A Simplified Approach to the Classical Laminate Theory of Composite Materials: Application of Bar and Beam Elements (Advanced Structured Materials, 192) [1st ed. 2023] 3031381912, 9783031381911

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Advanced Structured Materials

Andreas Öchsner

A Simplified Approach to the Classical Laminate Theory of Composite Materials Application of Bar and Beam Elements

Advanced Structured Materials Volume 192

Series Editors Andreas Öchsner, Faculty of Mechanical and Systems 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

A Simplified Approach to the Classical Laminate Theory of Composite Materials Application of Bar and Beam Elements

Andreas Öchsner Faculty of Mechanical and Systems Engineering Esslingen University of Applied Sciences Esslingen, Germany

ISSN 1869-8433 ISSN 1869-8441 (electronic) Advanced Structured Materials ISBN 978-3-031-38191-1 ISBN 978-3-031-38192-8 (eBook) https://doi.org/10.1007/978-3-031-38192-8 © 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 limitations of many classical engineering materials. Particularly the combination of a matrix with fibers provides far better properties than each component alone. Despite their importance, many engineering degree programs do not treat the mechanical behavior of this class of advanced structured materials in detail. Thus, some engineers are not able to thoroughly apply and introduce these modern engineering materials in their design process. 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 kinematics relationship, the constitutive law, and the equilibrium equation, differential equations are derived to describe the physical problem. The so-called classical 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. The procedure provides the solution of a 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. This laminate theory is typically provided for twodimensional plane problems, where the basic structural element is a simple superposition of a classical plane elasticity element with a thin plate element under the consideration of an orthotropic constitutive law. This two-dimensional approach and the underlying advanced continuum mechanical modeling might be very challenging for some students, particularly at universities of applied sciences. Thus, a reduced approach, the so-called simplified classical laminate theory, has been developed. The idea is to use solely isotropic one-dimensional elements, i.e., a superposition of bar and beam elements, to introduce the major calculation steps of the classical laminate theory. Understanding this simplified theory is much easier and the final step is to highlight the differences when moving to the general two-dimensional case. This monograph provides a systematic introduction to composite materials, which are obtained by a layer-wise stacking of one-dimensional bar/beam elements. Each v

vi

Preface

layer may have different mechanical properties but each single layer is considered as isotropic. The major idea is to provide a simplified theory to easily understand the classical two-dimensional laminate theory for composites based on laminae with unidirectional fibers. In addition to the elastic behavior, failure is investigated based on the maximum stress, maximum strain, Tsai–Hill and the Tsai–Wu criteria. Esslingen, Germany May 2023

Andreas Öchsner

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 5

2 Bar Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Constitutive Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 7 8 9 10 12 14

3 Euler–Bernoulli Beam Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Constitutive Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15 15 18 23 27 29 32

4 Combination of Bar and Beam Elements . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Constitutive Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Failure Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.1 Maximum Stress Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.2 Maximum Strain Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.3 Tsai–Hill Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.4 Tsai–Wu Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33 33 33 35 35 37 38 38 40 40 40 41

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Contents

5 Classical Laminate Theory for One-Dimensional Elements . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Generalized Stress–Strain Relationship . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Failure Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43 43 44 47 48

6 Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Problem 1: Stresses and Strains in a Symmetric Laminate . . . . . . . . 6.3 Problem 2: Stresses and Strains in an Asymmetric Laminate . . . . . . 6.4 Problem 3: Failure Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Problem 4: Ply-By-Ply Failure Loads . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49 49 49 57 66 78 88

7 Outlook to the Two-Dimensional Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Failure Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

89 89 92 96

Appendix A: Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

97

Appendix B: Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Appendix C: Units and Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

Symbols and Abbreviations

Latin Symbols (Capital Letters) A A A A A B B B B B C C C C C* D D D D D E F G I L M M N N

Area, cross-sectional area, matrix element of generalized elasticity matrix Matrix element of the inverted generalized elasticity matrix Matrix element Submatrix of the generalized elasticity matrix Submatrix of the inverted generalized elasticity matrix Matrix element of generalized elasticity matrix Matrix element of inverted generalized elasticity matrix Matrix element Submatrix of the generalized elasticity matrix Submatrix of the inverted generalized elasticity matrix Diameter, elasticity constant Element of the transformed elasticity matrix (to the 1–2 system) Elasticity matrix Transformed elasticity matrix (to the 1–2 system) Generalized elasticity matrix Bending stiffness matrix element of generalized elasticity matrix Matrix element of the inverted generalized elasticity matrix Matrix element Elastic compliance matrix submatrix of the generalized elasticity matrix Submatrix of the inverted generalized elasticity matrix Young’s modulus Force Shear modulus Second moment of area Element length Moment Column matrix of internal moments Normal force (internal) Column matrix of internal normal forces ix

x

Q R T U

Symbols and Abbreviations

Shear force (internal) Radius, radius of curvature, strength ratio Transformation matrix Perimeter

Latin Symbols (Small Letters) a b e f h k m n n p q q s s t t u u x y z

Geometric dimension (width), coefficient Volume-specific load, geometric dimension Column matrix of generalized strains Body force Geometric dimension Elastic embedding modulus, elastic foundation modulus, strength value Distributed moment Total number of layers Normal vector Distributed load in x-direction Distributed load in y-direction Column matrix of distributed loads Length Column matrix of generalized stresses Geometric dimension (height), traction force Tangential vector Displacement Column matrix of displacements Cartesian coordinate Cartesian coordinate Cartesian coordinate

Latin Numbers 1 2 3

Cartesian coordinate Cartesian coordinate Cartesian coordinate

Symbols and Abbreviations

Greek Symbols α γ ε ε0 ε0 ε k k σ σ ϕ

Rotation angle Shear strain (engineering definition) Normal strain Normal strain in neutral fiber Column matrix of strains in neutral fiber Column matrix of strain components Curvature Column matrix of curvatures Stress, normal stress Column matrix of stress components Rotation

Mathematical Symbols d… d( . . . ) dx L{. . . } L ( … )T |…|

Differential quantity First-order derivative with respect to the x-axis Differential operator Matrix of differential operators Transpose Absolute value

Indices, Superscripted …n …ε

Normalized (length-specific) Strain related

Indices, Subscripted …c …k …sym …t …ε …σ

Compression Layer number Symmetric Tensile Strain related Stress related

xi

xii

Symbols and Abbreviations

Abbreviations 1D 2D BEM CLT FDM FEM FVM Mat PDE SCLT

One dimensional Two dimensional Boundary element method Classical laminate theory Finite difference method Finite element method Finite volume method Material Partial differential equation Simplified classical laminate theory

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 much better properties than a single component for itself. The focus is many times on fiber-reinforced composites where a single layer has unidirectionally aligned reinforcing fibers. The basic equations of continuum mechanics to model such two-dimensional structures are introduced. A common approach is to avoid to solve the coupled system of differential equations. This is done by the so-called classical laminate theory for two-dimensional orthotropic composite materials. To facilitate the understanding, the idea of a simplified classical laminate theory is outlined. This reduced approach is based on one-dimensional bar/beam elements at which each single layer reveals isotropic properties.

A typical concept to design materials with improved 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, see Fig. 1.1. It is common to describe such a unidirectional lamina based on the (1, 2, 3) coordinate system where the 1-axis is aligned with the parallel fibers, the 2-axis is in-plane perpendicular to the fibers, and the 3-axis stands perpendicular to the 1–2 plane. The mechanical behavior of such a lamina is many times approximated as a two-dimensional superposition of a plane elasticity element (loading and deformation only in the 1–2 plane) and a classical plate element (only bending deformation). Furthermore, the constitutive description is based on an orthotropic Hooke’s law in the linear-elastic range. 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, see Fig. 1.2. The mechanical behavior is now described with reference to a global (x, y, z)-coordinate system. The orientation of a single

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, A Simplified Approach to the Classical Laminate Theory of Composite Materials, Advanced Structured Materials 192, https://doi.org/10.1007/978-3-031-38192-8_1

1

2

1 Introduction

Fig. 1.1 Schematic representation of a unidirectional lamina

Fig. 1.2 Unbonded view of single laminae forming a five-layer laminate

layer k with respect to the global coordinate system is expressed by the rotational angle αk , which is drawn from the global x-axis to the 1k -axis of layer k. The simplified approach, i.e., the simplified classical laminate theory (SCLT), will still consider a layer-wise composition of a mechanical member but is based on the following, see also Fig. 1.3: • The two-dimensional combination of a plane elasticity and classical plate element is replaced by a superposition of a two-dimensional bar and beam element. Bending occurs only in a single plane. • Each single layer k is considered as isotropic and homogeneous. The constitutive description is based on the one-dimensional Hooke’s law. It is obvious that this simplified approach disregards the transformations of stresses, strains, and stiffnesses between the (1, 2)- and (x, y)-coordinate systems.

1 Introduction

3

Fig. 1.3 Unbonded view of single bar/beam elements forming a five-layer composite

Nevertheless, the continuum mechanical modeling as well as the composition of a composite element by layers is well included. Typical thin laminates composed of orthotropic laminae (plies) can be modeled under particular assumptions as a simple superposition of a plane elasticity and a classical plate element. The same holds for our simplified composite model which consists of the superposition of one-dimensional bar and beam elements. These one-dimensional elements are described separately in Chaps. 2–3 and subsequently composed to a combined bar/beam element, see Chap. 4. 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 2023; Öchsner and Öchsner 2018) for general details on the method and Table 1.1 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

4

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 1.1 Some of the early textbooks which cover the finite element simulation of composite materials (no claim to completeness) Year (1st ed.) Author Title Refs. 1992

O.O. Ochoa, J.N. Reddy

1998

L.T. Tenek, J. Argyris

2000

F.L. Matthews, G.A.O. Davies, D. Hitchings, C. Soutis E.J. Barbero

2008

Finite element analysis of composite laminates Finite element analysis for composite structures Finite element modelling of composite materials and structures Finite element analysis of composite materials

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

Barbero (1996)

References

5

lot of experience and a solid foundation in the basics of applied mechanics, materials science, and mathematics. The Chap. 5 presents a simplified approach, the so-called (simplified) classical laminate theory (CLT) (Stravsky 1961; Dong et al. 1962), which is adopted to our one-dimensional bar/beam element. This approach aims to provide a stress and a subsequent strength/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 or generalized stresses and strains. It should be noted here that this simplified approach based on one-dimensional members has been already successfully applied in the context of the finite element method (Öchsner 2018), lightweight design (Öchsner 2022), and structural optimization (Öchsner and Makvandi 2020).

References Altenbach H, Öchsner A (eds) (2020) Encyclopedia of continuum mechanics. Springer, Berlin Barbero EJ (1996) Finite element analysis of composite materials. CRC Press, Boca Raton Dong SB, Pister KS, Taylor RL (1962) On the theory of laminated anisotropic shells and plates. J Aerosp Sci 29:696–975. https://doi.org/10.2514/8.9668 Matthews FL, Davies GAO, Hitchings D, Soutis C (2000) Finite element modelling of composite materials and structures. Woodhead Publishing Limited, Cambridge Ochoa OO, Reddy JN (1992) Finite element analysis of composite laminates. Kluwer Academic Publishers, Dordrecht Öchsner A, Merkel M (2018) One-Dimensional finite elements: an introduction to the FE method. Springer, Cham Öchsner A, Öchsner M (2018) A first introduction to the finite element analysis program MSC Marc/Mentat. Springer, Cham Öchsner A, Makvandi R (2020) Numerical engineering optimization: application of the computer algebra system maxima. Springer, Cham Öchsner A (2022) Stoff- und Formleichtbau: Leichter Einstieg mit eindimensionalen Strukturen. Springer Vieweg, Wiesbaden Öchsner A (2023) Computational statics and dynamics: an introduction based on the finite element method. Springer, Cham Stravsky Y (1961) Bending and stretching of laminated aeolotropic plates. J Eng Mech Div-ASCE 87:31–56. https://doi.org/10.1061/JMCEA3.0000267 Tenek LT, Argyris J (1998) Finite element analysis for composite structures. Springer, Dordrecht

Chapter 2

Bar Elements

Abstract This chapter provides the fundamental description of one-dimensional bar members. Based on the three basic equations of continuum mechanics, i.e., the kinematics relationship, the constitutive law (here: one-dimensional Hooke’s law for isotropic materials), and the equilibrium equation, the partial differential equation, which describes the physical problem, is derived. In addition to the classical notation, a formal notation is introduced. This allows to highlight the similarities between the different structural members.

2.1 Introduction A bar is defined as a prismatic body whose axial dimension is much larger than its transverse dimensions (Öchsner 2014). This structural member is only loaded in the direction of the main body axes, see Fig. 2.1a. As a result of this loading, the deformation occurs only along its main axis. The following derivations are restricted to some simplifications: • • • •

only applying to straight rods, displacements are (infinitesimally) small, strains are (infinitesimally) small, material is linear elastic (homogeneous and isotropic).

The ultimate goal of the classical modeling approach is to derive a partial differential equation, which describes at any location x the deformation (displacement) u x (x). For simple problems, e.g., constant properties (E, A) and constant loads, the analytical solution can be obtained from the differential equation by simple integration and adjusting the constants of integration based on the boundary conditions of the problem. It should be noted here that the alternatively nomenclature “rod” is also found in scientific literature to describe a bar member. Details on the continuum mechanical description of bars can be found in Gross et al. (2011); Hartsuijker and Welleman (2007) and the basic equations are derived in detail in the following sections. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, A Simplified Approach to the Classical Laminate Theory of Composite Materials, Advanced Structured Materials 192, https://doi.org/10.1007/978-3-031-38192-8_2

7

8

2 Bar Elements

Fig. 2.1 General configuration for bar problems: a example of boundary conditions and external loads; b cross-sectional area

2.2 Kinematics To derive the strain–displacement relation (kinematics relation), an axially loaded bar is considered as shown in Fig. 2.2. The length of the member is equal to L and the constant axial tensile stiffness is equal to E A. The load is either given as a single force Fx and/or as a distributed load px (x). This distributed load has the unit of force per unit length. In the case of a body force f x (unit: force per unit volume), the distributed load takes the form px (x) = f x (x)A(x) where A is the cross-sectional area of the bar. A typical example for a body force would be the dead weight, i.e., the mass under the influence of gravity. In the case of a traction force tx (unit: force per unit area), the distributed load can be written as px (x) = tx (x)U (x) where U (x) is the perimeter of the cross section. Typical examples are frictional resistance, viscous drag, and surface shear. Let us now consider a differential element dx of such a bar as shown in Fig. 2.3. Under an acting load, this element deforms as indicated in Fig. 2.3b where the initial point at the position x is displaced by u x and the end point at the position x + dx

Fig. 2.2 General configuration of an axially loaded bar: a geometry and material property; b prescribed loads

2.3 Constitutive Equation

9

Fig. 2.3 Elongation of a differential element of length dx: a undeformed configuration; b deformed configuration

is displaced by u x + du x . Thus, the differential element which has a length of dx in the unloaded state elongates to a length of dx + (u x + du x ) − u x . The engineering strain, i.e., the increase in length related to the original length, can be expressed as εx =

(dx + (u x + du x ) − u x ) − (dx) , dx

(2.1)

or finally as εx (x) =

du x (x) . dx

(2.2)

The last equation is often expressed in a less mathematical way (non-differential) as εx = ΔL where ΔL is the change in length of the entire bar element. L ) If we replace the common formulation of the first-order derivative, i.e., d(... , by a dx formal operator symbol, i.e., L1 (. . . ), the kinematics relation can be stated in a more formal way as εx (x) =

d u x (x) = L1 u x (x) . dx 

(2.3)

L1

2.3 Constitutive Equation The constitutive equation, i.e., the relation between the stress σx and the strain εx , is given in its simplest form as Hooke’s law1 σx (x) = Eεx (x),

1

Robert Hooke (1635–1703), English natural philosopher, architect, and polymath.

(2.4)

10

2 Bar Elements

Fig. 2.4 Axially loaded bar: a strain and b stress distribution

where Young’s modulus2 E is in the case of linear elasticity a material constant. For the considered bar element, the normal stress and strain are constant over the cross section as shown in Fig. 2.4. Using a more general notation, Hooke’s law can be written as σx (x) = Cεx (x) ,

(2.5)

where the scalar constant C = E transforms in the two- or three-dimensional case the so-called elasticity matrix.

2.4 Equilibrium The equilibrium equation between the external forces and internal reactions can be derived for a differential element of length dx as shown in Fig. 2.5. It is assumed for simplicity that the distributed load px and the cross-sectional area A are constant in this figure. The internal reactions N x are drawn in their positive directions, i.e., at the left-hand face in the negative and at the right-hand face in the positive x-direction. The force equilibrium in the x-direction for a static configuration requires that − N x (x) + px dx + N x (x + dx) = 0

(2.6)

holds. A first-order Taylor’s3 series expansion (cf. Appendix A.6) of the normal force N x (x + dx) around point x, i.e.,

2 3

Thomas Young (1773–1829), English polymath. Brook Taylor (1685–1731), English mathematician.

2.4 Equilibrium

11

Fig. 2.5 Differential element of a bar with internal reactions and constant external distributed load Table 2.1 Fundamental governing equations of a bar for deformation along the x-axis

Expression

Equation

Kinematics

εx (x) =

du x (x) dx

dN x (x) = − px (x) dx σx (x) = Eεx (x)

Equilibrium Constitution

 dN x  N x (x + dx) ≈ N x (x) +  dx , dx 

(2.7)

x

allows to finally express Eq. (2.6) as dN x (x) = − px (x) , dx

(2.8)

or with the operator symbol L as4 LT1 (N x (x)) = − px (x) .

(2.9)

The three fundamental equations to describe the behavior of a bar element are summarized in Table 2.1. A slightly different derivation of the equilibrium equation is obtained as follows: Equation (2.6) can be expressed based on the normal stresses as − σx (x)A + px dx + σx (x + dx)A = 0 .

(2.10)

A first-order Taylor’s series expansion of the stress σx (x + dx) around point x, i.e.,

 T ) It should be noted here that the transpose (“T”), i.e., LT1 = d(... , is only used to show later dx similar structures of the equations in the two-dimensional case.

4

12

2 Bar Elements

 dσx  σx (x + dx) ≈ σx (x) +  dx , dx 

(2.11)

x

allows to finally express Eq. (2.10) as dσx (x) px (x) + = 0. dx A The last equation with σx =

Nx A

(2.12)

immediately gives Eq. (2.8).

Finally, it should be noted here that the general definition of the internal normal force N x (x) is based on the normal stress distribution as N x (x) = σx (x)d A . (2.13) A

In the case of a rectangular cross section of width a and height t, this can be written based on Hooke’s law as t

2 N x (x) =

a Eεx (x)dz = ta Eεx (x) . −

(2.14)

t 2

2.5 Differential Equation To derive the governing partial differential equation, the three fundamental equations given in Table 2.1 must be combined. Introducing the kinematics relation (2.2) into Hooke’s law (2.4) gives σx (x) = E

du x . dx

(2.15)

Considering in the last equation that a normal stress is defined as an acting force N x over a cross-sectional area A: du x Nx =E . A dx

(2.16)

The last equation can be differentiated with respect to the x-coordinate to give

 dN x d du x = EA , (2.17) dx dx dx

2.5 Differential Equation

13

where the derivative of the normal force can be replaced by the equilibrium equation (2.8) to obtain in the general case:

 d du x (x) (2.18) E(x)A(x) = − px (x) . dx dx The general case in the formulation with the internal normal force distribution reads E(x)A(x)

du x (x) = N x (x) . dx

(2.19)

Thus, to obtain the displacement field u x (x), one may start from Eq. (2.18) or from Eq. (2.19). The first approach requires to state the distribution of the distributed load px (x) while for the second approach one requires the internal normal force distribution N x (x). If the axial tensile stiffness E A is constant, the formulation (2.18) can be simplified to EA

d2 u x (x) = − px (x) . dx 2

(2.20)

Some common formulations of the governing partial differential equation are collected in Table 2.2. It should be noted here that some of the different cases given in Table 2.2 can be combined. The last case in Table 2.2 refers to the case of elastic embedding of a bar where the embedding modulus k has the unit of force per unit

Table 2.2 Different formulations of the partial differential equation for a bar (x-axis: right facing) Configuration Partial Differential Equation d2 u x =0 dx 2

EA d dx

du x E(x)A(x) dx



EA

d2 u x = − px (x) dx 2

EA

d2 u x = k(x)u x dx 2

=0

14

2 Bar Elements

Table 2.3 Different formulations of the basic equations for a bar (x-axis along the principal rod axis). E: Young’s modulus; A: cross-sectional area; px : length-specific distributed normal load; ) L1 = d(... dx : first-order derivative; b: volume-specific distributed normal load Specific formulation

General formulation

Kinematics εx (x) =

du x (x) dx

εx (x) = L1 (u x (x))

Constitution σx (x) = Eεx (x)

σx (x) = Cεx (x)

Equilibrium dσx (x) px (x) + =0 dx A

LT1 (σx (x)) + b = 0

dN x (x) + px (x) = 0 dx

LT1 (N x (x)) + px (x) = 0

PDE (A = const.)

 d du x px (x) E(x) + =0 dx dx A

LT1 (C L1 (u x (x))) + b = 0

or LT1 (E AL1 (u x (x))) + px = 0

area. Analytical solutions for different loading and support conditions can be found, for example, in Öchsner (2014). ) , by If we replace the common formulation of the first-order derivative, i.e., d(... dx a formal operator symbol, i.e., L1 (. . . ), the basic equations can be stated in a more formal way as given in Table 2.3. Such a formulation is advantageous in the two- and three-dimensional cases (Altenbach and Öchsner 2020). It should be noted here that T  ) , is only used to show later similar structures the transpose (“T”), i.e., LT1 = d(... dx of the equations in the two- and three-dimensional case.

References Altenbach H, Öchsner A (eds) (2020) Encyclopedia of Continuum Mechanics. Springer, Berlin Gross D, Hauger W, Schröder J, Wall WA, Bonet J (2011) Engineering Mechanics 2: Mechanics of Materials. Springer, Berlin Hartsuijker C, Welleman JW (2007) Engineering Mechanics Volume 2: Stresses, Strains, Displacements. Springer, Dordrecht Öchsner A (2014) Elasto-Plasticity of Frame Structure Elements: Modeling and Simulation of Rods and Beams. Springer, Berlin

Chapter 3

Euler–Bernoulli Beam Elements

Abstract This chapter provides the fundamental description of one-dimensional thin beam members. Based on the three basic equations of continuum mechanics, i.e., the kinematics relationship, the constitutive law (here: one-dimensional Hooke’s law for isotropic materials), and the equilibrium equation, the partial differential equation, which describes the physical problem, is derived in different formulations. In addition to the classical notation, a formal notation is introduced. This allows to highlight the similarities between the different structural members.

3.1 Introduction A thin beam is defined as a long prismatic body as schematically shown in Fig. 3.1a. The following derivations are restricted to some simplifications: • • • • • • •

only applying to straight beams, no elongation along the x-axis, no torsion around the x-axis, no contribution of the shear force on the deformation, deformations in a single plane, i.e., symmetrical bending, small deformations, and simple cross sections.

The external loads, which are considered within this chapter, are single forces Fz , single moments M y , distributed loads qz (x), and distributed moments m y (x). These loads have in common that their line of action (force) or the direction of the momentum vector is orthogonal to the center line of the beam and cause its bending. This is a different type of deformation compared to the bar element from Chap. 2, see Table 3.1. It should be noted here that these basic types of deformation can be superposed to account for more complex loading conditions Boresi and Schmidt (2003).

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, A Simplified Approach to the Classical Laminate Theory of Composite Materials, Advanced Structured Materials 192, https://doi.org/10.1007/978-3-031-38192-8_3

15

16

3 Euler–Bernoulli Beam Elements

Fig. 3.1 General configuration for beam problems: a example of boundary conditions and external loads; b cross-sectional area

Table 3.1 Differentiation between bar and beam element; center line parallel to the x-axis Bar Beam Force Unknown

Along the bar axis Displacement along bar axis u x

Perpendicular to the beam axis displacement perpendicular and rotation perpendicular to the beam axis u z , ϕ y

The classic theories of beam bending distinguish between shear-rigid and shearflexible models. The shear-rigid beam, also called the Bernoulli1 beam,2 neglects the shear deformation from the shear forces. This theory implies that a cross-sectional plane which was perpendicular to the beam axis before the deformation remains in the deformed state perpendicular to the beam axis, see Fig. 3.2a. Furthermore, it is assumed that a cross-sectional plane stays plane and unwarped in the deformed state. These two assumptions are also known as Bernoulli’s hypothesis. Altogether one imagines that cross-sectional planes are rigidly fixed to the center line of the beam3 so that a change of the center line affects the entire deformation. Consequently, it is also assumed that the geometric dimensions4 of the cross-sectional planes do not change. In the case of a shear-flexible beam, also called the Timoshenko5 beam, the shear deformation is considered in addition to the bending deformation and cross-sectional planes are rotated by an angle γ compared to the perpendicular line, see Fig. 3.2b. For 1

Jakob I. Bernoulli (1655–1705), Swiss mathematician and physicist. More precisely, this beam is known as the Euler–Bernoulli beam. A historical analysis of the development of the classical beam theory and the contribution of different scientists can be found in Heyman (1998). 3 More precisely, this is the neutral fiber or the bending line. 4 Consequently, the width b and the height t of a, for example, rectangular cross section remains the same, see Fig. 3.1b. 5 Stepan Prokopovych Timoshenko (1878–1972), Ukrainian/US engineer. 2

3.1 Introduction

17

Fig. 3.2 Different deformation modes of a bending beam: a shear-rigid; b shear-flexible. Adapted from Hartmann and Katz (2007)

Fig. 3.3 Different stress distributions of a beam with rectangular cross section and linear-elastic material behavior: a normal stress and (b) shear stress

beams for which the length is 10–20 times larger than a characteristic dimension of the cross section, the shear fraction is usually disregarded in the first approximation. The different load types, meaning pure bending moment loading or shear due to shear force, lead to different stress fractions in a beam. In the case of a Bernoulli beam, deformation occurs solely through normal forces, which are linearly distributed over the cross section. Consequently, a tension—alternatively a compression maximum on the bottom face—maximum on the top face occurs, see Fig. 3.3a. In the case of symmetric cross sections, the zero crossing6 occurs in the middle of the cross section. The shear stress distribution for a rectangular cross section is parabolic at which the maximum occurs at the neutral axis and is zero at both the top and bottom surfaces, see Fig. 3.3b. This shear stress distribution can be calculated for the Bernoulli beam but is not considered for the derivation of the deformation. Finally, it needs to be noted that the one-dimensional beam theories have corresponding counterparts in two-dimensional space, see Table 3.2. In plate theories, the Bernoulli beam corresponds to the shear-rigid Kirchhoff7 plate and the Timoshenko beam corresponds to the shear-flexible Reissner8 –Mindlin9 plate, Timoshenko (1959), Gould (1988), Altenbach et al. (1998).

The sum of all points with σ = 0 along the beam axis is called the neutral fiber. Gustav Robert Kirchhoff (1824–1887), German physicist. 8 Eric Reissner (1913–1996), German/US engineer. 9 Raymond David Mindlin (1906–1987), US engineer. 6 7

18

3 Euler–Bernoulli Beam Elements

Table 3.2 Analogies between the beam and plate theories Beam theory Dimensionality Shear-rigid Shear-flexible

1D Bernoulli beam Timoshenko beam

Plate theory 2D Kirchhoff plate Reissner–Mindlin plate

Further details regarding the beam theory and the corresponding basic definitions and assumptions can be found in references Szabó (2003), Gross et al. (2009), Budynas (1999), Hibbeler (2008). In the following two chapters, only the Bernoulli beam is considered. Consideration of the shear part is, for example, treated in Öchsner (2014).

3.2 Kinematics For the derivation of the kinematics relation, a beam with length L is under constant moment loading M y (x) = const., meaning under pure bending, is considered, see Fig. 3.4. One can see that both external single moments at the left- and right-hand boundary lead to a positive bending moment distribution M y within the beam. The vertical position of a point with respect to the center line of the beam without action of an external load is described through the z-coordinate. The vertical displacement of a point on the center line of the beam, meaning for a point with z = 0, under action of the external load is indicated with u z . The deformed center line is represented by the sum of these points with z = 0 and is referred to as the bending line u z (x). In the case of a deformation in the x-z-plane, it is important to precisely distinguish between the positive orientation of the internal reactions, the positive rotational angle, and the slope, see Fig. 3.5. The internal reactions at a right-hand boundary are directed in the positive directions of the coordinate axes. Thus, a positive moment at a righthand boundary is clockwise oriented (as the positive rotational angle), see Fig. 3.5a. However, the slope is negative, see Fig. 3.5b. This difference requires some careful derivations of the corresponding equations. Only the center line of the deformed beam is considered in the following. Through the relation for an arbitrary point (x, u z ) on a circle with radius R around the center point (x0 , z 0 ), meaning (x − x0 )2 + (u z (x) − z 0 )2 = R 2 ,

(3.1)

one obtains through differentiation with respect to the x-coordinate 2(x − x0 ) + 2(u z (x) − z 0 )

du z (x) = 0, dx

(3.2)

3.2 Kinematics

19

Fig. 3.4 Beam under pure bending in the x-z-plane: a moment distribution; b deformed beam. Note that the deformation is exaggerated for better illustration. For the deformations considered in this chapter the following applies: R  L

Fig. 3.5 Positive definition of a internal reactions and b rotation (but negative slope)

alternatively after another differentiation: 2+2

du z du z d2 u z + 2(u z (x) − z 0 ) 2 = 0 . dx dx dx

(3.3)

20

3 Euler–Bernoulli Beam Elements

Equation (3.3) provides the vertical distance between an arbitrary point on the center line of the beam and the center point of a circle as  1+ (u z − z 0 ) = −

du z dx

2 ,

d2 u z dx 2

(3.4)

while the difference between the x-coordinates results from Eq. (3.2): (x − x0 ) = −(u z − z 0 )

du z . dx

(3.5)

If the expression according to Eq. (3.4) is used in Eq. (3.5) the following results:  du z (x − x0 ) = dx

1+

du z dx

2

d2 u z dx 2

.

(3.6)

Inserting both expressions for the x- and z-coordinate differences according to Eqs. (3.6) and (3.4) in the circle equation according to (3.1) leads to R 2 = (x − x0 )2 + (u z − z 0 )2    2 2  2 2 du z  2 1 + du z 1 + dx dx du z + =  2 2  2 2 dx d uz d uz dx 2

⎛ =⎝

d2 u z dx 2

2

dx 2

⎞ + 1⎠

  2 2 z 1 + du dx

2 ⎞3 du z ⎠ ⎝1 + dx = .  2 d2 u z dx 2 ⎛

(3.7)



d2 u z dx 2

2



(3.8)

3.2 Kinematics

21

Fig. 3.6 Definition of a negative curvature in the x-z-plane

Thus, the radius of curvature is obtained as ⎛



⎝1 +

du z dx

2 ⎞3/2 ⎠

d2 u z 2 dx

|R| =

.

(3.9)

To decide if the radius of curvature is positive or negative, let us have a look at Fig. 3.6 where a curve with its tangential and normal vectors is shown. Since the curve in 2 this configuration is bending away10 from the normal vector n, it holds that ddxu2z < 0 and the radius of curvature is obtained for a positive bending moment as ⎛ ⎝1 + R=−



du z dx

d2 u z dx 2

2 ⎞3/2 ⎠ .

(3.10)

Note that the expression curvature, which results as a reciprocal value from the curvature radius, i.e., κ = R1 , is used as well. z  1 results and Eq. (3.10) For small bending deflections, meaning u z  L, du dx simplifies to d2 u z 1 1 or κ = = − 2 . (3.11) R=− 2 R dx d uz dx 2 For the determination of the strain, one refers to its general definition, meaning elongation referring to initial length. Relating to the configuration shown in Fig. 3.7,

10

See Sect. A.10 for further details.

22

3 Euler–Bernoulli Beam Elements

Fig. 3.7 Segment of a beam under pure bending in the x-z-plane. Note that the deformation is exaggerated for better illustration

the longitudinal elongation of a fiber at distance z to the neutral fiber allows to express the strain as ds − dx . (3.12) εx = dx The lengths of the circular arcs ds and dx result from the corresponding radii and the enclosed angles in radian measure as dx = Rdϕ y ,

(3.13)

ds = (R + z)dϕ y .

(3.14)

If these relations for the circular arcs are used in Eq. (3.12), the following results εx =

dϕ y (R + z)dϕ y − Rdϕ y =z . dx dx dϕ

It results from Eq. (3.13) that dxy = can finally be expressed as follows: εx (x, z) = z

1 R

(3.15)

and together with relation (3.11) the strain

1 (3.11) d2 u z (x) (3.1.1) = −z = zκ . R dx 2

(3.16)

If we replace the common formulation of the second-order derivative, i.e., d dx(...2 ) , by a formal operator symbol, i.e., L2 (. . . ), the kinematics relation can be stated in a more formal way as (3.17) εx (x, z) = −zL2 u z (x) = zκ . 2

An alternative derivation of the kinematics relation results from consideration of Fig. 3.8. From the relation of the right-angled triangle 0 1 2 , this means11 sin ϕ y = ux , the following relation results for small angles (sin ϕ y ≈ ϕ y ): z u x = +zϕ y .

(3.18)

Note that according to the assumptions of the Bernoulli beam the lengths 01 and 0 1 remain unchanged.

11

3.3 Constitutive Equation

23

Fig. 3.8 Alternative configuration for the derivation of the kinematics relation. Note that the deformation is exaggerated for better illustration

Furthermore, it holds that the rotation angle of the slope equals the center line for small angles: − du z (x) ≈ ϕy . tan ϕ y = (3.19) dx If Eqs. (3.19) and (3.18) are combined, the following results: u x = −z

du z (x) . dx

(3.20)

The last relation equals (ds − dx) in Eq. (3.12) and differentiation with respect to the x-coordinate leads directly to Eq. (3.16).

3.3 Constitutive Equation The one-dimensional Hooke’s law according to Eq. (2.4) can also be assumed in the case of the bending beam, since, according to the requirement, only normal stresses are regarded in this section: (3.21) σx = Eεx .

24

3 Euler–Bernoulli Beam Elements

Fig. 3.9 a Schematic representation of the normal stress distribution σx = σx (z) of a bending beam; b Definition and position of an infinitesimal surface element for the derivation of the resulting moment action due to the normal stress distribution

Using a more general notation, Hooke’s law can be again written as σx (x) = Cεx (x),

(3.22)

where the scalar constant C = E transforms in the two- or three-dimensional case the so-called elasticity matrix. Through the kinematics relation according to Eq. (3.16), the stress results as a function of the deflection to σx (x, z) = −E z

d2 u z (x) . dx 2

(3.23)

The stress distribution shown in Fig. 3.9a generates the internal bending moment, which acts in this cross section. To calculate this internal moment, the stress is multiplied by a surface element, so that the resulting force is obtained. Multiplication with the corresponding lever arm then gives the internal moment. Since the stress is linearly distributed over the height, the evaluation is done for an infinitesimally small surface element: dM y = (+z)(+σx )dA = zσx dA .

(3.24)

Therefore, the entire moment results via integration over the entire surface in:

My =

(3.23)

zσx dA = − A

zEz

d2 u z (x) dA . dx 2

(3.25)

A

Assuming that Young’s modulus is constant, the internal moment around the y-axis results in

3.3 Constitutive Equation

25

d2 u z M y = −E 2 dx

z 2 dA =

I y σx . z

(3.26)

A

   Iy

The integral in Eq. (3.26) is the so-called axial second moment of area or axial surface moment of second order in the SI unit m4 . This factor is only dependent on the geometry of the cross section and is also a measure of the stiffness of a plane cross section against bending. The values of the axial second moment of area for simple geometric cross sections are collected in Table 3.3. Consequently the internal moment can also be expressed as M y = −E I y

d2 u z (3.11) E I y = E I y κ. = dx 2 R

(3.27)

Equation (3.27) describes the bending line u z (x) as a function of the bending moment and is therefore also referred to as the bending line–moment relation. The product E I y in Eq. (3.27) is also called the bending stiffness. If the result from Eq. (3.27) is used in the relation for the bending stress according to Eq. (3.23), the distribution of stress over the cross section results in σx (x, z) = +

M y (x) z(x) . Iy

(3.28)

The plus sign in Eq. (3.28) causes that a positive bending moment (see Fig. 3.4) leads to a tensile stress in the upper beam half (meaning for z > 0). The corresponding equations for a deformation in the x-y-plane can be found in Öchsner (2014). In the case of plane bending with M y (x) = const., the bending line can be approximated in each case locally through a circle of curvature, see Fig. 3.10. Therefore, the result for pure bending according to Eq. (3.27) can be transferred to the case of plane bending as d2 u z (x) = M y (x) . (3.29) − E Iy dx 2 Let us note at the end of this section that Hooke’s law in the form of Eq. (3.21) is not so easy to apply12 in the case of beams since the stress and strain are linearly changing over the height of the cross section, see Eq. (3.28) and Fig. 3.9. Thus, it might be easier to apply a so-called stress resultant or generalized stress, i.e., a simplified representation of the normal stress state13 based on the acting bending moment:

12

However, this formulation works well in the case of rod elements since stress and strain are constant over the cross section, i.e., σx = σx (x) and εx = εx (x), see Fig. 2.4. 13 A similar stress resultant can be stated for the shear stress based on the shear force: Q (x) = z  τx z (x, z) dA.

26

3 Euler–Bernoulli Beam Elements

Table 3.3 Axial second moment of area around the y- and z-axes Cross section Iy Iz π D4 π R4 = 64 4

π D4 π R4 = 64 4

πba 3 4

πab3 4

a4 12

a4 12

bh 3 12

hb3 12

bh 3 36

hb3 36

bh 3 36

bh 3 48

3.4 Equilibrium

27

Fig. 3.10 Deformation of a beam in the x-z-plane with M y (x) = const.

Fig. 3.11 Formulation of the constitutive law based on a stress and b stress resultant

M y (x) =

zσx (x, z) d A ,

(3.30)

which was already introduced in Eq. (3.24). Using in addition the curvature14 κ = κ(x) (see Eq. (3.16)) instead of the strain εx = εx (x, z), the constitutive equation can be easily expressed as shown in Fig. 3.11. The variables M y and κ have both the advantages that they are constant for any location x of the beam.

3.4 Equilibrium The equilibrium conditions are derived from an infinitesimal beam element of length dx, which is loaded by a constant distributed load qz , see Fig. 3.12. The internal reactions are drawn on both cut faces, i.e., at locations x and x + dx. One can see that a positive shear force is oriented in the positive z-direction at the right-hand face15 and that a positive bending moment has the same rotational direction as the positive

14

The curvature is then called a generalized strain. A positive cut face is defined by the surface normal on the cut plane which has the same orientation as the positive x-axis. It should be regarded that the surface normal is always directed outward.

15

28

3 Euler–Bernoulli Beam Elements

Fig. 3.12 Infinitesimal beam element in the x-z-plane with internal reactions and constant distributed load

y-axis (right-hand grip rule16 ). The orientation of shear force and bending moment is reversed at the left-hand face in order to cancel in sum the effect of the internal reactions at both faces. This convention for the direction of the internal directions is maintained in the following. Furthermore, it can be derived from Fig. 3.12 that an upward directed external force or alternatively a mathematically positive-oriented external moment at the right-hand face leads to a positive shear force or alternatively a positive internal moment. In a corresponding way, it results that a downward directed external force or alternatively a mathematically negative-oriented external moment at the left-hand face leads to a positive shear force or alternatively a positive internal moment. The equilibrium condition will be determined in the following for the vertical forces. Assuming that forces in the direction of the positive z-axis are considered positive, the following results: − Q z (x) + Q z (x + dx) + qz dx = 0 .

(3.31)

If the shear force on the right-hand face is expanded in a Taylor’s series of first order, meaning dQ z (x) dx , (3.32) Q z (x + dx) ≈ Q z (x) + dx Eq. (3.31) results in − Q z (x) + Q z (x) +

dQ z (x) dx + qz dx = 0 , dx

(3.33)

or alternatively after simplification finally to 16

If the axis is grasped with the right hand in a way so that the spread out thumb points in the direction of the positive axis, the bent fingers then show the direction of the positive rotational direction.

3.5 Differential Equation

29

dQ z (x) = −qz . dx

(3.34)

For the special case that no distributed load is acting (qz = 0), Eq. (3.34) simplifies to dQ z (x) = 0. dx

(3.35)

The equilibrium of moments around the reference point at x + dx gives M y (x + dx) − M y (x) − Q z (x)dx +

1 qz dx 2 = 0 . 2

(3.36)

If the bending moment on the right-hand face is expanded into a Taylor’s series of first order similar to Eq. (3.32) and consideration that the term 21 qz dx 2 as infinitesimal small size of higher order can be disregarded, finally the following results: dM y (x) = Q z (x) . dx

(3.37)

The combination of Eqs. (3.34) and (3.37) leads to the relation between the bending moment and the distributed load: d2 M y (x) dQ z (x) = −qz (x) , = dx 2 dx

(3.38)

or with the operator symbol L as17   LT2 M y (x) = −qz (x) .

(3.39)

Finally, the elementary basic equations for the bending of a beam in the x-z-plane for arbitrary moment loading M y (x) are summarized in Table 3.4.

3.5 Differential Equation Two-time differentiation of Eq. (3.27) and consideration of the relation between bending moment and distributed load according to Eq. (3.38) lead to the classical type of differential equation of the bending line,   d2 d2 u z (3.40) E I y 2 = qz , dx 2 dx  2 T It should be noted here that the transpose (“T”), i.e., LT2 = d dx(...2 ) , is only used to show later similar structures of the equations in the two-dimensional case.

17

30

3 Euler–Bernoulli Beam Elements

Table 3.4 Elementary basic equations for the bending of a Bernoulli beam in the x-z-plane. The differential equations are given under the assumption of constant bending stiffness E I y Name

Equation

Kinematics εx (x, z) = −z Equilibrium

Constitutive equation

d2 u z (x) dx 2

dQ z (x) dM y (x) = −qz (x) ; = Q z (x) dx dx σx (x, z) = Eεx (x, z)

Stress σx (x, z) = Diff’equation

M y (x) z(x) Iy

E Iy

d2 u z (x) = −M y (x) dx 2

E Iy

d3 u z (x) = −Q z (x) dx 3

E Iy

d4 u z (x) = qz (x) dx 4

which is also referred to as the bending line-distributed load relation. For a beam with constant bending stiffness E I y along the beam axis, the following results: E Iy

d4 u z = qz . dx 4

(3.41)

The differential equation of the bending line can of course also be expressed through the bending moment or the shear force as d2 u z = −M y or dx 2 d3 u z E I y 3 = −Q z . dx

E Iy

(3.42) (3.43)

Equations (3.42) and (3.43) can be also written in the more general form for variable bending stiffness: d2 u z = −M y (x) , 2 dx  d2 u z du z E(x)I y (x) 2 = −Q z (x) . dx dx E(x)I y (x)

(3.44) (3.45)

3.5 Differential Equation

31

Table 3.5 Different formulations of the partial differential equation for a Bernoulli beam in the x-z-plane (x-axis: right facing; z-axis: upward facing) Configuration Partial differential equation E Iy

d4 u z =0 dx 4 

d2 u z d2 E(x)I y (x) 2 2 dx dx

E Iy

d4 u z = qz (x) dx 4

E Iy

dm y (x) d4 u z = dx 4 dx

E Iy

d4 u z = −k(x)u z dx 4

 =0

Depending on the problem and the fact which distribution (qz (x), M y (x), or Q z (x)) is easier to state, one may start from one of the three formulations to derive the displacement field u z (x). Different formulations of the fourth-order differential equation are collected in Table 3.5 where different types of loadings, geometry, and bedding are differentiated. The last case in Table 3.5 refers to the elastic foundation of a beam which is also known in the literature as Winkler18 foundation Winkler (1867). The elastic foundation or Winkler foundation modulus k has in the case of beams19 the unit of force per unit area. 2 If we replace the common formulation of the second-order derivative, i.e., ddx...2 , by a formal operator symbol, i.e., L2 (. . . ), the basic equations can be stated in a more formal way as given in Table 3.6.

18

Emil Winkler (1835–1888), German engineer. In the general case, the unit of the elastic foundation modulus is force per unit area per unit length, i.e., mN2 /m = mN3 .

19

32

3 Euler–Bernoulli Beam Elements

Table 3.6 Different formulations of the basic equations for an Euler–Bernoulli beam (bending in the x-z-plane; x-axis along the principal beam axis). E: Young’s modulus; I y : second moment of area; qz : length-specific distributed force; L2 = Specific formulation

d2 (... ) : dx 2

second-order derivative

General formulation

Kinematics εx (x, z) = −z

d2 u z (x) dx 2

εx (x, z) = −z L2 (u z (x))

d2 u z (x) dx 2 Constitution

κ(x) = −L2 (u z (x))

σx (x, z) = Eεx (x, z)

σx (x, z) = Cεx (x, z)

M y (x) = E I y κ(x)

M y (x) = Dκ(x)

κ(x) = −

Equilibrium d2 M y (x) + qz (x) = 0 dx 2 PDE   d2 d2 u z (x) E I − qz (x) = 0 y dx 2 dx 2





LT2 M y (x) + qz (x) = 0

LT2 (D L2 (u z (x))) − qz (x) = 0

References Altenbach H, Altenbach J, Naumenko K (1998) Ebene Flächentragwerke: Grundlagen der Modellierung und Berechnung von Scheiben und Platten. Springer, Berlin Boresi AP, Schmidt RJ (2003) Advanced mechanics of materials. Wiley, New York Budynas RG (1999) Advanced strength and applied stress analysis. McGraw-Hill Book, Singapore Gould PL (1988) Analysis of shells and plates. Springer, New York Gross D, Hauger W, Schröder J, Wall WA (2009) Technische Mechanik 2: Elastostatik. Springer, Berlin Hartmann F, Katz C (2007) Structural analysis with finite elements. Springer, Berlin Heyman J (1998) Structural analysis: a historical approach. Cambridge University Press, Cambridge Hibbeler RC (2008) Mechanics of materials. Prentice Hall, Singapore Öchsner A (2014) Elasto-Plasticity of frame structure elements: modeling and simulation of rods and beams. Springer, Berlin Szabó I (2003) Einführung in die Technische Mechanik: Nach Vorlesungen István Szabó. Springer, Berlin Timoshenko S, Woinowsky-Krieger S (1959) Theory of plates and shells. McGraw-Hill Book Company, New York Winkler E (1867) Die Lehre von der Elasticität und Festigkeit mit besonderer Rücksicht auf ihre Anwendung in der Technik. H. Dominicus, Prag

Chapter 4

Combination of Bar and Beam Elements

Abstract This chapter focuses on the superposition of the bar and the thin beam element. Based on the combined basic equations of continuum mechanics, i.e., the kinematics relationship, the constitutive law (here: one-dimensional Hooke’s law for isotropic materials) and the equilibrium equation, the partial differential equations, which describe the combined structural element, are derived. In addition to the classical notation, a formal notation is introduced. Thus allows to highlight the similarities between the different structural members.

4.1 Introduction The simple superposition of a bar and thin beam element is illustrated in Fig. 4.1. It is assumed for this approach that there is no direct coupling between these two deformation modes, i.e., the tensile and bending mode. This combined bar/beam element is now described by the geometrical properties L , I y , A and the material property E. Further details on the approach and the total deformation can be found in Öchsner (2021b).

4.2 Kinematics The linear superposition of the normal strain from the tensile deformation (cf. Eq. (2.3)) and the bending deformation (cf. Eq. (3.17)) gives the total normal strain as εx (x) = L1 u x (x) − zL2 u z (x) = L1 u x +zκ y .   

(4.1) (4.2)

εx0

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, A Simplified Approach to the Classical Laminate Theory of Composite Materials, Advanced Structured Materials 192, https://doi.org/10.1007/978-3-031-38192-8_4

33

34

4 Combination of Bar and Beam Elements

Fig. 4.1 Superposition of a bar and a thin beam element

The last two equations can be differently arranged in matrix form as ⎤ d  0 ⎢ 0 ⎥  εx ⎥ ux ⎢ dx ⎥ =⎢ , 2 ⎥ ⎢ κy ⎣ 0 − d ⎦ uz dx 2 ⎡

or symbolically as

e = L u ,

(4.3)

(4.4)

where the column matrix of generalized strains is expressed as  0 ε e= x . κy

(4.5)

4.4 Equilibrium

35

4.3 Constitutive Equation The constitutive equation is the same for the bar and beam element (cf. Eqs. (2.4) and (3.21)) and does not require any adjustment: σx (x) = Eεx (x) = Cεx (x) .

(4.6)

4.4 Equilibrium The equilibrium relation for the rod (cf. Eq. (2.8)) and the thin beam (cf. Eq. (3.38)) can be combined in a single matrix equation to obtain: ⎡

⎤ d      0 ⎥ ⎢ px (x) 0 ⎢ dx ⎥ N x (x) ⎢ ⎥ + = . 2 ⎥ ⎢ q y (x) 0 ⎣ 0 d ⎦ M y (x) dx 2

(4.7)

In the context of two-dimensional composite layers, it is common to define the internal reactions per unit length (normalized with the corresponding side length of the plate element). The normalized (superscript “n”) internal reactions are obtained as N x (x) , a M y (x) M yn (x) = . a N xn (x) =

(4.8) (4.9)

Thus, the combined equilibrium equation can be written as ⎡

⎤ ⎤ ⎡ d 0 ⎥  n  ⎢ px (x)⎥   ⎢ 0 ⎢ dx ⎥ N x (x) ⎥ ⎢ ⎢ ⎥ +⎢ a ⎥= , 2 ⎥ ⎢ n ⎣ q y (x) ⎦ 0 ⎣ 0 d ⎦ M y (x) dx 2 a

(4.10)

or in symbolic notation: LT s n + q = 0 .

(4.11)

Let us note here that the derivation of the internal reactions or stress resultants, i.e., N x (x) and M y (x), must be revised for the combined bar/beam element (cf. the standalone derivations in Eqs. (2.14) and (3.25)). This comes from the fact that the total strain in the form of Eq. (4.1), i.e., εx (x) = ε0 + zκ, must be used in the derivation. The derivation for a rectangular cross section (see Fig. 4.2) reads

36

4 Combination of Bar and Beam Elements

Fig. 4.2 Particular configuration of a combined bar/beam element

t



t

2

N x (x) =

σx (x)dA =

2 σx (x)adz =

t − 2

A t

t

2 =

=

t 2

2 aCεx0 dz +

−

aCεx0

t 2

aC zκ y dz

−

 2t   2t z 2 z t + aCκ y − 2 − 2t 2

N x (x) = taCεx0 +

(4.12)

t

2 aC(εx0 + zκ y )dz =

−

aCεx (x)dz t − 2

(4.13)

t 2

(4.14)

t2 aCκ y . 4

(4.15)

In the same way, the evaluation of the internal bending moment gives t

t

2

 M y (x) =

zσx (x)dA = A

2 zσx (x)adz =

−

t 2

t

M y (x) =

t 2

t

aCεx0



z2 2

t

2

t − 2 3

 + aCκ y

t2 t aCεx0 + aCκ y . 4 12

2 aC zεx0 dz +

−

t 2

z3 3

(4.16)

t 2

2 aC z(εx0 + zκ y )dz =

−

=

− t

2 =

aC zεx (x)dz

−

t

2

−

t 2

aC z 2 κ y dz

(4.17)

t 2

(4.18) (4.19)

4.5 Differential Equations

37

Equations (4.15) and (4.19) can be combined to obtain a single matrix representation: ⎡ 

C ⎢ ta ⎢  ⎢ A N x (x) ⎢ =⎢ 2 ⎢t M y (x) ⎢ aC ⎣4    

⎤ t2 aC ⎥   4  ⎥ ⎥ εx0 ⎥ B , ⎥ ⎥ κy t3 aC ⎥ ⎦ 12 

B

(4.20)

D

or based on the internal reactions per unit length (normalized with the corresponding side length of the plate element, see Eqs. (4.8) and (4.9)) ⎡ 

N xn (x) M yn (x)



tC ⎢ ⎢ ⎢ A ⎢ =⎢ 2 ⎢t ⎢ C ⎣4 

⎤ t2 C⎥ 4 ⎥  0  ⎥ εx B ⎥ , ⎥ t 3 ⎥ κy ⎥ C⎦ 12  

B

or in symbolic notation:

(4.21)

D

sn = C ∗ e ,

(4.22)

where C ∗ is the generalized elasticity matrix.

4.5 Differential Equations Let us start the derivation from the combined equilibrium equation as given in Eq. (4.10) and introduce the representations of the normalized internal reactions according to Eq. (4.21). This procedure results in the following formulation: ⎡

⎤

⎡

tC d ⎢ 0 ⎥⎢ ⎢ ⎢ ⎥⎢ A ⎢ dx ⎥ ⎢ 2 ⎥⎢ 2 ⎢ t ⎣ 0 d ⎦⎢ ⎢ C 2 ⎣ dx 4  B

⎤ t2 ⎡ ⎤ C⎥ px (x) ⎥ 4      ⎥ ε0 ⎢ ⎥ 0 x ⎢ a ⎥ B ⎥ . +⎢ ⎥ ⎥= 3 ⎥ κy ⎣ q y (x) ⎦ t 0 ⎥ C⎦ a 12  

(4.23)

D

Consideration of the kinematics relationship according to Eq. (4.3) in the last equation gives

38

4 Combination of Bar and Beam Elements

⎡

⎤

⎡

tC d ⎢ 0 ⎥⎢ ⎢ ⎢ ⎢ dx ⎥⎢ A ⎢ ⎥ 2 ⎥⎢ 2 ⎢ d t ⎣0 ⎦⎢ ⎢ C dx 2 ⎣ 4 B

⎤ t2 ⎡ ⎤ ⎡ ⎤ C⎥ d px (x) ⎥ 4     0 ⎢ ⎥  ⎥ ⎢ dx ⎢ ⎥ 0 ⎥ ux ⎢ a ⎥ B ⎥⎢ ⎥ , +⎢ ⎥⎢ ⎥= 2 ⎥ u 3 ⎥⎣ d ⎦ z ⎣ q y (x) ⎦ t 0 ⎥ C⎦ 0 − 2 dx a 12  

(4.24)

D

or slightly simplified as ⎤⎡ d 0 ⎥⎢ A ⎢ ⎢ dx ⎥⎢ ⎢ ⎥⎢ 2 ⎥⎢ ⎢ ⎣ 0 d ⎦⎣ B dx 2 ⎡

⎤⎡

⎤ ⎡ ⎤ d 0 ⎥   ⎢ px (x)⎥   B ⎥⎢ 0 ⎥ ux ⎥ ⎢ dx ⎢ ⎥ ⎥ ⎥⎢ . +⎢ a ⎥= 2 ⎥ u ⎥⎢ d ⎦ z ⎣ q y (x) ⎦ 0 ⎣ ⎦ 0 − 2 C dx a

(4.25)

A symbolic representation of the last equation reads LT C ∗ L u + q = 0 .

(4.26)

Once the solution of the displacements u in Eq. (4.26) is obtained, for example, based on the finite element method (Öchsner, 2023) or the finite difference method (Öchsner, 2021a), the kinematics relationships (4.3) and (4.2) provide the strains from which the constitutive equation (4.6) allows the calculation of the stresses. Based on theses stress and/or strain values, a subsequent failure analysis can be performed. The basic equations of the combined bar/beam element are summarized in Table 4.1.

4.6 Failure Criteria The following subsections present some of the common orthotropic failure criteria for composite materials (Öchsner, 2023). However, the formulations were simplified to our one-dimensional problem. This means that the orthotropic nature is disregarded. The classical criteria for pure isotropic material behavior can be found, for example, in Öchsner (2016).

4.6.1 Maximum Stress Criterion This criterion assumes that there is no failure in the x-direction as long as the stress is in the following limits of the corresponding strength values k: kc < σx < kt .

(4.27)

4.6 Failure Criteria

39

Table 4.1 Different formulations of the basic equations for a combined one-dimensional bar/beam element (bending in the x-z-plane; x-axis along the principal element axis). C = E: elasticity modulus; C ∗ : generalized elasticity matrix; t: composite thickness; s: column matrix of normalized stress resultants; e: column matrix of generalized strains; q: column matrix of distributed loads; L: matrix of differential operators e = L u ⎤ ⎡ ⎡ ⎤ d ⎡ ⎤ 0 ⎥ ⎢ 0 ⎥ ⎢u x ⎥ ⎢εx ⎥ ⎢ dx ⎥⎢ ⎥ ⎢ ⎥ ⎢ , ⎥ ⎢ ⎥=⎢ 2 ⎥⎣ ⎦ ⎣ ⎦ ⎢ d ⎦ uz ⎣0 − κy dx 2

Kinematics

σx = Cεx

Constitution

sn = C ∗ e σx (x, z) = Eεx (x, z) ⎡

⎤ t2 ⎤ ⎢  ⎡ C ⎥⎡ ⎤ tC ⎢ 4 ⎥ n (x) ⎥ ⎢εx0 ⎥ ⎢ A  N ⎥ ⎢ ⎥ ⎢ x B ⎥⎢ ⎥ ⎥=⎢ ⎢ ⎥ ⎥⎢ ⎦ ⎢ 2 ⎣ 3 ⎥ ⎣κ ⎦ ⎢t t M yn (x) ⎥ ⎢ C y C⎦ ⎣4    12 B

Equilibrium

D

⎡

⎤

LT s n + q = 0

⎡ ⎤ d ⎤ ⎡ ⎡ ⎤ px (x) ⎢ dx 0 ⎥ n ⎢ ⎥ ⎢ ⎥ ⎢ N x (x) ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥⎢ ⎥+⎢ a ⎥=⎢ ⎥ ⎢ ⎥⎣ ⎥ ⎣ ⎦ ⎦ ⎢ 2 ⎢ ⎥ ⎣ q y (x) ⎦ ⎣ 0 d ⎦ M n (x) 0 y dx 2 a PDE ⎡

⎤

⎡

LT C ∗ L u + q = 0

⎤ ⎤ ⎡ t2 ⎡ ⎤ d ⎢tC ⎡ ⎤ ⎡ ⎤ C⎥ d  0 0 ⎥ ⎢ px (x) 4 ⎢ dx ⎥⎢ ⎥ ⎢ ⎢ ⎥  ⎥ ⎢ dx u ⎢ ⎥⎢ A ⎥ x⎥ ⎢ ⎥ ⎥ ⎢0 ⎥ ⎢ ⎥⎢ ⎥⎢ B ⎥⎢ ⎢ ⎥+⎢ a ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎥ ⎣ ⎦ 2 ⎥ 2 ⎥⎣ ⎦ ⎢ ⎣ q y (x) ⎦ t2 t3 ⎥ ⎢ ⎣ 0 d ⎦⎢ ⎣ 0 − d ⎦ uz 0 ⎢ C C⎥ 2 2 ⎦ ⎣ dx dx 4 a    12 B

D

40

4 Combination of Bar and Beam Elements

4.6.2 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: kcε < εx < ktε ,

(4.28)

where the k ε represent the corresponding failure strains.

4.6.3 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 laminae (Hill, 1950), (Tsai, 1968). This criterion can be stated for the one-dimensional case as follows: σx2 < 1, k2

(4.29)

where k = kt for σx > 0 or k = kc for σx < 0. This criterion reduces in this particular one-dimensional case to the maximum stress criterion, see Sect 4.6.1.

4.6.4 Tsai–Wu Criterion The Tsai–Wu criterion can be expressed for the particular one-dimensional case as follows (Tsai et al., 1971): (4.30) a1 σx + a11 σx2 < 1 , where the coefficients a can be related to the classical failure stresses as follows (Kaw, 2006), (Altenbach et al., 2018): 1 1 + , kt kc 1 =− . kt kc

a1 = a11

(4.31) (4.32)

References

41

References Altenbach H, Altenbach J, Kissing W (2018) Mechanics of composite structural elements. Springer Nature, Singapore Hill R (1950) The mathematical theory of plasticity. Oxford University Press, London Kaw AK (2006) Mechanics of composite materials. CRC Press, Boca Raton Öchsner A (2016) Continuum damage and fracture mechanics. Springer, Singapore Öchsner A (2021a) Structural mechanics with a pen: a guide to solve finite difference problems. Springer, Cham Öchsner A (2021b) Classical beam theories of structural mechanics. Springer, Cham Öchsner A (2023) Computational statics and dynamics: an introduction based on the finite element method. Springer, Cham Öchsner A (2023) Composite mechanics. Springer, Cham 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, pp 3–12 Tsai SW, Wu EM (1971) A general theory of strength for anisotropic materials. J Compos Mater 5:58–80 (1971). https://doi.org/10.1177/002199837100500106

Chapter 5

Classical Laminate Theory for One-Dimensional Elements

Abstract This chapter focuses on composite bar/beam elements with different layers. Based on a generalized stress–strain relationship, the calculation steps based on the simplified classical laminate theory are introduced. The simplifications come from the application of one-dimensional members and the treatment of each single layer as an isotropic material. Based on the internal reactions, i.e., the internal normal force and the internal bending moment, the strain and stress distribution in each single layer can be evaluated. These values may serve then for a subsequent failure analysis based on different criteria.

5.1 Introduction The procedure outlined in the following follows the description and notation provided in Öchsner (2022). For the earlier references, the reader may consult the following publications: Stravsky (1961), Dong et al. (1962). Let us consider in the following a composite bar/beam element, which is composed of n layers, see Fig. 5.1. In our simplified approach, each layer k is considered as an isotropic and homogeneous material. However, the properties can vary from layer to layer. The global coordinate system (x, y, z) is used to describe the entire composite bar/beam element. The height of a layer k (1 ≤ k ≤ n) can be expressed based on the thickness coordinate as (5.1) tk = z k − z k−1 , and the total height of the laminate results as t=

n 

tk .

(5.2)

k =1

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, A Simplified Approach to the Classical Laminate Theory of Composite Materials, Advanced Structured Materials 192, https://doi.org/10.1007/978-3-031-38192-8_5

43

44

5 Classical Laminate Theory for One-Dimensional Elements

Fig. 5.1 Geometry of a composite bar/beam element with n layers

5.2 Generalized Stress–Strain Relationship Let us focus in the following on the evaluation on the stress resultants as introduced in Eqs. (4.12) and (4.16) for a single layer. The internal normal force can be expressed in the laminate-specific coordinate system (x, y, z) as t/2 Nx = a

σx dz = a

k = 1z

−t/2

=a

zk

n  k = 1z

=a

n 

⎛

n z k  k = 1z

k−1

Ck εx,k dˆz

zk

zk Ck εx0 dˆz + z k−1

Ck (z k − 

z k−1 ) εx0 

Ak

(5.3)

k−1

  Ck εx0 + zˆ κ y dˆz

z k−1

n

 k =1

σx,k dˆz = a

k−1

⎝

k =1

=a

n z k 

(5.4) ⎞

Ck zˆ κ y dˆz ⎠

(5.5)

  1 2 2 + Ck (z k ) − (z k−1 ) κ y ,   2

(5.6)

Bk

or with N xn = N x /a as N xn (x)

=

n

 k =1

Ck (z k −  Ak

z k−1 ) εx0 

  1 2 2 + Ck (z k ) − (z k−1 ) κ y .   2 Bk

(5.7)

5.2 Generalized Stress–Strain Relationship

45

The corresponding derivation for the internal bending moment gives t/2 My = a

zσx dz = a

k = 1z

−t/2

=a

n z k  k = 1z

=a

n 

⎛

σx,k zˆ dˆz = a

⎝

k−1

k =1

Ck εx,k zˆ dˆz

zk

zk Ck εx0 zˆ dˆz +

(5.8)

k−1

  Ck εx0 zˆ + zˆ 2 κ y dˆz

z k−1

n



n z k  k = 1z

(5.9) ⎞

k−1

k =1

=a

n z k 

Ck zˆ 2 κ y dˆz ⎠

(5.10)

z k−1

  0  1 1 2 2 3 3 Ck (z k ) − (z k−1 ) εx + Ck (z k ) − (z k−1 ) κ y ,    2  3 Bk

(5.11)

Dk

or with M yn = M y /a as M yn (x)

=

n

 k =1

  0  1 1 2 2 3 3 Ck (z k ) − (z k−1 ) εx + Ck (z k ) − (z k−1 ) κ y .    2  3 Bk

(5.12)

Dk

Equations (5.7) and (5.12) can be combined in a single matrix form to give 

N xn



A

B



M yn B D     C∗

s

=

  εx0 κy  

,

(5.13)

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 scalar elements of C ∗ are given as follows: A= B= D=

n  k =1 n  k =1 n  k =1

Ak = Bk = Dk =

n 

Ck (z k − z k−1 ) ,

k =1 n 

(5.14)

  1 Ck (z k )2 − (z k−1 )2 , 2 k =1

(5.15)

n   1 Ck (z k )3 − (z k−1 )3 . 3 k =1

(5.16)

46

5 Classical Laminate Theory for One-Dimensional Elements

It should be noted here that the matrix element B represents a bending–tension coupling. Equation (5.13) can be inverted to obtain the strains and curvatures (generalized strains) as a function of the generalized stresses as   εx0 κy  

 =

A

B

B

D

−1 

N xn



M yn

 = 

A

B

B

D



(C ∗ )−1

e



N xn



M yn   

,

(5.17)

s

where the elements of the inverse are simply given for a 2 × 2 matrix as follows: D , AD − B B −B , B = AD − B B A D = . AD − B B A =

(5.18) (5.19) (5.20)

Based on the generalized strains obtained from Eq. (5.17), it is now possible to calculate the stresses in each layer k expressed in the x-y coordinate system:   σx,k (z) = Ck εx0 + zκ y ,

(5.21)

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. The obtained stress values may serve for a subsequent failure analysis of layer k (see Sects. 4.6 and 5.3). It should be mentioned at the end of this section that the generalized elasticity matrix C ∗ (see Eq. (5.13)) and the corresponding inverse (C ∗ )−1 (see (5.17)) take the following simplified forms for composite elements with a symmetric layer composition:  C ∗sym

=

A

0

0

D

 ,

⎤

⎡

(C ∗sym )−1

1 ⎢A =⎢ ⎣ 0

(5.22)

0⎥ ⎥. 1⎦ D

(5.23)

5.3 Failure Analysis

47

5.3 Failure Analysis Let us summarize here the recommended steps for a calculation according to the simplified classical laminate theory, see Table 5.1 for the case that the internal normal force and the internal bending moment are given and Table 5.2 for the case that the generalized strains are given.

Table 5.1 Recommended steps for a calculation according to the simplified classical laminate theory. Case: internal normal force N xn and internal bending moment M yn given Nr.

Steps to perform

①

Define for each layer k the material (E k ) and the geometrical (z k ; z k−1 ; αk ) properties

② ③ ④

⑤ ⑥

⑦

T  Define the column matrix of generalized stresses: s = N xn M yn Calculate the matrix elements A, B, and C according to Eqs. (5.14)–(5.16). Assemble the generalized elasticity matrix C ∗ according to Eq. (5.13) Calculate the generalized compliance matrix (C ∗ )−1 based on Eqs. (5.18)–(5.20)  T Calculate the generalized strains e = εx0 κ y according to Eq. (5.17) Calculate the stresses ineach layer k according to Eq. (5.21), σx,k (z) = Ck εx0 + zκ y , in the x-y laminate system (z k−1 ≤ z ≤ z k ).   The strains in the x-y laminate system are obtained from εx0 + zκ y Perform the failure analysis for each layer k, see Sect. 4.6

Table 5.2 Recommended steps for a calculation according to the simplified classical laminate T  given theory. Case: generalized strains e = εx0 κ y Nr.

Steps to perform

①

Define for each lamina k the material (E k ) and the geometrical (z k ; z k−1 ; αk ) properties

② ③

④ (⑤)

T  Define the column matrix of generalized strains: e = εx0 κ y Calculate the matrix elements A, B, and C according to Eqs. (5.14)–(5.16). Assemble the generalized elasticity matrix C ∗ according to Eq. (5.13)   Calculate the generalized stresses s = N xn M yn according to Eq. (5.13) In case that the stresses and strains in each layer are required (e.g., for a failure analysis), go to step ⑥ and ⑦ in Table 5.1

48

5 Classical Laminate Theory for One-Dimensional Elements

References Dong SB, Pister KS, Taylor RL (1962) On the theory of laminated anisotropic shells and plates. J Aerosp Sci 29:696–975. https://doi.org/10.2514/8.9668 Öchsner A (2022) Foundations of classical laminate theory. Springer, Cham Stravsky Y (1961) Bending and stretching of laminated aeolotropic plates. J Eng Mech Div-ASCE 87:31–56 (1961). https://doi.org/10.1061/JMCEA3.0000267

Chapter 6

Example Problems

Abstract This chapter presents the solution of some standard calculation problems based on the simplified classical laminate theory. Each solution is based on a stepby-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, as well as the application of failure criteria, and the ply-by-ply failure of laminates.

6.1 Introduction The following example problems are based on different materials, which are assumed to be isotropic, see Table 6.1. These material definitions were obtained from the real properties of an AS4 carbon/3501-6 epoxy prepreg system Soden et al. (1998) by considering the methods of the off-axis transformations outlined in Öchsner (2022). The last column in Table 6.1 indicates the assumed off-axis angle, i.e., the angle between the aligned fibers and the global x-axis. All the following example problems follow the steps of the simplified laminate theory as outlined in Table 5.1.

6.2 Problem 1: Stresses and Strains in a Symmetric Laminate Given is a symmetric eight-layer laminate of thickness t = 8 mm as shown in Fig. 6.1. Each layer has the same thickness (t/8) but different material properties, which can be taken from Table 6.2. Consider (a) a pure tensile load case with N xn = 1000 N/mm (the internal moment is equal to zero), a pure bending load case with M yn = 1000 Nmm/mm (the internal normal force is equal to zero), and (c) a combined load case with N xn = 1000 N/mm

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 A. Öchsner, A Simplified Approach to the Classical Laminate Theory of Composite Materials, Advanced Structured Materials 192, https://doi.org/10.1007/978-3-031-38192-8_6

49

50

6 Example Problems

Table 6.1 Selected mechanical properties of different materials (the angle α refers to the orientation of a uniaxial lamina with respect to the rod/beam x-axis) Material E x in MPa kt in MPa kc in MPa ktε in – kcε in – (α in deg) Mat1 Mat2 Mat3 Mat4

126000 11340 16380 26586

1950 48 96 182

−1450 −200 −158 −182

0.013800 0.004360 0.006153 0.008007

−0.011750 −0.020000 −0.016071 −0.011453

(0◦ ) (90◦ ) (±45◦ ) (30◦ )

Fig. 6.1 Cross section of a symmetric composite with eight layers Table 6.2 Geometrical properties (z k ; z k−1 ; tk ) of the composite shown in Fig. 6.1 Layer k Coordinate z k−1 in Coordinate z k in mm Thickness tk 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

t/8 t/8 t/8 t/8 t/8 t/8 t/8 t/8

and M yn = 1000 to determine the stresses and strains in each layer in the global x-y coordinate system. • Laminate loaded by N xn = 1000 N/mm The solution procedure will follow the recommended steps provided in Table 5.1. ① Define for each layer k the material (E k ) and the geometrical (z k ; z k−1 ; αk ) properties.

6.2 Problem 1: Stresses and Strains in a Symmetric Laminate

51

The material property (E x,k ) for each layer can be taken from Table 6.1. The geometrical properties for each layer are summarized in Table 6.2. T  ② Define the column matrix of generalized stresses: s = N xn M yn .

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

(6.1)

③ Calculate the matrix elements A, B, and C according to Eqs. (5.14)–(5.16). Assemble the generalized elasticity matrix C ∗ according to Eq. (5.13).

The evaluation of the matrix elements A, B, and C according to Eqs. (5.14)–(5.16) gives A=

8 

Ak = 340200.0

k =1

B=

8 

N , mm

(6.2)

Bk = 0.0 N ,

(6.3)

Dk = 1207080.0 Nmm ,

(6.4)

k =1

D=

8  k =1

from which the generalized elasticity matrix (written without units) can be assembled as follows:   340200.0 0.0 ∗ C = . (6.5) 0.0 1207080.0

④ Calculate the generalized compliance matrix (C ∗ )−1 based on Eqs. (5.18)–(5.20). The evaluation of the matrix elements A , B  , and C  according to Eqs. (5.18)–(5.20) gives

52

6 Example Problems

A = 29.394474 B  = 0.0

10−7 mm , N

1 , N

D  = 8.284455

(6.6) (6.7)

−7

10 , Nmm

(6.8)

from which the generalized compliance matrix (written without units) can be assembled as follows:   29.394474 0.0 ∗ −1 (C ) = 10−7 . (6.9) 0.0 8.284455

T  ⑤ Calculate the generalized strains e = εx0 κ y according to Eq. (5.17). The evaluation of Eq. (5.17) gives the generalized strains as follows:     εx0 2.939447 0 = (C ∗ )−1 s = /00 . 0.0 κy 

(6.10)

e k ⑥

0Calculate  the stresses in each layer k according to Eq. (5.21), σx (z) = Ck εx + zκ y , in the x-y laminate system (z

k−1 ≤ z ≤ z k ). The strains in the x-y laminate system are obtained from εx0 + zκ y .

Evaluation of Eq. (5.21) for each layer k, i.e.,

 σx,k (z) = Ck εx0 + zκ y ,

(6.11)

under consideration of the generalized strains given in Eq. (6.10) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 6.2 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 εx0 + zκ y provides these values. A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) is provided in Fig. 6.2 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. 6.2b) across the entire thickness of the laminate. The stress distribution (see Fig. 6.2a) reveals a layer-wise constant distri-

6.2 Problem 1: Stresses and Strains in a Symmetric Laminate

53

(a) z in mm 4 symmetric laminate 3 2 1 −400 −300 −200 −100

100

200

300

400

σx in MPa

−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. 6.2 Distribution of selected field quantities over the composite thickness for the tensile load case: a normal stress σx (z) and b normal strain εx (z)

54

6 Example Problems

bution with the characteristic that the stress is proportional to the stiffness value of each layer in the loading direction. • Laminate loaded by M yn = 1000 Nmm/mm The solution procedure will follow again the recommended steps provided in Table 5.1. However, subtasks ① and ③–④ 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.  T ② Define the column matrix of generalized stresses: s = N xn M yn .

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

(6.12)

 T ⑤ Calculate the generalized strains e = εx0 κ y according to Eq. (5.17). The evaluation of Eq. (5.17) gives the generalized strains as follows:     εx0 0.0 0 = (C ∗ )−1 s = / . 0.828446 00 κy 

(6.13)

e

k ⑥

0Calculate  the stresses in each layer k according to Eq. (5.21), σx (z) = Ck εx + zκ y , in the x-y laminate system (z ≤ z ≤ z ). The strains in the  k

k−1 x-y laminate system are obtained from εx0 + zκ y .

Evaluation of Eq. (5.21) for each layer k, i.e.,

 σx,k (z) = Ck εx0 + zκ y ,

(6.14)

6.2 Problem 1: Stresses and Strains in a Symmetric Laminate

55

under consideration of the generalized strains given in Eq. (6.13) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 6.2 gives the stresses for each layer in the x-y laminate system.

In casethat the strains in the x-y laminate system are required, the evaluation of εx0 + zκ y provides these values. A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) is provided in Fig. 6.3 as a function of the laminate thickness (z-coordinate). It can be seen that a pure bending load by M yn 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. 6.3b). The stress distribution (see Fig. 6.3a) 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. 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. • Laminate loaded by N xn = 1000 N/mm and M yn = 1000 Nmm/mm The solution procedure will follow again the recommended steps provided in Table 5.1. However, subtasks ① and ③–④ 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. T  ② Define the column matrix of generalized stresses: s = N xn M yn .

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

(6.15)

T  ⑤ Calculate the generalized strains e = εx0 κ y according to Eq. (5.17). The evaluation of Eq. (5.17) gives the generalized strains as follows:     εx0 2.939447 0 ∗ −1 = (C ) s = / . 0.828446 00 κy  e

(6.16)

56

6 Example Problems

Fig. 6.3 Distribution of selected field quantities over the composite thickness for the bending load case: a normal stress σx (z) and b normal strain εx (z)

6.3 Problem 2: Stresses and Strains in an Asymmetric Laminate

57

It should be noted here that this is a simple superposition of the results from case (a) and (b). k ⑥

0Calculate  the stresses in each layer k according to Eq. (5.21), σx (z) = Ck εx + zκ y , in the x-y laminate system (z

k−1 ≤ z ≤ z k ). The strains in the x-y laminate system are obtained from εx0 + zκ y .

Evaluation of Eq. (5.21) for each layer k, i.e.,

 σx,k (z) = Ck εx0 + zκ y ,

(6.17)

under consideration of the generalized strains given in Eq. (6.16) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 6.2 gives the stresses for each layer in the x-y laminate system.

In casethat the strains in the x-y laminate system are required, the evaluation of εx0 + zκ y provides these values. A graphical representation of the normal stress (σx ) and normal strain (εx ) in loading direction (x) is provided in Fig. 6.4 as a function of the laminate thickness (z-coordinate). It can be seen that combined load case with N xn and M yn 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. 6.4b). The stress distribution (see Fig. 6.4a) 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. Furthermore, it can be concluded that the stress and strain distribution is a simple linear superposition of the results from load cases (a) and (b).

6.3 Problem 2: Stresses and Strains in an Asymmetric Laminate Given is an asymmetric eight-layer laminate of thickness t = 8 mm as shown in Fig. 6.5. Each layer has the same thickness (t/8) but different material properties, which can be taken from Table 6.1. Consider (a) a pure tensile load case with N xn = 1000 N/mm (the internal moment is equal to zero), (b) a pure bending load case with M yn = 1000 Nmm/mm (the internal normal force is equal to zero), and (c) a combined load case with N xn = 1000 N/mm and M yn = 1000 to determine the stresses and strains in each layer in the global x-y coordinate system.

58

6 Example Problems

Fig. 6.4 Distribution of selected field quantities over the composite thickness for the combined load case: a normal stress σx (z) and b normal strain εx (z)

6.3 Problem 2: Stresses and Strains in an Asymmetric Laminate

59

Fig. 6.5 Cross section of an asymmetric composite with eight layers

Table 6.3 Geometrical properties (z k ; z k−1 ; tk ) of the composite shown in Fig. 6.5 Layer k Coordinate z k−1 in Coordinate z k in mm Thickness tk 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

t/8 t/8 t/8 t/8 t/8 t/8 t/8 t/8

• Laminate loaded by N xn = 1000 N/mm The solution procedure will follow the recommended steps provided in Table 5.1. ① Define for each layer k the material (E k ) and the geometrical (z k ; z k−1 ; αk ) properties. The material property (E x,k ) for each layer can be taken from Table 6.1. The geometrical properties for each layer are summarized in Table 6.3. T  ② Define the column matrix of generalized stresses: s = N xn M yn .

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

(6.18)

60

6 Example Problems

③ Calculate the matrix elements A, B, and C according to Eqs. (5.14)–(5.16). Assemble the generalized elasticity matrix C ∗ according to Eq. (5.13).

The evaluation of the matrix elements A, B, and C according to Eqs. (5.14)–(5.16) gives A=

8 

N , mm

(6.19)

Bk = −114660.0 N ,

(6.20)

Dk = 977760.0 Nmm ,

(6.21)

Ak = 340200.0

k =1

B=

8  k =1

D=

8  k =1

from which the generalized elasticity matrix (written without units) can be assembled as follows:   340200.0 −114660.0 C∗ = . (6.22) −114660.0 977760.0

④ Calculate the generalized compliance matrix (C ∗ )−1 based on Eqs. (5.18)– (5.20). The evaluation of the matrix elements A , B  , and C  according to Eqs. (5.18)–(5.20) gives 10−7 mm , N 10−7 B  =3.588878 , N 10−7 D  =10.648319 , Nmm A =30.604059

(6.23) (6.24) (6.25)

from which the generalized compliance matrix (written without units) can be assembled as follows:   30.604059 3.588878 (C∗ )−1 = 10−7 . (6.26) 3.588878 10.648319

6.3 Problem 2: Stresses and Strains in an Asymmetric Laminate

61

T  ⑤ Calculate the generalized strains e = εx0 κ y according to Eq. (5.17). The evaluation of Eq. (5.17) gives the generalized strains as follows:     εx0 3.060406 0 = (C ∗ )−1 s = / . 0.358888 00 κy 

(6.27)

e

k ⑥

0Calculate  the stresses in each layer k according to Eq. (5.21), σx (z) = Ck εx + zκ y , in the x-y laminate system (z

k−1 ≤ z ≤ z k ). The strains in the x-y laminate system are obtained from εx0 + zκ y .

Evaluation of Eq. (5.21) for each layer k, i.e.,

 σx,k (z) = Ck εx0 + zκ y ,

(6.28)

under consideration of the generalized strains given in Eq. (6.27) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 6.3 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. 6.6 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 z = 0. The strain distribution is linear and continuous (see Fig. 6.6b) across the entire thickness of the laminate. The stress distribution (see Fig. 6.6a) 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. 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 matrix element B where the bending–tension entry is now unequal to zero, see Eq. (6.22). • Laminate loaded by M yn = 1000 Nmm/mm. The solution procedure will follow again the recommended steps provided in Table 5.1. However, subtasks ① and ③–④ 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.

62

6 Example Problems

Fig. 6.6 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)

6.3 Problem 2: Stresses and Strains in an Asymmetric Laminate

63

T  ② Define the column matrix of generalized stresses: s = N xn M yn .

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

(6.29)

T  ⑤ Calculate the generalized strains e = εx0 κ y according to Eq. (5.17). The evaluation of Eq. (5.17) gives the generalized strains as follows:     εx0 0.358888 0 ∗ −1 = (C ) s = / . 1.064832 00 κy 

(6.30)

e

k ⑥

0Calculate  the stresses in each layer k according to Eq. (5.21), σx (z) = Ck εx + zκ y , in the x-y laminate system (z

k−1 ≤ z ≤ z k ). The strains in the x-y laminate system are obtained from εx0 + zκ y .

Evaluation of Eq. (5.21) for each layer k, i.e.,

 σx,k (z) = Ck εx0 + zκ y ,

(6.31)

under consideration of the generalized strains given in Eq. (6.30) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 6.3 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. 6.7 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. 6.7b). The stress distribution (see Fig. 6.7a) 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.

64

6 Example Problems

Fig. 6.7 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)

6.3 Problem 2: Stresses and Strains in an Asymmetric Laminate

65

This is also manifested in the matrix element B where the bending–tension entry is now unequal to zero, see Eq. (6.22). • Laminate loaded by N xn = 1000 N/mm and M yn = 1000 Nmm/mm. The solution procedure will follow again the recommended steps provided in Table 5.1. However, subtasks ① and ③–④ 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.  T ② Define the column matrix of generalized stresses: s = N xn M yn .

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

(6.32)

T  ⑤ Calculate the generalized strains e = εx0 κ y according to Eq. (5.17). The evaluation of Eq. (5.17) gives the generalized strains as follows:   εx0 κy 

 3.419294 0 = (C ) s = / . 1.423720 00 ∗ −1



(6.33)

e

k ⑥

0Calculate  the stresses in each layer k according to Eq. (5.21), σx (z) = Ck εx + zκ y , in the x-y laminate system (z

k−1 ≤ z ≤ z k ). The strains in the x-y laminate system are obtained from εx0 + zκ y .

Evaluation of Eq. (5.21) for each layer k, i.e.,

 σx,k (z) = Ck εx0 + zκ y ,

(6.34)

under consideration of the generalized strains given in Eq. (6.33) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 6.3 gives the stresses for each layer in the x-y laminate system. In case that the strains in the x-y laminate system

66

6 Example Problems

 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. 6.8 as a function of the laminate thickness (z-coordinate). It can be seen that combined load case with N xn and M yn 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. 6.8b). The stress distribution (see Fig. 6.8a) 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. Furthermore, it can be concluded that the stress and strain distribution is a simple linear superposition of the results from load cases (a) and (b).

6.4 Problem 3: Failure Criteria Given is an asymmetric four-layer laminate of thickness t = 8 mm as shown in Fig. 6.9. Each layer has the same thickness (t/4) but different material properties, which can be taken from Table 6.1. Consider (a) a tensile load case with N xn = 1000 N/mm (the internal moment is equal to zero), (b) a bending load case with M yn = 1000 Nmm/mm (the internal normal force is equal to zero), and (c) a combined load case with N xn = 1000 N/mm and M yn = 1000 Nmm/mm to determine the stresses and strains in each layer in the global x-y coordinate system. Use these stress and strain values to perform a failure analysis based on the maximum stress, maximum strain, and Tsai–Wu failure criteria. • Laminate loaded by N xn = 1000 N N/mm. The solution procedure will follow the recommended steps provided in Table 5.1. ① Define for each layer k the material (E k ) and the geometrical (z k ; z k−1 ; αk ) properties. The material property (E x,k ) for each layer can be taken from Table 6.1. The geometrical properties for each layer are summarized in Table 6.4. T  ② Define the column matrix of generalized stresses: s = N xn M yn .

6.4 Problem 3: Failure Criteria

67

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

68

6 Example Problems

Fig. 6.9 Cross section of an asymmetric composite with four layers Table 6.4 Geometrical properties (z k ; z k−1 ; tk ) of the composite shown in Fig. 6.9 Layer k Coordinate z k−1 in Coordinate z k in mm Thickness tk mm 1 2 3 4

−4 −2 0 2

−2 0 2 4

t/4 t/4 t/4 t/4

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

(6.35)

③ Calculate the matrix elements A, B, and C according to Eqs. (5.14)– (5.16). Assemble the generalized elasticity matrix C ∗ according to Eq. (5.13).

The evaluation of the matrix elements A, B, and C according to Eqs. (5.14)–(5.16) gives A=

8  k =1

B=

8 

N , mm

(6.36)

Bk = −198828.0 N ,

(6.37)

Dk = 1018416.0 Nmm ,

(6.38)

Ak = 370692.0

k =1

D=

8  k =1

from which the generalized elasticity matrix (written without units) can be assembled as follows:   370692.0 −198828.0 C∗ = . (6.39) −198828.0 1018416.0

6.4 Problem 3: Failure Criteria

69

④ Calculate the generalized compliance matrix (C ∗ )−1 based on Eqs. (5.18)– (5.20). The evaluation of the matrix elements A , B  , and C  according to Eqs. (5.18)–(5.20) gives 10−7 mm , N 10−7 B  =5.882727 , N 10−7 D  =10.967670 , Nmm A =30.131891

(6.40) (6.41) (6.42)

from which the generalized compliance matrix (written without units) can be assembled as follows:   30.131891 5.882727 (C∗ )−1 = (6.43) 10−7 . 5.882727 10.967670

T  ⑤ Calculate the generalized strains e = εx0 κ y according to Eq. (5.17). The evaluation of Eq. (5.17) gives the generalized strains as follows:   εx0 κy 

 3.013189 0 = (C ) s = / . 0.588273 00 ∗ −1



(6.44)

e

k ⑥

0Calculate  the stresses in each layer k according to Eq. (5.21), σx (z) = Ck εx + zκ y , in the x-y laminate system (z

k−1 ≤ z ≤ z k ). The strains in the x-y laminate system are obtained from εx0 + zκ y .

Evaluation of Eq. (5.21) for each layer k, i.e.,

 σx,k (z) = Ck εx0 + zκ y ,

(6.45)

70

6 Example Problems

under consideration of the generalized strains given in Eq. (6.44) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 6.4 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. 6.10 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 z = 0. The strain distribution is linear and continuous (see Fig. 6.10b) across the entire thickness of the laminate. The stress distribution (see Fig. 6.10a) 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 matrix element B where the bending–tension entry is now unequal to zero, see Eq. (6.39). Failure analysis based on the maximum stress, maximum strain, and Tsai–Wu failure criteria. To easily compare different failure criteria, we determine the limit loads by applying the load as N xn = R × 1000 N/mm, i.e., we multiply the load with the strength ratio R. In case of R > 1 there is no failure, whereas R ≤ 1 results in the failure of the respective layer. Considering the stresses from Fig. 6.10a in the condition of the maximum stress criterion according to Eq. (4.27) gives (kc 1). In addition, it can be seen that the Tsai–Wu criterion is identical to the maximum stress approach. • Laminate loaded by M yn = 1000 Nmm/mm. The solution procedure will follow again the recommended steps provided in Table 5.1. However, subtasks ① and ③–④ 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.  T ② Define the column matrix of generalized stresses: s = N xn M yn .

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

(6.49)

6.4 Problem 3: Failure Criteria

73

T  ⑤ Calculate the generalized strains e = εx0 κ y according to Eq. (5.17). The evaluation of Eq. (5.17) gives the generalized strains as follows:   εx0 κy 

  0.588273 0 = (C ) s = / . 1.096767 00 ∗ −1

(6.50)

e

k ⑥

0Calculate  the stresses in each layer k according to Eq. (5.21), σx (z) = Ck εx + zκ y , in the x-y laminate system (z

k−1 ≤ z ≤ z k ). The strains in the x-y laminate system are obtained from εx0 + zκ y .

Evaluation of Eq. (5.21) for each layer k, i.e.,

 σx,k (z) = Ck εx0 + zκ y ,

(6.51)

under consideration of the generalized strains given in Eq. (6.50) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 6.4 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. 6.11 as a function of the laminate thickness (z-coordinate). It can be seen that a pure bending load by M yn 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. 6.11b). The stress distribution (see Fig. 6.11a) 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 matrix element B where the bending–tension entry is now unequal to zero, see Eq. (6.39). Failure analysis based on the maximum stress, maximum strain, and Tsai–Wu failure criteria. To easily compare different failure criteria, we determine the limit loads by applying the load as M yn = R × 1000 Nmm/mm, i.e., we multiply the load with the strength

74

6 Example Problems

Fig. 6.11 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)

6.4 Problem 3: Failure Criteria

75

Table 6.6 Strength ratio R in the x-y coordinate systems for an external loading of M yn = R × 1000 Nmm/mm, asymmetric laminate Layer no. Position Direction Max. stress Max. strain Tsai–Wu 1

2

3

4

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

x x x x x x x x x x x x

6.009 3.570 2.539 26.308 22.631 7.169 2.461 4.063 11.637 1.178 1.511 2.107

10.011 5.948 4.231 23.459 23.107 7.320 2.878 4.752 13.611 1.237 1.586 2.212

6.009 3.570 2.539 26.308 22.631 7.169 2.461 4.063 11.637 1.178 1.511 2.107

ratio R. In case of R > 1 there is no failure, whereas R ≤ 1 results in the failure of the respective layer. As in the case of (a), application of Eqs. (6.46)–(6.48) allows the determination of the strength ratio R. The summary of the strength ratio is provided in Table 6.6. It can be seen from Table 6.6 that all failure criteria predict no failure of the lamina (R > 1). In addition, it can be seen that the Tsai–Wu criterion is identical to the maximum stress approach. • Laminate loaded by N xn = 1000 N/mm and M yn = 1000 Nmm/mm. The solution procedure will follow again the recommended steps provided in Table 5.1. However, subtasks ① and ③–④ 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. T  ② Define the column matrix of generalized stresses: s = N xn M yn .

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

(6.52)

76

6 Example Problems

T  ⑤ Calculate the generalized strains e = εx0 κ y according to Eq. (5.17). The evaluation of Eq. (5.17) gives the generalized strains as follows:   εx0 κy 

 3.601462 0 = (C ) s = / . 1.685040 00 ∗ −1



(6.53)

e

k ⑥

0Calculate  the stresses in each layer k according to Eq. (5.21), σx (z) = Ck εx + zκ y , in the x-y laminate system (z

k−1 ≤ z ≤ z k ). The strains in the x-y laminate system are obtained from εx0 + zκ y .

Evaluation of Eq. (5.21) for each layer k, i.e.,

 σx,k (z) = Ck εx0 + zκ y ,

(6.54)

under consideration of the generalized strains given in Eq. (6.53) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 6.4 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. 6.12 as a function of the laminate thickness (z-coordinate). It can be seen that the combined load case with N xn and M yn 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. 6.12b). The stress distribution (see Fig. 6.12a) 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 since the matrix element B (bending–tension coupling) is again unequal to zero, see Eq. (6.39). However, this coupling is not so obvious due to the combined load case. Failure analysis based on the maximum stress, maximum strain, and Tsai–Wu failure criteria.

6.4 Problem 3: Failure Criteria

77

Fig. 6.12 Distribution of selected field quantities over the laminate thickness for the combined load case: a normal stress σx (z) and b normal strain εx (z)

78

6 Example Problems

Table 6.7 Strength ratio R in the x-y coordinate systems for an external loading of N xn = R × 1000 N/mm and M yn = R × 1000 Nmm/mm, asymmetric laminate Layer no.

Position

Direction

Max. stress

Max. strain

Tsai–Wu

1

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

x x x x x x x x x x x x

25.330 6.636 3.073 4.297 8.076 66.886 0.982 1.295 1.901 0.567 0.677 0.841

26.592 11.056 5.120 3.832 7.201 59.642 1.149 1.515 2.223 0.595 0.711 0.883

25.330 6.636 3.073 4.297 8.076 66.886 0.982 1.295 1.901 0.567 0.677 0.841

2

3

4

To easily compare the different failure criteria, we determine the limit loads by applying the load as N xn = R × 1000 N/mm and M yn = R × 1000 Nmm/mm, i.e., we multiply the load with the strength ratio R. In case of R > 1 there is no failure, whereas R ≤ 1 results in the failure of the respective layer. As in the case of (a), application of Eqs. (6.46)–(6.48) allows the determination of the strength ratio R. The summary of the strength ratio is provided in Table 6.7. It can be seen from Table 6.7 that all failure criteria predict the failure of lamina 4 (R < 1). The maximum stress and the Tsai–Wu criteria predict in addition the failure of layer 3 (R < 1 for the top part of the layer). Finally, it can be stated that all criteria provide strength ratios in a similar order of magnitude.

6.5 Problem 4: Ply-By-Ply Failure Loads Given is a symmetric three-layer laminate of thickness t = 9 mm as shown in Fig. 6.13. Each layer has the same thickness (t/3) but material properties may vary from layer to layer, see Table 6.1. Consider a pure tensile load case with N xn > 0 (the internal bending moment is equal to zero) to investigate the ply-by-ply failure of this laminate based on the maximum stress criterion. The solution procedure will follow the recommended steps provided in Table 5.1.

6.5 Problem 4: Ply-By-Ply Failure Loads

79

Fig. 6.13 Cross section of a symmetric composite with three layers Table 6.8 Geometrical properties (z k ; z k−1 ; tk ) of the composite shown in Fig. 6.13 Layer k Coordinate z k−1 in Coordinate z k in mm Thickness tk mm 1 2 3

−4.5 −1.5 1.5

−1.5 1.5 4.5

t/3 t/3 t/3

① Define for each layer k the material (E k ) and the geometrical (z k ; z k−1 ; αk ) properties. The material property (E x,k ) for each layer can be taken from Table 6.1. The geometrical properties for each layer are summarized in Table 6.8. T  ② Define the column matrix of generalized stresses: s = N xn M yn .

The load matrix for case (a) is given as follows:  T s = 1 0 N/mm .

(6.55)

③ Calculate the matrix elements A, B, and C according to Eqs. (5.14)– (5.16). Assemble the generalized elasticity matrix C ∗ according to Eq. (5.13).

The evaluation of the matrix elements A, B, and C according to Eqs. (5.14)–(5.16) gives

80

6 Example Problems

A=

8 

Ak = 790020.0

k =1

B=

8 

N , mm

(6.56)

Bk = 0.0 N ,

(6.57)

Dk = 7396515.0 Nmm ,

(6.58)

k =1

D=

8  k =1

from which the generalized elasticity matrix (written without units) can be assembled as follows:   790020.0 0.0 ∗ . (6.59) C = 0.0 7396515.0

④ Calculate the generalized compliance matrix (C ∗ )−1 based on Eqs. (5.18)– (5.20). The evaluation of the matrix elements A , B  , and C  according to Eqs. (5.18)–(5.20) gives A =12.657907 B  =0.0

10−7 , N

D  =1.351988

10−7 mm , N

(6.60) (6.61)

10−7 , Nmm

(6.62)

from which the generalized compliance matrix (written without units) can be assembled as follows:   12.657907 0.0 (C∗ )−1 = 10−7 . (6.63) 0.0 1.351988

 T ⑤ Calculate the generalized strains e = εx0 κ y according to Eq. (5.17). The evaluation of Eq. (5.17) gives the generalized strains as follows:

6.5 Problem 4: Ply-By-Ply Failure Loads

  εx0 κy 

= (C ∗ )−1 s =

81



 1.265791 × 10−3 0 /00 . 0.0

(6.64)

e

k ⑥

0Calculate  the stresses in each layer k according to Eq. (5.21), σx (z) = Ck εx + zκ y , in the x-y laminate system (z

k−1 ≤ z ≤ z k ). The strains in the x-y laminate system are obtained from εx0 + zκ y .

Evaluation of Eq. (5.21) for each layer k, i.e.,

 σx,k (z) = Ck εx0 + zκ y ,

(6.65)

under consideration of the generalized strains given in Eq. (6.64) and the range of the z-coordinate (z k−1 ≤ z ≤ z k ) provided in Table 6.8 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. 6.14 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 z = 0. The strain distribution is constant (see Fig. 6.14b) across the entire thickness of the laminate. The stress distribution (see Fig. 6.14a) 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. Failure analysis based on the maximum stress criterion.

To easier apply the maximum stress criterion, we determine the limit loads by applying the load as N xn = R × 1 N/mm, i.e., we multiply the load with the strength ratio R. Considering the stresses from Fig. 6.14a in the condition of the maximum stress criterion according to Eq. (4.27) gives (kc