 5.7 Lamina k Coordinate z k−1 in Coordinate z k in mm Angle αk (◦ ) mm 1 2 3 4

−4 −2 0 2

−2 0 2 4

+ 45 0 + 30 − 45

under consideration of the generalized strains given in Eq. (5.50) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 5.12 gives the stresses for each layer in the x–y laminate system. ( In case) that the strains in the x–y laminate system are required, the evaluation of ε 0 + zκ provides these values. Graphical representations of the normal stresses (σx , σ y ) and the normal strains (εx , ε y ) in the x- and y-direction are provided in Figs. 5.8 and 5.9 as a function of the laminate thickness (z-coordinate). It can be seen that a biaxial load by N xn and N yn results in the case of this asymmetric laminate in an asymmetric distribution of the normal stress and strain with respect to to z = 0. The strain distributions are both linear and continuous (see Fig. 5.9) across the entire thickness of the laminate. The stress distributions (see Fig. 5.8) reveal a layer-wise linear distribution with the characteristic that the stress is proportional to the stiffness value of each layer in the corresponding loading direction. It can be seen that we have a clear tension–bending coupling, i.e., a superposition of a constant and linear distribution for the stresses and strains. This is also manifested in the B-matrix where the bending–tension entries are unequal to zero, see Eq. (5.42). ⑧ Transform the stresses in each layer to the 1–2 lamina system according to Eq. (4.21): σ k1,2 = T kσ σ kx,y . The strains in the 1–2 lamina system are obtained from Eq. (3.33): ε k1,2 = (C k )−1 σ k1,2 . The transformation of the stresses and strains into the 1–2 lamina systems is performed according to Eqs. (4.21) and (3.33). The corresponding numerical values for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Tables 5.13 and 5.14. These values will be used for the subsequent failure analysis in each lamina. Failure analysis based on the maximum stress, maximum strain, Tsai–Hill, and Tsai–Wu failure criteria.

5.4 Problem 3: Failure Criteria

123 z in mm

(a)

4 asymm. laminate 3 2 1

−200

−100

100

200

σx in MPa

−1 −2 −3 −4

z in mm

(b)

4 asymm. laminate 3 2 1

−200

σy in MPa

−100

100

200

−1 −2 −3 −4

Fig. 5.8 Distribution of selected field quantities over the laminate thickness for the biaxial tensile load case: a normal stress σx (z) and b normal stress σ y (z)

124

5 Example Problems (a)

z in mm

4 asymm. laminate 3 2 1

−5

−4

−3

−2

−1

1

2

3

4

5

εx in

−1 −2 −3 −4

(b)

z in mm

4 asymm. laminate 3 2 1

−5

−4

−3

−2

−1

1

2

3

4

5

εy in

−1 −2 −3 −4

Fig. 5.9 Distribution of selected field quantities over the laminate thickness for the biaxial tensile load case: a normal strain εx (z) and b normal strain ε y (z)

5.4 Problem 3: Failure Criteria

125

Table 5.13 Stresses in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm and N yn = 500 N/mm, asymmetric laminate Lamina No.

Position

σ1 in MPa

σ2 in MPa

σ12 in MPa

1

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

89.914 54.699 19.484 228.967 233.463 237.958 228.883 189.177 149.471 15.968 113.785 211.602

54.310 62.010 69.710 34.488 39.629 44.771 23.631 31.621 39.610 26.488 25.616 24.745

11.565 14.424 17.283 − 13.166 − 20.260 − 27.354 0.623 − 0.448 − 1.519 5.589 2.730 − 0.129

2

3

4

Table 5.14 Strains in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm and N yn = 500 N/mm, asymmetric laminate Lamina No.

Position

ε1 in 0/00

ε2 in 0/00

γ12 in 0/00

1

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

0.592913 0.296320 − 0.000274 1.740561 1.764812 1.789062 1.764021 1.431137 1.098254 0.067872 0.846133 1.624395

4.737439 5.515701 6.293962 2.626455 3.083873 3.541290 1.639661 2.454212 3.268763 2.372473 2.075880 1.779287

1.752228 2.185394 2.618561 − 1.994817 − 3.069672 − 4.144526 0.094386 − 0.067908 − 0.230202 0.846773 0.413606 − 0.019560

2

3

4

To easier compare the different failure criteria, we determine the limit loads by applying loads as N xn = R × 1000 N/mm and N yn = R × 500 N/mm, i.e., we multiply the loads with the strength ratio R. In case of R > 1 there is no failure, whereas R ≤ 1 results in the failure of the respective lamina. Considering the stresses from Table 5.13 in the conditions of the maximum stress criterion according to Eqs. (3.133)–(3.135) gives (k1c 0 (all other forces and moments are equal to zero) to investigate the ply-by-ply failure of this laminate based on the Tsai–Wu criterion. The solution procedure will follow the recommended steps provided in Table 4.1. ① Define for each lamina k the material (E 1,k ; E 2,k ; ν12,k ; G 12,k ) and the geometrical (z k ; z k−1 ; αk ) properties. The material properties (E 1,k ; E 2,k ; ν12,k ; G 12,k ) are the same for each lamina and can be taken from Table 5.1. The geometrical properties for each lamina are summarized in Table 5.19. ② Define the column matrix of generalized stresses: ⎤T ⎡ s = N xn N yn N xny Mxn M yn Mxny .

5.5 Problem 4: Ply-By-Ply Failure Loads

133

Table 5.18 Strength ratio R in the local 1–2 coordinate systems for an external loading of Mxn = R × 1000 Nmm/mm and M yn = R × 500 Nmm/mm, asymmetric laminate Lamina No. Position 1

Top

Center

Bottom

2

Top

Center

Bottom

3

Top

Center

Bottom

4

Top

Center

Bottom

Direction

Max. stress

Max. Strain Tsai–Hill

1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12

48.918 10.390 34.888 23.485 8.031 15.564 15.451 6.545 10.016 999.032 35.574 10.974 20.051 22.784 9.270 10.127 16.758 8.024 14.429 47.000 14.360 22.108 54.401 32.846 47.261 23.883 114.317 15.330 2.827 5.410 27.741 3.715 6.699 145.718 5.413 8.796

61.048 11.875 58.295 27.032 9.404 26.006 17.359 7.784 16.736 14164.355 39.380 18.336 21.193 31.382 15.489 10.587 26.084 13.408 12.893 96.395 23.994 19.486 37.720 54.882 39.876 23.448 191.010 14.200 3.458 9.039 26.079 4.281 11.194 159.533 5.616 14.698

Tsai–Wu

9.886

10.810

6.948

8.295

5.260

6.646

10.488

13.911

7.969

11.903

5.973

9.765

9.966

10.094

17.160

18.236

20.549

19.077

2.477

2.555

3.231

3.241

4.610

4.389

134

5 Example Problems

Fig. 5.12 Symmetric laminate: [0◦ /90◦ ]s

Table 5.19 Geometrical properties (z k ; z k−1 ; αk ) of the laminate shown in Fig. 5.12 Lamina k Coordinate z k−1 in Coordinate z k in mm Angle αk (◦ ) mm − 4.5 −1.5 1.5

1 2 3

− 1.5 1.5 4.5

0 90 0

Let us assume a unit load which facilitates the scaling of any failure loads: ⎡ ⎤T s = 1 0 0 0 0 0 N/mm .

(5.63)

③ Calculate for each layer k the elasticity matrix C k in the 1–2 lamina system according to Eq. (3.31). Transform each matrix according to Eq. (3.49) to obtain the elasticity matrix C k in the x–y laminate system. Since each lamina is composed of the same materials and the thickness is equal to t/3 = 3 mm, Eq. (3.31) yields for each lamina the same matrix: ⎤ E1 ν21 E 1 ⎤ 0 ⎥ ⎡126868.343 3101.226 ⎢1 − ν ν 1 − ν ν 0 12 21 12 21 ⎥ ⎢ N ⎥ = ⎣ 3101.226 11075.808 0 ⎦ E2 Ck = ⎢ . ⎥ ⎢ ν12 E 2 mm2 0 ⎦ ⎣1 − ν ν 1 − ν ν 0 0 6600.0 12 21 12 21 0 0 G 12 (5.64) ⎡

The transformation to the x–y laminate system according to Eq. (3.49) gives the following matrices: ⎡

C [0◦ ]

C [90◦ ]

⎤ 126868.343 3101.226 0.0 N = ⎣ 3101.226 11075.808 0.0 ⎦ , 2 0.0 0.0 6600.0 mm ⎡ ⎤ 11075.808 3101.226 0.0 N = ⎣ 3101.226 126868.343 0.0 ⎦ . 2 0.0 0.0 6600.0 mm

(5.65)

(5.66)

5.5 Problem 4: Ply-By-Ply Failure Loads

135

④ Calculate the submatrices A, B, and C according to Eqs. (4.13)–(4.15). Assemble the generalized elasticity matrix C ∗ according to Eq. (4.12).

The evaluation of the submatrices A, B, and C according to Eqs. (4.13)–(4.15) gives ⎡

⎤ 794437.483 27911.036 0.0 N A= , Ak = ⎣ 27911.036 447059.876 0.0 ⎦ 0.0 0.0 59400.0 mm k =1 ⎡ ⎤ 8 0.0 0.0 0.0 ∑ B= Bk = ⎣0.0 0.0 0.0⎦ N , 0.0 0.0 0.0 k =1 ⎡ ⎤ 8 7446718.652 188399.490 0.0 ∑ 0.0 ⎦ Nmm , D= Dk = ⎣ 188399.490 933388.526 0.0 0.0 400950.0 k =1 8 ∑

(5.67)

(5.68)

(5.69)

from which the generalized elasticity matrix (written without units) can be assembled as follows: ⎡ ⎤ 794437.483 27911.036 0.0 0.0 0.0 0.0 ⎢ 27911.036 447059.876 0.0 0.0 0.0 0.0 ⎥ ⎢ ⎥ ⎢ 0.0 59400.0 0.0 0.0 0.0 ⎥ 0.0 ⎥. C∗ = ⎢ ⎢ 0.0 0.0 0.0 7446718.652 188399.490 0.0 ⎥ ⎢ ⎥ ⎣ 0.0 0.0 0.0 188399.490 933388.526 0.0 ⎦ 0.0 0.0 0.0 0.0 0.0 400950.0 (5.70)

⑤ Calculate the generalized compliance matrix (C ∗ )−1 based on Eqs. (4.17)– (4.19). The evaluation of the submatrices A' , B' , and C ' according to Eqs. (4.17)–(4.19) gives ⎡

⎤ 12.615194 −0.787597 0.0 10−7 mm ⎦ 0.0 A' = ⎣−0.787597 22.417540 , N 0.0 0.0 168.350168 ⎡ ⎤ 0.0 0.0 0.0 1 B' = ⎣0.0 0.0 0.0⎦ , 0.0 0.0 0.0 N

(5.71)

(5.72)

136

5 Example Problems

⎡

⎤ 1.349766 −0.272443 0.0 10−7 ⎦ 0.0 D' = ⎣−0.272443 10.768643 , 0.0 0.0 24.940766 Nmm

(5.73)

from which the generalized compliance matrix (written without units) can be assembled as follows: ⎡

(C ∗ )−1

⎤ 0.0 0.0 0.0 0.0 12.61519 −0.787597 0.0 0.0 0.0 0.0 ⎢−0.787597 22.417540 ⎥ ⎢ ⎥ 0.0 0.0 168.350168 0.0 0.0 0.0 ⎢ ⎥ −7 =⎢ ⎥ 10 . 0.0 0.0 0.0 1.349766 −0.272443 0.0 ⎢ ⎥ ⎣ ⎦ 0.0 0.0 0.0 −0.272443 10.768643 0.0 0.0 0.0 0.0 0.0 0.0 24.940766

(5.74)

⎡ ⎤T ⑥ Calculate the generalized strains e = ε0 κ according to Eq. (4.16).

The evaluation of Eq. (4.16) gives the generalized strains as follows: ⎤ ⎡ ⎤ εx0 1.261519 × 10−3 ⎢ ε0 ⎥ ⎢−7.875972 × 10−5 ⎥ y ⎥ ⎢ ⎥ ⎡ 0⎤ ⎢ ⎢ 0⎥ ⎢ ⎥0 ε 0.0 ⎢γx y ⎥ ∗ −1 ⎢ ⎥ /00 . = ⎢ ⎥ = (C ) s = ⎢ ⎥ 0.0 κ ⎢ κx ⎥ ⎢ ⎥

  ⎢ ⎥ ⎣ ⎦ 0.0 ⎦ ⎣ κ y e 0.0 κx y ⎡

(5.75)

⑦ Calculate the stresses in each layer k according to Eq. (4.20), σ kx,y (z) = ( ) C k ε0 + zκ , in the x–y laminate system ) z ≤ z k ). The strains in the ( (z k−1 ≤ x–y laminate system are obtained from ε0 + zκ .

Evaluation of Eq. (4.20) for each layer k, i.e., ( ) σ kx,y (z) = C k ε0 + zκ ,

(5.76)

under consideration of the generalized strains given in Eq. (5.75) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 5.19 gives the stresses for each layer in the x–y laminate system.( In case )that the strains in the x–y laminate system are required, the evaluation of ε0 + zκ provides these values. A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) is provided in Fig. 5.13 as a function of the laminate thickness (z-coordinate). It can

5.5 Problem 4: Ply-By-Ply Failure Loads

(a)

137

z in mm 4.5 symmetric

1.5

−300

−200

−100

100

200

300

1

2

3

σx in 10−3 MPa

−1.5

−4.5

(b)

z in mm 4.5

1.5

−3

−2

−1

εx in 10−6

−1.5

−4.5

Fig. 5.13 Distribution of selected field quantities over the laminate thickness for the tensile load case: a normal stress σx (z) and b normal strain εx (z)

138

5 Example Problems

Table 5.20 Stresses in the local 1–2 coordinate systems for an external loading of N xn = 1 N/mm, symmetric laminate Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

Top Center Bottom Top Center Bottom Top Center Bottom

0.159803 0.159803 0.159803 − 0.006080 − 0.006080 − 0.006080 0.159803 0.159803 0.159803

0.003040 0.003040 0.003040 0.013728 0.013728 0.013728 0.003040 0.003040 0.003040

0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

be seen that a uniaxial load by N xn results in the case of this symmetric laminate in a symmetric distribution of the normal stress and strain with respect to z = 0. The strain distribution is constant (see Fig. 5.13b) across the entire thickness of the laminate. The stress distribution (see Fig. 5.13a) reveals a layer-wise constant distribution with the characteristic that the stress is proportional to the stiffness value of each layer in the loading direction. ⑧ Transform the stresses in each layer to the 1–2 lamina system according to Eq. (4.21): σ k1,2 = T kσ σ kx,y . The strains in the 1–2 lamina system are obtained from Eq. (3.33): ε k1,2 = (C k )−1 σ k1,2 . The transformation of the stresses into the 1–2 lamina systems is performed according to Eqs. (4.21) and (3.33). The corresponding numerical values for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Table 5.20. These values will be used for the subsequent failure analysis based on the Tsai–Wu criterion. Failure analysis based on the Tsai–Wu failure criterion.

To easier evaluate the Tsai–Wu failure criterion, we determine the limit loads by applying loads as N xn = R × 1 N/mm, i.e., we multiply the load with the strength ratio R. Considering the stresses from Table 5.20 in the Tsai–Wu criterion according to Eq. (3.140) gives

5.5 Problem 4: Ply-By-Ply Failure Loads

139

Table 5.21 Strength ratio R in the local 1–2 coordinate systems for an external loading of N xn = R × 1 N/mm, symmetric laminate Lamina No. Position Tsai–Wu 1

Top Center Bottom Top Center Bottom Top Center Bottom

2

3

10586.336 10586.336 10586.336 3468.294 3468.294 3468.294 10586.336 10586.336 10586.336

Fig. 5.14 Symmetric laminate: [0◦ /90◦ ]s after first ply failure

a1 (R × σ1 ) + a2 (R × σ2 ) + a11 (R × σ1 )2 + a22 (R × σ2 )2 + a12 (R × σ12 )2 + 2a1,2 (R × σ1 )(R × σ2 ) < 1 .

(5.77)

This is a classical quadratic equation in R, from which the value of R can be identified, see Table 5.21. It can be concluded from Table 5.21 that the minimum strength ratio R is obtained for lamina 2, i.e., the 90◦ layer. Thus, we obtain the maximum load as N xn = R × 1 N/mm = 3468.294 × 1 N/mm = 3468.294 N/mm ,

(5.78)

which corresponds to a maximum stress of N xn 3468.294 N/mm N = = 385.366 . t 9 mm mm2

(5.79)

The corresponding maximum normal strain in the x-direction for this load is obtained as R × εx0 = 3468.294 × 1.261519 × 10−3 0/00 = 4.3753190/00 .

(5.80)

For the next iteration, we consider the laminate without lamina 2, i.e., the 90◦ layer, see Fig. 5.14.

140

5 Example Problems

Steps ①–③ remain unchanged, and only the modified steps, i.e., without the 90◦ layer, will be presented in the following. ④ Calculate the submatrices A, B, and C according to Eqs. (4.13)–(4.15). Assemble the generalized elasticity matrix C ∗ according to Eq. (4.12).

The evaluation of the submatrices A, B, and C according to Eqs. (4.13)–(4.15) gives ⎡ ⎤ 761210.059966 18607.357021 0.0 N A= Ak = ⎣ 18607.357021 66454.846505 0.0 ⎦ , 0.0 0.0 39600.0 mm k =1 ⎡ ⎤ 8 0.0 0.0 0.0 ∑ B= Bk = ⎣0.0 0.0 0.0⎦ N , 0.0 0.0 0.0 k =1 ⎡ ⎤ 8 7421798.084668 181421.730959 0.0 ∑ 0.0 ⎦ Nmm , D= Dk = ⎣ 181421.730959 647934.753423 0.0 0.0 386100.0 k =1 8 ∑

(5.81)

(5.82)

(5.83)

from which the generalized elasticity matrix (written without units) can be assembled as follows: ⎡ ⎤ 761210.060 18607.357 0.0 0.0 0.0 0.0 ⎢ 18607.357 66454.847 0.0 0.0 0.0 0.0 ⎥ ⎢ ⎥ ⎢ ⎥ 0.0 0.0 39600.0 0.0 0.0 0.0 ∗ ⎢ ⎥. C =⎢ ⎥ 0.0 0.0 0.0 7421798.085 181421.731 0.0 ⎢ ⎥ ⎣ 0.0 0.0 0.0 181421.731 647934.753 0.0 ⎦ 0.0 0.0 0.0 0.0 0.0 386100.0 (5.84)

⑤ Calculate the generalized compliance matrix (C ∗ )−1 based on Eqs. (4.17)– (4.19). The evaluation of the submatrices A' , B' , and C ' according to Eqs. (4.17)–(4.19) gives ⎡

⎤ 13.227513 −3.703704 0.0 10−7 mm ⎦ 0.0 , A' = ⎣−3.703704 151.515152 N 0.0 0.0 252.525253

(5.85)

5.5 Problem 4: Ply-By-Ply Failure Loads

141

⎡

⎤ 0.0 0.0 0.0 1 B' = ⎣0.0 0.0 0.0⎦ , 0.0 0.0 0.0 N ⎡ ⎤ 1.356668 −0.379867 0.0 10−7 0.0 ⎦ D' = ⎣−0.379867 15.540016 , 0.0 0.0 25.900026 Nmm

(5.86)

(5.87)

from which the generalized compliance matrix (written without units) can be assembled as follows: ⎡

(C ∗ )−1

⎤ 0.0 0.0 0.0 0.0 13.227513 −3.703704 −3.703704 151.515152 0.0 0.0 0.0 0.0 ⎥ ⎢ ⎥ ⎢ 0.0 0.0 252.525253 0.0 0.0 0.0 ⎥ −7 ⎢ =⎢ ⎥ 10 . 0.0 0.0 0.0 1.356668 −0.379867 0.0 ⎥ ⎢ ⎦ ⎣ 0.0 0.0 0.0 −0.379867 15.540016 0.0 0.0 0.0 0.0 0.0 0.0 25.900026

(5.88)

⎡ ⎤T ⑥ Calculate the generalized strains e = ε0 κ according to Eq. (4.16).

The evaluation of Eq. (4.16) gives the generalized strains as follows: ⎤ ⎡ ⎤ εx0 1.322751 × 10−3 ⎢ 0⎥ ⎢−3.703704 × 10−4 ⎥ ⎢ ⎥ ⎡ 0 ⎤ ⎢ ε0y ⎥ ⎥ ⎢ ⎢ ⎥0 γx y ⎥ ε 0.0 ∗ −1 ⎢ ⎥ /00 . =⎢ ) s = = (C ⎥ ⎢ ⎢ ⎥ 0.0 κ κx ⎥ ⎢ ⎢ ⎥

  ⎣ κ ⎦ ⎣ ⎦ 0.0 y e 0.0 κx y ⎡

(5.89)

⑦ Calculate the stresses in each layer k according to Eq. (4.20), σ kx,y (z) = ) ( C k ε0 + zκ , in the x–y laminate system ) z ≤ z k ). The strains in the ( (z k−1 ≤ x–y laminate system are obtained from ε0 + zκ .

Evaluation of Eq. (4.20) for each layer k, i.e., ) ( σ kx,y (z) = C k ε0 + zκ ,

(5.90)

under consideration of the generalized strains given in Eq. (5.89) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 5.19 gives the stresses for each layer in the x–y laminate system. In case that the strains in the x–y laminate

142

5 Example Problems

Table 5.22 Stresses in the local 1–2 coordinate systems for an external loading of N xn = 1 N/mm after first ply failure, symmetric laminate Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

Top Center Bottom Top Center Bottom Top Center Bottom

0.166667 0.166667 0.166667 – – – 0.166667 0.166667 0.166667

0.0 0.0 0.0 – – – 0.0 0.0 0.0

0.0 0.0 0.0 – – – 0.0 0.0 0.0

) ( system are required, the evaluation of ε 0 + zκ provides these values. A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) is provided in Fig. 5.15 as a function of the laminate thickness (z-coordinate). It can be seen that a uniaxial load by N xn results in the case of this symmetric laminate in a symmetric distribution of the normal stress and strain with respect to to z = 0. The strain distribution is constant (see Fig. 5.15b) across the entire thickness of the laminate. The stress distribution (see Fig. 5.15a) reveals a layer-wise constant distribution with the characteristic that the stress is proportional to the stiffness value of each layer in the loading direction. ⑧ Transform the stresses in each layer to the 1–2 lamina system according to Eq. (4.21): σ k1,2 = T kσ σ kx,y . The strains in the 1–2 lamina system are obtained from Eq. (3.33): ε k1,2 = (C k )−1 σ k1,2 . The transformation of the stresses into the 1–2 lamina systems is performed according to Eqs. (4.21) and (3.33). The corresponding numerical values for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Table 5.22. These values will be used for the subsequent failure analysis based on the Tsai–Wu criterion. Failure analysis based on the Tsai–Wu failure criterion.

To easier evaluate the Tsai–Wu failure criterion, we determine the limit loads by applying loads as N xn = R × 1 N/mm, i.e., we multiply the load with the strength ratio R.

5.5 Problem 4: Ply-By-Ply Failure Loads

143

(a) z in mm 4.5 symmetric

1.5

−300

−200

−100

100

200

300

1

2

3

σx in 10−3 MPa

−1.5

−4.5

(b) z in mm 4.5

1.5

−3

−2

−1

εx in 10−6

−1.5

−4.5

Fig. 5.15 Distribution of selected field quantities over the laminate thickness for the tensile load case after the first ply failure: a normal stress σx (z) and b normal strain εx (z)

144

5 Example Problems

Table 5.23 Strength ratio R in the local 1–2 coordinate systems for an external loading of N xn = R × 1 N/mm after first ply failure, symmetric laminate Position Tsai–Wu Lamina No. 1

2

3

Top Center Bottom Top Center Bottom Top Center Bottom

11700.0 11700.0 11700.0 – – – 11700.0 11700.0 11700.0

Considering the stresses from Table 5.22 in the Tsai–Wu criterion according to Eq. (3.140) gives a1 (R × σ1 ) + a2 (R × σ2 ) + a11 (R × σ1 )2 + a22 (R × σ2 )2 + a12 (R × σ12 )2 + 2a1,2 (R × σ1 )(R × σ2 ) < 1 .

(5.91)

This is a classical quadratic equation in R, from which the value of R can be identified, see Table 5.23. It can be concluded from Table 5.23 that the minimum strength ratio R is obtained for lamina 1 and 3, i.e., the 0◦ layers. Thus, we obtain the maximum load as N xn = R × 1 N/mm = 11700.0 × 1 N/mm = 11700.0 N/mm ,

(5.92)

which corresponds to a maximum stress of 11700.0 N/mm N N xn = = 1300.0 . 9 mm mm2 t

(5.93)

The corresponding maximum normal strain in the x-direction for this load is obtained as R × εx0 = 11700.0 × 1.322751 × 10−3 0/00 = 15.4761870/00 .

(5.94)

The macroscopic stress–strain diagram in x-direction is shown in Fig. 5.16. It can be seen that a bilinear behavior is obtained in the elastic range for the entire laminate.

5.6 Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae

145

Nxn /t in MPa

last ply failure (0◦ ) 1250

1000

750

500 first ply failure (90◦ ) 250

2

4

6

8

10

12

14

16

εx in

Fig. 5.16 Stress–strain curve showing the ply-by-ply failure of the [0◦ /90◦ ]s laminate under uniaxial loading with N xn Fig. 5.17 Unidirectional lamina rotated under the angle α

5.6 Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae Given is a unidirectional lamina, i.e., a AS4 carbon/3501-6 epoxy prepreg, of 1 mm thickness, see Fig. 5.17. Calculate the apparent elastic properties E x , E y , G x y , νx y , and ν yx as a function of the angle of rotation. Represent the elastic properties in the form of pole diagrams (0◦ ≤ α ≤ 360◦ ). In addition, provide a classical representation in a Cartesian

146

5 Example Problems

coordinate system (0◦ ≤ α ≤ 90◦ ). The material properties of the lamina can be taken from Table 5.1. The apparent elastic properties can be obtained from the elements D i j of the transformed elastic compliance matrix according to Eq. (3.71) ⎤ ⎡ νx y ηx 1 − ⎥ ⎡ ⎤ ⎢ Ex Ex ⎥ ⎢ Ex D 11 D 12 D 14 ⎢ ν ηy ⎥ ⎥ ⎢ xy 1 D = ⎣ D 12 D 22 D 24 ⎦ = ⎢− (5.95) ⎥, ⎢ Ex E y E y ⎥ ⎥ ⎢ D 14 D 24 D 44 ηy 1 ⎦ ⎣ ηx Ex E y G x y where the compliance coefficients D i j can be evaluated based Eqs. (3.65)–(3.70) D 11 = D11 cos4 α + (2D12 + D44 ) sin2 α cos2 α + D22 sin4 α ,

(5.96)

D 12 = (D11 + D22 − D44 ) sin α cos +D12 (sin α + cos α) ,

(5.97)

2

2

4

4

D 22 = D11 sin4 α + (2D12 + D44 ) sin2 α cos2 α + D22 cos4 α ,

(5.98)

D 14 = (2D11 − 2D12 − D44 ) sin α cos α − (2D22 − 2D12 − D44 ) sin α cos α , (5.99) 3

3

D 24 = (2D11 − 2D12 − D44 ) sin3 α cos α − (2D22 − 2D12 − D44 ) sin α cos3 α , (5.100) D 44 = 2(2D11 + 2D22 − 4D12 − D44 ) sin2 α cos2 α + D44 (sin4 α + cos4 α) . (5.101) or after replacing the Di j [see Eq. (3.63)] as D 11 D 12 D 22 D 14

( ( ) ) 1 ν12 1 1 4 = cos α + 2 − sin4 α , + sin2 α cos2 α + E1 E2 E1 G 12 ( ( ) ) 1 1 1 ν12 2 2 = + − sin α cos + − (sin4 α + cos4 α) , E 1 E 2 G 12 E1 ( ( ) ) 1 1 ν12 1 4 = sin α + 2 − cos4 α , + sin2 α cos2 α + E1 G 12 E1 E2 ( ( ) ) 1 ν12 1 = 2 −2 − − sin α cos3 α E1 E1 G 12 ( ( ) ) ν12 1 1 − 2 −2 − − sin3 α cos α , E1 G 12 E2

(5.102) (5.103) (5.104)

(5.105)

5.6 Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae

(

D 24

D 44

( ) ) 1 ν12 1 = 2 −2 − − sin3 α cos α E1 E1 G 12 ( ( ) ) 1 ν12 1 − 2 −2 − − sin α cos3 α , E2 E1 G 12 ( ( ) ) 1 1 1 ν12 =2 2 +2 −4 − − sin2 α cos2 α E1 G 12 E1 E2

147

+

1 (sin4 α + cos4 α) . G 12

(5.106)

(5.107)

Thus, we can state the apparent elastic properties as follows2 ( ) 1 1 2ν12 1 1 4 = cos α + − sin4 α , (5.108) sin2 α cos2 α + E1 G 12 E2 E1 Ex ⎤ ⎡ ( ) ( ) 1 1 1 ν 12 νx y = E x − sin4 α + cos4 α , (5.109) + − sin2 α cos2 + E1 E 1 E 2 G 12 ( ) 1 1 2ν12 1 1 4 = sin α + − cos4 α, (5.110) sin2 α cos2 α + E1 G 12 E2 E1 Ey ( ) 2 2 4ν12 1 1 1 =2 + + − (sin4 α + cos4 α), sin2 α cos2 α + E1 E2 G 12 G 12 E1 Gxy (5.111) ηx

⎡(

( ⎤ ) ) 2 2ν12 1 2 2ν12 1 3 3 + − + − sin α cos α − sin α cos α , E1 E1 E1 G 12 E2 G 12 (5.112)

⎡(

( ⎤ ) ) 2ν12 1 2 2ν12 1 2 3 3 + − + − sin α cos α − sin α cos α . G 12 E2 G 12 E1 E1 E1 (5.113)

= Ex ηy = Ey

The graphical representations of the apparent elastic properties for the AS4 carbon/3501-6 epoxy prepreg are shown in Figs. 5.18, 5.19, 5.20, and 5.21 as pole diagrams and in Fig. 5.22 in a classical Cartesian representation. 2 It should be noted here that some textbooks present the right-hand sides of Eqs. (5.111) and (5.113) in a slightly different manner by considering E x → E 1 , E y → E 2 or E x , E y → E 1 see Altenbach et al. (2018) and Kaw (2006).

148

5 Example Problems

120◦

90◦ 60◦ Ey /E1

150◦

0

180◦

30◦

0.5

1

1.5 ◦ 0

Ex /E1

210◦

330◦

240◦

270◦

300◦

Fig. 5.18 Pole diagram of the elastic properties E x /E 1 and E y /E 1 for AS4 carbon/3501-6 epoxy 120◦

90◦ 60◦ Gxy /G12

150◦

0

180◦

1

210◦

30◦

2

0◦

330◦

240◦

270◦

300◦

Fig. 5.19 Pole diagram of the elastic property G x y /G 12 for AS4 carbon/3501-6 epoxy

5.6 Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae

120◦

90◦ 60◦

150◦

30◦ νxy

0

180◦

0.25

0.5 ◦ 0

210◦

330◦

240◦

300◦

270◦

Fig. 5.20 Pole diagram of the elastic property νx y for AS4 carbon/3501-6 epoxy 120◦

90◦ 60◦ −ηy

150◦

30◦

−ηx 0

180◦

1

2

210◦

3

0◦

330◦

240◦

270◦

300◦

Fig. 5.21 Pole diagram of the elastic properties −ηx and −η y for AS4 carbon/3501-6 epoxy

149

150

5 Example Problems

2

1.2 AS4 carbon / 3501-6 epoxy 1

Gxy /G12

1.5

0.8 1

0.6 0.4 Ey /E1

Ex /E1

0.5

0.2 0◦ 0

15◦

30◦

45◦

60◦

75◦

Elastic constant Gxy /G12 in –

Elastic constants Ex /E1 , Ey /E1 in –

(a)

0 90◦

Off-axis angle α (b)

Elastic constant νxy in –

AS4 carbon / 3501-6 epoxy 0.3 2 νxy

−ηx

−ηy

0.2 1 0.1

0◦ 0

15◦

30◦

45◦

60◦

75◦

Elastic constants −ηx , −ηy in –

3

0.4

0 90◦

Off-axis angle α

Fig. 5.22 Elastic constants as a function of the off-axis angle: a E x /E 1 , E y /E 1 , G x y /G 12 and b νx y , −ηx , −η y

5.7 Problem 6: Failure Envelopes for Unidirectional Laminae

151

Fig. 5.23 Unidirectional lamina loaded by a single normal stress σx

5.7 Problem 6: Failure Envelopes for Unidirectional Laminae Given is a unidirectional lamina, i.e., a AS4 carbon/3501-6 epoxy prepreg, of 1 mm thickness, which is loaded by a normal stress σx in the x-direction, see Fig. 5.23. Calculate and sketch the failure envelope, i.e., the failure stress in the x-direction as a function of the angle of rotation (0◦ ≤ α ≤ 90◦ ), based on the maximum stress, maximum strain, Tsai–Hill, and Tsai–Wu failure criteria. The material properties of the lamina can be taken from Table 5.1. The stresses in principal material coordinates (1–2 coordinate system) resulting from a single off-axis normal stress (σx ) can be calculated from transformation (3.35): ⎡

⎤ ⎡ ⎤⎡ ⎤ sin2 α 2 sin α cos α cos2 α σ1 σx ⎣ σ2 ⎦ = ⎣ sin2 α cos2 α −2 sin α cos α ⎦ ⎣ 0 ⎦ , σ12 0 − sin α cos α sin α cos α cos2 α − sin2 α

(5.114)

or based on this result the strains in principal material coordinates [see Eq. (3.28)] ⎡

⎤ ν21 1 ⎤ ⎢ E − E 0 ⎥⎡ ⎤ ⎡ 2 ⎢ 1 ⎥ σ1 (σx ) ε1 ⎢ ν12 1 ⎥ ⎦ ⎣ ε2 ⎦ = ⎢− ⎣ 0 ⎥ ⎢ E E ⎥ σ2 (σx ) . ⎢ ⎥ 1 2 2 ε12 σ (σ ) ⎣ 1 ⎦ 12 x 0 0 G 12

(5.115)

This means that the single off-axis stress (σx ) results in a plane stress state (σ1 , σ2 , σ12 ). To judge if this combined stress or strain state results in the failure of the lamina, the maximum stress, maximum strain, Tsai–Hill, and Tsai–Wu failure criteria will be applied in the following.

152

5 Example Problems

Table 5.24 Different failure modes of the unidirectional lamina (see Table 5.1 for details) according to the maximum stress criterion Failure mode Angle range Tension (σx > 0) 0 ◦ ≤ α ≤ 2◦ 3◦ ≤ α ≤ 30◦ 35◦ ≤ α ≤ 90◦ Compression (σx < 0) 0◦ ≤ α ≤ 3◦ 4◦ ≤ α ≤ 65◦ 70◦ ≤ α ≤ 90◦

Longitudinal tensile failure (k1t ) Shear failure (k12s ) Transverse tensile failure (k2t ) Longitudinal compressive failure (k1c ) Shear failure (k12s ) Transverse compressive failure (k2c )

To determine the limit load, one may apply a unit load, i.e., σx = 1 (or σx = −1 in the case of compression), and multiply this unit load with the strength ratio R, i.e., σx = R × 1. The minimum load factor is then equal to the limit load. • Maximum Stress Criterion: Considering the stresses from Eq. (5.114) in the conditions of the maximum stress criterion according to Eqs. (3.133)–(3.135) gives k1c < c1 R × (±1) < k1t ,

(5.116)

k2c < c2 R × (±1) < k2t , |c3 R × (±1)| < k12s ,

(5.117) (5.118)

from which the minimum value of R can be identified as the ultimate load [the ci are scalar values resulting from the evaluation of Eq. (5.114)]. The graphical representation of the failure (ultimate) load as a function of the off-axis angle is provided in Fig. 5.24 for the maximum stress criterion. It can be seen that small offaxis angles result in a significant reduction of the failure load (0◦ ≤ α ≤ 30◦ ). For larger angles (30◦ ≤ α ≤ 90◦ ), there is a clear difference between the tension and compression case. This difference is magnified in Fig. 5.24b based on a logarithmic stress axis. It should be noted that different failure modes occur depending on the off-axis angle, see Table 5.24. These different failure modes result in different sections of the failure envelope without a theoretical smooth transition. • Maximum Strain Criterion: Considering the strains from Eq. (5.115) in the conditions of the maximum strain criterion according to Eqs. (3.136)–(3.138) gives

5.7 Problem 6: Failure Envelopes for Unidirectional Laminae

153

(a)

Normal stress |σx | in MPa

2000 Maximum stress criterion AS4 carbon / 3501-6 epoxy 1500

tension (σx > 0) compression (σx < 0)

1000

500

0◦ 0

15◦

30◦

45◦

60◦

75◦

90◦

75◦

90◦

Off-axis angle α (b)

Normal stress | lg(σx )| in MPa

104 Maximum stress criterion AS4 carbon / 3501-6 epoxy

103

tension (σx > 0) compression (σx < 0)

102

0◦

15◦

30◦

45◦

60◦

Off-axis angle α

Fig. 5.24 Failure envelope for a unidirectional prepreg (see Table 5.1 for details) according to the maximum stress criterion: a linear division of the axes and b logarithmic vertical axis

154

5 Example Problems

Table 5.25 Different failure modes of the unidirectional lamina (see Table 5.1 for details) according to the maximum strain criterion Angle range Failure mode Tension (σx > 0) 0 ◦ ≤ α ≤ 4◦ 5◦ ≤ α ≤ 20◦ 25◦ ≤ α ≤ 90◦ Compression (σx < 0) 0◦ ≤ α ≤ 5◦ 6◦ ≤ α ≤ 55◦ 60◦ ≤ α ≤ 90◦

Longitudinal tensile failure (k1t ) Shear failure (k12s ) Transverse tensile failure (k2t ) Longitudinal compressive failure (k1c ) Shear failure (k12s ) Transverse compressive failure (k2c )

ε ε k1c < c1ε R × (±1) < k1t , ε ε ε k2c < c2 R × (±1) < k2t , | | ε |c R × (±1)| < k ε , 3

(5.119) (5.120) (5.121)

12s

from which the minimum value of R can be identified as the ultimate load [the ciε are scalar values resulting from the evaluation of Eq. (5.115)]. The graphical representation of the failure (ultimate) load as a function of the off-axis angle is provided in Fig. 5.25 for the maximum stress criterion. The characteristics of the failure envelopes are quite similar to the ones observed for the maximum stress criterion. As in the case of the maximum stress criterion, different failure modes can be distinguished, see Table 5.25 for details. • Tsai–Hill Criterion: Considering the stresses from Eq. (5.114) in the Tsai–Hill criterion according to Eq. (3.139) gives (c1 R × (±1))2 k12 +

−

(c1 R × (±1))(c2 R × (±1))

(c3 R × (±1))2 2 k12s

k12 < 1,

+

(c2 R × (±1))2 k22 (5.122)

from which the value of R can be identified as the ultimate load [the ci are scalar values resulting from the evaluation of Eq. (5.114)]. The graphical representation of the failure (ultimate) load as a function of the off-axis angle is provided in Fig. 5.26 for the maximum stress criterion. Different envelopes are again obtained for tension and compression, but the envelopes are much smoother.

5.7 Problem 6: Failure Envelopes for Unidirectional Laminae

155

(a)

Normal stress |σx | in MPa

2000 Maximum strain criterion AS4 carbon / 3501-6 epoxy tension (σx > 0) compression (σx < 0)

1500

1000

500

0◦ 0

15◦

30◦

45◦

60◦

75◦

90◦

75◦

90◦

Off-axis angle α (b)

Normal stress | lg(σx )| in MPa

104 Maximum strain criterion AS4 carbon / 3501-6 epoxy tension (σx > 0) compression (σx < 0)

103

102

0◦

15◦

30◦

45◦

60◦

Off-axis angle α

Fig. 5.25 Failure envelope for a unidirectional prepreg (see Table 5.1 for details) according to the maximum strain criterion: a linear division of the axes and b logarithmic vertical axis

156

5 Example Problems (a)

Normal stress |σx | in MPa

2000 Tsai-Hill criterion AS4 carbon / 3501-6 epoxy 1500

tension (σx > 0) compression (σx < 0)

1000

500

0◦ 0

15◦

30◦

45◦

60◦

75◦

90◦

75◦

90◦

Off-axis angle α (b)

Normal stress | lg(σx )| in MPa

104 Tsai-Hill criterion AS4 carbon / 3501-6 epoxy

103

tension (σx > 0) compression (σx < 0)

102

0◦

15◦

30◦

45◦

60◦

Off-axis angle α

Fig. 5.26 Failure envelope for a unidirectional prepreg (see Table 5.1 for details) according to the Tsai–Hill criterion: a linear division of the axes and b logarithmic vertical axis

5.7 Problem 6: Failure Envelopes for Unidirectional Laminae

157

(a)

Normal stress |σx | in MPa

2000 Tsai-Wu criterion AS4 carbon / 3501-6 epoxy 1500

1000

500

0◦ 0

15◦

30◦

45◦

60◦

75◦

90◦

75◦

90◦

Off-axis angle α (b)

Normal stress | lg(σx )| in MPa

104 Tsai-Wu criterion AS4 carbon / 3501-6 epoxy

103

102

0◦

15◦

30◦

45◦

60◦

Off-axis angle α

Fig. 5.27 Failure envelope for a unidirectional prepreg (see Table 5.1 for details) according to the Tsai–Wu criterion: a linear division of the axes and b logarithmic vertical axis

158

5 Example Problems

• Tsai–Wu Criterion: Considering the stresses from Eq. (5.114) in the Tsai–Wu criterion according to Eq. (3.140) gives a1 (c1 R × (±1)) + a2 (c2 R × (±1)) + a11 (c1 R × (±1))2 + a22 (c2 R × (±1))2 + a12 (c3 R × (±1))2 + 2a1,2 (c1 R × (±1))(c2 R × (±1)) < 1 ,

(5.123)

i.e., a classical quadratic equation in R, from which the value of R can be identified as the ultimate load [the ci are scalar values resulting from the evaluation of Eq. (5.114)]. A single and smooth failure envelope is obtained which facilitates the application compared to the previously mentioned criteria (Fig. 5.27).

References Altenbach H, Altenbach J, Kissing W (2018) Mechanics of composite structural elements. Springer, Singapore Buryachenko VA (2007) Micromechanics of heterogeneous materials. Springer, New York Dvorak GJ (2013) Micromechanics of composite materials. Springer, New York Herakovich CT (2012) Mechanics of composites: a historical review. Mech Res Commun 41:1–20 Kaw AK (2006) Mechanics of composite materials. CRC Press, Boca Raton MIL-HDBK-17/1F (2002) Composite materials handbook—volume 1. Polymer matrix composites guidelines for characterization of structural materials. Department of Defence, USA Soden PD, Hinton MJ, Kaddour AS (1998) Lamina properties, lay-up configurations and loading conditions for a range of fibre-reinforced composite laminates. Compos Sci Technol 58:1011– 1022

Chapter 6

Supplementary Problems

Abstract This chapter presents the short solutions of some additional calculation problems based on the classical laminate theory. The solution is based on the stepby-step approach presented in detail in Chap. 5. The presented problems comprise the calculation of stresses and strains in symmetric and asymmetric laminates, the application of failure criteria, and the ply-by-ply failure of laminates. In addition, the characterization of basic properties of unidirectional laminae based on pole diagrams for the elastic properties and failure envelopes is presented.

6.1 Problems All the supplementary problems are based on the same E-glass/epoxy lamina system; see Table 6.1 for the effective properties.

6.1.1 Supplementary Problem 1: Stresses and Strains in a Symmetric Laminate Given is a symmetric ten-layer laminate of thickness t = 10 mm as shown in Fig. 6.1. Each lamina has the same thickness (t/10) and identical material properties, which can be taken from Table 6.1. However, the orientation of each lamina changes from layer to layer as indicated in Fig. 6.1. Consider (a) a pure tensile load case with N xn = 1000 N/mm (all other forces and moments are equal to zero) and (b) a pure bending load case with Mxn = 1000 Nmm/mm (all other moments and forces are equal to zero) to determine the stresses and strains in each lamina in the global x–y-coordinate system as well as in the principal material coordinates (1–2 coordinate systems) of each lamina.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, Composite Mechanics, Advanced Structured Materials 184, https://doi.org/10.1007/978-3-031-32390-4_6

159

160

6 Supplementary Problems

Table 6.1 Selected mechanical properties of a unidirectional lamina (fiber: E-glass 21xK43 Gevetex; matrix: LY556/HT907/DY063 epoxy; fiber volume fraction of the lamina: 0.62) Value Property Elastic constants Longitudinal modulus, E 1 in MPa Transverse modulus, E 2 in MPa In-plane shear modulus, G 12 in MPa Major Poisson’s ratio, ν12 in – Strength parameters Longitudinal tensile strength, k1t in MPa Longitudinal compressive strength, k1c in MPa Transverse tensile strength, k2t in MPa Transverse compressive strength, k2c in MPa In-plane shear strength, k12s in MPa ε in % Longitudinal tensile failure strain, k1t ε in Longitudinal compressive failure strain, k1c % ε in % Transverse tensile failure strain, k2t ε in % Transverse compressive failure strain, k2c ε In-plane shear failure strain, k12s in %

53480 17700 5830 0.278 1140 − 570 35 − 114 72 2.132 − 1.065 0.197 − 0.644 3.8

Adapted from Soden et al. (1998) Fig. 6.1 Symmetric laminate: [0◦ / ± 45◦ /0◦ /90◦ ]s

6.1.2 Supplementary Problem 2: Stresses and Strains in an Asymmetric Laminate Given is an asymmetric ten-layer laminate of thickness t = 10 mm as shown in Fig. 6.2. Each lamina has the same thickness (t/10) and identical material properties, which can be taken from Table 6.1. However, the orientation of each lamina changes from layer to layer as indicated in Fig. 6.2. Consider (a) a pure tensile load case with N xn = 1000 N/mm (all other forces and moments are equal to zero) and (b) a pure bending load case with Mxn = 1000 Nmm/mm (all other moments and forces are equal to zero) to determine the

6.1 Problems

161

Fig. 6.2 Asymmetric laminate: [0◦ / ± 45◦ /0◦ /90◦ /0◦ /90◦ / ± 45◦ /90◦ ]

Fig. 6.3 Asymmetric laminate: [+ 45◦ / 0◦ / + 30◦ / − 45◦ ]

stresses and strains in each lamina in the global x–y-coordinate system as well as in the principal material coordinates (1–2 coordinate systems) of each lamina.

6.1.3 Supplementary Problem 3: Failure Criteria Given is an asymmetric four-layer laminate of thickness t = 10 mm as shown in Fig. 6.3. Each lamina has the same thickness (t/4) and identical material properties, which can be taken from Table 6.1. However, the orientation of each lamina changes from layer to layer as indicated in Fig. 6.3. Consider (a) a biaxial tensile load case with N xn = 1000 N/mm and N yn = 500 N/mm (all other forces and moments are equal to zero) and (b) a biaxial bending load case with Mxn = 1000 Nmm/mm and M yn = 500 Nmm/mm (all other moments and forces are equal to zero) to determine the stresses and strains in each lamina in the global x–y-coordinate system as well as in the principal material coordinates (1–2 coordinate systems) of each lamina. Use these stress and strain values to perform a failure analysis based on the maximum stress, maximum strain, Tsai–Hill, and Tsai–Wu failure criteria.

6.1.4 Supplementary Problem 4: Ply-by-Ply Failure Loads Given is a symmetric three-layer laminate of thickness t = 9 mm as shown in Fig. 6.4. Each lamina has the same thickness (t/3) and identical material properties, which can be taken from Table 6.1. However, the orientation of each lamina changes from layer to layer as indicated in Fig. 6.4.

162

6 Supplementary Problems

Fig. 6.4 Symmetric laminate: [0◦ /90◦ ]s

Fig. 6.5 Unidirectional lamina rotated under the angle α

Consider a pure tensile load case with N xn > 0 (all other forces and moments are equal to zero) to investigate the ply-by-ply failure of this laminate based on the Tsai–Wu criterion.

6.1.5 Supplementary Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae Given is a unidirectional lamina, i.e., an E-glass/epoxy lamina system, of 1 mm thickness; see Fig. 6.5. Calculate the apparent elastic properties E x , E y , G x y , νx y , and ν yx as a function of the angle of rotation. Represent the elastic properties in the form of pole diagrams (0◦ ≤ α ≤ 360◦ ). In addition, provide a classical representation in a Cartesian coordinate system (0◦ ≤ α ≤ 90◦ ). The material properties of the lamina can be taken from Table 6.1.

6.1.6 Supplementary Problem 6: Failure Envelopes for Unidirectional Laminae Given is a unidirectional lamina, i.e., an E-glass/epoxy lamina system, of 1 mm thickness, which is loaded by a normal stress σx in the x-direction; see Fig. 6.6. Calculate and sketch the failure envelope, i.e., the failure stress in the x-direction as a function of the angle of rotation (0◦ ≤ α ≤ 90◦ ), based on the maximum stress, maximum strain, Tsai–Hill, and Tsai–Wu failure criteria. The material properties of the lamina can be taken from Table 6.1.

6.2 Short Solutions

163

Fig. 6.6 Unidirectional lamina loaded by a single normal stress σx

6.2 Short Solutions 6.2.1 Supplementary Problem 1: Stresses and Strains in a Symmetric Laminate • Laminate loaded by N xn = 1000 N/mm A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) is provided in Fig. 6.7 as a function of the laminate thickness (zcoordinate). The numerical values of the stresses and strains in the 1–2 lamina systems for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Tables 6.2 and 6.3. These values could be used for a subsequent failure analysis in each lamina. • Laminate loaded by Mxn = 1000 Nmm/mm A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) is provided in Fig. 6.8 as a function of the laminate thickness (zcoordinate). The numerical values of the stresses and strains in the 1–2 lamina systems for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Tables 6.4 and 6.5. These values could be used for a subsequent failure analysis in each lamina.

164

6 Supplementary Problems (a) z in mm 5 symmetric laminate 4 3 2 1 −200

−100

100

σx in MPa

200

−1 −2 −3 −4 −5 (b) z in mm 5 4 3 2 1 −3

−2

−1

1

2

3

εx in

−1 −2 −3 −4 −5

Fig. 6.7 Distribution of selected field quantities over the laminate thickness for the tensile load case: a normal stress σx (z) and b normal strain εx (z)

6.2 Short Solutions

165

Table 6.2 Stresses in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm, symmetric laminate Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

4

5

6

7

8

9

10

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

159.503 159.503 159.503 61.691 61.691 61.691 61.691 61.691 61.691 159.503 159.503 159.503 − 36.122 − 36.122 − 36.122 − 36.122 − 36.122 − 36.122 159.503 159.503 159.503 61.691 61.691 61.691 61.691 61.691 61.691 159.503 159.503 159.503

− 1.846 − 1.846 − 1.846 23.895 23.895 23.895 23.895 23.895 23.895 − 1.846 − 1.846 − 1.846 49.636 49.636 49.636 49.636 49.636 49.636 − 1.846 − 1.846 − 1.846 23.895 23.895 23.895 23.895 23.895 23.895 − 1.846 − 1.846 − 1.846

0.0 0.0 0.0 − 22.886 − 22.886 − 22.886 22.886 22.886 22.886 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 22.886 22.886 22.886 − 22.886 − 22.886 − 22.886 0.0 0.0 0.0

6.2.2 Supplementary Problem 2: Stresses and Strains in an Asymmetric Laminate A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) is provided in Fig. 6.9 as a function of the laminate thickness (z-coordinate).

166

6 Supplementary Problems

Table 6.3 Strains in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm, symmetric laminate Position ε1 in 0/00 ε2 in 0/00 γ12 in 0/00 Lamina No. 1

2

3

4

5

6

7

8

9

10

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

2.992080 2.992080 2.992080 1.029319 1.029319 1.029319 1.029319 1.029319 1.029319 2.992080 2.992080 2.992080 − 0.933442 − 0.933442 − 0.933442 − 0.933442 − 0.933442 − 0.933442 2.992080 2.992080 2.992080 1.029319 1.029319 1.029319 1.029319 1.029319 1.029319 2.992080 2.992080 2.992080

− 0.933442 − 0.933442 − 0.933442 1.029319 1.029319 1.029319 1.029319 1.029319 1.029319 − 0.933442 − 0.933442 − 0.933442 2.992080 2.992080 2.992080 2.992080 2.992080 2.992080 − 0.933442 − 0.933442 − 0.933442 1.029319 1.029319 1.029319 1.029319 1.029319 1.029319 − 0.933442 − 0.933442 − 0.933442

0.0 0.0 0.0 − 3.925522 − 3.925522 − 3.925522 3.925522 3.925522 3.925522 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 3.925522 3.925522 3.925522 − 3.925522 − 3.925522 − 3.925522 0.0 0.0 0.0

The numerical values of the stresses and strains in the 1–2 lamina systems for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Tables 6.6 and 6.7. These values could be used for a subsequent failure analysis in each lamina.

6.2 Short Solutions

167

Table 6.4 Stresses in the local 1–2 coordinate systems for an external loading of Mxn = 1000 Nmm/mm, symmetric laminate Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

4

5

6

7

8

9

10

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

− 66.946 − 75.314 − 83.683 − 14.718 − 17.172 − 19.625 − 12.165 − 15.206 − 18.247 − 16.737 − 25.105 − 33.473 0.0 2.874 5.748 − 5.748 − 2.874 0.0 33.473 25.105 16.737 18.247 15.206 12.165 19.625 17.172 14.718 83.683 75.314 66.946

3.322 3.737 4.153 − 6.849 − 7.990 − 9.132 − 3.947 − 4.933 − 5.920 0.831 1.246 1.661 0.0 − 2.543 − 5.087 5.087 2.543 0.0 − 1.661 − 1.246 − 0.831 5.920 4.933 3.947 9.132 7.990 6.849 − 4.153 − 3.737 − 3.322

0.550 0.619 0.688 7.891 9.206 10.522 − 5.261 − 6.576 − 7.891 0.138 0.206 0.275 0.0 − 0.069 − 0.138 0.138 0.069 0.0 − 0.275 − 0.206 − 0.138 7.891 6.576 5.261 − 10.522 − 9.206 − 7.891 − 0.688 − 0.619 − 0.550

• Laminate loaded by Mxn = 1000 Nmm/mm A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) is provided in Fig. 6.10 as a function of the laminate thickness (zcoordinate). The numerical values of the stresses and strains in the 1–2 lamina systems for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Tables 6.8 and 6.9. These values could be used for a subsequent failure analysis in each lamina.

168

6 Supplementary Problems (a) z in mm 5 symmetric laminate 4 3 2 1 −200

−100

100

σx in MPa

200

−1 −2 −3 −4 −5 (b) z in mm 5 4 3 2 1 −3

−2

−1

1

2

3

εx in

−1 −2 −3 −4 −5

Fig. 6.8 Distribution of selected field quantities over the laminate thickness for the bending load case: a normal stress σx (z) and b normal strain εx (z)

6.2 Short Solutions

169

Table 6.5 Strains in the local 1–2 coordinate systems for an external loading of Mxn = 1000 Nmm/mm, symmetric laminate Position ε1 in 0/00 ε2 in 0/00 γ12 in 0/00 Lamina No. 1

2

3

4

5

6

7

8

9

10

Top Center Bottom Top Center Bottom Top Center Bottom Top Center bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

− 1.269065 − 1.4276987 − 1.586331 − 0.239613 − 0.279549 − 0.319484 − 0.206947 − 0.258684 − 0.310421 − 0.317266 − 0.475899 − 0.634532 0.0 0.066961 0.133921 − 0.133921 − 0.066961 0.0 0.634532 0.475899 0.317266 0.310421 0.258684 0.206947 0.319484 0.279549 0.239613 1.586331 1.427698 1.269065

0.535686 0.602646 0.669607 − 0.310421 − 0.362158 − 0.413895 0.159742 − 0.199678 − 0.239613 0.133921 0.200882 0.267843 0.0 − 0.158633 − 0.317266 0.317266 0.158633 0.0 − 0.267843 − 0.200882 − 0.133921 0.239613 0.199678 0.159742 0.413895 0.362158 0.310421 − 0.669607 − 0.602646 − 0.535686

0.094410 0.106212 0.118013 1.353563 1.579157 1.804750 − 0.902375 − 1.127969 − 1.353563 0.023603 0.035404 0.047205 0.0 − 0.011801 − 0.023603 0.023603 0.011801 0.0 − 0.047205 − 0.035404 − 0.023603 1.353563 1.127969 0.902375 − 1.804750 − 1.579157 − 1.353563 − 0.118013 − 0.106212 − 0.094410

6.2.3 Supplementary Problem 3: Failure Criteria • Laminate loaded by N xn = 1000 N/mm and N yn = 500 N/mm Graphical representations of the normal stresses (σx , σ y ) and the normal strains (εx , ε y ) in the x- and y-direction are provided in Figs. 6.11 and 6.12 as a function of the laminate thickness (z-coordinate).

170

6 Supplementary Problems (a) z in mm 5 asymmetric laminate 4 3 2 1 −200

−100

100

σx in MPa

200

−1 −2 −3 −4 −5 (b) z in mm 5 4 3 2 1 −4

−3

−2

−1

1

2

3

4

εx in

−1 −2 −3 −4 −5

Fig. 6.9 Distribution of selected field quantities over the laminate thickness for the tensile load case: a normal stress σx (z) and b normal strain εx (z)

6.2 Short Solutions

171

Table 6.6 Stresses in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm, asymmetric laminate Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

4

5

6

7

8

9

10

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

129.108 121.799 114.489 49.862 45.270 40.678 62.044 58.811 55.578 172.966 165.656 158.347 − 35.196 − 35.711 − 36.226 202.204 194.895 187.585 − 33.135 − 33.651 − 34.166 104.967 100.375 95.782 100.837 97.604 94.371 − 30.044 − 30.560 − 31.075

− 4.773 − 5.394 − 6.015 21.173 19.836 18.499 23.057 21.362 19.668 − 1.045 − 1.666 − 2.287 58.828 56.418 54.009 1.441 0.819 0.198 68.465 66.056 63.646 37.211 35.874 34.538 43.388 41.694 39.999 82.922 80.513 78.103

− 0.987 1.146 − 1.305 − 21.294 − 20.499 − 19.704 122.883 22.088 21.294 − 0.033 − 0.192 − 0.351 − 0.285 − 0.126 0.033 0.603 0.444 0.285 − 0.921 − 0.762 − 0.603 − 30.832 − 30.037 − 29.242 32.422 31.627 30.832 − 1.876 − 1.717 − 1.558

The numerical stress and strain values for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Tables 6.10 and 6.11. These values will be used for the subsequent failure analysis in each lamina.

172

6 Supplementary Problems

Table 6.7 Strains in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm, asymmetric laminate Position ε1 in 0/00 ε2 in 0/00 γ12 in 0/00 Lamina No. 1

2

3

4

5

6

7

8

9

10

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

2.438948 2.305498 2.172049 0.822290 0.743373 0.664455 1.040278 0.988637 0.936997 3.239645 3.106196 2.972746 − 0.963913 − 0.961021 − 0.958129 3.773443 3.639994 3.506544 − 0.975482 − 0.972590 − 0.969698 1.769295 1.690378 1.611460 1.659961 1.608321 1.556680 − 0.992835 − 0.989943 − 0.987051

− 0.940776 − 0.937884 − 0.934992 0.936997 0.885357 0.833716 0.980124 0.901207 0.822290 − 0.958129 − 0.955237 − 0.952345 3.506544 3.373095 3.239645 − 0.969698 − 0.966805 − 0.963913 4.040342 3.906893 3.773443 1.556680 1.505040 1.453400 1.927129 1.848212 1.769295 4.841039 4.707590 4.574140

− 0.169261 − 0.196538 − 0.223815 − 3.652407 − 3.516066 − 3.379724 3.925091 3.788749 3.652407 − 0.005600 − 0.032877 − 0.060154 − 0.048954 − 0.021677 0.005600 0.103507 0.076230 0.048954 − 0.158061 − 0.130784 − 0.103507 − 5.288507 − 5.152166 − 5.015824 5.561191 5.424849 5.288507 − 0.321722 − 0.294445 − 0.267168

6.2 Short Solutions

173

(a)

z in mm 5

asymmetric laminate 4 3 2 1 −100

−50

50

σx in MPa

100

−1 −2 −3 −4 −5

(b)

z in mm 5 4 3 2 1 −3

−2

−1

1

2

3

εx in

−1 −2 −3 −4 −5

Fig. 6.10 Distribution of selected field quantities over the laminate thickness for the bending load case: a normal stress σx (z) and b normal strain εx (z)

174

6 Supplementary Problems

Table 6.8 Stresses in the local 1–2 coordinate systems for an external loading of Mxn = 1000 Nmm/mm, asymmetric laminate Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

4

5

6

7

8

9

10

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

− 77.660 − 89.202 − 100.744 − 18.215 − 22.850 − 27.485 − 11.207 − 15.696 − 20.185 − 8.407 19.949 − 31.491 1.665 4.083 6.502 37.762 26.220 14.678 − 8.008 − 5.589 − 3.171 37.403 32.768 28.133 42.662 38.173 33.684 − 22.517 − 20.099 − 17.680

2.013 2.083 2.153 − 7.696 − 9.444 − 11.192 3.605 − 5.392 − 7.178 1.593 1.663 1.733 4.877 1.273 − 2.331 1.313 1.383 1.453 19.293 15.689 12.085 13.278 11.530 9.782 17.829 16.043 14.257 40.917 37.313 33.709

0.196 0.179 0.162 8.277 9.910 11.543 − 5.010 − 6.644 − 8.277 0.299 0.282 0.265 − 0.333 − 0.316 − 0.299 0.367 0.350 0.333 − 0.401 − 0.384 − 0.367 − 11.321 − 9.688 − 8.055 14.588 12.955 11.321 − 0.503 − 0.486 − 0.469

6.2 Short Solutions

175

Table 6.9 Strains in the local 1–2 coordinate systems for an external loading of Mxn = 1000 Nmm/mm, asymmetric laminate Position ε1 in 0/00 ε2 in 0/00 γ12 in 0/00 Lamina No. 1

2

3

4

5

6

7

8

9

10

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

− 1.462585 − 1.678771 − 1.894956 − 0.300596 − 0.378175 − 0.455755 − 0.190817 − 0.265471 − 0.340125 − 0.165472 − 0.381658 − 0.597843 0.005784 0.069736 0.133687 0.699270 0.483085 0.266899 − 0.250022 − 0.186070 − 0.122119 0.630360 0.552781 0.475201 0.705036 0.630382 0.555727 − 0.633730 − 0.569779 − 0.505828

0.517396 0.581348 0.645299 − 0.340125 − 0.414780 − 0.489434 − 0.145436 − 0.223016 − 0.300596 0.133687 0.197639 0.261590 0.266899 0.050714 − 0.165472 − 0.122119 − 0.058167 0.005784 1.131641 0.915456 0.699270 0.555727 0.481073 0.406419 0.785520 0.707940 0.630360 2.428754 2.212569 1.996383

0.033679 0.030754 0.027828 1.419707 1.699844 1.979981 − 0.859433 − 1.139570 − 1.419707 0.051231 0.048305 0.045380 − 0.057081 − 0.054156 − 0.051231 0.062932 0.060007 0.057081 − 0.068783 − 0.065857 − 0.062932 − 1.941937 − 1.661800 − 1.381663 2.502211 12.222074 1.941937 − 0.086334 − 0.083409 − 0.080484

176

6 Supplementary Problems (a) z in mm 5 asymm. laminate

2.5

−200

−100

100

200

σx in MPa

−2.5

−5 (b) z in mm 5 asymm. laminate

2.5

−200

σy in MPa

−100

100

200

−2.5

−5

Fig. 6.11 Distribution of selected field quantities over the laminate thickness for the biaxial tensile load case: a normal stress σx (z) and b normal stress σ y (z)

6.2 Short Solutions

177

(a) z in mm 5 asymm. laminate

2.5

−5

−4

−3

−2

−1

1

2

3

4

5

εx in

−2.5

−5 (b) z in mm 5 asymm. laminate

2.5

−5

−4

−3

−2

−1

1

2

3

4

5

εy in

−2.5

−5

Fig. 6.12 Distribution of selected field quantities over the laminate thickness for the biaxial tensile load case: a normal strain εx (z) and b normal strain ε y (z)

178

6 Supplementary Problems

Table 6.10 Stresses in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm and N yn = 500 N/mm, asymmetric laminate Lamina No.

Position

σ1 in MPa

σ2 in MPa

σ12 in MPa

1

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

69.137 58.286 47.436 148.826 146.572 144.318 134.841 123.165 111.489 67.851 84.995 102.140

59.818 64.721 69.623 34.752 37.392 40.033 34.339 39.459 44.578 47.875 45.410 42.945

− 5.707 − 4.443 − 3.179 − 5.333 − 8.608 − 11.884 − 8.712 − 9.255 − 9.798 13.289 12.026 10.762

2

3

4

Table 6.11 Strains in the local 1–2 coordinate systems for an external loading of N xn = 1000 N/mm and N yn = 500 N/mm, asymmetric laminate Lamina No.

Position

ε1 in 0/00

ε2 in 0/00

γ12 in 0/00

1

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

0.981817 0.753441 0.525064 2.602192 2.546314 2.490436 2.342830 2.097893 1.852956 1.019860 1.353244 1.686628

3.020163 3.353547 3.686930 1.189773 1.350659 1.511544 1.239121 1.589065 1.939009 2.352076 2.123699 1.895323

− 0.978892 − 0.762128 − 0.545364 − 0.914825 − 1.476585 − 2.038345 − 1.494289 − 1.587446 − 1.680603 2.279474 2.062710 1.845946

2

3

4

The values of the strength ratio R are summarized in Table 6.12 for considered four failure criteria. • Laminate loaded by Mxn = 1000 Nmm/mm and M yn = 500 Nmm/mm Graphical representations of the normal stresses (σx , σ y ) and the normal strains (εx , ε y ) in the x- and y-direction are provided in Figs. 6.13 and 6.14 as a function of the laminate thickness (z-coordinate).

6.2 Short Solutions

179

Table 6.12 Strength ratio R in the local 1–2 coordinate systems for an external loading of N xn = R × 1000 N/mm and N yn = R × 500 N/mm, asymmetric laminate (bold numbers indicate failure) Lamina No. Position 1

Top

Center

Bottom

2

Top

Center

Bottom

3

Top

Center

Bottom

4

Top

Center

Bottom

Direction

Max. stress

Max. Strain Tsai–Hill

1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12

16.489 0.585 12.616 19.559 0.541 16.205 24.033 0.503 22.645 7.660 1.007 13.500 7.778 0.936 8.364 7.899 0.874 6.059 8.454 1.019 8.265 9.256 0.887 7.780 10.225 0.785 7.349 16.801 0.731 5.418 13.412 0.771 5.987 11.161 0.815 6.690

21.715 0.652 38.819 28.297 0.587 49.860 40.605 0.534 69.678 8.193 1.656 41.538 8.373 1.459 25.735 8.561 1.303 18.643 9.100 1.590 25.430 10.163 1.240 23.938 11.506 1.016 22.611 20.905 0.838 16.671 15.755 0.928 18.422 12.641 1.039 20.586

Tsai–Wu

0.584

0.613

0.541

0.561

0.503

0.517

0.998

1.186

0.925

1.081

0.862

0.990

1.006

1.174

0.879

0.994

0.779

0.861

0.724

0.764

0.764

0.821

0.808

0.886

180

6 Supplementary Problems (a) z in mm 5 asymm. laminate

2.5

−200

−100

100

200

σx in MPa

−2.5

−5 (b) z in mm 5 asymm. laminate

2.5

−200

σy in MPa

−100

100

200

−2.5

−5

Fig. 6.13 Distribution of selected field quantities over the laminate thickness for the biaxial bending load case: a normal stress σx (z) and b normal stress σ y (z)

6.2 Short Solutions

181

(a) z in mm 5 asymm. laminate 2.5

−5

−4

−3

−2

−1

1

2

3

4

5

εx in

−2.5

−5 (b) z in mm 5 asymm. laminate

2.5

−5

−4

−3

−2

−1

1

2

3

4

5

εy in

−2.5

−5

Fig. 6.14 Distribution of selected field quantities over the laminate thickness for the biaxial bending load case: a normal strain εx (z) and b normal strain ε y (z)

182

6 Supplementary Problems

Table 6.13 Stresses in the local 1–2 coordinate systems for an external loading of Mxn = 1000 Nmm/mm and M yn = 500 Nmm/mm, asymmetric laminate Lamina No.

Position

σ1 in MPa

σ2 in MPa

σ12 in MPa

1

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

− 21.703 − 37.692 − 53.681 1.459 − 27.928 − 57.315 54.877 32.671 10.466 59.649 41.025 22.400

− 18.293 − 25.343 − 32.393 − 1.872 − 5.396 − 8.921 6.587 1.172 − 4.242 27.847 21.490 15.134

4.776 7.603 10.429 2.939 3.248 3.556 − 4.493 − 1.891 0.711 12.182 9.356 6.529

2

3

4

Table 6.14 Strains in the local 1–2 coordinate systems for an external loading of Mxn = 1000 Nmm/mm and M yn = 500 Nmm/mm, asymmetric laminate Lamina No.

Position

ε1 in 0/00

ε2 in 0/00

γ12 in 0/00

1

Top Center Bottom Top Center Bottom Top Center Bottom Top Center Bottom

− 0.310718 − 0.573041 − 0.835363 0.037016 − 0.494158 − 1.025332 0.991879 0.604810 0.217741 0.970600 0.655389 0.340177

− 0.920668 − 1.235880 − 1.551091 − 0.113334 − 0.159694 − 0.206054 0.086870 − 0.103595 − 0.294060 1.263217 1.000894 0.738572

0.819278 1.304092 1.788906 0.504172 0.557061 0.609950 − 0.770732 − 0.324426 0.121879 2.089606 1.604792 1.119978

2

3

4

The numerical stress and strain values for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Tables 6.13 and 6.14. These values will be used for the subsequent failure analysis in each lamina. The values of the strength ratio R are summarized in Table 6.15 for considered four failure criteria.

6.2 Short Solutions

183

Table 6.15 Strength ratio R in the local 1–2 coordinate systems for an external loading of Mxn = R × 1000 Nmm/mm and M yn = R × 500 Nmm/mm, asymmetric laminate Lamina No. Position 1

Top

Center

Bottom

2

Top

Center

Bottom

3

Top

Center

Bottom

4

Top

Center

Bottom

Direction

Max. stress

Max. Strain Tsai–Hill

1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12 1 2 12

26.264 6.232 15.074 15.123 4.498 9.470 10.618 3.519 6.904 781.227 60.905 24.495 20.410 21.126 22.170 9.945 12.779 20.247 20.774 5.314 16.024 34.893 29.854 38.067 108.929 26.874 101.329 19.112 1.257 5.910 27.788 1.629 7.696 50.893 2.313 11.027

34.275 6.995 46.382 18.585 5.211 29.139 12.749 4.152 21.242 575.974 56.823 75.371 21.552 40.327 68.215 10.387 31.254 62.300 21.495 22.678 49.304 35.251 62.165 117.130 97.914 21.900 311.784 21.966 1.560 18.185 32.530 1.968 23.679 62.673 2.667 33.929

Tsai–Wu

5.738

6.210

4.016

4.456

3.083

3.471

22.639

28.313

12.688

18.149

7.647

12.037

4.918

6.272

25.097

24.251

24.997

23.904

1.228

1.320

1.592

1.696

2.263

2.372

184

6 Supplementary Problems

6.2.4 Supplementary Problem 4: Ply-by-Ply Failure Loads • Original laminate (step 0) A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) for the original configuration (step 0) is provided in Fig. 6.15 as a function of the laminate thickness (z-coordinate). The numerical values for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Table 6.16. These values will be used for the subsequent failure analysis based on the Tsai–Wu criterion. The values of R for the classical Tsai–Wu criterion are summarized in Table 6.17. It can be concluded from Table 5.21 that the minimum strength ratio R is obtained for lamina 2, i.e., the 90◦ layer. Thus, we obtain the maximum load as N xn = R × 1 N/mm = 752.614 × 1 N/mm = 752.614 N/mm ,

(6.1)

which corresponds to a maximum stress of 752.614 N/mm N N xn = = 83.624 . mm2 t 9 mm

(6.2)

The corresponding maximum normal strain in the x-direction for this load is obtained as R × εx0 = 752.614 × 2.657817 × 10−3 0/00 = 2.0003100/00 .

(6.3)

For the next iteration, we consider the laminate without lamina 2, i.e., the 90◦ layer. • Laminate without lamina 2 (step 1) A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) for the reduced configuration (step 1) is provided in Fig. 6.16 as a function of the laminate thickness (z-coordinate). The numerical values for the bottom (z k−1 ), the middle ((z k−1 + z k )/2), and the top (z k ) of each lamina are summarized in Table 6.18. These values will be used for the subsequent failure analysis based on the Tsai–Wu criterion. The values of R for the classical Tsai–Wu criterion are summarized in Table 6.19. It can be concluded from Table 6.19 that the minimum strength ratio R is obtained for the remaining lamina 1 and 9, i.e., the 0◦ layers. Thus, we obtain the maximum load as N xn = R × 1 N/mm = 6840.0 × 1 N/mm = 6840.0 N/mm ,

(6.4)

6.2 Short Solutions

185

(a) z in mm 4.5 symmetric

1.5

−300

−200

−100

100

200

300

1

2

3

σx in 10−3 MPa

−1.5

−4.5

(b) z in mm 4.5

1.5

−3

−2

−1

εx in 10−6

−1.5

−4.5

Fig. 6.15 Distribution of selected field quantities over the laminate thickness for the tensile load case: a normal stress σx (z) and b normal strain εx (z), ply-by-ply failure (step 0)

186

6 Supplementary Problems

Table 6.16 Stresses in the local 1–2 coordinate systems for an external loading of N xn = 1 N/mm, symmetric laminate (step 0) Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

Top Center Bottom Top Center Bottom Top Center Bottom

0.143642 0.143642 0.143642 − 0.010806 − 0.010806 − 0.010806 0.143642 0.143642 0.143642

0.005403 0.005403 0.005403 0.046049 0.046049 0.046049 0.005403 0.005403 0.005403

0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

Table 6.17 Strength ratio R in the local 1–2 coordinate systems for an external loading of N xn = R × 1 N/mm, symmetric laminate (step 0) Position Tsai–Wu Lamina No. 1

2

3

Top Center Bottom Top Center Bottom Top Center Bottom

6889.844 6889.844 6889.844 752.614 752.614 752.614 6889.844 6889.844 6889.844

which corresponds to a maximum stress of N xn 6840.0 N/mm N = = 760.0 . t 9 mm mm2

(6.5)

The corresponding maximum normal strain in the x-direction for this load is obtained as R × εx0 = 6840.0 × 3.116430 × 10−3 0/00 = 21.3163810/00 .

(6.6)

The macroscopic stress–strain diagram in x-direction is shown in Fig. 6.17. It can be seen that a bilinear behavior is obtained in the elastic range for the entire laminate.

6.2 Short Solutions

187

(a) z in mm 4.5 symmetric

1.5

−300

−200

−100

100

200

300

1

2

3

σx in 10−3 MPa

−1.5

−4.5

(b) z in mm 4.5

1.5

−3

−2

−1

εx in 10−6

−1.5

−4.5

Fig. 6.16 Distribution of selected field quantities over the laminate thickness for the tensile load case after the first ply failure: a normal stress σx (z) and b normal strain εx (z), ply-by-ply failure (step 1)

188

6 Supplementary Problems

Table 6.18 Stresses in the local 1–2 coordinate systems for an external loading of N xn = 1 N/mm, symmetric laminate (step 1) Position σ1 in MPa σ2 in MPa σ12 in MPa Lamina No. 1

2

3

Top Center Bottom Top Center Bottom Top Center Bottom

0.166667 0.166667 0.166667 0.0 0.0 0.0 0.166667 0.166667 0.166667

0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

Table 6.19 Strength ratio R in the local 1–2 coordinate systems for an external loading of N xn = R × 1 N/mm, symmetric laminate (step 1) Position Tsai–Wu Lamina No. 1

2

3

Top Center Bottom Top Center Bottom Top Center Bottom

6840.0 6840.0 6840.0 – – – 6840.0 6840.0 6840.0

6.2.5 Supplementary Problem 5: Pole Diagrams of Elastic Properties for Unidirectional Laminae The graphical representations of the apparent elastic properties for the E-glass/epoxy lamina system are shown in Figs. 6.18, 6.19, 6.20, and 6.21 as pole diagrams and in Fig. 6.22 in a classical Cartesian representation.

6.2 Short Solutions

189

Nxn /t in MPa

800

last ply failure (0◦ )

600

400

200 first ply failure (90◦ ) 5

10

15

εx in

20

Fig. 6.17 Stress–strain curve showing the ply-by-ply failure of the [0◦ /90◦ ]s laminate under uniaxial loading with N xn 120◦

90◦ 60◦ Ey /E1

150

◦

30◦

0

180◦

0.5

1

1.5 ◦ 0

Ex /E1

210◦

330◦

240◦

270◦

300◦

Fig. 6.18 Pole diagram of the elastic properties E x /E 1 and E y /E 1 for E-glass/epoxy lamina system

190

6 Supplementary Problems

120◦

90◦ 60◦ Gxy /G12

150

◦

30◦

0

180◦

1

210◦

2

0◦

330◦

240◦

270◦

300◦

Fig. 6.19 Pole diagram of the elastic property G x y /G 12 for E-glass/epoxy lamina system 120◦

90◦ 60◦ νxy

150◦

0

180◦

30◦

0.25

0.5 ◦ 0

330◦

210◦

240◦

270◦

300◦

Fig. 6.20 Pole diagram of the elastic property νx y for E-glass/epoxy lamina system

6.2 Short Solutions

191

120◦

90◦ 60◦

150◦

30◦ −ηy

180

0

◦

1

−ηx

2

210◦

3

0◦

330◦

240◦

270◦

300◦

Fig. 6.21 Pole diagram of the elastic properties −ηx and −η y for E-glass/epoxy lamina system

6.2.6 Supplementary Problem 6: Failure Envelopes for Unidirectional Laminae Figures 6.23, 6.24, 6.25, and 6.26 present the different failure envelopes based on the four failure criteria.

192

6 Supplementary Problems

2.5

1.2 E-glass / epoxy lamina 1

2 Gxy /G12

0.8 0.6

Ey /E1

Ex /E1

1.5

1 0.4 0.5

0.2 0◦ 0

15◦

30◦

45◦

60◦

75◦

Elastic constant Gxy /G12 in –

Elastic constants Ex /E1 , Ey /E1 in –

(a)

0 90◦

Off-axis angle α (b)

Elastic constant νxy in –

E-glass / epoxy lamina 2 νxy

0.4

−ηy

−ηx

1

0.2 0

0◦ 0

15◦

30◦

45◦

60◦

75◦

Elastic constants −ηx , −ηy in –

3

0.6

−1 90◦

Off-axis angle α

Fig. 6.22 Elastic constants as a function of the off-axis angle: a E x /E 1 , E y /E 1 , G x y /G 12 and b νx y , −ηx , −η y

6.2 Short Solutions

193

(a)

Normal stress |σx | in MPa

2000 Maximum stress criterion E-glass / epoxy lamina 1500

tension (σx > 0) compression (σx < 0)

1000

500

0◦ 0

15◦

30◦

45◦

60◦

75◦

90◦

75◦

90◦

Off-axis angle α (b)

Normal stress | lg(σx )| in MPa

104 Maximum stress criterion E-glass / epoxy lamina

10

3

tension (σx > 0) compression (σx < 0)

102

0◦

15◦

30◦

45◦

60◦

Off-axis angle α

Fig. 6.23 Failure envelope for a unidirectional lamina (see Table 6.1 for details) according to the maximum stress criterion: a linear division of the axes and b logarithmic vertical axis

194

6 Supplementary Problems (a)

Normal stress |σx | in MPa

2000 Maximum strain criterion E-glass / epoxy lamina tension (σx > 0) compression (σx < 0)

1500

1000

500

0◦ 0

15◦

30◦

45◦

60◦

75◦

90◦

75◦

90◦

Off-axis angle α (b)

Normal stress | lg(σx )| in MPa

104 Maximum strain criterion E-glass / epoxy lamina

10

tension (σx > 0) compression (σx < 0)

3

102

0◦

15◦

30◦

45◦

60◦

Off-axis angle α

Fig. 6.24 Failure envelope for a unidirectional lamina (see Table 6.1 for details) according to the maximum strain criterion: a linear division of the axes and b logarithmic vertical axis

6.2 Short Solutions

195

(a)

Normal stress |σx | in MPa

2000 Tsai-Hill criterion E-glass / epoxy lamina 1500

tension (σx > 0) compression (σx < 0)

1000

500

0◦ 0

15◦

30◦

45◦

60◦

75◦

90◦

75◦

90◦

Off-axis angle α (b)

Normal stress | lg(σx )| in MPa

104 Tsai-Hill criterion E-glass / epoxy lamina

10

3

tension (σx > 0) compression (σx < 0)

102

0◦

15◦

30◦

45◦

60◦

Off-axis angle α

Fig. 6.25 Failure envelope for a unidirectional lamina (see Table 6.1 for details) according to the Tsai–Hill criterion: a linear division of the axes and b logarithmic vertical axis

196

6 Supplementary Problems (a)

Normal stress |σx | in MPa

2000 Tsai-Wu criterion E-glass / epoxy lamina 1500

1000

500

0◦ 0

15◦

30◦

45◦

60◦

75◦

90◦

75◦

90◦

Off-axis angle α (b)

Normal stress | lg(σx )| in MPa

104 Tsai-Wu criterion E-glass / epoxy lamina

103

102

0◦

15◦

30◦

45◦

60◦

Off-axis angle α

Fig. 6.26 Failure envelope for a unidirectional lamina (see Table 6.1 for details) according to the Tsai–Wu criterion: a linear division of the axes and b logarithmic vertical axis

Reference

197

Reference Soden PD, Hinton MJ, Kaddour AS (1998) Lamina properties, lay-up configurations and loading conditions for a range of fibre-reinforced composite laminates. Compos Sci Technol 58:1011– 1022

Chapter 7

Composite Laminate Analysis Tool—CLAT

Abstract This chapter provides a very brief introduction to a user-friendly computational tool, which allows the application of the classical laminate theory to different problems of composite laminates. The Python-based tool allows for classical stress and strain analyses, the generation of pole diagrams, and various approaches to the failure analysis.

7.1 Introduction The applications of the classical laminate theory as introduced in Chaps. 3 and 4 and the subsequent application to different problems such as stress and strain distributions, pole diagrams, and failure analyses based on different criteria (see the example problems in Chap. 5) require the handling of matrix equations and occasionally iterations depending on the problem at hand. Thus, a computational approach is appropriate to ease the application of the theory and to reduce the calculation time. A simple Internet search provides different approaches based on different programming languages or mathematical programs. The following section will briefly introduce the so-called composite laminate analysis tool (CLAT), a Python-based approach, which can be simply run as a web application in any common browser, see Öchsner and Makvandi (2023) for details.

7.2 Analysis Options Figure 7.1 shows the model definition interface of CLAT in a web browser. The definition of material and geometrical properties as well as load conditions can be either directly done in this interface or imported, as well as exported, as an Excel file.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, Composite Mechanics, Advanced Structured Materials 184, https://doi.org/10.1007/978-3-031-32390-4_7

199

200

7 Composite Laminate Analysis Tool—CLAT

Fig. 7.1 Model definition in CLAT for the laminate shown in Fig. 5.1, tensile load case

Fig. 7.2 Analysis options and layer representation in CLAT for a laminate

7.2 Analysis Options

201

4

4

3

3

2

2

1

1

0

0

1

1

2

2

3

3

4

4 200

0

200

200

0

200

Fig. 7.3 Distribution of selected field quantities over the laminate thickness for the tensile load case: a normal stress σ1 (z) and b normal stress σx (z)

Fig. 7.4 Numerical stress values in different coordinate systems (1–2 or x–y) and locations (bottom, center, or top) in the laminate

202

7 Composite Laminate Analysis Tool—CLAT

The different CLAT analysis options are indicated in Fig. 7.2. The options range from the classical stress and strain analysis of a single lamina or the stacking to comprehensive laminates. In addition to this elastic analysis, the failure can predicted based on the maximum stress, the maximum strain, the Tsai–Hill, as well as the Tsai– Wu criterion. Furthermore, particular representations of a single lamina, i.e., pole diagrams of the elastic parameters as well as failure envelopes for different angles can be generated. The lower part of Fig. 7.2 shows the layer representation in CLAT. Some selected results are represented in Figs. 7.3 and 7.4. Figure 7.3 shows the graphical stress distribution in the 1- and x-direction over the laminate thickness. The corresponding diagrams for 2 (or y) and 12 (or x y) are also available in CLAT. It might be required for post-processing purposes to have access to the numerical values of any graphical representation. Thus, Fig. 7.4 provides the numerical stress values, which can be exported in Excel format for further processing. This option is available for any diagrams provided in CLAT.

Reference Öchsner A, Makvandi R (2023) A numerical approach to the classical laminate theory of composite materials: the composite laminate analysis tool–CLAT v2.0. Springer, Cham

Index

A AS4 carbon/3501-6 epoxy properties, 96 B Beltrami equations, 80 Beltrami-Michell equations, 80 C Classical laminate theory, v, 8, 9 assumptions, 9 example problems, 95 literature, 7 steps for calculation, 93 supplementary problems, 159 Classical lamination theory, see classical laminate theory Classical plate element constitutive equation, 64 equilibrium, 73 kinematics, 59 CLAT, see composite laminate analysis tool Compliance matrix, 64, 65 rotated, 69 Composite laminate analysis tool, 199 Computational approach, 8, 199 finite element method, 8 Constitutive equation, 6 classical plate element, 64 combined element, 65 plane elasticity element, 63 Continuum mechanical modeling, 8

E E-glass / epoxy properties, 160 Elastic constants rotated lamina, 69, 146 Elasticity matrix, 64, 65 generalized, 87 Elasticity modulus, 63 Equilibrium equation, 6 classical plate element, 77 combined element, 78 plane elasticity element, 70 Euler-Bernoulli beam, 81

F Failure analysis of laminates, 93, 118, 132, 161, 162 Failure criterion maximum strain, 82 maximum stress, 82 Tsai-Hill, 83 Tsai-Wu, 83 Failure envelope, 151, 162 Fiber, 4

G Generalized elasticity matrix, 79, 87, 88 special cases, 88 Generalized strains, 62, 87 Generalized stresses, 72, 87

H Hooke’s law, 63 isotropic, 63

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, Composite Mechanics, Advanced Structured Materials 184, https://doi.org/10.1007/978-3-031-32390-4

203

204 orthotropic, 65

I Inverse rule of mixtures, 19 Isotropic material, 63, 64, 80

K Kinematics relation, 6, 58 classical plate element, 59 combined element, 61 plane elasticity element, 59

L Lamé-Navier equations, 79 Lamina, 4 failure envelope, 151, 162 pole diagram, 145, 162 Laminate, 4 angle-ply, 90 balanced, 91 cross-ply, 90 failure analysis, 93, 118, 132, 161, 162 orientation code, 4 quasi-isotropic, 92 symmetric, 88 symmetric with isotropic layers, 89 Laminate orientation code, 4

M Macromechanics lamina, 57 laminate, 85 Material properties, 2, 3, 96, 160 classical materials, 2 fibers, 3 lamina, 96, 160 matrices, 3 Matrix, 3

Index Mechanics of materials approach, 15 Micromechanics, 13, 95 concept, 3 elasticity solutions after Tsai, 25 experimental results, 34 Halpin-Tsai approximations, 31 mechanics of materials approach, 15 optimized predictions, 47

O Orthotropic material, 65 rotation, 65

P Plane elasticity element constitutive equation, 63 equilibrium, 70 kinematics, 59 Poisson’s ratio, 63, 70 Pole diagram, 145, 162 Prepreg, 95 Python, 199

R Reuss rule, 19 Rule of mixtures, 16

S Shear-extension coupling coefficient, 70 Shear modulus, 65 Steps laminate calculation, 93 Strength ratio, 125, 131, 138, 142, 152 Stress resultants, 72, 74, 78

V Voigt rule, 16