Advanced Mechanics
of Composite Materials
and Structures
Advanced Mechanics
of Composite Materials
and Structures
Fourth Edition
Valery V. Vasiliev
Professor of Structural Mechanics,
Institute for Problems in Mechanics Russian Academy of Sciences,
Moscow
Evgeny V. Morozov
Professor of Mechanical Engineering,
School of Engineering & IT, The University of New South Wales, Canberra,
Australia
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Preface
This book is concerned with describing, analyzing, and discussing topical problems in the mechanics of advanced composite materials and structures whose mechanical properties are controlled by
high-strength and/or high-stiffness continuous fibers embedded in a polymeric, metal, or ceramic
matrix. Although the idea of combining two or more components to produce materials with controlled properties has been known and used from time immemorial, modern composites have been
developed only over the last five or six decades and now have widespread applications in many different fields of engineering, but are of particular importance in aerospace structures for which high
strength-to-weight and stiffness-to-weight ratios are required.
Due to these widespread existing and potential applications, composite technology has been
developed very intensively over recent decades, and there exist publications that cover anisotropic
elasticity, mechanics of composite materials, design, analysis, fabrication, and application of composite structures. A particular feature of this book, which distinguishes it from the existing publications, is that it addresses a wide range of advanced problems in the mechanics of composite
materials, such as the physical statistical aspects of fiber strength, stress diffusion in composites
with damaged fibers, stress singularity, nonlinear elasticity, plasticity and creep of composite materials, hybrid and two-matrix composites, spatial fibrous structures, tensor strength criteria, progressive failure, environmental and manufacturing effects, as well as the traditional basic composite
material mechanics. In addition to classical lamination theory and its application to beams and
plates, this book covers the problems of consistency of nonclassical theories, buckling and postbuckling behavior of symmetric and nonsymmetrically laminated plates, and applied theories of
composite cylindrical shells. Since the advantages of composite materials are best demonstrated by
the optimal fibrous structures, we have particularly addressed the issues of optimality criteria and
the analysis of the most significant optimal composite structures: filament-wound pressure vessels,
rotating disks, and lattice cylindrical shells.
The authors of this book are experienced designers of composite structures who over the last 40
years have been involved in practically all the main Russian projects in composite technology. This
experience has led us to carefully select the problems addressed in our book, which can be referred
to as material problems challenging design engineers.
The primary aim of the book is the combined coverage of mechanics, technology, and analysis
of composite materials and structural elements at an advanced level. Such an approach enables an
engineer to take into account the essential mechanical properties of the material itself and the special features of practical implementation, including manufacturing technology, experimental results,
and design characteristics.
This book consists of 12 chapters and can be divided into 2 main parts: the first part including
Chapters 1 5 is devoted to composite materials, whereas the second part, Chapters 6 12, covers
the analysis and design of typical composite structural elements.
In comparison with the third edition, the book includes three new chapters—Chapter 8,
Equations of the Applied Theory of Thin-Walled Composite Structures, devoted to general equations of mechanics of composite structures, Chapter 11, Shells of Revolution, concerning composite
shells of revolution, and Chapter 12, Composite Pressure Vessels, about composite pressure vessels.
To retain a reasonable volume for the book, two chapters have been removed—Chapter 1,
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Preface
Introduction, including the general information about composite materials which can be found in
numerous books on composites and Chapter 2, Fundamentals of Mechanics of Solids, concerning
basic equations in the mechanics of solids that are presented in specialized books on the mechanics
of solids.
Chapter 1, Mechanics of a Unidirectional Ply, of the fourth edition is devoted to the basic structural element of a composite material, that is, unidirectional composite ply. In addition to the conventional description of micromechanical models and experimental results, the physical nature of
fiber strength, its statistical characteristics, and interaction of damaged fibers through the matrix are
discussed, and we show that fibrous composites comprise a special class of man-made materials
utilizing the natural potentials of material strength and structure.
Chapter 2, Mechanics of a Composite Layer, contains a description of typical composite layers
made of unidirectional, fabric, and spatially reinforced composite materials. Conventional linear
elastic models are supplemented in this chapter with nonlinear elastic and elastic plastic analyses
demonstrating specific types of the behavior of composites with metal and thermoplastic matrices.
Chapter 3, Mechanics of Laminates, is concerned with the mechanics of laminates and includes
a conventional description of the laminate stiffness matrix, coupling effects in typical laminates
and procedures for stress calculations for in-plane and interlaminar stresses.
Chapter 4, Failure Criteria and Strength of Laminates, presents a practical approach to the evaluation of laminate strength. Three main types of failure criteria, that is, structural criteria indicating
the modes of failure, approximation polynomial criteria treated as formal approximations of experimental data, and tensor-polynomial criteria are discussed and compared with available experimental
results for unidirectional and fabric composites. A combined elastoplastic damage model and a
strain-driven implicit integration procedure for fiber reinforced composite materials and structures
that involve a consideration of their mechanical response before the initiation of damage, prediction
of damage initiation, and modeling of postfailure behavior are discussed.
Chapter 5, Environmental, Special Loading, and Manufacturing Effects, dealing with environmental and special loading effects includes analysis of thermal conductivity, hydrothermal elasticity, material aging, creep, and durability under long-term loading, fatigue, damping, and impact
resistance of typical advanced composites. The effect of manufacturing factors on material properties and behavior is demonstrated for filament winding accompanied with nonuniform stress distribution between the fibers and ply waviness and laying-up processing of nonsymmetric laminate
exhibiting warping after curing and cooling.
Composite beams are discussed in Chapter 6, Laminated Composite Beams and Columns.
Along with the general solutions for static and buckling problems, refined beam theories are considered and evaluated.
Chapter 7, Laminated Composite Plates, is devoted to laminated composite plates and covers
traditional and specific problems of the plate theory, particularly, Kirchhoff and Thomson—Tait
transformation of boundary conditions in classical plate theory and the interaction of penetrating
and boundary-layer solutions in the theory of shear deformable plates. Exact solution of the problem of bending of a clamped plate and the problem of the plate torsion by corner forces are presented. Buckling and postbuckling problems are discussed with application to composite plates
with symmetrically and unsymmetrically laminated structure. The problem of optimization of laminates under in-plane loading and bending is considered in the closure of the chapter.
Preface
xv
Chapter 8, Equations of the Applied Theory of Thin-Walled Composite Structures, contains a
brief derivation of the general equations of applied theory of thin-walled composite structures
allowing us to consider the problems of statics, stability, and dynamics of composite structures.
Chapter 9, Thin-Walled Composite Beams, is concerned with laminated thin-walled composite
beams. Free and restrained bending and torsion problems are considered for composite beams with
closed, open, and multicell contours of the beam cross section. Coupling effects in anisotropic
beams are discussed with application to composite beams with controlled properties. Practical
methods for the analysis of beams stiffened with ribs and loaded with surface forces are presented.
Chapter 10, Circular Cylindrical Shells, deals with composite cylindrical shells and covers
applied theories and practical methods of analysis, particularly the problems of engineering and
semimembrane shell theories. A new approach to the study of shell buckling under axial compression, which results in simple and boundary-condition-independent analytical solutions, is presented.
Optimal lattice composite structures are discussed at the end of this chapter.
In Chapter 11, Shells of Revolution, analytical and numerical methods are described with application to the problem of axisymmetric deformation of composite shells of revolution.
Chapter 12, Composite Pressure Vessels, is devoted to optimal composite pressure vessels—the
structures that demonstrate excellent weight efficiency of composites as structural materials in a
most pronounced way.
This fourth edition is a revised, updated, and extended version of the third edition, with new
chapters on composite shells and structures.
A new title “Advanced Mechanics of Composite Materials and Structures” has been adopted for
the fourth edition, because this provides a better reflection of the overall contents and improvements, extensions and revisions introduced in the present version.
This book offers a comprehensive coverage of the topic over a wide range: from basics and fundamentals to the advanced modeling and analysis including practical design and engineering applications and can be used as an up-to-date introductory text book aimed at senior undergraduates and
graduate students. At the same time, it includes a detailed and comprehensive coverage of the contemporary theoretical models at the micro- and macrolevels of material structure, practical methods
and approaches, experimental results, and optimization of composite material properties and component performance that can be used by researchers and engineers.
The authors would like to thank several people for their time and effort in making the book a
reality. Specifically, we would like to thank our Elsevier editors who have encouraged and participated in the preparation of the fourth edition. These include Matthew Deans, Senior Publisher and
Peter Jardim, Editorial Project Manager. Special thanks are due to Prof. Leslie Henshall, for his
work on the text improvements.
Valery V. Vasiliev and Evgeny V. Morozov
Introduction
Materials are the basic elements of all natural and man-made structures. Figuratively speaking,
these materialize the structural conception. Technological progress is associated with continuous
improvement of existing material properties as well as with the expansion of structural material
classes and types. Usually, new materials emerge due to the necessity to improve structural efficiency and performance. In addition, new materials themselves as a rule, in turn provide new
opportunities to develop updated structures and technology, while the latter challenges materials
science with new problems and tasks. One of the best manifestations of this interrelated process in
the development of materials, structures, and technology is associated with composite materials and
structural elements to which this book is devoted.
Composite materials emerged in the middle of the 20th century as a promising class of engineering materials providing new prospects for modern technology. Generally speaking, any material
consisting of two or more components with different properties and distinct boundaries between the
components can be referred to as a composite material. Moreover, the idea of combining several
components to produce a material with properties that are not attainable with the individual components has been used by man for thousands of years. Correspondingly the majority of natural materials that have emerged as a result of a prolonged evolution process can be treated as composite
materials.
With respect to the problems covered in this book, we can classify existing composite materials
(composites) into two main groups.
The first group comprises composites that are known as “filled materials.” The main feature of
these materials is the existence of some basic or matrix material whose properties are improved by
filling it with some particles. Usually the matrix volume fraction is more than 50% in such materials, and material properties, being naturally modified by the fillers, are governed mainly by the
matrix. As a rule, filled materials can be treated as homogeneous and isotropic, i.e., traditional
models of the mechanics of materials developed for metals and other conventional materials can be
used to describe their behavior. This group of composites is not touched on in this book.
The second group of composite materials that is under study here involves composites that are
called “reinforced materials.” The basic components of these materials (sometimes referred to as
“advanced composites”) are long thin fibers. The fibers are bound with a matrix material whose
volume fraction in a composite is usually less than 50%. The main properties of advanced composites, due to which these materials find a wide application in engineering, are provided by fibers
possessing high strength and stiffness. At present, glass, mineral, carbon, organic, boron, ceramic,
and metal fibers are used in reinforced composite materials.
Continuous glass fibers (the first type of fibers used in advanced composites) are made by pulling molten glass (at a temperature about 1300 C) through 0.83.0 mm diameter dies and further
high-speed stretching to a diameter of 319 µm. Usually, glass fibers have solid circular crosssections. However, there exist fibers with rectangular (square or plane), triangular, and hexagonal
cross-sections, as well as hollow circular fibers. The important properties of glass fibers as components of advanced composites for engineering applications are their high strength, which is maintained in humid environments but degrades under elevated temperatures, relatively low stiffness
(about 40% of the stiffness of steel), high chemical and biological resistance, and low cost. Being
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Introduction
actually elements of monolithic glass, the fibers do not absorb water and do not change their
dimensions in water. For the same reason, they are brittle and sensitive to surface damage.
Quartz fibers are similar to glass fibers and are obtained by high-speed stretching of quartz rods
made of (at temperatures of about 2200 C) fused quartz crystals or sand. The original process
developed for manufacturing glass fibers cannot be used because the viscosity of molten quartz is
too high to make thin fibers directly. However, this more complicated process results in fibers with
higher thermal resistance than glass fibers.
The same process that is used for glass fibers can be employed to manufacture mineral fibers,
e.g., basalt fibers made of molten basalt rocks. Having relatively low strength and high density,
basalt fibers are not used for high performance, e.g., aerospace structures, but are promising reinforcing elements for prestressed reinforced concrete structures in civil engineering.
Substantial improvement of a fiber’s stiffness in comparison with glass fibers has been achieved
with the development of carbon (or graphite) fibers. Modern high-modulus carbon fibers have a
modulus that is a factor of about 4 higher than the modulus of steel, whereas the fiber’s density is
lower by the same factor. Modern high-strength fibers have a 40% higher tensile strength compared
with the strength of the best glass fibers, whereas the density of carbon fibers is 30% less than that
of glass fibers. Carbon fibers are made by pyrolysis of organic fibers of which there exist two main
types—polyacrylonitrile (PAN)-based and pitch-based fibers. For PAN-based fibers the process
consists of three stages—stabilization, carbonization, and graphitization. In the first step (stabilization) a system of PAN filaments is stretched and heated up to about 400 C in an oxidation furnace,
while in the subsequent step (carbonization at 900 C in an inert gas media), most elements of the
filaments other than carbon are removed or converted into carbon. During the successive heat treatment at a temperature reaching 2800 C (graphitization) a crystalline carbon structure oriented along
the fiber’s length is formed, resulting in PAN-based carbon fibers. The same process is used for
rayon organic filaments (instead of PAN), but results in carbon fibers with lower modulus and
strength because rayon contains less carbon than PAN. For pitch-based carbon fibers the initial
organic filaments are made in approximately the same manner as for glass fibers from molten
petroleum or coal pitch and pass through carbonization and graphitization processes. Since pyrolysis is accompanied by a loss of material, carbon fibers have a porous structure and their specific
gravity (about 1.8) is less than that of graphite (2.26). The properties of carbon fibers are affected
by the crystallite size, crystalline orientation, porosity and purity of the carbon structure. As components of advanced composites for engineering applications, carbon fibers are characterized by very
high modulus and strength, high chemical and biological resistance, electrical conductivity, and
very low coefficient of thermal expansion. The strength of carbon fibers practically does not change
with temperature up to 1500 C (in an inert media preventing oxidation of the fibers). The exceptional strength of 7.06 GPa is reached in Toray T-1000 carbon fibers, whereas the highest modulus
of 850 GPa is obtained in Carbonic HM-85 fibers. Carbon fibers are anisotropic, very brittle, and
sensitive to damage. They do not absorb water and do not change their dimensions in humid environments. There exist more than 50 types of carbon fibers with a broad spectrum of strength, stiffness, and cost, and the process of fiber advancement is not over—one may expect fibers with
strength up to 10 GPa and modulus up to 1000 GPa within a few years.
Organic fibers commonly encountered in textile applications can be employed as reinforcing
elements of advanced composites. Naturally, only high-performance fibers, i.e., fibers possessing
high stiffness and strength, can be used for this purpose. The most widely used organic fibers that
Introduction
xix
satisfy these requirements are known as aramid (aromatic polyamide) fibers. They are extruded
from a liquid crystalline solution of the corresponding polymer in sulfuric acid with subsequent
washing in a cold water bath and stretching under heating. As components of advanced composites
for engineering applications, aramid fibers are characterized by low density providing high specific
strength and stiffness, low thermal conductivity resulting in high heat insulation, and a negative
thermal expansion coefficient allowing us to construct hybrid composite elements that do not
change their dimensions under heating. Consisting actually of a system of very thin filaments
(fibrils), aramid fibers have very high resistance to damage. Their high strength in the longitudinal
direction is accompanied by relatively low strength under tension in the transverse direction.
Aramid fibers are characterized by pronounced temperature and time dependence for stiffness and
strength. Unlike the inorganic fibers discussed earlier, they absorb water resulting in moisture content up to 7% and degradation of material properties by 15%20%. The list of organic fibers has
been supplemented recently with extended chain polyethylene fibers demonstrating outstanding low
density (less than that of water) in conjunction with relatively high stiffness and strength.
Polyethylene fibers are extruded from the corresponding polymer melt in a similar manner to glass
fibers. They do not absorb water and have high chemical resistance, but demonstrate relatively low
temperature and creep resistance.
Boron fibers were developed to increase the stiffness of composite materials when glass fibers
were mainly used to reinforce composites of the day. Being followed by high-modulus carbon
fibers with higher stiffness and lower cost, boron fibers have now rather limited application. Boron
fibers are manufactured by chemical vapor deposition of boron onto about 12 µm diameter tungsten
or carbon wire (core). As a result, boron fibers have a relatively large diameter, 100200 µm. They
are extremely brittle and sensitive to surface damage. Being mainly used in metal matrix composites, boron fibers degrade on the contact with aluminum or titanium matrices at the temperature
that is necessary for processing (above 500 C). To prevent this degradation, chemical vapor deposition is used to cover the fiber surface with about 5 µm-thick layer of silicon carbide, SiC, (such
fibers are called Borsic) or boron carbide, B4 C.
There exists a special class of ceramic fibers for high-temperature applications composed of
various combinations of silicon, carbon, nitrogen, aluminum, boron, and titanium. The most commonly used are silicon carbide (SiC) and alumina (Al2 O3 ) fibers. Silicon carbide is deposited on a
tungsten or carbon core-fiber by the reaction of a gas mixture of silanes and hydrogen. Thin
(815 µm in diameter) SiC fibers can be made by pyrolysis of polymeric (polycarbosilane) fibers
at temperatures of about 1400 C in an inert atmosphere. Silicon carbide fibers have high strength
and stiffness, moderate density and very high melting temperature (2600 C). Alumina ðAl2 O3 Þ
fibers are fabricated by the sintering of fibers extruded from the viscous alumina slurry with rather
complicated composition. Alumina fibers, possessing approximately the same mechanical properties as SiC fibers, have relatively large diameter and high density. The melting temperature is about
2000 C. Silicon carbide and alumina fibers are characterized by relatively low reduction in strength
at elevated temperatures. Promising ceramic fibers for high-temperature applications are boron carbide ðB4 CÞ fibers that can be obtained either as a result of reaction of a carbon fiber with a mixture
of hydrogen and boron chloride at high temperature (around 1800 C) or by pyrolysis of cellulosic
fibers soaked with boric acid solution. Possessing high stiffness and strength and moderate density,
boron carbide fibers have very high thermal resistance (up to 2300 C).
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Introduction
Metal fibers (thin wires) made of steel, beryllium, titanium, tungsten, and molybdenum are used
for special, e.g., low-temperature and high-temperature applications.
In advanced composites, fibers provide not only high strength and stiffness but also a possibility
to tailor the material so that the directional dependence of its mechanical properties matches that of
the loading environment. The principle of directional properties can be traced in all natural materials that have emerged as a result of a prolonged evolution and, in contrast to man-made metal
alloys, are neither isotropic nor homogeneous. Many natural materials have fibrous structures and
utilize the high strength and stiffness of natural fibers. Natural fibers have lower strength and stiffness than man-made fibers.
Before being used as reinforcing elements of advanced composites, the fibers are normally subjected to special finish surface treatments, undertaken to prevent any fiber damage under contact
with processing equipment, to provide surface wetting when the fibers are combined with matrix
materials, and to improve the interface bond between fibers and matrices. The most commonly
used surface treatments are chemical sizing performed during the basic fiber formation operation
and resulting in a thin layer applied to the surface of the fiber, surface etching by acid, plasma, or
corona discharge, and coating of the fiber surface with thin metal or ceramic layers.
With only a few exceptions (e.g., metal fibers), individual fibers, being very thin and sensitive
to damage, are not used in composite manufacturing directly, but in the form of tows (rovings),
yarns, and fabrics. A unidirectional tow (roving) is a loose assemblage of parallel fibers consisting
usually of thousands of elementary fibers. The two main designations are used to indicate the size
of the tow, namely the K-number that gives the number of fibers in the tow (e.g., 3K tow contains
3000 fibers) and the tex-number which is the mass in grams of 1000 m of the tow. The tow texnumber depends not only on the number of fibers but also on the fiber diameter and density. For
example, AS4-6K tow consisting of 6000 AS4 carbon fibers has 430 tex. A yarn is a fine tow (usually it includes hundreds of fibers) slightly twisted (about 40 turns per meter) to provide the integrity of its structure necessary for textile processing. Yarn size is indicated in tex-numbers or in
textile denier-numbers (den) such that 1 tex 5 9 den. Continuous yarns are used to make fabrics
with various weave patterns. There exists a wide variety of glass, carbon, aramid, and hybrid fabrics that are widely used in composite technology.
To utilize the high strength and stiffness of fibers in a monolithic composite material
suitable for engineering applications, fibers are bound with a matrix material whose strength and
stiffness are, naturally, much lower than those of fibers (otherwise, no fibers would be necessary).
Matrix materials provide the final shape of the composite structure and govern the parameters of
the manufacturing process. The optimal combination of fiber and matrix properties should satisfy a
set of operational and manufacturing requirements that are sometimes of a contradictory nature,
and have not been completely met yet in existing composites.
First of all the stiffness of the matrix should correspond to the stiffness of the fibers and be sufficient to provide uniform loading of fibers. The fibers are usually characterized by relatively high
scatter in strength that may be increased due to damage of the fibers caused by the processing
equipment. Naturally, fracture of the weakest or damaged fiber should not result in material failure.
Instead the matrix should evenly redistribute the load from the broken fiber to the adjacent ones
and then load the broken fiber at a distance from the cross-section at which it failed. The higher the
matrix stiffness, the smaller is the distance, and less is the influence of damaged fibers on the composite material strength and stiffness (which should be the case). Moreover, the matrix should
Introduction
xxi
provide the proper stress diffusion (this is the term traditionally used for this phenomenon in the
analysis of stiffened structures) in the material at a given operational temperature. That is why this
temperature is limited, as a rule, by the matrix rather than by the fibers. But on the other hand, to
provide material integrity up to the failure of the fibers, the matrix material should possess high
compliance. Obviously, for a linear elastic material, a combination of high stiffness and high ultimate strain results in high strength which is not the case for modern matrix materials. Thus close to
optimal (with respect to the foregoing requirements) and realistic matrix material should have a
nonlinear stressstrain diagram and possess high initial modulus of elasticity and high ultimate
strain.
However, matrix properties, even though being optimal for the corresponding fibers, do not
manifest in the composite material if the adhesion (the strength of fibermatrix interface bonding)
is not high enough. Strong adhesion between fibers and matrices, providing material integrity up to
the failure of the fibers, is a necessary condition for high-performance composites. Proper adhesion
can be reached for appropriately selected combinations of fiber and matrix materials under some
additional conditions. First a liquid matrix should have viscosity low enough to allow the matrix to
penetrate between the fibers of such dense systems of fibers as tows, yarns, and fabrics. Second the
fiber surface should have good wettability with the matrix. Third the matrix viscosity should be
high enough to retain the liquid matrix in the impregnated tow, yarn, or fabric in the process of fabrication of a composite part. Finally the manufacturing process providing the proper quality of the
resulting material should not require high temperature and pressure to make a composite part.
At present, typical matrices are made from polymeric, metal, carbon, and ceramic materials.
Polymeric matrices are divided into two main types: thermoset and thermoplastic. Thermoset
polymers, which are the most widely used matrix materials for advanced composites, include polyester, epoxy, polyimide, and other resins cured under elevated or room temperature. Being cured
(polymerized) a thermoset matrix cannot be reset, dissolved, or melted. Heating of a thermoset
material results first in degradation of its strength and stiffness, and then in thermal destruction.
In contrast to thermoset resins, thermoplastic matrices do not require any curing reaction. They
melt under heating and convert to a solid state under cooling. The possibility to remelt and dissolve
thermoplastic matrices allows us to reshape composite parts forming them under heating and simplifies their recycling, which is a problem for thermoset materials.
Polymeric matrices can be combined with glass, carbon, organic, or boron fibers to yield a wide
class of polymeric composites with high strength and stiffness, low density, high fatigue resistance,
and excellent chemical resistance. The main disadvantage of these materials is their relatively low
(in comparison with metals) temperature resistance limited by the matrix. The so-called thermomechanical curves are plotted to determine this important (for applications) characteristic of the
matrix. These curves show the dependence of some stiffness parameter on the temperature and
allow us to find the so-called glass transition temperature, Tg , which indicates a dramatic reduction
in material stiffness. Naturally, to retain the complete set of properties of polymeric composites,
the operating temperature, in general, should not exceed Tg . However, the actual material behavior
depends on the type of loading. As a rule, heating above the glass transition temperature only
slightly influences material properties under tension in the fiber direction and dramatically reduces
its strength in longitudinal compression and transverse bending. The glass transition temperature
depends on the processing temperature, Tp , at which a material is fabricated, and higher Tp results,
as a rule, in higher Tg . Thermoset epoxy matrices cured at a temperature in the range 120160 C
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Introduction
have Tg 5 60 2 1403 C. There also exist a number of high-temperature thermoset matrices (e.g.,
organosilicone, polyimide, and bismaleimide resins) with Tg 5 250 2 3003 C and curing temperatures up to 400 C. Thermoplastic matrices are also characterized by a wide range of glass transition
temperatures—from 100 to 270 C. The processing temperature for different thermoplastic matrices
varies from 300 to 4003 C.
Further enhancement to the temperature resistance of composite materials is associated with the
use of metal matrices in combination with high-temperature boron, carbon, ceramic fibers, and
metal wires. The most widespread metal matrices are aluminum, magnesium, and titanium alloys
possessing high plasticity, whereas for special applications nickel, copper, niobium, cobalt, and
lead matrices can be used. Fiber reinforcement essentially improves the mechanical properties of
such metals. For example, carbon fibers increase the strength and stiffness of such a soft metal as
lead by an order of magnitude.
Metal matrices allow us to increase the operational temperatures for composite structures. In
polymeric composites the matrix materials play an important but secondary role of holding the
fibers in place and providing good load dispersion into the fibers, whereas material strength and
stiffness are controlled by the reinforcements. In contrast the mechanical properties of metal matrix
composites are controlled by the matrix to a considerably larger extent, though the fibers still provide the major contribution to the strength and stiffness of the material.
The next step in the development of composite materials that can be treated as matrix materials
reinforced with fibers rather than fibers bonded with matrix (which is the case for polymeric composites) is associated with ceramic matrix composites, possessing very high thermal resistance. The
stiffnesses of the fibers that are usually metal (steel, tungsten, molybdenum, and niobium), carbon,
boron, or ceramic (SiC, Al2 O3 ) and the ceramic matrices (oxides, carbides, nitrides, borides, and
silicides) are not very different, and the fibers do not carry the main fraction of the load in ceramic
composites. The function of the fibers is to provide strength and mainly toughness (resistance to
cracks) of the composite, because nonreinforced ceramic materials are very brittle. Ceramic composites can operate under very high temperatures depending on the melting temperature of the matrix
that varies from 1200 to 3500 C. Naturally the higher the temperature, the more complicated is the
manufacturing process. The main shortcoming of ceramic composites is associated with a low ultimate tensile elongation of the ceramic matrix resulting in cracks appearing in the matrix under relatively low tensile stress applied to the material.
An outstanding combination of high mechanical characteristics and temperature resistance is
demonstrated by carboncarbon composites in which both components—fibers and matrix—are
made from one and the same material but with different structure. A carbon matrix is formed as a
result of carbonization of an organic resin (phenolic and furfural resin or pitch) with which carbon
fibers are impregnated, or of chemical vapor deposition of pyrolytic carbon from a hydrocarbon
gas. In an inert atmosphere or in a vacuum, carboncarbon composites can withstand very high
temperatures (more than 3000 C). Moreover, their strength increases under heating up to 2200 C
while the modulus degrades at temperatures above 1400 C. However in an oxygen atmosphere,
they oxidize and sublime at relatively low temperatures (about 600 C). To use carboncarbon
composite parts in an oxidizing atmosphere, they must have protective coatings, made usually from
silicon carbide. Manufacturing of carboncarbon parts is a very energy and time consuming process. To convert an initial carbonphenolic composite into carboncarbon, it should receive a
thermal treatment at 250 C for 150 h, carbonization at about 800 C for about 100 h and several
Introduction
xxiii
cycles of densification (one-stage pyrolysis results in high porosity of the material) each including
impregnation with resin, curing, and carbonization. To refine the material structure and to provide
oxidation resistance, a further high-temperature graphitization at 2700 C and coating (at 1650 C)
can be required. Vapor deposition of pyrolytic carbon is also a time consuming process performed
at 900 2 12003 C under a pressure of 1502000 kPa.
Composite materials do not exist apart from composite structures and are formed while the
structure is fabricated. Being a heterogeneous media a composite material has two levels of heterogeneity. The first level represents a microheterogeneity induced by at least two phases (fibers and
matrix) that form the material microstructure. At the second level the material is characterized by a
macroheterogeneity caused by the laminated or more complicated macrostructure of the material
which consists usually of a set of layers with different orientations. A number of technologies have
been developed by now to manufacture composite structures, all these technologies involve two
basic processes during which the material’s microstructure and macrostructure are formed.
The first basic process yielding material microstructure involves the application of a matrix
material to the fibers. The simplest way to do this, normally utilized in the manufacturing of composites with thermosetting polymeric matrices, is a direct impregnation of tows, yarns, fabrics, or
more complicated fibrous structures with liquid resins. Thermosetting resin has relatively low viscosity (10100 Pa s), which can be controlled using solvents or heating, and good wetting ability
for the majority of fibers. There exist two versions of this process. According to the so-called
“wet” process, impregnated fibrous material (tows, fabrics, etc.) is used to fabricate composite parts
directly, without any additional treatment or interruption of the process. In contrast to that, in “dry”
or “prepreg” processes, impregnated fibrous material is dried (partially cured) and thus preimpregnated tapes (prepregs) are stored for further utilization (usually under low temperature to prevent
uncontrolled premature polymerization of the resin). Both processes, having similar advantages and
shortcomings, are widely used for composites with thermosetting matrices. For thermoplastic matrices, application of direct impregnation (“wet” processing) is limited by the relatively high viscosity
(about 1012 Pa s) of thermoplastic polymer solutions or melts. For this reason, “prepreg” processes
with preliminary fabricated tapes or sheets in which fibers are already combined with the thermoplastic matrix are used to manufacture composite parts. There also exist other processes that
involve application of heat and pressure to hybrid materials including reinforcing fibers and a thermoplastic polymer in the form of powder, films, or fibers. A promising process (called fibrous technology) utilizes tows, tapes, or fabrics with two types of fibers: reinforcing and thermoplastic.
Under heat and pressure, the thermoplastic fibers melt and form the matrix of the composite material. Metal and ceramic matrices are applied to fibers by means of casting, diffusion welding, chemical deposition, plasma spraying, processing by compression molding or with the aid of powder
metallurgy methods.
The second basic process provides the appropriate macrostructure of a composite material corresponding to the loading and operational conditions of the composite part that is fabricated. There
exist the three main types of material macrostructure—linear structure which is appropriate for
bars, profiles, and beams, plane-laminated structure suitable for thin-walled plates and shells, and
spatial structure which is necessary for thick-walled and bulk-solid composite parts.
A linear structure is formed by pultrusion, table rolling, or braiding and provides high strength
and stiffness in one direction coinciding with the axis of a bar, profile, or a beam. Pultrusion results
in a unidirectionally reinforced composite profile made by pulling a bundle of fibers impregnated
xxiv
Introduction
with resin through a heated die to cure the resin and, to provide the desired shape of the profile
cross-section. Table rolling is used to fabricate small diameter tapered tubular bars (e.g., ski poles
or fishing rods) by rolling preimpregnated fiber tapes in the form of flags around the metal mandrel
which is pulled out of the composite bar after the resin is cured. Fibers in the flags are usually oriented along the bar axis or at an angle to the axis thus providing more complicated reinforcement
than the unidirectional one typical of pultrusion. Even more complicated fiber placement with orientation angle varying from 5 to 853 along the bar axis can be achieved using two-dimensional
(2D) braiding which results in a textile material structure consisting of two layers of yarns or tows
interlaced with each other while they are wound onto the mandrel.
A plane-laminated structure consists of a set of composite layers providing the necessary stiffness and strength in at least two orthogonal directions in the plane of the laminate. Such a plane
structure would be formed by hand or machine lay-up, fiber placement, or filament winding.
Lay-up and fiber placement technology provides fabrication of thin-walled composite parts of
practically arbitrary shape by hand or automated placing of preimpregnated unidirectional or fabric
tapes onto a mold. Layers with different fiber orientations (and even with different fibers) are combined to result in the laminated composite material exhibiting the desired strength and stiffness in
given directions. Lay-up processes are usually accompanied by pressure applied to compact the
material and to remove entrapped air. Depending on the required quality of the material, as well as
on the shape and dimensions of a manufactured composite part, compacting pressure can be provided by rolling or vacuum bags, in autoclaves, or by compression molding.
Filament winding is an efficient automated process of placing impregnated tows or tapes onto a
rotating mandrel that is removed after curing of the composite material. Varying the winding angle,
it is possible to control the material strength and stiffness within the layer and through the thickness
of the laminate. Preliminary tension applied to the tows in the process of winding induces pressure
between the layers providing compaction of the material. Filament winding is the most advantageous in manufacturing thin-walled shells of revolution though it can also be used in building composite structures with more complicated shapes.
Structural elements of rather complicated shapes can be fabricated by braiding which is a sort
of combination of weaving and filament winding. In braiding, two or more tows come from spools
that rotate in the opposite directions around a mandrel that moves in the axial direction. The tows
interlace with one another making flat, tubular, shell-like, or solid structural elements. Braiding
over shaped mandrels allows us to fabricate the structures with curvature and cross-sections varying
along the axes. The spatial macrostructure of a composite material that is specific for thick-walled
and solid members requiring fiber reinforcement in at least three directions (not lying in one plane)
can be formed by 3D braiding (with three interlaced yarns) or using such textile processes as weaving, knitting, or stitching. Spatial (3D, 4D, etc.) structures used in carboncarbon technology are
assembled from thin carbon composite rods fixed in different directions.
There are two specific manufacturing procedures that have an inverse sequence of the basic processes described earlier, i.e., first, the macrostructure of the material is formed and then the matrix
is applied to fibers. The first of these procedures is the aforementioned carboncarbon technology
that involves chemical vapor deposition of a pyrolytic carbon matrix on preliminary assembled and
sometimes rather complicated structures made from dry carbon fabric. The second procedure is the
resin transfer molding (RTM). Fabrication of a composite part starts with a preform that is assembled in the internal cavity of a mold from dry fabrics, tows, yarns, etc., and forms the
Introduction
xxv
macrostructure of a composite part. The shape of this part is governed by the shape of the mold
cavity into which liquid resin is transferred under pressure through injection ports. Vacuum-assisted
RTM or resin infusion process in contrast to RTM required a single-sided tool on which a dry fiber
preform is placed and covered with a vacuum bag. Inlet and exit resin feed pipes are going through
the bag, and vacuum is drawn at the exit to facilitate an infusion of the preform with resin.
The basic processes described earlier are normally accompanied by a thermal treatment resulting in the solidification of the matrix. Heating is applied to cure thermosetting resins, cooling is
used to transfer thermoplastic, metal, and ceramic matrices to a solid phase, whereas a carbon
matrix is made by pyrolysis. The final stages of the manufacturing procedure involve removal of
mandrels, molds, or other tooling and machining of a composite part as required.
The specific features of fibers, matrices, and manufacturing processes briefly described earlier
are utilized to provide the desired microstructure and macrostructure of composite materials which
require the material models and methods of analysis of composite structures that are discussed in
depth in this book.
CHAPTER
MECHANICS OF A
UNIDIRECTIONAL PLY
1
A ply or lamina is the simplest element of a composite material, an elementary layer of unidirectional fibers in a matrix (see Fig. 1.1) formed when a unidirectional tape impregnated with resin is
placed onto the surface of a tool, thus providing the shape of a composite part.
1.1 PLY ARCHITECTURE
As the tape consists of tows (bundles of fibers), the ply thickness (whose minimum value is about
0.1 mm for modern composites) is much greater than the fiber diameter (about 0.005 mm). In an
actual ply, the fibers are randomly distributed, as in Fig. 1.2. Since the actual distribution is not
known and can hardly be predicted, some typical idealized regular distributions, i.e., square
(Fig. 1.3), hexagonal (Fig. 1.4), and layer-wise (Fig. 1.5) are used for the analysis.
A composite ply is generally taken to consist of two constituents: fibers and a matrix whose
quantities in the materials are specified by volume, v, and mass, m, fractions
vf 5
Vf
;
Vc
vm 5
Vm
Vc
(1.1)
mf 5
Mf
;
Mc
mm 5
Mm
Mc
(1.2)
Here, V and M are volume and mass, whereas subscripts f, m, and c correspond to fibers, matrix,
and composite material, respectively. Since Vc 5 Vf 1 Vm and Mc 5 Mf 1 Mm , we have
vf 1 vm 5 1;
mf 1 mm 5 1
(1.3)
There exist the following relationships between volume and mass fractions
vf 5
ρc
mf ;
ρf
vm 5
ρc
mm
ρm
(1.4)
where ρf , ρm , and ρc are the densities of the fibers, the matrix, and the composite respectively. In
analysis, volume fractions are used because they enter the stiffness coefficients for a ply, whereas
mass fractions are usually measured directly during processing or experimental study of the
fabricated material.
Two typical situations usually occur. The first situation assumes that we know the mass of the
fibers used to fabricate a composite part and the mass of the part itself. The mass of the fibers can
be found if we weigh the spools with fibers before and after they are used or calculate the total
length of tows and multiply it by the tow tex-number that is the mass in grams of a 1000 m long
Advanced Mechanics of Composite Materials and Structures. DOI: https://doi.org/10.1016/B978-0-08-102209-2.00001-3
© 2018 Elsevier Ltd. All rights reserved.
1
2
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
3
2
1
FIGURE 1.1
A unidirectional ply.
FIGURE 1.2
Actual carbon fiber distribution in the cross section of a ply.
tow. So, we know the values of Mf and Mc and can use the first equations of Eqs. (1.2) and (1.4) to
calculate vf .
The second situation arises if we have a sample of a composite material and know the densities
of the fibers and the matrix used for its fabrication. Then, we can find the experimental value of
material density, ρec , and use the following equation for theoretical density
ρc 5 ρf vf 1 ρm vm
(1.5)
Putting ρc 5 ρec and taking into account Eq. (1.3), we obtain
vf 5
ρec 2 ρm
ρf 2 ρm
(1.6)
Consider, for example, a carbonepoxy composite material with fibers AS4 and matrix EPON
DPL-862, for which ρf 5 1:79 g/cm3 and ρm 5 1:2 g/cm3. Let ρec 5 1:56 g/cm3. Then, Eq. (1.6)
yields vf 5 0:61.
1.1 PLY ARCHITECTURE
FIGURE 1.3
Square fiber distribution in the cross section of a ply.
FIGURE 1.4
Hexagonal fiber distribution in the cross section of a ply.
3
4
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
FIGURE 1.5
Layer-wise fiber distribution in the cross section of a ply.
This result is approximate because it ignores possible material porosity. To determine the actual
fiber fraction, we should remove the resin using matrix destruction, solvent extraction, or burning the
resin out in an oven. As a result, we get Mf , and having Mc , can calculate mf and vf with the aid of
Eqs. (1.2) and (1.4). Then we find ρc using Eq. (1.5) and compare it with ρec . If ρc . ρec , the material
includes voids whose volume fraction (porosity) can be calculated using the following equation:
vp 5 1 2
ρec
ρc
(1.7)
For the carbonepoxy composite material considered above as an example, assume that the
foregoing procedure results in mf 5 0:72. Then, Eqs. (1.4), (1.5), and (1.7) give vf 5 0:63,
ρc 5 1:58 g/cm3, and vp 5 0:013, respectively.
For real unidirectional composite materials, we normally have vf 5 0:50 2 0:65. Lower fiber
volume content results in lower ply strength and stiffness under tension along the fibers, whereas
higher fiber content, close to the ultimate value, leads to reduction of the ply strength under longitudinal compression and in-plane shear due to poor bonding of the fibers.
Since the fibers usually have uniform circular cross sections, there exists the ultimate fiber volume fraction, vuf which is less than unity and depends on the fiber arrangement. For typical arrangements shown in Figs. 1.31.5, the ultimate arrays are presented in Fig. 1.6, and the corresponding
ultimate fiber volume fractions
are:
1 πd 2
π
u
Square array
vf 5 2
5 5 0:785
d
4
4
2
2
πd
π
Hexagonal array vuf 5 pffiffiffi
5 pffiffiffi 5 0:907
d2 3 4
2 3
1 πd 2
π
u
Layer-wise array vf 5 2
5 5 0:785
d
4
4
1.2 FIBERMATRIX INTERACTION
(A)
5
(C)
(B)
d
d
d
d
d
d
FIGURE 1.6
Ultimate fiber arrays for square (A), hexagonal (B), and layer-wise (C) fiber distributions.
1.2 FIBERMATRIX INTERACTION
Composite materials consist of fibers and matrix acting together, and the material properties are
governed by the properties of fibers and matrix as the result of their interaction.
1.2.1 THEORETICAL AND ACTUAL STRENGTH
The most important property of advanced composite materials is associated with the very high
strength of a unidirectional ply, accompanied with relatively low density. This strength of the material is provided mainly by the fibers. Correspondingly, a natural question arises as to how such traditional lightweight materials like glass or graphite, which were never utilized as primary loadbearing structural materials, can be used to make fibers with a strength exceeding the strength of
such traditional structural materials as aluminum or steel. The general answer is well known: The
strength of a thin wire is usually much higher than the strength of the corresponding bulk material.
This is demonstrated in Fig. 1.7 showing that the wire strength increases as the wire diameter is
reduced.
In connection with this, two questions arise. First, what is the upper limit of strength that can
be predicted for an infinitely thin wire or fiber, and second, what is the nature of this
phenomenon?
The answer to the first question is given in The Physics of Solids. Consider an idealized model
of a solid, namely a regular system of atoms located as shown in Fig. 1.8 and find the tensile stress,
σ, that causes the two planes of atoms to completely separate. The dependence of σ on the atomic
spacing is presented in Fig. 1.9. Point O of the curve corresponds to the equilibrium of the
unloaded system, whereas point U specifies the ultimate theoretical stress, σt . The initial tangent
angle, α, characterizes the material’s modulus of elasticity, E. To evaluate σt , we can use the following sine approximation (Gilman, 1959) for the OU segment of the curve
σ , GPa
4
3
2
1
0
0.4
0.8
1.2
d, mm
1.6
FIGURE 1.7
Dependence of high-carbon steel wire strength on the wire diameter.
σ
a
σ
FIGURE 1.8
Material microstructural model.
σ
U
σt
a0
α
a
0
FIGURE 1.9
Atoms’ interaction curve (ss) and its sine approximation ( ).
1.2 FIBERMATRIX INTERACTION
σ 5 σt sin2π
7
a 2 a0
a0
Since the strain
ε5
a 2 a0
a0
we arrive at
σ 5 σt sin2πε
Now, we can calculate the modulus as
E5
dσ
jε50 5 2πσt
dε
Thus,
σt 5
E
2π
(1.8)
This equation yields a very high value for the theoretical strength. For example, for a steel wire,
σt 5 33:4 GPa. At present, the highest strength reached in 2μm-diameter monocrystals of iron
(whiskers) is about 12 GPa.
The model under study allows us to introduce another important characteristic of the material.
The specific energy that should be spent to break the material can be presented in accordance with
Fig. 1.9 as
N
ð
2γ 5
σðaÞda
(1.9)
a0
As material fracture results in the formation of two new free surfaces, γ can be referred to as
the specific surface energy (energy spent to form a new surface of unit area).
The answer to the second question (why the fibers are stronger than the corresponding bulk
materials) was in fact given by Griffith (1921) whose results have formed the basis of fracture
mechanics.
Consider a fiber loaded in tension and having a thin circumferential crack as shown in
Fig. 1.10. The crack length, l, is much less than the fiber diameter, d.
For a linear elastic fiber, σ 5 Eε, and the density of elastic energy can be presented as
1
σ2
U 5 σε 5
2E
2
When the crack appears, the strain energy is released in a material volume adjacent to the crack.
Suppose that this volume is comprised of a conical ring whose generating lines are represented in Fig. 1.10
by dashed lines and whose heights are proportional to the crack length, l. Then, the total released energy,
which is the density of the elastic energy multiplied by the volume of the conical ring, becomes
W5
1 σ2 2
kπ l d
2
E
(1.10)
where k is some constant coefficient of proportionality. On the other hand, the formation of new
surfaces consumes the energy
S 5 2π γ ld
(1.11)
8
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
σ
l
l
dl
dl
d
σ
FIGURE 1.10
A fiber with a crack.
where γ is the surface energy, Eq. (1.9). Now assume that the crack length is increased by an
infinitesimal increment, dl. Then, if for some value of acting stress, σ
dW
dS
.
dl
dl
(1.12)
the crack will propagate, and the fiber will fail. Substituting W and S in the inequality, Eq. (1.12),
with their expressions as per Eqs. (1.10) and (1.11), we arrive at
rffiffiffiffiffiffiffiffiffi
2γE
σ . σc 5
kl
(1.13)
The most important result that follows from this condition specifying some critical stress, σc ,
beyond which the fiber with a crack cannot retain structural integrity, is the fact that σc depends on
the absolute value of the crack length (not on the ratio l/d). Clearly, for a continuous fiber, 2l , d
so, the thinner the fiber, the smaller the length of the crack that can exist in this fiber and the higher
is the critical stress, σc . More rigorous analysis shows that, reducing l to a in Fig. 1.8, we arrive at
σc 5 σt .
Consider for example, glass fibers that are widely used as reinforcing elements in composite
materials and have been studied experimentally to support the fundamentals of fracture mechanics
1.2 FIBERMATRIX INTERACTION
9
(Griffith, 1921). The theoretical strength of glass, Eq. (1.8), is about 14 GPa, whereas the actual
strength of 1-mm diameter glass fibers is only about 0.2 GPa, and for 5-mm diameter fibers, this
value is much lower (about 0.05 GPa). The fact that such low actual strength is caused by surface
cracks can be readily proved if the fiber surface is smoothed by etching the fiber with acid. Then
the strength of 5-mm diameter fibers can be increased up to 2 GPa. If the fiber diameter is reduced
by heating and stretching the fibers to a diameter of about 0.0025 mm, the strength is increased to
6 GPa. Theoretical extrapolation of the experimental curve, showing the dependence of the fiber
strength on the fiber diameter for very small fiber diameters, yields σ 5 11 GPa, which is close to
σt 5 14 GPa.
Thus, we arrive at the following conclusion, clarifying the nature of the high performance of
advanced composites and their place among modern structural materials. The actual strength of
advanced structural materials is much lower than their theoretical strength. This difference is
caused by defects in the material microstructure (e.g., crystalline structure) or macrocracks inside
the material and on its surface. Using thin fibers, we reduce the influence of cracks and thus
increase the strength of materials reinforced with these fibers. So, advanced composites comprise a
special class of structural materials in which we try to utilize the natural potential properties of the
material, rather than the capabilities of technology as we do when developing high-strength alloys.
1.2.2 STATISTICAL ASPECTS OF FIBER STRENGTH
Fiber strength, being relatively high, is still less than the corresponding theoretical strength, which
means that fibers of advanced composites have microcracks or other defects randomly distributed
along the fiber length. This is supported by the fact that the fiber strength depends on the length of
the tested fiber. The dependence of strength on length for boron fibers (Mikelsons & Gutans, 1984)
is shown in Fig. 1.11. The longer the fiber, the higher is the probability of a deleterious defect to
exist within this length, and the lower the fiber strength. The tensile strengths of fiber segments
with the same length but taken from different parts of a long continuous fiber, or from different
fibers, also demonstrate the strength deviation. A typical strength distribution for boron fibers is
presented in Fig. 1.12.
The first important characteristic of the strength deviation is the strength scatter,
Δσ 5 σmax 2 σmin . For the case corresponding to Fig. 1.12, σmax 5 4:2 GPa, σmin 5 2 GPa, and
Δσ 5 2:2 GPa. To plot the diagram presented in Fig. 1.12, Δσ is divided into a set of increments,
and a normalized number of fibers n 5 Nσ =N (Nσ is the number of fibers failing at that stress within
the increment, and N is the total number of tested fibers) is calculated and shown on the vertical
axis. Thus, the so-called frequency histogram can be plotted. This histogram allows us to determine
the mean value of the fiber strength as
σm 5
and the strength dispersion as
N
1X
σi
N i51
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
N
u 1 X
dσ 5 t
ðσm 2σi Þ2
N 2 1 i51
(1.14)
(1.15)
10
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
σ , GPa
4
3
2
1
L, mm
0
10
20
30
40
FIGURE 1.11
Dependence of strength of boron fibers on the fiber length.
n
0.25
0.2
0.15
0.1
0.05
0
1
2
3
4
5
σ , GPa
FIGURE 1.12
Strength distribution for boron fibers.
The deviation of fiber strength is characterized by the coefficient of the strength variation,
which is presented as follows:
rσ 5
dσ
100%
σm
(1.16)
For the boron fibers under consideration, Eqs. (1.14)(1.16) yield σm 5 3:2 GPa, dσ 5 0:4 GPa,
and rσ 5 12:5%.
1.2 FIBERMATRIX INTERACTION
11
j
L
FIGURE 1.13
Tension of a bundle of fibers.
σ , GPa
2
2
1.5
1
1
0.5
0
0
0.5
1
1.5
2
2.5
ΔL
L
3
FIGURE 1.14
Stressstrain diagrams for bundles of carbon (1) and aramid (2) fibers.
To demonstrate the influence of fiber strength deviation on the strength of a unidirectional ply,
consider a bundle of fibers, i.e., a system of approximately parallel fibers with different strengths
and slightly different lengths as in Fig. 1.13. Typical stressstrain diagrams for fibers tested under
tension in a bundle are shown in Fig. 1.14 (Vasiliev & Tarnopol’skii, 1990). As can be seen, the
diagrams have two nonlinear segments. The nonlinearity in the vicinity of zero stresses is associated with different lengths of fibers in the bundles, whereas the nonlinear behavior of the bundle
under stresses close to the ultimate values is caused by fracture of the fibers with lower strength.
Useful qualitative results can be obtained if we consider model bundles consisting of five fibers
with different strengths. Five such bundles are presented in Table 1.1 showing the normalized
strength of each fiber. As can be seen, the deviation of fiber strength is such that the mean strength,
σm 5 1, is the same for all the bundles, whereas the variation coefficient, rσ , changes from 31.6%
12
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
Table 1.1 Strength of Bundles Consisting of Fibers
With Different Strength
Bundle No.
Fiber No.
1
2
3
4
5
1
2
3
4
5
σm
rσ ð %Þ
0.6
0.8
1.0
1.2
1.4
1.0
31.6
3.2
0.7
0.9
1.0
1.1
1.3
1.0
22.4
3.6
0.85
0.9
1.0
1.1
1.15
1.0
12.8
4.25
0.9
0.95
1.0
1.05
1.1
1.0
7.8
4.5
1.0
1.0
1.0
1.0
1.0
1.0
0
5.0
F
Table 1.2 Strength of Bundles Consisting of Fibers
With Different Strength
Bundle No.
Fiber No.
1
2
3
4
5
1
2
3
4
5
σm
rσ ð %Þ
0.6
0.8
1.0
1.6
3.0
1.4
95.0
3.2
0.7
0.9
1.2
1.4
1.6
1.16
66.0
3.6
0.85
0.9
1.1
1.15
1.4
1.08
22.0
4.25
0.9
0.85
1.0
1.05
1.1
1.0
7.8
4.5
0.95
0.95
0.95
0.95
0.95
0.95
0
4.75
F
for bundle No. 1 to zero for bundle No. 5. The last row in the table shows the effective (observed)
ultimate force, F, for a bundle. Consider, for example, the first bundle. When the force is increased
to F 5 3, the stresses in all the fibers become σj 5 0:6, and fiber No. 1 fails. After this happens,
the force, F 5 3, is taken by four fibers, and σj 5 0:75 (j 5 2, 3, 4, 5). When the force reaches
the value F 5 3.2, the stresses become σj 5 0:8, and fiber No. 2 fails. After that, σj 5 1:07 (j 5 3,
4, 5). This means that fiber No. 3 also fails at force F 5 3.2. Then, for the two remaining fibers,
σ4 5 σ5 5 1:6, and they also fail. Thus, F 5 3:2 for bundle No. 1. In a similar way, F can be calculated for the other bundles in the table. As can be seen, the lower the fiber strength variation, the
higher the F, which reaches its maximum value, F 5 5, for bundle No. 5 consisting of fibers with
the same strength.
Table 1.2 demonstrates that strength variation can be more important than the mean strength. In
fact, while the mean strength, σm , goes down for bundle Nos. 15, the ultimate force, F, increases.
So, it can be better to have fibers with relatively low strength and low strength variation than highstrength fibers with high strength variation.
1.2 FIBERMATRIX INTERACTION
13
1.2.3 STRESS DIFFUSION IN FIBERS INTERACTING THROUGH THE MATRIX
The foregoing discussion concerned individual fibers or bundles of fibers that are not joined
together. This is not the case for composite materials in which the fibers are embedded in the
matrix material. Usually, the stiffness of the matrix is much lower than that of the fibers and the
matrix practically does not take any significant load in the fiber direction. However, the fact that
the fibers are bonded with the matrix even with relatively low stiffness changes the mechanism of
fiber interaction and considerably increases their effective strength. To show this, the strength of
dry fiber bundles can be compared with the strength of the same bundles after impregnating them
with epoxy resin and curing. The results are listed in Table 1.3. As can be seen, composite bundles
in which fibers are joined together by the matrix demonstrate significantly higher strength, and the
higher the fiber sensitivity to damage, the higher the difference in strength of dry and composite
bundles. The influence of a matrix on the variation of strength is even more significant. As follows
from Table 1.4, the variation coefficients of composite bundles are lower by an order of magnitude
than those of individual fibers.
To clarify the role of a matrix in composite materials, consider the simple model of a unidirectional ply shown in Fig. 1.15 and apply the method of analysis developed for stringer panels
(Goodey, 1946).
Let the ply of thickness δ consist of 2k fibers symmetrically distributed on both sides of the central fiber n 5 0. The fibers are joined with layers of the matrix material, and the fiber volume fraction is
Table 1.3 Strength of Dry Bundles and Composite Bundles
Ultimate Tensile Load F (N)
Fibers
Sensitivity of Fibers to Damage
Dry Bundle
Composite Bundle
Strength Increase (%)
Carbon
Glass
Aramid
High
Moderate
Low
14
21
66
26
36
84
85.7
71.4
27.3
Table 1.4 Variation Coefficient for Fibers and Unidirectional Composites
Variation Coefficient rσ (%)
Fibers
Fibers
Composite
Glass
Carbon
Aramid
Boron
29
30
24
23
2.0
4.7
5.0
3.0
14
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
y
δ
σ
k
(n + 1)
σ
σ
σ
(n + 1)
n
n
(n − 1)
1
am
a
af
σ
σ
1
0
x,u
1'
sym.
sym.
FIGURE 1.15
Model of a unidirectional ply with a broken fiber.
vf 5
af
;
a
a 5 af 1 am
(1.17)
Let the central fiber have a crack induced by fiber damage or by the shortage of this fiber’s
strength. At a distance from the crack, the fibers are uniformly loaded with stress σ (see
Fig. 1.15).
First, derive the set of equations describing the ply under study. Since the stiffness of the matrix
is much less than that of fibers, we neglect the stress in the matrix acting in the x direction and
assume that the matrix works only in shear. We also assume that there are no displacements in the
y direction.
Considering equilibrium of the last (n 5 k) fiber, an arbitrary fiber, and the central (n 5 0)
fiber shown in Fig. 1.16, we arrive at the following equilibrium equations:
af σ0k 2 τ k 5 0
af σ0n 1 τ n11 2 τ n 5 0
af σ00 1 2τ 1 5 0
(1.18)
0
in which ð:::Þ 5 dð:::Þ=dx.
Constitutive equations for fibers and the matrix can be written as
σn 5 Ef εn ;
τ n 5 Gm γ n
(1.19)
Here, Ef is the fiber elasticity modulus and Gm is the matrix shear modulus, whereas
εn 5 u0n
(1.20)
is the fiber strain expressed in terms of the displacement in the x direction. The shear strain in the
matrix follows from Fig. 1.17, i.e.,
γn 5
1
ðun 2 un21 Þ
am
(1.21)
This set of equations, Eqs. (1.18)(1.21), is complete—it includes 10k 1 3 equations and contains the same number of unknown stresses, strains, and displacements.
1.2 FIBERMATRIX INTERACTION
σk
(σk +
k
k
15
dσk
dx )
dx
τk
(A)
n+1
τ n+1
n
σn
n
(σ n +
dσ n
dx)
dx
(σ 0 +
dσ 0
dx)
dx
τn
(B)
1
σ0
τ1
0
τ1
(C)
FIGURE 1.16
Stresses acting in fibers and matrix layers for the last (A), arbitrary n-th fiber (B), and the central n 5 0 fiber (C).
n
un
am
γn
n−1
FIGURE 1.17
Shear strain in the matrix layer.
un−1
16
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
Consider the boundary conditions. If there is no crack in the central fiber, the solution of the
problem is evident and has the form σn 5 σ, τ n 5 0. Assuming that the perturbation of the stressed
state induced by the crack vanishes at a distance from the crack, we take
σn ðx-NÞ 5 σ;
τ n ðx-NÞ 5 0
(1.22)
As a result of the crack in the central fiber, we have
σ0 ðx 5 0Þ 5 0
(1.23)
For the remaining fibers, symmetry conditions yield
un ðx 5 0Þ 5 0 ðn 5 1; 2; 3 . . . kÞ
(1.24)
To solve the problem, we use the stress formulation and, in accordance with the stress formulation of the problem in The Theory of Elasticity, should consider equilibrium equations in conjunction with compatibility equations expressed in terms of stresses.
First, transform equilibrium equations introducing the stress function, FðxÞ, such that
τ n 5 F 0n ;
Fn ðx-NÞ 5 0
(1.25)
Substituting Eq. (1.25) for τ n in the equilibrium equations, Eq. (1.18), integrating the resulting
equations from x to N, and taking into account Eqs. (1.22) and (1.25), we obtain
1
Fk
af
1
σn 5 σ 2 ðFn11 2 Fn Þ
af
2
σ0 5 σ 2 F1
af
σk 5 σ 1
(1.26)
The compatibility equations follow from Eqs. (1.20) and (1.21), i.e.,
γ 0n 5
1
ðεn 2 εn21 Þ
am
Using constitutive equations, Eq. (1.19), we can write them in terms of stresses, i.e.,
τ 0n 5
Gm
ðσn 2 σn21 Þ
am Ef
Substituting stresses from Eqs. (1.25) and (1.26) and introducing the dimensionless coordinate
x 5 x=a (see Fig. 1.15), we finally arrive at the following set of governing equations:
Fvk 2 μ2 ð2Fk 2 Fk21 Þ 5 0
Fvn 1 μ2 ðFn11 2 2Fn 1 Fn21 Þ 5 0
(1.27)
Fv1 1 μ ðF2 2 3F1 Þ 5 0
2
in which, in accordance with Eq. (1.17),
μ2 5
Gm a2
Gm
5
af am Ef
vf ð1 2 vf ÞEf
(1.28)
Taking into account the second equation in Eq. (1.25), we can present the general solution of
Eq. (1.27) in the form
1.2 FIBERMATRIX INTERACTION
Fn ðxÞ 5 An e2λx
17
(1.29)
The corresponding substitution in Eq. (1.27) yields:
λ2
Ak 2 2 2 2 Ak21 5 0
μ
λ2
An11 2 An 2 2 2 1 An21 5 0
μ
λ2
A2 2 A1 3 2 2 5 0
μ
(1.30)
(1.31)
(1.32)
The finite-difference equation, Eq. (1.31), can be reduced to the following form:
An11 2 2An cosθ 1 An21 5 0
(1.33)
where
cosθ 5 1 2
λ2
2μ2
(1.34)
As can be readily checked, the solution for Eq. (1.33) is
An 5 Bcosnθ 1 Csinnθ
(1.35)
Then, Eq. (1.34) yields after some transformation
λ 5 2μsin
θ
2
(1.36)
Substituting the solution given by Eq. (1.35) for Ak and Ak21 in Eq. (1.30), we obtain, after
some transformation,
B 5 2 Ctanðk 1 1Þθ
Thus, Eq. (1.35) can be written as
An 5 C½sinnθ 2 cosnθUtanðk 1 1Þθ
(1.37)
Substituting Eq. (1.37) for A1 and A2 in Eq. (1.32) and performing rather cumbersome trigonometric transformations, we arrive at the following equation for θ:
tankθ 5 2 tan
θ
2
(1.38)
The periodic properties of the tangent function in Eq. (1.38) mean that it has k 11 different
roots corresponding to intersection points of the curves z 5 tankθ and z 5 2 tanθ=2. For the case k
5 4, considered below as an example, these points are shown in Fig. 1.18. Further transformation
allows us to reduce Eq. (1.38) to
sin
2k 1 1
θ50
2
from which it follows that
θi 5
2πi
2k 1 1
ði 5 0; 1; 2. . .kÞ
(1.39)
18
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
z = tan kθ
0
0
1
2
3
4
θ , rad
–2
–4
–6
z = − tan
θ
2
z
FIGURE 1.18
Geometric interpretation of Eq. (1.38) for k 5 4.
The first root, θ0 5 0, corresponds to λ 5 0 and Fn 5 const, i.e., to a ply without a crack in the
central fiber. So, Eq. (1.39) specifies k roots (i 5 1, 2, 3, . . ., k) for the ply under study, and the
solution in Eqs. (1.29) and (1.37) can be generalized as
Fn ðxÞ 5
k
X
Ci ½sinnθi 2 cosnθi Utanðk 1 1Þθi e2λi x
(1.40)
θi
2
(1.41)
i51
where, in accordance with Eq. (1.36),
λi 5 2μsin
and θi are determined by Eq. (1.39).
Using Eq. (1.38), we can transform Eq. (1.40) to the following final form:
Fn ðxÞ 5
k
X
Ci Sn ðθi Þe2λi x
(1.42)
2n 2 1
θi
2
θi
cos
2
(1.43)
i51
where
Sn ðθi Þ 5
sin
Applying Eqs. (1.25) and (1.26), we can find shear and normal stresses, i.e.,
τ n ðxÞ 5 2
k
1X
λi Ci Sn ðθi Þe2λi x
a i51
σk ðxÞ 5 σ 1
σn ðxÞ 5 σ 2
ðn 5 1; 2; 3. . .kÞ
k
1X
Ci Sk ðθi Þe2λi x
af i51
k
1X
Ci ½Sn11 ðθi Þ 2 Sn ðθi Þe2λi x
af i51
ðn 5 1; 2; 3. . .k 2 1Þ
(1.44)
(1.45)
(1.46)
1.2 FIBERMATRIX INTERACTION
σ0 ðxÞ 5 σ 2
k
2X
Ci S1 ðθi Þe2λi x
af i51
19
(1.47)
Displacements can be determined with the aid of Eqs. (1.19), (1.21), and (1.25). Changing x for
x 5 x=a, we get
am 0
F ðxÞ 1 un21 ðxÞ
aGm n
un ðxÞ 5
For the first fiber (n 5 1), we have
u1 ðxÞ 5
am 0
F ðxÞ 1 u0 ðxÞ
aGm 1
Substituting Fn in these equations with Eq. (1.42), we arrive at
un ðxÞ 5 2
k
am X
Ci λi Sn ðθi Þe2λi x 1 un21 ðxÞ
aGm i51
u1 ðxÞ 5 2
ðn 5 2; 3; 4. . .kÞ
k
am X
Ci λi S1 ðθi Þe2λi x 1 u0 ðxÞ
aGm i51
(1.48)
(1.49)
To determine the coefficients Ci , we should apply the boundary conditions and write
Eqs. (1.23) and (1.24) in the explicit form using Eqs. (1.47)(1.49). Substituting Sn from
Eq. (1.43) and λi from Eq. (1.41), we have
k
X
Ci tan
i51
k
X
Ci tan
i51
θi
σaf
5
2
2
θi
2n 2 1
θi 5 0
sin
2
2
k
X
Ci tan
i51
ðn 5 2; 3; 4. . .kÞ
θi
θi
aGm
sin 5
u0 ð0Þ
2
2
2μam
Introducing new coefficients
Di 5 Ci tan
θi
2
(1.50)
we arrive at the final form of the boundary conditions, i.e.,
k
X
σaf
2
(1.51)
2n 2 1
θi 5 0 ðn 5 2; 3; 4. . .kÞ
2
(1.52)
i51
k
X
i51
Di sin
k
X
i51
Di sin
Di 5
θi
aGm
5
u0 ð0Þ
2
2μam
(1.53)
20
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
This set contains k 11 equations and includes k unknown coefficients Di and displacement
u0 ð0Þ.
The foregoing set of equations allows us to obtain the exact analytical solution for any number
of fibers, k. To find this solution, some transformations are required. First, multiply Eq. (1.52) by
sin½ð2n 2 1Þθs =2 and sum up all the equations from n 5 2 to n 5 k. Adding Eq. (1.53) for n 5 1
multiplied by sinðθs =2Þ, we obtain
k X
k
X
Di sin
n51 i51
2n 2 1
2n 2 1
aGm
θs
θi sin
θs 5
u0 ð0Þsin
2
2
2μam
2
Now the sequence of summation can be changed, as follows
k
X
Di
i51
k
X
sin
n51
2n 2 1
2n 2 1
aGm
θs
u0 ð0Þsin
θi sin
θs 5
2μam
2
2
2
(1.54)
Using the following known series
k
X
cosð2n 2 1Þθ 5
n51
sin2kθ
2sinθ
we get in several steps
k
X
2n 2 1
2n 2 1
θi sin
θs
sin
2
2
n51
k 1X
2n 2 1
2n 2 1
ðθi 2 θs Þ 2 cos
ðθi 1 θs Þ
5
cos
2 n51
2
2
2
3
Ris 5
5
7
16
6 sinkðθi 2 θs Þ 2 sinkðθi 1 θs Þ 7
5
1
44 1
sin ðθi 2 θs Þ sin ðθi 1 θs Þ
2
2
cos
5
θi
θs
coskθi cos coskθs θi
θs
2
2
tankθi tan 2 tankθs tan
cosθs 2 cosθi
2
2
Using Eq. (1.38), we can conclude that Ris 5 0 for i 6¼ s. For the case i 5 s, we have
Rss 5
k
X
sin
2 2n 2 1
2
n51
k
1X
1
sin2kθs
θs 5
k2
½1 2 cosð2n 2 1Þθs 5
2 n51
2
2sinθs
As a result, Eq. (1.54) yields
θs
sinθs
2
Ds 5
μam ð2ksinθs 2 sin2kθs Þ
2aGm u0 ð0Þsin
ðs 5 1; 2; 3 . . .kÞ
(1.55)
Substituting these expressions for the coefficientsDi in Eq. (1.51), we can find u0 ð0Þ, i.e.,
0
121
θi
k
sin sinθi
σμaf am BX
C
2
u0 ð0Þ 5
@
A
4aGm
2ksinθ
2sin2kθ
i
i
i51
(1.56)
1.2 FIBERMATRIX INTERACTION
21
σn σ
1.5
1
1.25
2
1
3,4
0.75
0.5
0.25
0
li
0
0
5
10
15
20
25
30
35
40
45
50
x
FIGURE 1.19
Distribution of normal stresses along the fibers.
n 5 0, 1, 2, 3, 4 for k 5 4, Ef 5 250 GPa, Gm 5 1 GPa.
Thus, the solution for the problem under study is specified by Eqs. (1.44)(1.50), (1.55),
and (1.56).
For example, consider a carbonepoxy ply with the following parameters: Ef 5 250 GPa,
Gm 5 1 GPa, vf 5 0:6, and k 5 4. The distribution of the normalized stresses in the fibers along
the ply is shown in Fig. 1.19, whereas the same distribution of shear stresses in the matrix is presented in Fig. 1.20. As can be seen, in the vicinity of the crack in the central fiber, the load carried
by this fiber is transmitted by shear through the matrix to adjacent fibers. At a distance from the
end of the fiber, greater than li , the stress in the broken fiber becomes close to σ, and for x . li , the
fiber behaves as if there is no crack. A portion of the broken fiber corresponding to 0 # x # li is not
fully effective in resisting the applied load, and li 5 li a is referred to as the fiber ineffective length.
Since the fiber defects are randomly distributed along its length, their influence on the strength of
the ply is minimal if there are no other defects in the central fiber and its adjacent fibers within distance li from the crack. To minimize the probability of such defects, we should minimize li which
depends on fiber and matrix stiffnesses and material microstructure.
To evaluate li , consider Eq. (1.47) and assume that σ0 ðxÞ becomes close to σ if
e2λi li 5 k
where k is some small parameter indicating how close σ0 ðxÞ should be to σ to neglect the difference
between them (as a matter of fact, this difference vanishes only for x-N). Taking approximately
λi 5 2μ in accordance with Eq. (1.41) and using Eq. (1.28) specifying μ, we arrive at
1
li 5 2 lnk U
2
For k 5 0.01, we get
li 5 2:3a U
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Ef
vf ð1 2 vf Þ
Gm
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Ef
vf ð1 2 vf Þ
Gm
(1.57)
For a typical carbonepoxy ply (see Fig. 1.19) with a 5 0.016 mm and vf 5 0.6, Eq. (1.57)
yields 0.29 mm.
22
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
τn σ
0
–0.02
–0.04
–0.06
–0.08
–0.1
0
5
10
15
20
25
30
35
40
45
50
x
FIGURE 1.20
Distribution of shear stresses along the fibers.
for k 5 4, Ef 5 250 GPa, Gm 5 1 GPa.
Numbers of the matrix layers:
n51
n52
n53
n54
σn σ
1.5
1
1.25
2
1
3,4
0.75
0.5
0.25
0
li
0
0
5
10
15
20
25
30
35
40
45
50
x
FIGURE 1.21
Distribution of normal stresses along the fibers.
n 5 0, 1, 2, 3, 4 for k 5 4, E f 5 250 GPa, Gm 5 0.125 GPa.
Thus, for real composites, the length li is very small, and this explains why a unidirectional
composite demonstrates much higher strength in longitudinal tension than a dry bundle of fibers
(see Table 1.3). Reducing Gm , i.e., the matrix stiffness, we increase the fiber ineffective length
which becomes infinitely large for Gm -0. This effect is demonstrated in Fig. 1.21 which
1.2 FIBERMATRIX INTERACTION
23
τn σ
0
–0.02
–0.04
–0.06
–0.08
–0.1
0
5
10
15
20
25
30
35
40
45
50
x
FIGURE 1.22
Distribution of shear stresses along the fibers.
for k 5 4, E f 5 250 GPa, Gm 5 0.125 GPa
Numbers of the matrix layers:
n51
n52
n53
n54
corresponds to a material whose matrix shear stiffness is much lower than that in the foregoing
example (see Fig. 1.19). For this case, li 5 50, and Eq. (1.57) yields li 5 0.8 mm. The distribution
of shear stresses in this material is shown in Fig. 1.22. Experiments with unidirectional
glassepoxy composites (Ef 5 86:8 GPa, vf 5 0:68, a 5 0.015) have shown that reduction of the
matrix shear modulus from 1.08 GPa (li 5 0.14 mm) to 0.037 GPa (li 5 0.78 mm) results in reduction of longitudinal tensile strength from 2010 to 1290 MPa, i.e., by 35.8% (Chiao, 1979).
The ineffective length of a fiber in a matrix can be found experimentally by using the singlefiber fragmentation test. For this test, a fiber is embedded in a matrix, and tensile load is applied to
the fiber through the matrix until the fiber breaks. Further loading results in fiber fragmentation,
and the length of the fiber fragment that no longer breaks is the fiber ineffective length. For a carbon fiber in epoxy matrix, li 5 0.3 mm (Fukuda, Miyazawa, & Tomatsu, 1993).
According to the foregoing reasoning, it looks as though the stiffness of the matrix should be as
high as possible. However, there exists an upper limit of this stiffness. Comparing Figs. 1.20 and
1.22, we can see that the higher the value of Gm , the higher is the shear stress concentration in the
matrix in the vicinity of the crack. If the maximum shear stress, τ m , acting in the matrix reaches its
ultimate value, τ m , delamination will occur between the matrix layer and the fiber, and the matrix
will not transfer the load from the broken fiber to the adjacent ones. This maximum shear stress
depends on the fiber stiffness—the lower the fiber modulus, the higher the value of τ m . This is
shown in Figs. 1.23 and 1.24, where shear stress distributions are presented for aramid fibers
(Ef 5 150 GPa) and glass fibers (Ef 5 90 GPa), respectively.
24
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
τn σ
0
–0.02
–0.04
–0.06
–0.08
–0.1
–0.12
0
5
10
15
20
25
30
35
40
45
50
x
FIGURE 1.23
Distribution of shear stresses along the fibers.
for k 5 4, E f 5 150 GPa, Gm 5 1 GPa.
Numbers of the matrix layers:
n51
n52
n53
n54
Finally, it should be emphasized that the plane model of a ply, considered in this section (see
Fig. 1.15), provides only qualitative results concerning fibers and matrix interaction. In real materials, a broken fiber is surrounded with more than two fibers (at least five or six as can be seen in
Fig. 1.2), so the stress concentration in these fibers and in the matrix is much lower than would be
predicted by the foregoing analysis. For a hexagonal fiber distribution (see Fig. 1.4), the stress concentration factor for the fibers does not exceed 1.105 (Tikhomirov & Yushanov, 1980). The effect
of fiber breakage on the tensile strength of unidirectional composites has been studied by AbuFarsakh, Abdel-Jawad, and Abu-Laila (2000).
1.2.4 FRACTURE TOUGHNESS
Fracture toughness is an important characteristic of a structural material indicating its resistance to
cracks and governed by the work needed to fracture a material (work of fracture). It is well known
that there exist brittle and ductile metal alloys, whose typical stressstrain diagrams are shown in
Fig. 1.25. Comparing alloys with one and the same basic metal (e.g. steel alloys), we can see that
while brittle alloys have higher strength, σ, ductile alloys have higher ultimate elongation, ε, and,
as a result, a higher work of fracture since this is proportional to the area under the stressstrain
diagram. Though brittle materials have, in general, higher strength, they are sensitive to cracks that,
by propagating, can cause material failure for a stress that is much lower than the static strength.
1.2 FIBERMATRIX INTERACTION
25
τn σ
0
–0.02
–0.04
–0.06
–0.08
–0.1
–0.12
–0.14
0
5
10
15
20
25
30
35
40
45
50
x
FIGURE 1.24
Distribution of shear stresses along the fibers.
for k 5 4, E f 5 90 GPa, Gm 5 1 GPa.
Numbers of the matrix layers:
n51
n52
n53
n54
σ
1
σ
2
ε
ε
FIGURE 1.25
Typical stressstrain diagrams of brittle (1) and ductile (2) metal alloys.
That is why design engineers usually prefer ductile materials with lower strength but higher fracture
toughness. A typical dependence of fracture toughness on static strength for metals is shown in
Fig. 1.26 (line 1). For composites, this dependence is entirely different (line 2)—a higher static
strength corresponds usually to higher fracture toughness (Mileiko, 1982). This phenomenon is
26
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
K
2
1
σ
FIGURE 1.26
Typical relations between fracture toughness (K) and strength (σ) for metals (1) and composites (2).
6
5
4
3
3
1
2
2
1
0
0
0.1
0.2
0.3
0.4
0.5
vf
FIGURE 1.27
Dependence of static strength (1), work of fracture (2), and fatigue strength (3) on fiber volume fraction for a
boronaluminum composite material.
demonstrated for a unidirectional boronaluminum composite in Fig. 1.27 (Mileiko, 1982). As can
be seen, an increase in fiber volume fraction, vf , results not only in higher static strength along the
fibers (line 1), which is quite natural, but also is accompanied by an increase in the work of fracture
(curve 2) and, consequently, results in an increase in the material fatigue strength (bending under
106 cycles, line 3), which shows a material’s sensitivity to cracks.
The reason for such a specific behavior in composite materials is associated with their inhomogeneous microstructure, particularly with fibermatrix interfaces that restrain free propagation of a
crack (see Fig. 1.28). Of some importance also are fiber defects, local delaminations, and fiber
1.3 MICROMECHANICS OF A PLY
σ
σ
σ
σ
σ
σ
27
FIGURE 1.28
Mechanism of the crack stopping at the fibermatrix interface.
strength deviation, which reduce the static strength but increase the fracture toughness. As a result,
by combining brittle fibers and brittle matrix, we usually arrive at a composite material whose fracture toughness is higher than that of its constituents.
Thus, we can conclude that composites comprise a new class of structural materials that are
entirely different from traditional man-made materials for several reasons. First, using thin fibers,
we make an attempt to utilize the high-strength capacity that is naturally inherent in all the materials. Second, this utilization is provided by the matrix material, which increases the fiber performance and makes it possible to manufacture composite structures. Third, the combination of fibers
and matrices can result in new qualities of composite materials that are not inherent either in individual fibers or in the matrices, and are not described by the laws of mechanical mixtures. For
example, as noted above, brittle fiber and matrix materials, both having low fracture toughness on
their own, can provide a heterogeneous composite material with higher fracture toughness.
1.3 MICROMECHANICS OF A PLY
Consider a unidirectional composite ply under the action of in-plane normal and shear stresses as in
Fig. 1.29. Since the normal stresses do not change the right angle between axes 1 and 2, and shear
stresses do not cause elongations in the longitudinal and transverse directions 1 and 2, the ply is
orthotropic, and the corresponding constitutive equations yield for the case under study
σ1
σ2
2 ν 12
E1
E2
σ2
σ1
ε2 5
2 ν 21
E2
E1
ε1 5
γ 12 5
1
τ 12
G12
(1.58)
28
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
σ1
τ 12
τ 12
3
σ2
1
2
σ2
τ 12
τ 12
σ1
FIGURE 1.29
A unidirectional ply under in-plane loading.
FIGURE 1.30
Specimens of matrix material.
The inverse form of these equations is
σ1 5 E1 ðε1 1 ν 12 ε2 Þ
σ2 5 E2 ðε2 1 ν 21 ε1 Þ
τ 12 5 G12 γ 12
(1.59)
where
E1;2 5
E1;2
1 2 ν 12 ν 21
and the following symmetry condition is valid
E1 ν 12 5 E2 ν 21
(1.60)
The constitutive equations, Eqs. (1.58) and (1.59), include effective or apparent longitudinal,
E1 , transverse, E2 , and shear, G12 , moduli of a ply and Poisson’s ratios ν 12 and ν 21 , only one of
which is independent, since the second one can be found from Eq. (1.60).
The elastic constants, E1 , E2 , G12 , and ν 12 or ν 21 , are governed by the fibers and matrix properties and the ply microstructure, i.e., the shape and size of the fibers’ cross sections, fiber volume
fraction, distribution of fibers in the ply, etc. The task of micromechanics is to derive the corresponding governing relationships, i.e., to establish the relation between the properties of a unidirectional ply and those of its constituents.
To achieve this, we should first know the mechanical characteristics of the fibers and the matrix
material of the ply. To determine the matrix modulus, Em , its Poisson’s ratio, ν m , and strength, σm ,
conventional material specimens and testing procedures can be used (see Figs. 1.30 and 1.31). The
shear modulus, Gm , can be calculated as Gm 5 Em =2ð1 1 ν m Þ. To find the fibers’ properties is a
1.3 MICROMECHANICS OF A PLY
29
FIGURE 1.31
Testing of the matrix specimen.
more complicated problem. There exist several methods to test elementary fibers by bending or
stretching 1030 mm long fiber segments. All of them are rather specific because of the small
(about 0.005 mm) fiber diameter, and, more importantly, the fiber properties in a composite material can be different from those of individual fibers (see Section 1.2.3) with the preassigned lengths
provided by these methods.
It is worth knowing a fiber’s actual modulus and strength, not only for micromechanics but also
to check the fibers’ quality before they are used to fabricate a composite part. For this purpose, a
simple and reliable method has been developed to test the fibers in simulated actual conditions.
According to this method, a fine tow or an assembly of fibers is carefully impregnated with resin,
slightly stretched to avoid fiber waviness, and cured to provide a specimen of the so-called microcomposite material. The microcomposite strand is overwrapped over two discs as in Fig. 1.32, or
fixed in special friction grips as in Fig. 1.33, and tested under tension to determine the ultimate tensile force F and strain ε corresponding to some force F , F. Then, the resin is removed by burning
it out, and the mass of fibers being divided by the strand length and fiber density yields the crosssectional area of fibers in the strand, Af . Fiber strength and modulus can be calculated as
30
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
FIGURE 1.32
Testing of a microcomposite specimen overwrapped over discs.
σf 5
F
;
Af
Ef 5
F
Af ε
In addition to fiber and matrix mechanical properties, micromechanical analysis requires information about the ply microstructure. Depending on the level of this information, there exist micromechanical models of different levels of complexity.
The simplest or zero-order model of a ply is a monotropic model ignoring the strength and stiffness of the matrix and assuming that the ply works only in the fiber direction. Taking E2 5 0 and
G12 5 0 in Eq. (1.59) and putting ν 12 5 0 in accordance with Eq. (1.60), we arrive at the following
equations describing this model:
σ1 5 E1 ε1 ;
σ2 5 0;
τ 12 5 0
(1.61)
in which E1 5 Ef vf . Being very simple and too approximate to be used in the stressstrain analysis
of composite structures, Eq. (1.61) are extremely efficient for the design of optimal composite
structures in which the loads are carried mainly by the fibers.
1.3 MICROMECHANICS OF A PLY
31
FIGURE 1.33
Testing of a microcomposite specimen gripped at the ends.
First-order models allow for the matrix stiffness but require only one structural parameter to be
specified: fiber volume fraction, vf . Since the fiber distribution in the ply is not important for these
models, the ply can be modeled as a system of strips as shown in Fig. 1.34 where the fibers and
matrix are represented by the shaded and light areas, respectively. The structural parameters of the
model can be expressed in terms of fiber and matrix volume fractions only, i.e.,
af
5 vf ;
a
am
5 vm ;
a
vf 1 vm 5 1
(1.62)
Suppose that the ply is under in-plane loading with some effective stresses σ1 , σ2 , and τ 12 as in
Fig. 1.34, and determine the corresponding effective elastic constants E1 , E2 , G12 , ν 12 , and ν 21 to
be inserted into Eq. (1.58). The constitutive equations for isotropic fiber and matrix strips can be
written as
ε1f ;m 5
1
ðσf ;m 2 ν f ;m σf2;m Þ
Ef ;m 1
ε2f ;m 5
1
ðσf ;m 2 ν f ;m σf1;m Þ
Ef ;m 2
γ f12;m 5
1 f ;m
τ
Gf ;m 12
(1.63)
32
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
σ1
τ 12
τ 12
3
2
σ2
σ2
1
τ 12
τ 12
σ1
am af
a
FIGURE 1.34
A first-order model of a unidirectional ply.
Here, as previously, the indices f and m correspond to fibers and matrix, respectively.
Let us make some assumptions concerning the model behavior. First, it can be assumed that the
effective stress resultant σ1 a is distributed between fiber and matrix strips and that the longitudinal
strains of these strips are the same as the effective (apparent) longitudinal strain of the ply, ε1 , i.e.,
σ1 a 5 σf1 af 1 σm
1 am
(1.64)
εf1 5 εm
1 5 ε1
(1.65)
Second, as can be seen in Fig. 1.34, under transverse tension the stresses in the strips are the
same and are equal to the effective stress σ2 , whereas the ply elongation in the transverse direction
is the sum of the fiber and matrix strips’ elongations, i.e.,
σf2 5 σm
2 5 σ2
(1.66)
Δa 5 Δaf 1 Δam
(1.67)
Introducing transverse strains
ε2 5
Δa
;
a
εf2 5
Δaf
;
af
εm
2 5
Δam
am
we can write Eq. (1.67) in the following form:
ε2 a 5 εf2 af 1 εm
2 am
(1.68)
The same assumptions can be made for shear stresses and strains, so that
τ f12 5 τ m
12 5 τ 12
(1.69)
γ 12 a 5 γ f12 af 1 γ m
12 am
(1.70)
Taking into account Eqs. (1.65), (1.66), and (1.69), constitutive equations, Eq. (1.63) can be
reduced to
ε1 5
1 f
ðσ 2 ν f σ2 Þ;
Ef 1
ε1 5
1 m
ðσ 2 ν m σ2 Þ
Em 1
(1.71)
1.3 MICROMECHANICS OF A PLY
εf2 5
1
ðσ2 2 ν f σf1 Þ;
Ef
εm
2 5
1
τ 12 ;
Gf
γm
12 5
γ f12 5
33
1
ðσ2 2 ν m σm
1Þ
Em
(1.72)
1
τ 12
Gm
(1.73)
The first two equations, Eq. (1.71), allow us to find longitudinal stresses, i.e.,
σf1 5 Ef ε1 1 ν f σ2 ;
σm
1 5 Em ε1 1 ν m σ 2
(1.74)
Equilibrium equation, Eq. (1.64), can be rearranged with the aid of Eq. (1.62) in the form
σ1 5 σf1 vf 1 σm
1 vm
(1.75)
Substituting σf1 and σm
1 in this equation with Eq. (1.74), we can express ε1 in terms of σ 1 and
σ2 . Combining this result with the first constitutive equation in Eq. (1.58), we arrive at
E1 5 Ef vf 1 Em vm
(1.76)
ν 12
ν f vf 1 ν m vm
5
E2
Ef vf 1 Em vm
(1.77)
The first of these equations specifies the apparent longitudinal modulus of the ply and corresponds to the so-called rule of mixtures, according to which the property of a composite can be calculated as the sum of its constituent material properties multiplied by the corresponding volume
fractions.
Now consider Eq. (1.68) that can be written as
ε2 5 εf2 vf 1 εm
2 vm
f
m
Substituting strains εf2 and εm
2 from Eq. (1.72), stresses σ1 and σ1 from Eq. (1.74), and ε1 from
Eq. (1.58) and taking into account Eqs. (1.76) and (1.77), we can express ε2 in terms of σ1 and σ2 .
Comparing this expression with the second constitutive equation in Eq. (1.58), we get
1
vf
vm
vf vm ðEf ν m 2Em ν f Þ2
5
1
2
Ef
Em
Ef Em ðEf vf 1 Em vm Þ
E2
(1.78)
ν 21
ν f vf 1 ν m vm
5
E1
Ef vf 1 Em vm
(1.79)
Using Eqs. (1.76) and (1.79), we have
ν 21 5 ν f vf 1 ν m vm
(1.80)
This result corresponds to the rule of mixtures. The second Poisson’s ratio can be found from
Eqs. (1.77) and (1.78). Finally, Eqs. (1.58), (1.70), and (1.73) yield the apparent shear modulus
1
vf
vm
5
1
G12
Gf
Gm
(1.81)
This expression can be derived from the rule of mixtures if we use compliance coefficients
instead of stiffnesses, as in Eq. (1.76).
34
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
Since the fiber modulus is typically many times greater than the matrix modulus, Eqs. (1.76),
(1.78), and (1.81) can be simplified, neglecting small terms, and presented in the following approximate form:
E1 5 Ef vf ;
E2 5
Em
;
vm ð1 2 ν 2m Þ
G12 5
Gm
vm
Only two of the foregoing expressions, namely Eq. (1.76) for E1 and Eq. (1.80) for ν 21 , both
following from the rule of mixtures, demonstrate good agreement with experimental results.
Moreover, expressions analogous to Eqs. (1.76) and (1.80) follow from practically all the numerous
studies based on different micromechanical models. A comparison of predicted and experimental
results is presented in Figs. 1.351.37, where the theoretical dependencies of normalized moduli
on the fiber volume fraction are shown with lines. The dots correspond to the test data for epoxy
composites reinforced with different fibers that have been measured by the authors or taken from
the publications of Tarnopol’skii and Roze (1969), Kondo and Aoki (1982), and Lee, Jeong, and
Kim (1995). As can be seen in Fig. 1.35, not only the first-order model, Eq. (1.76), but also the
zero-order model, Eq. (1.61), provide fair predictions for E1 , whereas Figs. 1.36 and 1.37 for E2
and G12 call for an improvement to the first-order model (the corresponding results are shown with
solid lines).
Second-order models allow for the fiber shape and distribution, but in contrast to higher-order
models, ignore the complicated stressed state in the fibers and matrix under loading of the ply as
shown in Fig. 1.38. To demonstrate this approach, consider a layer-wise fiber distribution (see
Fig. 1.5) and assume that the fibers are absolutely rigid and the matrix is in the simplest uniaxial
E1 E f
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
FIGURE 1.35
Dependence of the normalized longitudinal modulus on fiber volume fraction.
zero-order model, Eq. (1.61);
ss first-order model, Eq. (1.76);
K experimental data.
vf
E2 Em
10
8
6
4
2
0
0
0.2
0.4
0.6
0.8
vf
FIGURE 1.36
Dependence of the normalized transverse modulus on fiber volume fraction.
first-order model, Eq. (1.78);
second-order model, Eq. (1.89);
higher-order model (elasticity solution) (Van Fo Fy, 1966);
the upper bound;
experimental data.
G12 Gm
10
8
6
4
2
0
0
0.2
0.4
0.6
0.8
vf
FIGURE 1.37
Dependence of the normalized in-plane shear modulus on fiber volume fraction.
first-order model, Eq. (1.81);
second-order model, Eq. (1.90);
higher-order model (elasticity solution) (Van Fo Fy, 1966);
experimental data.
36
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
x3
l(α )
α
σ2
3
R
A
A
σ2
x2
2
1
σm
Δa
a
FIGURE 1.38
Microstructural model of the second order.
stressed state under transverse tension. The typical element of this model is shown in Fig. 1.38
from which we can obtain the following equation:
vf 5
πR2
πR
5
2a
2Ra
(1.82)
Since 2R , a, vf , π=4 5 0:785. The equilibrium condition yields
ðR
2Rσ2 5
σm dx3
(1.83)
2R
where x3 5 Rcosα and σ2 is some average transverse stress that induces average strain
ε2 5
Δa
a
(1.84)
such that the effective (apparent) transverse modulus is calculated as
E2 5
σ2
ε2
(1.85)
The strain in the matrix can be determined with the aid of Fig. 1.38 and Eq. (1.84), i.e.,
εm 5
Δa
Δa
5
5
lðαÞ a 2 2Rsinα
ε2
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x 2ffi
3
12λ 12
R
(1.86)
where in accordance with Eq. (1.82),
λ5
2R 4vf
5
π
a
(1.87)
Assuming that there is no strain in the matrix in the fiber direction and there is no stress in the
matrix in the x3 direction, we have
σm 5
Em εm
1 2 ν 2m
(1.88)
1.3 MICROMECHANICS OF A PLY
37
Substituting σ2 and σm in Eq. (1.83) with their expressions given by Eq. (1.85) and Eq. (1.88),
and using Eq. (1.86) to express εm , we arrive at
Em
E2 5
2Rð1 2 ν 2m Þ
ðR
dx3
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
2
2R 1 2 λ 1 2 x3 =R
Calculating the integral and taking into account Eq. (1.87), we finally get
E2 5
πEm rðλÞ
2vf ð1 2 ν 2m Þ
where
1
rðλÞ 5 pffiffiffiffiffiffiffiffiffiffiffiffiffi tan21
1 2 λ2
(1.89)
rffiffiffiffiffiffiffiffiffiffiffi
11λ π
2
12λ 4
A similar derivation for an in-plane shear yields
G12 5
πGm
rðλÞ
2vf
(1.90)
The dependencies of E2 and G12 on the fiber volume fraction corresponding to Eqs. (1.89) and
(1.90) are shown in Figs. 1.36 and 1.37 (dotted lines). As can be seen, the second-order model of a
ply provides better agreement with the experimental results than the first-order model. This agreement can be further improved if we take a more realistic microstructure of the material. Consider
the actual microstructure shown in Fig. 1.2 and single out a typical square element with size a as in
Fig. 1.39. The dimension a should provide the same fiber volume fraction for the element as for
the material under study. To calculate E2 , we divide the element into a system of thin (h ,, a)
strips parallel to axis x2 . The ith strip is shown in Fig. 1.39. For each strip, we measure the lengths,
lij , of the matrix elements, the jth of which is shown in Fig. 1.39. Then, equations analogous to
Eqs. (1.83), (1.88), and (1.86) take the form
σ2 a 5 h
X
σðiÞ
m;
σðiÞ
m 5
i
ε2 a
ðiÞ
εm
5P
j lij
Em ðiÞ
ε ;
1 2 ν 2m m
a
x3
j
i
h
l ij
FIGURE 1.39
Typical structural element.
a
x2
38
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
and the final result is
E2 5
Em h X X
lij
1 2 ν 2m i
j
!21
where h 5 h=a, lij 5 lij =a. The second-order models considered above can be readily generalized to
account for the fiber transverse stiffness and matrix nonlinearity.
Numerous higher-order microstructural models and descriptive approaches have been proposed,
including:
•
•
•
•
•
Analytical solutions in the problems of elasticity for an isotropic matrix having regular
inclusions: fibers or periodically spaced groups of fibers
Numerical (finite element, finite difference methods) stress analysis of the matrix in the vicinity
of fibers
Averaging of stress and strain fields for a media filled in with regularly or randomly distributed
fibers
Asymptotic solutions of elasticity equations for inhomogeneous solids characterized by a small
microstructural parameter (fiber diameter)
Photoelasticity methods
An exact elasticity solution for a periodical system of fibers embedded in an isotropic matrix
(Van Fo Fy (Vanin), 1966) is shown in Figs. 1.36 and 1.37. As can be seen, due to the high scatter
in experimental data, the higher-order model does not demonstrate significant advantages with
respect to elementary models.
Moreover, all the micromechanical models are not realistically usable for the practical analysis
of composite materials and structures. The reason for this is that irrespective of how rigorous the
micromechanical model is, it cannot describe adequately enough a real material microstructure governed by a particular manufacturing process, taking into account voids, microcracks, randomly
damaged or misaligned fibers, and many other effects that cannot be formally reflected in a mathematical model. As a result of this, micromechanical models are mostly used for qualitative analysis,
providing us with the understanding of how material microstructural parameters affect the mechanical properties rather than with quantitative information about these properties. In particular, the
foregoing analysis should result in two main conclusions. First, the ply stiffness along the fibers is
governed by the fibers and linearly depends on the fiber volume fraction. Second, the ply stiffness
across the fibers and in shear is determined not only by the matrix (which is natural), but by the
fibers as well. Although the fibers do not directly take the load applied in the transverse direction,
they significantly increase the ply transverse stiffness (in comparison with the stiffness of a net
matrix), acting as rigid inclusions in the matrix. Indeed, as can be seen in Fig. 1.34, the higher the
fiber fraction, af , the lower the matrix fraction, am , for the same a, and the higher stress σ2 should
be applied to the ply to cause the same transverse strain ε2 because only matrix strips are deformable in the transverse direction.
Due to the aforementioned limitations of micromechanics, only the basic models were considered in the previous examples. A historical overview of micromechanical approaches and more
detailed description of the corresponding results can be found elsewhere (Bogdanovich & Pastore,
1996; Jones, 1999).
1.3 MICROMECHANICS OF A PLY
39
To analyze the foregoing micromechanical models, we used the traditional approach based on
direct derivation and solution of the system of equilibrium, constitutive, and strain-displacement
equations. Alternatively, the same problems can be solved with the aid of variational principles. In
their application to micromechanics, these principles allow us not only to determine the apparent
stiffnesses of the ply, but also to establish the upper and the lower bounds on them.
Consider, for example, the problem of transverse tension of a ply under the action of some average stress σ2 (see Fig. 1.29) and apply the principle of minimum strain energy. According to this
principle, the actual stress field provides the value of the body strain energy, which is equal to or
less than that of any statically admissible stress field. Equality takes place only if the admissible
stress state coincides with the actual one. Excluding this case, i.e., assuming that the class of admissible fields under study does not contain the actual field, we can write the following strict
inequality
Wσadm . Wσact
(1.91)
In the case of transverse tension, the fibers can be treated as absolutely rigid, and only the
matrix strain energy needs to be taken into account. We can also neglect the energy of shear strain
and consider the energy corresponding to normal strains only. Taking into account these assumptions, we get
W5
ððð
UdVm
(1.92)
Vm
where Vm is the volume of the matrix, and
U5
1 m m
m
m m
ðσ ε 1 σm
2 ε2 1 σ 3 ε3 Þ
2 1 1
(1.93)
is the density of the elastic energy. To find the value of the energy Wσ in inequality (1.91), we
should express strains in terms of stresses with the aid of constitutive equations, i.e.
1 m
m
ðσ 2 ν m σm
2 2 ν m σ3 Þ
Em 1
1 m
m
εm
ðσ 2 ν m σm
1 2 ν m σ3 Þ
2 5
Em 2
1 m
m
εm
ðσ 2 ν m σm
3 5
1 2 ν m σ2 Þ
Em 3
εm
1 5
(1.94)
Consider first the actual stress state. Let the ply in Fig. 1.29 be loaded with stress σ2 inducing
apparent strain ε2 such that
ε2 5
σ2
E2act
(1.95)
Here, E2act is the actual apparent modulus, which is not known. Taking into account Eqs. (1.92)
and (1.93), we get
1
W 5 σ2 ε2 V;
2
Wσact 5
σ22
V
2E2act
(1.96)
40
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
where V is the volume of the material. As an admissible field, we can take any state of stress that
satisfies the equilibrium equations and force boundary conditions. Using the simplest first-order
model shown in Fig. 1.34, we assume that
m
σm
1 5 σ 3 5 0;
σm
2 5 σ2
Then, Eqs. (1.92)(1.94) yield
Wσadm 5
σ22
Vm
2Em
(1.97)
Substituting Wσadm and Wσact in the inequality (1.91) with their expressions given by Eqs. (1.96)
and (1.97), we arrive at
E2act . E2l
where, in accordance with Eq. (1.62) and Fig. 1.34,
E2l 5
Em V
Em
5
Vm
vm
This result, specifying the lower bound on the apparent transverse modulus, follows from
Eq. (1.78) if we put Ef -N. Thus, the lower (solid) line in Fig. 1.36 actually represents the lower
bound on E2 .
To derive the expression for the upper bound, we should use the principle of minimum total
potential energy, according to which (we again assume that the admissible field does not include
the actual state)
Tadm . Tact
(1.98)
where T 5 Wε 2 A. Here, Wε is determined by Eq. (1.92), in which stresses are expressed in terms
of strains with the aid of Eq. (1.94), and A, for the problem under study, is the product of the force
acting on the ply and the ply extension induced by this force. Since the force is the resultant of
stress σ2 (see Fig. 1.29) which induces strain ε2 , the same for actual and admissible states, A is also
the same for both states, and we can present inequality (1.98) as
Wεadm . Wεact
(1.99)
For the actual state, we can write equations similar to Eq. (1.96), i.e.,
W5
1
σ2 ε2 V;
2
1
Wεact 5 E2act ε22 V
2
(1.100)
in which V 5 2Ra in accordance with Fig. 1.38. For the admissible state, we use the second-order
model (see Fig. 1.38) and assume that
εm
1 5 0;
εm
2 5 εm ;
εm
3 50
where εm is the matrix strain specified by Eq. (1.86). Then, the equations given by Eq. (1.94) yield
m
σm
1 5 μm σ 2 ;
m
σm
3 5 μm σ 2 ;
σm
2 5
where
μm 5
ν m ð1 1 ν m Þ
1 2 ν 2m
Em εm
1 2 2ν m μm
(1.101)
1.3 MICROMECHANICS OF A PLY
41
Substituting Eq. (1.101) for the stresses in Eq. (1.93) and performing integration in accordance
with Eq. (1.92), we have
Wεadm 5
Here,
Em ε22
U
1 2 2ν m μm
a
y
2ð
ðR
dx3
2R
0
dx2
πRaEm ε22 rðλÞ
5
2
y
2vf ð1 2 2ν m μm Þ
(1.102)
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x 2ffi
3
y512λ 12
R
and rðλÞ is given above, see also Eq. (1.89). Applying Eqs. (1.100) and (1.102) in conjunction with
inequality (1.99), we arrive at
E2act , E2u
where
E2u 5
πEm
2vf ð1 2 2ν m μm Þ
is the upper bound on E2 shown in Fig. 1.36 with a dashed curve.
Taking statically and kinematically admissible stress and strain fields that are closer to the
actual state of stress and strain, one can increase E2l and decrease E2u making the difference between
the bounds smaller (Hashin & Rosen, 1964).
It should be emphasized that the bounds established thus are not the bounds imposed on the
modulus of a real composite material but on the result of calculation corresponding to the adopted
material model. Indeed, we can return to the first-order model shown in Fig. 1.34 and consider inplane shear with stress τ 12 . As can be readily proved, the actual stressstrain state of the matrix in
this case is characterized by the following stresses and strains
m
m
σm
1 5 σ 2 5 σ 3 5 0;
τm
12 5 τ 12 ;
m
τm
13 5 τ 23 5 0
m
m
εm
1 5 ε2 5 ε3 5 0;
γm
12 5 γ 12 ;
m
γm
13 5 γ 23 5 0
(1.103)
Assuming that the fibers are absolutely rigid and considering stresses and strains in Eq. (1.103)
as statically and kinematically admissible, we can readily find that
l
u
Gact
12 5 G12 5 G12 5
Gm
vm
Thus, we have found the exact solution, but its agreement with experimental data is rather poor
(see Fig. 1.37) because the material model is not sufficiently adequate.
As follows from the foregoing discussion, micromechanical analysis provides only qualitative prediction of the ply stiffness. The same is true for the ply strength. Although the
micromechnical approach can be used in principle for strength analysis (Skudra, Bulavs,
Gurvich, & Kruklinsh, 1989), it mainly provides better understanding of the failure mechanism rather than the values of the ultimate stresses for typical loading cases. For practical
applications, these stresses are determined by experimental methods described in the next
section.
42
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
1.4 MECHANICAL PROPERTIES OF A PLY UNDER TENSION, SHEAR, AND
COMPRESSION
As shown in Fig. 1.29, a ply can experience five types of elementary loading:
•
•
•
•
•
Tension along the fibers
Tension across the fibers
In-plane shear
Compression along the fibers
Compression across the fibers
The actual mechanical properties of a ply under these loading cases are determined experimentally by testing of specially fabricated specimens. Since the thickness of an elementary ply is very
small (0.10.2 mm), the specimen usually consists of tens of plies having the same fiber
orientations.
Mechanical properties of composite materials depend on the processing method and parameters.
So, to obtain the adequate material characteristics that can be used for the analysis of structural elements, the specimens should be fabricated by the same processes that are used to manufacture the
structural elements. In connection with this, there exist two standard types of specimens— flat ones
that are used to test materials made by hand or machine lay-up and cylindrical (tubular or ring) specimens that represent materials made by winding.
Typical mechanical properties of unidirectional advanced composites are presented in Table 1.5
and in Figs. 1.401.43. More data relevant to the various types of particular composite materials
could be found in Peters (1998).
We now consider typical loading cases.
1.4.1 LONGITUDINAL TENSION
The stiffness and strength of unidirectional composites under longitudinal tension are determined
by the fibers. As follows from Fig. 1.35, material stiffness linearly increases with increase in the
fiber volume fraction. The same law following from Eq. (1.75) is valid for the material strength. If
the fiber’s ultimate elongation, εf , is less than that of the matrix (which is normally the case), the
longitudinal tensile strength is determined as
σ1
1 5 ðEf vf 1 Em vm Þεf
(1.104)
However, in contrast to Eq. (1.76) for E1 , this equation is not valid for very small and very high
fiber volume fractions. The dependence of σ1
1 on vf is shown in Fig. 1.44. For very low vf , the
fibers do not restrain the matrix deformation. Being stretched by the matrix, the fibers fail because
their ultimate elongation is less than that of the matrix and the induced stress concentration in the
matrix can reduce material strength below the strength of the matrix (point B). Line BC in
Fig. 1.44 corresponds to Eq. (1.104). At point C, the amount of matrix reduces below the level necessary for a monolithic material, and the material strength at point D approximately corresponds to
the strength of a dry bundle of fibers, which is less than the strength of a composite bundle of fibers
bound with matrix (see Table 1.3).
Table 1.5 Typical Properties of Unidirectional Composites
Property
Glass
Epoxy
Carbon
Epoxy
Carbon
PEEK
Aramid
Epoxy
Boron
Epoxy
BoronAl
Carbon
Carbon
Al2 O3 Al
Fiber volume fraction,vf
Density, ρ (g/cm3)
Longitudinal modulus, E1 (GPa)
Transverse modulus, E2 (GPa)
Shear modulus, G12 (GPa)
Poisson’s ratio, ν 21
Longitudinal tensile strength, σ1
1 (MPa)
Longitudinal compressive strength, σ2
1 (MPa)
(MPa)
Transverse tensile strength, σ1
2
Transverse compressive strength, σ2
2 (MPa)
In-plane shear strength, τ 12 (MPa)
0.65
2.1
60
13
3.4
0.3
1800
650
40
90
50
0.62
1.55
140
11
5.5
0.27
2000
1200
50
170
70
0.61
1.6
140
10
5.1
0.3
2100
1200
75
250
160
0.6
1.32
95
5.1
1.8
0.34
2500
300
30
130
30
0.5
2.1
210
19
4.8
0.21
1300
2000
70
300
80
0.5
2.65
260
140
60
0.3
1300
2000
140
300
90
0.6
1.75
170
19
9
0.3
340
180
7
50
30
0.6
3.45
260
150
60
0.24
700
3400
190
400
120
44
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
(A)
σ 1 , MPa
2000
σ +1
1600
1200
800
σ −1
400
0
(B)
0
1
2
3
ε1, %
σ 2 ; τ12 , MPa
90
σ 2−
60
0
τ 12
σ 2+
30
0
1
2
3
4
5
ε 2 ; γ12, %
FIGURE 1.40
Stressstrain curves for unidirectional glassepoxy composite material under longitudinal tension and
compression (A), transverse tension and compression (B), and in-plane shear (B).
The strength and stiffness under longitudinal tension are determined using unidirectional strips
or rings. The strips are cut out of unidirectionally reinforced plates, and their ends are made thicker
(usually glassepoxy tabs are bonded onto the ends) to avoid specimen failure in the grips of the
testing machine (Jones, 1999; Lagace, 1985). Rings are cut out of a circumferentially wound cylinder or wound individually on a special mandrel, as shown in Fig. 1.45. The strips are tested using
traditional approaches, whereas the rings should be loaded with internal pressure. There exist several methods to apply the pressure (Tarnopol’skii & Kincis, 1985), the simplest of which involves
the use of mechanical fixtures with various numbers of sectors as in Figs. 1.46 and 1.47. The
1.4 MECHANICAL PROPERTIES OF A PLY UNDER TENSION
(A)
45
σ 1 , MPa
2000
σ +1
1600
1200
σ −1
800
400
0
0.5
0
1.5
1
ε1, %
(B) σ 2 ; τ12 , MPa
200
σ 2−
150
100
σ 2+
50
0
0
τ 12
1
2
3
ε 2 ; γ 12, %
FIGURE 1.41
Stressstrain curves for unidirectional carbonepoxy composite material under longitudinal tension and
compression (A), transverse tension and compression (B), and in-plane shear (B).
failure mode is shown in Fig. 1.48. Longitudinal tension yields the following mechanical properties
of the material:
•
•
•
Longitudinal modulus, E1
Longitudinal tensile strength, σ1
1
Poisson’s ratio, ν 21
Typical values of these characteristics for composites with various fibers and matrices are listed
in Table 1.5. It follows from Figs. 1.401.43 that the stressstrain diagrams are linear practically
up to failure.
46
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
(A)
σ 1 , MPa
2800
σ +1
2400
2000
1600
1200
800
400
0
σ −1
0
0.5
1
1.5
2
2.5
3
ε1, %
(B) σ 2; τ 12 , MPa
150
σ 2−
100
50
σ 2+
τ 12
ε2
0
0
1
2
12, %
3
FIGURE 1.42
Stressstrain curves for unidirectional aramidepoxy composite material under longitudinal tension and
compression (A), transverse tension and compression (B), and in-plane shear (B).
1.4.2 TRANSVERSE TENSION
There are three possible modes of material failure under transverse tension with stress σ2 shown in
Fig. 1.49: failure of the fibermatrix interface (adhesion failure), failure of the matrix (cohesion
failure), and fiber failure. The last failure mode is specific for composites with aramid fibers, which
consist of thin filaments (fibrils) and have low transverse strength. As follows from the micromechanical analysis (Section 1.3), the material stiffness under tension across the fibers is higher than
that of a net matrix (see Fig. 1.36).
For qualitative analysis of transverse strength, consider again the second-order model in
Fig. 1.38. As can be seen, the stress distribution σm ðx3 Þ is not uniform, and the maximum stress in
the matrix corresponds to α 5 903 . Using Eqs. (1.85), (1.86), and (1.88), we obtain
1.4 MECHANICAL PROPERTIES OF A PLY UNDER TENSION
(A)
47
σ 1 , MPa
2400
σ −1
2000
1600
1200
σ +1
800
400
0
(B)
0
0.4
0.8
1.2
1.6
ε1, %
σ 2; τ 12, MPa
300
σ 2−
200
100
τ 12
σ 2+
0
0
1
2
3
ε2
12, %
FIGURE 1.43
Stressstrain curves for unidirectional boronepoxy composite material under longitudinal tension and
compression (A), transverse tension and compression (B), and in-plane shear (B).
σmax
m 5
Em σ2
ð1 2 ν 2m ÞE2 ð1 2 λÞ
1
1
Taking σmax
m 5 σm and σ2 5 σ2 , where σ m and σ2 are the ultimate stresses for the matrix and
composite material, respectively, and substituting for λ and E2 their expressions given by
Eqs. (1.87) and (1.89), we arrive at
σ1
2 5 σm
rðλÞ
ðπ 2 4vf Þ
2vf
(1.105)
48
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
σ 1+
1
C
0.8
D
0.6
0.4
0.2
A
B
0
0
vf
0.2
0.4
0.6
0.8
FIGURE 1.44
Dependence of normalized longitudinal strength on fiber volume fraction (x—experimental results).
FIGURE 1.45
A mandrel for test rings.
The variation of the ratio σ1
2 =σm for epoxy composites is shown in Fig. 1.50. As can be seen,
the transverse strength of a unidirectional material is considerably lower than the strength of the
matrix. It should be noted that for the first-order model, which ignores the shape of the fiber cross
1
sections (see Fig. 1.34), σ1
2 is equal to σm . Thus, the reduction of σ 2 is caused by stress concentration in the matrix induced by cylindrical fibers.
However, both polymeric and metal matrices exhibit elasticplastic behavior, and it is known
that plastic deformation reduces the effect of stress concentration. Nevertheless, the stressstrain
FIGURE 1.46
Two-, four-, and eight-sector test fixtures for composite rings.
FIGURE 1.47
A composite ring on an eight-sector test fixture.
FIGURE 1.48
Failure modes of unidirectional rings.
50
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
1
σ2
σ2
2
3
FIGURE 1.49
Modes of failure under transverse tension.
1 Adhesion failure
2 Cohesion failure
3 Fiber failure
σ 2+ σ m
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
vf
FIGURE 1.50
Dependence of material strength under transverse tension on fiber volume fraction.
———Eq. (1.105); (K) Experimental data.
diagrams σ1
2 2 ε2 shown in Figs. 1.401.43 are linear up to the failure point, which means that the
plastic deformation does not affect the material behavior. To explain this phenomenon, consider
element A of the matrix located in the vicinity of a fiber as in Fig. 1.38. Assuming that the fiber is
absolutely rigid, we can conclude that the matrix strains in directions 1 and 3 are close to zero.
m
Taking εm
1 5 ε3 5 0 in Eq. (1.94), we arrive at Eq. (1.101) for stresses, according to which
m
m
σ1 5 σ3 5 μm σm
2 . The dependence of parameter μm on the matrix Poisson’s ratio is presented in
Fig. 1.51. As follows from this figure, in the limiting case ν m 5 0:5, we have μm 5 1 and
1.4 MECHANICAL PROPERTIES OF A PLY UNDER TENSION
51
μm
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.3
0.35
0.4
0.45
0.5
υm
FIGURE 1.51
Dependence of parameter μm on the matrix Poisson’s ratio.
m
m
σm
1 5 σ2 5 σ 3 , i.e., the state of stress under which all the materials behave as absolutely brittle. For
epoxy resin, ν m 5 0:35 and μm 5 0:54 which, as can be expected, does not allow the resin to demonstrate its rather limited plastic properties.
The strength and stiffness under transverse tension are experimentally determined using flat
strips (see Fig. 1.52) or tubular specimens (see Fig. 1.53). These tests allow us to determine the
transverse modulus, E2 , and transverse tensile strength, σ1
2 . For typical composite materials, these
properties are given in Table 1.5.
1.4.3 IN-PLANE SHEAR
The failure modes in unidirectional composites under in-plane pure shear with stress τ 12 shown in
Fig. 1.29 are practically the same as those for the case of transverse tension (see Fig. 1.49).
However, there is a significant difference in material behavior. As follows from Figs. 1.401.43,
the stressstrain curves τ 12 2 γ 12 are not linear, and τ 12 exceeds σ1
2 . This means that the fibers do
not restrict the free shear deformation of the matrix, and the stress concentration in the vicinity of
the fibers does not significantly influence material strength because of the matrix plastic
deformation.
Strength and stiffness under in-plane shear are determined experimentally by testing plates and
thin-walled cylinders. A plate is reinforced at 453 to the loading direction and is fixed in a square
frame consisting of four hinged members, as shown in Fig. 1.54. Simple equilibrium consideration
and geometric analysis yield the following equations:
P
τ 12 5 pffiffiffi ;
2ah
γ12 5 εy 2 εx ;
G12 5
τ 12
γ 12
in which h is the plate thickness. Thus, for a given P and measured strains in the x and y directions,
we can determine τ 12 and G12 . More accurate and reliable results can be obtained if we induce
pure shear in a twisted tubular specimen reinforced in the circumferential direction (Fig. 1.55).
Again, using simple equilibrium and geometric analysis, we get
52
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
FIGURE 1.52
Test fixture for transverse tension and compression of unidirectional strips.
τ 12 5
M
;
2πR2 h
γ 12 5
ϕR
;
l
G12 5
τ 12
γ 12
Here, M is the torque, R and h are the cylinder radius and thickness, and ϕ is the twist angle
between two cross sections located at some distance l from each other. Thus, for given M and measured ϕ, we can find τ 12 and G12 .
1.4 MECHANICAL PROPERTIES OF A PLY UNDER TENSION
FIGURE 1.53
Test fixture for transverse tension or compression of unidirectional tubular specimens.
P
y
1
2
45°
a
τ 12
P
FIGURE 1.54
Simulation of pure shear in a square frame.
x
53
54
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
FIGURE 1.55
A tubular specimen for shear test.
τ
σ1
α
τ
σ1
FIGURE 1.56
Shear failure under compression.
1.4.4 LONGITUDINAL COMPRESSION
Failure under compression along the fibers can occur in different modes, depending on the material
microstructural parameters, and is difficult to predict using micromechanical analysis because of
the rather complicated interaction of these modes. Nevertheless, useful qualitative results allowing
us to understand the material behavior and hence to improve its properties can be obtained from
microstructural models.
Consider typical compression failure modes. The usual failure mode under compression is associated with shear in some oblique plane as in Fig. 1.56. The shear stress can be calculated as
τ 5 σ1 sin α cos α
and reaches its maximum value at α 5 453 . Shear failure under compression is usually typical for
unidirectional composites that demonstrate the highest strength under longitudinal compression. On
the other hand, materials showing the lowest strength under compression exhibit a transverse extension failure mode typical of wood compressed along the fibers, and shown in Fig. 1.57. This failure
is caused by tensile transverse strain, whose absolute value is
ε2 5 ν 21 ε1
(1.106)
where ν 21 is Poisson’s ratio and ε1 5 σ1 =E1 is the longitudinal strain. Consider Table 1.6, showing
some data taken from Table 1.5 and the results of calculations for epoxy composites. The fourth
1
column displays the experimental ultimate transverse strains ε1
2 5 σ2 =E2 calculated with the aid of
1.4 MECHANICAL PROPERTIES OF A PLY UNDER TENSION
55
2
1
σ1
σ1
FIGURE 1.57
Transverse extension failure mode under longitudinal compression.
Table 1.6 Characteristics of Epoxy Composites
Characteristic
Material
σ2
1 (MPa)
ε2
1 (%)
ν 21
ε1
2 (%)
ε2 5 ν 21 ε2
1
Glassepoxy
Carbonepoxy
Aramidepoxy
Boronepoxy
600
1200
300
2000
1.00
0.86
0.31
0.95
0.30
0.27
0.34
0.21
0.31
0.45
0.59
0.37
0.30
0.23
0.11
0.20
data presented in Table 1.5, whereas the last column shows the results following from Eq. (1.106).
As can be seen, the failure mode associated with transverse tension under longitudinal compression
is not appropriate for the composites under consideration because ε1
2 . ε2 . However, this is true
only for fiber volume fractions vf 5 0:50 2 0:65, to which the data presented in Table 1.6 correspond. To see what happens for higher fiber volume fractions, let us use the second-order micromechanical model and the corresponding results in Figs. 1.36 and 1.50. We can plot the strain
concentration factor kε (which is the ratio of the ultimate matrix elongation, εm , to ε1
2 for the composite material) versus the fiber volume fraction. As can be seen in Fig. 1.58, this factor, being
about 6 for vf 5 0:6, becomes as high as 25 for vf 5 0:75. This means that ε1
2 dramatically
decreases for higher vf , and the fracture mode shown in Fig. 1.57 becomes significant for composites with high fiber volume fractions.
Both fracture modes shown in Figs. 1.56 and 1.57 are accompanied with fibers bending induced
by local buckling of fibers. According to N.F. Dow and B.W. Rosen (Jones, 1999), there can exist
two modes of fiber buckling, as shown in Fig. 1.59, i.e., a shear mode and a transverse extension
mode. To study the fiber’s local buckling (or microbuckling, which means that the material specimen is straight, whereas the fibers inside the material are curved), consider a plane model of a unidirectional ply, shown in Figs. 1.15 and 1.60, and take am 5 a and af 5 δ 5 d, where d is the fiber
diameter. Then, Eq. (1.17) yields
vf 5
d
;
11d
d5
d
a
(1.107)
Due to the symmetry conditions, consider two fibers 1 and 2 in Fig. 1.60 and the matrix
between these fibers. The buckling displacement, v, of the fibers can be represented with a sine
function as
56
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
kε
25
20
15
10
5
0
0
0.2
0.4
0.6
0.8
vf
FIGURE 1.58
Dependence of strain concentration factor on the fiber volume fraction.
(A)
(B)
FIGURE 1.59
Shear (A) and transverse extension (B) modes of fiber local buckling.
y , uy
c
v2 ( x)
2
v1 ( x)
σ1
a
1
d
α
FIGURE 1.60
Local buckling of fibers in unidirectional ply.
ln
σ1
x, ux
1.4 MECHANICAL PROPERTIES OF A PLY UNDER TENSION
57
d
d 2
σ1
1
a
σ1
d 2
2
ln
FIGURE 1.61
A typical ply element.
v1
dv1
δ
x
dx
ln
FIGURE 1.62
Deformation of a fiber.
v1 ðxÞ 5 Vsinλn x; v2 ðxÞ 5 Vsinλn ðx 2 cÞ
(1.108)
where V is an unknown amplitude value, the same for all the fibers, λn 5 π=ln , ln is a half of a fiber
wavelength (see Fig. 1.60), and c 5 ða 1 dÞcotα is a phase shift. Taking c 5 0, we can describe the
shear mode of buckling (Fig. 1.59a), whereas c 5 ln corresponds to the extension mode
(Fig. 1.59b). To find the critical value of stress σ1 , we use the Timoshenko energy method
(Timoshenko & Gere, 1961), yielding the following buckling condition:
A5W
(1.109)
Here, A is the work of external forces, and W is the strain energy accumulated in the material
while the fibers undergo buckling. Work A and energy W are calculated for a typical ply element
consisting of two halves of fibers 1 and 2 and the matrix between them (see Fig. 1.61). The work,
A, can be calculated as
A 5 σ1 ða 1 dÞdUδ
(1.110)
with displacement δ following from Fig. 1.62, i.e.,
δ 5 ln 2
lnð
2δ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
dv1
11
dx
dx
0
Using conventional assumptions, i.e., taking ðdv1 =dxÞ , , 1 and δ , , l and substituting v1
from Eq. (1.108), we arrive at
58
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
δ5
1
2
ðln 2
dv1
1
dx 5 V 2 λ2n ln
4
dx
0
Thus, Eq. (1.110) yields
π2
σ1 V 2 adð1 1 dÞ
4ln
A5
(1.111)
The strain energy consists of three parts, i.e.,
W 5 Wf 1 Wms 1 Wme
(1.112)
where Wf is the energy of buckled fibers, whereas Wms and Wme correspond to shear strain and transverse extension of the matrix that supports the fibers. The strain energy of fibers deformed in accordance with Eq. (1.108) and shown in Fig. 1.61 has the form
1
Wf 5 D f
4
ðln " 2 2 2 2 #
d v1
d v2
dx
1
dx2
dx2
0
where Df is the fiber bending stiffness. Substituting Eq. (1.108) and calculating the integrals, we
get
Wf 5
π4
Df V 2
4l3n
(1.113)
To determine the strain energy of the matrix, we assume that the matrix element shown in
Fig. 1.61 is in a state of plane stress (nonzero stresses are σx , σy , and τ xy ), and the equilibrium
equations can be written as
@σx
@τ xy
1
5 0;
@x
@y
@σy
@τ xy
1
50
@y
@x
(1.114)
To simplify the solution, we assume that the longitudinal stress, σx , acting in the matrix can be
neglected in comparison with the corresponding stress acting in the fibers. Thus, we can set σx 5 0.
Then, Eq. (1.114) can be integrated and yield
τ xy 5 τðxÞ;
σy 5 σðxÞ 2 τ 0 ðxÞy
(1.115)
0
Here, τðxÞ and σðxÞ are arbitrary functions of integration and ð Þ 5 dð Þ=dx. Also neglecting
the Poisson effects, we can express the strains as follows
γ xy 5
τðxÞ
;
Gm
εy 5
1
½σðxÞ 2 τ 0 ðxÞy
Em
(1.116)
which in turn can be expressed in terms of displacements, i.e.,
γ xy 5
@ux
@uy
1
;
@y
@x
εy 5
@uy
@y
(1.117)
Substituting strains in Eq. (1.117) with their expressions given by Eq. (1.116) and integrating,
we can determine the displacements as
ux 5 uðxÞ 1
τðxÞ
1
1
2 v0 ðxÞ y 2
σ0 ðxÞy2 2 τvðxÞy3
Gm
2Em
3
1.4 MECHANICAL PROPERTIES OF A PLY UNDER TENSION
uy 5 vðxÞ 1
59
1
1
σðxÞy 2 τ 0 ðxÞy2
Em
2
Here, uðxÞ and vðxÞ are functions of integration that, in addition to the functions τðxÞand σðxÞ,
should be found using compatibility conditions at the fibermatrix interfaces. Using Fig. 1.63, we
can write these conditions in the following form:
d
ux ðy 5 0Þ 5 2 v01 ðxÞ;
2
ux ðy 5 aÞ 5
d 0
v ðxÞ
2 2
uy ðy 5 0Þ 5 v1 ðxÞ;
uy ðy 5 aÞ 5 v2 ðxÞ
Satisfying them, we can find uðxÞ and vðxÞdirectly as
d
uðxÞ 5 2 Vλn cosλn x;
2
vðxÞ 5 Vsinλn x
and derive the following equations for σðxÞ and τðxÞ:
σðxÞ 5
Em
1
V ½sinλn ðx 2 cÞ 2 sinλn x 1 τ 0 ðxÞa
2
a
a2
2
τvðxÞ 2
τðxÞ 5 2 Vλn ð1 1 dÞ½cosλn ðx 2 cÞ 1 cosλn x
Gm
6Em
(1.118)
(1.119)
We need a periodic solution of Eq. (1.119) and can find it in the following form:
τðxÞ 5 C½cosλn ðx 2 cÞ 1 cosλn x
(1.120)
Substituting τðxÞ in Eq. (1.119) with its expression given by Eq. (1.120) and taking into account
that λn 5 π=ln , we have
C5V
πGm ð1 1 dÞ
;
2ln ð1 1 β n Þ
βn 5
π2 a2 Gm
12l2n Em
d 2
1
d 2
v2 ( x )
v '1 ( x )
v '2 ( x )
v1 ( x )
d 2
2
FIGURE 1.63
Compatible fibermatrix deformation.
d 2
(1.121)
60
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
Now, using Eqs. (1.115), (1.118), and (1.120), we can write the final expressions for the stresses
acting in the matrix
τ xy 5 C½cosλn ðx 2 cÞ 1 cosλn x;
a
Em
2y 2
σy 5 2 Cλn
V sinλn ðx 2 cÞ
2
a
a
Em
2y 1
2 Cλn
V sinλn x
2
a
(1.122)
in which C is specified by Eq. (1.121). The corresponding strain energies of the typical element in
Fig. 1.61 are
Wms 5
ð
ad ln 2
τ dx;
2Gm 0 xy
Wme 5
ð
ad ln 2
σ dx
2Em 0 y
Substituting Eq. (1.122) and integrating, we arrive at
Wms 5
Wme 5
adln 2
C ð1 1 cosλn cÞ
2Gm
adln π2 a2 2
Em2 2
C
ð1
1
cosλ
cÞ
1
V
ð1
2
cosλ
cÞ
n
n
2Em 12l2n
a2
In conjunction with these results, Eqs. (1.109), (1.111)(1.113), and (1.121) allow us to determine σ1 , which takes the following final form:
π2 Df
σ1 5 2
1
ln dð1 1 dÞa4
2 πc
2Em ln
πc
! 1 1 cos
1 2
1 2 cos
ln
π ð1 1 dÞ
ln
π2 Gm
Gm ð1 1 dÞ
2 11
(1.123)
2
12ln Em
where d 5 d=a, ln 5 ln =a, and c 5 c=a. The critical value of σ1 can be found by minimization of the
right-hand part of Eq. (1.123) with respect to ln and c. However, having in mind only qualitative
analysis, we can omit this cumbersome procedure and use Eq. (1.123) for qualitative assessments
and estimates.
As follows from this equation, the strength of a unidirectional composite under longitudinal compression should increase with an increase in the fiber bending stiffness. This prediction is definitely
supported by the experimental data presented in Table 1.6. The highest strength is demonstrated by
composites reinforced with boron fibers that have relatively high diameter and high modulus, providing very high fiber bending stiffness. Carbon fibers, also having high modulus but smaller diameter
than boron fibers, provide compressive strength that is 40% lower than that of boron composites, but
is twice the strength of a composite reinforced with glass fibers having the same diameter but lower
modulus. The lowest strength in compression is demonstrated by composites with aramid fibers. As
was already noted, these fibers, although having high tensile stiffness, consist of a system of poorly
bonded thin filaments and possess low bending stiffness. As can be seen in Eq. (1.123), compressive
strength also increases with an increase in the matrix stiffness. The available experimental results
(Woolstencroft, Haresceugh, & Curtis, 1982; Crasto & Kim, 1993) show that the strength of carbon
composites in compression increases linearly, while the matrix shear modulus rises up to Gm 5 1500
MPa, which is the value typical for epoxy resins. For higher values of Gm , the compressive strength
1.4 MECHANICAL PROPERTIES OF A PLY UNDER TENSION
61
FIGURE 1.64
Failure mode of a unidirectional carbonepoxy composite under longitudinal compression.
does not change, and we can expect that there exists some maximum value of Gm , beyond which the
matrix does not allow fibers to buckle, and the material strength is controlled by the fiber strength in
compression. Results listed in Table 1.5 support this conclusion. As can be seen, changing an epoxy
matrix for an aluminum one with higher stiffness does not increase the compressive strength of boron
fiber composites. Moreover, by increasing the matrix stiffness, we usually reduce its ultimate elongation. As a result, the material can fail under relatively low stress because of delamination (see
Fig. 1.57). An example of such a material can also be found in Table 1.5. Carboncarbon unidirectional composites with brittle carbon matrix possessing very high stiffness demonstrate very low
strength under longitudinal compression.
Fracture of actual unidirectional composites usually occurs as a result of interaction of the fracture modes discussed above. Such fracture is shown in Fig. 1.64. The ultimate stress depends on
material structural and manufacturing parameters, has considerable scatter, and can hardly be predicted theoretically. For example, the compressive strength of composites with the same fibers and
matrices having the same stiffness but different nature (thermoset or thermoplastic) can be different
(Crasto & Kim, 1993).
The strength of composites under longitudinal compression is determined experimentally using
ring or flat specimens and special methods to prevent the specimen buckling (Tarnopol’skii &
Kincis, 1985). The most accurate results are provided by compression of sandwich specimens with
composite facings made from the material under study (Crasto & Kim, 1993).
1.4.5 TRANSVERSE COMPRESSION
Under compression across the fibers, unidirectional composites exhibit conventional shear mode of
fracture of the type shown in Fig. 1.65. The transverse compression strength is higher than in-plane
62
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
σ2
σ2
FIGURE 1.65
Failure under transverse compression.
l/2
P
b
l/2
h
FIGURE 1.66
Three-point bending of a laminated composite beam with a rectangular cross section.
shear strength (see Table 1.5) due to two main reasons. First, the area of the oblique failure plane
is larger than the area of the orthogonal longitudinal ply cross section in which the ply fails under
in-plane shear. Second, additional compression across the oblique failure plane (see Fig. 1.65)
increases the shear strength. Strength under transverse compression is measured using flat or tubular specimens as shown in Figs. 1.52 and 1.53.
1.5 INTERLAMINAR STIFFNESS AND STRENGTH
Composite structures are usually fabricated by placing plies one by one on top of the layup. The
resulting laminated structure is normally compacted applying pressure to its surface. For these reasons, the stiffness and strength of a unidirectional composite material under interlaminar shear can
differ from the corresponding in-plane characteristics of a ply discussed in Section 1.4.3. The interlaminar shear stiffness and strengths are usually determined using the so-called three-point bending
test shown in Fig. 1.66. The maximum deflection of the beam can be found using the results presented in Chapter 6, Laminated Composite Beams and Columns, of this book concerning beams
and has the following form:
w5
Pl3
D
1 1 12 2
Sl
48D
(1.124)
Here, D and S are the bending and the transverse shear stiffnesses of a beam. For a beam with a
rectangular cross section (see Fig. 1.66), we have
D5
1
Ebh3 ; S 5 kGbh
12
(1.125)
Using the three-point bending test, we can determine the material modulus E and the shear
modulus G. The parameter k in the second equation of Eq. (1.125) is the so-called shear correction
factor discussed in Section 6.1. This factor takes into account a nonuniform distribution of shear
1.5 INTERLAMINAR STIFFNESS AND STRENGTH
63
stress over the beam height and is usually taken in accordance with the following condition:
5=6 # k # 1. Substituting Eq. (1.125) for D and S in Eq. (1.124), we get
w5
1 Pl3
Pl
1
4b Eh3
kGh
(1.126)
Introduce normalized variables as
l
λ5 ;
h
w5
4bw
Pλ
w5
λ2
1
1
kG
E
and reduce Eq. (1.126) to
(1.127)
Testing two beams with parameters Pi , li , and bi (i 5 1; 2Þ; and measuring the beam deflections
wi (i 5 1; 2Þ; we arrive at two equations analogous to Eq. (1.127) written in terms of E and G.
Solving these equations for E and G, we get
E5
λ21 2 λ22
;
w1 2 w2
kG 5
λ22 2 λ21
2
λ2 w 1 2 λ21 w 2
(1.128)
To obtain reliable results, we should test the beams with different values of the parameter λ
(e.g., one beam could have λ1 equal to about 20 whereas another beam could have λ2 equal to
about 10). Nevertheless, even in this case the accuracy of the measured beam dimensions and
deflections must be rather high because of the subtraction of close numbers in Eq. (1.128).
The test can be simplified if we determine the material modulus E using the tensile test described
in Section 1.4.1. In this case, the shear modulus can be found using Eq. (1.127). It should be taken
into account that parameter λ in this test must not be too high (the value normally approximately
ranges from 5 to 10). Three-point bending test can also be used to determine the interlaminar shear
strength τ . For a beam with rectangular cross section, the maximum shear stress can be calculated as
τ5
3P
4bh
where P is the load that causes beam delamination. To increase the test accuracy, we can use a
beam with a cross section having side cutouts as shown in Fig. 1.67. For such a beam, we get
d2
b0
1
2
3P
3h2
b
τ5
4bh
d3
b0
12 3 12
4h
b
12
The multiplicand in this equation is usually close to unity so the following approximate equation which does not depend on the shape of the cutout can be used:
τD
3P
4b0 h
This test allows us to control the location of delamination and to increase the value of τ by
10%—20% in comparison with the test shown in Fig. 1.66 as demonstrated by Mikhailov,
Gunyaev, Ivanova, and Kuznetsova (1985).
64
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
b
h
d
b0
FIGURE 1.67
Cross section of a beam with cutouts.
1.6 HYBRID COMPOSITES
The foregoing sections of this chapter are concerned with the properties of unidirectional plies reinforced with fibers of a certain type: glass, carbon, aramid, etc. In hybrid composites, the plies can
include fibers of two or maybe more types, e.g., carbon and glass, glass and aramid, and so on.
Hybrid composites provide greater opportunities to control material stiffness, strength, and cost. A
promising application of these materials is associated with the so-called thermostable structures,
which do not change their dimensions under heating or cooling. For some composites, e.g., with glass
of boron fibers, the longitudinal coefficient of thermal expansion is positive, whereas for other materials, e.g., with carbon or aramid fibers, it is negative (see Table 5.1 and Section 5.1.2 of Chapter 5:
Environmental, Special Loading, and Manufacturing Effects). So, the appropriate combination of
fibers with positive and negative coefficients can result in material with zero thermal expansion.
Consider the problem of micromechanics for a unidirectional ply reinforced with two types of
fibers. Normally, the stiffness of these fibers would be different, and we assume that Efð1Þ . Efð2Þ .
The first-order model of the ply that generalizes the model in Fig. 1.34 is presented in Fig. 1.68.
For tension in the fiber direction, the apparent stress and strain, σ1 and ε1 , are linked by Hooke’s
law
σ1 5 E1 ε1
(1.129)
in which the effective modulus is specified by the following equation, generalizing Eq. (1.76)
ð2Þ ð2Þ
E1 5 Efð1Þ vð1Þ
f 1 Ef vf 1 Em vm
(1.130)
ð2Þ
Here, vð1Þ
f and vf are volume fractions of the fibers of the first and second type, and vm is the
matrix volume fraction, so that
ð2Þ
vð1Þ
f 1 vf 1 vm 5 1
We also introduce the total volume fraction of the fibers
ð2Þ
vf 5 vð1Þ
f 1 vf
1.6 HYBRID COMPOSITES
65
σ1
3
2
1
σ1
FIGURE 1.68
A first-order microstructural model of a hybrid unidirectional ply.
and normalized volume fractions of fibers as
wð1Þ
f 5
vð1Þ
f
vf
;
wð2Þ
f 5
vfð2Þ
vf
Obviously,
ð2Þ
wð1Þ
f 1 wf 5 1
Then, Eq. (1.130) can be written in the form
ð2Þ
ð1Þ
E1 5 vf ½Efð1Þ wð1Þ
f 1 Ef ð1 2 wf Þ 1 Em ð1 2 vf Þ
(1.131)
The linear dependence of E1 on wð1Þ
predicted by Eq. (1.131) is in good correlation with the
f
experimental data reported by Zabolotskii and Varshavskii (1984), and is presented in Fig. 1.69.
Since the fibers of hybrid composites have different stiffnesses, they are characterized as a rule
with different ultimate elongations. As follows from Fig. 1.70, plotted using the data listed in
Table 1.5, there exists an inverse linear dependence between the ply longitudinal modulus and the
ð2Þ
ultimate elongation ε1 . So, assuming Efð1Þ . Efð2Þ , we should take into account that εð1Þ
f , εf . This
ð1Þ
ð1Þ
means that Eq. (1.129) is valid for ε1 # εf and the strain ε1 5 εf is accompanied by failure of
fibers of the first type. The corresponding part of a possible stressstrain diagram is shown in
ð1Þ
Fig. 1.71 with the line OA. The stress at point A is σð1Þ
1 5 E1 εf . After the fibers of the first type
fail, the material modulus reduces to
E1 5 Efð2Þ vf ð1 2 wð1Þ
f Þ 1 Em ð1 2 vf Þ
This modulus determines the slope of line OC in Fig. 1.71.
Since E1 , E1 , the ply experiences a jump in strain under constant stress σ1 5 σð1Þ
1 . As follows
from Fig. 1.71, the final strain is
ε1 5
σð1Þ
1
E1
There are two possible scenarios for the further material behavior, depending on the relation
ð2Þ
between strain ε1 and the ultimate strain of the fibers of the second type, εð2Þ
f . If ε1 $ εf , these
ð1Þ
fibers will also fail under stress σ1 , and the material stressstrain diagram corresponds to the
66
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
E1 , GPa
300
250
200
1
150
2
3
100
50
4
5
6
wf(1)
0
0
0.2
0.4
0.6
0.8
1
FIGURE 1.69
Experimental dependencies of longitudinal modulus on the volume fraction of the higher modulus fibers in hybrid
unidirectional composites.
1 - boroncarbon, 2 - boronaramid, 3 - boronglass,
4 - carbonaramid, 5 - carbonglass, 6 - aramidglass.
E1 , GPa
250
Boron
200
150
Carbon
Aramid
100
50
Glass
ε 1,%
0
0
1
2
3
FIGURE 1.70
Longitudinal modulus versus ultimate tensile strain for advanced epoxy unidirectional composites.
1.6 HYBRID COMPOSITES
67
σ1
C
σ 1(1)
0
A
B
ε f(1)
ε1∗
ε1
ε f(2)
FIGURE 1.71
Typical stressstrain diagrams for hybrid unidirectional composites.
σ 1 , MPa
6
1200
5
800
1 2
4
3
400
ε1 ,%
0
0
1
2
3
FIGURE 1.72
Experimental stressstrain diagrams for hybrid carbonglass epoxy unidirectional composite with various
volume fraction of glass fibers vg and carbon fibers vc.
1 2 vg 5 0;
4 2 vg 5 0:25;
2 2 vg 5 0:07;
5 2 vg 5 0:5;
3 2 vg 5 0:14;
6 2 vc 5 0:
dashed line OA in Fig. 1.71. If εð2Þ
f . ε1 , the material would retain integrity to point C in this figure.
Experimental diagrams supporting this prediction are shown in Fig. 1.72 (Gunyaev, 1981).
The threshold value of wð2Þ
f , indicating the minimum amount of the fibers of second type that is
sufficient to withstand the load after the failure of the first type fibers, can be found from the condition ε1 5 εð2Þ
f (Skudra et al., 1989). The final result is as follows
wð2Þ
f 5
ð1Þ
Efð1Þ vf εfð1Þ 2 ð1 2 vf ÞEm ðεð2Þ
f 2 εf Þ
ð2Þ ð2Þ
ð1Þ
vf ½Efð1Þ εð1Þ
f 1 Ef ðεf 2 εf Þ
ð2Þ
ð1Þ
ð2Þ
ð2Þ
For wð2Þ
f , wf , material strength can be calculated as σ1 5 E1 εf , whereas for wf . wf ,
ð2Þ
σ1 5 E1 εf . The corresponding theoretical prediction of the dependence of material strength on wð2Þ
f
is shown in Fig. 1.73 (Skudra et al., 1989).
68
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
σ 1 , MPa
1200
800
400
w (2)
f
0
0
0.2
0.4
w (2)
f
0.8
1
FIGURE 1.73
Dependence of the longitudinal strength of unidirectional carbonglass epoxy composite on the volume fraction
of glass fibers.
1.7 COMPOSITES WITH HIGH FIBER FRACTION
We now return to Fig. 1.44, which shows the dependence of the tensile longitudinal strength of unidirectional composites on the fiber volume fraction vf . As follows from this figure, the strength
increases up to vf , which is close to 0.7 and becomes lower for higher fiber volume fractions. This
is a typical feature of unidirectional fibrous composites (Andreevskaya, 1966). However, there are
some experimental results (e.g., Roginskii & Egorov, 1966) showing that the material strength can
increase up to vf 5 0:88, which corresponds to the maximum theoretical fiber volume fraction discussed in Section 1.1. The reason that the material strength usually starts to decrease at higher fiber
volume fractions is associated with material porosity, which becomes significant for materials with
a shortage of resin. By reducing the material porosity, we can increase material tensile strength for
high fiber volume fractions.
Moreover, applying the correct combination of compacting pressure and temperature to composites with organic (aramid or polyethylene) fibers, we can deform the fiber cross sections and reach
a value of vf that would be close to unity. Such composite materials studied by Golovkin (1985),
Kharchenko (1999), and other researchers are referred to as composites with high fiber fraction
(CHFF). The cross section of a typical CHFF is shown in Fig. 1.74.
The properties of aramidepoxy CHFF are listed in Table 1.7 (Kharchenko, 1999). Comparing
traditional composites (vf 5 0:65) with CHFF, we can conclude that CHFF have significantly higher
longitudinal moduli (up to 50%) and longitudinal tensile strength (up to 30%), whereas the density
is only 6% higher. However, the transverse and shear strengths of CHFF are lower than those of
traditional composites. As a result of this, CHFF can be efficient in composite structures whose
loading induces high tensile stresses acting mainly along the fibers, e.g., in cables, pressure vessels,
etc.
1.7 COMPOSITES WITH HIGH FIBER FRACTION
FIGURE 1.74
Cross section of aramidepoxy composite with high fiber fraction:
(A) initial structure,
(B) structure with delaminated fibers.
Table 1.7 Properties of AramidEpoxy Composites with High Fiber Fraction
Fiber Volume Fraction, vf
Property
Density, ρ (g/cm )
Longitudinal modulus, E1 (GPa)
Transverse modulus, E2 (GPa)
Shear modulus, G12 (GPa)
Longitudinal tensile strength, σ1
1 (MPa)
Longitudinal compressive strength, σ2
1 (MPa)
Transverse tensile strength, σ1
2 (MPa)
Transverse compressive strength, σ2
2 (MPa)
In-plane shear strength,τ 12 (MPa)
3
0.65
0.92
0.96
1.33
85
3.3
1.6
2200
293
22
118
41
1.38
118
2.1
1.7
2800
295
12
48
28
1.41
127
4.5
2800
310
18
69
70
CHAPTER 1 MECHANICS OF A UNIDIRECTIONAL PLY
1.8 LIMITATIONS OF THE PHENOMENOLOGICAL HOMOGENEOUS MODEL
OF A PLY
It follows from the foregoing discussion that micromechanical analysis provides rather approximate
predictions for the ply stiffness and only qualitative information concerning the ply strength.
However, the design and analysis of composite structures requires reasonably accurate and reliable
information about the properties of the ply as the basic element of composite structures. This information is provided by experimental methods as discussed above. As a result, the ply is presented as
an orthotropic homogeneous material possessing some apparent (effective) mechanical characteristics determined experimentally. This means that, on the ply level, we use a phenomenological
model of a composite material that ignores its actual microstructure.
It should be emphasized that this model, being quite natural and realistic for the majority of
applications, sometimes does not allow us to predict actual material behavior. To demonstrate this,
consider a problem of biaxial compression of a unidirectional composite in the 23-plane as shown
in Fig. 1.75. Testing a glassepoxy composite material described by Koltunov, Pleshkov,
Kanovich, Roginskii, and Natrusov (1977) shows a surprising result—its strength is about
σ 5 1200MPa which is quite close to the level of material strength under longitudinal tension, and
material failure is accompanied by fiber breakage typical for longitudinal tension.
The phenomenological model fails to predict this mode of failure. Indeed, the average stress in
the longitudinal direction specified by Eq. (1.75) is equal to zero under loading (shown in
Fig. 1.75), i.e.,
σ1 5 σf1 vf 1 σm
1 vm 5 0
(1.132)
To apply the first-order micromechanical model considered in Section 1.3, we generalize constitutive equations, Eq. (1.63), for the three-dimensional stress state of the fibers and the matrix as
εf1; m 5
i
1 h f; m
σ1 2 ν f ; m ðσf2; m 1 σ3f ; m Þ ð1; 2; 3Þ
Ef ; m
(1.133)
Changing 1 for 2, 2 for 3, and 3 for 1, we can write the corresponding equations for ε2 and ε3 .
σ
1
3
σ
σ
FIGURE 1.75
Biaxial compression of a unidirectional composite.
2
σ
REFERENCES
71
Suppose that the stresses acting in the fibers and in the matrix in the plane of loading are the
same, i.e.,
m
σf2 5 σf3 5 σm
2 5 σ3 5 2 σ
(1.134)
f
m
and that εf1 5 εm
1 . Substituting ε1 and ε1 from Eq. (1.133), we get (taking into account Eq. (1.134))
1 f
1 m
ðσ 1 2ν f σÞ 5
ðσ 1 2ν m σÞ
Ef 1
Em 1
In conjunction with Eq. (1.132), this equation allows us to find σf1 , which has the form
σf1 5
2σðEf ν m 2 Em ν f Þvm
Ef vf 1 Em vm
Simplifying this result for the situation Ef . . Em , we arrive at
σf1 5 2σ
ν m vm
vf
Thus, the loading shown in Fig. 1.75 indeed induces tension in the fibers as can be revealed
using the micromechanical model. The ultimate stress can be expressed in terms of the fibers’
strength σf as
σ5
1
vf
σf
2 ν m vm
The actual material strength is not as high as follows from this equation, which is derived using
the condition that the adhesive strength between the fibers and the matrix is infinitely high.
Tension in the fibers is induced by the matrix which expands in the 1 direction (see Fig. 1.75) due
to Poisson’s effect and interacts with fibers through shear stresses whose maximum value is limited
by the fibermatrix adhesion strength. Under high shear stress, debonding of the fibers can occur,
reducing the material strength, which is, nevertheless, very high. This effect is utilized in composite
shells with radial reinforcement designed to withstand an external pressure of high intensity
(Koltunov et al., 1977).
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tensile strength of fiber composite materials. Mechanics of Composite Materials and Structures, 7(1),
105122.
Andreevskaya, G. D. (1966). High-strength Oriented Fiberglass Plastics. Moscow (in Russian): Nauka.
Bogdanovich, A. E., & Pastore, C. M. (1996). Mechanics of Textile and Laminated Composites. London:
Chapman & Hall.
Chiao, T.T. (1979). Some interesting mechanical behaviors of fiber composite materials. In G. C. Sih, & V.P.
Tamuzh (Eds.), Proc. of 1st USA-USSR Symposium on Fracture of Composite Materials, Riga, USSR,
47 September, 1978 (pp. 385392). Alphen aan den Rijn: Sijthoff and Noordhoff.
Crasto, A.S. & Kim, R.Y. (1993). An improved test specimen to determine composite compression strength. In
Proc. 9th Int. Conf. on Composite Materials (ICCM/9), Madrid, 1216 July 1993, Vol. 6, Composite
Properties and Applications (pp. 621630). Woodhead Publishing Ltd.
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Fukuda, H., Miyazawa, T., & Tomatsu, H. (1993). Strength distribution of monofilaments used for advanced
composites. In Proc. 9th Int. Conf. on Composite Materials (ICCM/9), Madrid, 1216 July 1993, Vol. 6,
Composite Properties and Applications (pp. 687694). Woodhead Publishing Ltd.
Gilman, J. J. (1959). Cleavage, Ductility and Tenacity in Crystals. In Fracture, New York: Wiley.
Golovkin, G. S. (1985). Manufacturing parameters of the formation process for ultimately reinforced organic
plastics. Plastics, 4, 3133. (in Russian).
Goodey, W. J. (1946). Stress diffusion problems. Aircraft Engineering, June 1946, 195-198; July 1946, 227234; August 1946, 271-276; September 1946, 313-316; October 1946, 343-346; November 1946, 385-389.
Griffith, A.A. (1921). The phenomenon of rupture and flow in solids. Philosophical Transactions of the Royal
Society. Series A221, 163198.
Gunyaev, G. M. (1981). Structure and properties of polymeric fibrous composites. Moscow (in Russian):
Khimia.
Hashin, Z., & Rosen, B. W. (1964). The elastic moduli of fiber reinforced materials. Journal of Applied
Mechanics., 31E, 223232.
Jones, R. M. (1999). Mechanics of composite materials (2nd ed). Philadelphia, PA: Taylor & Francis.
Kharchenko, E.F. (1999). High-strength Ultimately Reinforced Organic Plastics. Moscow (in Russian).
Koltunov, M. A., Pleshkov, L. V., Kanovich, M. Z., Roginskii, S. L., & Natrusov, V. I. (1977). High-strength
glass-reinforced plastic shells with radial orientation of the reinforcement. Polymer Mechanics/Mechanics
of Composite Materials, 13(6), 928930.
Kondo, K., & Aoki, T. (1982). Longitudinal shear modulus of unidirectional composites. In Kawata Hayashi,
& Umeka. (Eds.), Proc. 4th Int. Conf. on Composite Materials (ICCM-IV), Vol. 1, Progr. in Sci. and Eng.
of Composites (pp. 357364), Tokyo, 1982.
Lagace, P. A. (1985). Nonlinear stressstrain behavior of graphite/epoxy laminates. AIAA J, 223(10),
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Lee, D.J., Jeong, T.H., & Kim, H.G. (1995). Effective longitudinal shear modulus of unidirectional composites.
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Technology Reviews, Composite Materials, Part 2, Moscow.
CHAPTER
MECHANICS OF A COMPOSITE
LAYER
2
A typical composite laminate consists of individual layers (see Fig. 2.1), which are usually made of
unidirectional plies with the same or regularly alternating orientations. A layer can also be made
from metal, thermosetting or thermoplastic polymer, or fabric, or can have a spatial threedimensionally reinforced structure. In contrast to a ply as considered in Chapter 1, Mechanics of a
Unidirectional Ply, a layer is generally referred to the global coordinate frame x, y, and z of the
structural element rather than to coordinates 1, 2, and 3 associated with the ply orientation.
Usually, a layer is much thicker than a ply and has a more complicated structure, but this structure
does not change through its thickness, or this change is ignored. Thus, a layer can be defined as a
three-dimensional structural element that is uniform in the transverse (normal to the layer plane)
direction.
FIGURE 2.1
Laminated structure of a composite pipe.
Advanced Mechanics of Composite Materials and Structures. DOI: https://doi.org/10.1016/B978-0-08-102209-2.00002-5
© 2018 Elsevier Ltd. All rights reserved.
75
2.1 ISOTROPIC LAYER
The simplest layer that can be observed in composite laminates is an isotropic layer of metal or thermoplastic polymer that is used to protect the composite material (Fig. 2.2) and to provide tightness. For
example, filament-wound composite pressure vessels usually have a sealing metal (Fig. 2.3) or thermoplastic (Fig. 2.4) internal liner, which can also be used as a mandrel for winding. Since the layer is isotropic, we need only one coordinate system such as the global coordinate frame shown in Fig. 2.5.
2.1.1 LINEAR ELASTIC MODEL
The explicit form of Hooke’s law for an isotropic material can be written as
1
τ xy
ðσx 2 νσy 2 νσz Þ; γ xy 5
E
G
1
τ xz
εy 5 ðσy 2 νσx 2 νσz Þ; γ xz 5
G
E
1
τ yz
εz 5 ðσz 2 νσx 2 νσy Þ; γ yz 5
;
G
E
εx 5
FIGURE 2.2
Composite drive shaft with external metal protection layer.
(2.1)
2.1 ISOTROPIC LAYER
77
FIGURE 2.3
Aluminum liner for a composite pressure vessel.
where E is the modulus of elasticity, ν is Poisson’s ratio, and G is the shear modulus, which can be
expressed in terms of E and ν as G 5 E=2ð1 1 νÞ: Adding equations in Eq. (2.1) for normal strains,
we get
ε0 5
1
σ0
K
(2.2)
where
ε0 5 εx 1 εy 1 εz
(2.3)
is the volume deformation (dilatation). For small strains, the volume dV1 of an infinitesimal material element after deformation can be found knowing the volume dV before the deformation and ε0
as
dV1 5 ð1 1 ε0 ÞdV
Volume deformation is related to the mean stress
1
σ0 5 ðσx 1 σy 1 σz Þ
3
(2.4)
through the volume or bulk modulus
K5
E
:
3ð1 2 2νÞ
(2.5)
It follows from Eq. (2.5) that for ν 5 1=2 we have K-N. Then, Eq. (2.2) yields ε0 5 0 and
dV1 5 dV for any stress. Such materials are called incompressible—they do not change their volume under deformation and can only change their shapes.
The foregoing equations correspond to the general three-dimensional stress state of a layer.
However, working as a structural element of a thin-walled composite laminate, a layer is usually
loaded with a system of stresses one of which, namely, transverse normal stress σz , is much less
78
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
FIGURE 2.4
Thermoplastic liners for composite pressure vessels.
than the other stresses. Bearing this in mind, we can neglect the terms in Eq. (2.1) that include σz
and write these equations in a simplified form
1
1
ðσx 2 νσy Þ; εy 5 ðσy 2 νσx Þ
E
E
τ xy
τ xz
τ yz
; γ xz 5
; γ yz 5
γ xy 5
G
G
G
(2.6)
σx 5 Eðεx 1 νεy Þ; σy 5 Eðεy 1 νεx Þ
(2.7)
εx 5
or
2.1 ISOTROPIC LAYER
79
z
x
σz
τyz
y
τxz
σx
τxy
τyz
σy
τxz
τxy
FIGURE 2.5
An isotropic layer.
τ xy 5 Gγ xy ; τ xz 5 Gγ xz ; τ yz 5 Gγ yz
where E 5 E=ð1 2 ν Þ.
2
2.1.2 NONLINEAR MODELS
Materials of metal and polymeric layers considered in this section demonstrate linear response only
under moderate stresses. Further loading results in nonlinear behavior, which to describe adequately
we need to apply one of the nonlinear material models.
A relatively simple nonlinear constitutive theory suitable for polymeric layers can be constructed using a nonlinear elastic material model. In the strictest sense, this model can be applied to
materials whose stressstrain curves are the same for active loading and unloading. However, normally structural analysis is undertaken only for active loading. If unloading is not considered, an
elastic model can be formally used for materials that are not perfectly elastic.
There exist a number of models developed to describe the nonlinear behavior of highly deformable elastomers such as rubber (Green & Adkins, 1960). Polymeric materials used to form isotropic
layers of composite laminates allowing, in principal, high strains usually do not demonstrate them
in composite structures whose deformation is governed by fibers with relatively low ultimate elongation (1%3%). So, creating this model, we can restrict ourselves to the case of small strains, i.e.,
to materials whose typical stressstrain diagram is shown in Fig. 2.6.
A standard approach is to apply the energy method allowing us to express the stresses as the partial derivatives of the density of the strain energy (elastic potential) U with respect to strains (we use
tensor notations for stresses and strains and the rule of summation over repeated subscripts), i.e.,
σij 5
@U
; dU 5 σij dεij
@εij
(2.8)
Approximation of the elastic potential U as a function of εij with some unknown parameters
allows us to write constitutive equations directly using the first equation of Eq. (2.8). However, the
80
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
σ , MPa
50
40
30
20
10
0
0
0.5
1
1.5
2
2.5
ε, %
FIGURE 2.6
A typical stressstrain diagram (circles) for a polymeric film and its cubic approximation (solid line).
polynomial approximation, which is the most simple and natural, results in a constitutive equation
of the type σ 5 Sεn , in which S is some stiffness coefficient and n is an integer. As can be seen in
Fig. 2.7, the resulting stressstrain curve is not typical for the materials under study. A better
agreement with nonlinear experimental diagrams presented, e.g., in Fig. 2.6 is demonstrated by the
curve specified by the equation ε 5 Cσn , in which C is some compliance coefficient. To arrive at
this form of a constitutive equation, we need to have relationships similar to Eq. (2.8), but allowing
us to express strains in terms of stresses. To derive them, we introduce the complementary elastic
potential Uc in accordance with the following equation:
dUc 5 εij dσij
(2.9)
The term “complementary” becomes clear if we consider a stressstrain curve shown in
Fig. 2.8. The area 0BC below the curve represents U, whereas the area 0AC above the curve is
equal to Uc . It can be shown that, dU in Eq. (2.8) is an exact differential. To prove the same for
dUc , consider the following sum:
dU 1 dUc 5 σij dεij 1 εij dσij 5 dðσij εij Þ
which is obviously an exact differential. Since dU in this sum is also an exact differential, dUc
should have the same property and can be expressed as
dUc 5
@Uc
dσij
@σij
2.1 ISOTROPIC LAYER
σ
81
σ = Sε n
ε = Cσ n
ε
FIGURE 2.7
Two forms of approximation of the stressstrain curve.
σ
A
B
Uc
dσ
U
σ
ε
C
0
ε
dε
FIGURE 2.8
Geometric interpretation of elastic potential, U, and complementary potential, Uc.
Comparing this result with Eq. (2.9), we arrive at
εij 5
@Uc
@σij
(2.10)
which are valid for any elastic solid (for a linear elastic solid, Uc 5 U).
The complementary potential, Uc , in general, depends on stresses. For an isotropic material,
Eq. (2.10) should yield invariant constitutive equations that do not depend on the directions of
82
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
the coordinate axes. This means that Uc should depend on stress invariants I1 , I2 , and I3
which are
I1 5 σ11 1 σ22 1 σ33
I2 5 2 σ11 σ22 2 σ11 σ33 2 σ22 σ33 1 σ212 1 σ213 1 σ223
I3 5 σ11 σ22 σ33 1 2σ12 σ13 σ23 2 σ11 σ223 2 σ22 σ213 2 σ33 σ212
Using different approximations for the function Uc ðI1 ; I2 ; I3 Þ, we can construct different classes
of nonlinear elastic models. Existing experimental verification of such models shows that the
dependence of Uc on I3 can be neglected. Thus, we can present the complementary potential in a
simplified form Uc ðI1 ; I2 Þ and expand this function as a Taylor series as follows:
1
1
1
Uc 5 c0 1 c11 I1 1 c12 I12 1 c13 I13 1 c14 I14 1 . . .
2
3!
4!
1
1
1
2
3
1c21 I2 1 c22 I2 1 c23 I2 1 c24 I24 1 . . .
2
3!
4!
1
1
1
1 c1121 I1 I2 1 c1221 I12 I2 1 c1122 I1 I22 1 . . .
2
3!
3!
1
1
1
1 c1321 I13 I2 1 c1222 I12 I22 1 c1123 I1 I23 1 . . .
4!
4!
4!
where
@n Uc cin 5
@ Iin σij 50
@n1m Uc ; cinjm 5 n m @ Ii @ Ij (2.11)
σij 50
Constitutive equations following from Eq. (2.10) can be written in the form
εij 5
@Uc @I1
@Uc @I2
1
@I1 @σij
@I2 @σij
(2.12)
Assuming that for zero stresses Uc 5 0 and εij 5 0 we should take c0 5 0 and c11 5 0 in
Eq. (2.11).
Consider a plane stress state with stresses σx , σy , and τ xy . The stress invariants to be substituted
in Eq. (2.12) are
I1 5 σx 1 σy ; I2 5 2 σx σy 1 τ 2xy
(2.13)
A linear elastic material model is described by Eq. (2.11) if we take
1
Uc 5 c12 I12 1 c21 I2
2
Using Eqs. (2.12)(2.14) and engineering notations for stresses and strains, we arrive at
εx 5 c12 ðσx 1 σy Þ 2 c21 σy ; εy 5 c12 ðσx 1 σy Þ 2 c21 σx ;
γ xy 5 2c21 τ xy
These equations coincide with the corresponding equations in Eq. (2.6) if we take
c12 5
1
11ν
; c21 5
E
E
(2.14)
2.1 ISOTROPIC LAYER
83
To describe a nonlinear stressstrain diagram of the type shown in Fig. 2.6, we can generalize
Eq. (2.14) as
1
1
1
Uc 5 c12 I12 1 c21 I2 1 c14 I14 1 c22 I22
2
4!
2
Then, Eq. (2.12) yields the following cubic constitutive law
1
εx 5 c12 ðσx 1 σy Þ 2 c21 σy 1 c14 ðσx 1σy Þ3 1 c22 ðσx σy 2 τ 2xy Þσy
6
1
εy 5 c12 ðσx 1 σy Þ 2 c21 σx 1 c14 ðσx 1σy Þ3 1 c22 ðσx σy 2 τ 2xy Þσx
6
γ xy 5 2 ½c21 2 c22 ðσx σy 2 τ 2xy Þτ xy
The corresponding approximation is shown in Fig. 2.6 with a solid line. Retaining more higherorder terms in Eq. (2.11), we can describe the nonlinear behavior of any isotropic polymeric material.
To describe the nonlinear elasticplastic behavior of metal layers, we should use constitutive
equations from the theory of plasticity. There exist two basic versions of this theory—deformation
theory and flow theory which are briefly described below.
According to the deformation theory of plasticity, the strains are decomposed into two components—elastic strains (with superscript “e”) and plastic strains (superscript “p”), i.e.,
εij 5 εeij 1 εpij
(2.15)
We again use the tensor notations of strains and stresses (i.e., εij and σij ). Elastic strains are
related to stresses by Hooke’s law, Eq. (2.1), which can be written with the aid of Eq. (2.10) in the
form
εeij 5
@Ue
@σij
(2.16)
where Ue is the elastic potential that for a linear elastic solid coincides with the complementary
potential Uc in Eq. (2.10). An explicit expression for Ue is
Ue 5
1 2
σ 1 σ222 1 σ233 2 2νðσ11 σ22 1 σ11 σ33 1 σ22 σ33 Þ
2E 11
1 2
ðσ 1 σ213 1 σ223 Þ
1
2G 12
(2.17)
Now represent the plastic strains in Eq. (2.15) in a form similar to Eq. (2.16)
εpij 5
@Up
@σij
(2.18)
where Up is the plastic potential. To approximate the dependence of Up on stresses, a special generalized stress characteristic, i.e., the so-called stress intensity σ, is introduced in the classical theory
of plasticity as
2
#12
1 4
2
2
2
2
2
2
σ 5 pffiffiffi ðσ11 2σ22 Þ 1 ðσ22 2σ33 Þ 1 ðσ11 2σ33 Þ 16ðσ12 1σ13 1σ23 Þ
2
(2.19)
84
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
Transforming Eq. (2.19) with the aid of equations for the stress invariants, we can reduce it to
the following form:
σ5
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
I12 1 3I2
This means that σ is an invariant characteristic of a stress state, i.e., that it does not depend on
the orientation of a coordinate frame. For unidirectional tension, we have only one nonzero stress,
e.g., σ11 . Then, Eq. (2.19) yields σ 5 σ11 . In a similar way, the strain intensity ε can be
introduced as
pffiffiffi "
12
2
ε5
ðε11 2ε22 Þ2 1 ðε22 2ε33 Þ2 1 ðε11 2ε33 Þ2 16ðε212 1ε213 1ε223 Þ
3
(2.20)
The strain intensity is also an invariant characteristic. For uniaxial tension, with stress σ11 and
strain ε11 in the loading direction, we have ε22 5 ε33 5 2 ν p ε11 , where ν p is the elasticplastic
Poisson’s ratio which, in general, depends on σ11 . For this case, Eq. (2.20) yields
2
ε 5 ð1 1 ν p Þε11
3
(2.21)
For an incompressible material (see Section 2.1.1), ν p 5 1=2 and ε 5 ε11 . Thus, the numerical
coefficients in Eqs. (2.19) and (2.20) provide σ 5 σ11 and ε 5 ε11 for uniaxial tension in an incompressible material. The stress and strain intensities in Eqs. (2.19) and (2.20) have an important
physical meaning. As known from experiments, metals do not demonstrate plastic properties under
loading with stresses σx 5 σy 5 σz 5 σ0 resulting only in a change of material volume. Under such
loading, materials exhibit only elastic volume deformation specified by Eq. (2.2). Plastic strains
occur in metals if we change material shape. For a linear elastic material, the elastic potential U
can be reduced after rather cumbersome transformation with the aid of Eqs. (2.3) and (2.4), and
(2.19) and (2.20) to the following form:
1
1
U 5 σ0 ε0 1 σε
2
2
(2.22)
The first term in the right-hand part of this equation is the strain energy associated with the volume change, whereas the second term corresponds to the change of material shape. Thus, σ and ε
in Eqs. (2.19) and (2.20) are stress and strain characteristics associated with the change of a material’s shape when it demonstrates plastic behavior.
In the theory of plasticity, the plastic potential Up is assumed to be a function of stress intensity
σ, and according to Eq. (2.18), the plastic strains are given by
εpij 5
dUp @σ
dσ @σij
(2.23)
Consider further a plane stress state with stresses σx , σy , and τ xy in Fig. 2.5. For this case,
Eq. (2.19) takes the form
σ5
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
σ2x 1 σ2y 2 σx σy 1 3τ 2xy
(2.24)
2.1 ISOTROPIC LAYER
85
Using Eqs. (2.15)(2.17), (2.23), and (2.24), we finally arrive at the following constitutive
equations
1
1
ðσx 2 νσy Þ 1 ωðσÞ σx 2 σy
E
2 1
1
εy 5 ðσy 2 νσx Þ 1 ωðσÞ σy 2 σx
E
2
1
γ xy 5 τ xy 1 3ωðσÞτ xy
G
εx 5
(2.25)
in which
ωðσÞ 5
1 dUp
σ dσ
(2.26)
To find ωðσÞ, we need to specify the dependence of Uc on σ. The most simple and suitable for
practical applications is the power law approximation
Up 5 Cσn
(2.27)
where C and n are experimental constants. As a result, Eq. (2.26) yields
ωðσÞ 5 Cnσn22
(2.28)
To determine the coefficients C and n, we introduce the basic assumption of plasticity theory
concerning the existence of a universal stressstrain diagram (master curve). According to this
assumption, for any particular material, there exists a relationship between stress and strain intensities, i.e., σ 5 ϕðεÞ (or ε 5 f ðσÞ), that is one and the same for all loading cases. This assumption
enables us to find coefficients C and n from a test under uniaxial tension and thus extend the results
obtained to an arbitrary state of stress.
Indeed, consider uniaxial tension with stress σ11 . For this case, σ 5 σx , and Eq. (2.25) yield
σx
1 ωðσx Þσx
E
ν
1
εy 5 2 σx 2 ωðσx Þσx
E
2
γ xy 5 0
εx 5
(2.29)
(2.30)
Solving Eq. (2.29) for ωðσx Þ, we get
ωðσx Þ 5
1
1
2
Es ðσx Þ E
(2.31)
where Es 5 σx =εx is the so-called secant modulus. Using now the assumption regarding the existence of a universal diagram for stress intensity σ and taking into account that σ 5 σx for uniaxial
tension, we can generalize Eq. (2.31) and write it for an arbitrary state of stress as
ωðσÞ 5
1
1
2
Es ðσÞ E
(2.32)
To determine Es ðσÞ 5 σ=ε, we need to plot the universal stressstrain curve. For this purpose,
we can use an experimental diagram σx ðεx Þ for the case of uniaxial tension, e.g., the one shown in
Fig. 2.9 for an aluminum alloy with a solid line. To plot the universal curve σðεÞ, we should put
86
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
σ x , σ , MPa
250
200
150
100
50
0
0
1
2
3
4
ε x , ε, %
FIGURE 2.9
Experimental stressstrain diagram for an aluminum alloy under uniaxial tension (solid line); the universal
stressstrain curve (dashed line), and its power approximation (dots).
σ 5 σx and change the scale on the strain axis in accordance with Eq. (2.21). To do this, we need
to know the plastic Poisson’s ratio ν p , which can be found as ν p 5 2 εy =εx . Using Eqs. (2.29) and
(2.30), we arrive at
νp 5
1 Es 1
2
2ν
2
E 2
It follows from this equation that ν p 5 ν if Es 5 E and ν p -1=2 for Es -0: The dependencies of
Es and ν p on ε for the aluminum alloy under consideration are presented in Fig. 2.10. With the aid
of this figure and Eq. (2.21) in which we should take ε11 5 εx , we can calculate ε and plot the
universal curve shown in Fig. 2.9 with a dashed line. As can be seen, this curve is slightly different
from that in the diagram corresponding to a uniaxial tension. For the power law approximation in
Eq. (2.27), we get from Eqs. (2.26) and (2.32) the following equations:
ωðσÞ 5 Cnσn22 ; ωðσÞ 5
ε
1
2
σ
E
Matching these results, we find
ε5
σ
1 Cnσn21
E
(2.33)
This is a traditional approximation for a material with a power hardening law. Now, we can
find C and n using Eq. (2.33) to approximate the dashed line in Fig. 2.9. The results of this
2.1 ISOTROPIC LAYER
87
νp
E
0.5
100
νp
80
0.4
60
0.3
40
0.2
0.1
20
Es
Et
0
0
1
2
3
4
0
ε x, %
FIGUER 2.10
Dependencies of the secant modulus (Es), tangent modulus (Et), and the plastic Poisson’s ratio (vp) on strain for
an aluminum alloy.
approximation are shown in this figure with dots that correspond to E 5 71:4 GPa, n 5 6, and
C 5 6:23U10215 ðMPaÞ25 .
Thus, constitutive equations of the deformation theory of plasticity are specified by Eqs. (2.25)
and (2.32). These equations are valid only for active loading that can be identified by the condition
dσ . 0. Applied for unloading (i.e., for dσ , 0), Eq. (2.25) corresponds to a nonlinear elastic material that has the same stressstrain diagrams for active loading and unloading. For an elasticplastic material, the unloading curve does not coincide with the curve for the active loading and is
linear. So, if we reduce the stresses by some decrements Δσx , Δσy , and Δτ xy , the corresponding
decrements of strains will be
1
1
ðΔσx 2 νΔσy Þ; Δεy 5 ðΔσy 2 νΔσx Þ;
E
E
1
Δγ xy 5 Δτ xy
G
Δεx 5
Direct application of the nonlinear equations (Eq. (2.25)) substantially hinders the stressstrain
analysis because these equations include the function ωðσÞ in Eq. (2.32) which, in turn, contains the
unknown secant modulus Es ðσÞ. For the power law approximation corresponding to Eq. (2.33), Es
can be expressed analytically, i.e.,
1
1
5 1 Cnσn22
Es
E
88
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
However, in many cases Es is given graphically as in Fig. 2.10 or numerically in the form of a
table. Thus, Eq. (2.25) sometimes cannot be even written in an explicit analytical form. This
implies application of numerical methods in conjunction with iterative linearization of Eq. (2.25).
There exist several methods of such linearization that will be demonstrated using the first equation in Eq. (2.25), i.e.,
εx 5
1
1
ðσx 2 νσy Þ 1 ωðσÞ σx 2 σy
E
2
(2.34)
In the method of elastic solutions (Ilyushin, 1948), Eq. (2.34) is used in the following form:
εsx 5
1 s
ðσ 2 νσsy Þ 1 ηs21
E x
(2.35)
where s is the number of the iteration step and
1 s21
σ
ηs21 5 ωðσs21 Þ σs21
2
x
2 y
For the first step (s 5 1), we take η0 5 0 and solve the problem of linear elasticity with
Eq. (2.35) in the form
ε1x 5
1 1
ðσ 2 νσ1y Þ
E x
(2.36)
Finding the stresses, we calculate η1 and write Eq. (2.35) as
ε2x 5
1 2
ðσ 2 νσ2y Þ 1 η1
E x
where the first term is linear, whereas the second term is a known nonlinear function of coordinates. Thus we have another linear problem, resolving which we find stresses, calculate η2 , and
switch to the third step. This process is continued until the strains corresponding to some step
become sufficiently close within the stipulated accuracy to the results found at the previous step.
Thus, the method of elastic solutions reduces the initial nonlinear problem to a sequence of linear problems of the theory of elasticity for the same material, but with some initial strains that can
be transformed into initial stresses or additional loads. This method readily provides a nonlinear
solution for any problem that has a linear solution, analytical or numerical. The main shortcoming
of the method is its poor convergence. Graphical interpretation of this process for the case of uniaxial tension with stress σ is presented in Fig. 2.11A. This figure shows a simple way to improve the
convergence of the process. If we need to find the strain at the point of the curve that is close to
point A, it is not necessary to start the process with initial modulus E. Taking E0 , E in Eq. (2.36)
we can reach the result with substantially fewer steps.
According to the method of elastic variables (Birger, 1951), we should present Eq. (2.34) as
εsx 5
1 s
1
ðσx 2 νσsy Þ 1 ωðσs21 Þ σsx 2 σsy
E
2
(2.37)
In contrast to Eq. (2.35), stresses σsx and σsy in the second term correspond to the current step
rather than to the previous one. This enables us to write Eq. (2.37) in a form analogous to Hooke’s
law, i.e.,
2.1 ISOTROPIC LAYER
σ
A
σ
89
A
ε
ε
(A)
(B)
σ
A
σ
A
ε
(C)
ε
(D)
FIGURE 2.11
Geometric interpretation of (A) the method of elastic solutions, (B) the method of variable elasticity parameters,
(C) Newton’s method, and (D) method of successive loading.
1
ðσs 2 ν s21 σsy Þ
Es21 x
(2.38)
21
1
ν
1
1ωðσs21 Þ ; ν s21 5 Es21
1 ωðσs21 Þ
E
E
2
(2.39)
εsx 5
where
Es21 5
are elastic variables corresponding to the step with number s 2 1. The iteration procedure is similar
to that described previously. For the first step we take E0 5 E and ν 0 5 ν in Eq. (2.38). We then
find σ1x , σ1y and σ1 , determine E1 , and ν 1 , switch to the second step, and so on. Graphical interpretation of the process is shown in Fig. 2.11B. Convergence of this method is by an order faster than
that of the method of elastic solutions. However, the elastic variables in the linear constitutive
equation of this method, Eq. (2.38), depend on stresses and hence, on coordinates whence the
method has obtained its name. This method can be efficiently applied in conjunction with the finite
element method, according to which the structure is modeled with the system of elements with constant stiffness coefficients. Being calculated for each step with the aid of Eq. (2.39), the stiffness
will change only with transition from one element to another, which clearly would not practically
hinder the calculation procedure for the finite element method.
The iteration process having the best convergence is provided by the classical Newton method,
requiring the following form of Eq. (2.34)
s
s21
εsx 5 εs21
1 cs21
x
11 ðσ x 2 σ x Þ
s21 s
s21
s
s21
1 c12 ðσy 2 σy Þ 1 cs21
13 ðτ xy 2 τ xy Þ
(2.40)
90
where
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
1
1 s21
@
1 ωðσs21 Þ 1 σs21
σ
2
ωðσs21 Þ
x
y
s21
E
2
@σ
x
ν
1
1
@
2 ωðσs21 Þ 1 σs21
2 σs21
ωðσs21 Þ
cs21
12 5 2
x
E
2
2 y
@σs21
y
1
@
s21
2 σs21
ωðσs21 Þ
cs21
13 5 σ x
2 y
@τ s21
xy
cs21
11 5
Since coefficients c are known from the previous step ðs 2 1Þ, Eq. (2.40) is linear with respect
to stresses and strains corresponding to step number s. Graphical interpretation of this method is
presented in Fig. 2.11C. In contrast to the methods discussed previously, Newton’s method has no
physical interpretation, and being characterized with very high convergence, is rather cumbersome
for practical applications.
The iteration methods discussed above are used to solve direct problems of stress analysis, i.e.,
to find stresses and strains induced by a given load. However, there exists another class of problems
requiring us to evaluate the load carrying capacity of the structure. To solve these problems, we
need to trace the evolution of stresses while the load increases from zero to some ultimate value.
To do this, we can use the method of successive loading. According to this method, the load is
applied with some increments, and for each s-step of loading the strain is determined as
εsx 5 εs21
1
x
1
ðΔσsx 2 ν s21 Δσsy Þ
Es21
(2.41)
where Es21 and ν s21 are specified by Eq. (2.39) and correspond to the previous loading step.
Graphical interpretation of this method is shown in Fig. 2.11D. To obtain reliable results, the load
increments should be as small as possible, because the error of calculation is cumulative in this
method. To avoid this effect, the method of successive loading can be used in conjunction with the
method of elastic variables. Being applied after several loading steps (black circles in Fig. 2.11D),
the latter method allows us to eliminate the accumulated error and to start the process of loading
again from a “correct” initial state (light circles in Fig. 2.11D).
Returning to the constitutive equations of the deformation theory of plasticity, Eq. (2.25), it is
important to note that these equations are algebraic. This means that the strains corresponding to
some combination of loads are determined by the stresses induced by these loads and do not
depend on the history of loading, i.e., on what happened to the material before this combination of
loads was reached.
However, existing experimental data show that, in general, strains should depend on the history
of loading. This means that constitutive equations should be differential, rather than algebraic as
they are in the deformation theory. Such equations are provided by the flow theory of plasticity.
According to this theory, decomposition in Eq. (2.15) is used for infinitesimal increments of stresses, i.e.,
dεij 5 dεeij 1 dεpij
(2.42)
Here, increments of elastic strains are related to the increments of stresses by Hooke’s law, e.g.,
for the plane stress state
2.1 ISOTROPIC LAYER
dεex 5
1
1
ðdσx 2 νdσy Þ; dεey 5 ðdσy 2 νdσx Þ;
E
E
1
dγ xy 5 dτ xy
G
91
(2.43)
whereas increments of plastic strains
dεpij 5
@Up
dλ
@σij
are expressed in the form of Eq. (2.18) but include a parameter λ which characterizes the loading
process.
Assuming that Up 5 Up ðσÞ, where σ is the stress intensity specified by Eq. (2.19) or (2.24), we get
dεpij 5
dUp @σ
dλ
dσ @σij
The explicit form of these equations for the plane stress state is
1
dεpx 5 dωðσÞ σx 2 σy
2 1
dεpy 5 dωðσÞ σy 2 σx
2
dγ pxy 5 3dωðσÞτ xy
(2.44)
where
dωðσÞ 5
1 dUp
dλ
σ dσ
(2.45)
To determine the parameter λ, assume that the plastic potential Up , being on the one hand a
function of σ, can be treated as the work performed by stresses on plastic strains, i.e.,
@Up
dσ
@σ
5 σx dεpx 1 σy dεpy 1 τ xy dγ pxy
dUp 5
Substituting strain increments from Eq. (2.44) and taking into account Eq. (2.24) for σ, we have
@Up
dσ 5 σ2 dωðσÞ
@σ
Taking into account Eq. (2.45), we arrive at the following straightforward relationship,
dλ 5 dσ=σ. Thus, Eq. (2.45) takes the form
dωðσÞ 5
dσ dUp
σ2 dσ
(2.46)
and Eqs. (2.42)(2.44) result in the following constitutive equations in the flow theory:
1
1
ðdσx 2 νdσy Þ 1 dωðσÞ σx 2 σy
E
2 1
1
dεy 5 ðdσy 2 νdσx Þ 1 dωðσÞ σy 2 σx
E
2
1
dγ xy 5 dτ xy 1 3dωðσÞτ xy
G
dεx 5
(2.47)
92
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
As can be seen, in contrast to the deformation theory, stresses govern the increments of plastic
strains rather than the strains themselves.
In the general case, irrespective of any particular approximation of plastic potential Up , we can
obtain for function dωðσÞ in Eq. (2.47) an expression similar to Eq. (2.32). Consider uniaxial tension, for which Eq. (2.47) yield
dεx 5
dσx
1 dωðσx Þσx
E
Repeating the derivation of Eq. (2.32), we finally have
dωðσÞ 5
dσ
1
1
2
σ Et ðσÞ E
(2.48)
where Et ðσÞ 5 dσ=dε is the so-called tangent modulus. The dependence of Et on strain for an aluminum alloy is shown in Fig. 2.10. Consider the power law approximation for plastic potential
Up 5 Bσn
(2.49)
Then, matching Eqs. (2.46) and (2.48), we arrive at the following equation:
dε
1
5 1 Bnσn22
dσ
E
Upon integration, we get
ε5
σ
Bn n21
1
σ
E
n21
(2.50)
As can be seen, this equation has the same form as Eq. (2.33). The only difference is in the
form of coefficients C and B. As in the theory of deformation, Eq. (2.50) can be used to approximate the experimental stressstrain curve and to determine the coefficients B and n. Thus, the constitutive equations for the flow theory of plasticity are specified by Eqs. (2.47) and (2.48).
For a plane stress state, introduce the stress space shown in Fig. 2.12 and refer it to a Cartesian
coordinate frame with stresses as coordinates. In this space, any loading can be presented as a curve
τ xy
A
0
σy
FIGURE 2.12
Loading path (0A) in the stress space.
σx
2.2 UNIDIRECTIONAL ORTHOTROPIC LAYER
93
specified by the parametric equations σx 5 σx ðpÞ; σy 5 σy ðpÞ, and τ xy 5 τ xy ðpÞ, in which p is the
loading parameter. To find strains corresponding to point A on the curve, we should integrate
Eq. (2.47) along this curve, thus taking into account the whole history of loading. In the general
case, the result obtained will be different from what follows from Eq. (2.25) of the deformation theory for point A. However, there exists one loading path (the straight line 0A in Fig. 2.12) that is
completely determined by the location of its final point A. This is the so-called proportional loading, during which the stresses increase in proportion to parameter p, i.e.,
σx 5 σ0x p; σy 5 σ0y p; τ xy 5 τ 0xy p
(2.51)
where stresses with superscript “0” can depend on coordinates only. For such loading, σ 5 σ0 p,
dσ 5 σo dp, and Eqs. (2.46) and (2.49) yield
n23
dωðσÞ 5 Bnσn23 dσ 5 Bnσn22
dp
0 p
(2.52)
Consider, e.g., the first equation of Eq. (2.47). Substituting Eqs. (2.51) and (2.52), we have
1 0
1 0 n22
0
n22
0
dεx 5 ðσx 2 νσy Þdp 1 Bnσ0
σx 2 σy p dp
E
2
This equation can be integrated with respect to p. Using again Eq. (2.51), we arrive at the constitutive equation of the deformation theory
εx 5
1
n
1
ðσx 2 νσy Þ 1 B
σn22 σx 2 σy
E
n21
2
Thus, for a proportional loading, the flow theory reduces to the deformation theory of plasticity.
Unfortunately, before the problem is solved and the stresses are found we do not know whether the
loading is proportional or not and which particular theory of plasticity should be used. There exists
a theorem of proportional loading (Ilyushin, 1948), according to which the stresses increase proportionally and the deformation theory can be used if:
1. External loads increase in proportion to one loading parameter.
2. The material is incompressible and its hardening can be described with the power law σ 5 Sεn .
In practice, both conditions of this theorem are rarely met. However, existing experience shows
that the second condition is not very important and that the deformation theory of plasticity can be
reliably (but approximately) applied if all the loads acting on the structure increase in proportion to
one parameter.
2.2 UNIDIRECTIONAL ORTHOTROPIC LAYER
A composite layer with the simplest structure consists of unidirectional plies whose material coordinates, 1, 2, and 3, coincide with coordinates of the layer, x, y, and z, as shown in Fig. 2.13. An
example of such a layer is presented in Fig. 2.14: The principal material axes of an outer circumferential unidirectional layer of a pressure vessel coincide with global (axial and circumferential) coordinates of the vessel.
94
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
z,3
x,1
σ3
τ 23
σ2
FIGURE 2.13
An orthotropic layer.
FIGURE 2.14
Filament-wound composite pressure vessel.
τ 13
τ 12
τ 23
y,2
τ13
τ 12
σ1
2.2 UNIDIRECTIONAL ORTHOTROPIC LAYER
95
2.2.1 LINEAR ELASTIC MODEL
For the layer under study, the constitutive equations are
σ1
σ2
σ3
2 ν 12
2 ν 13
E1
E2
E3
σ2
σ1
σ3
ε2 5
2 ν 21
2 ν 23
E2
E1
E3
σ3
σ1
σ2
ε3 5
2 ν 31
2 ν 32
E3
E1
E2
τ 12
τ 13
τ 23
γ 12 5
; γ 5
; γ 5
G12 13 G13 23 G23
ε1 5
(2.53)
where
ν 12 E1 5 ν 21 E2 ; ν 13 E1 5 ν 31 E3 ; ν 23 E2 5 ν 32 E3
The inverse form of Eq. (2.53) is
σ1 5 A1 ðε1 1 μ12 ε2 1 μ13 ε3 Þ
σ2 5 A2 ðε2 1 μ21 ε1 1 μ23 ε3 Þ
σ3 5 A3 ðε3 1 μ31 ε1 1 μ32 ε2 Þ
τ 12 5 G12 γ 12 ; τ 13 5 G13 γ 13 ; τ 23 5 G23 γ 23
(2.54)
where
E1
E2
E3
ð1 2 ν 23 ν 32 Þ; A2 5
ð1 2 ν 13 ν 31 Þ; A3 5
ð1 2 ν 12 ν 21 Þ
D
D
D
D 5 1 2 ν 12 ν 23 ν 31 2 ν 13 ν 21 ν 32 2 ν 13 ν 31 2 ν 12 ν 21 2 ν 23 ν 32
ν 12 1 ν 13 ν 32
ν 21 1 ν 23 ν 31
μ12 5
; μ21 5
1 2 ν 23 ν 32
1 2 ν 13 ν 31
ν 13 1 ν 12 ν 23
ν 31 1 ν 21 ν 32
μ13 5
; μ31 5
1 2 ν 23 ν 32
1 2 ν 12 ν 21
ν 23 1 ν 13 ν 21
ν 32 1 ν 12 ν 31
μ23 5
; μ32 5
1 2 ν 13 ν 31
1 2 ν 12 ν 21
A1 5
As for an isotropic layer considered in Section 2.1, the terms including the transverse normal
stress σ3 can be neglected for a thin layer in Eqs. (2.53) and (2.54), and they can be written in the
following simplified forms:
σ1
σ2
σ2
σ1
2 ν 12 ; ε2 5
2 ν 21
E1
E2
E2
E1
τ 12
τ 13
τ 23
γ 12 5
; γ 5
; γ 5
G12 13 G13 23 G23
(2.55)
σ1 5 E1 ðε1 1 ν 12 ε2 Þ; σ2 5 E2 ðε2 1 ν 21 ε1 Þ
τ 12 5 G12 γ 12 ; τ 13 5 G13 γ 13 ; τ 23 5 G23 γ 23
(2.56)
ε1 5
and
where
E1;2 5
E1;2
1 2 ν 12 ν 21
96
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
The constitutive equations presented above include elastic constants for a layer that are determined experimentally. For in-plane characteristics E1 , E2 , G12 , and ν 12 , the corresponding test
methods are discussed in Chapter 1, Mechanics of a Unidirectional Ply. The transverse modulus E3
is usually found by testing the layer under compression in the z-direction. The transverse shear
moduli G13 and G23 can be obtained by various methods (see Section 1.5) or by inducing pure
shear in two symmetric specimens shown in Fig. 2.15 and calculating the shear modulus as
G13 5 P=ð2AγÞ, where A is the in-plane area of the specimen.
For unidirectional composites, G13 5 G12 (see Table 1.5) whereas typical values of G23 are
listed in Table 2.1 (Herakovich, 1998).
Poisson’s ratios ν 31 and ν 32 can be determined by measuring the change in the layer thickness
under in-plane tension in directions 1 and 2.
2.2.2 NONLINEAR MODELS
Consider Figs. 1.401.43, showing typical stressstrain diagrams for unidirectional advanced composites. As can be seen, the materials demonstrate linear behavior only under tension. The curves
corresponding to compression are slightly nonlinear, whereas the shear curves are definitely nonlinear. It should be emphasized that this does not mean that the linear constitutive equations presented
in Section 2.2.1 are not valid for these materials. First, it should be taken into account that the
deformations of properly designed composite materials are controlled by the fibers, and they do not
allow the shear strain to reach the values at which the shear stressstrain curve is highly nonlinear.
Second, the shear stiffness is usually very small in comparison with the longitudinal one, and so is
its contribution to the apparent material stiffness. The material behavior is usually close to linear
γ
3
1
P
3
1
FIGURE 2.15
A test to determine transverse shear modulus.
Table 2.1 Transverse Shear Moduli of Unidirectional Composites (Herakovich, 1998)
Material
GlassEpoxy
CarbonEpoxy
AramidEpoxy
BoronAluminum
G23 (GPa)
4.1
3.2
1.4
49.1
2.2 UNIDIRECTIONAL ORTHOTROPIC LAYER
97
even if the shear deformation is nonlinear. Thus, a linear elastic model provides, as a rule, a reasonable approximation to the actual material behavior. However, there exist problems to in which we
need to allow for material nonlinearity and apply one of the nonlinear constitutive theories
discussed below.
First, note that material behavior under elementary loading (pure tension, compression, and
shear) is specified by experimental stressstrain diagrams of the type shown in Figs. 1.401.43,
and we do not need any theory. The necessity for a theory arises if we are to study the interaction
of simultaneously acting stresses. Since for the layer under study this interaction usually takes place
for in-plane stresses σ1 , σ2 , and τ 12 (see Fig. 2.13), we consider further the plane state of stress.
In the simplest (but quite useful for practical engineering analysis) approach, the stress interaction is ignored completely, and the linear constitutive equations, Eq. (2.55), are generalized as
ε1 5
σ1
σ2
σ2
σ1
τ 12
2 ν s12 s ; ε2 5 s 2 ν s21 s ; γ 12 5 s
E1s
E2
E2
E1
G12
(2.57)
where the superscript “s” indicates the corresponding secant modulus, which depends on stresses
and is determined using experimental diagrams similar to those presented in Figs. 1.401.43. In
particular, the diagrams σ1 ðε1 Þ and ε2 ðε1 Þ plotted under uniaxial longitudinal loading yield E1s ðσ1 Þ
and ν s21 ðσ1 Þ, and the secant moduli E2s ðσ2 Þ and Gs12 ðτ 12 Þ are determined from experimental curves
for σ2 ðε2 Þ and τ 12 ðγ 12 Þ, respectively, whereas ν s12 is found from the symmetry condition in
Eq. (2.53). In a more rigorous model (Jones, 1977), the secant characteristics of the material in
Eq. (2.57) are also functions, but in this case they are functions of strain energy U rather than of
individual stresses. Models of this type provide adequate results for unidirectional composites with
moderate nonlinearity.
To describe pronounced nonlinear elastic behavior of a unidirectional layer, we can use
Eq. (2.10). Expanding the complementary potential Uc into a Taylor series with respect to stresses,
we have
1
1
1
Uc 5 c0 1 cij σij 1 cijkl σij σkl 1 cijklmn σij σkl σmn 1 cijklmnpq σij σkl σmn σpq
2
3!
4!
1
1
1 cijklmnpqrs σij σkl σmn σpq σrs 1 cijklmnpqrstw σij σkl σmn σpq σrs σtw 1 :::
5!
6!
where
@Uc c0 5 Uc ðσij 5 0Þ; cij 5
@σij σij 50
@2 Uc ; cijkl 5
@σij @σkl (2.58)
; etc:
σij 50;σkl 50
A sixth-order approximation with the terms presented in Eq. (2.58) (where summation over
repeated subscripts is implied) allows us to construct constitutive equations including stresses to the
fifth power. The coefficients “c” should be determined from experiments with material specimens.
Since these coefficients are partial derivatives that do not depend on the sequence of differentiation,
the sequence of their subscripts is not important. As a result, the sixth-order polynomial in
Eq. (2.58) includes 84 “c”-coefficients. This is clearly far too many for the practical analysis of
composite materials. To reduce the number of coefficients, we can first use some general considerations. Namely, assume that Uc 5 0 and εij 5 0 if there are no stresses ðσij 5 0Þ. Then, c0 5 0 and
cij 5 0. Second, we should take into account that the material under study is orthotropic. This
98
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
means that normal stresses do not induce shear strain, and shear stresses do not cause normal
strains. Third, the direction of shear stresses should influence only shear strains, i.e., shear stresses
should have only even powers in constitutive equations for normal strains, whereas the corresponding equation for shear strain should include only odd powers of shear stresses. As a result, the constitutive equations will contain 37 coefficients and take the following form (in new notations for
coefficients and stresses):
ε15 a1 σ1 1 a2 σ21 1 a3 σ31 1 a4 σ41 1 a5 σ51 1 d1 σ1
1 2d2 σ1 σ2 1 d3 σ22 1 3d4 σ21 σ2 1 d5 σ32 1 d6 σ1 σ22 1 4d7 σ31 σ2 1 3d8 σ21 σ22 1 2d9 σ1 σ32
1 d10 σ42 1 5d11 σ41 σ2 1 4d12 σ31 σ22 1 3d13 σ21 σ32 1 2d14 σ1 σ42 1 d15 σ52 1 k1 σ1 τ 212 1 k2 σ2 τ 212
1 3k3 σ21 τ 212 1 4k4 σ31 τ 212 1 2k5 σ1 τ 412
ε25 b1 σ2 1 b2 σ22 1 b3 σ32 1 b4 σ42 1 b5 σ52
1 d1 σ1 1 d2 σ21 1 2d3 σ1 σ2 1 d4 σ31 1 3d5 σ1 σ22 1 d6 σ21 σ2 1 d7 σ41 1 2d8 σ31 σ2
(2.59)
1 3d9 σ21 σ22 1 4d10 σ1 σ32 1 d11 σ51 1 2d12 σ41 σ2 1 3d13 σ31 σ22 1 2d14 σ21 σ32 1 5d15 σ1 σ42
1 m1 σ2 τ 212 1 k2 σ1 τ 212 1 3m2 σ22 τ 212 1 4m3 σ32 τ 212 1 2m4 σ2 τ 412
γ 12 5 c1 τ 12 1 c2 τ 312 1 c3 τ 512 1 k1 τ 12 σ21 1 m1 τ 12 σ22
1 2k2 τ 12 σ1 σ2 1 2k3 τ 12 σ31 1 2m2 τ 12 σ32
1 2k4 τ 12 σ41 1 4k5 τ 312 σ21 1 2m3 τ 12 σ42 1 4m4 τ 312 σ22
For unidirectional composites, the dependence ε1 ðσ1 Þ is linear which means that we should put
d2 5 ::: d15 5 0, k1 5 ::: k5 5 0. Then, the foregoing equations reduce to
ε1 5 a1 σ1 1 d1 σ2
ε2 5 b1 σ2 1 b2 σ22 1 b3 σ32 1 b4 σ42 1 b5 σ52
1 d1 σ1 1 m1 σ2 τ 212 1 3m2 σ22 τ 212 1 4m3 σ32 τ 212 1 2m4 σ2 τ 412
(2.60)
γ 12 5 c1 τ 12 1 c2 τ 312 1 c3 τ 512 1 m1 τ 12 σ22
12m2 τ 12 σ32 1 2m3 τ 12 σ42 1 4m4 τ 312 σ22
As an example, consider a specific unidirectional two-matrix fiberglass composite with high inplane transverse and shear deformation (see Section 2.4.4 for further details). The stressstrain
curves corresponding to transverse tension, compression, and in-plane shear are shown in Fig. 2.16.
The solid lines are used to approximate the experimental results (circles in Fig. 2.16) with the aid
of Eq. (2.60). The coefficients a1 and d1 in Eq. (2.60) are found using diagrams ε1 ðσ1 Þ and ε2 ðσ2 Þ,
which are linear and not shown here. The coefficients b1 :::b5 and c1 , c2 , and c3 are determined
2
using the least-squares method to approximate curves σ1
2 ðε2 Þ, σ 2 ðε2 Þ, and τ 12 ðγ 12 Þ. The other coefficients, i.e., m1 :::m4 , should be determined with the aid of a more complicated experiment involving loading that induces both stresses σ2 and τ 12 acting simultaneously. This experiment is
described in Section 2.3.
As follows from Figs. 1.401.43, unidirectional composites demonstrate pronounced nonlinearity only under shear. Assuming that the dependence ε2 ðσ2 Þ is also linear, we can reduce Eq. (2.60)
to
ε1 5 a1 σ1 1 d1 σ2 ; ε2 5 b1 σ2 1 d1 σ1 ; γ 12 5 c1 τ 12 1 c2 τ 312 1 c3 τ 512
2.2 UNIDIRECTIONAL ORTHOTROPIC LAYER
99
σ 2 , τ12 , MPa
20
σ 2– ( ε 2)
16
σ 2+ ( ε 2)
12
τ 12( γ12 )
8
4
0
0
2
4
6
8
10
ε 2 , γ 12
FIGURE 2.16
Calculated (solid lines) and experimental (circles) stressstrain diagrams for a two-matrix unidirectional
2
composite under in-plane transverse tension (σ1
2 ), compression (σ2 ) and shear (τ 12 ).
For practical analysis, an even simpler form of these equations (with c3 5 0) can be used (Hann
& Tsai, 1973).
Nonlinear behavior in composite materials can also be described with the aid of the theory of
plasticity, which can be constructed as a direct generalization of the classical plasticity theory
developed for metals and described in Section 2.1.2.
To construct such a theory, we decompose the strains in accordance with Eq. (2.15) and use
Eqs. (2.16) and (2.18) to determine elastic and plastic strains as
εeij 5
@Ue p @Up
; ε 5
@σij ij @σij
(2.61)
where Ue and Up are elastic and plastic potentials. For the elastic potential, elasticity theory yields
U 5 cijkl σij σkl
(2.62)
where cijkl are compliance coefficients, and summation over repeated subscripts is implied. The
plastic potential is assumed to be a function of stress intensity, σ, which is constructed for a plane
stress state as a direct generalization of Eq. (2.24), i.e.,
σ 5 aij σij 1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
aijkl σij σkl 1 3 aijklmn σij σkl σmn 1 :::
(2.63)
where the coefficients “a” are material constants characterizing its plastic behavior. Finally, we use
the power law in Eq. (2.27) for the plastic potential.
To write constitutive equations for a plane stress state, we return to engineering notations for
stresses and strains and use conditions that should be imposed on an orthotropic material and
100
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
which are discussed earlier in application to Eq. (2.59). Finally, Eqs. (2.15), (2.27) and (2.61)
(2.63) yield
1
1
ðb11 σ1 1 c12 σ2 Þ 1 2 ðd11 σ21 1 2e12 σ1 σ2 1 e21 σ22 Þ
R1
R2
1
1
2
2
n21
ε2 5 b1 σ2 1 d1 σ1 1 nσ
ðb22 σ2 1 c12 σ1 Þ 1 2 ðd22 σ2 1 2e21 σ2 σ1 1 e12 σ1
R1
R2
b
12
γ 12 5 c1 τ 12 1 2nσn21
τ 12
R1
ε1 5 a1 σ1 1 d1 σ2 1 nσn21
(2.64)
where
σ 5 R1 1 R2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
3
R1 5 b11 σ21 1 b22 σ22 1 b12 τ 212 1 2c12 σ1 σ2 ; R2 5 d11 σ31 1 d22 σ32 1 3e12 σ21 σ2 1 3e21 σ1 σ22
In deriving Eq. (2.64), we use new notations for coefficients and restrict ourselves to the threeterm approximation for σ as in Eq. (2.63).
For independent uniaxial loading along the fibers, across the fibers, and in pure shear,
Eq. (2.64) reduce to
ε1 5 a1 σ1 1 nð
pffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffi
pffiffiffiffiffiffiffi σ1
b11 σ21 1σ1 3 d11 Þn21
b11 pffiffiffiffiffi2 1 3 d11
σ1
qffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffin21 pffiffiffiffiffiffiffi σ2
pffiffiffiffiffiffiffi
ε2 5 b1 σ2 1 n
b22 pffiffiffiffiffi2 1 3 d22
b22 σ22 1σ2 3 d22
σ2
pffiffiffiffiffiffiffi qffiffiffiffiffiffiffi n21
n
2
τ 12
γ 12 5 c1 1 2n b12
τ 12
(2.65)
If nonlinear material behavior does not depend on the sign of normal stresses, then
d11 5 d22 5 0 in Eq. (2.65). In the general case, Eq. (2.65) allows us to describe materials with high
nonlinearity and different behaviors under tension and compression.
As an example, consider a boronaluminum unidirectional composite whose experimental
stressstrain diagrams (Herakovich, 1998) are shown in Fig. 2.17 (circles) along with the corresponding approximations (solid lines) plotted with the aid of Eq. (2.65).
2.3 UNIDIRECTIONAL ANISOTROPIC LAYER
Consider now a unidirectional layer studied in the previous section and assume that its principal
material axis 1 makes some angle φ with the x-axis of the global coordinate frame (see Fig. 2.18).
An example of such a layer is shown in Fig. 2.19.
2.3 UNIDIRECTIONAL ANISOTROPIC LAYER
101
σ 2 , MPa
(A)
150
100
50
0
–2
–1.8 –1.6 –1.4 –1.2
–1
–0.8 –0.6 –0.4 –0.2
0
– 50
0.2
ε 2 ,%
– 100
– 150
– 200
–250
– 300
(B)
τ12 , MPa
140
120
100
80
60
40
20
γ12 ,%
0
0
1
2
3
4
5
6
7
8
FIGURE 2.17
Calculated (solid lines) and experimental (circles) stressstrain diagrams for a boronaluminum composite under
transverse loading (A) and in-plane shear (B).
2.3.1 LINEAR ELASTIC MODEL
Constitutive equations of the layer under study referred to the principal material coordinates are
given by Eqs. (2.55) and (2.56). We now need to derive such equations for the global coordinate
frame x, y, and z (see Fig. 2.18). To do this, we should transform stresses σ1 , σ2 , τ 12 , τ 13 , and τ 23
acting in the layer and the corresponding strains ε1 , ε2 , γ 12 , γ 13 , and γ 23 into stress and strain components σx , σy , τ xy , τ xz , and τ yz , and εx , εy , γ xy , γ xz , and γ yz . The resulting expressions are
(Vasiliev, 1993)
102
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
z
3
2
y
x
φ1
τ23
τ13
τxz
τyz
τxy
σ2
τ12 σ1
σx
σy
FIGURE 2.18
A composite layer consisting of a system of unidirectional plies with the same orientation.
FIGURE 2.19
An anisotropic outer layer of a composite pressure vessel.
σ1 5 σx c2 1 σy s2 1 2τ xy cs
σ2 5 σx s2 1 σy c2 2 2τ xy cs
τ 12 5 ðσy 2 σx Þcs 1 τ xy ðc2 2 s2 Þ
(2.66)
τ 13 5 τ xz c 1 τ yz s
τ 23 5 2 τ xz s 1 τ yz c
(2.67)
2.3 UNIDIRECTIONAL ANISOTROPIC LAYER
103
where c 5 cosφ and s 5 sinφ: The inverse form of these equations is
σx 5 σ1 c2 1 σ2 s2 2 2τ 12 cs
σy 5 σ1 s2 1 σ2 c2 1 2τ 12 cs
τ xy 5 ðσ1 2 σ2 Þcs 1 τ 12 ðc2 2 s2 Þ
τ xz 5 τ 13 c 2 τ 23 s
τ yz 5 τ 13 s 1 τ 23 c
(2.68)
The corresponding transformation for strains is (Vasiliev, 1993)
ε1 5 εx c2 1 εy s2 1 γ xy cs
ε2 5 εx s2 1 εy c2 2 γ xy cs
γ 12 5 2ðεy 2 εx Þcs 1 γ xy ðc2 2 s2 Þ
γ 13 5 γ xz c 1 γ yz s
γ 23 5 2 γ xz s 1 γ yz c
(2.69)
εx 5 ε1 c2 1 ε2 s2 2 γ 12 cs
εy 5 ε1 s2 1 ε2 c2 1 γ 12 cs
γ xy 5 2ðε1 2 ε2 Þcs 1 γ 12 ðc2 2 s2 Þ
γ xz 5 γ 13 c 2 γ 23 s
γ yz 5 γ 13 s 1 γ 23 c
(2.70)
or
To derive constitutive equations for an anisotropic unidirectional layer, we substitute the strains
in Hooke’s law, Eq. (2.56) with their expressions given by Eq. (2.69), and the resulting expressions
substitute for stresses in Eq. (2.68). The final result is as follows:
σx 5 A11 εx 1 A12 εy 1 A14 γ xy
σy 5 A21 εx 1 A22 εy 1 A24 γ xy
τ xy 5 A41 εx 1 A42 εy 1 A44 γ xy
(2.71)
τ xz 5 A55 γ xz 1 A56 γ yz
τ yz 5 A65 γ xz 1 A66 γ yz
The stiffness coefficients in these equations are
A11 5 E1 c4 1 E2 s4 1 2E12 c2 s2
A12 5 A21 5 E1 ν 12 1 ðE1 1 E2 2 2E12 Þc2 s2
A14 5 A41 5 ½E1 c2 2 E2 s2 2 E12 ðc2 2 s2 Þcs
A22 5 E1 s4 1 E2 c4 1 2E12 c2 s2
A24 5 A42 5 ½E1 s2 2 E2 c2 1 E12 ðc2 2 s2 Þcs
A44 5 ðE1 1 E2 2 2E1 ν 12 Þc2 s2 1 G12 ðc2 2s2 Þ2
A55 5 G13 c2 1 G23 s2
A56 5 A65 5 ðG13 2 G23 Þcs
A66 5 G13 s2 1 G23 c2
where
E1;2 5
E1;2
; E12 5 E1 ν 12 1 2G12 ; c 5 cosφ; s 5 sinφ
1 2 ν 12 ν 21
(2.72)
104
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
The dependence of the stiffness coefficients Amn in Eq. (2.72) on φ has been studied by Tsai
and Pagano (see, e.g., Tsai, 1987; Verchery, 1999). Changing the powers of sinφ and cosφ in
Eq. (2.72) for multiple-angle trigonometric functions, we can reduce these equations to the following form:
A11 5 S1 1 S2 1 2S3 cos2φ 1 S4 cos4φ
A12 5 2 S1 1 S2 2 S4 cos4φ
A14 5 S3 sin2φ 1 S4 sin4φ
A22 5 S1 1 S2 2 2S3 cos2φ 1 S4 cos4φ
A24 5 S3 sin2φ 2 S4 sin4φ
(2.73)
A44 5 S1 2 S4 cos4φ
A55 5 S5 1 S6 cos2φ
A56 5 4S6 sin2φ
A66 5 S5 2 S6 cos2φ
(Verchery, 1999). In these equations,
1
S1 5 ðA011 1 A022 2 2A012 1 4A044 Þ
8
1
S2 5 ðA011 1 A022 1 2A012 Þ
4
1
S3 5 ðA011 2 A022 Þ
4
1
S4 5 ðA011 1 A022 2 2A012 2 4A044 Þ
8
1
S5 5 ðA055 1 A066 Þ
2
1
S6 5 ðA055 2 A066 Þ
2
where A0n are stiffness coefficients corresponding to φ 5 0. It follows from Eq. (2.72) that,
A011 5 E1 ; A012 5 E1 ν 12 ; A014 5 A024 5 A056 5 0
A022 5 E2 ; A044 5 G12 ; A055 5 G13 ; A066 5 G23
As can be seen in Eq. (2.73), there exist the following differential relationships between tensile
and coupling stiffnesses (Verchery & Gong, 1999)
dA11
dA22
5 2 4A14 ;
5 4A24
dφ
dφ
It can be directly checked that Eq. (2.73) provide three invariant stiffness characteristics whose
forms do not depend on φ, i.e.,
A11 ðφÞ 1 A22 ðφÞ 1 2A12 ðφÞ 5 A011 1 A022 1 2A012
A44 ðφÞ 2 A12 ðφÞ 5 A044 2 A012
A55 ðφÞ 1 A66 ðφÞ 5 A055 1 A066
(2.74)
2.3 UNIDIRECTIONAL ANISOTROPIC LAYER
105
Any linear combination of these equations is also an invariant combination of stiffness
coefficients.
The strains in Eq. (2.71) can be obtained if we substitute for stresses in Hooke’s law, Eq. (2.55)
their expressions given by Eqs. (2.66) and (2.67), and substitute the resulting equations for the
strains in Eq. (2.70). As a result, we arrive at
σx
σy
τ xy
σy
σx
τ xy
2 ν xy
1 ηx;xy
; εy 5
2 ν yx
1 ηy;xy
Ex
Ey
Gxy
Ey
Ex
Gxy
τ xy
σx
σy
γ xy 5
1 ηxy;x
1 ηxy;y
Gxy
Ex
Ey
τ xz
τ yz
τ yz
τ xz
γ xz 5
1 λxz;yz
; γ 5
1 λyz;xz
Gxz
Gyz yz Gyz
Gxz
εx 5
(2.75)
in which the compliance coefficients are
1
c4
s4
1
2ν 21 2 2
c s
5
1
1
2
Ex
G12
E1
E2
E1
ν xy
ν yx
ν 21
1
1
2ν 21
1
c2 s2
5
5
2
1
1
2
E1
E2
G12
Ey
Ex
E1
E1
ηx;xy
ηxy;x
c2
s2
1
ν 21
ðc2 2 s2 Þ cs
5
52
2
2
2
2G12
Gxy
Ex
E1
E2
E1
1
s4
c4
1
2ν 21 2 2
c s
5
1
1
2
E1
E2
E1
Ey
G12
ηy;xy
ηxy;y
s2
c2
1
ν 21
ðc2 2 s2 Þ cs
5
52
2
1
2
2G12
Gxy
Ey
E1
E2
E1
1
1
1
2ν 21 2 2
1 2 22
c s 1
54
1
1
ðc 2s Þ
Gxy
E1
E2
G12
E1
1
c2
s2 λxz;yz
λyz;xz
1
1
1
s2
c2
cs;
5
1
;
5
5
2
5
1
Gxz
G13
G23
Gyz
G13
G23 Gyz
Gxz
G13
G23
There exist the following dependencies between the coefficients of Eqs. (2.71) and (2.75)
1
1
ν xy
ν yx
1
5
ðA22 A44 2 A224 Þ;
5
5
ðA12 A44 2 A14 A24 Þ
Ex
D1
D1
Ey
Ex
ηx;xy
ηxy;x
1
1
1
5
5
ðA12 A24 2 A22 A14 Þ;
5
ðA11 A44 2 A214 Þ
D1
Ey
D1
Gxy
Ex
ηy;xy
ηxy;y
1
1
1
5
5
ðA12 A14 2 A11 A24 Þ;
5
ðA11 A22 2 A212 Þ
D1
Gxy
D1
Gxy
Ey
1
A66 1
A55 λxz;yz
λyz;xz
A56
5
;
5
;
5
52
Gxz
D2 Gyz
D2 Gyz
Gxz
D2
Here,
D1 5 A11 A22 A44 2 A11 A224 2 A22 A214 2 A44 A212 1 2A12 A14 A24 ; D2 5 A55 A66 2 A256
(2.76)
106
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
and
1 2 ηy;xy ηxy;y
ν xy 2 ηx;xy ηxy;y
; A12 5
D3 Ey Gxy
D3 Ey Gxy
ηx;xy 1 ν xy ηy;xy
1 2 ηx;xy ηxy;y
A14 5 2
; A22 5
D3 Ey Gxy
D3 Ex Gxy
ηy;xy 1 ν yx ηx;xy
1 2 ν xy ν yx
A24 5 2
; A44 5
D3 Ex Gxy
D3 Ex Ey
1
λxz;yz
1
A55 5
; A56 5 2
; A66 5
D4 Gyz
D4 Gxz
D4 Gyz
A11 5
where
1
ð1 2 ν xy ν yx 2 ηx;xy ηxy;x 2 ηy;xy ηxy;y
Ex Ey Gxy
2 ν xy ηy;xy ηxy;x 2 ν yx ηx;xy ηxy;y Þ
1
D4 5
ð1 2 λxz;yz λyz;xz Þ
Gxz Gyz
D3 5
As can be seen in Eqs. (2.71) and (2.75), the layer under study is anisotropic in the plane xy because
the constitutive equations include shearextension and shearshear coupling coefficients η and λ. For
φ 5 0, the foregoing equations degenerate into Eqs. (2.55) and (2.56) for an orthotropic layer.
The dependencies of stiffness coefficients on the orientation angle for a carbonepoxy composite with properties listed in Table 1.5 are presented in Figs. 2.20 and 2.21.
Uniaxial tension of the anisotropic layer (the so-called off-axis test of a unidirectional composite) is often used to determine material characteristics that cannot be found in tests with orthotropic
specimens or to evaluate constitutive and failure theories. Such a test is shown in Fig. 2.22. To
study this loading case, we should take σy 5 τ xy 5 0 in Eq. (2.75). Then,
εx 5
σx
σx
σx
; εy 5 2 ν xy ; γ xy 5 ηxy;x
Ex
Ex
Ex
(2.77)
As can be seen in these equations, tension in the x-direction is accompanied not only by transverse contraction, as in orthotropic materials, but also by shear. This results in the deformed shape
of the sample shown in Fig. 2.23. This shape is expected because the material stiffness in the fiber
direction is much higher than that across the fibers.
Such an experiment, in cases where it can be performed, allows us to determine the in-plane
shear modulus, G12 , in principal material coordinates using a simple tensile test as opposed to the
much more complicated tests described in Section 1.4.3 and shown in Figs. 1.54 and 1.55. Indeed,
if we know Ex from the tensile test in Fig. 2.23 and find E1 , E2 , and ν 21 from tensile tests along
and across the fibers (see Sections 1.4.1 and 1.4.2), we can use the first equation of Eq. (2.76) to
determine
G12 5
sin2 φcos2 φ
1
cos φ sin4 φ 2ν 21 2
2
2
1
sin φcos2 φ
E1
Ex
E1
E2
4
(2.78)
In connection with this, a question arises as to what angle should be substituted into this equation to provide the most accurate result. The answer is given in Fig. 2.24, which displays the strains
2.3 UNIDIRECTIONAL ANISOTROPIC LAYER
107
Amn, GPa
140
120
A11
A22
100
80
60
40
A44
20
φ°
0
0
15
30
45
60
75
90
FIGURE 2.20
Dependencies of tensile (A11, A22) and shear (A44) stiffnesses of a unidirectional carbonepoxy layer on the
orientation angle.
Amn, GPa
40
A14
30
20
A12
A24
10
0
φ°
0
15
30
45
60
75
90
FIGURE 2.21
Dependencies of coupling stiffnesses of a unidirectional carbonepoxy layer on the orientation angle.
108
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
FIGURE 2.22
An off-axis test.
σx
γ
γ
σx
FIGURE 2.23
Deformation of a unidirectional layer loaded at an angle to fiber orientation.
in principal material coordinates for a carbonepoxy layer calculated with the aid of Eqs. (2.69)
and (2.77). As can be seen in this figure, the most appropriate angle is about 10 . At this angle, the
shear strain γ 12 is much higher than normal strains ε1 and ε2 , so that material deformation is associated mainly with shear. An off-axis test with φ 5 10 can also be used to evaluate the material
strength in shear τ 12 (Chamis, 1979). Stresses acting under off-axis tension in the principal material
coordinates are statically determinate and can be found directly from Eq. (2.67) as
σ1 5 σx cos2 φ; σ2 5 σx sin2 φ; τ 12 5 2 σx sinφcosφ
(2.79)
Thus, applying stress σx and changing φ we can induce proportional loading with different combinations of stresses σ1 , σ2 , and τ 12 to evaluate putative constitutive or failure theories for a material under study.
2.3 UNIDIRECTIONAL ANISOTROPIC LAYER
3
109
ε 1 ε x , ε 2 ε x ,γ 12 εx
2.5
γ12 ε x
2
1.5
1
ε1 ε x
ε 2 εx
0.5
φ°
0
0
15
30
45
60
75
90
FIGURE 2.24
Dependencies of normalized strains in the principle material coordinates on the angle of the off-axis test.
However, the test shown in Fig. 2.23 can hardly be performed because the test fixture (see
Fig. 2.22) restrains the shear deformation of the specimen and induces a corresponding shear stress.
The constitutive equations for the specimen loaded with uniaxial tension as in Fig. 2.23 and fixed
as in Fig. 2.22 follow from Eq. (2.75) if we take σy 5 0, i.e.,
σx
τ xy
1 ηx;xy
Ex
Gxy
τ xy
σx
1 ηxy;x
γ xy 5
Gxy
Ex
εx 5
(2.80)
(2.81)
in which the elastic constants are specified by Eq. (2.76). The shear stress, being of a reactive
nature, can be found from Eq. (2.81) if we put γ xy 5 0. Then,
τ xy 5 2 ηxy;x
Gxy
σx
Ex
Substituting this result into Eq. (2.80), we arrive at
εx 5
Here,
σx
Exa
(2.82)
110
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
Exa 5
Ex
1 2 ηx;xy ηxy;x
(2.83)
is the apparent elastic modulus that can be found from the test shown in Fig. 2.22. As follows from
Eq. (2.83), Exa , in general, does not coincide with Ex as used in Eq. (2.78) for G12 .
Thus, measuring σx and εx we can determine Ex from Eq. (2.82) only under the condition Exa 5 Ex ,
which means that the shearextension coupling coefficient η must be zero. Applying Eq. (2.76)
and assuming that φ 6¼ 0 and φ ¼
6 90 , we arrive at the following condition providing η 5 0:
sin2 φ0 5
e1
e2
(2.84)
in which
e1 5
1 1 ν 21
1
1 1 ν 21
1 1 ν 12
1
2
; e2 5
1
2
2G12
G12
E1
E1
E2
Since 0 # sin2 φ # 1, there exist two cases in which Eq. (2.84) is valid. The first case corresponds to the following set of inequalities
e1 $ 0; e2 . 0; e2 $ e1
(2.85)
e1 # 0; e2 , 0; e2 # e1
(2.86)
whereas for the second case,
To be specific, suppose that E1 . E2 . Then, taking into account the symmetry condition
ν 12 E1 5 ν 21 E2 we have
1 1 ν 12
1 1 ν 21
.
E2
E1
(2.87)
Consider the first set of inequalities in Eq. (2.85) and assume that the first of them, which has
the following explicit form
1 1 ν 21
1
$
2G12
E1
(2.88)
is valid. Then, Eq. (2.87) yields
1 1 ν 12
1 1 ν 21
1
.
$
2G12
E2
E1
or
1 1 ν 12
1
.
2G12
E2
Matching this result with the last inequality in Eq. (2.85) presented in the form
1 1 ν 12
1
$
2G12
E2
(2.89)
we can conclude that if the first condition in Eq. (2.85) is valid, then the last of these conditions
holds too.
Consider the second condition in Eq. (2.85) and write it in explicit form, i.e.,
2.3 UNIDIRECTIONAL ANISOTROPIC LAYER
111
1 1 ν 12
1 1 ν 21
1
1
$
G12
E2
E1
(2.90)
Transforming Eq. (2.87) and using Eq. (2.89), we have
1 1 ν 12
1 1 ν 21
1 1 ν 21
1
1
.2
$
G12
E2
E1
E1
which means that the condition in Eq. (2.90) is valid.
So, the set of conditions in Eq. (2.85) can be reduced to one inequality in Eq. (2.88), which can
be written in a final form as
G12 $
E1
2ð1 1 ν 21 Þ
(2.91)
Consider the conditions given by Eq. (2.86) and assume that the last of them, which can be presented in the following explicit form
1 1 ν 12
1
#
E2
2G12
(2.92)
is valid. Using Eqs. (2.87) and (2.92), we get
1 1 ν 21
1 1 ν 12
1
,
#
E1
E2
2G12
or
1 1 ν 21
1
,
E1
2G12
Since the first condition in Eq. (2.86) can be presented as
1 1 ν 21
1
#
2G12
E1
we can conclude that it is satisfied.
Consider the second inequality in Eq. (2.86) and write it in an explicit form, i.e.,
1 1 ν 21
1 1 ν 12
1
1
,
E1
E2
G12
(2.93)
Using Eqs. (2.87) and (2.92), we get
1 1 ν 21
1 1 ν 12
1 1 ν 12
1
1
,2
#
G12
E1
E2
E2
which means that the condition in Eq. (2.93) is satisfied.
So, the set of conditions in Eq. (2.86) is reduced to one inequality in Eq. (2.92), which can be
written in the following final form:
G12 #
E2
2ð1 1 ν 12 Þ
(2.94)
Thus, Eq. (2.84) determines the angle φ0 for orthotropic materials whose mechanical characteristics satisfy the conditions in Eqs. (2.91) or (2.94). Such materials, being loaded at an angle
φ 5 φ0 , do not experience shearstretching coupling. The shear modulus can be found from
112
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
Eq. (2.78) in which Ex 5 σx =εx where σx and εx are the stress and the strain determined in the offaxis tension test shown in Fig. 2.22.
Consider as examples unidirectional composites with typical properties (Table 1.5).
For fiberglassepoxy composite, we have E1 5 60 GPa, E2 5 13 GPa, G12 5 3.4 GPa,
ν 12 5 0.065, ν 21 5 0.3. Calculation in accordance with Eqs. (2.91) and (2.94) yields
E1
E2
5 23:08 GPa;
5 6:1 GPa
2ð1 1 ν 21 Þ
2ð1 1 ν 12 Þ
Thus, the condition in Eq. (2.94) is satisfied, and Eq. (4.84) gives φ0 5 54:313 .
For aramidepoxy composite E1 5 95 GPa, E2 5 5:1 GPa, G12 5 1:8 GPa, ν 12 5 0:018,
ν 21 5 0:34, we have
E1
E2
5 36:45 GPa;
5 2:5 GPa; and φ0 5 61:453
2ð1 1 ν 21 Þ
2ð1 1 ν 12 Þ
For carbonepoxy composite with E1 5 140 GPa, E2 5 11 GPa, G12 5 5:5 GPa, ν 12 5 0:021,
ν 21 5 0:27, we have
E1
E2
5 55:12 GPa;
5 5:39 GPa
2ð1 1 ν 21 Þ
2ð1 1 ν 12 Þ
As can be seen, the conditions in Eqs. (2.91) and (2.94) are not satisfied, and the angle φ0 does
not exist for this material.
As can be directly checked with the aid of Eq. (2.76), there exists the following relationship
between the elastic constants of anisotropic materials (Verchery & Gong, 1999)
ηx;xy
d 1
522
dφ Ex
Gxy
This equation means that ηx;xy 5 0 for materials whose modulus Ex reaches the extremum in the
interval 0 , φ , 90 . The dependencies of Ex =E1 on φ for the materials considered above as examples are shown in Fig. 2.25.
As can be seen, curves 1 and 2 corresponding to glass and aramid composites reach the minimum value at φ0 5 54:31 and φ0 5 61:45 , respectively, whereas curve 3 for carbon composite
does not have a minimum at 0 , φ , 90 .
The dependence Ex ðφÞ with the minimum value of Ex reached at φ 5 φ0 , where 0 , φ , 90 , is
typical for composites reinforced in two orthogonal directions. For example, for a fabric composite
having E1 5 E2 and ν 12 5 ν 21 , Eq. (2.84) yields the well-known result φ0 5 45 . For a typical fiberglass fabric composite with E1 5 26 GPa, E2 5 22 GPa, G12 5 7:2 GPa, ν 12 5 0:11, ν 21 5 0:13, we
have
E1
E2
5 11:5 GPa;
5 9:9 GPa; and φ0 5 49:13
2ð1 1 ν 21 Þ
2ð1 1 ν 12 Þ
In conclusion, it should be noted that the actual application of Eq. (2.78) is hindered by the fact
that the angle φ0 specified by Eq. (2.84) depends on G12 , which is not known and needs to be
determined from Eq. (2.78). To find G12 , we actually need to perform several tests for several
values of G12 in the vicinity of the expected value and the corresponding values of φ0 following
2.3 UNIDIRECTIONAL ANISOTROPIC LAYER
Ex
113
E1
1
0.9
0.8
0.7
0.6
1
0.5
0.4
3
0.3
2
0.2
0.1
φ°
0
0
10
20
30
40
50
60
°
70
80
90
φ = 61.45°
φ = 54.31
FIGURE 2.25
Dependencies of Ex/E1 on φ for fiberglass(1), aramid(2), and carbon(3) epoxy composites.
y
σ
σ
V
x
a
l
M
FIGURE 2.26
Off-axis tension of a strip fixed at the ends.
from Eq. (2.84), and select the correct value of G12 , which satisfies in conjunction with the corresponding value of φ0 , both equations: Eqs. (2.78) and (2.84) (Morozov & Vasiliev, 2003).
Consider the general case of an off-axis test (see Fig. 2.22) for a composite specimen with an
arbitrary fiber orientation angle φ (see Fig. 2.26). To describe this test, we need to study the coupled problem for an anisotropic strip in which shear is induced by tension but is restricted at the
strip ends by the jaws of a test fixture as in Figs 2.22 and 2.26. As follows from Fig. 2.26, the
action of the grip can be simulated if we apply a bending moment M and a transverse force V such
that the rotation of the strip ends (γ in Fig. 2.23) will become zero. As a result, bending normal
and shear stresses appear in the strip that can be analyzed with the aid of composite beam theory
(see Chapter 6: Laminated Composite Beams and Columns).
114
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
To derive the corresponding equations, introduce the conventional assumptions of beam theory
according to which axial, ux , and transverse, uy , displacements can be presented as
ux 5 uðxÞ 1 yθ; uy 5 vðxÞ
where u and θ are the axial displacement and the angle of rotation of the strip cross-section,
x 5 constant, and v is the strip deflection in the xy-plane (see Fig. 2.26). The strains corresponding to these displacements are
@ux
5 u0 1 yθ0 5 ε 1 yθ0
@x
@ux
@uy
1
5 θ 1 v0
γ xy 5
@y
@x
εx 5
(2.95)
where ð. . .Þ0 5 dð. . .Þ=dx and ε is the deformation of the strip axis. These strains are related to stresses by Eq. (2.75) which reduces to
σx
τ xy
1 ηx;xy
Ex
Gxy
τ xy
σx
γ xy 5
1 ηxy;x
Gxy
Ex
εx 5
(2.96)
The inverse form of these equations is
σx 5 B11 εx 1 B14 γ xy ; τ xy 5 B41 εx 1 B44 γ xy
(2.97)
Ex
Gxy
; B44 5
1 2 ηx;xy ηxy;x
1 2 ηx;xy ηxy;x
Ex ηx;xy
Gxy ηxy;x
B14 5 B41 5 2
52
1 2 ηx;xy ηxy;x
1 2 ηx;xy ηxy;x
(2.98)
where
B11 5
Now, decompose the strip displacements, strains, and stresses into two components corresponding to
1. free tension (see Fig. 2.23), and
2. bending.
For free tension, we have τ xy 5 0 and v 5 0. So, Eqs. (2.95) and (2.96) yield
0
ð1Þ
εð1Þ
x 5 ε1 1 yθ1 ; γ xy 5 θ1
ð1Þ
σx
σð1Þ
x
εð1Þ
;
γ ð1Þ
x 5
xy 5 ηxy;x
Ex
Ex
(2.99)
Here, ε1 5 u01 and σð1Þ
x 5 σ 5 F=ah, where F is the axial force applied to the strip, a the strip
width, and h is its thickness. Since σð1Þ
x 5 constant, Eq. (2.99) give
θ1 5 ηxy;x
σ
σ
F
5 constant; ε1x 5 ε1 5
5
Ex
Ex
ah
(2.100)
Adding components corresponding to bending (with index 2), we can write the total displacements and strains as
2.3 UNIDIRECTIONAL ANISOTROPIC LAYER
ux 5 u1 1 u2 1 yðθ1 1 θ2 Þ;
εx 5 ε1 1 ε2 1 yθ02 ;
115
uy 5 v2
γ xy 5 θ1 1 θ2 1 v02
The total stresses can be expressed with the aid of Eq. (2.97), i.e.,
σx 5 B11 ðε1 1 ε2 1 yθ02 Þ 1 B14 ðθ1 1 θ2 1 v02 Þ
τ xy 5 B41 ðε1 1 ε2 1 yθ02 Þ 1 B44 ðθ1 1 θ2 1 v02 Þ
Transforming these equations with the aid of Eqs. (2.98) and (2.100), we arrive at
σx 5 σ 1 B11 ðε2 1 yθ02 Þ 1 B14 ðθ2 1 v02 Þ
τ xy 5 B41 ðε2 1 yθ02 Þ 1 B44 ðθ2 1 v02 Þ
(2.101)
These stresses are statically equivalent to the axial force P, the bending moment M, and the
transverse force V, which can be introduced as
a=2
ð
P5h
a=2
ð
σx dy; M 5 h
2a=2
a=2
ð
σx ydy; V 5 h
2a=2
τ xy dy
2a=2
Substitution of Eq. (2.101) and integration yields
P 5 ah½σ 1 B11 ε2 1 B14 ðθ2 1 v02 Þ
M 5 B11 h
(2.102)
3
a 0
θ
12 2
(2.103)
V 5 ah½B41 ε2 1 B44 ðθ2 1 v02 Þ
(2.104)
These forces and this moment should satisfy the equilibrium equation that follows from
Fig. 2.27, i.e.,
P0 5 0; V 0 5 0; M 0 5 V
(2.105)
Solution of the first equation is P 5 F 5 σah. Then, Eq. (2.102) gives
ε2 5 2
B14
ðθ2 1 v02 Þ
B11
(2.106)
The second equation of Eq. (2.105) shows that V 5 C1 , where C1 is a constant of integration.
Then, substituting this result into Eq. (2.104) and eliminating ε2 with the aid of Eq. (2.106), we have
θ2 1 v02 5
C1
ahB44
(2.107)
where B44 5 B44 2 B14 B41 .
V + V' dx
P
P + P' dx
M
V
FIGURE 2.27
Forces and moments acting on the strip element.
M + M' dx
dx
116
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
Taking in the third equation of Eq. (2.105), V 5 C1 , and substituting M from Eq. (2.103), we
arrive at the following equation for θ2
θv2 5
12C1
a3 hB11
Integration yields
θ2 5
6C1
1 C2 x 1 C3
a3 hB11
The total angle of rotation θ 5 θ1 1 θ2 , where θ1 is specified by Eq. (2.100), should be zero at
the ends of the strip, i.e., θðx 5 6 l=2Þ 5 0. Satisfying these conditions, we have
θ2 5
3C1
l2
σ
2
2 ηxy;x
2x
2
Ex
a3 hB11
2
(2.108)
Substitution into Eq. (4.107) and integration allows us to find the deflection
C1 x
3C1 x 2x2
l2
σ0 x
2
1 ηxy;x
v2 5
2 3
1 C4
a hB11 3
Ex
2
ahB44
(2.109)
This expression includes two constants, C1 and C4 , which can be determined from the boundary
conditions v2 ðx 5 6 l=2Þ 5 0. The final result, following from Eqs. (2.100), (2.108), and (2.109), is
"
#
2
σx
B11 1 l B44 ð3=2 2 2x2 Þ
v 5 lηxy;x
12
2
Ex
B11 1 l B44
2
3σl B44 ð2x2 2 1=2Þ
θ 5 ηxy;x
Ex B11 1 l2 B44
(2.110)
where l 5 l=a and x 5 x=l. The deflection of a carbonepoxy strip having φ 5 45 and l 5 10 is
shown in Fig. 2.28.
Now, we can write the relationship between modulus Ex corresponding to the ideal test shown
in Fig. 2.23 and apparent modulus Exa that can be found from the real test shown in Figs. 2.22 and
2.26. Using Eqs. (2.98), (2.100), (2.106), and (2.110), we finally get
σ 5 Exa ε
where
Exa 5
12
Ex
Ex ηx;xy ηxy;x
2
Ex 1 l Gxy ð1 2 ηx;xy ηxy;x Þ
Consider two limiting cases. For an infinitely long strip (l-N), we have Exa 5 Ex . This result
corresponds to the case of free shear deformation specified by Eq. (2.77). For an infinitely short
strip (l-0), taking into account Eq. (2.98), we get
Exa 5
Ex
5 B11
1 2 ηx;xy ηxy;x
2.3 UNIDIRECTIONAL ANISOTROPIC LAYER
0.1
117
vE x
lσ
0.05
–0.5 –0.4 –0.3 –0.2 –0.1 0
–0.05
0.1
0.2
0.3
0.4
0.5
x
l
–0.1
FIGURE 2.28
Normalized deflection of a carbonepoxy strip (φ 5 45 ; l 5 10).
In accordance with Eq. (2.97), this result corresponds to a restricted shear deformation
(γ xy 5 0). For a strip with finite length, Ex , Exa , B11 . The dependence of the normalized apparent
modulus on the length-to-width ratio for a 45 carbonepoxy layer is shown in Fig. 2.29. As can
be seen, the difference between Exa and Ex becomes less than 5% for l . 3a.
2.3.2 NONLINEAR MODELS
Nonlinear deformation of an anisotropic unidirectional layer can be studied rather straightforwardly
because stresses σ1 , σ2 , and τ 12 in the principal material coordinates (see Fig. 2.18) are statically
determinate and can be found using Eq. (2.67). Substituting these stresses into the nonlinear constitutive equations, Eqs. (2.60) or (2.64), we can express strains ε1 , ε2 , and γ 12 in terms of stresses σx ,
σy , and τ xy . Further substitution of the strains ε1 , ε2 , and γ 12 into Eq. (2.70) yields constitutive
equations that link strains εx ,εy , and γ xy with stresses σx , σy , and τ xy , thus allowing us to find
strains in the global coordinates x, y, and z if we know the corresponding stresses.
As an example of the application of a nonlinear elastic material model described by Eq. (2.60),
consider a two-matrix fiberglass composite (see Section 2.4.4) whose stressstrain curves in the
principal material coordinates are presented in Fig. 2.16. These curves allow us to determine coefficients “b” and “c” in Eq. (2.60). To find the coupling coefficients “m,” we use a 45 off-axis test.
Experimental results (circles) and the corresponding approximation (solid line) are shown in
Fig. 2.30. Thus, the constructed model can be used now to predict material behavior under tension
at any other angle (different from 0, 45, and 90 —the corresponding results are given in Fig. 2.31
for 60 ) or to study more complicated material structures and loading cases (see Section 2.5).
As an example of the application of the elasticplastic material model specified by Eq. (2.64),
consider a boronaluminum composite whose stressstrain diagrams in principal material coordinates are shown in Fig. 2.17. Theoretical and experimental curves (Herakovich, 1998) for 30 and
45 off-axis tension of this material are presented in Fig. 2.32.
118
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
E xa E x
1.4
1.3
1.2
1.1
1
l
0
2
4
8
6
FIGURE 2.29
Dependence of the normalized apparent modulus on the strip length-to-width ratio for a 45 carbonepoxy layer.
σ x , MPa
16
12
8
4
0
0
2
4
6
ε x, %
FIGURE 2.30
Calculated (solid line) and experimental (circles) stressstrain diagram for 45 off-axis tension of a two-matrix
unidirectional composite.
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
119
σ x , MPa
16
12
8
4
0
εx , %
0
1
2
3
4
FIGURE 2.31
Theoretical (solid line) and experimental (dashed line) stressstrain diagrams for 60 off-axis tension of a twomatrix unidirectional composite.
σx , MPa
200
30°
160
45°
120
80
40
0
εx , %
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
FIGURE 2.32
Theoretical (solid lines) and experimental (dashed lines) stressstrain diagrams for 30 and 45 off-axis tension
of a boronaluminum composite.
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
The simplest layer reinforced in two directions is the so-called cross-ply layer that consists of alternating plies with 0 and 90 orientations with respect to the global coordinate frame x, y, z as in
120
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
z
3
1
2
3
1
x
σz
τ yz
h
τ xz
τ xz
2
y
σx
τ xy
τ yz
τ xy
σy
FIGURE 2.33
A cross-ply layer.
Fig. 2.33. Actually, this is a laminated structure, but being formed with a number of plies, it can be
treated as a homogeneous orthotropic layer.
2.4.1 LINEAR ELASTIC MODEL
Let the layer consist of m longitudinal (0 ) plies with thicknesses hðiÞ
0 (i 5 1; 2; 3; :::; m) and n
transverse (90 ) plies with thicknesses hðjÞ
(j
5
1;
2;
3;
:::;
n)
made
of one and the same com90
posite material. Then, stresses σx , σy , and τ xy that comprise the plane stress state in the global
coordinate frame can be expressed in terms of stresses in the principal material coordinates of the
plies as
m
n
X
X
ðiÞ
ðjÞ
σðiÞ
σðjÞ
1 h0 1
2 h90
σx h 5
i51
j51
i51
m
X
j51
n
X
i51
j51
m
n
X
X
ðiÞ
ðjÞ
σy h 5
σðiÞ
σðjÞ
2 h0 1
1 h90
τ xy h 5
ðiÞ
τ ðiÞ
12 h0 1
(2.111)
ðjÞ
τ ðjÞ
12 h90
Here, h is the total thickness of the layer (see Fig. 2.33), i.e.,
h 5 h0 1 h90
where
h0 5
m
n
X
X
hðiÞ
hðjÞ
0 ; h90 5
90
i51
j51
are the total thicknesses of the longitudinal and transverse plies.
The stresses in the principal material coordinates of the plies are related to the corresponding
strains by Eqs. (1.59) or (2.56)
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
121
ði;jÞ
ði;jÞ
σði;jÞ
1 5 E1 ðε1 1 ν 12 ε2 Þ
ði;jÞ
ði;jÞ
σði;jÞ
2 5 E2 ðε2 1 ν 21 ε1 Þ
(2.112)
ði;jÞ
τ ði;jÞ
12 5 G12 γ 12
in which, as earlier, E1;2 5 E1;2 =ð1 2 ν 12 ν 21 Þ and E1 ν 12 5 E2 ν 21 . Now assume that the deformation
of all the plies is the same as that of the deformation of the whole layer, i.e., that
ðjÞ
ðiÞ
ðjÞ
ðiÞ
ðjÞ
εðiÞ
1 5 ε2 5 εx ; ε2 5 ε1 5 εy ; γ 12 5 γ 12 5 γ xy
Then, substituting the stresses, Eq. (2.112), into Eq. (2.111), we arrive at the following constitutive equations
σx 5 A11 εx 1 A12 εy
σy 5 A21 εx 1 A22 εy
τ xy 5 A44 γ xy
(2.113)
A11 5 E1 h0 1 E2 h90 ; A22 5 E1 h90 1 E2 h0
A12 5 A21 5 E1 ν 12 5 E2 ν 21 ; A44 5 G12
(2.114)
in which the stiffness coefficients are
and
h0 5
h0
h90
; h90 5
h
h
The inverse form of Eq. (2.113) is
εx 5
σx
σy
σy
σx
τ xy
2 ν xy ; εy 5
2 ν yx ; γ xy 5
Ex
Ey
Ey
Ex
Gxy
(2.115)
where
A212
A2
; Ey 5 A22 2 12 ; Gxy 5 A44
A22
A21
A12
A12
ν xy 5
; ν yx 5
A11
A22
Ex 5 A11 2
(2.116)
To determine the transverse shear moduli Gxz and Gyz , consider, e.g., pure shear in the xz-plane
(see Fig. 2.34). It follows from the equilibrium conditions for the plies that
ðjÞ
ðiÞ
ðjÞ
τ ðiÞ
13 5 τ 23 5 τ xz ; τ 23 5 τ 13 5 τ yz
The total shear strains can be found as
m
n
X
1 X
γ xz 5
γ ðiÞ
γ ðjÞ
13 h0 1
23 h90
h i51
j51
γ yz 5
where in accordance with Eq.(4.56)
m
n
X
1 X
γ ðiÞ
γ ðjÞ
23 h0 1
13 h90
h i51
j51
(2.117)
!
!
(2.118)
122
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
τ xz
τ13
τ 23
τ xz
τ 13
τ xz
τ 23
τ xz
FIGURE 2.34
Pure transverse shear of a cross-ply layer.
ði;jÞ
γ 13
5
ði;jÞ
τ 13
τ ði;jÞ
ði;jÞ
; γ 23
5 23
G13
G23
(2.119)
Substituting Eq. (2.119) into Eq. (2.118) and using Eq. (2.117), we arrive at
γ xz 5
τ xz
τ yz
; γ 5
Gxz yz Gyz
where
1
h0
h90
1
h0
h90
5
1
;
5
1
Gxz
G13
G23 Gyz
G23
G13
2.4.2 NONLINEAR MODELS
The nonlinear behavior of a cross-ply layer associated with nonlinear material response under loading in the principal material coordinates (see, e.g., Figs. 2.16 and 2.17) can be described using nonlinear constitutive equations, Eqs. (2.60) or (2.64), instead of linear equations, Eq. (2.113).
However, this layer can demonstrate nonlinearity which is entirely different from that studied in
the previous sections. This nonlinearity is observed in a cross-ply layer composed of linear elastic
plies and is caused by microcracking of the matrix.
To study this phenomenon, consider a cross-ply laminate consisting of three plies as in
Fig. 2.35. Equilibrium conditions yield the following equations:
2ðσx1 h1 1 σx2 h2 Þ 5 σ
2ðσy1 h1 1 σy2 h2 Þ 5 0
in which
h1 5 h1 =h; h2 5 h2 =h; h 5 2ðh1 1 h2 Þ
The constitutive equations are
(2.120)
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
123
z
b
σ
x
1
h1
2
2h2
1
h1
σ
y
FIGURE 2.35
Tension of a cross-ply laminate.
σx1;2 5 E1;2 ðεx 1 ν 12;21 εy Þ
σy1;2 5 E2;1 ðεy 1 ν 21;12 εx Þ
(2.121)
in which E1;2 5 E1;2 =ð1 2 ν 12 ν 21 Þ. We assume that the strains εx and εy do not change through the
laminate thickness. Substituting Eq. (2.121) into Eq. (2.120), we can find strains and then stresses
using again Eq. (2.121). The final result is
σx1;2 5
σE1;2 ½E2 h1 1 E1 h2 2 E1;2 ν 212;21 ðh1 1 h2 Þ
2½ðE1 h1 1 E2 h2 ÞðE2 h1 1 E1 h2 Þ 2 E12 ν 212 ðh1 1h2 Þ2 To simplify the analysis, neglect Poisson’s effect, i.e., take ν 12 5 ν 21 5 0. Then
σx1 5 σ01 5
σE1
σE2
; σx2 5 σ02 5
2ðE1 h1 1 E2 h2 Þ
2ðE1 h1 1 E2 h2 Þ
(2.122)
Consider, e.g., the case h1 5 h2 5 0:5 and find the ultimate stresses corresponding to the failure
0
1
of longitudinal plies or to the failure of the transverse ply. Putting σ01 5 σ1
1 and σ2 5 σ2 , we get
E2
E1
1
ð2Þ
1
;
σð1Þ
5
σ
1
1
σ
5
σ
1
1
x
1
x
2
E1
E2
The results of calculation for the composites listed in Table 1.5 are presented in Table 2.2.
ð2Þ
As can be seen, σð1Þ
x . . σ x . This means that the first failure occurs in the transverse ply under
the stress
σ 5 σ 5 2σ1
2
E1
h2 1
h1
E2
(2.123)
This stress does not cause failure of the whole laminate, because the longitudinal plies can carry
the load, but the material behavior becomes nonlinear. Actually, the effect under consideration is
the result of the difference between the ultimate elongations of the unidirectional plies along and
across the fibers. From the data presented in Table 2.2 we can see that for all the materials listed in
this table ε1 . . ε2 . As a result, transverse plies drawn under tension by longitudinal plies that
have much higher stiffness and elongation fail because their ultimate elongation is smaller. This
failure is accompanied by a system of cracks parallel to the fibers which can be observed not only
in cross-ply layers but also in many other laminates that include unidirectional plies experiencing
transverse tension caused by interaction with the adjacent plies (see Fig. 2.36).
ð2Þ
Table 2.2 Ultimate Stresses Causing the Failure of Longitudinal ðσð1Þ
x ) or Transverse ðσx Þ Plies and Deformation Characteristics of Typical
Advanced Composites
σ(MPa);ε (%)
GlassEpoxy
CarbonEpoxy
CarbonPEEK
AramidEpoxy
BoronEpoxy
BoronAluminum
CarbonCarbon
Al2 O3 Al
σxð1Þ
2190
2160
2250
2630
1420
2000
890
1100
σxð2Þ
225
690
1125
590
840
400
100
520
ε1
ε2
ε1 =ε2
3
0.31
9.7
1.43
0.45
3.2
1.5
0.75
2
2.63
0.2
13.1
0.62
0.37
1.68
0.50
0.1
5
0.47
0.05
9.4
0.27
0.13
2.1
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
125
FIGURE 2.36
Cracks in the circumferential layer of a failed pressure vessel induced by transverse (for the vessel, axial) tension
of the layer.
z
σ
h1
1
2h2
2
h1
3
x
σ
FIGURE 2.37
A cross-ply layer with a crack in the transverse ply.
Now assume that the acting stress σ $ σ, where σ is specified by Eq. (2.123) and corresponds
to the load causing the first crack in the transverse ply as in Fig. 2.37. To study the stress state in
the vicinity of the crack, decompose the stresses in the three plies shown in Fig. 2.37 as
σx1 5 σx3 5 σ01 1 σ1 ; σx2 5 σ02 2 σ2
(2.124)
and assume that the crack also induces transverse through-the-thickness shear and normal stresses
τ xzi 5 τ i ; σzi 5 si ; i 5 1; 2; 3
(2.125)
The stresses σ01 and σ02 in Eq. (2.124) are specified by Eq. (2.122) with σ 5 σ, corresponding to
the applied stress at which the first crack appears in the transverse ply. Stresses σ1 and σ2 should
be self-balanced, i.e.,
σ1 h1 5 σ2 h2
(2.126)
126
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
The total stresses in Eqs. (2.124) and (2.125) should satisfy the equilibrium equations, which
yield for the problem under study
@σxi
@τ xzi
@σzi
@τ xzi
1
5 0;
1
50
@x
@z
@z
@x
(2.127)
where i 5 1; 2; 3:
To simplify the problem, suppose that the additional stresses σ1 and σ2 do not depend on z, i.e.,
that they are uniformly distributed through the thickness of the longitudinal plies. Then,
Eq. (2.127), upon substitution of Eqs. (2.124) and (2.125), can be integrated with respect to z. The
resulting stresses should satisfy the following boundary and interface conditions (see Fig. 2.37)
τ 1 ðz 5 h1 1 h2 Þ 5 0;
τ 1 ðz 5 h2 Þ 5 τ 2 ðz 5 h2 Þ;
τ 2 ðz 5 2 h2 Þ 5 τ 3 ðz 5 2 h2 Þ;
s1 ðz 5 h1 1 h2 Þ 5 0
s1 ðz 5 h2 Þ 5 s2 ðz 5 h2 Þ
s2 ðz 5 2 h2 Þ 5 s3 ðz 5 2 h2 Þ
τ 3 ðz 5 2 h1 2 h2 Þ 5 0;
s3 ðz 5 2 h1 2 h2 Þ 5 0
Finally, using Eq. (2.126) to express σ1 in terms of σ2 , we arrive at the following stress distribution (Vasiliev, Dudchenko, & Elpatievskii, 1970):
h2
; σx2 5 σ02 2 σ2 ðxÞ
h1
h2
h2
τ 1 5 2 σ02 ðxÞz1 ; τ 2 5 σ02 ðxÞz; τ 3 5 2 σ02 ðxÞz2
h1
h1
h2
1
s1 5
σv2 ðxÞz21 ; s2 5 2 σv2 ðxÞðz2 2 h1 h2 2 h22 Þ
2
2h1
h2
2
σv2 ðxÞz2
s3 5
2h1
σx1 5 σx3 5 σ01 1 σ2 ðxÞ
(2.128)
where
z1 5 z 2 h1 2 h2 ; z2 5 z 1 h1 1 h2 ; and ð Þ0 5 dð Þ=dx
Thus, we need to find only one unknown function: σ2 ðxÞ. To do this, we can use the principle
of minimum strain energy, according to which the function σ2 ðxÞ should deliver the minimum value
of the energy
Wσ 5
ðl 3 ð 2
σ2
τ2
1 X
σxi
ν xzi
1 zi 2 2
σxi σzi 1 xzi dx
Exi
Ezi
Ezi
Gxzi
2 i51
0
(2.129)
hi
where Ex1 5 Ex3 5 E1 , Ex2 5 E2 , Ezi 5 E2 , Gxz1 5 Gxz3 5 G13 , Gxz2 5 G23 , ν xz1 5 ν xz3 5 ν 13 , and
ν xz2 5 ν 23 , whereas E1 ; E2 ; G13 ; G23 ; ν 13 , and ν 23 are elastic constants of a unidirectional ply.
Substituting stresses, Eq. (2.128), into the functional in Eq. (2.129), integrating with respect to z,
and using the traditional procedure of variational calculus providing δWσ 5 0, we arrive at the following equation for σ2 ðxÞ:
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
127
d 4 σ2
d 2 σ2
2 2a2 2 1 b4 σ2 5 0
dx4
dx
in which
1 h2
h1
ν 23
2
h1 ν 23
1
1
1
; b4 5
1
2
ðh1 1 h2 Þ 1
1
3G13
E2
3E2
d 3G23
3
d h1 E1
h2 E2
1 1 3
2
ðh 1 h32 Þ 2 h22 ðh1 1 h2 Þ 1 h2 ðh1 1h2 Þ2
d5
2E2 5 1
3
a2 5
The general solution for this equation is
σ2 5 e2k1 x ðc1 sink2 x 1 c2 cosk2 xÞ 1 ek1 x ðc3 sink2 x 1 c4 cosk2 xÞ
where
(2.130)
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 2
1 2
2
ða 1 b Þ; k2 5
ðb 2 a2 Þ
k1 5
2
2
Suppose that the strip shown in Fig. 2.37 is infinitely long in the x-direction. Then, we should
have σ1 -0 and σ2 -0 for x-N in Eq. (2.124). This means that we should put c3 5 c4 5 0 in
Eq. (2.130). The other two constants, c1 and c2 , should be determined from the conditions on the
crack surface (see Fig. 2.37), i.e.,
σx2 ðx 5 0Þ 5 0; τ xz2 ðx 5 0Þ 5 0
Satisfying these conditions, we obtain the following expressions for stresses
h2
k1
σx1 5 σx3 5 σ01 1 σ02 e2k1 x
sink2 x 1 cosk2 x
h1
k2
k
1
0
2k1 x
sink2 x 1 cosk2 x
σx2 5 σ2 1 2 e
k2
σ0
τ xz2 5 2 2 ðk12 1 k22 Þze2k1 x sink2 x
k2
σ02 2
ðk1 1 k22 Þ z2 2 h2 ðh1 1 h2 Þ e2k1 x ðk1 sink2 x 2 k2 cosk2 xÞ
σz2 5 2
2k2
(2.131)
As an example, consider a glassepoxy sandwich layer with the following parameters:
h1 5 0:365 mm, h2 5 0:735 mm, E1 5 56 GPa, E2 5 17 GPa, G13 5 5:6 GPa, G23 5 6:4 GPa,
ν 13 5 0:095, ν 23 5 0:35, and σ1
2 5 25:5 MPa. The distributions of stresses normalized to the acting
stress σ are presented in Fig. 2.38. As can be seen, there is a stress concentration in the longitudinal
plies in the vicinity of the crack, whereas the stress in the transverse ply, being zero on the crack
surface, practically reaches σ02 at a distance of about 4 mm (or about twice the thickness of the laminate) from the crack. The curves have the expected forms for this problem of stress diffusion.
However, analysis of the second equation of Eq. (2.131) allows us to reveal an interesting phenomenon which can be demonstrated if we increase the vertical scale of the graph in the vicinity of
points A and B (see Fig. 2.38). It follows from this analysis that stress σx2 becomes equal to σ02 at
point A with coordinate
128
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
(σ x1, σ x2) / σ
3
2.5
σ x1
2
σ 10
1.5
B
σ 20
A
1
σ 20
0.5
A B
σ x2
x, mm
0
0
4
8
12
16
FIGURE 2.38
Variation of normalized normal stresses in longitudinal ðσx1 Þ and transverse ðσx2 Þ plies with distance from the
crack.
z
1
h1
2
2h2
3
h1
x
lc
FIGURE 2.39
A system of cracks in the transverse ply.
xA 5
1
k2
π 2 tan21
k2
k1
and reaches a maximum value at point B with coordinate xB 5 π=k2 . This maximum value
0
2π
0B
σmax
x2 5 σ 2 @1 1 e
1
k1
k2 C
A
is higher than stress σ02 which causes failure of the transverse ply. This means that a single crack
cannot exist. When stress σ02 reaches its ultimate value σ1
2 , a regular system of cracks located at a
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
129
distance of lc 5 π=k2 from one another appears in the transverse ply (see Fig. 2.39). For the example considered above, lc 5 12:6 mm.
To study the stress state of a layer with cracks shown in Fig. 2.39, we can use solution (2.130)
but should write it in a different form, i.e.,
σ2 5 C1 sinhk1 x sink2 x 1 C2 sinhk1 x cosk2 x
1 C3 coshk1 x sink2 x 1 C4 coshk1 x cosk2 x
(2.132)
Since the stress state of an element 2lc =2 # x # lc =2 is symmetric with respect to coordinate x,
we should put C2 5 C3 5 0 and find constants C1 and C4 from the following boundary conditions:
σx2 ðx 5 lc =2Þ 5 0; τ xz2 ðx 5 lc =2Þ 5 0
(2.133)
where lc 5 π=k2 .
The final expressions for stresses are
h2 k1
σx1 5 σx3 5 σ01 1 σ02
coshk1 x cosk2 x 1 sinhk1 x sink2 x
h1 c k2
1 k1
0
coshk1 x cosk2 x 1 sinhk1 x sink2 x
σx2 5 σ2 1 2
c k2
σ02 2
τ xz2 5
ðk 1 k22 Þz sinhk1 x cosk2 x
k2 c 1
σ0
σz2 5 2 2 ðk12 1 k22 Þ z2 2 h2 ðh1 1 h2 Þ ðk1 coshk1 x cosk2 x 2 k2 sinhk1 x sink2 xÞ
2k2 c
(2.134)
in which c 5 sinhðπk1 =2k2 Þ.
For the layer under consideration, stress distributions corresponding to σ 5 σ 5 44:7MPa are
shown in Figs. 2.40 and 2.41. Under further loading (σ . σ), two modes of layer failure are possible. The first one is the formation of another transverse crack dividing the block with length lc
in Fig. 2.39 into two pieces. The second one is a delamination in the vicinity of the crack
caused by stresses τ xz and σz (see Fig. 2.41). Usually, the first situation takes place because
stresses τ xz and σz are considerably lower than the corresponding ultimate stresses, whereas the
maximum value of σx2 is close to the ultimate stress σ02 5 σ1
2 . Indeed, the second equation of
Eq. (2.134) yields
0
σmax
x2 5 σ x2 ðx 5 0Þ 5 σ 2 ð1 2 kÞ
0
where k 5 k1 =ðk2 cÞ. For the foregoing example, k 5 3:85 3 1024 . So, σmax
x2 is so close to σ2 that we
can assume that under practically the same load, another crack occurs in the central cross-section
x 5 0 of the central block in Fig. 2.39 (as well as in all the other blocks). Thus, the distance
between the cracks becomes lc 5 π=2k2 (6.4 mm for the example under study). The corresponding
stress distribution can be determined with the aid of Eqs. (2.128) and (2.132), and boundary conditions (2.133), in which we should take lc 5 π=2k2 . The next crack will again appear at the block
center and this process will be continued until failure of the longitudinal plies.
To plot the stressstrain diagram of the cross-ply layer with allowance for the cracks in the
transverse ply, we introduce the mean longitudinal strain
(σ x1, σ x2 ) / σ
3
σx2
2
σ 10
lc
1
σ 20
σx2
x, mm
–8
–4
0
4
8
FIGURE 2.40
Distribution of normalized stresses in longitudinal ðσx1 Þ and transverse ðσx2 Þ plies between the cracks.
lc
(τxz2 ,σz2 ) / σ
0.3
0.2
τxz2
σ z2
0.1
x, mm
–8
–4
0
4 σ
z2
8
–0.1
τ xz2
–0.2
–0.3
FIGURE 2.41
Distribution of normalized shear ðτ xz2 Þ and transverse normal stresses ðσz2 Þ at the ply interface ðz 5 h2 Þ between
the cracks.
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
2
εx 5
h2 lc
lcð=2
hð2
εx2 dz
dx
0
131
0
where
εx2 5
1
ðσx2 2 ν 23 σz2 Þ
E2
For a layer with the properties given above, such a diagram is shown in Fig. 2.42 with a solid
line, and it is in good agreement with experimental results (circles). The formation of cracks is
accompanied by horizontal jumps and reduction in material stiffness. The stressstrain diagram for
the transverse layer that is formally singled out of the diagram in Fig. 2.42 is presented in
Fig. 2.43.
To develop a nonlinear phenomenological model of the cross-ply layer, we need to approximate
the diagram in Fig. 2.43. As follows from this figure and numerous experiments, the most
suitable and simple approximation is that shown by the dashed line. It implies that the ply is linear
1
elastic until its transverse stress σ2 reaches its ultimate value σ1
2 , and after that σ2 5 σ2 , i.e., σ2
remains constant up to failure of the longitudinal plies. This means that under transverse tension, a
unidirectional ply is in a state of permanent failure and takes from the longitudinal plies the necessary load to support this state (Vasiliev & Elpatievskii, 1967). The stressstrain diagram of the
cross-ply layer corresponding to this model is shown in Fig. 2.42 with a dashed line.
Now consider a general plane stress state with stresses σx , σy , and τ xy as in Fig. 2.44. As can be
seen, stress σx induces cracks in the inner ply and stress σy causes cracks in the outer orthogonal
σ , MPa
200
150
100
50
εx , %
0
0
0.2
0.4
FIGURE 2.42
Stressstrain diagram for a glassepoxy cross-ply layer:
x experiment;
sss theoretical prediction;
model.
0.6
0.8
σ x2 , MPa
30
σ 2+
20
10
εx, %
0
0
0.2
0.4
0.6
0.8
FIGURE 2.43
Stressstrain diagram for a transverse ply.
σx
τ xy
τ xy
σy
FIGURE 2.44
A cross-ply layer in a plane stress state.
σ 1, σ 2,τ 12
σ1
σ1
τ 12
∗
τ 12
σ 2∗
σ2
σ2
ε1, ε 2, γ 12
σ2
σ2
σ1
σ1
FIGURE 2.45
Stressstrain diagrams of a unidirectional ply simulating its behavior in the laminate and allowing for cracks in
the matrix.
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
133
plies, whereas the shear stress τ xy can give rise to cracks in all the plies. The ply model that generalizes the model introduced above for a uniaxial tension is demonstrated in Fig. 2.45. To determine
strains corresponding to a given combination of stresses σx , σy , and τ xy , we can use the following
procedure:
1. For the first stage of loading (before the cracks appear), the strains are calculated with the aid
ð1Þ
ð1Þ
of Eqs. (2.114) and (2.115) providing εð1Þ
x ðσÞ, εy ðσÞ, and γ xy ðσÞ, where σ 5 ðσ x ; σy ; τ xy Þ is the
given combination of stresses. Using Eq. (2.112), we find stresses σ1 , σ2 , and τ 12 in principal
material coordinates for all the plies.
2. We determine the combination of stresses σ1k , σ2k and τ 12k which induce the first failure of the
matrix in some ply and indicate the number of this ply, say k, applying the appropriate strength
criterion (see Section 4.2). Then, the corresponding stresses σ 5 ðσx ; σy ; τ xy Þ and strains
ð1Þ ð1Þ εð1Þ
x ðσ Þ, εy ðσ Þ, and γ xy ðσ Þ are calculated.
3. To proceed, i.e., to study the material behavior for σ . σ , we need to consider two possible
cases for the layer stiffnesses. For this purpose, we should write Eq. (2.114) for stiffness
coefficients in a more general form, i.e.,
A11 5
m
n
X
X
ðiÞ ðiÞ
ðjÞ ðjÞ
E1 h0 1
E2 h90 ;
i51
j51
m
n
X
X
ðiÞ ðiÞ
ðjÞ ðjÞ
E2 h0 1
E1 h90
i51
m
n
X
X
ðiÞ ðiÞ
ðjÞ ðjÞ
A12 5
ν ðiÞ
ν ðjÞ
12 E1 h0 1
12 E 1 h90 ;
i51
A22 5
j51
j51
m
n
X
X
ðiÞ
ðjÞ
A44 5
GðiÞ
GðjÞ
12 h0 1
12 h90
i51
(2.135)
j51
ðiÞ
ðjÞ
ðjÞ
where h0 5 hðiÞ
0 =h and h90 5 h90 =h:
a. If σ2k . 0 in the kth ply, it can work only along the fibers, and we should calculate the
stiffnesses of the degraded layer taking E2k 5 0, Gk12 5 0, and ν k12 5 0 in Eq. (2.135).
b. If σ2k , 0 in the kth ply, it cannot work only in shear, so we should take Gk12 5 0 in Eq. (2.135).
Thus, we find coefficients Að2Þ
st (st 5 11, 12, 22, 44) corresponding to the second stage of loading
ð2Þ
(with one degraded ply). Using Eqs. (2.116) and (2.115), we can determine Exð2Þ , Eyð2Þ , Gð2Þ
xy , ν xy , and
ð2Þ
ð2Þ
ð2Þ
ð2Þ
ν yx , and express the strains in terms of stresses, i.e., εx ðσÞ, εy ðσÞ, and γ xy ðσÞ. The final strains
corresponding to the second stage of loading are calculated as
ð2Þ
f
ð1Þ ð2Þ
εfx 5 εð1Þ
x ðσ Þ 1 εx ðσ 2 σ Þ; εy 5 εy ðσ Þ 1 εy ðσ 2 σ Þ
ð2Þ
γ fxy 5 γ ð1Þ
xy ðσ Þ 1 γ xy ðσ 2 σ Þ
To study the third stage, we should find σ1 , σ2 , and τ 12 in all the plies, except the kth one, identify the next degraded ply and repeat step 3 of the procedure which is continued up to failure of the
fibers. The resulting stressstrain curves are multisegmented broken lines with straight segments
and kinks corresponding to degradation of particular plies.
The foregoing procedure was described for a cross-ply layer consisting of plies with different
properties. For the layer made of one and the same material, there are only three stages of loading:
First, before any ply degradation; second, after the degradation of the longitudinal or the transverse
ply only; and third, after the degradation of all the plies.
As a numerical example, consider a carbonepoxy cylindrical pressure vessel consisting of
axial plies with total thickness h0 and circumferential plies with total thickness h90 . The vessel has
134
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
x
σx
y
σy
FIGURE 2.46
Element of a composite pressure vessel.
the following parameters: radius R 5 500 mm, total thickness of the wall h 5 7:5 mm, h0 5 2:5 mm,
h90 5 5 mm. The mechanical characteristics of a carbonepoxy unidirectional ply are E1 5 140 GPa,
1
E2 5 11 GPa, ν 12 5 0:0212, ν 21 5 0:27, σ1
1 5 2000 MPa, and σ2 5 50 MPa. The axial, σx , and
circumferential, σy , stresses are expressed as (see Fig. 2.46)
σx 5
pR
pR
; σy 5
2h
h
(2.136)
where p is the internal pressure.
Using Eqs. (2.114) and (2.116), we calculate first the stiffness coefficients. The result is as
follows:
A11 5 54:1 GPa;
Ex 5 54 GPa;
A12 5 3 GPa;
Ey 5 97 GPa;
A22 5 97:1 GPa
ν xy 5 0:055; ν yx 5 0:031
(2.137)
Substituting the stresses, Eq. (2.136) into the constitutive equations, Eq. (2.115), we obtain
εxð1Þ ðpÞ 5
pR 1
ν xy
pR 1
ν yx
2
ðpÞ
5
2
5 0:58U1023 p; εð1Þ
5 0:66U1023 p
y
h 2Ex
h Ey
Ey
2Ex
where p is measured in megapascals. For axial plies, εx 5 ε1;0 and εy 5 ε2;0 . The corresponding
stresses are
ð1Þ
σð1Þ
1;0 ðpÞ 5 E1 ðε1;0 1 ν 12 ε2;0 Þ 5 83:2p; σ 2;0 ðpÞ 5 E 2 ðε2;0 1 ν 21 ε1;0 Þ 5 9:04p
For circumferential plies, εx 5 ε2;90 , εy 5 ε1;90 and
ð1Þ
σð1Þ
1;90 ðpÞ 5 E1 ðε1;90 1 ν 12 ε2;90 Þ 5 94:15p; σ 2;90 ðpÞ 5 E2 ðε2;90 1 ν 21 ε1;90 Þ 5 8:4p
ð1Þ
As can be seen, σð1Þ
2;0 . σ2;90 . This means that the cracks appear first in the axial plies under the
1
pressure p that can be found from the equation σð1Þ
2;0 ðp Þ 5 σ 2 . The result is p 5 5:53 MPa.
To study the second stage of loading for p . p , we should put E2 5 0, and ν 12 5 0 in
Eq. (2.135) for the axial plies. Then, the stiffness coefficients and elastic constants become
A11 5 54:06 GPa; A12 5 2 GPa; A22 5 93:4 GPa
Ex 5 54 GPa; Ey 5 93:3 GPa; ν xy 5 0:037; ν yx 5 0:021
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
135
The strains and stresses in the plies are
23
23
ð2Þ
εð2Þ
x ðpÞ 5 0:59U10 p; εy ðpÞ 5 0:7U10 p
ð2Þ
σð2Þ
1;0 ðpÞ 5 82:6p; σ 1;90 ðpÞ 5 99:8p
σð2Þ
2;90 ðpÞ 5 8:62p
The total transverse stress in the circumferential plies can be calculated as
σ2;90 5 σð1Þ
2;90 ðpÞ 1 8:62ðp 2 pÞ
Using the condition σ2;90 ðp Þ 5 σ1
2 , we find the pressure p 5 5:95 MPa at which cracks
appear in the matrix of the circumferential plies.
For p $ p , we should take E2 5 0 and ν 12 5 0 for all the plies. Then
A11 5 46:2 GPa; A12 5 0; A22 5 93:4 GPa
Ex 5 46:2 GPa; Ey 5 93:4 GPa; ν xy 5 ν yx 5 0
23
εxð3Þ ðpÞ 5 0:72U1023 p; εð3Þ
y ðpÞ 5 0:71U10 p
(2.138)
ð3Þ
σð3Þ
1;0 ðpÞ 5 100:8p; σ1;90 ðpÞ 5 99:4p
The total stresses acting along the fibers are
ð2Þ ð3Þ
σ1;0 ðpÞ 5 σð1Þ
1;0 ðp Þ 1 σ1;0 ðp 2 pÞ 1 σ 1;0 ðp 2 p Þ 5 100:8p 2 105
ð2Þ
ð3Þ
σ1;90 ðpÞ 5 σð1Þ
1;90 ðp Þ 1 σ 1;90 ðp 2 p Þ 1 σ 1;90 ðp 2 p Þ 5 99:4p 2 28:9
To determine the ultimate pressure, we can use two possible strength conditions: for axial fibers,
and for circumferential fibers. The criterion σ1;0 ðpÞ 5 σ1
1 yields p 5 20:9 MPa, whereas the criterion
σ1;90 ðpÞ 5 σ1
gives
p
5
20:4
MPa.
Thus,
the
burst
pressure
governed by failure of the fibers in the
1
circumferential plies, is p 5 20:4 MPa.
The strains can be calculated for all three stages of loading using the following equations:
•
•
•
for p # p
εx;y ðpÞ 5 εð1Þ
x;y ðpÞ
for p , p # p
ð2Þ
εx;y ðpÞ 5 εð1Þ
x;y ðp Þ 1 εx;y ðp 2 p Þ
for p , p # p
ð2Þ ð3Þ
εx;y ðpÞ 5 εð1Þ
x;y ðp Þ 1 εx;y ðp 2 p Þ 1 εx;y ðp 2 p Þ
For the pressure vessel under study, the dependency of the circumferential strain on pressure is
shown in Fig. 2.47 (solid line). The circles correspond to failure of the matrix and fibers.
For comparison, consider two limiting cases. First, assume that no cracks occur in the matrix,
and the material stiffness is specified by Eq. (2.137). The corresponding diagram is shown in
Fig. 2.47 with a dashed line. Second, suppose that the load is taken by the fibers only, i.e., use the
monotropic model of a ply introduced in Section 1.3. Then, the material stiffnesses are given by
Eq. (2.138). The corresponding result is also presented in Fig. 2.47. It follows from this figure that
1
all three models give close results for the burst pressure (which is expected since σ1
2 , , σ1 ), but
provide different strains.
136
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
Ρ, MPa
25
20
15
10
5
0
0
0.5
1
εy, %
1.5
FIGURE 2.47
Dependence of the circumferential strain of the carbonepoxy pressure vessel on pressure:
model allowing for cracks in the matrix;
model ignoring cracks in the matrix;
model ignoring the matrix.
2.4.3 COMPOSITES WITH CONTROLLED CRACKS
As follows from the previous section, for the case when the transverse ultimate elongation of a ply,
ε2 , is less than the corresponding longitudinal elongation, ε1 , (see Table 2.2), the stress σ in
Eq. (2.123) induces a system of cracks in the matrix of the transverse ply as shown in Fig. 2.39. As
has been already noted, these cracks do not cause laminate failure because its strength is controlled
by the longitudinal plies. What is actually not desirable is matrix failure in the process of laminate
loading. So, since the cracks shown in Fig. 2.39 will occur anyway at some stress σ, suppose that
the material has these cracks before loading, i.e., that the transverse ply consists of individual strips
with width lc as in Fig. 2.39. The problem is to find the width lc for which no other cracks will
appear in the transverse ply up to failure of the fibers in the longitudinal plies.
Consider the solution in Eq. (2.132), take C2 5 C3 5 0 and find the constants C1 and C4 from
the boundary conditions in Eq. (2.133) in which lc is some unknown width. The resulting expression for the stress in the transverse ply is
(
σx2
5 σ02
12
1
[ðk2 coshλ1 sinλ2 2 k1 sinhλ1 cosλ2 Þsinhk1 x sink2 x
k1 sinλ2 cosλ2 1 k2 sinhλ1 coshλ1
1 ðk1 coshλ1 sinλ2 1 k2 sinhλ1 cosλ2 Þcoshk1 x cosk2 x
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
137
in which λ1 5 k1 lc =2 and λ2 5 k2 lc =2. The maximum stress acts at x 5 0 (see Fig. 2.40) and can be
presented as
0
σm
2 5 σ2 ½1 2 Fðlc Þ
(2.139)
where
Fðlc Þ 5
2ðk1 coshλ1 sinλ2 1 k2 sinhλ1 cosλ2 Þ
k1 sin2λ2 1 k2 sinh2λ1
(2.140)
The stress σ02 in Eq. (2.139) is specified by the second equation of Eq. (2.122). Taking into
account the first equation, we have
σ02 5
E2 0
σ
E1 1
where σ01 is the stress in the longitudinal plies. So, Eq. (2.139) can be written as
σm
2 5
E2 0
σ ½1 2 Fðlc Þ
E1 1
(2.141)
Now suppose that σ1 5 σ1 , i.e., that the longitudinal stress reaches the corresponding ultimate
value. The cracks in the matrix of the transverse ply do not appear if σm
2 # σ2 , where σ2 is the
transverse tensile strength of the ply. Then, Eq. (2.141) yields
Fðlc Þ $ t
(2.142)
where t 5 1 2 E1 σ2 =E2 σ1 .
As an example, consider a cross-ply (see Fig. 2.35) carbonepoxy composite with the following
parameters
E1 5 140 GPa; E2 5 11 GPa; G13 5 5:5 GPa; G23 5 4:1 GPa
ν 23 5 0:3; σ1 5 2000 MPa; σ2 5 50 MPa
for which t 5 0:68. Introduce normalized thicknesses of the plies as
h1 5
2h1
2h2
; h2 5
h
h
where h 5 2ðh1 1 h2 Þ (see Fig. 2.37). Let h1 5 1 2 α, h2 5 α, where the parameter α specifies the
relative thickness of the transverse ply. The dependencies of the coefficients k1 5 k1 =h and
k2 5 k2 =h (in which k1 and k2 are given in the notations to Eq. (2.130)) on the parameter α are
shown in Fig. 2.48. The dependence of function F in Eq. (2.140) on the normalized distance
between the cracks lc 5 lc =h is presented in Fig. 2.49 for α 5 0:1; 0:5; and 0:9 . The intersections
of the horizontal line F 5 t 5 0:68 give the values of lc for which no new cracks appear in the transverse ply up to the fibers’ failure. The final dependence of lc on α is shown in Fig. 2.50. As can be
seen, lc varies from about two up to four thicknesses of the laminate. For h1 5 h2 5 δ, where
δ 5 0:15 mm is the thickness of the unidirectional ply, we get h 5 4δ, α 5 0:5, and lc 5 1:9 mm. A
yarn of such width is typical for carbon fabrics made of 3K carbon tows. Clearly, there will be no
cracks between the yarns in the fabric layer, since the gaps between the yarns are filled with resin
characterized by high elongation. Experiments with such fabric composites show that the tensile
stressstrain diagram of the material is linear up to failure, and no cracks are observed in the
matrix.
138
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
k1 , k2
2
1.81
1.8
1.6
1.55
1.4
k1
1.41
1.32
1.2
1.2
1.15
1.16
1.25
1
0.8
1.17
1.15
0.1
0.05 0.02
0.79
0.59
0.6
k2
0.45
0.4
0.33
0.22
0.2
α
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
FIGURE 2.48
Dependencies of the coefficients k 1 and k 2 on the relative thickness of the transverse ply α.
2.4.4 TWO-MATRIX COMPOSITES
The problem of the analysis of a cracked cross-ply composite laminate has been studied by Tsai
and Azzi (1966), Vasiliev and Elpatievskii (1967), Vasiliev et al. (1970), Hahn and Tsai (1974),
Reifsnaider (1977), Hashin (1987), Lungren and Gudmundson (1999), and many other authors. In
spite of this, the topic is still receiving repeated attention in the literature. Taking into account that
matrix degradation leads to reduction of material stiffness and fatigue strength, absorption of moisture, and many other consequences that are difficult to predict but are definitely undesirable, it is
surprising how many attempts have been made to study this phenomenon rather than try to avoid it.
At first glance, the problem looks simple; all we need is to synthesize a unidirectional composite
whose ultimate elongations along and across the fibers, i.e., ε1 and ε2 , are the same. Actually, the
problem is even simpler, because ε2 can be less than ε1 by a factor that is equal to the safety factor
of the structure. This means that matrix degradation can occur, but at the load that exceeds the
operational level (the safety factor is the ratio of the failure load to the operational load and can
vary from 1.25 up to 3 or more depending on the application of a particular composite structure).
Returning to Table 2.2, in which ε1 and ε2 are given for typical advanced composites, we can see
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
F
1
0.995
0.99
0.94
0.95
0.9
0.92
0.8
0.81
0.7
t = 0.68
0.79
0.74
0.68
0.68
0.68
0.59
0.6
0.58
0.5
0.49
0.4
0.4
α = 0.1
0.3
0.3
0.2
0.175
0.1
1.8
0
1
0.03
3.15
2
0.26
α = 0.5
0.19
0
α = 0.9
3
3.6
0.16
0.1
0.1
0
4
5
6
7
0.06
0.034
0.014
0.05 0.03
8
9
10
FIGURE 2.49
Dependencies of function F on the normalized width of the strip for α 5 0.1, 0.5, and 0.9.
lc
4
3.6
3
3.15
2.6
2
1.8
1
0
α
0
0.2
0.4
0.6
0.8
1
FIGURE 2.50
Normalized width of the strip l c as a function of the relative thickness of transverse ply α.
lc
139
140
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
that ε1 . ε2 for all the materials and that for polymeric matrices the problem could be, in principle,
solved if we could increase ε2 up to about 1%.
Two main circumstances hinder the direct solution of this problem. The first is that being locked
between the fibers, the matrix does not show the high elongation that it has under uniaxial tension
and behaves as a brittle material (see Section 1.4.2). To study this effect, epoxy resins were modified to have different ultimate elongations. The corresponding curves are presented in Fig. 2.51
(only the initial part of curve 4 is shown in this figure; the ultimate elongation of this resin is
60%). Fiberglass composites that have been fabricated with these resins were tested under transverse tension. As can be seen in Fig. 2.52, the desired value of ε2 (that is, about 1%) is reached if
σm, MPa
120
2
100
1
3
80
60
40
20
4
60 %
0
0
4
8
12
16
20
εm,%
FIGURE 2.51
Stressstrain curves for epoxy matrices modified for various ultimate elongations.
σ2, MPa
2
50
3
40
1
30
4
20
10
ε2,%
0
0
0.2
0.4
0.6
0.8
1
1.2
FIGURE 2.52
Stressstrain curves for transverse tension of unidirectional fiberglass composites with various epoxy matrices
(numbers on the curves correspond to Fig. 2.51).
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
141
the matrix elongation is about 60%. However, the stiffness of this matrix is relatively low, and the
second circumstance arises; matrix material with low stiffness cannot provide sufficient stress diffusion in the vicinity of damaged or broken fibers (see Section 1.2.3). As a result, the main material
characteristic—its longitudinal tensile strength—decreases. Experimental results corresponding to
composites with resins 1, 2, 3, and 4 are presented in Fig. 2.53. Thus, a significant increase in
transverse elongation is accompanied by an unacceptable drop in longitudinal strength (see also
Chiao, 1979).
One of the possible ways for synthesizing composite materials with high transverse elongation
and high longitudinal strength is to combine two matrix materials: one with high stiffness to bind
the fibers and the other with high elongation to provide the appropriate transverse deformability
(Vasiliev & Salov, 1984). The manufacturing process involves two-stage impregnation. At the first
stage, a fine tow is impregnated with a high-stiffness epoxy resin (of the type 2 in Fig. 2.51) and
cured. The properties of fiberglass composite fabricated in this way are as follows:
•
•
•
•
•
•
•
•
Number of elementary glass fibers in the cross-section—500
Mean cross-sectional area—0.15 mm2
Fiber volume fraction—0.75
Density—2.2 g=cm3
Longitudinal modulus—53.5 GPa
Longitudinal strength—2100 MPa
Longitudinal elongation—4.5%
Transverse modulus—13.5 GPa
σ1, MPa
1500
1
2
1400
3
1300
1200
4
1100
εm,%
1000
0
20
40
60
FIGURE 2.53
Dependence of the longitudinal strength on the matrix ultimate elongation (numbers on the curve correspond to
Figs. 2.51 and 2.52).
142
•
•
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
Transverse strength—400 MPa
Transverse elongation—0.32%
At the second stage, a tape formed of composite fibers is impregnated with a highly deformable
epoxy matrix whose stressstrain diagram is presented in Fig. 2.54. The microstructure of the
resulting two-matrix unidirectional composite is shown in Fig. 2.55 (the dark areas are crosssections of composite fibers; the magnification is not sufficient to see the elementary glass fibers).
Stressstrain diagrams corresponding to transverse tension, compression, and in-plane shear of this
material are presented in Fig. 2.16.
The main mechanical characteristics of the two-matrix fiberglass composite are listed in
Table 2.3 (material No.1). As can be seen, two-stage impregnation results in relatively low fiber
volume content (about 50%). Material No.2, composed of composite fibers and a conventional
epoxy matrix, has also a low fiber fraction; however, its transverse elongation is 10 times lower
than that of material No.1. Material No.3 is a conventional glassepoxy composite that has the
highest longitudinal strength and the lowest transverse strain. Comparing materials Nos.1 and 3, we
σm, MPa
20
16
12
8
4
εm,%
0
0
20
40
60
FIGURE 2.54
Stressstrain diagram of a deformable epoxy matrix.
FIGURE 2.55
Microstructure of a unidirectional two-matrix composite.
80
100
120
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
143
Table 2.3 Properties of GlassEpoxy Unidirectional Composites
No
1
2
3
4
Material
Components
Composite fibers and
deformable matrix
Composite fibers and
high-stiffness matrix
Glass fibers and highstiffness matrix
Glass fibers and
deformable matrix
Fiber
Volume
Fraction
Longitudinal
Strength σ1
1
(MPa)
Ultimate
Transverse Strain
ε1
2 (%)
Density
ρ (g/cm3)
Specific Strength
3
σ1
1 =ρ 3 10 (m)
0.51
1420
3.0
1.83
77.6
0.52
1430
0.3
1.88
76.1
0.67
1470
0.2
2.07
71.0
0.65
1100
1.2
2.02
54.4
σ , MPa
500
1
2
400
300
200
100
0
0
0.4
0.8
1.2
1.6
2
εx , %
FIGURE 2.56
Stressstrain diagrams of a conventional (1) and two-matrix (2) cross-ply glassepoxy layers under tension:
sss theoretical prediction;
x experiment.
can see that although the fiber volume fraction of the two-matrix composite is lower by 24%, its
longitudinal strength is less than that of a traditional composite by only 3.4% (because the composite fibers are not damaged in the processing of composite materials), whereas its specific strength is
a bit higher (due to its lower density). Material No.4 demonstrates that direct application of a
highly deformable matrix allows us to increase transverse strains but results in a 23% reduction in
longitudinal specific strength.
Thus, two-matrix glassepoxy composites have practically the same longitudinal strength as
conventional materials but their transverse elongation is greater by an order of magnitude.
Comparison of a conventional cross-ply glassepoxy layer and a two-matrix one is presented in
Fig. 2.56. Line 1 corresponds to a traditional material and, typically for this material, has a kink
corresponding to matrix failure in the transverse plies (see also Fig. 2.37). A theoretical diagram
144
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
FIGURE 2.57
Intensity of acoustic emission for a cross-ply two-matrix composite (upper) and a conventional fiberglass
composite (lower).
was plotted using the procedure described above. Line 2 corresponds to a two-matrix composite.
As can be seen, there is no kink on the stressstrain diagram. To prove that no cracks appear in
the matrix of this material under loading, the intensity of acoustic emission was recorded during
loading. The results are shown in Fig. 2.57.
Composite fibers of two-matrix materials can be also made from fine carbon or aramid tows,
and the deformable thermosetting resin can be replaced with a thermoplastic matrix (Vasiliev,
Salov, & Salov, 1997). The resulting hybrid thermosetthermoplastic unidirectional composite tape
(“prepreg”) is characterized by high longitudinal strength and transverse strain exceeding 1%.
Having high strength, composite fibers are not damaged in the process of laying-up or winding,
and the tapes formed from these fibers are readily impregnated even with high-viscosity thermoplastic polymers. Another way to fabricate twomatrix thermoset—thermoplastic composites is
based on three-dimensional printing technology.
2.4.5 TWO-MATRIX COMPOSITE MADE BY THREE-DIMENSIONAL PRINTING
The basic process of additive manufacturing technology using three-dimensional (3D) printing
involves building of three-dimensional geometry by deposition of successive layers of molten thermoplastic polymer. Due to the relatively low mechanical properties of thermoplastics, this process
is mainly used to produce scale models rather than functional load-carrying structural elements. To
2.4 ORTHOGONALLY REINFORCED ORTHOTROPIC LAYER
145
increase the stiffness of 3D printed thermoplastic parts, the polymer can be prefilled with short carbon fibers. However, these fibers do not significantly affect the material properties which are only
slightly higher than those of the base polymer.
Dramatic improvement can be achieved if 3D printed thermoplastic parts are reinforced by continuous fibers. However, existing approaches do not demonstrate significant progress in this direction. For example, modern 3D printers are capable of reinforcing polymer with continuous aramid
fibers that provide strength up to 100 MPa and modulus of elasticity up to 9 GPa (see Melenka,
Cheung, Schofield, Dawson, & Carey, 2016). Obviously, these values are much less than the corresponding characteristics of traditional laminated composites. The reason for this is associated with
high viscosity of molten thermoplastic polymer which considerably hinders the impregnation of the
reinforcing fibers. For high fiber content composites, the high viscosity of polymer melt usually
requires application of high temperature, pressure, and time (Peters, 1998). Three-dimensional
printing cannot provide the required parameters in the manufacturing process and cannot yield
proper fiber impregnation and thus high mechanical properties in the fabricated composite material.
Application of the two-matrix composite technology in conjunction with three-dimensional printing
offers an opportunity to overcome these difficulties. The approach is based on utilization of composite fibers described in Section 2.4.4. These fibers are made of fine fibrous tows impregnated
with high stiffness thermosetting resin and completely cured. Composite fibers made of 1 or 3K
carbon tows have relatively large diameters (0.3 and 0.6 mm, respectively, in contrast to 57 μm
for the elementary filaments) and are readily impregnated with molten thermoplastic polymer.
Composite fibers taken from the spool mounted on the 3D printer (see Fig. 2.58) pass through the
printer head in which they are covered with molten thermoplastic polymer and placed on a heated
FIGURE 2.58
3D printer.
146
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
FIGURE 2.59
Microstructure of the 3D printed composite material.
plate forming the designed structural element (loop-shaped structure shown in Fig. 2.58). The
microstructure of the final material is shown in Fig. 2.59 (Azarov et al., 2016).
Using the first-order micromechanical model described in Section 1.3, we can calculate the density of the two-matrix composite as
ρ 5 ρf vf 1 ρms vms 1 ρmp vmp
where vf , vms , and vmp are the volume fractions of fibers, thermosetting matrix, and thermoplastic
matrix, respectively, such that
vf 1 vms 1 vmp 5 1
The equations derived in Section 1.3 for the material elastic constants, i.e., Eqs. (1.76), (1.78),
(1.79), and (1.81) can be generalized for a two-matrix composite as follows (Golubev, 2016):
E1 5 Ef vf 1 Ems vms 1 Emp vmp
vmp ν 2mp
1
vf
vms
vmp
vms ν 2ms
ðvf ν f 1vms ν ms 1vmp ν mp Þ2
5
1
1
2
2
2
Ef
Ems
Emp
Ems
Emp
Ef vf 1 Ems vms 1 Emp vmp
E2
ν 21
ν f vf 1 ν ms vms 1 vmp ν mp
5
E1
Ef vf 1 Ems vms 1 Emp vmp
1
vf
vms
vmp
5
1
1
Gf
Gms
Gmp
G12
Experimental samples of unidirectional two-matrix composite material were fabricated using
T-300-3K carbon tows, epoxy resin and polyamide (PLA) thermoplastic polymer. The properties of
the components are: Ef 5 230 GPa, ρf 5 1770 kg=m3 , Ems 5 3:9 GPa, ρms 5 1270 kg=m3 ,
2.5 ANGLE-PLY ORTHOTROPIC LAYER
147
Emp 5 3 GPa, ρmp 5 1100 kg=m3 . The operating melting temperature of PLA is 250 C. The volume
fractions of the constituents of the two-matrix composite material are about vf 5 0:3, vms 5 0:3, and
vmp 5 0:4. The fabricated unidirectional material had a longitudinal modulus E1 5 65 GPa, strength
2
under longitudinal tension σ1
1 5 950 MPa, strength under longitudinal compression σ 1 5 300 MPa,
3
density ρ 5 1290 kg=m and the ultimate elongation under tension across the fibers was about 2%.
Comparing these results with the properties of unidirectional composites fabricated by conventional methods, we can conclude that stiffness and strength of a two-matrix composite are relatively
low. The main reason for that is associated with relatively low fiber volume fraction. The maximum fiber volume fraction that can be achieved in the process with two-stage impregnation is
about 0.45 which is less than that for traditional process with one-stage impregnation (about 0.65).
For this reason, two-matrix composites cannot compete with traditional composite materials.
However, in combination with 3D printing, they allow us to manufacture structures with complicated fiber trajectories which cannot be made by traditional methods.
2.5 ANGLE-PLY ORTHOTROPIC LAYER
The angle-ply layer is a combination of an even number of alternating plies with angles 1φ and
2φ as in Fig. 2.60. The structure of this layer is typical for the process of filament winding (see
Fig. 2.61). As for the cross-ply layer considered in the previous section, an angle-ply layer is actually a laminate, but for a large number of plies it can be approximately treated as a homogeneous
orthotropic layer (see Section 3.6.3).
2.5.1 LINEAR ELASTIC MODEL
Consider two symmetric systems of unidirectional anisotropic plies consisting of the same number
of plies, made of one and the same material and having alternating angles 1φ and 2φ. Then, the
+φ
σ x+
+
τ xy
τ xz
σy
+
τ yz
−φ
σ x−
−
τ xy
σ y−
FIGURE 2.60
Two symmetric plies forming an angle-ply layer.
σy
σx
τ xy
148
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
FIGURE 2.61
Angle-ply layer of a filament-wound shell.
total stresses σx , σy , and τ xy acting on the layer can be expressed in terms of the corresponding
stresses acting in the 1φ and 2φ plies as
h
h
h
h
1 σ2
; σ y h 5 σ1
1 σ2
x
y
y
2
2
2
2
2 h
1h
τ xy h 5 τ xy 1 τ xy
2
2
σ x h 5 σ1
x
(2.143)
where h is the total thickness of the layer. Stresses with superscripts “ 1 ” and “ 2 ” are related to
strains εx , εy , and γ xy (which are assumed to be the same for all the plies) by Eq. (2.71), i.e.,
6
6
6
6
6
6
σx6 5 A11
εx 1 A12
εy 1 A14
γ xy ; σy6 5 A21
εx 1 A22
εy 1 A24
γ xy
6
6
6
6
τ xy 5 A41 εx 1 A42 εy 1 A44 γ xy
(2.144)
2
1
2
1
2
1
2
1
2
in which A1
11 5 A11 5 A11 , A12 5 A12 5 A12 , A22 5 A22 5 A22 , A14 5 2 A14 5 A14 , A24 5 2 A24 5 A24 ,
1
2
and A44 5 A44 5 A44 , where Amn (mn 5 11, 12, 22, 14, 24, 44) are specified by Eq. (2.72).
Substituting Eq. (2.144) into Eq. (2.143), we arrive at the following constitutive equations for an
angle-ply layer
σx 5 A11 εx 1 A12 εy
σy 5 A21 εx 1 A22 εy
τ xy 5 A44 γ xy
(2.145)
The inverse form of these equations is
εx 5
σx
σy
σy
σx
τ xy
2 ν xy ; εy 5
2 ν yx ; γ xy 5
Ex
Ey
Ey
Ex
Gxy
(2.146)
where
A212
A2
; Ey 5 A22 2 12 ; Gxy 5 A44
A22
A11
A12
A12
ν xy 5
; ν yx 5
A11
A22
Ex 5 A11 2
(2.147)
It follows from Eqs. (2.145) and (2.146) that the layer under study is orthotropic.
Now derive constitutive equations relating transverse shear stresses τ xz and τ yz and the corresponding shear strains γ xz and γ yz . Let the angle-ply layer be loaded by stress τ xz . Then for all the
2
1
2
1
2
plies,τ 1
xz 5 τ xz 5 τ xz , and because the layer is orthotropic, γ xz 5 γ xz 5 γ xz , and γ yz 5 γ yz 5 γ yz 5 0. In
2.5 ANGLE-PLY ORTHOTROPIC LAYER
149
2
1
2
1
2
a similar way, applying stress τ yz we have τ 1
yz 5 τ yz 5 τ yz , γ yz 5 γ yz 5 γ yz , and γ xz 5 γ xz 5 γ xz 5 0.
Writing the last two constitutive equations of Eq. (2.71) for these two cases, we arrive at
τ xz 5 A55 γ xz ; τ yz 5 A66 γ yz
(2.148)
where the stiffness coefficients A55 and A66 are specified by Eq. (2.72).
The dependencies of Ex and Gxy on φ, plotted using Eq. (2.147), are shown in Fig. 2.62 with
solid lines. The theoretical curve for Ex is in very good agreement with experimental data shown
with circles (Lagace, 1985). For comparison, the same moduli are presented for the 1φ anisotropic
layer considered in Section 2.3.1. As can be seen, Ex ð 6 φÞ $ Ex1 . To explain this effect, consider
uniaxial tension of both layers in the x-direction. Whereas tension of the 1φ and 2φ individual
plies shown in Fig. 2.63 is accompanied with shear strain, the system of these plies does not demonstrate shear under tension and, as a result, has higher stiffness. Working as plies of a symmetric
angle-ply layer, individual anisotropic 1φ and 2φ plies are loaded not only with a normal stress
σx that is applied to the layer, but also with shear stress τ xy that restricts the shear of individual
plies (see Fig. 2.63). In order to find the reactive shear stress, which is balanced between the plies,
we can use Eq. (2.75). Taking σy 5 0, we can simulate the stressstrain state of the ply in the
angle-ply layer putting γ xy 5 0. Then, the third equation yields
Ex, Gxy, GPa
140
120
E x (±φ )
100
80
Ex+
60
Gxy (±φ )
40
20
+
Gxy
φ°
0
0
15
30
45
60
75
90
FIGURE 2.62
Dependencies of the moduli of a carbonepoxy layer on the orientation angle:
sss orthotropic angle-ply 6 φ layer;
anisotropic 6 φ layer;
x experiment for an angle-ply layer.
150
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
(A)
(B)
FIGURE 2.63
Deformation and stresses induced in individual plies (A) and bonded symmetric plies (B) by uniaxial tension.
τ xy 5 2 η1
xy;x
G1
xy
σx
Ex1
Superscript “ 1 ” indicates that elastic constants correspond to an individual 1φ ply.
Substituting this shear stress into the first equation of Eq. (2.75), we arrive at σx 5 Ex εx , where
Ex 5
Ex1
5
1
1 2 η1
x;xy ηxy;x
Ex1
G1
xy
1 2 1 ðη1
Þ2
Ex xy;x
(2.149)
is the modulus of the 6 φ angle-ply layer.
Under pure shear of an angle-ply layer, its plies are loaded with additional normal stresses.
These stresses can be found if we take εx 5 0 and εy 5 0 in the first two equations of Eq. (2.75).
The result is
σx 5 2 τ xy
1 1
1 1
Ex1 ðη1
Ey1 ðη1
x;xy 2 ν xy ηy;xy Þ
y;xy 2 ν yx ηx;xy Þ
;
σ
5
2
τ
y
xy
1 1
1 1
G1
G1
xy ð1 2 ν xy ν yx Þ
xy ð1 2 ν xy ν yx Þ
Substituting these expressions for the stresses in the third equation, we get τ xy 5 Gxy γ xy , where
Gxy 5
1 1
G1
xy ð1 2 ν xy ν yx Þ
1
1
1
1
1
1
1
1
1 1
1
1 2 ν xy ν yx 2 ηx;xy ηxy;x 2 ηy;xy ηxy;y 2 ν 1
xy ηy;xy ηxy;x 2 ν yx ηx;xy ηxy;y
is the shear modulus of an angle-ply layer which is much higher than G1
xy (see Fig. 2.62).
Tensile loading of a 6 45 angle-ply specimen provides a simple way to determine the inplane shear modulus of a unidirectional ply, G12. Indeed, for this layer, Eqs. (2.72) and (2.147)
yield
1
1
45
45
A45
11 5 A22 5 ðE 1 1 E2 1 2E1 ν 12 1 4G12 Þ; A12 5 E1 ν 12 1 ðE1 1 E2 2 2E1 ν 12 2 4G12 Þ
4
4
and
E45 5
1
1
45
45
45
45
45
ðA45
11 1 A12 ÞðA11 2 A12 Þ; 1 1 ν 45 5 45 ðA11 1 A12 Þ
A45
A11
11
2.5 ANGLE-PLY ORTHOTROPIC LAYER
151
45
Taking into account that A45
11 2 A12 5 2G12 , we have
G12 5
E45
2ð1 1 ν 45 Þ
(2.150)
Thus, to find G12, we can test a 6 45 specimen under tension, measure εx and εy , determine
E45 5 σx =εx , ν 45 5 2 εy =εx , and use Eq. (2.150) rather than perform the cumbersome tests
described in Chapter 1, Mechanics of a Unidirectional Ply.
2.5.2 NONLINEAR MODELS
To describe the nonlinear behavior of an angle-ply layer associated with material nonlinearity in its
plies, we can use nonlinear constitutive equations, Eqs. (2.60) or (2.64) instead of Hooke’s law.
Indeed, assuming that the ply behavior is linear under tension or compression along the fibers, we
can write these equations in the following general form
ε1 5 c11 σ1 1 c12 σ2 ; ε2 5 c12 σ1 1 c22 σ2 1 ω2 ðσ2 ; τ 12 Þ; γ 12 5 c44 τ 12 1 ω12 ðσ2 ; τ 12 Þ
The functions ω2 and ω12 include all the nonlinear terms. The inverse form of these equations is
σ1 5 C11 ε1 1 C12 ε2 2 C12 ω2 ; σ2 5 C12 ε1 1 C22 ε2 2 C22 ω2 ;
τ 12 5 C44 γ 12 2 C44 ω12
(2.151)
in which
C11 5
c22
c11
1
c12
; C22 5
; C44 5
; c 5 c11 c22 2 c212 :
; C12 5 2
c44
c
c
c
Repeating the derivation of Eq. (2.145) and using this time Eq. (2.151) as the constitutive equations for the ply, we arrive at
σx 5 A11 εx 1 A12 εy 2 Aω11 ; σy 5 A21 εx 1 A22 εy 2 Aω22 ;
τ xy 5 A44 γ xy 2 Aω44
where s 5 sinφ, and c 5 cosφ.
Aω11 5 ðC22 s2 1 C12 c2 Þω2 2 2C44 csω12 ; Aω22 5 ðC22 c2 1 C12 s2 Þω2 1 2C44 csω12 ;
Aω44 5 ðC12 2 C22 Þcsω2 1 C44 ðc2 2 s2 Þω12
These equations can be used in conjunction with the method of elastic solutions described in
Section 2.1.2.
As an example, consider the two-matrix glassepoxy composite described in Section 2.4.4
(see also Figs. 2.16, 2.30, and 2.31). Theoretical (solid lines) and experimental (dashed lines)
stressstrain diagrams for 6 30 , 6 45 , and 6 75 angle-ply layers under tension along the
x-axis are shown in Fig. 2.64.
Angle-ply layers demonstrate a specific type of material nonlinearity: structural nonlinearity
that can occur in the layers composed of linear elastic plies due to a change in the plies’ orientations caused by loading. Since this effect manifests itself at high strains, consider a geometrically
nonlinear problem of the ply deformation. This deformation can be described with the longitudinal,
ε1 , transverse, ε2 , and shear, γ 12 , strains that follow from Fig. 2.65 and can be expressed as
152
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
(A) σ x , MPa
80
(B) σ x , MPa
60
12
40
8
20
4
0
0
16
0.4
0.8
(C)
εx, %
1.2
0
εx, %
0
1
2
3
4
5
σ x , MPa
12
8
4
εx, %
0
0
0.4
0.8
1.2
1.6
FIGURE 2.64
Theoretical (solid lines) and experimental (dashed lines) stressstrain diagrams for 6 30 (A), 6 45 (B), and
6 75 (C) angle-ply two-matrix composites under uniaxial tension.
ε1 5
1
1
π
ðds0 2 ds1 Þ; ε2 5
ðds0 2 ds2 Þ; γ 12 5 2 ψ
ds1 1
ds2 2
2
(2.152)
In addition to this, we introduce strain εv2 in the direction normal to the fibers
εv2 5
1
ðdsv2 2 ds2 Þ
ds2
(2.153)
and the angle of rotation of the element as a solid in the 12-plane
1
ω12 5 ðω1 2 ω2 Þ
2
π
where ω1 5 φ0 2 φ; ω2 5 1 φ 2 ðφ0 1 ψÞ are the angles of rotation of axes 10 and 20 (see
2
Fig. 2.65). Thus,
2.5 ANGLE-PLY ORTHOTROPIC LAYER
153
y
2′′ 2
ω2
2′ ds1′
ds2′′
ψ
2
ds2′
1′
ω1
1
φ′
ds1
ds2
1
φ
x
FIGURE 2.65
Ply element before and after deformation.
y
dsα′
α′
ux
dy ′
d x′
dsα
α
uy
dy
x
dx
FIGURE 2.66
Linear element before and after deformation.
ω12 5 φ0 2 φ 1
ψ π
2
2
4
(2.154)
Consider some arbitrary element dsα , shown in Fig. 2.66, and introduce its strain
dsα 5
1
ðds0 2 dsα Þ
dsα α
(2.155)
Taking into account that as a result of deformation, the elements dx and dy become
dx0 5 dx 1 dux and dy0 5 dy 1 duy ; we get
ds2α 5 dx2 1 dy2
ðds0α Þ2 5 ðdx0 Þ2 1 ðdy0 Þ2 5 ðdx1dux Þ2 1 ðdy1duy Þ2
5 ð11εx Þ2 dx2 1 ð11εy Þ2 dy2 1 2εxy dxdy
154
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
where
"
#
@ux
1 @ux 2 1 @uy 2
1
ð11εx Þ 5 1 1 2
1
2 @x
2 @x
@x
"
2
#
@uy
1 @ux
1 @uy 2
2
1
1
ð11εy Þ 5 1 1 2
@y
2 @y
2 @y
2
εxy 5
(2.156)
@ux
@uy
@ux @ux
@uy @uy
1
1
1
@y
@x
@x @y
@x @y
Using Eq. (2.155), we arrive at
ð11εα Þ2 5 ð11εx Þ2 cos2 α 1 ð11εy Þ2 sin2 α 1 εxy sin2α
(2.157)
where cosα 5 dx=dsα and sinα 5 dy=dsα .
In a similar way, we can find the angle α0 after the deformation, i.e.,
dy0
1
@uy
@uy
sinα 1
cosα
5
11
1 1 εα dsα
@y @x
dx0
1
@ux
@ux
0
cosα 1
sinα
cosα 5
5
11
dsα
@x
@y
1 1 εα
sinα0 5
(2.158)
Now return to the ply element in Fig. 2.65. Taking α 5 φ in Eqs. (2.157) and (2.158), we obtain
Putting α 5
2
2
2
ð11ε1 Þ2 5 ð11εx Þ2 cos
φ 1 ð11ε
y Þ sin φ 1 εxy sin2φ
1
@u
@uy
y
sinφ0 5
sinφ 1
cosφ
11
1 1 ε1 @y @x
1
@u
@ux
x
cosφ 1
sinφ
cosφ0 5
11
1 1 ε1
@x
@y
(2.159)
ð11ε2 Þ2 5 ð11εx Þ2 sin2 φ1 ð11εy
Þ2 cos2 φ 2 2εxy sin2φ
1
@u
@uy
y
0
sinðφ 1 ψÞ 5
cosφ 2
sinφ
11
@y @x
1 1 ε2 1
@ux
@ux
cosφ 1
sinφ
cosðφ0 1 ψÞ 5
2 11
1 1 ε2
@x
@y
(2.160)
π
1 φ, we have
2
Using the last equation of Eq. (2.152), we can find the shear strain as sinγ 12 5 cosψ. After some
rearrangement, with the aid of Eqs. (2.159) and (2.160), we arrive at
sinγ 12 5
1
½ð11εy Þ2 2 ð11εx Þ2 sinφcosφ 1 εxy cos2φ
ð1 1 ε1 Þð1 1 ε2 Þ
(2.161)
For φ 5 0, axes 1 and 2 coincide, respectively, with axes x and y (see Fig. 2.65), and
Eq. (2.161) yields
sinγ xy 5
εxy
ð1 1 εx Þð1 1 εy Þ
(2.162)
2.5 ANGLE-PLY ORTHOTROPIC LAYER
155
Using this result to express εxy , we can write Eqs. (2.159)(2.161) in the following final form:
ð11ε1 Þ2 5 ð11εx Þ2 cos2 φ 1 ð11εy Þ2 sin2 φ 1 ð1 1 εx Þð1 1 εy Þsinγ xy sin2φ
ð11ε2 Þ2 5 ð11εx Þ2 sin2 φ 1 ð11εy Þ2 cos2 φ 2 ð1 1 εx Þð1 1 εy Þsinγ xy sin2φ
n
o
1
sinγ 12 5
ð11εy Þ2 2 ð11εx Þ2 sinφ cosφ 1 ð1 1 εx Þð1 1 εy Þsinγ xy cos2φ
ð1 1 ε1 Þð1 1 ε2 Þ
(2.163)
It follows from Fig. 2.65 and the last equation of Eqs. (2.152), that dsv2 5 ds02 sinψ 5 ds02 cosγ 12 .
So, in accordance with Eqs. (2.152) and (2.153),
1 1 εv2 5 ð1 1 ε2 Þcosγ 12
Using Eq. (2.163) to transform this equation, we get
1 1 εv2 5
ð1 1 εx Þð1 1 εy Þ
cosγ xy
1 1 ε1
(2.164)
To express φ0 in terms of φ and strains referred to the global coordinate frame x, y, consider
Eq. (2.154). After rather cumbersome transformation with the aid of Eqs. (2.159) and (2.160), we
obtain
sin2ω12
5
1
@uy
@ux
@uy @uy
@ux @ux
2
1
2
cos2 2φ
@x
@y
@x @y
@x @y
ð1 1 ε1 Þð1 1 ε2 Þ
"
2 #
@uy
@ux
@ux @uy
@ux @uy
1 @ux @uy 2
@uy @ux
2
2
1
2
sin 2φ 1
2
2
1
sin4φ
1
@x @y
@x @y
4
@x
@y
@x @x
@y @y
Taking φ 5 0, we can express the rotation angle ωz around the z-axis of the global coordinate
frame, i.e.,
sin2ωz 5
1
@uy
@ux
@uy @uy
@ux @ux
2
1
2
ð1 1 εx Þð1 1 εy Þ @x
@y
@x @y
@x @y
(2.165)
Consider now Eqs. (2.156), (2.162), and (2.165) which form a set of four algebraic equations
with respect to the derivatives of the displacements. Omitting the solution procedure, we can write
the final outcome as
γ xy
@ux
5 ð1 1 εx Þ cos
1 ωz 2 1;
@x
2
γ xy
@uy
5 ð1 1 εx Þ sin
1 ωz ;
@x
2
γ xy
@ux
5 ð1 1 εy Þ sin
2 ωz ;
@y
2
γ xy
@uy
5 ð1 1 εy Þ cos
2 ωz 2 1
@y
2
Substituting these expressions into Eq. (2.159), we have
γ xy
γ xy
1
1 ωz cosφ 1ð1 1 εy Þ cos
2 ωz sinφ
ð1 1 εx Þ sin
2
2
1 1 ε1
γ xy
γ xy
1
1 ωz cosφ 1ð1 1 εy Þ sin
2 ωz
sinφ
ð1 1 εx Þ cos
cosφ0 5
1 1 ε1
2
2
sinφ0 5
(2.166)
156
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
The derived nonlinear equations, Eq. (2.163), generalize Eq. (2.69) for the case of large strains,
whereas Eq. (2.166) allow us to find the fiber orientation angle after deformation.
The equilibrium equations, Eq. (2.68), retain their form but should be written for the deformed
state, i.e.,
σx 5 σ01 cos2 φ0 1 σv2 sin2 φ0 2 τ 012 sin2φ0
σy 5 σ01 sin2 φ0 1 σv2 cos2 φ0 1 τ 012 sin2φ0
τ xy 5 ðσ01 2 σv2 Þsinφ0 cosφ0 1 τ 012 cos2φ0
(2.167)
where σ01 , σv2 , and τ 012 are stresses referred to coordinate frame 10 2v (see Fig. 2.65) and to the current thickness of the ply.
Consider the problem of uniaxial tension of a 6 φ angle-ply layer with stress σx . For this case,
γ xy 5 0, ωz 5 0, and Eqs. (2.163), (2.164), (2.166) take the form
ð11ε1 Þ2 5 ð11εx Þ2 cos2 φ 1 ð11εy Þ2 sin2 φ
ð11ε2 Þ2 5 ð11εx Þ2 sin2 φ 1 ð11εy Þ2 cos2 φ
sinφcosφ
sinγ 12 5
ð11εy Þ2 2 ð11εx Þ2
ð1 1 ε1 Þð1 1 ε2 Þ
ð1 1 εx Þð1 1 εy Þ
1 1 εv2 5
1 1 ε1
1
1
ε
1 1 εx
y
sinφ0 5
sinφ; cosφ0 5
cosφ
1 1 ε1
1 1 ε1
For composite materials, the longitudinal strain ε1 is usually small, and these equations can be
further simplified as follows:
1 2 2
ðε cos φ 1 ε2y sin2 φÞ
2 x
ð11ε2 Þ2 5 ð11εx Þ2 sin2 φ 1 ð11εy Þ2 cos2 φ
1 ð11εy Þ2 2 ð11εx Þ2 sinφcosφ
sinγ 12 5
1 1 ε2
1 1 εv2 5 ð1 1 εx Þð1 1 εy Þ
1 1 εy
tanφ
tanφ0 5
1 1 εx
ε1 5 εx cos2 φ 1 εy sin2 φ 1
(2.168)
As an example, consider a specially synthesized highly deformable composite material made
from glass composite fibers and thermoplastic matrix as discussed in Section 2.4.4. Neglecting
interaction of strains, we take constitutive equations for the unidirectional ply as
σ01 5
E1 ε1
; σ0 5 ω2 ðεv2 Þ; τ 012 5 ω12 ðγ 12 Þ
1 1 εv2 2
(2.169)
where E1 in the first equation is the longitudinal elasticity modulus, whereas εv2 in the denominator takes account of the decrease of the ply stiffness due to the increase in the fiber spacing. The
constant E1 and functions ω2 and ω12 are determined from the experimental stressstrain
diagrams for 0, 90, and 45 specimens that are shown in Fig. 2.67. The results of calculations
with the aid of Eqs. (2.167)(2.169) are presented together with the corresponding experimental
data in Fig. 2.68.
2.5 ANGLE-PLY ORTHOTROPIC LAYER
157
σ x , MPa
1
φ = 0°
0.8
0.6
0.4
± 45°
0.2
φ = 90°
εx , %
0
0
10
20
30
40
50
FIGURE 2.67
Experimental stressstrain diagrams for 0 , 6 45 , and 6 90 angle-ply layers.
σ , MPa
1
±15°
0.8
0.6
± 30°
0.4
0.2
± 60°
± 75°
0
0
20
40
60
80
ε, %
100
FIGURE 2.68
Calculated (circles) and experimental (solid lines) stressstrain diagrams for 6 15 , 6 30 , 6 60 , and 6 75
angle-ply layers.
The foregoing equations comprise the analytical background for a promising manufacturing process that allows us to fabricate composite parts with complicated shapes by deforming partially
cured preforms of simple shapes made by winding or laying-up (see, e.g., Cherevatsky, 1999). An
example of such a part is presented in Fig. 2.69. The curved composite pipe shown in this
figure was fabricated from a straight cylinder that was partially cured, loaded with preassigned
internal pressure and end forces and moments, and cured completely in this state. The desired
deformation of the part under loading is provided by the appropriate change of the fibers’ orientation angles governed by Eqs. (2.163), (2.166), and (2.167).
158
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
FIGURE 2.69
A curved angle-ply pipe made by deformation of a filament-wound cylinder.
Angle-ply layers can also demonstrate nonlinear behavior caused by the matrix cracking
described in Section 2.4.2. To illustrate this type of nonlinearity, consider carbonepoxy 6 15,
6 30, 6 45, and 6 75 angle-ply specimens studied experimentally by Lagace (1985). The unidirectional ply has the following mechanical properties: E1 5 131 GPa, E2 5 11 GPa, G12 5 6 GPa,
1
2
ν 21 5 0:28, σ1
1 5 1770 MPa, σ 2 5 54 MPa, σ 2 5 230 MPa, and τ 12 5 70 MPa. The dependencies
σ1 ðε1 Þ and σ2 ðε2 Þ are linear, whereas for in-plane shear, the stressstrain diagram is not linear and
is shown in Fig. 2.70. To take into account the material nonlinearity associated with shear, we use
the constitutive equation derived in Section 2.2.2, i.e.,
γ 12 5 c1 τ 12 1 c2 τ 312
in which c1 5 1=G12 and c2 5 5:2U1028 ðMPaÞ23 .
The specimens were tested in uniaxial tension in the x-direction. To calculate the applied stress
σx that causes failure of the matrix, we use the simplest maximum stressstrength criterion (see
Chapter 4: Failure Criteria and Strength of Laminates) which ignores the interaction of stresses,
i.e.,
1
2σ2
2 # σ 2 # σ2 ; jτ 12 j # τ 12
The nonlinear behavior associated with ply degradation is predicted applying the procedure
described in Section 2.4.2. Stressstrain diagrams are plotted using the method of successive loading (see Section 2.1.2).
Consider a 6 15 angle-ply layer. Point 1 on the theoretical diagram, shown in Fig. 2.71, corresponds to cracks in the matrix caused by shear. These cracks do not result in complete failure of
the matrix because the transverse normal stress σ2 is compressive (see Fig. 2.72) and does not reach
σ2
2 before the failure of fibers under tension (point 2 on the diagram). As can be seen, the theoretical prediction of the material stiffness is rather good, whereas the predicted material strength (point
2) is much higher than the experimental (dot on the solid line). The reasons for this are discussed
in the next section.
2.5 ANGLE-PLY ORTHOTROPIC LAYER
159
σ 2 , τ 12 , MPa
80
60
1
2
40
20
0
0
1
2
3
ε 2 , γ 12 , %
FIGURE 2.70
Experimental stressstrain diagrams for transverse tension (1) and in-plane shear (2) of a carbonepoxy
unidirectional ply.
σ x , MPa
2000
1600
2
1200
1
0°
± 15°
800
2
400
± 30°
1
εx, %
0
0
0.4
0.8
1.2
1.6
2
2.4
FIGURE 2.71
Experimental (solid lines) and calculated (dashed lines) stressstrain diagrams for 0 , 6 15 , and 6 30 angleply carbonepoxy layers.
The theoretical diagram corresponding to the 6 303 layer (see Fig. 2.71) also has two specific
points. Point 1 again corresponds to cracks in the matrix induced by the shear stress τ 12 , whereas
point 2 indicates complete failure of the matrix caused by the compressive stress σ2 which reaches
σ2
2 at this point. After the failure of matrix, the fibers of an angle-ply layer cannot take the load.
Indeed, putting E2 5 G12 5 ν 12 5 0 in Eq. (2.72), we obtain the following stiffness coefficients:
A11 5 E1 cos4 φ; A22 5 E1 sin4 φ; A12 5 E1 sin2 φcos2 φ
160
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
1.2
σ1 σ x
1
σ2 σ x
0.8
0.6
0.4
τ 12 σ x
0.2
φ°
0
15
30
45
60
75
90
–0.2
FIGURE 2.72
Dependencies of the normalized stresses in the plies on the ply orientation angle.
FIGURE 2.73
A failure mode of 6 30 angle-ply specimen.
With these coefficients, the first equation of Eq. (2.147) yields Ex 5 0, which means that the
system of fibers becomes a mechanism, and the stresses in the fibers, no matter how high they are,
cannot balance the load. A typical failure mode for a 6 30 angle-ply specimen is shown in
Fig. 2.73.
Angle-ply layers with fiber orientation angles exceeding 45 demonstrate a different type of
behavior. As can be seen in Fig. 2.72, the transverse normal stress σ2 is tensile for φ $ 45 . This
means that the cracks induced in the matrix by normal, σ2 , or shear, τ 12 , stresses cause failure of
the layer. The stressstrain diagrams for 6 60 and 6 75 layers are shown in Fig. 2.74. As
follows from this figure, the theoretical curves are linear and are close to the experimental
ones, whereas the predicted ultimate stresses (circles) are again higher than the experimental
values (dots).
2.5 ANGLE-PLY ORTHOTROPIC LAYER
161
σ x , MPa
100
80
± 60°
60
± 75°
40
20
0
εx, %
0
0.2
0.4
0.6
0.8
FIGURE 2.74
Experimental (solid lines) and calculated (dashed lines) stressstrain diagrams for 6 60 and 6 75 angle-ply
carbonepoxy layers.
Now consider the 6 45 angle-ply layer, which demonstrates a very specific behavior. For this
layer, the transverse normal stress, σ2 , is tensile but not high (see Fig. 2.72), and the cracks in the
matrix are caused by the shear stress, τ 12 . According to the ply model we use, to predict the material response after the cracks’ appearance, we should take G12 5 0 in the stiffness coefficients.
Then, Eq. (2.72) yield
1
1
A11 5 A12 5 A22 5 ðE1 1 E2 Þ 1 E1 ν 12
4
2
whereas Eqs. (2.146) and (2.147) give
εx 5
A22 σx
A12 σx
; εy 5 2
A11 A22 2 A212
A11 A22 2 A212
The denominator of both expressions is zero, so it looks as though the material becomes a
mechanism and should fail under the load that causes cracks in the matrix. However, this is not the
case. To explain why, consider the last equation of Eq. (2.168), i.e.,
tanφ0 5
1 1 εy
tanφ
1 1 εx
For the layer under study, tanφ 5 1, εy , 0 and εx . 0, so tanφ0 , 1 and φ0 , 45 . However, in
the plies with φ , 45 the transverse normal stresses, σ2 , become compressive (see Fig. 2.72) and
close the cracks. Thus, the load exceeding the level at which the cracks appear due to shear locks
the cracks and induces compression across the fibers, thus preventing material failure. Since φ0 is
only slightly less than 45 , the material stiffness, Ex , is very low and slightly increases with the
increase in strains and decrease of φ0 . For the material under study, the calculated and experimental
162
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
σ x , MPa
160
120
80
40
0
0
0.4
0.8
1.2
1.6
2
ε x, %
FIGURE 2.75
Experimental (solid line) and calculated (dashed line) stressstrain diagrams for 6 45 angle-ply carbonepoxy
layer.
σ x , MPa
(φ ′ )°
300
45
2
200
40
1
35
100
0
0
2
4
6
8
30
εx , %
FIGURE 2.76
Experimental dependencies of stress (1) and ply orientation angle (2) on strain for 6 45 angle-ply glassepoxy
composite.
diagrams are shown in Fig. 2.75. The circle on the theoretical curve indicates the stress σx that
causes cracks in the matrix. More pronounced behavior of this type is demonstrated by the
glassepoxy composites whose stressstrain diagram is presented with curve 1 in Fig. 2.76
(Alfutov & Zinoviev, 1982). A specific plateau on the curve and material hardening at high strain
are the result of the angle variation as is also shown in Fig. 2.76 (line 2).
2.5 ANGLE-PLY ORTHOTROPIC LAYER
163
2.5.3 FREE-EDGE EFFECTS
As shown in the previous section, there is a significant difference between the predicted and measured strength of an angle-ply specimen loaded in tension. This difference is associated with the
stress concentration that occurs in the vicinity of the specimen longitudinal edges and was not taken
into account in the analysis.
To study a free-edge effect in an angle-ply specimen, consider a strip whose initial width a is
much smaller than the length l. Under tension with longitudinal stress σ, symmetric plies with orientation angles 1φ and 2φ tend to deform as shown in Fig. 2.77. As can be seen, deformation of
the plies in the y-direction is the same, whereas the deformation in the x-direction tends to be different. This means that symmetric plies forming the angle-ply layer interact through the interlaminar shear stress τ xz acting between the plies in the longitudinal direction. To describe the ply
interaction, introduce the model shown in Fig. 2.78, according to which the in-plane stresses in the
plies are applied to their middle surfaces, whereas transverse shear stresses act in some hypothetical
layers introduced between these surfaces.
To simplify the problem, we further assume that the transverse stress can be neglected, i.e.,
σy 5 0, and that the axial strain in the middle part of the long strip is constant, i.e.,
εx 5 ε 5 constant. Thus, the constitutive equations, Eq. (2.75), for a 1φ ply have the form
εx 5
σx
τ xy
1 η1
x;xy 1
Ex1
Gxy
εy 5 2 ν 1
yx
(2.170)
σx
τ xy
1 η1
y;xy 1
Ex1
Gxy
(2.171)
y
σ
x
a
l
FIGURE 2.77
Deformation of symmetric plies under tension.
FIGURE 2.78
A model simulating the plies’ interaction.
σ
164
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
γ xy 5 η1
xy;x
σx
τ xy
1 1
Ex1
Gxy
(2.172)
where the elastic constants for an individual ply are specified by Eq. (2.76). The straindisplacement equations for the problem under study are
εx 5 ε 5
@ux
@uy
@ux
@uy
; εy 5
; γ xy 5
1
@x
@y
@y
@x
(2.173)
Integration of the first equation yields
2φ
u1φ
x 5 εUx 1 uðyÞ; ux 5 εUx 2 uðyÞ
(2.174)
for the 1φ and 2φ plies where uðyÞ is the displacement shown in Fig. 2.78. This displacement
results in the following transverse shear deformation and transverse shear stress
2
γ xz 5 uðyÞ; τ xz 5 Gxz γ xz ;
δ
(2.175)
where Gxz is the transverse shear modulus of the ply specified by Eq. (2.76). Consider the equilibrium state of the 1φ ply element shown in Fig. 2.79. Equilibrium equations can be written as
δ
@τ xy
@τ xy
5 0; δ
2 2τ xz 5 0
@x
@y
(2.176)
The first of these equations shows that τ xy does not depend on x. Since the axial stress, σx , in
the middle part of a long specimen also does not depend on x, Eqs. (2.171) and (2.173) allow us to
conclude that εy , and hence uy , do not depend on x. As a result, the last equation of Eq. (2.173)
yields, in conjunction with the first equation of Eq. (2.174),
γ xy 5
@ux
du
5
dy
@y
Substituting this expression into Eq. (2.172) and taking into account the equation for ε,
Eq. (2.170), we arrive at
G1
du
xy
1
2 ηxy;x ε
τ xy 5
1 2 η dy
(2.177)
1
where η 5 η1
x;xy ηxy;x .
(τ xy +
∂y
dy )δ
(τ xy +
σ xδ
τ xz
dy
τ xyδ
∂τ xy
dx
τxyδ
FIGURE 2.79
Forces acting on the infinitesimal element of a ply.
σ xδ
∂τ xy
∂x
dx)δ
2.5 ANGLE-PLY ORTHOTROPIC LAYER
165
Substitution of Eqs. (2.175) and (2.177) into the second equation of Eq. (2.176) provides the following governing equation for the problem under study
d2 u
2 k2 u 5 0
dy2
(2.178)
in which
k2 5
4Gxz ð1 2 ηÞ
2
G1
xy δ
Using the symmetry conditions, we can obtain the solution of Eq. (2.178) as
u 5 Csinh ky
The constant C can be determined from the boundary conditions for the free longitudinal edges
of the specimen (see Fig. 2.77), according to which τ xy ðy 5 6 a=2Þ 5 0. Satisfying these conditions
and using Eqs. (2.170), (2.171), (2.175), and (2.177), we finally obtain
εx 5 ε
ε
coshλy
coshλy
1
1
η1
2
1
1
ν
2
1
η
η
yx
1 2 η y;xy xy;x coshλ
coshλ
1 coshλy
γ xy 5 εηxy;x
;
coshλ
η
coshλy
21
σx 5 εEx1 1 2
1 2 η coshλ
1 G1
η
xy xy;x coshλy
21
τ xy 5
coshλ
12η
2ε
sinhλy
τ xz 5 Gxz η1
xy;x
kδ
coshλ
εy 5
where
ka a
λ5
5
2
δ
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Gxz
2y
ð1 2 ηÞ 1 ; y 5
Gxy
a
(2.179)
(2.180)
The axial stress, σx , should provide the stress resultant equal to σa (see Fig. 2.77), i.e.,
a=2
ð
σx dy 5 σ a
2a=2
This condition allows us to determine the axial strain as
ε5
where
Ex 5 Ex1 1 1
σ
Ex
η
1
1 2 tanhλ
12η
λ
is the apparent modulus of an angle-ply specimen.
(2.181)
166
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
Consider two limiting cases. First, suppose that Gxz 5 0, i.e., that the plies are not bonded.
Then, λ 5 0 and because
1
lim tanhλ 5 1
λ
λ-0
Ex 5 Ex1 . Second, assume that Gxz -N, i.e., that the interlaminar shear stiffness is infinitely
high. Then λ-N and Eq. (2.181) yields
Ex 5
Ex1
12η
(2.182)
This result coincides with Eq. (2.149), which specifies the modulus of an angle-ply layer.
For finite values of Gxz , the parameter λ in Eq. (2.180) is rather large because it includes the
ratio of the specimen width, a, to the ply thickness, δ, which is, usually, a large number. Taking
into account that tanhλ # 1, we can neglect the last term in Eq. (2.181) in comparison with unity.
Thus, this equation reduces to Eq. (2.182). This means that the tensile loading of angle-ply specimens allows us to measure material stiffness with good accuracy despite the fact that the fibers are
cut on the longitudinal edges of the specimens.
However, this does not hold for the strength. The distribution of stresses over the half-width
of the carbonepoxy specimen with the properties given above and a=δ 5 20, φ 5 45 is shown
in Fig. 2.80. The stresses σx , τ xy , and τ xz were calculated with the aid of Eq. (2.179), whereas
stresses σ1 , σ2 , and τ 12 in the principal material directions of the plies were found using Eq. (2.69)
for the corresponding strains and Hooke’s law for the plies. As can be seen in Fig. 2.80, there exists
a significant concentration of stress σ2 that causes cracks in the matrix. Moreover, the interlaminar
σx σ
1
σ1 σ
0.8
0.6
τ12 σ
τ xy σ
0.4
0.2
0
τ xz σ
σ2 σ
y
0
0.2
0.4
0.6
0.8
1
FIGURE 2.80
Distribution of normalized stresses over the width of a 6 45 angle-ply carbonepoxy specimen.
2.6 LAYER MADE BY ANGLE-PLY CIRCUMFERENTIAL WINDING
167
σ , MPa
800
600
400
200
0
0
40
80
120
a, mm
FIGURE 2.81
Experimental dependence of strength of a 6 35 angle-ply layer on the width of the specimen.
shear stress τ xz that appears in the vicinity of the specimen edge can induce delamination of the
specimen. The maximum value of stress σ2 is
2
2
1
σmax
5 σ2 ðy 5 1Þ 5 E2 ε½ð1 2 ν 21 ν 1
2
yx Þ sin φ 1 ðν 21 2 ν yx Þ cos φ 2 ð1 2 ν 21 Þηxy;x sinφ cosφ
Using the modified strength condition, i.e., σmax
5 σ1
2
2 to evaluate the strength of the 6 60
specimen, we arrive at the result shown with a triangular symbol in Fig. 2.74. As can be seen,
the allowance for the stress concentration results in fair agreement with the experimental
strength (dot).
Thus, the strength of angle-ply specimens is reduced by the free-edge effects, which causes a
dependence of the observed material strength on the width of the specimen. Such dependence is
shown in Fig. 2.81 for 105 mm diameter and 2.5 mm thick fiberglass rings made by winding at
6 35 angles with respect to the axis and loaded with internal pressure by two half-discs as in
Fig. 1.46 (Fukui et al., 1966).
It should be emphasized that the free-edge effect occurs in laboratory test specimens only and is
not relevant to composite structures which, being properly designed, must not have free edges of
such a type.
2.6 LAYER MADE BY ANGLE-PLY CIRCUMFERENTIAL WINDING
Composite materials with a special spatial structure can be fabricated using the so-called angle-ply
circumferential winding process which is a combination of winding and weaving shown in
Fig. 2.82. The schematic of the process is shown in Fig. 2.83. A system of tows is placed on the
168
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
FIGURE 2.82
Angle-ply circumferential winding.
w
φ
Axial
yarns
ωm
Mandrel
Rotating
ring
ωr
Vc
Circumferential
tows (yarns)
FIGURE 2.83
Parameters of angle-ply circumferential winding.
2.6 LAYER MADE BY ANGLE-PLY CIRCUMFERENTIAL WINDING
169
rotating mandrel and moves in the axial direction. On the way to the mandrel, the tows are overwrapped with fine yarns drawn from the spools placed on the ring which rotates around the system
of tows. The resulting composite layer is reinforced in four directions. First, it has hoop tows; second, the 6 φ angle-ply structure is formed by the yarns drawn from the rotating ring; and finally,
the yarns also provide through-the-thickness reinforcement of the layer. Composite materials made
by angle-ply winding may have at least two applications. First, in composite pipes and pressure
vessels made by continuous hoop filament winding (Vasiliev, Krikanov, & Razin, 2003). Second,
they can be utilized to fabricate wing and fuselage skin panels for composite aircraft structures
using circular mandrels of large diameter or mandrels with rectangular cross-sections. For panels,
the primary tows should be oriented along the major loading direction of the panel. It is important
that the panel with the material structure under consideration can be made as thick as required.
This is achieved by controlling (reducing) the speed of the carriage Vc (see Fig. 2.83). Additional
advantage is the higher resistance of the panel to delamination caused by impact. This is achieved
due to the inherent through-the-thickness reinforcement introduced in the process of fabrication and
the resulting nonlaminated structure of the material.
The mechanical properties of a composite material made by angle-ply circumferential winding
depend on the material structural parameters and normally could be determined experimentally.
However, some theoretical predictions can also be made.
Assume that the circumferential tape (primary system of tows) consists of Nc fibers with crosssectional areas fc , modulus of elasticity Ec , and strength under tension σc . Then, the total area of
the fiber cross-section is
Ac 5 Nc fc 5 nt nft fc
where nt is the number of circumferential tows and nft is the number of fibers in the tow. The effective thickness of the resulting circumferential layer can be found by dividing the area Ac by the
tape width w (see Fig. 2.83), i.e.,
hc 5
Ac
fc
5 nt nft
w
w
To find the corresponding parameter for the axial yarns, assume that the mandrel makes rm rotations per minute, so that the time of one rotation is tm 5 1=rm . Let the ring with nr spools (see
Fig. 2.82) make rr rotations per minute. Taking into account that the ring places 2nr yarns onto the
hoop tape during one rotation, we can find the number of axial fibers Na that are placed onto the
mandrel while it makes this rotation, i.e.,
Na 5 2nr nfa rr tm 5 2nr nfa
rr
rm
where nfa is the number of fibers in the yarn. Then, the total area of the fiber cross-sections
becomes Aa 5 Na fa , where fa is the area of the fiber cross-section, and the effective thickness of the
axial layer is
ha 5
Aa
nr nfa rr fa
5
2πR
πRrm
where R is the radius of the mandrel cross-section. Introduce the total effective thickness of the
layer h and the normalized thickness as
170
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
h 5 hc 1 ha ; hc 5
hc
;
h
ha 5
ha
; hc 1 ha 5 1
h
Introducing the parameter
λ5
ha
nr nfa rr fa w
5
πRrm fc nt nft
hc
(2.183)
we arrive at
hc 5
1
;
11λ
ha 5
λ
11λ
(2.184)
The axial yarns make angles 6 φ with the tows’ direction. The linear velocity of the circumferential tow is vc 5 ωm R, where ωm is the mandrel angular velocity ωm 5 πrm =30; where, as earlier,
rm is the number of the mandrel rotations per minute. The time, tr , over which the ring makes one
complete rotation, is tr 5 2π=ωr , where ωr 5 πrr =30 is the angular velocity of the ring. It follows
from Fig. 2.83 that
a 5 vc tr 5 2πR
rm
;
rr
tanφ 5
2w
wrr
5
πRrm
a
The effects of manufacturing parameters on the mechanical properties of composite material
have been studied experimentally. The materials under consideration have been manufactured using
the angle-ply circumferential winding process shown in Fig. 2.82. The mandrel radius is
R 5 75 mm and the tape width is w 5 200 mm: Fiberglass yarns are used to reinforce material in
the circumferential and axial directions, so that nfa 5 nft and fa 5 fc . Correspondingly, the parameter
λ in Eq. (2.183) becomes
λ5
nr rr w
πRrm nt
The manufacturing parameters of five different versions of the fabrication process are presented
in Table 2.4. Experimental mechanical properties are listed in Table 2.5. The dependencies of Ec ,
Ea and σc and σa on the relative effective thickness of circumferential yarns hc (see Eq. (2.184))
Table 2.4 Manufacturing Parameters of the Process
Process Version
Process Parameters
1
2
3
4
5
Number of circumferential yarns, nt
Number of spools on the ring, nr
Mandrel rotational speed, rm (rpm)
Ring rotational speed, rr (rpm)
Parameter λ
154
5
31
220
0.196
0.836
132
7
34
220
0.291
0.774
132
9
33
220
0.386
0.721
132
3
10
220
0.463
0.683
132
11
32
220
0.487
0.673
0.164
0.226
0.279
0.317
0.327
80.6
79.7
80
87.2
80.3
hc
ha
φ
2.6 LAYER MADE BY ANGLE-PLY CIRCUMFERENTIAL WINDING
171
Table 2.5 Mechanical Properties of the Composite Material
Process Version
Property
1
2
3
4
5
Circumferential modulus, Ec (GPa)
Axial modulus, Ea (GPa)
Poisson’s ratio, ν ac
Circumferential strength, σc (MPa)
Axial strength, σa (MPa)
39.5
9.8
0.15
1070
290
36.3
11.7
0.13
1010
350
33.4
16.8
0.13
960
390
30.1
19.5
0.11
930
430
29.6
20.1
0.11
900
440
E, GPa
60
Ec + Ea
40
Ec
20
0
Ea
0.6
0.7
hc
0.8
0.9
experiment,
analysis.
FIGURE 2.84
Dependencies of material moduli on parameter hc ,
are given in Figs. 2.84 and 2.85. It follows from the graphs that the sums Ec 1 Ea 5 E and
σc 1 σa 5 σ do not depend on the material structure. For the material under study, E 5 49:5 GPa
and σ 5 1350 MPa. Compare these characteristics with the corresponding properties of the unidirectional composite. The glass fibers used to fabricate the material with the properties as per
Figs. 2.84 and 2.85 have the following modulus of elasticity and strength:
Ef 5 85 GPa;
σf 5 2350 MPa
The fiber volume fraction in the composite material made by the angle-ply circumferential
winding is usually lower than that typical for conventional composites fabricated by winding. For
the material under consideration, it is vf 5 0:59. The products Ef vf 5 50:15 GPa and
σf vf 5 1386 MPa are slightly higher than E (by 1.3%) and σ (by 2.7%), respectively. This means
that angle-ply circumferential winding, as opposed to weaving, allows us to utilize completely the
original fiber stiffness and strength.
172
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
σ , MPa
1600
σc + σa
1200
σc
800
σa
400
0
0.6
0.7
0.8
0.9
hc
FIGURE 2.85
Dependencies of material tensile strength on parameter hc ,
experiment,
analysis.
To evaluate approximately the stiffness and strength of the material under consideration, assume
that it consists of two layers of fibers with relative thicknesses hc and ha in Eq. (2.184) and apply
the approach described in Sections 2.4 and 2.5. Neglecting the contribution of the matrix, we can
determine the stiffness coefficients of the layers using Eq. (2.72), i.e.,
Ac11 5 vf Efc ;
Ac12 5 Ac22 5 Ac33 5 0
Aa11 5 vf Efa cos4 φ;
Aa12 5 Aa33 5 vf Efa sin2 φcos2 φ;
Aa22 5 vf Efa sin4 φ
Here, vf is the fiber volume fraction which can be found only experimentally and Efc and Efa are
the fiber moduli of the circumferential and axial yarns. The stiffness coefficients of the material
can be calculated using Eq. (2.114), i.e.,
A11 5 vf ðEfc hc 1 Efa ha cos4 φÞ
A12 5 A33 5 vf Efa sin2 φcos2 φ
A22 5 vf Efa sin4 φ
whereas the elastic constants can be found from Eq. (2.116)
Ec 5 A11 2
ν ca 5
A12
;
A11
A212
;
A22
Ea 5 A22 2
ν ac 5
A12
A22
A212
;
A11
Gca 5 A33
(2.185)
For vf 5 0:59, the results of calculations are presented in Fig. 2.84 and qualitatively correlate
with the experimental data.
2.7 FABRIC LAYERS
173
To predict the strength, apply Eq. (2.69) in conjunction with Hooke’s law to find the stresses in
the fibers
σcf 5 Efc εc ; σaf 5 Efa εc cos2 φ 1 εa sin2 φ
(2.186)
The strains, εc and εa , specify the corresponding stresses
σc 5 Ec ðεc 1 ν ca εa Þ; σa 5 Ea ðεa 1 ν ac εc Þ
(2.187)
where E 5 E=ð1 2 ν ca ν ac Þ, whereas E and ν are given by Eq. (2.185). Substituting Eq. (2.186) into
Eq. (2.187), we arrive at
"
#
σcf ν ca σaf
2
σc 5 Ec c 1 2 ν ca cotan φ 1 2 a
Ef
sin φ Ef
"
#
a
c
σf 1 σf
2
1 c ν ac 2 cotan φ
σa 5 E a
Ef
sin2 φ Efa
For the ultimate state, taking σcf 5 σcf and σaf 5 σaf , we can find the material strength, i.e., σc and
σa . The results of calculations are presented in Fig. 2.85 and are in fair agreement with experimental data.
2.7 FABRIC LAYERS
Textile preforming plays an important role in composite technology, providing glass, aramid, carbon (see Fig. 2.86), and hybrid fabrics that are widely used as reinforcing materials. The main
advantages of woven composites are their cost efficiency and high processability, particularly, in
FIGURE 2.86
A carbonfabric tape.
174
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
FIGURE 2.87
A composite hull of a boat made of fiberglass fabric by lay-up method.
FIGURE 2.88
A composite leading edge of an aeroplane wing made of carbon fabric by lay-up method.
lay-up manufacturing of large-scale structures (see Figs. 2.87 and 2.88). However, processing of
the fibers and their bending in the process of weaving results in substantial reduction in material
strength and stiffness. As can be seen in Fig. 2.89, in which a typical woven structure is shown, the
warp (lengthwise) and fill (crosswise) yarns forming the fabric make angle α $ 0 with the plane of
the fabric layer.
To demonstrate how this angle influences material stiffness, consider tension of the structure
shown in Fig. 2.89 in the warp direction. The apparent modulus of elasticity can be expressed as
2.7 FABRIC LAYERS
fill
warp
3
175
h
2
h
2
α
2
1
t1
t2
t1
FIGURE 2.89
Unit cell of a fabric structure.
Ea Aa 5 Ef Af 1 Ew Aw
(2.188)
where Aa 5 hð2t1 1 t2 Þ is the apparent cross-sectional area and
Af 5
h
h
ð2t1 1 t2 Þ ; Aw 5 ð4t1 1 t2 Þ
2
4
are the areas of the fill and warp yarns in the cross-section. Substitution in Eq. (2.188) yields
Ea 5
1
Ew ð4t1 1 t2 Þ
Ef 1
2
2ð2t1 1 t2 Þ
Since the fibers of the fill yarns are orthogonal to the loading direction, we can take Ef 5 E2 ,
where E2 is the transverse modulus of a unidirectional composite. The compliance of the warp yarn
can be decomposed into two parts corresponding to t1 and t2 in Fig. 2.89, i.e.,
2t1 1 t2
2t1
t2
5
1
Ew
E1
Eα
where E1 is the longitudinal modulus of a unidirectional composite, whereas Eα can be determined
with the aid of the first equation of Eq. (2.76) if we change φ for α, i.e.,
1
cos4 α sin4 α
1
2ν 21
sin2 α cos2 α
5
1
1
2
E1
Eα
E1
E2
G12
(2.189)
The final result is as follows:
Ea 5
E2
1
2
E1 ð4t1 1 t2 Þ
E
E1
1
4 2t1 1 t2 cos4 α 1
sin4 α 1
2 2ν 21 sin2 α cos2 α
E2
G12
(2.190)
For example, consider a glass fabric with the following parameters: α 5 12 ; t2 5 2t1 . Taking
elastic constants for a unidirectional material from Table 1.5, we get Ea 5 23:5 GPa for the fabric
composite. For comparison, a cross-ply ½0=90 laminate made of the same material has
E 5 36:5 GPa. Thus, the modulus of a woven structure is lower by 37% than the modulus of the
same material reinforced with straight fibers. Typical mechanical characteristics of fabric composites are listed in Table 2.6.
176
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
Table 2.6 Typical Properties of Fabric Composites
Property
Glass FabricEpoxy
Aramid FabricEpoxy
Carbon FabricEpoxy
Fiber volume fraction
Density (g / cm3)
Longitudinal Modulus, GPa
Transverse modulus (GPa)
Shear modulus (GPa)
Poisson’s ratio
Longitudinal tensile strength (MPa)
Longitudinal compressive strength (MPa)
Transverse tensile strength (MPa)
Transverse compressive strength (MPa)
In-plane shear strength (MPa)
0.43
1.85
26
22
7.2
0.13
400
350
380
280
45
0.46
1.25
34
34
5.6
0.15
600
150
500
150
44
0.45
1.40
70
70
5.8
0.09
860
560
850
560
150
(A)
(B)
(C)
(D)
FIGURE 2.90
Plain (A), twill (B) and (C), and triaxial (D) woven fabrics.
The stiffness and strength of fabric composites depend not only on the yarns and matrix properties, but on the material structural parameters, i.e., on fabric count and weave. The fabric count specifies the number of warp and fill yarns per inch (25.4 mm), whereas the weave determines how
the warp and the fill yarns are interlaced. Typical weave patterns are shown in Fig. 2.90 and
2.7 FABRIC LAYERS
177
include plain, twill, and triaxial woven fabrics. In the plain weave (see Fig. 2.90A) which is the
most common and the oldest, the warp yarn is repeatedly woven over the fill yarn and under the
next fill yarn. In the twill weave, the warp yarn passes over and under two or more fill yarns (as in
Figs. 2.90B and C) in a regular way.
Being formed from one and the same type of yarns, plain and twill weaves provide approximately the same strength and stiffness of the fabric in the warp and the fill directions. Typical
stressstrain diagrams for a fiberglass fabric composite of such a type are presented in Fig. 2.91.
As can be seen, this material demonstrates relatively low stiffness and strength under tension at an
angle of 45 with respect to the warp or fill directions. To improve these properties, multiaxial
woven fabrics, one of which is shown in Fig. 2.90D, can be used.
Fabric materials whose properties are closer to those of unidirectional composites are made by
weaving a greater number of larger yarns in the longitudinal direction and fewer and smaller yarns
in the orthogonal direction. Such a weave is called unidirectional. It provides materials with high
stiffness and strength in one direction, which is specific for unidirectional composites, and high
processability typical of fabric composites.
Being fabricated as planar structures, fabrics can be shaped on shallow surfaces using the material’s high stretching capability under tension at 45 to the yarns’ directions. Many more possibilities for such shaping are provided by the implementation of knitted fabrics whose strain to failure
exceeds 100%. Moreover, knitting allows us to shape the fibrous preform in accordance with the
shape of the future composite part. There exist different knitting patterns, some of which are shown
in Fig. 2.92. Relatively high curvature of the yarns in knitted fabrics, and possible fiber breakage in
the process of knitting, result in materials whose strength and stiffness are less than those of woven
fabric composites but whose processability is better and the cost is lower. Typical stress-strain diagrams for composites reinforced by knitted fabrics are presented in Fig. 2.93.
σ , MPa
400
0°
10°
300
20°
30°
200
45°
100
0
0
1
2
3
4
5
ε, %
FIGURE 2.91
Stressstrain curves for fiberglass fabric composite loaded in tension at different angles with respect to the warp
direction.
178
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
FIGURE 2.92
Typical knitted structures.
σ , MPa
250
0°
200
45°
150
90°
100
50
0
0
1
2
3
4
ε, %
FIGURE 2.93
Typical stressstrain curves for fiberglass knitted composites loaded in tension at different angles with respect to
direction indicated by the arrow in Fig. 2.92.
Material properties close to those of woven composites are provided by braided structures
which, being usually tubular in shape, are fabricated by mutual intertwining, or twisting of yarns
about each other. Typical braided structures are shown in Fig. 2.94. The biaxial braided fabrics in
Fig. 2.94 can incorporate longitudinal yarns forming a triaxial braid whose structure is similar to
that shown in Fig. 2.90D. Braided preforms are characterized by very high processability providing
near net-shape manufacturing of tubes and profiles with various cross-sectional shapes.
Although microstructural models of the type shown in Fig. 2.89 which lead to equations similar
to Eq. (2.190) have been developed to predict the stiffness and even strength characteristics of fabric composites (e.g., Skudra, Bulavs, Gurvich, & Kruklinsh, 1989), for practical design and
2.7 FABRIC LAYERS
(A)
179
(B)
FIGURE 2.94
Diamond (A) and regular (B) braided fabric structures.
analysis, these characteristics are usually determined by experimental methods. The elastic constants entering the constitutive equations written in principal material coordinates, e.g., Eq. (2.55),
are determined by testing strips cut out of fabric composite plates at different angles with respect to
the orthotropy axes. The 0 and 90 specimens are used to determine moduli of elasticity E1 and E2
and Poisson’s ratios ν 12 ; ν 21 (or parameters for nonlinear stressstrain curves), whereas the inplane shear stiffness can be obtained with the aid of off-axis tension described in Section 2.3.1. For
fabric composites, the elastic constants usually satisfy conditions in Eqs. (2.85) and (2.86), and
there exists the angle φ specified by Eq. (2.84) such that off-axis tension under this angle is not
accompanied by shearextension coupling.
Since Eq. (2.84) specifying φ includes the shear modulus G12 , which is not known, we can
transform the results presented in Section 2.3.1. Using Eq. (2.76) and assuming that there is no
shearextension coupling ðηx;xy 5 0Þ, we can write the following equations:
1
1 1 ν 21
1 1 ν 12 4
ν 21
1
5
cos4 φ 1
sin φ 2
1
sin2 φ cos2 φ
Ex
G
E1
E
E
2
1
12
ν yx
ν 21
1 1 ν 21
1 1 ν 12
1
sin2 φ cos2 φ
5
2
1
2
G12
Ex
E1
E1
E2
1 1 ν 21
1 1 ν 12 2
1
cos2 φ 2
sin φ 2
cos2φ 5 0
2G12
E1
E2
(2.191)
Summing up the first two of this equation, we get
1 1 ν yx
1 1 ν 21
1 1 ν 12 2
2
2
5
cos φ 2
sin φ cos2φ 1
sin2 φ cos2 φ
Ex
E1
E2
G12
Using the third equation, we arrive at the following remarkable result:
G12 5
Ex
2ð1 1 ν yx Þ
(2.192)
similar to the corresponding formula for isotropic materials. It should be emphasized that
Eq. (2.192) is valid for off-axis tension in the x-direction making some special angle φ with the
180
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
principal material axis 1. This angle is given by Eq. (2.84). Another form of this expression follows
from the last equation of Eqs. (2.191) and (2.192), i.e.,
1 1 ν yx
1 1 ν 21
2
Ex
E1
sin2 φ 5
1 1 ν yx
1 1 ν 21
1 1 ν 12
2
2
2
Ex
E1
E2
(2.193)
For fabric composites whose stiffness in the warp and the fill directions is the same ðE1 5 E2 Þ,
Eq. (2.193) yields φ 5 45 .
2.8 SPATIALLY REINFORCED LAYERS AND BULK MATERIALS
The layers considered in the previous sections and formed of unidirectionally reinforced plies and
tapes (Sections 2.22.5) or fabrics reinforced in the layer plane (Section 2.7) suffer from a serious
shortcoming: their transverse (normal to the layer plane) stiffness and strength are substantially
lower than the corresponding in-plane characteristics. To improve the material properties under tension or compression in the z-direction and in shear in the xz- and the yz-planes (see, e.g., Fig. 2.18),
the material should be additionally reinforced with fibers or yarns directed along the z-axis or making some angles (less than a right angle) with this axis.
A simple and natural way of such triaxial reinforcement is provided by the implementation of
three-dimensionally woven or braided fabrics. Three-dimensional weaving or braiding is a variant
of the corresponding planar process wherein some yarns are going in the thickness direction. An
alternative method involves assembling elementary fabric layers or unidirectional plies into a threedimensionally reinforced structure by sewing or stitching. Depending on the size of the additional
yarn and frequency of sewing or stitching, the transverse mechanical properties of the twodimensionally reinforced composite can be improved to a greater or lesser extent. A third way is
associated with the introduction of composite or metal pins parallel to the z-axis that can be
inserted in the material before or after it is cured. A similar effect can be achieved by so-called needle punching. The needles puncture the fabric, break the fibers that compose the yarns, and direct
the broken fibers through the layer thickness. Short fibers (or whiskers) may also be introduced
into the matrix with which the fabrics or the systems of fibers are impregnated.
Another class of spatially reinforced composites, used mainly in carboncarbon technology, is
formed by bulk materials multidimensionally reinforced with fine rectilinear yarns composed of carbon
fibers bound with a polymeric or carbon matrix. The basic structural element of these materials is a
parallelepiped shown in Fig. 2.95. The simplest spatial structure is the so-called 3D (three-dimensionally
reinforced), in which reinforcing elements are placed along the ribs AA1 ; AB, and AD of the basic
parallelepiped in Fig. 2.95. This structure is shown in Fig. 2.96 (Vasiliev & Tarnopol’skii, 1990). A
more complicated 4D structure with reinforcing elements directed along the diagonals AC1 ; A1 C; BD1 ,
and B1 D (see Fig. 2.95) is shown in Fig. 2.97 (Tarnopol’skii, Zhigun, & Polyakov, 1987). An example
of this structure is presented in Fig. 2.98. A cross-section of a 5D structure reinforced along diagonals
AD1 ; A1 D, and ribs AA1 ; AB; and AD is shown in Fig. 2.99 (Vasiliev & Tarnopol’skii, 1990). There
exist structures with a greater number of reinforcing directions. For example, combination of a 4D structure (Fig. 2.97) with reinforcements along the ribs AB and AD (see Fig. 2.95) results in a 6D structure;
2.8 SPATIALLY REINFORCED LAYERS AND BULK MATERIALS
B1
181
C1
B
C
A1
D1
A
D
FIGURE 2.95
The basic structural element of multidimensionally reinforced materials.
FIGURE 2.96
3D spatially reinforced structure.
addition of reinforcements in the direction of the rib AA1 gives a 7D structure; and so on, up to 13D,
which is the most complicated of the spatial structures under discussion.
The mechanical properties of multidimensional composite structures can be qualitatively predicted with the microstructural models discussed, e.g., by Tarnopol’skii, Zhigun, & Polyakov
(1992). However, for practical applications these characteristics are usually obtained by experimental methods. Being orthotropic in the global coordinates of the structure x, y, and z, spatially
182
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
FIGURE 2.97
4D spatially reinforced structure.
FIGURE 2.98
A 4D spatial structure.
reinforced composites are described within the framework of a phenomenological model ignoring
their microctructure by three-dimensional constitutive equations analogous to Eqs. (2.53) or (2.54),
in which 1 should be changed for x, 2 for y, and 3 for z. These equations include nine independent
2.8 SPATIALLY REINFORCED LAYERS AND BULK MATERIALS
183
FIGURE 2.99
Cross-section of a 5D spatially reinforced structure.
3
τ 32
τ 23
2
σ
τ
τ
σy
τ
σ2
τ
τ
τxy
τ 21
σ3
τ 31
τ 13
τ 12
σ1
1
σx
x
y
FIGURE 2.100
Material elements referred to the global structural coordinate frame x, y, z and to the principal material axes 1, 2, 3.
elastic constants. Stiffness coefficients in the basic plane, i.e., Ex ; Ey ; Gxy ; and ν xy , are determined
using traditional tests developed for unidirectional and fabric composites as discussed in Sections
1.4, 2.2, and 2.7. The transverse modulus Ez and the corresponding Poisson’s ratios ν xz and ν yz can
be determined using material compression in the z-direction. Transverse shear moduli Gxz and Gyz
can be calculated using the results of a three-point beam bending test described in Section 1.5.
The last spatially reinforced structure that is considered here is formed by a unidirectional composite material whose principal material axes 1, 2, and 3 make some angles with the global structural axes x, y, and z (see Fig. 2.100). In the principal material coordinates, the constitutive
184
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
equations have the form of Eqs. (2.53) or (2.54). Introducing directional cosines lxi ; lyi ; and lzi
which are cosines of the angles that the i-axis ði 5 1; 2; 3Þ makes with axes x, y, and z, respectively, transforming stresses and strains in coordinates 1, 2, and 3 to stresses and strains referred to
coordinates x, y, and z, and using the procedure described in Section 2.3.1, we finally arrive at the
following constitutive equations in the global structural coordinate frame:
8
8
9
9
εx >
σx >
>
>
>
>
>
>
>
>
>
>
>
>
εy >
σy >
>
>
>
>
>
>
>
<
<ε >
=
=
σz
z
5 ½S
>
>
> τ xy >
> γ xy >
>
>
>
>
>
>
>
>
γ >
τ xz >
>
>
>
>
>
>
>
:
: xz >
;
;
γ yz
τ yz
in which
2
6
6
6
½S 5 6
6
6
4
S1111
S1122
S2222
S1133
S2233
S3333
sym
S1112
S2212
S3312
S1212
(2.195)
S1113
S2213
S3313
S1213
S1313
3
S1123
S2223 7
7
S3323 7
7
S1223 7
7
S1323 5
S2323
is the stiffness matrix where
S1111 5 A1 l4x1 1 A2 l4x2 1 A3 l4x3 1 2A112 l2x1 l2x2 1 2A113 l2x1 l2x3 1 2A223 l2x2 l2x3
S1122 5 A1 l2x1 l2y1 1 A2 l2x2 l2y2 1 A3 l2x3 l2y3 1 A1 μ12 ðl2x1 l2y2 1 l2x2 l2y1 Þ
1 A1 μ13 ðl2x1 l2y3 1 l2x3 l2y1 Þ 1 A2 μ23 ðl2x2 l2y3 1 l2x3 l2y3 Þ
1 4ðG12 lx1 lx2 ly1 ly2 1 G13 lx1 lx3 ly1 ly3 1 G23 lx2 lx3 ly2 ly3 Þ
ð1; 2; 3Þ
ð1; 2; 3Þ
S1112 5 A1 l3x1 ly1 1 A2 l3x2 ly2 1 A3 l3x3 ly3 1 A112 ðlx1 ly2 1 lx2 ly1 Þlx1 l2x
1 A113 ðlx1 ly3 1 lx3 ly1 Þlx1 lx3 1 A223 ðlx2 ly3 1 lx3 ly2 Þlx2 lx3
S1113 5 A1 l3x3 lz1 1 A2 l3x2 lz2 1 A3 l3x3 lz3 1 A112 ðlx1 lz2 1 lx2 lz1 Þlx1 lx2
1 A113 ðlx1 lz2 1 lx3 lz1 Þlx1 lx3 1 A223 ðlx2 lz3 1 lx3 lz2 Þlx2 lx3
ð1; 2; 3Þ
ð1; 2; 3Þ
S1123 5 A1 l2x1 ly1 lz1 1 A2 l2x2 ly2 lz2 1 A3 l2x3 ly3 lz3
1 A1 μ12 ðl2x1 ly2 lz2 1 l2x2 ly1 lz1 Þ 1 A1 μ13 ðl2x1 ly3 lz3 1 l2x3 ly1 lz1 Þ
1 A2 μ23 ðl2x2 ly3 lz3 1 l2x3 ly2 lz2 Þ 1 2½G12 ðly1 lz2 1 lz1 ly2 Þlx1 lx2
1 G13 ðly1 lz2 1 lz1 ly2 Þlx1 lx3 1 G23 ðly3 lz2 1 ly2 lz3 Þlx2 lx3 ð1; 2; 3Þ
S1212 5 A1 l2x1 l2y1 1 A2 l2x2 l2y2 1 A3 l2x3 l2y3
1 2ðA1 μ12 lx1 lx2 ly1 ly2 1 A1 μ13 lx1 lx3 ly1 ly3 1 A2 μ23 lx2 lx3 ly2 ly3 Þ
1 G12 ðlx1 ly2 1lx2 ly1 Þ2 1 G13 ðlx1 ly3 1lx3 ly1 Þ2 1 G23 ðlx2 ly3 1lx3 ly2 Þ2
(2.196)
ð1; 2; 3Þ
S1213 5 A1 l2x1 ly1 lz1 1 A2 l2x2 ly2 lz2 1 A3 l2x3 ly3 lz3
1 A1 μ12 ðly1 lz2 1 ly2 lz1 Þlx1 lx2 1 A1 μ13 ðly1 lz3 1 ly2 lz1 Þlx1 lx3
1 A2 μ23 ðly2 lz3 1 ly3 lz2 Þlx1 lx3 1 G12 ðlx1 ly2 1 lx2 ly1 Þðlx1 ly2 1 lx2 lz1 Þ
1 G13 ðlx1 ly3 1 lx3 ly1 Þðlx1 lz3 1 lx3 lz2 Þ 1 G23 ðlx2 ly3 1 lx3 ly2 Þðlx2 lz3 1 lx3 lz2 Þ
ð1; 2; 3Þ
2.8 SPATIALLY REINFORCED LAYERS AND BULK MATERIALS
185
It should be noted that stiffness coefficients are symmetric with respect to the couples of subscripts ðSijkl 5 Sklij Þ and that notation (1, 2, 3) means that performing permutation, i.e., changing 1
for 2, 2 for 3, and 3 for 1, we can use Eq. (2.196) to write the expressions for all the stiffness coefficients entering Eq. (2.195). The coefficients Ai and μij in Eq. (2.196) are given in the notations to
Eq. (2.54) and
A112 5 A1 μ12 1 2G12 ; A113 5 A1 μ13 1 2G13 ;
A223 5 A2 μ23 1 2G23
Solving Eq. (2.195) for strains, we arrive at the following constitutive equations:
ηx;xy
ηx;yz
η
σx
ν xy
ν xz
2
σy 2
σz 1
τ xy 1 x;xz τ xz 1
τ yz ðx; y; zÞ
Ex
Ey
Ez
Gxy
Gxz
Gyz
η
η
η
τ xy
λxy;xz
λxy;yz
xy;x
xy;y
xy;z
γ xy 5
1
τ xz 1
τ yz 1
σx 1
σy 1
σz ðx; y; zÞ
Gxy
Gxz
Gyz
Ex
Ey
Ez
εx 5
in which the compliance coefficients are
l4y1
l4
1
l4
5 x1 1
1 z1 1 C122 l2x1 l2y1 1 C133 l2x1 l2z1 1 C233 l2y1 l2z1
Ex
E1
E2
E3
ν xy
ν yx
ν 12 2 2
ν 13 2 2
5
5
ðl l 1 l2 l2 Þ 1
ðl l 1 l2 l2 Þ
Ey
Ex
E2 x1 y2 x2 y1
E3 x1 z2 x2 z1
ð1; 2; 3Þ; ðx; y; zÞ
l2y1 l2y2
l2 l2
ν 23 2 2
l2 l2
ðly1 lz2 1 l2y2 l2z1 Þ 2 x1 x2 2
2 z1 z2
E3
E1
E2
E3
1
1
1
2
lx1 lx2 ly1 ly2 2
lx1 lx2 lz1 lz2 2
ly1 ly2 lz1 lz2 ð1; 2; 3Þ; ðx; y; zÞ
G12
G13
G23
!
l3y1 ly2
ηx;xy
ηxy;x
l3 lz2
l3 lx2
1 C122 ðlx1 ly2 1 lx2 ly1 Þlx1 ly1
5
5 2 x1 1
1 z1
Gxy
Ex
E1
E2
E3
(2.197)
1
1 C133 ðlx1 lz2 1 lx2 lz1 Þlx1 lz1 1 C233 ðly1 lz2 1 ly2 lz1 Þly1 lz1 ð1; 2; 3Þ; ðx; y; zÞ
!
l3y1 ly3
ηx;xz
η
l3 lz3
l3 lx3
5 xz;x 5 2 x1 1
1 z1
Gxz
Ex
E1
E2
E3
"
1 C122 ðlx1 ly3 1 ly1 lx3 Þlx1 ly1
1 C133 ðlx1 lz3 1 lx3 lz1 Þlx1 lz1 1 C233 ðly1 lz3 1 ly3 lz1 Þly1 lz1 ð1; 2; 3Þ; ðx; y; zÞ
l2y1 ly2 ly3
ηx;yz
ηyz;x
l2 lz2 lz3
l2 lx2 lx3
5
5 2 x1
1
1 z1
Gyz
Ex
E1
E2
E3
#
ν 12 2
ν 13 2
ν 23 2
lx1 ly1
2
2
2
2
ðl ly2 ly3 1 ly1 lx2 lx3 Þ 2
ðl lz2 lz3 1 lz1 lx2 lx3 Þ 2
ðl lz2 lz3 1 lz1 ly2 ly3 Þ 1
ðlx2 ly3 1 lx3 ly2 Þ
E2 x1
E3 x1
E3 y1
G12
lx1 lz1
ly1 lz1
ðlx2 lz3 1 lx3 lz2 Þ 1
ðly2 lz3 1 ly3 lz2 Þð1; 2; 3Þ; ðx; y; zÞ
1
G13
G23
"
#
l2y1 l2y2
l2 l2
1
l2 l2
ν 12
ν 13
ν 23
5 4 x1 x2 1
1 z1 z2 2 2
lx1 lx2 ly1 ly2 1
lx1 lx2 lz1 lz2 1
ly1 ly2 lz1 lz2
E1
E2
E3
E2
E3
E3
Gxy
1
1
2
2
1
ðlx1 ly2 1lx2 ly1 Þ 1
ðlx1 lz2 1lx2 lz1 Þ
G12
G13
1
1
ðly1 lz2 1ly2 lz1 Þ2 ð1; 2; 3Þ; ðx; y; zÞ
G23
186
CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
"
l2y1 ly2 ly3
l2 lz2 lz3
λxy;xz
λxz;xy
l2 lx2 lx3
5
5 4 x1
1
1 z1
Gxz
Gxy
E1
E2
E3
ν 12
ν 13
2
ðlx3 ly2 1 lx2 ly3 Þlx1 ly1 2
ðlx2 lz3 1 lz2 lx3 Þlx1 lz1
E2
E3
#
ν 23
1
2
ðly2 lz3 1 ly3 lz2 Þly1 lz1 1
ðlx1 ly3 1 lx3 ly1 Þðlx1 ly2 1 lx2 ly1 Þ
G12
E3
1
ðlx1 lz3 1 lx3 lz1 Þðlx1 lz2 1 lx2 lz1 Þ
G13
1
1
ðly1 lz3 1 ly3 lz1 Þðly1 lz2 1 ly2 lz1 Þ ð1; 2; 3Þ; ðx; y; zÞ
G23
1
in which
C122 5
1
2ν 12
1
2ν 13
2
; C133 5
2
;
G12
G13
E2
E3
1
2ν 23
C233 5
2
E3
G23
Consider a special spatial structure (Pagano & Whitford, 1985) formed by a fabric composite in
which the plies reinforced at angle φ (warp direction) with respect to the x-axis make angles α and
β with the x-axis and the y-axis, respectively, as in Fig. 2.101. The directional cosines for this
structure are
lx1 5 cosλcosψ;
lx3 5 2 sinψ;
ly2 5 cosλcosβ 1 sinλsinβsinψ;
lz1 5 sinλsinβ 1 cosλcosβsinψ
lz2 5 cosλsinβ 2 sinλcosβsinψ;
21
lx2 5 2 sinλcosψ
ly1 5 sinλcosβ 2 cosλsinβsinψ
ly3 5 2 sinβcosψ
lz3 5 cosβcosψ
21
where λ 5 φ 1 tan ðtanβsinψÞ, ψ 5 tan ðtanαcosβÞ
The dependencies of elastic constants Ex ; Ey ; Gxz , and Gyz calculated with the aid of
Eq. (2.197) for the material with E1 5 12:9 GPa, E2 5 5:2 GPa, E3 5 3 GPa, G12 5 G13 5 1:5 GPa,
G23 5 1 GPa, ν 21 5 0:15; ν 31 5 0:2; and ν 32 5 0:2 are presented in Fig. 2.102 (Vasiliev &
Morozov, 1988).
z
y
3
β
1
2
φ
α
FIGURE 2.101
Orientation angles in a spatial composite structure.
x
REFERENCES
E , GPa
16
G, GPa
1.6
1
1
12 2
Ex
Ey
3
8
187
2
1.2 3
3
2
1
0.8
3
Gyz
2
1
Gxz
1
4 2
0.4
3
0
0
15
30
45
60
75
φ°
90
0
0
15
30
45
60
75
φ°
90
FIGURE 2.102
Dependencies of the elastic constants of a spatially reinforced composite on the orientation angles:
ð1Þ α 5 β 5 0 ; ð2Þ α 5 β 5 8 ; and ð3Þ α 5 β 5 16 .
For planar structures ðα 5 β 5 0Þ, Eqs. (2.196) and (2.197) generalize Eqs. (2.72) and (2.76) for
a three-dimensional stress state of an orthotropic layer.
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Chiao, T. T. (1979). Some interesting mechanical behaviors of fiber composite materials. In G. C. Sih, & V. P.
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Golubev, M. (2016). Evaluation of effective elastic constants in unidirectional three-component composite
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CHAPTER 2 MECHANICS OF A COMPOSITE LAYER
Green, A. E., & Adkins, J. E. (1960). Large elastic deformations and nonlinear continuum mechanics.
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Hahn, H. T., & Tsai, S. W. (1974). On the behavior of composite laminates after initial failures. Journal of
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Hashin, Z. (1987). Analysis of orthogonally cracked laminates under tension. Journal of Appilied Mechanics,
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CHAPTER
MECHANICS OF LAMINATES
3
A typical composite structure consists of a system of layers bonded together. The layers can be
made of different isotropic or anisotropic materials, and have different structures (see Chapter 2:
Mechanics of a Composite Layer), thicknesses, and mechanical properties. In contrast to the typical
layers which are described in Chapter 2, Mechanics of a Composite Layer and whose basic
properties are determined experimentally, the laminate characteristics are usually calculated using information concerning the number of layers, their stacking sequence, and geometric and mechanical
properties, which must be known. A finite number of layers can be combined to form so many
different laminates that the concept of studying them using experimental methods does not seem realistic.
Whereas the most complicated typical layer is described by nine stiffness coefficients Amn (mn 5 11, 22,
12, 14, 24, 44, 55, 56, and 66), some of which can be calculated, a laminate is characterized by 21 coefficients and demonstrates coupling effects that are difficult to simulate in experiments.
Thus, the topic of this chapter is to provide equations allowing us to predict the behavior of a
laminate as a system of layers with given properties. The only restriction that is imposed on the
laminate as an element of a composite structure concerns its total thickness, which is assumed to be
much smaller than the other dimensions of the structure.
3.1 STIFFNESS COEFFICIENTS OF A NONHOMOGENEOUS ANISOTROPIC
LAYER
Consider a thin nonhomogeneous layer, which is anisotropic in the xy-plane and whose mechanical
properties are some functions of the normal coordinate z (see Fig. 3.1). A typical example of such a
generalized layer is a laminate composed of different plies. Another practical example is a plate
made of functionally graded material (FGM), which is characterized by gradual variation in
z
b
b
τ xz
a
y
τ yz
σy
s h
e
a
τ xy
x
σx
τ xy
FIGURE 3.1
An element of a nonhomogeneous layer.
Advanced Mechanics of Composite Materials and Structures. DOI: https://doi.org/10.1016/B978-0-08-102209-2.00003-7
© 2018 Elsevier Ltd. All rights reserved.
191
192
CHAPTER 3 MECHANICS OF LAMINATES
composition and/or in microstructure over the material thickness, resulting in corresponding
changes in the material properties. Coordinate axes x and y belong to some plane which is referred
to as a reference plane such that z 5 0 on this plane and 2 e # z # s for the layer under study.
There exist some special locations of the reference plane discussed below, but in this section its
coordinates e and s are not specified. We introduce two assumptions, both based on the fact that
the thickness h 5 e 1 s is small.
First, it is assumed that the layer thickness, h, does not change under the action of stresses
shown in Fig. 3.1. Actually, the thickness does change, but because it is small, this change is negligible. This means that there is no strain in the z-direction, i.e.,
εz 5
@uz
5 0;
@z
uz 5 wðx; yÞ
(3.1)
Here, wðx; yÞ is the so-called normal deflection which is a translational displacement of a normal
element a-b (see Fig. 3.1) as a solid in the z-direction.
Second, we assume that in-plane displacements ux and uy are linear functions of the thickness
coordinate z, i.e.,
ux ðx; y; zÞ 5 uðx; yÞ 1 zθx ðx; yÞ
uy ðx; y; zÞ 5 vðx; yÞ 1 zθy ðx; yÞ
(3.2)
where u and v are the displacements of the points of the reference plane z 5 0 or, which is the
same, the translational displacements of the normal element a-b (see Fig. 3.1) as a solid in the xand y-directions, whereas θx and θy are the angles of rotations (usually referred to as “rotations”) of
the normal element a-b in the xz- and yz-planes. Geometric interpretation of the first expression in
Eq. (3.2) is presented in Fig. 3.2.
In-plane strains of the layer, εx , εy , and γ xy , can be found using Eqs. (3.1) and (3.2) as
@ux
5 ε0x 1 zκx
@x
@uy
5 ε0y 1 zκy
εy 5
@y
@ux
@uy
1
5 γ 0xy 1 zκxy
γ xy 5
@y
@x
εx 5
ux
u
z
b
A
θx z
(3.3)
b′
A′
a′
v
x
z
a
FIGURE 3.2
Decomposition of displacement ux of point A into translational (u) and rotation (zθx) components.
3.1 STIFFNESS COEFFICIENTS OF A NONHOMOGENEOUS ANISOTROPIC
193
where
@u
@v
@u @v
; ε0y 5 ; γ 0xy 5
1
@x
@y
@y
@x
@θx
@θy
@θx
@θy
; κy 5
; κxy 5
1
κx 5
@x
@y
@y
@x
ε0x 5
These generalized strains correspond to the following four basic deformations of the layer
shown in Fig. 3.3:
•
•
•
•
In-plane tension or compression ðε0x ; ε0y Þ
In-plane shear ðγ 0xy Þ
Bending in the xz- and yz-planes ðκx ; κy Þ
Twisting ðκxy Þ
(A)
ε x0 , ε y0
(B)
γ xy0
(C)
κx
(D)
κ xy
FIGURE 3.3
Basic deformations of the layer: (A) in-plane tension and compression ðε0x ; ε0y Þ, (B) in-plane shear ðγ 0xy Þ, (C)
bending ðκx Þ, and (D) twisting ðκxy Þ.
194
CHAPTER 3 MECHANICS OF LAMINATES
The constitutive equations for an anisotropic layer, Eq. (2.71), upon substitution of
Eq. (3.3), yield
σx 5 A11 ε0x 1 A12 ε0y 1 A14 γ 0xy 1 zðA11 κx 1 A12 κy 1 A14 κxy Þ
σy 5 A21 ε0x 1 A22 ε0y 1 A24 γ 0xy 1 zðA21 κx 1 A22 κy 1 A24 κxy Þ
(3.4)
τ xy 5 A41 ε0x 1 A42 ε0y 1 A44 γ 0xy 1 zðA41 κx 1 A42 κy 1 A44 κxy Þ
where Amn 5 Anm are the stiffness coefficients of the material that depend, in general, on the
coordinate z.
It follows from Eq. (3.4) that the stresses depend on six generalized strains ε, γ, and κ, which
are functions of coordinates x and y only. To derive the constitutive equations for the layer under
study, we introduce the corresponding force functions as stress resultants and couples shown in
Fig. 3.4 and specified as
ðs
Nx 5
ðs
σx dz;
Ny 5
2e
ðs
Mx 5
ðs
σy dz;
Nxy 5
2e
ðs
σx zdz;
My 5
2e
τ xy dz
2e
ðs
σy zdz; Mxy 5
2e
τ xy zdz
2e
(see also Fig. 3.1). Substituting the stresses, Eq. (3.4), into these equations, we arrive at the following constitutive equations that relate stress resultants and couples to the corresponding generalized strains, i.e.,
Nx 5 B11 ε0x 1 B12 ε0y 1 B14 γ 0xy 1 C11 κx 1 C12 κy 1 C14 κxy
Ny 5 B21 ε0x 1 B22 ε0y 1 B24 γ 0xy 1 C21 κx 1 C22 κy 1 C24 κxy
Nxy 5 B41 ε0x 1 B42 ε0y 1 B44 γ 0xy 1 C41 κx 1 C42 κy 1 C44 κxy
(3.5)
Mx 5 C11 ε0x 1 C12 ε0y 1 C14 γ 0xy 1 D11 κx 1 D12 κy 1 D14 κxy
My 5 C21 ε0x 1 C22 ε0y 1 C24 γ 0xy 1 D21 κx 1 D22 κy 1 D24 κxy
Mxy 5 C41 ε0x 1 C42 ε0y 1 C44 γ 0xy 1 D41 κx 1 D42 κy 1 D44 κxy
M xy
Mx
My
Ny
N xy
M xy
Mx
N xy
M xy
Nx
N xy
M xy
My
Ny
N xy
FIGURE 3.4
Stress resultants and couples applied to the reference plane of the layer.
Nx
3.1 STIFFNESS COEFFICIENTS OF A NONHOMOGENEOUS ANISOTROPIC
195
These equations include membrane stiffness coefficients
ðs
Bmn 5 Bnm 5
Amn dz
(3.6)
2e
which specify the layer stiffness under in-plane deformation (Figs. 3.3A and B), bending stiffness
coefficients
ðs
Dmn 5 Dnm 5
Amn z2 dz
(3.7)
2e
which are associated with the layer bending and twisting (Figs. 3.3C and D), and membranebending coupling coefficients
ðs
Cmn 5 Cnm 5
Amn zdz
(3.8)
2e
through which in-plane stress resultants are related to bending deformations, and stress couples are
linked with in-plane strains.
Coefficients with subscripts 11, 12, 22, and 44 compose the basic set of the layer stiffnesses
associated with in-plane extension, contraction, and shear (B11 , B12 , B22 , and B44 ), bending and
twisting (D11 , D12 , D22 , and D44 ), and coupling effects (C11 , C12 , C22 , and C44 ). For an anisotropic
layer there also exists coupling between extension (A) and shear (B) in Fig. 3.3 (coefficients B14
and B24 ), extension (A) and twisting (D) in Fig. 3.3 (coefficients C14 and C24 ), and bending (C)
and twisting (D) in Fig. 3.3 (coefficients D14 and D24 ).
The forces and moments N and M specified by Eq. (3.5) are resultants and couples of in-plane
stresses σx , σy , and τ xy (see Fig. 3.1). However, there also exist transverse shear stresses τ xz and
τ yz , which should be expressed in terms of the corresponding shear strains. Unfortunately, we cannot apply the direct approach that was used above to derive Eq. (3.5) for this purpose. This direct
approach involves straindisplacement equations, i.e.,
@uy
@uz
1
@z
@y
(3.9)
τ xz 5 A55 γ xz 1 A56 γ yz ;
τ yz 5 A65 γ xz 1 A66 γ yz
(3.10)
γ xz 5 a55 τ xz 1 a56 τ yz ;
γ yz 5 a65 τ xz 1 a66 τ yz
(3.11)
γ xz 5
@ux
@uz
1
;
@z
@x
γ yz 5
in conjunction with Hooke’s law
or
where Amn and amn are stiffness and compliance coefficients, respectively. The problem is associated with Eq. (3.2), which specify only approximate dependence of displacements ux and uy on
coordinate z (the actual distribution of ux and uy through the layer thickness is not known) and
must not be differentiated with respect to z. So, we cannot substitute Eq. (3.2) into Eq. (3.9), which
includes derivatives of ux and uy with respect to z. To see what can happen if we violate this
well-known mathematical restriction, consider a sandwich laminate shown in Fig. 3.5. It can be
seen that while linear approximation of uðzÞ (dashed line) looks reasonable, the derivatives of the
actual displacements and the approximate ones have little in common.
196
CHAPTER 3 MECHANICS OF LAMINATES
∂u
∂z
z
u (z)
x
(A)
(B)
FIGURE 3.5
Actual (solid lines) and approximate (dashed lines) distributions of a displacement (A) and its derivative (B)
through the thickness of a sandwich laminate.
z
b
b
τ yz
s
τ xz
Vx
Vy
e
a
a
FIGURE 3.6
Reduction of transverse shear stresses to stress resultants (transverse shear forces).
To derive constitutive equations for transverse shear, consider Fig. 3.6. The actual distribution
of shear stresses τ xz and τ yz across the layer thickness is not known, but we can assume that it is
not important. Indeed, as follows from Eq. (3.1), elements a-b (see Fig. 3.6) along which the shear
stresses act are absolutely rigid. This means (in accordance with the corresponding theorem of
Statics of Solids) that the displacements of these elements in the z- direction depend only on the
resultants of the shear stresses, i.e., on transverse shear forces
ðs
Vx 5
ðs
τ xz dz;
Vy 5
2e
τ yz dz
(3.12)
2e
Since the particular distributions of τ xz and τ yz do not influence the displacements, we can introduce some average stresses having the same resultants as the actual ones, i.e.,
τx 5
Vx
1
5
h
h
ðs
τ xz dz;
τy 5
2e
Vy
1
5
h
h
ðs
τ yz dz
2e
However, according to Eq. (3.11), shear strains are linear combinations of shear stresses. So, we
can use the same law to introduce average shear strains as
γx 5
1
h
ðs
γ xz dz;
2e
γy 5
1
h
ðs
γ yz dz
2e
(3.13)
3.1 STIFFNESS COEFFICIENTS OF A NONHOMOGENEOUS ANISOTROPIC
197
Average shear strains γ x and γ y can be readily expressed in terms of displacements if we substitute Eq. (3.9) into Eq. (3.13), i.e.,
2
3
ðs
14
@uz 5
dz
γx 5
ux ðsÞ 2 ux ð2 eÞ 1
@x
h
2e
2
3
ðs
14
@uz 5
uy ðsÞ 2 ux ð2 eÞ 1
dz
γy 5
h
@y
2e
These equations, in contrast to Eq. (3.9), do not include derivatives with respect to z. So, we
can substitute Eqs. (3.1) and (3.2) to get the final result
γ x 5 θx 1
@w
;
@x
γ y 5 θy 1
@w
@y
(3.14)
Consider Eqs. (3.10) and (3.11). Integrating them over the layer thickness and using Eqs. (3.12)
and (3.13), we get
ðs
ðs
Vx 5
ðA55 γ xz 1 A56 γ yz Þdz;
2e
1
γx 5
h
Vy 5
ðA65 γ xz 1 A66 γ yz Þdz
2e
ðs
1
γy 5
h
ða55 τ xz 1 a56 τ yz Þdz;
2e
ðs
ða65 τ xz 1 a66 τ yz Þdz
2e
Since the actual distribution of stresses and strains according to the foregoing reasoning is
not significant (note that this is strictly true for the stresses only), we can change them for the
corresponding average stresses and strains:
Vx 5 S55 γ x 1 S56 γ y ;
Vy 5 S65 γ x 1 S66 γ y
(3.15)
γ x 5 s55 Vx 1 s56 Vy ;
γ y 5 s65 Vx 1 s66 Vy
(3.16)
where
ðs
Smn 5 Snm 5
Amn dz
(3.17)
2e
smn 5 snm 5
1
h2
ðs
amn dz
(3.18)
2e
It should be emphasized that Eq. (3.16) are not the inverse form of Eq. (3.15). Indeed, solving
Eq. (3.16), using Eq. (3.18), and taking into account that
a55 5 A66 ;
a56 5 2 A56 ; a66 5 A55 ;
Amn
Amn 5
A55 A66 2 A256
198
CHAPTER 3 MECHANICS OF LAMINATES
we arrive at Eq. (3.15) in which Smn should be changed to
Ðs
h2
Amn dz
2e
Smn 5 Ðs
Ðs
Ðs
ð A55 dzÞð A66 dzÞ 2 ð A56 dzÞ2
2e
2e
(3.19)
2e
These expressions, in general, do not coincide with Eq. (3.17).
Thus, the constitutive equations for transverse shear are specified by Eq. (3.15), and there exist
two, in general, different approximate forms of stiffness coefficients: Eqs. (3.17) and (3.19). The
fact that equations obtained in this way are approximate is quite natural because the assumed
displacement field, Eqs. (3.1) and (3.2), is also approximate.
To compare two possible forms of constitutive equations for transverse shear, consider for the
sake of brevity an orthotropic layer for which
A56 5 0;
a56 5 0; A55 5 Gxz ; A66 5 Gyz ;
1
1
; a66 5 A55 5
a55 5 A66 5
Gxz
Gyz
For transverse shear in the xz-plane, Eq. (3.15) yield
Vx 5 S55 γ x
Vx 5 S55 γ x
(3.20)
Gxz dz
(3.21)
or
in which, in accordance with Eq. (3.17)
ðs
S55 5
2e
whereas Eq. (3.19) yields
h2
dz
G
2e xz
S55 5 Ðs
(3.22)
If the shear modulus does not depend on z Eqs. (3.21) and (3.22) both, give the same result,
S55 5 S55 5 Gxz h. The same, of course, holds for the transverse shear in the yz-plane.
Using the energy method applied in Section 1.3, we can show that Eqs. (3.21) and (3.22)
provide the upper and the lower bounds for the exact transverse shear stiffness. Indeed, consider a
strip with unit width experiencing transverse shear induced by force Vx as in Fig. 3.7. Assume that
z
Δ
h
y
FIGURE 3.7
Transverse shear of a strip with unit width.
s
e
γx
1
l
Vx
x
3.1 STIFFNESS COEFFICIENTS OF A NONHOMOGENEOUS ANISOTROPIC
199
Eq. (3.20) links the actual force Vx with the exact angle γ x 5 Δ=l through the exact shear stiffness
Se55 , which we do not know, and which we would like to evaluate. To do this, we can use the two
basic variational principles. According to the principle of minimum total potential energy
Text # Tadm
(3.23)
where
ε
Text 5 Uext
2 Aext ;
ε
Tadm 5 Uadm
2 Aadm
are the total energies of the exact state and some admissible kinematic state expressed in terms of the
strain energy, U, and work A performed by force Vx on displacement Δ (see Fig. 3.7). For both states
Aext 5 Aadm 5 Vx Δ
and the condition in Eq. (3.23) reduces to
ε
ε
Uext
# Uadm
(3.24)
For the exact state, with due regard to Eq. (3.20), we get
Uext 5
l
l
Vx γ x 5 Se55 γ 2x
2
2
(3.25)
For the admissible state, we should use the following general equation:
U5
ðl
ðl
ðs
ðs
1
1
dx τ xz γ xz dz 5
dx Gxz γ 2xz dz 5 U ε
2
2
0
2e
2e
0
and use some approximation for γ xz . The simplest one is γ xz 5 γ x , so that
l
ε
Uadm
5 γ 2x
2
ðs
Gxz dz
(3.26)
2e
Then, Eqs. (3.24)(3.26) yield
ðs
Se55 #
Gxz dz
2e
Comparing this inequality with Eq. (3.21), we can conclude that this equation specifies the
upper bound for Se55 .
To determine the lower bound, we should apply the principle of minimum strain energy,
according to which
σ
Uext # Uadm
where
l
l V2
Uext 5 Vx γ x 5 U ex
2
2 S55
(3.27)
200
CHAPTER 3 MECHANICS OF LAMINATES
For the admissible state we should apply
U5
ðl
ðl
ðs
ðs 2
τ xz
1
1
dx τ xz γ xz dz 5
dx
dz 5 U σ
2
2
Gxz
2e
0
2e
0
and use some admissible distribution for τ xz . The simplest approximation is τ xz 5 Vx =h so that
σ
Uadm
5
l 2
V
2h2 x
ðs
2e
dz
Gxz
Substitution in the condition (3.27) yields
h2
dz
G
2e xz
Se55 $ Ðs
Thus, Eq. (3.22) provides the lower bound for Se55 , and the exact stiffness satisfies the following
inequality:
ðs
h2
e
#
S
#
Gxz dz
55
Ðs dz
2e
2e Gxz
It should be emphasized that Se55 in this analysis is not the actual (experimental) shear stiffness
coefficient of the laminate. It is the exact value of the stiffness coefficient which can be found
using the exact stress and strain fields following from three-dimensional elasticity equations.
So, constitutive equations for the laminate under study are specified by Eqs. (3.5) and (3.15).
Stiffness coefficients, which are given by Eqs. (3.6)(3.8), and (3.17) or (3.19), can be written in a
form more suitable for calculations. To do this, introduce the new coordinate t 5 z 1 e such that
0 # t # h (see Fig. 3.8). Transforming the integrals to this new variable, we have
ð0Þ
ð1Þ
ð0Þ
Bmn 5 Imn
; Cmn 5 Imn
2 eImn
;
ð2Þ
ð1Þ
ð0Þ
Dmn 5 Imn
2 2eImn
1 e2 Imn
(3.28)
where mn 5 11, 12, 22, 14, 24, and 44 and
ðrÞ
Imn
5
ðh
r 5 0; 1; 2
Amn tr dt;
0
z
A
h
s
e
FIGURE 3.8
Coordinates of an arbitrary point A.
x z
t
x
(3.29)
3.2 STIFFNESS COEFFICIENTS OF A HOMOGENEOUS LAYER
201
The transverse shear stiffnesses, Eqs. (3.17) and (3.19), take the form
ð0Þ
Smn 5 Imn
(3.30)
ð0Þ
h2 I mn
ð0Þ ð0Þ
ð0Þ
I 55 I 66 2 ðI 56 Þ2
(3.31)
and
Smn 5
where mn 5 55, 56, 66 and
ð0Þ
ðh
I mn 5
Amn dt
(3.32)
0
The coefficients Amn are specified by the expression given in notations to Eqs. (3.19).
3.2 STIFFNESS COEFFICIENTS OF A HOMOGENEOUS LAYER
Consider a layer whose material stiffness coefficients Amn do not depend on coordinate z. Then
ðrÞ
Imn
5
Amn r11
h ;
r11
ð0Þ
I mn 5 Amn h
(3.33)
and Eqs. (3.28), (3.30), and (3.31) yield the following stiffness coefficients for the layer:
Bmn 5 Amn h;
3
h
2 eh 1 e2 ;
Dmn 5 Amn
3
Cmn 5 Amn
h
2e ;
2
(3.34)
Smn 5 Amn h
Both Eqs. (3.30) and (3.31) give the same result for Smn . It follows from the second of
Eq. (3.34) that the membranebending coupling coefficients Cmn become equal to zero if we take
e 5 h=2, i.e., if the reference plane coincides with the middle-plane of the layer shown in Fig. 3.9.
In this case, Eqs. (3.5) and (3.15) take the following decoupled form:
z
h/2
h/2
y
FIGURE 3.9
Middle-plane of a laminate.
x
202
CHAPTER 3 MECHANICS OF LAMINATES
Nx 5 B11 ε0x 1 B12 ε0y 1 B14 γ 0xy ;
Ny 5 B21 ε0x 1 B22 ε0y 1 B24 γ 0xy ;
0
Nxy 5 B41 εx 1 B42 ε0y 1 B44 γ 0xy ;
Mx 5 D11 κx 1 D12 κy 1 D14 κxy ; My 5 D21 κx 1 D22 κy 1 D24 κxy
Mxy 5 D41 κx 1 D42 κy 1 D44 κxy
Vx 5 S55 γ x 1 S56 γ y ;
(3.35)
Vy 5 S65 γ x 1 S66 γ y
As can be seen, we have arrived at three independent groups of constitutive equations, i.e., for
the in-plane stressed state of the layer, bending and twisting, and transverse shear. The stiffness
coefficients, Eq. (3.34), become
Bmn 5 Amn h;
Dmn 5
Amn 3
h ;
12
Smn 5 Amn h
(3.36)
For an orthotropic layer, there are no in-plane stretchingshear coupling ðB14 5 B24 5 0Þ and
transverse shear coupling ðS56 5 0Þ. Then, Eq. (3.35) reduces to
Nx 5 B11 ε0x 1 B12 ε0y ; Ny 5 B21 ε0x 1 B22 ε0y ; Nxy 5 B44 γ 0xy
Mx 5 D11 κx 1 D12 κy ; My 5 D21 κx 1 D22 κy ; Mxy 5 D44 κxy
Vx 5 S55 γ x ; Vy 5 S66 γ y
(3.37)
In terms of engineering elastic constants, the material stiffness coefficients of an orthotropic
layer can be expressed as
A11 5 Ex ;
A12 5 ν xy Ex ; A22 5 Ey ;
A55 5 Gxz ; A66 5 Gyz
A44 5 Gxy
(3.38)
where Ex;y 5 Ex;y =ð1 2 ν xy ν yx Þ. Then, Eq. (3.36) yields
B11 5 Ex h; B12 5 ν xy Ex h; B22 5 Ey h; B44 5 Gxy h
1
ν xy
1
1
Ex h3 ; D22 5 Ey h3 ; D44 5 Gxy h3
D11 5 Ex h3 ; D12 5
12
12
12
12
S55 5 Gxz h; S66 5 Gyz h
(3.39)
Finally, for an isotropic layer, we have
Ex 5 Ey 5 E;
ν xy 5 ν yx 5 ν;
Gxy 5 Gxz 5 Gyz 5 G 5
E
2ð1 1 νÞ
and
B11 5 B22 5 Eh; B12 5 νEh; B44 5 S55 5 S66 5 Gh
1
ν
1
Gh3
Eh3 ; D12 5 Eh3 ; D44 5
D11 5 D22 5
12
12
12
(3.40)
where E 5 E=ð1 2 ν 2 Þ:
3.3 STIFFNESS COEFFICIENTS OF A LAMINATE
Consider the general case, i.e., a laminate consisting of an arbitrary number of layers with different
thicknesses hi and stiffnesses AðiÞ
mn ði 5 1; 2; 3; ?; kÞ. The location of an arbitrary ith layer of
3.3 STIFFNESS COEFFICIENTS OF A LAMINATE
203
z
tk = h
s
k
x
hi
i
tk
y
t i −1 t i
t2
t1
e
2
1
t0 = 0
FIGURE 3.10
Structure of the laminate.
the laminate is specified by the coordinate ti , which is the distance from the bottom plane of the
laminate to the top plane of the ith layer (see Fig. 3.10). Assuming that the material stiffness coefficients do not change within thickness of the layer, and using piecewise integration, we can write
the parameter Imn in Eqs. (3.29) and (3.32) as
ðrÞ
Imn
5
k
1 X
r11
AðiÞ ðtr11 2 ti21
Þ;
r 1 1 i51 mn i
ð0Þ
I mn 5
k
X
ðiÞ
Amn ðti 2 ti21 Þ
(3.41)
i51
where r 5 0, 1, 2 and t0 5 0, tk 5 h (see Fig. 3.10). For thin layers, Eq. (3.41) can be reduced to the
following form, which is more suitable for calculations
ð0Þ
Imn
5
k
X
AðiÞ
mn hi ;
ð0Þ
I mn 5
k
X
ðiÞ
Amn hi
i51
i51
k
1X
ð1Þ
Imn
5
AðiÞ hi ðti 1 ti21 Þ
2 i51 mn
k
1X
2
ð2Þ
5
AðiÞ hi ðt2 1 ti ti21 1 ti21
Þ
Imn
3 i51 mn i
(3.42)
in which hi 5 ti 2 ti21 is the thickness of the ith layer.
The membrane, coupling, and bending stiffness coefficients of the laminate are specified by
Eqs. (3.28) and (3.42).
Consider transverse shear stiffnesses that have two different forms determined by Eqs. (3.30)
and (3.31) in which
ð0Þ
Imn
5
k
X
i51
AðiÞ
mn hi ;
ð0Þ
I mn 5
k
X
ðiÞ
Amn hi
(3.43)
i51
A particular case, important for practical applications, is an orthotropic laminate for which
Eq. (3.5) takes the form
204
CHAPTER 3 MECHANICS OF LAMINATES
Nx 5 B11 ε0x 1 B12 ε0y 1 C11 κx 1 C12 κy
Ny 5 B21 ε0x 1 B22 ε0y 1 C12 κx 1 C22 κy
Nxy 5 B44 γ 0xy 1 C44 κxy
Mx 5 C11 ε0x 1 C12 ε0y 1 D11 κx 1 D12 κy
(3.44)
My 5 C21 ε0x 1 C22 ε0y 1 D21 κx 1 D22 κy
Mxy 5 C44 γ 0xy 1 D44 κxy
Here, membrane, coupling, and bending stiffnesses, Bmn , Cmn , and Dmn , are specified by
Eq. (3.28), i.e.,
ð0Þ
ð1Þ
ð0Þ
Bmn 5 Imn
; Cmn 5 Imn
2 eImn
;
ð2Þ
ð1Þ
ð0Þ
2 2eImn
1 e2 Imn
Dmn 5 Imn
(3.45)
where mn 5 11, 12, 22, and 44.
Transverse shear forces Vx and Vy are specified by equations similar to Eq. (3.20)
Vx 5 S55 γ x ;
Vy 5 S66 γ y
in which the corresponding stiffness coefficients, Eqs. (3.30) and (3.31) reduce to (mn 5 55, 66)
Smn 5
k
X
AðiÞ
mn hi ;
i51
Smm 5 P
h2
k
i51
hi
(3.46)
AðiÞ
mm
Laminates composed of unidirectional plies have special stacking-sequence notations. For
example, notation ½02 = 1 45 = 2 45 =902 means that the laminate consists of a 0 layer having two
plies, a 6 45 angle-ply layer, and a 90 layer also having two plies. The notation ½0 =90 5 means
that the laminate has five cross-ply layers.
3.4 SYMMETRIC LAMINATES
Symmetric laminates are composed of layers that are symmetrically arranged with respect to the
laminate’s middle plane as shown in Fig. 3.11. Introduce the layer coordinate zi (see Fig. 3.11).
k 2
i
z i −1
i
k 2
FIGURE 3.11
Layer coordinates of a symmetric laminate.
zi
h
2
h
2
3.4 SYMMETRIC LAMINATES
205
Since for any layer which is above the middle surface z 5 0 and has the coordinate zi there is a
similar layer, which is located under the middle surface and has the coordinate (2zi ), the integration over the laminate thickness can be performed from z 5 0 to z 5 h=2 (see Fig. 3.11). Then, the
integrals for Bmn and Dmn similar to Eqs. (3.6) and (3.7) must be doubled, whereas the integral for
Cmn similar to Eq. (3.8) is equal to zero. Thus, the stiffness coefficients entering Eq. (3.5) become
Bmn 5 2
ð h=2
Amn dz;
ð h=2
Dmn 5 2
Amn z2 dz;
0
Cmn 5 0
(3.47)
0
For a symmetric laminate shown in Fig. 3.11, we get
k=2
k=2
X
X
Bmn 5 2 AðiÞ
AðiÞ
mn ðzi 2 zi21 Þ 5 2
mn hi
i51
i51
Cmn 5 0
(3.48)
k=2
k=2
2 X ðiÞ 3
2 X ðiÞ
Dmn 5
Amn ðzi 2 z3i21 Þ 5
A hi ðz2i 1 zi zi21 1 z2i21 Þ
3 i51
3 i51 mn
where hi 5 zi 2 zi21 . The transverse shear stiffness coefficients are given by Eqs. (3.30) and (3.31)
in which
ð0Þ
Imn
52
k
X
AðiÞ
mn hi ;
ð0Þ
I mn 5 2
i51
k=2
X
ðiÞ
Amn hi ;
ðiÞ
Amn 5
i51
AðiÞ
mn
ðiÞ ðiÞ
2
A55 A66 2 ðAðiÞ
56 Þ
(3.49)
To indicate symmetric laminates, a contracted stacking-sequence notation is used, e.g.,
½0 =90 =45 s instead of ½0 =90 =45 =45 =90 =0 . Symmetric laminates are characterized by a specific feature: their bending stiffness is higher than the bending stiffness of any asymmetric laminate
composed of the same layers. To show this property of symmetric laminates, consider Eqs. (3.28)
and (3.29) and apply them to calculate stiffness coefficients with some combination of subscripts,
e.g., m 5 1 and n 5 1. Since the coordinate of the reference plane, e, is an arbitrary parameter, we
can find it from the condition C11 5 0. Then,
I ð1Þ
e 5 11
ð0Þ
I11
and
"
ð2Þ
D11 5 I11
2
(3.50)
ð1Þ 2
ðI11
Þ
#
ð0Þ
I11
(3.51)
Introduce a new coordinate for an arbitrary point A in Fig. 3.12 as z 5 t 2 ðh=2Þ. Changing t to
z, we can present Eq (3.29) in the form
ðrÞ
I11
5
h=2
ð
2h=2
h
A11 ð 1zÞr dz
2
206
CHAPTER 3 MECHANICS OF LAMINATES
z
A
h/2
h/2
t
FIGURE 3.12
Coordinate of point A referred to the middle plane.
Substituting these integrals into Eqs. (3.50) and (3.51), we have
e5
ð1Þ
h J11
1 ð0Þ
2 J11
and
"
ð2Þ
D11 5 J11
2
ð1Þ 2
ðJ11
Þ
(3.52)
#
(3.53)
ð0Þ
J11
where
ðrÞ
J11
5
h=2
ð
A11 zr dz
(3.54)
2h=2
and r 5 0, 1, 2.
Now decompose A11 as a function of z into symmetric and antisymmetric components, i.e.,
A11 ðzÞ 5 As11 ðzÞ 1 Aa11 ðzÞ
Then, Eq. (3.54) yields
ð0Þ
J11
5
h=2
ð
As11 dz;
2h=2
ð1Þ
J11
5
h=2
ð
Aa11 zdz;
2h=2
ð2Þ
J11
5
h=2
ð
As11 z2 dz
2h=2
ð1Þ
As can be seen from Eq. (3.53), D11 reaches its maximum value if J11
5 0. Then, Aa11 5 0 and
s
A11 5 A11 , which means that the laminate is symmetric. In this case, Eq. (3.52) gives e 5 h=2.
Thus, symmetric laminates provide the maximum bending stiffness for a given number and for
mechanical properties of layers and, being referred to the middle-plane, do not have membranebending coupling effects. This essentially simplifies the behavior of the laminate under loading
and constitutive equations, which have the form specified by Eq. (3.35).
3.5 ENGINEERING STIFFNESS COEFFICIENTS OF ORTHOTROPIC
LAMINATES
It follows from Eq. (3.28) that the laminate stiffness coefficients depend, in the general case, on the
coordinate of the reference surface e. By changing e, we can change the bending stiffness coefficient Dmn . Naturally, the result of the laminate analysis undertaken with the aid of the constitutive
3.5 ENGINEERING STIFFNESS COEFFICIENTS OF ORTHOTROPIC
207
z
M
x
y
M
e
N
h
N
FIGURE 3.13
Laminated element under tension and bending.
equations, Eqs. (3.5), does not depend on the particular preassigned value of the coordinate e
because of the coupling coefficients Cmn which also depend on e. To demonstrate this, consider an
orthotropic laminated element loaded with axial forces N and bending moments M uniformly
distributed over the element width as in Fig. 3.13. Suppose that the element displacement does not
depend on coordinate y. Then, taking Nx 5 N, Mx 5 M, ε0y 5 0 and κy 5 0 in Eq. (3.44), we get
N 5 B11 ε0x 1 C11 κx ;
M 5 C11 ε0x 1 D11 κx
(3.55)
where, in accordance with Eq. (3.28),
ð0Þ
ð1Þ
ð0Þ
B11 5 I11
; C11 5 I11
2 eI11
;
ð2Þ
ð1Þ
2 ð0Þ
D11 5 I11 2 2eI11 1 e I11
(3.56)
k
1 X
r11
AðiÞ ðtr11 2 ti21
Þ
r 1 1 i51 11 i
(3.57)
Here, as follows from Eq. (3.41)
ðrÞ
I11
5
(r 5 0, 1, 2) are the coefficients which do not depend on the coordinate of the reference plane e.
It is important to emphasize that forces N in Fig. 3.13 act in the reference plane z 5 0, and the
strain ε0x in Eq. (3.55) is the strain of the reference plane. Solving Eq. (3.55) for ε0x and κx , we have
ε0x 5
1
ðD11 N 2 C11 MÞ;
D1
κx 5
1
ðB11 M 2 C11 NÞ
D1
(3.58)
where
2
D1 5 B11 D11 2 C11
(3.59)
Substituting B, D, and C with their expressions given by Eq. (3.56), we find
ð0Þ ð2Þ
ð1Þ 2
D1 5 I11
I11 2 ðI11
Þ
As can be seen, the parameter D1 does not depend on e.
Consider now the same element but loaded with forces P applied to the middle plane of the
element as in Fig. 3.14. As follows from Fig. 3.15 showing the element cross-section, the forces
and the moments in Fig. 3.13 induced by the forces shown in Fig. 3.14 are
208
CHAPTER 3 MECHANICS OF LAMINATES
P
h/2
h/2
FIGURE 3.14
Laminated element under tension.
z
A
h/2
t
h/2
e
FIGURE 3.15
Cross-section of the element.
N 5 P;
h
2e
M5P
2
(3.60)
Substitution of Eq. (3.60) into Eq. (3.58) yields
ε0x 5
P ð2Þ
h ð1Þ
ð1Þ
ð0Þ
I11 2 eI11
2 ðI11
2 eI11
Þ
D1
2
P h ð0Þ
ð1Þ
I11 2 I11
κx 5
D1 2
(3.61)
(3.62)
It follows from Eq. (3.62), that κx does not depend on e, which is expected because the curvature induced by forces P in Fig. 3.14 is the same for all the planes z 5 constant of the element.
However, Eq. (3.61) includes e which is also expected because ε0x is the strain in the plane z 5 0
located at the distance e from the lower plane of the element (see Fig. 3.15). Let us find the strain
εtx at some arbitrary point A of the cross-section for which z 5 t 2 e (see Fig. 3.15). Using the first
equation of Eq. (3.3), we have
εtx 5 ε0x 1 ðt 2 eÞκx 5
P ð2Þ h ð0Þ
ð1Þ
ð1Þ
I11 2 ðtI11 2 I11
Þ 2 tI11
D1
2
This equation includes the coordinate of point A and does not depend on e. Thus, taking an
arbitrary coordinate of the reference plane, and applying Eq. (3.56) for the stiffness coefficients, we
arrive at values of C11 and D11 , the combination of which provides the final result that does not
depend on e. However, the stiffness coefficient D11 is not the actual bending stiffness of the laminate which cannot depend on e.
To determine the actual stiffness of the laminate, return to Eq. (3.58) for ε0x and κx . Suppose
that C11 5 0, which means that the laminate has no bendingstretching coupling effects. Then,
Eq. (3.59) yields D 5 B11 D11 and Eq. (3.58) becomes
3.5 ENGINEERING STIFFNESS COEFFICIENTS OF ORTHOTROPIC
ε0x 5
N
;
B11
κx 5
M
D11
209
(3.63)
It is obvious that now B11 is the actual axial stiffness and D11 is the actual bending stiffness of
the laminate. However, Eq. (3.63) is valid only if C11 5 0. Using the second equation of Eq. (3.56),
we get
I ð1Þ
e 5 11
ð0Þ
I11
(3.64)
Substituting this result into Eq. (3.56) and introducing new notations Bx 5 B11 and Dx 5 D11 for
the actual axial and bending stiffness of the laminate in the x-direction, we arrive at
ð0Þ
Bx 5 I11
;
ð2Þ
Dx 5 I11
2
ð1Þ 2
ðI11
Þ
ð0Þ
I11
(3.65)
ðrÞ
Here, coefficients I11
(r 5 0, 1, 2) are specified by Eq. (3.57). The corresponding stiffnesses in
the y-direction (see Fig. 3.13) are determined from similar equations, i.e.,
ð0Þ
By 5 I22
;
ð2Þ
Dy 5 I22
2
ð1Þ 2
ðI22
Þ
ð0Þ
I22
(3.66)
in which
ðrÞ
I22
5
k
1 X
r11
AðiÞ ðtr11 2 ti21
Þ
r 1 1 i51 22 i
For symmetric laminates, as discussed in Section 3.4, Cmn 5 0 and coefficients Dmn in
Eq. (3.48) specify the actual bending stiffnesses of the laminate, i.e.,
2 X ðiÞ
A hi ðz2i 1 zi zi21 1 z2i21 Þ
3 i51 11
k=2
Dx 5
2 X ðiÞ
A hi ðz2i 1 zi zi21 1 z2i21 Þ
Dy 5
3 i51 22
k=2
(3.67)
where coordinates zi and zi21 are shown in Fig. 3.11. Note, that if the number of layers k is not
even, the central layer is divided by the plane z 5 0 into two identical layers so k becomes even.
To find the shear stiffness, consider the element in Fig. 3.13 but loaded with shear forces, S,
and twisting moments H, uniformly distributed along the element edges as shown in Fig. 3.16. It
should be recalled that forces and moments are applied to the element reference plane z 5 0 (see
Fig. 3.13). Taking Nxy 5 S and Mxy 5 H in the corresponding Eq. (3.44), we get
S 5 B44 γ 0xy 1 C44 κxy ;
H 5 C44 γ 0xy 1 D44 κxy
(3.68)
in which, in accordance with Eqs. (3.28) and (3.41),
ð0Þ
B44 5 I44
;
ð1Þ
ð0Þ
C44 5 I44
2 eI44
;
ð2Þ
ð1Þ
ð0Þ
D44 5 I44
2 2eI44
1 e2 I44
(3.69)
210
CHAPTER 3 MECHANICS OF LAMINATES
x
y
H
S
FIGURE 3.16
Shear and torsion of the element.
where
ðrÞ
I44
5
k
1 X
r11
AðiÞ ðtr11 2 ti21
Þ
r 1 1 i51 44 i
The solution of Eq. (3.68) is
γ 0xy 5
1
ðD44 S 2 C44 HÞ;
D4
κxy 5
1
ðB44 M 2 C44 SÞ
D4
(3.70)
ð0Þ ð2Þ
ð1Þ 2
in which D4 5 I44
I44 2 ðI44
Þ .
A further transformation is used similar to that for Eqs. (3.58) and (3.59). Taking the coordinate
of the reference plane as
I ð1Þ
e 5 44
ð0Þ
I44
(3.71)
we get C44 5 0, and Eq. (3.70) reduce to
S
;
B44
γ 0xy 5
κxy 5
H
D44
(3.72)
Using the new notations B44 5 Bxy and D44 5 Dxy and applying Eqs. (3.69) and (3.71), we arrive
at
ð0Þ
Bxy 5 I44
;
ð2Þ
Dxy 5 I44
2
ð1Þ 2
ðI44
Þ
ð0Þ
I44
(3.73)
where Bxy is the actual in-plane shear stiffness of the laminate, whereas Dxy needs some comments.
The second equation of Eq. (3.72) yields
H 5 Dxy κxy
(3.74)
where κxy is given in notations to Eq. (3.3), i.e.,
κxy 5
@θx
@θy
1
@y
@x
(3.75)
The deformed state of the laminated element (see Fig. 3.16) loaded with twisting moments only
is shown in Fig. 3.17. Consider the deflection of point A with coordinates x and y. It follows from
Fig. 3.17 that w 5 xθx or w 5 yθy . Introduce the gradient of the torsional angle
3.5 ENGINEERING STIFFNESS COEFFICIENTS OF ORTHOTROPIC
211
x
y
θx
x
y
A
θy
w
FIGURE 3.17
Deformation of the element under torsion.
θ0 5
@θx
@θy
5
@y
@x
Since θ0 does not depend on x and y, θx 5 yθ0 , θy 5 xθ0 , and w 5 xyθ0 . Using Eq. (3.75), we have
κxy 5 2θ0 . Then, Eq. (3.74) yields
H 5 Dpt θ0
(3.76)
Dpt 5 2Dxy
(3.77)
where
is the plate torsional stiffness specifying the stiffness of the element which is loaded with twisting
moments applied to all four edges of the element, as shown in Fig. 3.16.
However, in practice we usually need the torsional stiffness of the element loaded with twisting
moments applied to only two opposite edges of the element, whereas the two other edges are free.
Since such loading induces not only twisting moments but also transverse shear forces V (see
Fig. 3.6), we must first determine the actual transverse (through-the-thickness) stiffnesses of a
laminate.
Consider an orthotropic laminated element loaded with transverse shear forces Vx 5 V uniformly
distributed over the element edge as in Fig. 3.18. From Eq. (3.20), we have two possible constitutive
equations, i.e.,
V 5 S55 γ x ;
V 5 S55 γ x
(3.78)
in which, in accordance with Eq. (3.46),
S55 5
k
X
AðiÞ
55 hi ;
i51
S55 5 P
h2
k
i51
hi
(3.79)
ðiÞ
A55
ðiÞ
ðiÞ
For the orthotropic material, AðiÞ
55 5 Gxz , where Gxz is the transverse shear modulus of the ith
layer. Thus, Eq. (3.79) takes the form
S55 5
k
X
i51
GðiÞ
xz hi ;
S55 5 P
h2
k
i51
hi
GðiÞ
xz
(3.80)
212
CHAPTER 3 MECHANICS OF LAMINATES
z
V
x
y
h
FIGURE 3.18
Laminated element loaded with transverse shear forces.
As shown in Section 3.1, S55 gives the upper bound and S55 gives the lower bound of the actual
transverse shear stiffness of the laminate. For a laminate consisting of identical layers, i.e., for the
case GðiÞ
xz 5 Gxz for all the layers, both equations of Eq. (3.80) give the same result,
S55 5 S55 5 Gxz h. However, in some cases, the results following from Eq. (3.80) can be dramatically
different, whereas for engineering applications we must have a unique constitutive equation instead
of Eq. (3.78), i.e.,
V 5 Sx γ x
(3.81)
and the question arises whether S55 or S55 should be taken as Sx in this equation. Since for a homogeneous material there is no difference between S55 and S55 , we can expect that this difference
shows itself in those laminates consisting of layers with different transverse shear moduli.
Consider, for example, sandwich structures composed of high-stiffness thin facing layers
(facings) and low-stiffness light foam core (Fig. 3.19A). The facings (2 in Fig. 3.19A) are made of
aluminum alloy with modulus Ef 5 70 GPa and shear modulus Gf 5 26:9 GPa. The foam core (1 in
Fig. 3.19A) has Ec 5 0:077 GPa and Gc 5 0:0385 GPa. The geometric and stiffness parameters of
two sandwich beams studied experimentally (Aleksandrov, Brukker, Kurshin, & Prusakov, 1960)
are presented in Table 3.1. The beams with length l 5 280 mm have been tested under transverse
bending. The coefficient Sa in the table corresponds to the actual shear stiffness found from the
experimental results. Actually, experimental study allows us to determine the shear parameter
(Vasiliev, 1993)
kG 5
D
Sa l2
(3.82)
which is presented in the third column of the table and depends on the bending stiffness, D, and the
beam length, l. Since the sandwich structure is symmetric, we can use Eq. (3.67) for Dx in which
2k 5 2 (the core is divided into two identical layers as in Fig. 3.19A)
h1 5
Að1Þ
11 5 Ec ;
Að2Þ
11 5 Ef ;
hc
z0 5 0; z1 5 ;
2
hc
;
2
h2 5 hf ;
Dx 5
2 1
3 2 3
Ec h3c 1 Ef hf
hc 1 hc hf 1 h2f
3 8
4
2
The final expression is
z2 5
hc
1 hf
2
3.5 ENGINEERING STIFFNESS COEFFICIENTS OF ORTHOTROPIC
2
hf
2
hf
1
hc
1
hc
213
hf
(B)
(A)
FIGURE 3.19
Three-layered (sandwich) and two-layered laminates.
Table 3.1 Parameters of Sandwich Structures
Shear Stiffness (GPa mm)
hf (mm)
hc (mm)
kG
Sa
S55
S55
Bending Stiffness (GPa mm3)
2.4
1.0
18.8
17.0
0.444
0.184
1.09
0.79
1.14
0.82
130
54.5
37,960
11,380
The results of the calculation are listed in the last column of Table 3.1. The shear stiffness
coefficients S55 and S55 can be found from Eq. (3.80), which for the structure in Fig. 3.19A take
the form
S55 5 hc Gc 1 2hf Gf
S55 5
ðhc 12hf Þ2
hc
2hf
1
Gc
Gf
The results of the calculation are presented in Table 3.1. As can be seen, coefficients S55 are in
good agreement with the corresponding experimental data, whereas coefficients S55 are higher by an
order of magnitude. Note, that S55 , providing the lower boundary for the exact shear stiffness, is higher than the actual stiffness Sa . The reason for this effect has been discussed in Section 3.1. The coefficient S55 specifies the lower boundary for the theoretical exact stiffness corresponding to the applied
model of the laminate, but not for the actual stiffness following from experiment. For example, the
actual shear stiffness of the sandwich beams described above can be affected by the compliance of
adhesive layers which bond the facings and the core and are not allowed for in the laminate model.
So, it can be concluded that the shear stiffness coefficient S55 specified by the corresponding
equation of Eq. (3.79) can be used to describe the transverse shear stiffness of composite laminates.
However, there are special structures for which coefficient S55 provides a better approximation of
shear stiffness than coefficient S55 . Consider, for example, a two-layered structure shown in
Fig. 3.19B and composed of a high-stiffness facing (2 in Fig. 3.19B) and a low-stiffness core (1 in
Fig. 3.19B). Assume, as for the sandwich structure considered above, that Gf 5 26:9 GPa and
Gc 5 0:0385 GPa, so that Gf =Gc 5 699, and take hc 5 9:9 mm, and hf 5 2:4 mm. It is obvious that
the core, having such a low shear modulus, does not work, and the transverse shear stiffness of the
laminate is governed by the facing layer. For this layer only, we get
S55 5 S55 5 Gf hf 5 64:6 GPa mm
214
CHAPTER 3 MECHANICS OF LAMINATES
whereas for the laminate, Eq. (3.80) yield
S55 5 hc Gc 1 hf Gf 5 65 GPa mm
ðhc 1hf Þ2
S55 5
5 0:59 GPa mm
hc
hf
1
Gc
Gf
As can be seen, coefficient S55 is too far from the value that would be expected. However,
structures of the type for which the stiffness coefficient S55 is more appropriate than the coefficient
S55 are not typical in composite technology and, being used, they usually do not require the calculation of transverse shear stiffnesses. For laminated composites it can be recommended to use the
coefficient S55 (Chen & Tsai, 1996). Thus, the transverse shear stiffness coefficient in Eq. (3.81)
can be taken in the following form:
Sx 5 P
h2
k
i51
(3.83)
hi
ðiÞ
Gxz
For shear in the yz-plane (see Fig. 3.18), we get a similar expression, i.e.,
Sy 5 P
h2
k
i51
(3.84)
hi
ðiÞ
Gyz
In the engineering analysis of laminated composites, transverse shear stiffnesses are mainly used
to study the problems of transverse bending of composite beams and plates. Note that the so-called
classical theory of laminated beams and plates ignores the transverse shear deformation of the laminate. Consider the constitutive equations for the shear forces and write them in the following form:
γx 5
Vx
;
Sx
γy 5
Vy
Sy
Taking Sx -N and Sy -N, we get γ x 5 0 and γ y 5 0. Applying Eq. (3.14) for γ x and γ y , we
can express the rotation angles in terms of the deflection as
θx 5 2
@w
;
@x
θy 5 2
@w
@y
Then, the expressions for curvatures entering Eq. (3.3) take the form
κx 5 2
@2 w
;
@x2
κy 5 2
@2 w
;
@y2
κxy 5 2 2
@2 w
@x@y
For actual laminates, the transverse shear stiffness coefficients are not infinitely high, but nevertheless, the classical theories ignoring the corresponding deformation are widely used in the analysis of composite structures. To evaluate the possibility of neglecting transverse shear deformation,
we can use the parameter kG specified by Eq. (3.82) and compare it with unity. The effect of the
transverse shear deformation is demonstrated in Table 3.2 for the problem of transverse bending of
simply supported sandwich beams with various parameters kG listed in the table. The right-hand
column of the table shows the ratio of the maximum deflections of the beam, w, found with
allowance for transverse shear deformation (wG ) and those corresponding to classical beam theory
3.5 ENGINEERING STIFFNESS COEFFICIENTS OF ORTHOTROPIC
215
Table 3.2 The Effect of Transverse Shear Deformation
on the Deflection of Sandwich Beams
kG 5
1
2
3
4
5
0.444
0.184
0.015
0.0015
0.0004
Mt
wG
wN
D
Sl2
Beam No.
5.386
2.805
1.152
1.014
1.002
e
z
b
τ xz
Mt
y
x
τ xy
h
FIGURE 3.20
Torsion of a laminated strip.
z
N xy
Vx
Mt
y
M xy
e
FIGURE 3.21
Forces and moments acting in the strip cross-section.
(wN ). As can be seen, for beams number 4 and 5, having parameter kG which is negligible in comparison with unity, the shear deformation practically does not affect the beams’ deflections.
Returning to the problem of torsion, we consider an orthotropic laminated strip with width b
loaded with a torque Mt as in Fig. 3.20. In contrast to the laminate shown in Fig. 3.6, the strip in
Fig. 3.20 is loaded only at the transverse edges, whereas the longitudinal edges y 5 6 b=2 are free.
The shear stresses τ xz and τ yz induced by torsion give rise to the shear forces Nxy , twisting moment
Mxy and transverse shear force Vx shown in Fig. 3.21. Applying the corresponding constitutive
equations, Eqs. (3.44) and (3.81), we get
Nxy 5 B44 γ 0xy 1 C44 κxy ;
Mxy 5 C44 γ 0xy 1 D44 κxy
Vx 5 Sx γ x
(3.85)
(3.86)
216
CHAPTER 3 MECHANICS OF LAMINATES
where the stiffness coefficients B, C, D, and S are specified by Eqs. (3.69) and (3.82). Preassign the
coordinate of the reference plane e in accordance with Eq. (3.71). Then, C44 5 0 and Eq. (3.85)
reduce to
Nxy 5 Bxy γ 0xy
(3.87)
Mxy 5 Dxy κxy
(3.88)
where Bxy and Dxy are given by Eq. (3.73). Since the strip is loaded with a torque Mt only (see
Fig. 3.20), Nxy 5 0, and as follows from Eq. (3.87), γ 0xy 5 0. So, we have only two constitutive
equations, i.e., Eqs. (3.86) and (3.88) for Vx and Mxy which are expressed in terms of the transverse
shear strain γ x and the twisting deformation κxy . Applying Eqs. (3.14) and (3.75), we have
γ x 5 θx 1
@w
;
@x
κxy 5
@θx
@θy
1
@y
@x
(3.89)
Consider the deformation of the strip. Assume that the strip cross-section rotates around the longitudinal axis x through an angle θ which depends only on x (Fig. 3.22). Then, as follows from
Fig. 3.22,
w 5 2 yθ;
θy 5 θ
and the corresponding substitution in Eq. (3.89) yields
γ x 5 θx 2 θ0 y;
κxy 5
@θx
1 θ0
@y
where θ0 5 dθ=dx. Using the first of these equations to transform the second one, we get
κxy 5
@γ x
1 2θ0
@y
Thus, the constitutive equations, Eqs. (3.86) and (3.88) take the following final form:
Vx 5 Sx γ x ;
@γ x
0
1 2θ
Mxy 5 Dxy
@y
(3.90)
Consider the equilibrium of the strip element shown in Fig. 3.23. The equilibrium equations in
this case are
@Vx
5 0;
@x
@Mxy
50
@x
z, w
y
θy
FIGURE 3.22
Rotation of the strip cross-section.
w
θ
(3.91)
3.5 ENGINEERING STIFFNESS COEFFICIENTS OF ORTHOTROPIC
217
M xy
Vx +
M xy
dy
dx
M xy +
h
Vx
M xy +
∂M xy
∂y
∂Vx
dx
∂x
∂M xy
∂x
dy
dy
FIGURE 3.23
Forces and moments acting on the strip element.
@Mxy
2 Vx 5 0
@y
(3.92)
The first two equations, Eq. (3.91) show that Vx 5 Vx ðyÞ and Mxy 5 Mxy ðyÞ. Then, as follows from
Eq. (3.90) for Vx , γ x 5 γ x ðyÞ. Substituting Mxy and Vx from Eq. (3.90) into Eq. (3.92) and taking into
account that θ0 does not depend on y, we arrive at the following ordinary differential equation for γ x :
d2 γx
2 k2 γ x 5 0
dy2
in which k2 5 Sx =Dxy . The general solution of this equation is
γ x 5 C1 sinh ky 1 C2 cosh ky
Substitution in Eq. (3.90) for Mxy yields
Mxy 5 Dxy ½2θ0 1 kðC1 cosh ky 1 C2 sinh kyÞ
The constants of integration C1 and C2 can be found from the boundary conditions according to
which Mxy ðy 5 6 b=2Þ 5 0 (see Fig. 3.20). The final solution is
Vx 5 2
2Sx sinh ky 0
θ;
kcosh λ
cosh ky
Mxy 5 2Dxy θ0 1 2
cosh λ
in which
1
b
λ 5 kb 5
2
2
sffiffiffiffiffiffiffi
Sx
Dxy
(3.93)
(3.94)
Consider Fig. 3.21 and express the applied torque Mt in terms of internal forces and moments
Vx and Mxy as
b=2
ð
Mt 5
ðMxy 2 Vx yÞdy
2b=2
218
CHAPTER 3 MECHANICS OF LAMINATES
Substituting Mxy and Vx from Eq. (3.93), we arrive at
Mt 5 Dt θ0
where
1
Dt 5 4Dxy b 1 2 tanh λ
λ
(3.95)
is the torsional stiffness of the strip. For a homogeneous orthotropic laminate discussed in
Section 3.2,
Dxy 5
and Eq. (3.95) reduces to
1
Gxy h3 ;
12
Sx 5 Gxz h
1
1
Dt 5 bh3 Gxy 1 2 tanh λ
3
λ
where
b
λ5
h
(3.96)
sffiffiffiffiffiffiffiffiffi
3Gxz
Gxy
The stiffness coefficient in Eq. (3.96) is in good agreement with the exact elasticity theory solutions (Vasiliev, 1993). Particularly, for b=h $ 3 the difference between Dt given by Eq. (3.96) and
the exact result is less than 2%. For a wide strip with relatively large b, the parameter λ in
Eq. (3.94) is also large, and Eq. (3.95) can be approximately reduced to
Dt 5 4Dxy b
(3.97)
Dividing Dt by b, we can find the stiffness of a laminate with unit width, i.e.,
Dbt 5 4Dxy
(3.98)
This is a beam torsional stiffness which is twice as high as the plate stiffness specified by
Eq. (3.77). The difference between Eqs. (3.77) and (3.98) is expected because Eq. (3.77)
corresponds to torsion with the moments acting on all four edges of the element (see Fig. 3.16),
whereas Eq. (3.98) describes torsion with only two moments applied at the transverse edges (see
Fig. 3.20).
Thus, the laminate membrane, bending, transverse shear, and torsional stiffness coefficients are
specified by Eqs. (3.65), (3.66), (3.82), (3.83), and (3.95).
3.6 QUASI-HOMOGENEOUS LAMINATES
Some typical layers considered in Chapter 2, Mechanics of a Composite Layer were actually
quasi-homogeneous laminates (see Sections 2.4 and 2.5), but being composed of a number of
identical plies, they were treated as homogeneous layers. The accuracy of this assumption is
evaluated below.
3.6 QUASI-HOMOGENEOUS LAMINATES
219
3.6.1 LAMINATE COMPOSED OF IDENTICAL HOMOGENEOUS LAYERS
Consider a laminate composed of layers with different thicknesses but the same stiffnesses,
i.e., AðiÞ
mn 5 Amn for all i 5 1, 2, 3, . . ., k. Then, Eqs. (3.29) and (3.32) yield
ðrÞ
Imn
5
Amn r11
h ;
r11
ð0Þ
I mn 5 Amn h
This result coincides with Eq. (3.33), which means that a laminate consisting of layers with the
same mechanical properties is a homogeneous laminate (layer) as studied in Section 3.2.
3.6.2 LAMINATE COMPOSED OF INHOMOGENEOUS ORTHOTROPIC LAYERS
Let the laminate have the following structure½0 =90 p , where p 5 1, 2, 3, . . . specifies the number
of elementary cross-ply couples of 0 and 90 plies. In Section 2.4, this laminate was treated as a
homogeneous layer with material stiffness coefficients specified by Eq. (2.114). Taking
h0 5 h90 5 0:5 in these equations, we have
1
A11 5 A22 5 ðE1 1 E2 Þ;
2
A12 5 E1 ν 12 ;
A44 5 G12
(3.99)
In accordance with Eq. (3.36), the stiffness coefficients of this layer should be
B0mn 5 Amn h;
0
Cmn
5 0;
D0mn 5
1
Amn h3
12
(3.100)
To calculate the actual stiffnesses of the laminate, we should put hi 5 δ, ti 5 iδ, k 5 2p, e 5 h=2,
and h 5 2pδ (see Fig. 3.10), where δ is the thickness of a unidirectional ply. Then, Eqs. (3.28) and
(3.42) yield
ð0Þ
Bmn 5 Imn
;
ð1Þ
ð0Þ
Cmn 5 Imn
2 pδ Imn
ð2Þ
ð1Þ
ð0Þ
Dmn 5 Imn
2 2pδ Imn
1 p2 δ2 Imn
(3.101)
Here,
h
ð0Þ
ð0Þ
I11
5 I22
5 pδ E1 ð1 1 αÞ 5 E1 ð1 1 αÞ;
2
ð0Þ
I12
5 2pδ E1 ν 12 5 E1 ν 12 h;
ð0Þ
5 2pδ G12 5 G12 h;
I44
ð1Þ
I11
5
p
δ2 X
E1 ½4jð1 1 αÞ 2 ð3 1 αÞ;
2
j51
p
δ2 X
1
ð1Þ
E1 ½4jð1 1 αÞ 2 ð3α 1 1Þ;
I12
5 E1 ν 12 h2 ;
2
2
j51
1
ð1Þ
I44
5 G12 h2 ;
2
p
δ3 X
ð2Þ
I11 5 E1 ½12j2 ð1 1 αÞ 2 6jð3 1 αÞ 1 7 1 α;
3
j51
p
δ3 X
ð2Þ
I22 5 E1 ½12j2 ð1 1 αÞ 2 6jð3α 1 1Þ 1 7α 1 1;
3
j51
1
1
ð2Þ
ð2Þ
I12 5 E1 ν 12 h3 ; I44
5 G12 h3
3
3
ð1Þ
5
I22
where α 5 E2 =E1 .
(3.102)
220
CHAPTER 3 MECHANICS OF LAMINATES
Matching Eqs. (3.99), (3.100), (3.101), and (3.102), we can see that Bmn 5 B0mn , i.e., the membrane stiffnesses are the same for both models of the laminate. The coupling and bending stiffnesses are also the same for mn 5 12, 44. There is no difference between the models for α 5 1
because the laminate reduces in this case to a homogeneous layer.
Performing further analysis, by summing the series in Eq. (3.102) and using Eq. (3.101), we
arrive at
C11 5 2 C22 5
1
E1 δ2 pðα 2 1Þ;
2
1
D11 5 D22 5 E1 δ3 p3 ð1 1 αÞ;
3
C12 5 C44 5 0;
D12 5 D012 ;
D44 5 D044
(3.103)
Taking into account that in accordance with Eq. (3.100) and accepted notations
1
D011 5 D022 5 E1 δ3 p3 ð1 1 αÞ
3
we can conclude that the only difference between the homogeneous and the laminated models is
associated with the coupling coefficients C11 and C22 which are equal to zero for the homogeneous
model and are specified by Eq. (3.103) for the laminated one. Since pδ 5 h=2, we can write these
coefficients in the form
1
C11 5 2 C22 5 E1 hδ ð1 1 αÞ
4
showing that Cmn -0 for δ-0:
3.6.3 LAMINATE COMPOSED OF ANGLE-PLY LAYERS
Consider a laminate with the following structure ½1φ =2φ p , where p is the number of layers
each consisting of 1φ and 2φ unidirectional plies. Taking the coordinate of the reference surface
e 5 h/2, we can write the constitutive equations, Eq. (3.5), as
Nx 5 B11 ε0x 1 B12 ε0y 1 C14 κxy
Ny 5 B21 ε0x 1 B22 ε0y 1 C24 κxy
Nxy 5 B44 γ 0xy 1 C41 κx 1 C42 κy
(3.104)
Mx 5 C14 γ 0xy 1 D11 κx 1 D12 κy
My 5 C24 γ 0xy 1 D21 κx 1 D22 κy
Mxy 5 C41 ε0x 1 C42 ε0y 1 D44 κxy
in which
Bmn 5 Amn h;
1
Cmn 5 2 Amn hδ;
2
Dmn 5
1
Amn h3
12
where h is the laminate thickness, δ is the ply thickness, and Amn are material stiffness coefficients
specified by Eq. (2.72). As can be seen, the laminate is anisotropic because the 1φ and 2φ plies
are located in different planes. The homogeneous model of the laminate ignores this fact and yields
3.6 QUASI-HOMOGENEOUS LAMINATES
221
C14 5 C24 5 0. Calculations show that these coefficients, although not actually equal to zero, have
virtually no practical influence on the laminate behavior for h=δ $ 20.
Laminates in which any ply or layer with orientation angle 1φ is accompanied by the same ply
or layer but with angle 2φ are referred to as balanced laminates. Being composed of only angleply layers, these laminates have no shearextension ðB14 5 B24 5 0Þ, bendingstretching, and
sheartwisting coupling ðC11 5 C12 5 C22 5 C44 5 0Þ. As follows from Eq. (3.104), only stretchingtwisting and bendingshear coupling can exist in balanced laminates. Such laminates can
include also 0 and 90 layers, however membranebending coupling can appear in such
laminates.
3.6.4 FIBER METAL LAMINATES
Fiber metal laminates (FMLs) are composed of thin (0.30.5 mm thick) metal layers alternating
with composite plies. To provide an appropriate adhesion, the metal layers are subjected to a special surface treatment, whereas the preimpregnated with resin, high-strength fibers are used to form
the composite layers having a unidirectional, cross-ply, or fabric structure. FMLs are
normally made by hot pressing, during which the metal layers are formed to a required shape and
the composite plies are cured. To date, there exists a number of FMLs, e.g. ARALL (aramid fibers
and aluminum), ALOR (aluminum and organic fibers), GLARE and SIAL (glass fibers and aluminum), and TIGR (titanium and graphite fibers), described elsewhere (Bucci & Mueller 1988; Bucci,
Mueller, Schulte, & Prohaska, 1987; Fridlyander et al., 1997; Vlot, Bulters, & Fredell, 1992; Vlot
& Gunnik, 2001; Vermeeren, 2003). Being originally developed to increase the strength of bolted
joints for composite laminates (Kolesnikov, Herbech, & Fink, 2008; Sirotkin, 1973; Vorobey &
Sirotkin, 1985), FMLs are now used to fabricate the airframe skin of modern airplanes (Ohrloff &
Horst, 1992). Compared to monolithic metals, FMLs have the following advantages, partially combining the benefits of metals and composites:
•
•
•
•
•
•
•
•
Considerably (by about 60%) higher tensile strength due to the higher fiber strength
Lower (by about 15%20%) density, since the density of composite materials is usually lower
than that of the metals included into the laminate
Ability to demonstrate elasticplastic behavior in contrast to pure composites which are
normally linear elastic
Higher (by an order of magnitude) fatigue strength, since transverse (through the thickness)
cracks in metal layers are arrested by fibers of cross-ply or fabric composite layers
Considerably higher damping properties which provide acoustic fatigue strength
FML skin able to be efficiently joined with mechanical fasteners, such as bolts or rivets
Impact damage resistance of FML usually higher than that of pure composite laminates
Parts with variable thickness readily fabricated by terminating the layers.
On the other hand, FMLs suffer from the following shortcomings:
•
Since Young’s and shear moduli of cross-ply or fabric glass or aramid plies are lower than the
corresponding characteristics of aluminum, the stiffness of FML is also lower than the stiffness
of aluminum. The same holds for FMLs composed of titanium and carbonepoxy plies.
222
•
•
•
•
CHAPTER 3 MECHANICS OF LAMINATES
In the metal layers, yielding starts once the strain exceeds the yield limit, and the lower
modulus of FMLs results in lower effective yield stress.
Hot pressing of FMLs induces residual interlaminar stresses.
Delamination caused by an impact impedes the application of FMLs in compressed areas of the
structure.
The cost of FMLs is an order of magnitude higher than the cost of metals.
Consider an FML shown in Fig. 3.24. Since the thickness of individual layers is very small, the
laminate can be treated as quasi-homogeneous and composed of the isotropic metal layer with total
thickness of hm and the orthotropic composite layer of thickness hc . The stiffness coefficients of the
metal (m) and composite (c) layers are
Ac11 5 E 1 ;
Ac12 5 E 1 ν 12 ;
m
Am
11 5 A22 5 E;
Ac22 5 E 2 ;
Am
12 5 Eν;
Ac44 5 G12 ;
Ac55 5 Ac66 5 Gcxz 5 Gcyz
m
m
Am
44 5 A55 5 A66 5 G 5 E=ð2ð1 1 ν ÞÞ
(the subscripts 1, 2, 3 correspond to the x, y, and z coordinate axes shown in Fig. 3.24). The membrane stiffness coefficients of the laminate become
B11 5 E 1 hc 1 Ehm ;
B22 5 E 2 hc 1 Ehm ;
B12 5 E 1 ν 12 hc 1 Eνhm
B44 5 G12 hc 1 Ghm
where, as earlier, E 1;2 5 E1;2 =ð1 2 ν 12 ν 21 Þ and E 5 E=ð1 2 ν 2 Þ. The corresponding bending stiffness
coefficients are expressed in terms of Bmn in accordance with Eq. (3.100), i.e.,
Dmn 5
h2
Bmn
12
where h 5 hc 1 hm is the total thickness of the laminate. Transverse shear stiffness coefficients are
specified by Eq. (3.83) and (3.84), which yield
z
y
metal
x
FIGURE 3.24
Fiber metal laminate.
composite
3.7 QUASI-ISOTROPIC LAMINATES IN THE PLANE STRESS STATE
Sx 5 Sy 5
223
h2
hc
hm
1
Gxz
G
3.7 QUASI-ISOTROPIC LAMINATES IN THE PLANE STRESS STATE
The layers of a laminate can be arranged in such a way that the laminate will behave as an isotropic layer under in-plane loading. Actually, the laminate is not isotropic (that is why it is
called a quasi-isotropic laminate) because under transverse (normal to the laminate plane) loading and under interlaminar shear its behavior is different from that of an isotropic (e.g., metal)
layer. The stresses in the layers of quasi-isotropic laminates are also different from those in isotropic materials.
To derive the conditions that should be met by the structure of a quasi-isotropic laminate, consider in-plane loading with stresses σx , σy , and τ xy that are shown in Fig. 3.1 and induce only inplane strains ε0x , ε0y , and γ 0xy . Taking κx 5 κy 5 κxy 5 0 in Eq. (3.5) and introducing average (through
the laminate thickness h) stresses as
σx 5 Nx =h;
σy 5 Ny =h;
τ xy 5 Nxy =h
we can write the first three equations of Eq. (3.5) in the following form:
σx 5 B11 ε0x 1 B12 ε0y 1 B14 γ 0xy
σy 5 B21 ε0x 1 B22 ε0y 1 B24 γ 0xy
(3.105)
τ xy 5 B41 ε0x 1 B42 ε0y 1 B44 γ 0xy
in which, in accordance with Eq. (3.28) and (3.42)
Bmn 5
k
X
AðiÞ
mn hi ;
hi 5 hi =h
(3.106)
i51
where hi is the thickness of the ith layer normalized to the laminate thickness and Amn are the stiffness coefficients specified by Eq. (2.72). For an isotropic layer, the constitutive equations analogous to Eq. (3.105) are
σx 5 Eðε0x 1 νε0y Þ;
σy 5 Eðε0y 1 νε0x Þ;
τ xy 5 Gγ 0xy
(3.107)
where
E5
E
;
1 2 ν2
G5
E
1
5 ð1 2 νÞE
2ð1 1 νÞ 2
(3.108)
Comparing Eqs. (3.105) and (3.107), we can see that the shearstretching coefficients of the
laminate, i.e., B14 5 B41 and B24 5 B42 , should be equal to zero. As follows from Eq. (2.72) and
Section 3.6.3, this means that the laminate should be balanced, i.e., it should be composed of 0 ,
6 φi (or φi and π 2 φi ), and 90 layers only. Since the laminate stiffness in the x- and the y- directions must be the same, we require that B11 5 B22 . Using Eq. (2.72), taking hi 5 h for all i, and performing the appropriate transformation, we arrive at the following condition:
224
CHAPTER 3 MECHANICS OF LAMINATES
k
X
cos2φi 5 0
i51
As can be checked by direct substitutions, for k 5 1 this equation is satisfied if φ1 5 45 and
for k 5 2 if φ1 5 0 and φ2 5 90 . Naturally, such one- and two-layered materials cannot be isotropic even in one plane. So, consider the case k $ 3, for which the solution has the form
π
φi 5 ði 2 1Þ ;
k
i 5 1; 2; 3; . . .; k
(3.109)
Using the sums that are valid for angles specified by Eq. (3.109), i.e.,
k
k
X
X
k
sin2 φi 5
cos2 φi 5
2
i51
i51
k
k
X
X
3k
sin4 φi 5
cos4 φi 5
8
i51
i51
k
X
k
sin2 φi cos2 φi 5
8
i51
and calculating stiffness coefficients from Eqs. (3.106) and (2.72), we get
1
B11 5 B22 5 3ðE1 1 E2 Þ 1 2ðE1 ν 12 1 2G12 Þ
8
1
B12 5 E1 1 E2 1 2ð3E1 ν 12 2 2G12 Þ
8
1
B44 5 E1 1 E2 2 2ðE1 ν 12 2 2G12 Þ
8
These stiffnesses provide constitutive equations in the form of Eq. (3.107) and satisfy the conditions in Eq. (3.108), which can be written as
B11 5 B22 5
E
;
1 2 ν2
B44 5 G
if
ðE1 1 E2 1 2E1 ν 12 ÞðE1 1 E2 2 2E1 ν 12 1 4G12 Þ
3ðE1 1 E2 Þ 1 2ðE1 ν 12 1 2G12 Þ
E1 1 E2 1 2ð3E1 ν 12 2 2G12 Þ
E
; G5
ν5
2ð1 1 νÞ
3ðE1 1 E2 Þ 1 2ðE1 ν 12 1 2G12 Þ
E5
(3.110)
Possible solutions to Eq. (3.109) providing quasi-isotropic properties of the laminates with
different number of layers are listed in Table 3.3 for k # 6.
All quasi-isotropic laminates, having different structures determined by Eq. (3.109) for a given
number of layers, k, possess the same apparent modulus and Poisson’s ratio specified by
Eq. (3.110). For typical advanced composites with the properties listed in Table 1.5, these characteristics are presented in Table 3.4.
It follows from Table 3.4 that the specific stiffness of quasi-isotropic composites with carbon
and boron fibers exceeds the corresponding characteristic of traditional isotropic structural materials: steel, aluminum, and titanium.
3.7 QUASI-ISOTROPIC LAMINATES IN THE PLANE STRESS STATE
225
Table 3.3 Angles Providing Quasi-Isotropic Properties of the Laminates
Orientation Angle of the ith layer
Number of Layers, k
φ1
φ2
φ3
φ4
φ5
φ6
3
4
5
6
0
0
0
0
60
45
36
30
120
90
72
60
135
108
90
144
120
150
Table 3.4 Modulus of Elasticity and Poisson’s Ratio of Quasi-Isotropic Laminates Made of
Typical Advanced Composites
Property
GlassEpoxy
CarbonEpoxy
AramidEpoxy
BoronEpoxy
BoronAl
Modulus, E (GPa)
Poisson’s ratio, ν
Specific modulus
kE 3 103 (m)
27.0
0.34
1290
54.8
0.31
3530
34.8
0.33
2640
80.3
0.33
3820
183.1
0.28
6910
Consider in more detail the behavior of the most popular quasi-isotropic laminate composed
of layers with angles 0 , 90 , 145 , and 245 (second row in Table 3.3) and equal thicknesses.
The laminate under study is made by filament winding of carbon tows HTS 5131 impregnated
with epoxy resin on a flat mandrel. The material density is 1510 kg/m3 , fiber volume fraction
55%, porosity 1.65%. The mechanical properties of the unidirectional plies are: longitudinal modulus
E1 5 105 GPa, transverse modulus E2 5 8:9 GPa, shear modulus G12 5 5 GPa, Poisson’s ratios
ν 12 5 0:025, ν 21 5 0:3, tensile and compressive longitudinal ultimate stresses σ 1
1 5 1700 MPa,
σ2
tensile and compressive transverse ultimate stresses σ 1
1 5 1300 MPa,
2 5 35 MPa,
σ2
2 5 170 MPa, and in-plane shear ultimate stress τ 12 5 35 MPa.
The experimental stressstrain diagram for this laminate is shown in Fig. 3.25 by the solid line.
As can be seen, the tensile diagram has two knees corresponding to matrix failure (the so-called
first and second ply failures) in the plies with angles 90 and 6 45 . To describe analytically the
laminate deformation under uniaxial tension in the x-direction (see, e.g., Fig. 2.33), apply the model
presented in Section 2.4. Stiffness coefficients of the plies specified by Eq. (2.72) are
A011 5 E1 ;
A012 5 E1 ν 12 ; A022 5 E2
1
45
A45
11 5 A22 5 ðE1 1 E 2 1 2E12 Þ
4
1
45
A12 5 E1 ν 12 1 ðE1 1 E2 2 2E12 Þ
4
90
A90
5
E
;
A
A90
2
11
12 5 E1 ν 12 ;
22 5 E1
in which, as earlier, E 1;2 5 E1;2 =ð1 2 ν 12 ν 21 Þ, E12 5 E 1 ν 12 1 2G12 . Taking h 0 5 h 90 5 0:25,
h 45 5 0:5, we can calculate the normalized membrane stiffness coefficients of the laminate using
Eq. (3.106), i.e.,
226
CHAPTER 3 MECHANICS OF LAMINATES
600
σ x , MPa
(1.52, 601.4)
3
500
3
(1.18, 439)
400
300
(0.474, 211.4)
2
(0.48, 203)
200
1
(0.4, 179.5)
(0.39, 171)
100
ε x,%
– 1.2
– 0.8
0
– 0.4
0.4
0.8
1.2
1.6
– 100
(−0.516, − 231.1)
4
– 200
– 300
(−1.156, − 489.2)
– 400
(−1.13, − 490)
5
– 500
FIGURE 3.25
Stressstrain diagram for the quasi-isotropic carbonepoxy laminate: analysis (x ), experiment (K ▬▬▬).
1
1
B11 5 B22 5
E1 1 E1 1 E2 1 2E12 1 E2
2
4
1
1
1
B12 5
E1 ν 12 1 2E1 ν 12 1 E1 1 E2 2 2E12 1 E1 ν 12 5 E1 ν 12 1 E1 1 E2 2 2E12
4
2
8
For uniaxial tension in the x-direction,
σx 5 B11 εx 1 B12 εy ;
σy 5 B12 εx 1 B22 εy 5 0;
τ xy 5 0
(3.111)
3.7 QUASI-ISOTROPIC LAMINATES IN THE PLANE STRESS STATE
227
Thus,
σx 5 Ex εx ;
εy 5 2 ν yx εx
(3.112)
where
2
Ex 5 B11 2
B12
;
B22
ν yx 5
B12
B22
(3.113)
Strains in the principal ply directions are specified by Eq. (2.69) according to which
ε01 5 εx ε02 5 εy ; ε012 5 0
1
45
ε45
ε45
12 5 ðεy 1 εx Þ
1 5 ε2 5 ðεx 1 εy Þ;
2
90
90
ε1 5 εy ; ε2 5 εx ; ε90
12 5 0
(3.114)
The corresponding stresses can be found using Hooke’s law, i.e.,
σ1 5 E1 ðε1 1 ν 12 ε2 Þ;
σ2 5 E2 ðε2 1 ν 21 ε1 Þ;
τ 12 5 G12 ε12
(3.115)
For the laminate under study, Eqs. (3.111) and (3.113) yield
B11 5 B22 5 49:39 GPa;
Ex 5 44:82 GPa;
B12 5 15:02 GPa
ν yx 5 0:304
(3.116)
Then, it follows from Eqs. (3.112) and (3.114) that
εx 5 2:23 3 1023 σx ;
εy 5 2 0:68 3 1023 σx
(3.117)
Here and further in this section, stresses are measured in MPa, whereas strains are expressed
in %.
The theoretical stressstrain diagram is shown in Fig. 3.25 by the dashed line. For the initial
loading (before the first ply failure) the dashed line corresponding to the first equation of
Eq. (3.117) closely coincides with the experimental solid line.
To find the stress which causes failure in the first ply, we apply the strength criterion,
Eq. (4.17) derived in Section 4.1.2, i.e.,
2 2
σ2
τ 12
1
51
σ2
τ 12
(3.118)
For the stresses in the plies, Eqs. (3.114) and (3.115) yield
σ01 5 2:51σx ;
σ45
1 5 0:9σ x ;
90
σ1 5 2 0:71σx ;
σ02 5 2 0:0009σx ;
τ 012 5 0
σ45
τ 45
2 5 0:097σ x ;
12 5 2 0:145σ x
90
σ2 5 0:195σx ;
τ 90
12 5 0
Substitution of these stresses into Eq. (3.118) allows us to conclude that the first ply failure
occurs in the 90 - layer at the stress σx 5 179:5 MPa. The corresponding strains which can be found
from Eq. (3.112) are: εx 5 0:4% and εy 5 0:122%. The experimental results are σx 5 171 MPa and
εx 5 0:39%. First ply failure is indicated in Fig. 3.25 by circle 1 (analysis) and dot 1 (experiment).
Now we consider further loading corresponding to σx $ σx . According to the model described
in Section 2.4, we should take in Eq. (3.111) E2 5 0, G12 5 0, and ν 12 5 0 for the 90 -layer. Then,
we arrive at the stiffness coefficients and elastic constants
228
CHAPTER 3 MECHANICS OF LAMINATES
B 11 5 46:98 GPa; B12 5 14:3 GPa; B22 5 49:39 GPa
Ex 5 42:83 GPa; ν yx 5 0:29
Comparing Ex with Eq. (3.116), we can conclude that the first ply failure reduces the laminate
stiffness. For σx $ σx , the strains can be found as
εx 5 εx 1 2:33ðσx 2 σx Þ U 1023 ;
εy 5 εy 2 0:68ðσx 2 σx Þ U 1023
(see Section 2.4). The stresses acting in the plies are
σ01 5 2:51σx 1 2:63ðσx 2 σx Þ; σ02 5 2 0:009σx 2 0:0023ðσx 2 σx Þ; τ 012 5 0
σ45
σ45
1 5 0:9σ x 1 0:96ðσ x 2 σ x Þ;
2 5 0:097σx 1 0:1ðσ x 2 σ x Þ
τ 45
12 5 2 0:145σ x 2 0:151ðσx 2 σx Þ
Substitution in Eq. (3.118) yields the stress σ
x 5 211:4 MPa which induces matrix failure in
the 45 plies. The corresponding strain is ε
x 5 0:474% (circle 2 in Fig. 3.25). The experimental
results are σx 5 203 MPa and εx 5 0:48% (dot 2 in Fig. 3.25).
To describe further loading (σx $ σ
x ), we should take E2 5 0, G12 5 0, and ν 12 5 0 for the 45
layer and arrive at
B 11 5 42:57 GPa; B12 5 14:9 GPa; B22 5 44:97 GPa
Ex 5 37:63 GPa; ν yx 5 0:33
The strains become
23
εx 5 ε
x 1 2:66ðσ x 2 σx ÞU10 ;
23
εy 5 ε
y 2 0:88ðσ x 2 σ x Þ U10
and the stress along the fibers of the 0 layer is
σ01 5 2:51σx 1 2:63 σ
x 2 σ x 1 2:99ðσ x 2 σ x Þ
The ultimate stress can be found using the strength condition σ01 5 σ 1
1 which yields
σ x 5 601:4 MPa and ε x 5 1:52%. This result corresponds to circle 3 in Fig. 3.25. It follows from this
figure that the experimental ultimate stress (dot 3) is only 439 MPa, i.e., some 27% lower. The reason for this difference is discussed in Section 2.5 (see Figs. 2.69 and 2.72). Whereas the theoretical
results correspond to a laminate reinforced with continuous fibers, the experimental data are obtained
using 38 mm wide and 2.6 mm thick samples cut out of a 250 3 250 mm plate. Thus, the fibers of
and 45 and 90 layers are not continuous and the edge effects discussed in Section 2.5 reduce the
effective strength of the laminate. Intralaminar and interlaminar cracks in the matrix result in low
efficiency of 45 plies. Indeed, if we assume that the load is taken by 0 plies only, we arrive at
σ x 5 425 MPa, which is fairly close to the experimental strength (439 MPa).
Under compression, the strains are specified by Eq. (3.117) and the strength criterion,
Eq. (3.118), in which σ 2 5 σ 2
2 shows that the first ply failure occurs in 45 layer under stress
σx 5 231:3 MPa and strain εx 5 0:516% (circle 4 on the diagram shown in Fig. 3.25). To describe
further loading (σx . σx ), we should take G12 5 0 for the 45 layer (see Section 2.4). Then, the stiffness coefficients become
B11 5 B22 5 46:9 GPa;
Ex 5 40:34 GPa;
B12 5 17:52 GPa
ν yx 5 0:37
3.8 ANTISYMMETRIC LAMINATES
229
The modulus obtained is close to the initial one and the knee on the experimental diagram is
usually not observed. For σx . σx , the strains become
εx 5 εx 1 2:48ðσx 2 σx Þ U 1023 ;
εy 5 εy 2 0:93ðσx 2 σx Þ U1023
These strains correspond to the dashed line between points 4 and 5 in Fig. 3.25. No other cracks
in the matrix appear up to the failure of the 0 ply fibers. The corresponding stress is
σ01 5 2:51σx 1 2:79ðσx 2 σx Þ
Putting σ01 5 σ 2
1 , we get σ x 5 489:2 MPa and ε x 5 1:156% (circle 5 in Fig. 3.25). The corresponding experimental results, i.e., σ x 5 490 MPa and ε x 5 1:13% (dot 5 in Fig. 3.25) are very
close to the theoretical ones.
The quasi-isotropic laminate under discussion is widely used as the skin of modern composite airplanes.
Demonstrating relatively high tensile and compressive strength and low density, this material is, on the
other hand, limited by relatively low stresses, inducing cracks in the matrix. Clearly, such cracks are not
acceptable under limit loads for commercial airplanes. That is why the weight savings that such composite
structures demonstrate in comparison with aluminum equivalents are relatively low (about 10%15%).
3.8 ANTISYMMETRIC LAMINATES
In antisymmetric laminates, symmetrically located layers have mutually reversed orientations. For
example, whereas laminates ½0 =90 =90 =0 and ½ 1 φ = 2 φ = 2 φ = 1 φ are symmetric, lami
nates ½0 =90 =0 =90 or ½0 =0 =90 =0 and ½ 1 φ= 2 φ= 1 φ= 2 φ are antisymmetric. In contrast
to symmetric laminates which have maximum bending and zero coupling stiffness coefficients,
antisymmetric laminates demonstrate pronounced coupling that can be important for some particular applications (robotic parts undergoing complicated deformation under simple loading, rotor
blades that twist under centrifugal forces, airplane wings twisting under bending, etc.).
The simplest antisymmetric laminate is a cross-ply layer consisting of two plies with angles
0 and 90 , and the same thickness h/2 (see Fig. 3.26). Taking e 5 h=2and using Eqs. (3.28) and
(3.41), we arrive at the following stiffness coefficients entering Eq. (3.44):
h
B11 5 B22 5 ðE1 1 E2 Þ;
2
h2
C11 5 2 C22 5 ðE2 2 E1 Þ;
8
h3
D11 5 D22 5 ðE1 1 E2 Þ;
24
B12 5 E1 ν 12 h;
B44 5 G12 h;
C12 5 0;
C44 5 0;
D12 5
3
h
E1 ν 12 ;
12
D44 5
h3
G12
12
Comparing these results with Eqs. (3.99) and (3.100), corresponding to a quasi-homogeneous
cross-ply laminate, we can see that the antisymmetric cross-ply laminate has the same membrane
and bending stiffnesses, but nonzero coupling coefficients C11 and C22 . This fact shows, in accordance with Eq. (3.44), that in-plane tension or compression of this laminate induces bending.
As another typical example of an antisymmetric laminate, consider an angle-ply structure consisting of two plies with the same thickness h/2 and orientation angles 1φ and 2φ, respectively
(see Fig. 3.27). The plies (or layers) are characterized by the following stiffness coefficients
230
CHAPTER 3 MECHANICS OF LAMINATES
z
h2
h2
x
y
FIGURE 3.26
An antisymmetric cross-ply laminate.
z
h 2
−φ
x
h 2
+φ
y
FIGURE 3.27
Unbonded view of an antisymmetric angle-ply laminate.
ð2Þ
Að1Þ
11 5 A11 5 A11 ;
ð2Þ
Að1Þ
12 5 A12 5 A12 ;
ð2Þ
Að1Þ
22 5 A22 5 A22 ;
ð2Þ
ð1Þ
ð2Þ
ð1Þ
ð2Þ
Að1Þ
14 5 2 A14 5 A14 ; A24 5 2 A24 5 A24 ; A44 5 A44 5 A44
where coefficients Amn are specified by Eq. (2.72). Taking again e 5 h=2, we arrive at constitutive
equations in Eq. (3.104) in which
Bmn 5 Amn h;
Cmn 5 2
h2
Amn ;
4
Dmn 5
h3
Amn
12
(3.119)
Comparing these coefficients with those in Eq. (3.104) and corresponding to a quasihomogeneous angle-ply laminate, we can conclude that the antisymmetric laminate has much larger
coupling coefficients C14 and C24 , and thus a much more pronounced extensiontwisting coupling
effect.
In composite technology, an antisymmetric 6 φ angle-ply laminate is usually fabricated by a
continuous filament winding process. A typical structure made by filament winding is shown in
Fig. 2.61 of Chapter 2, Mechanics of a Composite Layer. As can be seen in this figure, the angleply layer is composed from two plies with 1φ and 2φ orientation of the fibers and these plies
are interlaced in the process of filament winding. As a result, the structure of the layer is characterized by a distinctive regular mosaic pattern consisting of triangular-shaped, two-ply segments
(T-segments) repeating in chessboard fashion, with alternating 6 φ and 7 φ reinforcement. The
T-segments are arranged in a regular geometric pattern around the circumference and along the
axis forming the so-called cross-over circles (see Fig. 2.61). Depending on the parameters of
3.8 ANTISYMMETRIC LAMINATES
(A)
(B)
(C)
231
(D)
FIGURE 3.28
Filament-wound cylinders with various numbers nT of T-segments: nT 5 2 (A), 4 (B), 8 (C), and 16 (D).
the winding process, various numbers nT of T-segments located along the circumference can be
obtained. For a cylindrical shell, the structures corresponding to nT 5 2; 4; 8; and 16 are shown
in Fig. 3.28.
Each T-segment consists of two plies with either [1φ/2φ] or [2φ/1φ] structure, and the
plies are not interlaced within the T-segment area. If, for instance, a T-segment consists of the
top curvilinear-triangular-shaped ply, reinforced with fibers oriented at angle 1φ, and the bottom one reinforced with an angle 2φ, then the neighboring adjacent T-segments have an
inverse structure: their top plies are reinforced at angle 2φ, and the bottom ones are reinforced at 1φ.
The traditional approach used to analyze the laminates under consideration is based on
the model discussed in Section 2.5, according to which the laminate is treated as a homogeneous
orthotropic layer with stiffness coefficients specified by Eqs. (2.72) and (2.147). The constitutive
equations are taken in accordance with Eq. (3.44), i.e.,
Nx 5 B11 ε0x 1 B12 ε0y ;
Mx 5 D11 κx 1 D12 κy ;
Ny 5 B21 ε0x 1 B22 ε0y ;
Nxy 5 B44 γ 0xy ;
My 5 D21 κx 1 D22 κy ; Mxy 5 D44 κxy
where
Bmn 5 Amn h;
Dmn 5
h3
Amn
12
and Amn are specified by Eq. (2.72). The approach based on these constitutive equations corresponds to an infinite number of T-segments, i.e., to nT -N.
Considering a T-segment as an antisymmetric laminate, we must apply a more general version
of Eq. (3.44) including the coupling stiffness coefficients, i.e.,
232
CHAPTER 3 MECHANICS OF LAMINATES
Nx 5 B11 ε0x 1 B12 ε0y 1 C14 κxy ;
Ny 5 B21 ε0x 1 B22 ε0y 1 C24 κxy ;
Nxy 5 B44 γ 0xy 1 C41 κx 1 C42 κy ;
Mx 5 C14 γ 0xy 1 D11 κx 1 D12 κy ;
My 5 C24 γ 0xy 1 D21 κx 1 D22 κy ;
Mxy 5 C41 ε0x 1 C42 ε0y 1 D44 κxy ;
where the stiffness coefficients are specified by Eq. (3.119). It is important that, whereas for
the laminate with [1φ/2φ] structure shown in Fig. 3.27 the coupling stiffness coefficient is
negative, for the adjacent T-segment having [2φ/1φ], this coefficient is positive. This difference results in the specific behavior of the two different laminate structures of T-segments that
exhibit antisymmetric opposite anisotropic stretchingtwisting and bendingshear coupling
effects alternating along the circumference and axis of rotation of the shell. Due to the general
alternating pattern of the T-segments (chessboard structure) and their interactions within a layer
the anisotropic effects are balancing each other inducing, at the same time, additional stresses
in the plies.
To study the effect of the filament-wound mosaic pattern, the stress analysis of cylindrical shells
has been performed (Morozov, 2006). The shells under consideration consist of one filamentwound 6 φ angle-ply layer and are loaded with internal pressure. The solid modeling (Solid Edge)
and finite-element analysis (MSC NASTRAN) techniques have been employed to model the shells
with different mosaic pattern structures. Each shell is partitioned into triangular-shaped T-segments
according to the particular filament-wound pattern. Correspondingly, the finite elements are also
combined into the respective alternating groups. The material structure of the finite elements for
each of these groups is defined as either [1φ/2φ] or [2φ/1φ] laminate. The cylindrical shells
under consideration are reinforced with a winding angle φ 5 6 60 and loaded with internal pressure 1 MPa. The mechanical properties of the unidirectional glassepoxy composite ply correspond
to Table 1.5. The ends of the shells are clamped and the distance between the ends (length of the
cylinder) is fixed and equal to 140 mm. The diameter of the cylinder is 60 mm and total thickness
of the wall is h 5 1.4 mm (with the thickness of the unidirectional ply being 0:7 mm). The stress
analysis was performed for four types of shells.
The first cylinder is modeled with a homogeneous orthotropic angle-ply layer and analyzed
using finite-element models available within the MSC NASTRAN software. This model corresponds to nT -N. The other three cylinders have 2, 4, and 8 triangular-shaped segments around
the circumference (nT 5 2; 4; 8) and are analyzed using finite-element modeling of the shells with
allowance for their mosaic structure. The finite-element models and the deformed shapes for
nT -N, nT 5 2, and nT 5 4 are shown in Fig. 3.29.
As can be seen, the deformation of the shells distinctively reflects the corresponding filamentwound mosaic texture. The calculated maximum values of stresses along and across fibers, σ1 , σ2 ,
and shear stresses,τ 12 , acting in the plies are presented in Table 3.5. It can be noted from this
table that the maximum stresses strongly depend on the laminate structure. The traditional model
(nT -N) significantly underestimated the stresses. With an increase in the structural parameter nT ,
the stresses acting along the fibers reduce and approach the value obtained from the traditional
laminate model.
3.8 ANTISYMMETRIC LAMINATES
233
FIGURE 3.29
Finite-element models (A) and deformed shapes (B) of the cylinders with, nT -N, nT 5 2, and nT 5 4.
Table 3.5 Maximum Stresses in the Plies of the Shells With Various
Filament-Wound Structures
Structural Parameter, nT
σ1 (MPa)
σ2 (MPa)
τ 21 (MPa)
N
2
4
8
24.9
40.99
33.2
27.30
3.79
17.7
20.3
18.2
1.98
4.82
5.33
4.94
Thus, it can be expected that the higher the parameter nT , the higher the strength of 6 φ angleply filament-wound structures. This prediction is confirmed by the test results presented in
Fig. 3.30 (Vorobey, Morozov, & Tatarnikov, 1992). Carbon-phenolic cylindrical shells with the
geometrical parameters given above have been loaded with internal pressure up to the failure. As
234
CHAPTER 3 MECHANICS OF LAMINATES
p, MPa
ε x ⋅ 10 −5
–150
–100
p, MPa
1.5
1.5
1.0
1.0
0.5
0.5
–50
0
50
ε y ⋅ 10 −5
ε x ⋅ 10 −5 –200
(A)
–150
–100
–50
0
50
ε y ⋅ 10 −5
(B)
p, MPa
2.0
1.5
1.0
0.5
ε x ⋅ 10 −5 –200
–100
0
100
ε y ⋅ 10 −5
(D)
(C)
FIGURE 3.30
Dependencies of the axial (εx ) and the circumferential (εy ) strains on internal pressure (p) for cylindrical shells
with nT 5 2 (A), nT 5 4 (B), nT 5 16 (C), and the corresponding failure modes (D).
follows from Fig. 3.30, the increase of parameter nT from 2 (Fig. 3.30A) to 16 (Fig. 3.30C) results
in a significant increase in the burst pressure.
In conclusion, it should be noted that the effect under discussion shows itself mainly in 6 φ
angle-ply structures consisting of two symmetric plies. For laminated structures consisting of a
system of 6 φ angle-ply layers, the coupling stiffness coefficient which causes the specific behavior discussed above is given in notations to Eq. (3.104) and has the form
1
Cmn 5 2 Amn hδ;
2
(3.120)
in which h is the laminate thickness and δ is the thickness of the ply. Since δ is relatively small,
the coefficient Cmn in Eq. (3.120) is smaller than the corresponding coefficient in Eq. (3.119), and
the coupling effect caused by this coefficient is less pronounced.
3.9 SANDWICH STRUCTURES
235
FIGURE 3.31
Composite sandwich panel with honeycomb core.
FIGURE 3.32
Composite sandwich rings with foam core.
3.9 SANDWICH STRUCTURES
Sandwich structures are three-layered laminates consisting of thin facings and a light-weight honeycomb or foam core as in Figs. 3.31 and 3.32. Since the in-plane stiffnesses of the facings are much
higher than those of the core, whereas their transverse shear compliance is much lower than that of
the core, the stiffness coefficients of sandwich structures are usually calculated presuming that the
in-plane stiffnesses of the core are equal to zero. The transverse shear stiffnesses of the facings are
assumed to be infinitely high. For the laminate shown in Fig. 3.33 this means that
Að2Þ
mn 5 0;
ð1;3Þ
Amn
-N;
mn 5 11; 12; 14; 24; 44;
mn 5 55; 56; 66
As a result, the coefficients in Eq. (3.29) become
ðrÞ
Imn
5
1 ð1Þ r11
r11
2 t2r11 Þ ;
A t 1 Að3Þ
mn ðt3
r 1 1 mn 1
(3.121)
236
CHAPTER 3 MECHANICS OF LAMINATES
z
h3
h2
3
t3 = h
t2
x
e
h1
2
1
t1
y
FIGURE 3.33
Sandwich laminate with two laminated facings (1 and 3) and foam core (2).
where mn 5 11, 12, 22, 14, 24, 44. The membrane, bending, and coupling stiffness coefficients can
be determined using Eq. (3.28). Transverse shear stiffnesses, Eqs. (3.83) and (3.84), for an orthotropic core, can be presented in the form
Sx 5
h2
Gxz ;
h2
Sy 5
h2
Gyz
h2
where Gxz and Gyz are the shear moduli of the core.
3.10 COORDINATE OF THE REFERENCE PLANE
The stiffness coefficients specified by Eq. (3.28) include the coordinate of the reference plane e
(see Fig. 3.1) that, being appropriately preassigned, allows us to simplify the constitutive equations
for the laminate. As was shown in Sections 3.2 and 3.4, taking the middle plane as the reference
plane, i.e., putting e 5 h=2 for a homogeneous and symmetric laminate, we have Cmn 5 0, and the
constitutive equations take their simplest form with zero membranebending coupling terms.
Now, a natural question as to whether it is possible to reduce Eq. (3.5) to this form in the general case arises. Taking Cmn 5 0 in Eq. (3.28), we have
I ð1Þ
e 5 mn
ð0Þ
Imn
(3.122)
It is important that the reference plane should be one and the same for all mn 5 11, 12, 22, 14,
24, and 44, and these six equations should give the same value of e. In the general case, this is not
possible, so a universal reference plane providing Cmn 5 0 cannot exist.
However, there are some other particular laminates (in addition to homogeneous and symmetric
structures) for which this condition can be met. For example, consider a laminate composed of isotropic layers (see Sections 2.1 and 3.2). For such laminates,
3.10 COORDINATE OF THE REFERENCE PLANE
ðiÞ
AðiÞ
11 5 A22 5
Ei
;
1 2 ν 2i
ðiÞ
A12
5
Ei ν i
;
1 2 ν 2i
AðiÞ
44
237
Ei
2ð1 1 ν i Þ
and in accordance with Eq. (3.42)
ð0Þ
ð0Þ
I11
5 I22
5
ð1Þ
ð1Þ
5 I22
5
I11
ð1Þ
5
I44
k
X
Ei hi
1 2 ν 2i
i51
;
ð0Þ
I12
5
k
X
Ei ν i hi
1 2 ν 2i
i51
k
1X
Ei hi
ðti 1 ti21 Þ;
2 i51 1 2 ν 2i
;
ð1Þ
I12
5
ð0Þ
I44
5
k
X
Ei hi
;
2ð1
1 νiÞ
i51
k
1X
Ei ν i hi
ðti 1 ti21 Þ;
2 i51 1 2 ν 2i
k
1X
Ei hi
ðti 1 ti21 Þ
2 i51 2ð1 1 ν i Þ
As can be seen, these parameters, when substituted into Eq. (3.122), do not provide one and the
same value of e. However, if Poisson’s ratio is the same for all the layers, i.e., ν i 5 ν (i 5 1, 2,
3,. . ., k), we get
k
P
Ei hi ðti 1 ti21 Þ
e 5 i51
2
k
P
Ei hi
i51
For practical analysis, this result is often used even if the Poisson’s ratios of the layers are different. In these cases, it is assumed that all the layers can be approximately characterized with
some average value of Poisson’s ratio, i.e.,
ν5
k
1X
ν i hi
h i51
As another example, consider the sandwich structure described in Section 3.9 (see Fig. 3.33). In the
general case, we again fail to find the desired reference plane. However, if we assume that the facings are
made of one and the same material (only the thicknesses are different), Eqs. (3.121) and (3.122) yield
e5
h21 1 h3 ðh3 1 2h1 1 2h2 Þ
2ðh1 1 h3 Þ
Returning to the general case, we should emphasize that the reference plane providing Cmn 5 0
for all the mn values cannot be found in the general case only if the laminate structure is given. If
the stacking-sequence of the layers is not preassigned and there is a sufficient number of layers,
they can be arranged in such a way that Cmn 5 0. Indeed, consider a laminate in Fig. 3.34 and suppose that its structure is, in general, not symmetric, i.e., that z0i 6¼ zi and k0 6¼ k. Using plane z 5 0
as the reference plane, we can write the membranebending coupling coefficients as
0
k=2
k =2
1 X ðiÞ
1 X ði0 Þ 0 0
Cmn 5
Amn hi ðzi 1 zi21 Þ 2
A h ðz 1 z0i21 Þ
2 i51
2 i0 51 mn i i
where zi $ 0 and z0i $ 0. Introduce a new layer coordinate zi 5 ðzi 1 zi21 Þ=2, which is the distance
between the reference plane of the laminate and the middle plane of the ith layer. Then, the condition Cmn 5 0 yields
238
CHAPTER 3 MECHANICS OF LAMINATES
z
k
i
zi−1
zi′−1
zi
zi′
x
i′
e
k′
FIGURE 3.34
Layer coordinates with respect to the reference plane.
k=2
X
0
AðiÞ
mn hi zi 5
i51
k =2
X
i0 51
0
AðimnÞ h0i z0i
Now assume that we have a group of identical layers or plies with the same stiffness coeffi
cients Amn and thicknesses. For example, the laminate could include a 1.5 mm thick 0 unidirectional layer which consists of 10 plies (the thickness of an elementary ply is 0.15 mm). Arranging
these plies above ðzi Þ and below ðz0i Þ the reference plane in such a way that
10
X
ðzj 2 z0j Þ 5 0
(3.123)
j51
we have no coupling for this group of plies. Doing the same with the other layers, we arrive at a
laminate with no coupling. Naturally, some additional conditions following from the fact that the
laminate is a continuous structure should be satisfied. However even with these conditions,
Eq. (3.123) can be met with several systems of ply coordinates, and a symmetric arrangement of
the plies ðzj 5 z0j Þ is only one of these systems. The general analysis of the problem under discussion
has been presented by Verchery (1999).
Returning to laminates with preassigned stacking-sequences for the layers, it follows from
Eq. (3.122) that we can always make one of coupling stiffness coefficients equal to zero, e.g.,
taking e 5 est where
I ð1Þ
est 5 stð0Þ
Ist
(3.124)
we get Cst 5 0 (the rest of the coupling coefficients are not zero).
Another way to simplify the equations for stiffnesses is to take e 5 0, i.e., to take the surface of
the laminate as the reference plane. In this case, Eq. (3.28) take the form
ð0Þ
Bmn 5 Imn
;
ð1Þ
Cmn 5 Imn
;
ð2Þ
Dmn 5 Imn
In practical analysis, the constitutive equations for laminates with arbitrary structure are often
approximately simplified using the method of reduced or minimum bending stiffnesses described,
e.g., by Ashton (1969), Karmishin (1974), and Whitney (1987). To introduce this method, consider
the corresponding equation of Eq. (3.28) for bending stiffnesses, i.e.,
3.11 STRESSES IN LAMINATES
ð2Þ
ð1Þ
ð0Þ
Dmn 5 Imn
2 2eImn
1 e2 Imn
239
(3.125)
and find the coordinate e delivering the minimum value of Dmn . Using the minimum conditions
d
Dmn 5 0;
de
d2
Dmn . 0
de2
we have
I ð1Þ
e 5 emn 5 mn
ð0Þ
Imn
(3.126)
ð1Þ
ð0Þ
This result coincides with Eq. (3.124) and yields Cmn 5 0. Thus, calculating Imn
and Imn
, we use
for each mn 5 11; 12; 22; 14; 24; and 44 the corresponding value emn specified by Eq. (3.126).
Substitution yields
ð2Þ
Drmn 5 Imn
2
ð1Þ 2
ðImn
Þ
ð0Þ
Imn
;
r
Cmn
50
(3.127)
and the constitutive equations, Eq. (3.5), become uncoupled. Naturally, this approach is only
approximate because the reference plane coordinate should be the same for all stiffnesses, but it is
not in the method under discussion. It follows from the foregoing derivation that the coefficients
Drmn specified by Eq. (3.127) do not exceed the actual values of bending stiffnesses, i.e.,
Drmn # Dmn . So, the method of reduced bending stiffnesses leads to underestimation of the laminate
bending stiffness. In conclusion, it should be noted that this method is not formally grounded and
can yield both good and poor approximation of the laminate behavior, depending on the laminate
structure.
3.11 STRESSES IN LAMINATES
The constitutive equations derived in the previous sections of this chapter relate forces and
moments acting on the laminate to the corresponding generalized strains. For composite structures,
forces and moments should satisfy equilibrium equations, whereas strains are expressed in terms of
displacements. As a result, a complete set of equations is formed allowing us to find forces,
moments, strains, and displacements corresponding to a given system of loads acting on the structure. We assume that this problem has already been solved, i.e., we know either the generalized
strains ε, γ, and κ entering Eq. (3.5) or the forces and moments N and M. If the latter is the case,
we can use Eq. (3.5) to find ε, γ, and κ. Now, to complete the analysis, we need to determine the
stresses acting in each layer of the laminate.
To do this, we should first find strains in any ith layer using Eq. (3.3), which yield
0
εðiÞ
x 5 εx 1 zi κx ;
0
εðiÞ
y 5 εy 1 zi κy ;
0
γ ðiÞ
xy 5 γ xy 1 zi κxy
(3.128)
where zi is the layer normal coordinate changing over the thickness of the ith layer. If the ith layer
is orthotropic with principal material axes coinciding with axes x and y, (e.g., made of fabric),
Hooke’s law provides the stresses we need, i.e.,
240
CHAPTER 3 MECHANICS OF LAMINATES
ðiÞ
ðiÞ
ðiÞ ðiÞ
σðiÞ
x 5 Ex ðεx 1 ν xy εy Þ;
ðiÞ
ðiÞ
ðiÞ ðiÞ
σðiÞ
y 5 Ey ðεy 1 ν yx εx Þ;
ðiÞ ðiÞ
τ ðiÞ
xy 5 Gxy γ xy
(3.129)
ðiÞ
ðiÞ
ðiÞ
ðiÞ
ðiÞ
ðiÞ
ðiÞ
ðiÞ
where Ex;y 5 Ex;y
=ð1 2 ν ðiÞ
xy ν yx Þ and Ex , Ey , Gxy , ν xy , and ν yx are the elastic constants of the layer
referred to the principal material axes. For an isotropic layer (e.g., metal or polymeric), we should
ðiÞ
i
take in Eq. (3.129) ExðiÞ 5 EyðiÞ 5 Ei , ν ðiÞ
xy 5 ν yx 5 ν i , and Gxy 5 Gi 5 Ei =2ð1 1 ν i Þ.
If the layer is reinforced at some angle with respect to axes x and y; we need to use a different
approach. Consider a layer composed of unidirectional plies with orientation angle φi . Using
Eq. (2.69), we can express strains in the principal material coordinates as
2
ðiÞ
2
ðiÞ
ðiÞ
εðiÞ
1 5 εx cos φi 1 εy sin φi 1 γ xy sinφi cosφi ;
2
ðiÞ
ðiÞ
2
ðiÞ
εðiÞ
2 5 εx sin φi 1 εy cos φi 2 γ xy sinφi cosφi ;
(3.130)
ðiÞ
ðiÞ
ðiÞ
γ ðiÞ
12 5 2ðεy 2 εx Þsinφi cosφi 1 γ xy cos2φi ;
and find the corresponding stresses, i.e.,
ðiÞ
ðiÞ
ðiÞ
ðiÞ ðiÞ
ðiÞ
ðiÞ ðiÞ
σðiÞ
σðiÞ
1 5 E 1 ðε1 1 ν 12 ε2 Þ;
2 5 E2 ðε2 1 ν 21 ε1 Þ;
ðiÞ
ðiÞ ðiÞ
τ 12 5 G12 γ 12
(3.131)
ðiÞ
ðiÞ
ðiÞ
ðiÞ
ðiÞ
ðiÞ
ðiÞ
ðiÞ
where E1;2 5 E1;2
=ð1 2 ν ðiÞ
12 ν 21 Þ and E1 , E2 , G12 , ν 12 , ν 21 are the elastic constants of a unidirectional ply. Thus, Eq. (3.128)(3.131) allow us to find in-plane stresses acting in each layer or in an
elementary composite ply.
Compatible deformation of the layers is provided by interlaminar stresses τ xz , τ yz , and σz . To
find these stresses, we need to use the three-dimensional equilibrium equations
@σx
@τ yx
@τ zx
@σy
@τ xy
@τ yz
1
1
5 0;
1
1
5 0;
@x
@y
@z
@y
@x
@z
@σz
@τ xz
@τ yz
1
1
50
@z
@x
@y
(3.132)
The stresses σx , σy , and τ xy entering these equations are specified by Eq. (3.4). Integrating the first
two equations of Eq. (3.132) with respect to z from the laminate surface z 5 2 e (see Fig. 3.8) to
some current surface z 5 const (Fig. 3.35), we can calculate the transverse shear stresses τ xz , τ yz , as
z
σ zz
τ zyz
qy
σy
τ xy
y
FIGURE 3.35
Transverse shear (τ zzx , τ zzy ) and normal (σzz ) stresses.
x
τ
qz
σx
z
zx
qx
τ xy
REFERENCES
τ zzx 5 2
ðz @σx
@τ yx
1
dz 1 qx ;
@x
@y
2e
τ zzy 5 2
ðz @σy
@τ xy
1
dz 1 qy
@y
@x
241
(3.133)
2e
Here, q are the forces that can be applied at the surface z 5 2 e: Substitution of the shear
stresses in Eq. (3.133) into the last equation of Eq. (3.132) and integration yields the following
expression for the normal stress:
ðz
σzz 5
2e
ðz 2
@ σx
@2 σy
@2 τ xy
@qx
@qy
dz
1
1
2
dz
2
1
ðz 1 eÞ 1 qz
@x2
@y2
@x@y
@x
@y
2e
REFERENCES
Aleksandrov, A. Ya, Brukker, L. E., Kurshin, L. M., & Prusakov, A. P. (1960). Analysis of Sandwich Plates.
Moscow: Mashinostroenie. (in Russian).
Ashton, J. E. (1969). Approximate solutions for unsymmetrically laminated plates. Journal of Composite
Materials, 3, 189191.
Bucci, R.J., & Mueller, L.N. (1988). ARALLt laminates: properties and design update. In Proc. 33rd Int.
SAMPE Symp. March 7-10, 1988 (pp. 1237-1248). Anaheim, CA, USA.
Bucci, R.J., Mueller, L.N., Schulte, R.W., & Prohaska, J.L. (1987). ARALL laminates—results from a cooperative test program. In Proc. 32nd Int. SAMPE Symp. April 6 -9, 1987 (pp. 902-916). Anaheim, CA, USA.
Chen, H.-J., & Tsai, S. W. (1996). Three-dimensional effective moduli of symmetric laminates. Journal of
Composite Materials, 30(8), 906917.
Fridlyander, I. N., Senatorova, O. G., Anihovskaya, L. I., Shokin, G. I., Gorushkin, V. A., Dementyeva, L. A.,
. . . Sandler, V. S. (1997). Laminated structural Al/Glass Epoxy SIAL composites. Composite Materials,
Technologies and Automation of Products Manufacturing (pp. 7885). Moscow: SAMPE.
Karmishin, A. V. (1974). Equations for nonhomogeneous thin-walled elements based on minimum stiffnesses.
Applied Mechanics (Prikladnaya Mekhanika), 10(6), 3442. (in Russian).
Kolesnikov, B., Herbech, L., & Fink, A. (2008). GFRP/Titanium hybrid material for improving composite
bolted joints. Composite Structures, 83, 368380.
Morozov, E. V. (2006). The effect of filament-winding mosaic patterns on the strength of thin-walled composite shells. Composite Structures, 76, 123129.
Ohrloff, N., & Horst, P. (1992). Feasibility study of the application of GLARE materials in wide body aircraft
fuselage. In Proc. 13th Int. SAMPE European Conf., May 11-13, 1992 (pp. 131-142). Hamburg, Germany.
Sirotkin, O. S. (1973). Design of mechanical joints for composite parts. Design, Analysis and Testing of
Composite Structures Issue 1 (pp. 144151). Moscow: TSAGI. (in Russian).
Vasiliev, V. V. (1993). Mechanics of Composite Structures. Washington: Taylor & Francis.
Verchery, G. (1999). Designing with anisotropy. Part 1: Methods and general results for laminates. In Proc.
12th Int. Conf. on Composite Materials (ICCM-12), July 5-9, 1999 (11p). Paris, France(CD-ROM).
Vermeeren, C. A. J. R. (2003). An historic overview of the development of fibre metal laminates. Applied
Composite Materials, 10(4-5)), 189205.
Vlot, A., Bulters, P.B.M., & Fredell, R.S. (1992). Low and high velocity impact loading on fiber/metal laminates, carbon/PEEK and monolithic aluminum 2024-T3. In Proc. 13th Int. SAMPE European Conf., May
11-13, 1992 (pp. 347-360). Hamburg, Germany.
242
CHAPTER 3 MECHANICS OF LAMINATES
Vlot, A., & Gunnik, J. W. (2001). Fibre Metal Laminates: An Introduction. Dordrecht-Boston-London: Kluwer
Academic Publishers.
Vorobey, V. V., Morozov, E. V., & Tatarnikov, O. V. (1992). Analysis of Thermostressed Composite
Structures. Moscow: Mashinostroenie (in Russian).
Vorobey, V. V., & Sirotkin, O. S. (1985). Joining of Composite Structures. Moscow: Machinostroenie (in
Russian).
Whitney, J. M. (1987). Structural Analysis of Laminated Anisotropic Plates. Lancaster, PA, USA: Technomic
Publishing Co., Inc.
CHAPTER
FAILURE CRITERIA AND
STRENGTH OF LAMINATES
4
Consider a laminate consisting of orthotropic layers or plies whose principal material axes 1, 2, and
3, in general, do not coincide with the global coordinates of the laminate (x, y, z) and assume that
this layer or ply is in a state of plane stress as depicted in Fig. 4.1. It should be emphasized that, in
contrast to a laminate that can be anisotropic and demonstrate coupling effects, the layer under consideration is orthotropic and is referred to its principal material axes. Using the procedure that is
described in Section 3.11, we can find stresses σ1 , σ2 , and τ 12 corresponding to a given system of
loads acting on the laminate. The problem that we approach now is to evaluate the laminate’s
load-carrying capacity, that is, to calculate the loads that cause failure of the individual layers
and of the laminate as a whole. For a layer, this problem can be readily solved if we have a failure
or strength criterion
Fðσ1 ; σ2 ; τ 12 Þ 5 1
(4.1)
specifying the combination of stresses that causes layer fracture. In other words, the layer functions
while F , 1, fails if F 5 1, and does not exist as a load-carrying structural element if F . 1. In the
relevant stress space, that is, σ1 , σ2 , and τ 12 , Eq. (4.1) specifies the so-called failure surface (or
failure envelope) shown in Fig. 4.2. Each point in this space corresponds to a particular stress state,
and if a point is inside the surface, the layer withstands the corresponding combination of stresses
without failure.
Thus, the problem of strength analysis is reduced to the construction of a failure criterion in its
analytical, Eq. (4.1), or graphical (Fig. 4.2) form. Up to the present time, numerous variants of
these forms have been proposed for traditional and composite structural materials and discussed in
specialized monographs and reviews (Ashkenazi, 1966; Gol’denblat & Kopnov, 1968; Hashin,
1980; Rowlands, 1975; Tsai & Hahn, 1975; Vicario & Toland, 1975; Wu, 1974; and others). The
comprehensive analysis of the existing strength criteria is presented by Hinton, Kaddour, and
Soden (2004). Omitting the history and comparative analysis of particular criteria, we mainly discuss here the principle and the practical aspects of the problem.
In addition, in many applications, a progressive failure analysis of composite laminates is
required to predict their mechanical behavior under various loading conditions. We present a combined elastoplastic damage model and a strain-driven implicit integration procedure for fiber reinforced composite materials and structures that involve a consideration of their mechanical response
before the initiation of damage, prediction of damage initiation, and modeling of postfailure
behavior.
Advanced Mechanics of Composite Materials and Structures. DOI: https://doi.org/10.1016/B978-0-08-102209-2.00004-9
© 2018 Elsevier Ltd. All rights reserved.
243
244
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
z
3
τ12
σ1
2
σ2
y
x
φ
σ2
τ12
τ12
1
τ12
σ1
FIGURE 4.1
An orthotropic layer or ply in a plane stressed state.
τ12
τ12
C
D
σ1+
0
A
σ1
σ2+
B
σ2
FIGURE 4.2
Failure surface in the stress space.
4.1 FAILURE CRITERIA FOR AN ELEMENTARY COMPOSITE LAYER OR PLY
There exist, in general, two approaches to construct the failure surface, the first of which can be
referred to as the microphenomenological approach. The term “phenomenological” means that the
actual physical mechanisms of failure at the microscopic material level are not considered and that
4.1 FAILURE CRITERIA FOR AN ELEMENTARY COMPOSITE LAYER OR PLY
245
we deal with stresses and strains, that is, with conventional and not actually observed state variables. In the microapproach, we evaluate the layer strength using microstresses acting in the fibers
and in the matrix and failure criteria proposed for homogeneous materials. To a certain extent (see,
e.g., Skudra, Bulavs, Gurvich, & Kruklinsh, 1989), this approach requires the minimum number of
experimental material characteristics, that is, only those determining the strengths of fibers and
matrices. As a result, the coordinates of all the points of the failure surface shown in Fig. 4.2
including points A, B, and C corresponding to uniaxial and pure shear loading are found by calculation. To do this, we should simulate the layer or the ply with a suitable microstructural model
(see, e.g., Section 1.3), apply a preassigned system of average stresses σ1 , σ2 , and τ 12 (e.g., corresponding to vector 0D in Fig. 4.2), find the stresses acting in the material components, specify the
failure mode that can be associated with the fibers or with the matrix, and determine the ultimate
combination of average stresses corresponding, for example, to point D in Fig. 4.2. Thus, the whole
failure surface can be constructed. However, the uncertainty and approximate character of the existing micromechanical models discussed in Section 1.3 result in relatively poor accuracy of this
method which, being in principle rather promising, has not found wide practical application at the
present time.
The second basic approach, which can be referred to as the macrophenomenological one, deals
with the average stresses σ1 , σ2 , and τ 12 shown in Fig. 4.1 and ignores the ply’s microstructure.
For a plane stress state in an orthotropic ply, this approach requires at least five experimental
results specifying material strength under:
•
•
•
•
•
Longitudinal tension, σ1
1 (point A in Fig. 4.2)
Longitudinal compression, σ2
1
Transverse tension, σ1
2 (point B in Fig. 4.2)
Transverse compression, σ2
2
In-plane shear, τ 12 (point C in Fig. 4.2)
Obviously, these data are insufficient to construct the complete failure surface, and two possible
ways leading to two types of failure criteria can be used.
The first type, referred to as structural failure criteria, involves some assumptions concerning
the possible failure modes that can help us to specify the shape of the failure surface. According to
the second type, which provides a failure surface of an approximate form, experiments simulating a
set of complicated stress states (such that two or all three stresses σ1 , σ2 , and τ 12 are induced
simultaneously) are undertaken. As a result, a system of points, similar to point D in Fig. 4.2, is
determined and approximated with some suitable surface.
The experimental data that are necessary to construct the failure surface are usually obtained by
testing thin-walled tubular specimens such as those shown in Figs. 4.3 and 4.4. These specimens
are loaded with internal or external pressure p, tensile or compressive axial forces P, and end torques T, providing a prescribed combination of axial stress σx , circumferential stress σy , and shear
stress τ xy that can be calculated as
σx 5
P
;
2πRh
σy 5
pR
;
h
τ xy 5
T
2πR2 h
246
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
FIGURE 4.3
Glass-epoxy fabric test tubular specimens.
FIGURE 4.4
Carbon-epoxy test tubular specimens made by circumferential winding (the central cylinder failed under axial
compression and the right one under torsion).
4.1 FAILURE CRITERIA FOR AN ELEMENTARY COMPOSITE LAYER OR PLY
247
Here, R is the cylinder radius and h is its thickness. For the tubular specimens shown in
Fig. 4.4, which were made from unidirectional carbon-epoxy composite by circumferential winding,
σx 5 σ2 , σy 5 σ1 , and τ xy 5 τ 12 (see Fig. 4.1).
We shall now consider the classical structural and approximation strength criteria applied to
composite layers and plies.
4.1.1 MAXIMUM STRESS AND STRAIN CRITERIA
These criteria belong to a structural type and are based on the assumption that there can exist three
possible modes of failure caused by stresses σ1 , σ2 , and τ 12 or strains ε1 , ε2 , and γ 12 when they
reach the corresponding ultimate values.
The maximum stress criterion can be presented in the form of the following inequalities:
σ 1 # σ1
σ 2 # σ1
if σ1 . 0 σ2 . 0
1;
2
2
jσ 1 j # σ 1 ; jσ 2 j # σ 2
if σ1 , 0 σ2 , 0
2
jτ 12 j # τ 12
(4.2)
It should be noted here, and subsequently, that all the ultimate stresses σ and τ including compressive strength values are taken as positive quantities. The failure surface corresponding to the
criterion in Eqs. (4.2) is shown in Fig. 4.5. As can be seen, according to this criterion failure is
associated with independently acting stresses, and any possible stress interaction is ignored.
It can be expected that the maximum stress criterion adequately describes the behavior of those
materials in which stresses σ1 , σ2 , and τ 12 are taken by different structural elements. A typical
example of such a material is the fabric composite layer discussed in Section 2.7. Indeed, warp and
filling yarns (see Fig. 2.89) working independently provide material strength under tension and
compression in two orthogonal directions (1 and 2), whereas the polymeric matrix controls the
layer strength under in-plane shear. A typical failure envelope in the plane ðσ1 ; σ2 Þ for a glassepoxy fabric composite is shown in Fig. 4.6. The corresponding results in the plane ðσ1 ; τ 12 Þ, but
τ12
τ12
– σ2–
σ1+
–σ1–
σ2+
σ2
FIGURE 4.5
Failure surface corresponding to maximum stress criterion.
σ1
248
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
σ2, MPa
300
200
100
σ1, MPa
0
100
200
300
400
500
600
–100
–200
–300
FIGURE 4.6
Failure envelope for glass-epoxy fabric composite in plane (σ1 ; σ2 ). ( sss ) maximum stress criterion,
Eqs. (4.2); ( x ) experimental data.
for a different glass fabric experimentally studied by Annin and Baev (1979), are presented in
Fig. 4.7. It follows from Figs. 4.6 and 4.7 that the maximum stress criterion provides a satisfactory
prediction of the strength for fabric composites within the accuracy determined by the scatter of
experimental results. As has been already noted, this criterion ignores the interaction of stresses.
However, this interaction occurs in fabric composites that are loaded with compression in two
orthogonal directions because compression of the filling yarns increases the strength in the warp
direction and vice versa. The corresponding experimental results from Belyankin, Yatsenko, and
Margolin (1971) are shown in Fig. 4.8. As can be seen, there is a considerable discrepancy between
the experimental data and the maximum stress criterion shown with solid lines. However, even in
such cases, this criterion is sometimes used to design composite structures because it is simple and
conservative, that is, it underestimates material strength, thus increasing the safety factor for the
structure under design. There exist fabric composites for which the interaction of normal stresses is
exhibited in tension as well. An example of such a material is presented in Fig. 4.9 (experimental
data from Gol’denblat and Kopnov (1968)). Naturally, the maximum stress criterion (solid lines in
Fig. 4.9) should not be used in this case because it overestimates the material strength, and the
structure can fail under loads that are lower than those predicted by this criterion.
The foregoing discussion concerns fabric composites. Now consider a unidirectional ply and try
to apply the maximum stress criterion in this situation. First of all, since the longitudinal strength
of the ply is controlled by the fibers whose strength is much higher than that of the matrix, it is
4.1 FAILURE CRITERIA FOR AN ELEMENTARY COMPOSITE LAYER OR PLY
249
τ12, MPa
100
80
60
40
20
0
σ1, MPa
0
40
80
120
160
200
240
280
FIGURE 4.7
Failure envelope for glass-epoxy fabric composite in plane (σ1 ; τ 12 ). ( sss ) maximum stress criterion,
Eqs. (4.2); ( x ) experimental data.
σ2, MPa
–120
–100
–80
–60
–40
–20
0
0
–20
–40
–60
–80
–100
–120
σ1, MPa
FIGURE 4.8
Failure envelope for glass-phenolic fabric composite loaded with compression in plane (σ1 ; σ2 ). ( sss )
maximum stress criterion, Eqs. (4.2); ( s s s s ) polynomial criterion, Eqs. (4.16); ( x ) experimental data.
250
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
σ2, MPa
300
200
100
–300
–200
–100
100
200
300
500
σ1, MPa
–100
–300
FIGURE 4.9
Failure envelope for glass-epoxy fabric composite in plane (σ1 ; σ2 ). ( sss ) maximum stress criterion, Eqs. (4.2);
( s s s s ) approximation criterion, Eqs. (4.11) and (4.12); (………) approximation criterion, Eqs. (4.15); ( x )
experimental data.
reasonable to neglect the interaction of stress σ1 on one side and stresses σ2 and τ 12 on the other
side. In other words, we can apply the maximum stress criterion to predict material strength under
tension or compression in the fiber direction and, hence, use the first part of Eqs. (4.2), that is,
σ1 # σ1
if σ1 . 0
1
jσ 1 j # σ 2
if σ1 , 0
1
(4.3)
Actually, there exist unidirectional composites with a very brittle matrix (carbon or ceramic) for
which the other conditions in Eqs. (4.2) can be also applied. As an example, Fig. 4.10 displays the
failure envelope for a carboncarbon unidirectional material (experimental data from Vorobey,
Morozov, & Tatarnikov, 1992). However, for the majority of unidirectional composites, the interaction of transverse normal and shear stresses is essential and should be taken into account. This
means that we should apply Eq. (4.1) but can simplify it as follows
Fðσ2 ; τ 12 Þ 5 1
(4.4)
The simplest way to induce a combined stress state for a unidirectional ply is to use the off-axis
tension or compression test as discussed in Section 2.3.1. Applying stress σx as in Figs. 2.22 and
2.23, we have stresses σ1 , σ2 , and τ 12 specified by Eqs. (2.79). Then, Eqs. (4.2) yield the following
ultimate stresses:
For σx . 0,
σx 5
σ1
1
;
cos2 φ
σx 5
σ1
2
;
sin2 φ
σx 5
τ 12
sinφcosφ
(4.5)
4.1 FAILURE CRITERIA FOR AN ELEMENTARY COMPOSITE LAYER OR PLY
251
σ2, MPa
10
τ12, MPa
0
5
10
15
20
25
–10
–20
–30
–40
FIGURE 4.10
Failure envelope for carboncarbon unidirectional composite in plane (σ2 ; τ 12 ). ( sss ) maximum stress
criterion, Eqs. (4.2); ( x ) experimental data.
For σx , 0,
σx 5
σ2
1
;
cos2 φ
σx 5
σ2
2
;
sin2 φ
σx 5
τ 12
sinφcosφ
(4.6)
The actual ultimate stress is the minimum σx value of the three values provided by Eqs. (4.5)
for tension or Eqs. (4.6) for compression. The experimental data of S.W. Tsai (taken from Jones,
1999) and corresponding to a glass-epoxy unidirectional composite are presented in Fig. 4.11. As
can be seen, the maximum stress criterion (solid lines) demonstrates fair agreement with experimental results for angles close to 03 and 903 only. An important feature of this criterion belonging
to a structural type is its ability to predict the failure mode. Curves 1, 2, and 3 in Fig. 4.11 correspond to the first, the second, and the third equations of Eqs. (4.5) and (4.6). It follows from
Fig. 4.11A that fiber failure occurs only for φ 5 03 . For 03 , φ , 303 , material failure is associated
with in-plane shear, whereas for 303 , φ # 903 , it is caused by the transverse normal stress σ2 .
The maximum strain failure criterion is similar to the maximum stress criterion discussed earlier, but is formulated in terms of strains, that is,
ε # ε1
ε2 # ε1
if ε1 . 0 ε2 . 0
1;
2
2
jε1 j # ε1 ; jε2 j # ε2
if ε1 , 0 ε2 , 0
2
γ 12 # γ 12
(4.7)
where
ε1 5
σ1
σ2
2 ν 12 ;
E1
E2
ε2 5
σ2
σ1
2 ν 12 ;
E2
E1
γ 12 5
τ 12
G12
(4.8)
252
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
σx, MPa
σx, MPa
1000
1000
1
1
500
500
3
3
200
200
100
100
2
50
50
2
0
0
15
30
45
(A)
60
75
φ°
90
0
0
15
30
45
(B)
60
75
φ°
90
FIGURE 4.11
Dependence of the stress on the fiber orientation angle for off-axis tension (A) and compression (B) of glassepoxy unidirectional composite. ( ss s ) maximum stress criterion, Eqs. (4.2); (………) approximation
criterion, Eqs. (4.3) and (4.17); (s s s s) approximation criterion, Eqs. (4.3) and (4.18).
The maximum strain criterion ignores the strain interaction but allows for the stress interaction
caused by Poisson’s effect. This criterion provides results that are generally closely similar to those
following from the maximum stress criterion.
However, there exists a unique stress state which can only be studied using the maximum strain
criterion. This is longitudinal compression of a unidirectional ply as discussed in Section 1.4.4.
Under this type of loading, only longitudinal stress σ1 is induced, whereas σ2 5 0 and τ 12 5 0.
Nevertheless, fracture is accompanied with cracks parallel to the fibers (see Fig. 4.12, showing tests
performed by Katarzhnov (1982)). These cracks are caused by the transverse tensile strain ε2
induced by Poisson’s effect. The corresponding strength condition follows from Eqs. (4.7) and
(4.8) and can be written as
jσ1 j # ε1
2
E1
ν 21
It should be emphasized that the test shown in Fig. 4.12 can be misleading because transverse
deformation of the ply is not restricted in this test, whereas it is normally restricted in actual laminated composite structural elements. Indeed, a long cylinder with material structure ½0311 being
tested under compression yields a material strength σ2
1 5 300 MPa, whereas the same cylinder with
material structure ½0310 =903 gives σ2
1 5 505 MPa (Katarzhnov, 1982). Thus, if we change one
4.1 FAILURE CRITERIA FOR AN ELEMENTARY COMPOSITE LAYER OR PLY
253
FIGURE 4.12
Failure modes of a unidirectional glass-epoxy composite under longitudinal compression.
longitudinal ply for a circumferential ply which practically does not bear any of the load in compression along the cylinder axis, but restricts its circumferential deformation, we increase the
material strength in compression by 68.3%. Correspondingly, the failure mode is quite different in
this case (see Fig. 4.13).
4.1.2 APPROXIMATION STRENGTH CRITERIA
In contrast to structural strength criteria, approximation criteria do not indicate the mode of failure
and are constructed by approximation of available experimental results with some appropriate function depending on stresses σ1 ; σ2 , and τ 12 . The simplest and the most widely used criterion is a
second-order polynomial approximation, typical forms of which are presented in Fig. 4.14. In the
stress space shown in Fig. 4.2, the polynomial criterion corresponding to Fig. 4.14A can be written as
Fðσ1 ; σ2 ; τ 12 Þ 5 R11 σ21 1 R22 σ22 1 S12 τ 212 5 1
(4.9)
To determine the coefficients R and S, we need to perform three tests providing material
strength under uniaxial loading in 1 and 2 directions and in shear. Then, applying the following
conditions:
Fðσ1 5 σ1 ; σ2 5 0; τ 12 5 0Þ 5 1
Fðσ1 5 0; σ2 5 σ2 ; τ 12 5 0Þ 5 1
Fðσ1 5 0; σ2 5 0; τ 12 5 τ 12 Þ 5 1
(4.10)
254
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
FIGURE 4.13
Failure mode of a glass-epoxy tubular specimen with 10 longitudinal plies and 1 outside circumferential ply: (A)
inside view and (B) outside view.
y
b
(A)
x
a
a
2
2
+
y
=1
b
+
y
b
+
y
b
x
y
x
a
2
2
–
x
=1
c
+
xy
=1
d2
x
(B)
y
x
a
(C)
x
2
2
FIGURE 4.14
Typical shapes of the curves corresponding to the second-order polynomials specified by equations
(A), (B), and (C).
4.1 FAILURE CRITERIA FOR AN ELEMENTARY COMPOSITE LAYER OR PLY
255
we can find R and S and write Eq. (4.9) in its final form
2 2 2
σ1
σ2
τ 12
1
1
51
σ1
σ2
τ 12
(4.11)
It appears as though this criterion yields the same strength estimate in tension and compression.
However, it can be readily made specific for tension or compression. It is important to realize that
when evaluating a material’s strength we usually know the stresses acting in this material. Thus,
we can take in Eq. (4.10)
σ1 5 σ1
1
σ2 5 σ1
2
if
if
σ1 . 0 or
σ2 . 0 or
σ 1 5 σ2
1
σ 2 5 σ2
2
if
if
σ1 , 0
σ2 , 0
(4.12)
describing in this way the cases of tension and compression. The failure criterion given by
Eqs. (4.11) and (4.12) is demonstrated in Fig. 4.9 with application to a fabric composite loaded
with stresses σ1 and σ2 ðτ 12 5 0Þ. Naturally, this criterion requires different equations for different
quadrants in Fig. 4.9.
For some problems, for example, for the problem of design, for which we usually do not know
the signs of the stresses, we may need to use a universal form of the polynomial criterion valid
both for tension and compression. In this case, we should apply an approximation of the type
shown in Fig. 4.14B and generalize Eq. (4.9) as
Fðσ1 ; σ2 ; τ 12 Þ 5 R1 σ1 1 R2 σ2 1 R11 σ21 1 R22 σ22 1 S12 τ 212 5 1
(4.13)
Using criteria similar to Eqs. (4.10), that is,
Fðσ1 5 σ1
σ1 . 0
1 ; σ 2 5 0; τ 12 5 0Þ 5 1 if
Fðσ1 5 2 σ2
;
σ
5
0;
τ
5
0Þ
5
1
if
σ1 , 0
2
12
1
σ2 . 0
Fðσ1 5 0; σ2 5 σ1
2 ; τ 12 5 0Þ 5 1 if
Fðσ1 5 0; σ2 5 2 σ2
;
τ
5
0Þ
5
1
if
σ2 , 0
12
2
Fðσ1 5 0; σ2 5 0; τ 12 5 τ 12 Þ 5 1 if
(4.14)
we arrive at
σ1
2
1
1
1
1
σ21
σ22
τ 12
51
1 2 2 1 σ2
1 2 2 1 1 2 1 1 2 1
σ1
σ1
σ2
σ2
σ1 σ1
σ2 σ2
τ 12
(4.15)
Comparison of this criterion with the criteria discussed earlier and with experimental results is
presented in Fig. 4.9. As can be seen, the criteria specified by Eqs. (4.11), (4.12), and (4.15) provide results that are in fair agreement with the experimental data for all the stress states except,
possibly, biaxial compression for which there are practically no experimental results shown in
Fig. 4.9. Such results are presented in Fig. 4.8 and allow us to conclude that the failure envelope
can be approximated in this case by a polynomial of the type shown in Fig. 4.14C, that is,
Fðσ1 ; σ2; τ 12 Þ 5 R11 σ21 1 R12 σ1 σ2 1 R22 σ22 1 S12 τ 212 5 1
The coefficients R11 , R22 , and S12 can be found as earlier from Eqs. (4.10), and we need to use
an additional strength condition to determine the coupling coefficient, R12 . A reasonable form of
2
2
this condition is Fðσ1 5 2 σ2
1 ; σ 2 5 2 σ2 ; τ 12 5 0Þ 5 1. This means that whereas for jσ 1 j , σ1
256
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
and jσ2 j , σ2
2 the interaction of stresses increases material strength under compression, the combi2
nation of compressive failure stresses jσ1 j 5 σ2
1 and jσ2 j 5 σ2 results in material failure. Then,
2 2
σ1 2 σ1 σ2
σ2
τ 12
2 2 21 2 1
51
2
σ1
σ1 σ2
σ2
τ 12
(4.16)
Comparison of this criterion with experimental data is presented in Fig. 4.8. The term allowing
for the interaction of stresses σ1 and σ2 can be added to the strength criterion in Eq. (4.15), which
is generalized as
σ1
2
1
1
1
1
σ21
σ1 σ2
σ22
τ 12
p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
1
1
σ
1
2
2
2
1
51
2
2
1
2
1 2
1 2
1 2 1 2
σ1
σ
σ
σ
σ
σ
σ
σ
τ 12
σ
σ
σ
σ
1
1
2
2
1 1
2 2
1 1 2 2
(Tsai & Wu, 1971). Now consider unidirectional composites and return to Fig. 4.11. As can be
seen, the maximum stress criterion (solid lines), ignoring the interaction of stresses σ2 and τ 12 ,
demonstrates rather poor agreement with the experimental data. The simplest approximation
criterion, Eqs. (4.11) and (4.12), takes, for the case under study, the form
2 2
σ2
τ 12
Fðσ2 ; τ 12 Þ 5
1
51
σ2
τ 12
(4.17)
and the corresponding failure envelope is shown in Fig. 4.11 with dotted lines. Although providing
fair agreement with experimental results for tension (Fig. 4.11A), this criterion fails to predict material strength under compression (Fig. 4.11B). Moreover, for this case, the approximation criterion
yields worse results than those demonstrated by the maximum stress criterion. There are simple physical reasons for this discrepancy. In contrast to the maximum stress criterion, Eq. (4.17) allows for
stress interaction, but in such a way that the transverse stress σ2 reduces the material strength under
shear. However, this holds true only if the transverse stress is tensile. As can be seen in Fig. 4.15, in
τ12, MPa
80
40
20
–160
–120
–80
–40
0
40
σ2, MPa
FIGURE 4.15
Failure envelope for glass-epoxy unidirectional composite in plane (σ2 ; τ 12 ). ( sss ) approximation criterion,
Eqs. (4.12) and (4.17); ( s s s ) approximation criterion, Eqs. (4.18); ( x ) experimental data.
4.1 FAILURE CRITERIA FOR AN ELEMENTARY COMPOSITE LAYER OR PLY
257
which the experimental results taken from Barbero’s (1998) book are presented, a compressive
stress σ2 increases the ultimate value of shear stress τ 12 . As a result, the simplest polynomial criterion in Eq. (4.17), being, as it has been already noted, quite adequate for σ2 . 0, significantly
underestimates material strength for σ2 , 0 (solid line in Fig. 4.15). As also follows from
Fig. 4.15, a reasonable approximation to the experimental results can be achieved if we use a
curve of the type shown in Fig. 4.14B (but moved to the left with respect to the y-axis), that is, if
we apply for this case the criterion presented by Eq. (4.15) which can be written as
Fðσ2 ; τ 12 Þ 5 σ2
2
1
1
σ22
τ 12
2 2 1 1 21
51
σ1
σ
σ
σ
τ 12
2
2
2 2
(4.18)
The corresponding approximations are shown in Figs 4.11 and 4.15 by dashed lines.
In conclusion it should be noted that there exist more complicated polynomial strength criteria
than those considered above, for example, the fourth-order criterion of Ashkenazi (1966) and cubic
criterion proposed by Tennyson, Nanyaro, and Wharram (1980).
4.1.3 TENSOR STRENGTH CRITERIA
The polynomial approximation strength criteria discussed in Section 4.1.2 have been introduced
as some formal approximations of the experimental data in the principal material coordinates.
When written in some other coordinate frame, these criteria become much more complicated.
Consider, for example, an orthotropic material shown in Fig. 4.16 and referred to the principal
material axes 1 and 2, and to some axes 10 and 20 which make an angle φ 5 453 with the principal
axes. For the principal material axes 1 and 2, apply a generalized form of the criterion in
Eq. (4.13), that is,
Fðσ1 ; σ2 ; τ 12 Þ 5 R1 σ1 1 R2 σ2 1 R11 σ21 1 R12 σ1 σ2 1 R22 σ22 1 S12 τ 212 5 1
2
1′
2′
φ = 45°
1
FIGURE 4.16
An orthotropic material referred to coordinates (1, 2) and ð10 ; 20 Þ.
(4.19)
258
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
Using the strength conditions in Eqs. (4.14) to determine the coefficients R and S, we arrive at
Fðσ1 ; σ2 ; τ 12 Þ 5
1
1
1
1
σ21
σ
2
1
2
1
1
2
1
2 σ 2 1 1 2 1 R12 σ 1 σ 2
σ1
σ1
σ2
σ2
σ1 σ1
2
σ2
1 1 2 2 1 τ 12 5 1
τ 12
σ2 σ2
(4.20)
This criterion is similar to the criterion in Eq. (4.15); however, it includes the coefficient R12 ,
which cannot be found from simple tests using Eqs. (4.14). Treating Eq. (4.20) as the approximation strength criterion, we can undertake some additional testing or adopt additional assumptions
similar to those used for the derivation of Eq. (4.16) to determine the coefficient R12 . We can also
simplify the problem and take R12 5 0 arriving at Eq. (4.15), that is,
1
1
1
1
σ2
Fðσ1 ; σ2 ; τ 12 Þ 5 1 2 2 σ1 1 1 2 2 σ2 1 1 1 2
σ1
σ1
σ2
σ2
σ1 σ1
2
σ2
1 1 2 2 1 τ 12 5 1
τ 12
σ2 σ2
(4.21)
which is in good agreement with experimental results (see Fig. 4.9).
However, if we assume that the system of ultimate stresses form the so-called strength tensor,
we can determine the coefficient R12 in Eq. (4.20) analytically. The main property of the tensor
strength criterion is that its form is invariant with respect to coordinate transformation, which
means that for both coordinate frames 1; 2 and 10 ; 20 in Fig. 4.16 the form of the strength criterion
is the same. This property of the tensor criterion is an advantage because it has a form which is universal and suitable for any coordinate frame. However, this criterion has a serious shortcoming. If
the ultimate stresses compose a tensor, they must satisfy special invariant conditions according to
which some combinations of the ultimate stresses must be the same for all coordinate frames. Since
the ultimate stresses are found experimentally, it can happen that the invariant conditions are not
met and the tensor criterion cannot be applied to such materials.
Consider first the polynomial criterion in Eq. (4.21). To simplify the analysis, assume that
the material strength in tension and compression is the same for both principal directions 1 and 2,
that is,
2
1
2
σ1
1 5 σ1 5 σ2 5 σ2 5 σ0 ;
τ 12 5 τ 0
(4.22)
Then, Eq. (4.21) reduces to
Fðσ1 ; σ2 ; τ 12 Þ 5
2
ðσ21 1 σ22 Þ
τ 12
1
51
τ0
σ20
(4.23)
Now, let us write Eq. (4.23) in coordinates 10 and 20 (see Fig. 4.16). To transform the stresses σ1 , σ2 ,
45
45
0
0
and τ 12 to the stresses σ45
1 , σ2 , and τ 12 corresponding to coordinates 1 and 2 , we can use Eqs. (2.68).
3
45
45
Taking φ 5 45 , σx 5 σ1 , σy 5 σ2 , and τ xy 5 τ 12 , and σ1 5 σ1 , σ2 5 σ2 , and τ 12 5 τ 45
12 , we get
1
45
σ1 5 ðσ45
1 σ45
2 Þ 2 τ 12
2 1
1
45
σ2 5 ðσ45
1 σ45
2 Þ 1 τ 12
2 1
1
τ 12 5 ðσ45
2 σ45
2 Þ
2 1
(4.24)
4.1 FAILURE CRITERIA FOR AN ELEMENTARY COMPOSITE LAYER OR PLY
259
Substitution in Eq. (4.23) yields
45
45
Fðσ45
1 ; σ 2 ; τ 12 Þ 5
1 2
1 45 2
1 2
1
45 2
1
ðσ
σ45 σ45
1
Þ
1
ðσ
Þ
2
1
2
4 σ20
2 σ20
τ 20
τ 20 1 2
(4.25)
2
1 2 ðτ 45
Þ2 5 1
σ0 12
For tension in the directions of axes 10 and 20 in Fig. 4.16 and for shear in plane 10 20 , we can
write Eq. (4.25) in the following forms similar to Eqs. (4.10):
Fðσ45
σ45
τ 45
1 5 σ 45 ;
2 5 0;
12 5 0Þ 5 1
45
45
Fðσ1 5 0; σ2 5 σ45 ; τ 45
12 5 0Þ 5 1
45
45
Fðσ45
5
0;
σ
5
0;
τ
1
2
12 5 τ 45 Þ 5 1
(4.26)
Here, σ45 and τ 45 determine the material strength in coordinates 10 and 20 (see Fig. 4.16). Then,
Eq. (4.25) can be reduced to
45
45
Fðσ45
1 ; σ 2 ; τ 12 Þ 5
where σ45 and τ 45 are given by
45 1 45 2
1
2
τ 12
45 2
45 45
1
σ
51
ðσ
Þ
1
ðσ
Þ
2
σ
1
1
2
τ 45
σ245
τ 245
σ245 1 2
1
1 2
1
;
5
1
4 σ20
σ245
τ 20
(4.27)
1
τ 245 5 σ20
2
Comparing Eq. (4.27) with Eq. (4.23), we can see that Eq. (4.27), in contrast to Eq. (4.23),
45
includes a term with the product of stresses σ45
1 and σ 2 . So, the strength criterion under study
changes its form with a transformation of the coordinate frame (from 1 and 2 to 10 and 20 in
Fig. 4.16) which means that the approximation polynomial strength criterion in Eq. (4.23) and,
hence, the original criterion in Eq. (4.21), is not invariant with respect to rotation of the coordinate
frame and, correspondingly, is not a tensor criterion.
Consider the class of invariant strength criteria that are formulated in a tensor-polynomial form
as linear combinations of mixed invariants of the stress tensor σij and the strength tensors of different ranks Sij ; Sijkl ; etc., that is,
X
i;k
Sik σik 1
X
Sikmn σik σmn 1 ? 5 1
(4.28)
i;k;m;n
Using the standard transformation for tensor components, we can readily write this equation for
an arbitrary coordinate frame. However, as has been already mentioned, the fact that the strength
components form a tensor induces some invariant conditions that should be imposed on these components which do not necessarily correlate with experimental data.
To be specific, consider a second-order tensor criterion. Introducing contracted notations for
tensor components and restricting ourselves to the consideration of orthotropic materials referred to
the principal material coordinates 1 and 2 (see Fig. 4.16), we can present Eq. (4.22) as
Fðσ1 ; σ2 ; τ 12 Þ 5 R01 σ1 1 R02 σ2 1 R011 σ21 1 2R012 σ1 σ2 1 R022 σ22 1 4S012 τ 212 5 1
(4.29)
260
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
which corresponds to Eq. (4.28) if we put
σ11 5 σ1 ; σ12 5 τ 12 ; σ22 5 σ2 and S11 5 R1 ; S22 5 R2 ;
S1111 5 R011 ; S1122 5 S2211 5 R012 ; S2222 5 R022 ; S1212 5 S2121 5 S1221 5 S2112 5 S012
The subscript “0” indicates that the components of the strength tensors are referred to the principal material coordinates. Applying the strength conditions in Eqs. (4.14), we can reduce Eq. (4.29)
to the following form:
1
1
1
1
1
σ
2
2
2
σ1
σ2
σ1
σ2
1
1
2
2
Fðσ1 ; σ2 ; τ 12 Þ 5 σ1
(4.30)
2
σ2
σ2
1 1 1 2 1 2R012 σ1 σ2 1 1 2 2 1 τ 12 5 1
τ 12
σ1 σ1
σ2 σ2
This equation looks similar to Eq. (4.20); however, there is a difference in principle between
them. Whereas Eq. (4.20) is only an approximation to the experimental results, and we can take
any suitable value of coefficient R12 (in particular, we can put R12 5 0), the criterion in Eq. (4.30)
has an invariant tensor form, and the coefficient R012 should be determined using this characteristic
of the criterion.
Following Gol’denblat and Kopnov (1968), consider two cases of pure shear in coordinates
1
2
2
10 and 20 shown in Fig. 4.17 and assume that τ 1
45 5 τ 45 and τ 45 5 τ 45 , where the overbar denotes, as
2
earlier, the ultimate value of the corresponding stress. In the general case, τ 1
45 6¼ τ 45 . Indeed, for a
1
2
unidirectional composite, stress τ 45 induces tension in the fibers, whereas τ 45 causes compression
of the fibers, and the corresponding ultimate values can be different. Thus, we can conclude that
for the loading case shown in Fig. 4.17A, σ1 5 τ 1
σ2 5 2 τ 1
45 ;
45 , and τ 12 5 0, whereas for the case
2
2
in Fig. 4.17B, σ1 5 2 τ 45 ; σ2 5 τ 45 , and τ 12 5 0. Applying the strength criterion in Eq. (4.30) for
these loading cases, we arrive at
2
2
+
τ45
–
τ45
1′
2′
1′
2′
45°
45°
1
(A)
1
(B)
FIGURE 4.17
Two cases of pure shear in coordinates (10 ; 20 ) rotated by 453 with respect to the principal material coordinates (1,
2
2): (A) τ 1
45 and (B) τ 45 .
4.1 FAILURE CRITERIA FOR AN ELEMENTARY COMPOSITE LAYER OR PLY
261
Fðσ1 5 τ 1
σ2 5 2 τ 1
τ 12 5 0Þ
45 ;
45 ;
1
1
1
1
1
1
0
1 2
1
ðτ
2
2
1
Þ
1
2
2R
5 τ1
45
45
12 5 1
2
2
σ1
σ2
σ1
σ2
σ1
σ1
1
1
2
2
1 σ1
2 σ2
Fðσ1 5 2 τ 2
; σ2 5 τ 2
τ 12 5 0Þ
45 ;
45
1
1
1
1
1
1
0
2
2 2
5 τ 45 2 2 1 1 1 2 2 1 ðτ 45 Þ
2 1 1 2 2 2R12 5 1
σ1
σ1
σ2
σ2
σ1
σ2 σ2
1 σ1
In general, these two equations give different solutions for R012 . A unique solution exists if the
following compatibility condition is valid
1
1
1
1
1
1
2 22 11 25 1 2 2
σ1
σ
σ
σ
τ
τ
1
1
2
2
45
45
(4.31)
If the actual material strength characteristics do not satisfy this equation, the strength criteria in
Eq. (4.30) cannot be applied to this material. If they do, the coefficient R012 can be found as
R012 5
1
1
1
1
1
2
2
2
2
2 σ1
σ1
τ1
1 σ1
2 σ2
45 τ 45
(4.32)
For further analysis, consider for the sake of brevity a particular orthotropic material shown in
2
1
2
2
Fig. 4.16 for which, in accordance with Eqs. (4.22), σ1
τ1
1 5 σ1 5 σ2 5 σ2 5 σ0 ;
45 5 τ 45 5 τ 45 ,
and τ 12 5 τ 0 . As can be seen, Eq. (4.31) is satisfied in this case, and the strength criterion,
Eq. (4.30), referred to the principal material coordinates (1, 2) in Fig. 4.16 takes the form
2
1 2
τ 12
2
0
ðσ
1
σ
Þ
1
2R
σ
σ
1
51
1
2
12
1
2
τ0
σ20
(4.33)
where, in accordance with Eq. (4.32),
R012 5
1
1
2 2
2τ 45
σ20
(4.34)
Substituting Eq. (4.34) into Eq. (4.33), we arrive at the final form of the criterion under
consideration
Fðσ1 ; σ2 ; τ 12 Þ 5
2
ðσ21 1 σ22 Þ
2
1
τ 12
σ
1
2
σ
1
51
1
2
2
2
2
τ0
σ0
σ0
τ 45
(4.35)
Now, present Eq. (4.33) in the following matrix form:
fσgT R0 fσg 5 1
where
9
8
2 0
R11
>
=
< σ1 >
0
6
R 5 4 R012
fσg 5 σ2 ;
>
>
;
:
τ 12
0
1
1
1
0
0
R11 5 2 ;
R12 5 2 2 2 ;
σ0
σ0
2τ 45
(4.36)
R012
0
3
7
0 5
4S012
1
S012 5 2
4τ 0
R011
0
(4.37)
The superscript “T” means transposition, that is, converting the column vector fσg into the row
vector fσgT .
262
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
Let us transform stresses referred to axes (1, 2) into stresses corresponding to axes ð10 ; 20 Þ
shown in Fig. 4.16. Such a transformation can be performed with the aid of Eqs. (4.24). The matrix
form of this transformation is
fσg 5 ½T σ45
(4.38)
where
21
62
61
6
½T 5 6
62
41
2
1
2
1
2
2
21
1
2
3
7
7
7
1 7
7
5
0
Substitution of the stresses in Eq. (4.38) into Eq. (4.36) yields
σ45
T
½T T R0 ½T σ45 5 1
This equation, being rewritten as
σ45
T 45
R
σ45 5 1
(4.39)
0
0
specifies the strength criterion for the same material but referred to coordinates ð1 ; 2 Þ. The
strength matrix has the following form:
2
R45
11
45
0
6 45
T
R 5 ½T R ½T 5 4 R12
0
where
R45
12
R45
11
0
3
0
7
0 5
4S45
12
1
1 1
1
1
2 2
2
2
4 τ0
σ0
τ 45
1
1 1
1
R45
5
2
1
12
4 τ 20
σ20
τ 245
1
S45
12 5
4τ 245
(4.40)
45 2
1
1 1
1
2
ðσ1 Þ 1 ðσ45
1
2 2
2 Þ
2
2
4 τ0
σ0
τ 45
45 2
1
1 1
1
τ
45 45
σ1 σ2 1 12 5 1
1 2
12 2 2
τ 45
4 τ 20
σ0
τ 45
(4.41)
R45
11 5
The explicit form of Eq. (4.39) is
Now apply the strength conditions in Eqs. (4.26) which yield
1
1
1 1
1
5
1
2
4 τ 20
σ245
σ20
τ 245
(4.42)
4.1 FAILURE CRITERIA FOR AN ELEMENTARY COMPOSITE LAYER OR PLY
263
Then, the strength criterion in Eq. (4.41) can be presented as
45
45
Fðσ45
1 ; σ 2 ; τ 12 Þ 5
45 2
1 h 45 2 45 2 i
2
1
τ 12
45 45
1
σ
σ
1
σ
2
σ
1
51
1
2
1
2
τ 45
σ245
σ245
τ 20
(4.43)
Thus, we have two formulations of the strength criterion under consideration which are specified by Eq. (4.35) for coordinates (1, 2) and by Eq. (4.43) for coordinates (10 ; 20 ) (see Fig. 4.16).
As can be seen, Eqs. (4.35) and (4.43) have similar forms and follow from each other if we change
the stresses in accordance with the following rule:
σ1 2σ45
1 ;
σ2 2σ45
2 ;
τ 12 2τ 45
12 ;
σ0 2σ45 ;
τ 0 2τ 45
However, such correlation is possible under the condition imposed by Eq. (4.42) which can be
written in the form
Is 5
1
1
1
1
1 25 2 1 2
4τ 0
4τ 45
σ20
σ45
(4.44)
This result means that Is is the invariant of the strength tensor, that is, its value does not depend
on the coordinate frame for which the strength characteristics entering Eq. (4.44) have been found.
If the actual material characteristics do not satisfy Eq. (4.44), the tensor strength criterion cannot
be applied to this material. However, if this equation is consistent with the experimental data, the
tensor criterion offers considerable possibilities to study material strength. Indeed, restricting ourselves to two terms presented in Eq. (4.28), let us write this equation in coordinates ð10 ; 20 Þ shown
in Fig. 4.16 and assume that φ 6¼ 453 . Then
X φ φ
X φ
Sik σik 1
Sikmn σφik σφmn 5 1
i;k
(4.45)
i;k;m;n
Here, Sφik and Sφikmn are the components of the second and the fourth rank strength tensors which
are transformed in accordance with tensor calculus as
Sφik 5
X
lip lkq S0pq
p;q
Sφikmn 5
X
lip lkq lmr lns S0pqrs
(4.46)
p;q;r;s
Here, l are directional cosines of axes 10 and 20 on the plane referred to coordinates 1 and 2,
that is, l11 5 cosφ; l12 5 sinφ; l21 5 2 sinφ; l22 5 cosφ. Substitution of Eqs. (4.46) into
Eq. (4.45) yields the strength criterion in coordinates (10 , 20 ) but written in terms of strength components corresponding to coordinates (1, 2), that is,
XX
i;k
lip lkq S0pq σφik
p;q
1
X X
lip lkq lmr lns S0pqrs σφik σφmn 5 1
(4.47)
i;k;m;n p;q;r;s
Apply Eq. (4.47) to the special orthotropic material studied earlier (see Fig. 4.16) and for which,
in accordance with Eq. (4.22),
Spq 5 0;
S1111 5 S2222 5 R011 5 R022 5
S1122 5 S2211 5 R012 5
1
1
2 2
2τ 45
σ20
S1212 5 S2121 5 S1221 5 S2112 5 S012 5
1
σ20
(4.48)
1
4τ 20
264
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
Following Gol’denblat and Kopnov (1968), consider the material strength under tension in the
10 direction and in shear in plane ð10 ; 20 Þ. Taking first σφ11 5 σφ ; σφ22 5 0; τ φ12 5 0 and then
τ φ12 5 τ φ ; σφ11 5 0; σφ22 5 0, we get from Eq. (4.47)
σ2φ 5 X
1
l1p l1q l1r l1s S0pqrs
p;q;r;s
;
τ 2φ 5 X
1
l1p l2q l1r l2s S0pqrs
p;q;r;s
or in explicit form
σ2φ 5 ½R011 ðcos4 φ1sin4 φÞ12ðS012 12R012 Þsin2 φcos2 φ21
τ 2φ 5 4½2ðR011 2R012 Þsin2 φcos2 φ1S012 cos2 2φ21
(4.49)
These equations allow us to calculate the material strength in any coordinate frame whose axes
make angle φ with the corresponding principal material axes. Taking into account Eqs. (4.44) and
(4.48), we can derive the following relationship from Eqs. (4.49)
1
1
1
1
1 2 5 2 1 2 5 Is
4τ φ
4τ 0
σ2φ
σ0
(4.50)
So, Is is indeed the invariant of the strength tensor whose value for a given material does not
depend on φ.
Thus, tensor-polynomial strength criteria provide universal equations that can be readily written
in any coordinate frame, but on the other hand, the material mechanical characteristics, particularly
material strength in different directions, should follow the rules of tensor transformation, that is,
compose invariants (such as Is ) which must be the same for all coordinate frames.
4.1.4 INTERLAMINAR STRENGTH
The failure of composite laminates can also be associated with interlaminar fracture caused by
transverse normal and shear stresses σ3 and τ 13 ; τ 23 or σz and τ xz ; τ yz (see Fig. 2.18). Since
σ3 5 σz and shear stresses in coordinates (1, 2, 3) are linked with stresses in coordinates (x, y, z) by
simple relationships in Eqs. (2.67) and (2.68), the strength criterion is formulated here in terms of
stresses σz ; τ xz ; τ yz which can be found directly from Eqs. (3.132). Since the laminate strength in
tension and compression across the layers is different, we can use the polynomial criterion similar
to Eq. (4.15). For the stress state under study, we get
σz
where
τr 5
2
1
1
τr
2
51
2 1
σ1
σ
τi
3
3
(4.51)
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
τ 213 1 τ 223 5 τ 2xz 1 τ 2yz
is the resultant transverse shear stress and τ i determines the interlaminar shear strength of the material.
In thin-walled structures, the transverse normal stress is usually small and can be neglected in
comparison with the shear stress. Then, Eq. (4.51) can be simplified and written as
τr 5 τi
(4.52)
4.2 PRACTICAL RECOMMENDATIONS
265
P, kN
2
1.6
1.2
0.8
0.4
w/R
0
0
0.004
0.008
0.012
0.016
0.02
FIGURE 4.18
Experimental dependence of the normalized maximum deflection of a fiberglass-epoxy cylindrical shell on the
radial concentrated force.
As an example, Fig. 4.18 displays the dependence of the normalized maximum deflection w/R
on the force P for a fiberglass-epoxy cross-ply cylindrical shell of radius R loaded with a radial
concentrated force P (Vasiliev, 1970). Failure of the shell is caused by delamination. The shaded
interval shows the possible values of the ultimate force calculated with the aid of Eq. (4.52) (this
value is not unique because of the scatter in interlaminar shear strength).
4.2 PRACTICAL RECOMMENDATIONS
It follows from the foregoing analysis that for the practical strength evaluation of fabric composites,
we can use either the maximum stress criterion, Eqs. (4.2) or second-order polynomial criterion in
Eq. (4.15) in conjunction with Eq. (4.16) for the case of biaxial compression. For unidirectional
composites with polymeric matrices, we can apply Eqs. (4.3) and (4.4) in which the function F is
specified by Eq. (4.18). It should be emphasized that the experimental data usually have rather high
scatter, and the accuracy of more complicated and rigorous strength criteria can be more apparent
than real.
Comparing the tensor-polynomial and approximation strength criteria, we can conclude the following. The tensor criteria should be used if our purpose is to develop a theory of material strength
because a consistent physical theory must be covariant, that is, the constraints that are imposed on
material properties within the framework of this theory should not depend on a particular coordinate frame. For practical applications, the approximation criteria are more suitable, however, in the
266
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
forms, they are presented here they should be used only for orthotropic unidirectional plies or fabric
layers in coordinates whose axes coincide with the fibers’ directions.
To evaluate the laminate strength, we should first determine the stresses acting in the plies or
layers (see Section 3.11), identify the layer that is expected to fail first and apply one of the foregoing strength criteria. The fracture of the first ply or layer may not necessarily result in failure of the
whole laminate. Then, simulating the failed element with a suitable model (see, e.g., Section 2.4.2),
the strength analysis is repeated and continued up to failure of the last ply or layer (i.e., a progressive failure analysis is undertaken).
In principle, failure criteria can be constructed for the whole laminate as a quasi-homogeneous
material. This is not realistic for design problems, since it would be necessary to compare the solutions for numerous laminate structures which cannot practically be tested. However, this approach
can be used successfully for structures that are well developed and in mass production. For example, the segments of two structures of composite drive shafts, one made of fabric and the other of
unidirectional composite, are shown in Fig. 4.19. Testing these segments in tension, compression,
FIGURE 4.19
Segments of composite drive shafts with test fixtures.
4.3 EXAMPLES
267
and torsion, we can plot the strength envelope on the plane (M, T), where M is the bending moment
and T is the torque, and thus evaluate the shaft strength for different combinations of M and T with
high accuracy and reliability.
4.3 EXAMPLES
For the first example, consider a problem of torsion of a thin-walled cylindrical drive shaft (see
Fig. 4.20) made by winding a glass-epoxy fabric tape at angles 6 453 . The material properties are
2
E1 5 23:5 GPa, E2 5 18:5 GPa, G12 5 7:2 GPa, ν 12 5 0:16, ν 21 5 0:2, σ1
1 5 510 MPa, σ1 5 460 MPa,
1
2
σ2 5 280 MPa, σ2 5 260 MPa, and τ 12 5 85 MPa. The shear strain induced by torque T is
γ xy 5
T
2πR2 B44
(Vasiliev, 1993). Here, T is the torque, R 5 0.05 m is the shaft radius, and B44 is the shear stiffness
of the wall. According to Eqs. (3.38) and (3.39), B44 5 A44 h, where h 5 5 mm is the wall thickness,
and A44 is specified by Eqs. (2.72) and can be written as
A44 5
1
ðE1 1 E2 2 2E1 ν 12 Þ
4ð1 2 ν 12 ν 21 Þ
ðφ 5 453 Þ. Using Eqs. (3.130), we can determine the strains in the principal material coordinates
1 and 2 of 6 453 layers (see Fig. 4.20)
ε16 5 6
1
γ ;
2 xy
1
ε26 5 7 γ xy ;
2
6
γ 12
50
Applying Eqs. (3.131) and the foregoing results, we can express stresses in terms of T as
TE1 ð1 2 ν 12 Þ
πR2 hðE1 1 E2 2 2E1 ν 12 Þ
TE2 ð1 2 ν 21 Þ
σ26 5 7
πR2 hðE1 1 E2 2 2E1 ν 12 Þ
6
50
τ 12
σ16 5 6
The task is to determine the ultimate torque, T u .
x
2
y
h
1
+45°
R
T
–45°
T
2
FIGURE 4.20
Torsion of a drive shaft.
1
268
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
Use the maximum stress criterion, Eqs. (4.2), which gives the following four values of the ultimate torque corresponding to tensile or compressive failure of 6 453 layers
1
σ1
Tu 5 34 kNm
1 5 σ1 ;
2
2
σ1 5 σ1 ; Tu 5 30:7 kNm
1
σ1
Tu 5 25:5 kNm
2 5 σ2 ;
2
σ 2 5 σ2
Tu 5 23:7 kNm
2;
The actual ultimate torque is the lowest of these values, that is, T u 5 23:7 kNm.
As a second example, consider progressive failure analysis of a two-layered cylindrical section
of a composite pressure vessel (Fig. 4.21) which consists of a 6 363 angle-ply layer with total
thickness h1 5 0:62 mm and a 903 unidirectional layer with thickness h2 5 0:60 mm . A 200 mm
diameter vessel is made by filament winding from glass-epoxy composite with the following
mechanical properties: E1 5 44 GPa , E2 5 9:4 GPa , G12 5 4 GPa , and ν 21 5 0:26, and loaded
with internal pressure p. Axial Nx and circumferential Ny stress resultants can be found as
Nx 5
1
pR;
2
Ny 5 pR
where R 5 100 mm is the shell radius. Applying the constitutive equations, Eqs. (3.44), and neglecting the change in the cylinder curvature ðκx 5 κy 5 0Þ, we arrive at the following equations
for strains
1
B11 ε0x 1 B12 ε0y 5 pR ;
2
FIGURE 4.21
Composite pressure vessel.
B12 ε0x 1 B22 ε0y 5 pR
(4.53)
4.3 EXAMPLES
269
Using Eqs. (3.130) and (3.131) to determine strains and stresses in the principal material coordinates of the plies, we have
σðiÞ
1 5
pR E1 ðB22 2 2B12 Þðcos2 φi 1 ν 12 sin2 φi Þ 1 ð2B11 2 B12 Þðsin2 φi 1 ν 12 cos2 φi Þ
2B
σðiÞ
2 5
pR E2 ðB22 2 2B12 Þðsin2 φi 1 ν 21 cos2 φi Þ 1 ð2B11 2 B12 Þðcos2 φi 1 ν 21 sin2 φi Þ
2B
τ ðiÞ
12 5
pR
G12 ð2B11 1 B12 2 B22 Þsin2φi
2B
(4.54)
Here, B 5 B11 B22 2 B212 , and the membrane stiffnesses Bmn for the shell under study are:
ð0Þ
ð0Þ
ð0Þ
B11 5 I11
5 21:2 GPa mm, B12 5 I12
5 7:7 GPa mm, and B22 5 I22
5 35:6 GPa mm. Subscript “i” in
Eqs. (4.54) denotes the helical plies for which i 5 1; φ1 5 φ 5 363 and circumferential plies for
which i 5 2 and φ2 5 903 .
The task that we consider is to find the ultimate pressure pu . For this purpose, we use the
strength criteria in Eqs. (4.3), (4.4), and (4.17), and the following material properties:
1
σ1
1 5 1300 MPa, σ2 5 27 MPa, and τ 12 5 45 MPa.
Calculation with the aid of Eqs. (4.54) yields
σð1Þ
1 5 83:9p;
σð1Þ
2 5 24:2p;
ð1Þ
τ 12
5 1:9p
σð2Þ
1 5 112p;
σð2Þ
2 5 19:5p;
τ ð2Þ
12 5 0
Applying Eqs. (4.3) to evaluate the plies’ strength along the fibers, we get
1
σð1Þ
1 5 σ1 ;
pu 5 14:9 MPa
1
σð2Þ
1 5 σ1 ;
pu 5 11:2 MPa
The failure of the matrix can be identified using Eq. (4.17), that is,
ð1Þ 2 ð1Þ 2
σ2
τ
1 12 5 1;
τ 12
σ1
2
ð2Þ 2 ð2Þ 2
σ2
τ
1 12 5 1;
τ 12
σ1
2
pu 5 1:1 MPa
pu 5 1:4 MPa
Thus, we can conclude that failure occurs initially in the matrix of helical plies and takes place
at the pressure pð1Þ
u 5 1:1 MPa. This pressure fractures only the matrix of the helical plies, whereas
the fibers are not damaged and continue to work. According to the model of a unidirectional layer
with failed matrix discussed in Section 2.4.2, we should take E2 5 0; G12 5 0; and ν 12 5 0 in the
helical layer. Then, the stiffness coefficients, Eqs. (2.72) for this layer, become
4
Að1Þ
11 5 E1 cos φ ;
2
2
Að1Þ
12 5 E1 sin φcos φ ;
4
Að1Þ
22 5 E1 sin φ
(4.55)
Calculating again the membrane stiffnesses Bmn (see Eq. (3.28)) and using Eqs. (4.53), we get
for p $ pð1Þ
u
σð1Þ
1 5 92:1p;
σð2Þ
1 5 134:6p;
ð1Þ
σð1Þ
2 5 24:2pu 5 26:6 MPa;
σ2ð2Þ 5 22:6p;
τ ð2Þ
12 5 0
ð1Þ
τ ð1Þ
12 5 1:9pu 5 2:1 MPa
270
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
For a pressure p $ pð1Þ
u , three modes of failure are possible. The pressure causing failure of the
helical plies under longitudinal stress σð1Þ
1 can be calculated from the following equation:
ð1Þ
ð1Þ
1
σð1Þ
1 5 83:9pu 1 92:1ðpu 2 pu Þ 5 σ 1
which yields pu 5 14:2 MPa. The analogous value for the circumferential ply is determined by the
following condition
ð1Þ
ð1Þ
1
σð2Þ
1 5 112pu 1 134:6ðpu 2 pu Þ 5 σ 1
which gives pu 5 9:84 MPa. Finally, the matrix of the circumferential layer can fail under tension
across the fibers. Since τ ð2Þ
12 5 0, we put
ð1Þ
ð1Þ
1
σð2Þ
2 5 19:5pu 1 22:6ðpu 2 pu Þ 5 σ 2
and find pu 5 1:4 MPa.
Thus, the second failure stage takes place at pð2Þ
u 5 1:4 MPa and is associated with cracks in the
matrix of the circumferential layer (see Fig. 2.36).
For p $ pð2Þ
u , we should put E2 5 0; G12 5 0; and ν 12 5 0 in the circumferential layer whose
stiffness coefficients become
Að2Þ
11 5 0 ;
Að2Þ
12 5 0 ;
Að2Þ
22 5 E1
(4.56)
The membrane stiffnesses of the structure now correspond to the monotropic model of a composite unidirectional ply (see Section 1.3) and can be calculated as
ð2Þ
Bmn 5 Að1Þ
mn h1 1 Amn h2
where Amn are specified by Eqs. (4.55) and (4.56), and h1 5 0:62 mm and h2 5 0:6 mm are the
thicknesses of the helical and the circumferential layers. Again using Eqs. (4.54), we have for
p $ pð2Þ
u
σð1Þ
1 5 137:7p ;
σð2Þ
1 5 122:7p
The vessel’s failure can now be caused by fracture of either helical fibers or circumferential
fibers. The corresponding values of the ultimate pressure can be found from the following
equations:
ð1Þ
ð2Þ
ð1Þ
ð2Þ
1
σð1Þ
1 5 83:9pu 1 92:1ðpu 2 pu Þ 1 137:7ðpu 2 pu Þ 5 σ1
ð1Þ
ð2Þ
ð1Þ
ð2Þ
1
σð2Þ
1 5 112pu 1 134:6ðpu 2 pu Þ 1 122:7ðpu 2 pu Þ 5 σ 1
ð2Þ
in which pð1Þ
u 5 1:1 MPa and pu 5 1:4 MPa. The first of these equations yields pu 5 10 MPa,
whereas the second gives pu 5 10:7 MPa.
Thus, failure of the structure under study occurs at pu 5 10 MPa as a result of fiber fracture in
the helical layer.
The dependencies of the strains, which can be calculated using Eqs. (4.53) and the appropriate
values of Bmn , are shown in Fig. 4.22 (solid lines). As can be seen, the theoretical prediction is in
fair agreement with the experimental results. The same conclusion can be drawn for the burst pressure which is listed in Table 4.1 for two types of filament-wound fiberglass pressure vessels. A typical example of the failure mode for the vessels presented in Table 4.1 is shown in Fig. 4.23.
4.3 EXAMPLES
271
P, MPa
10
8
6
4
2
0
0
0.4
0.8
1.2
1.6
2
2.4
εx,%
2.8
1.6
2
2.4
2.8
(A)
P, MPa
10
8
6
4
2
0
εy,%
0
0.4
0.8
1.2
(B)
FIGURE 4.22
Dependence of the axial (A) and the circumferential (B) strains on internal pressure. ( sss ) analysis and ( x )
experimental data.
Table 4.1 Burst Pressure for the Filament-Wound Fiberglass Pressure Vessels
Layer
Thickness (mm)
Diameter of
the Vessel
(mm)
h1
200
200
0.62
0.92
Experimental Burst Pressure
h2
Calculated
Burst Pressure
(MPa)
Number
of Tested
Vessels
Mean Value
(MPa)
Variation
Coefficient (%)
0.60
0.93
10
15
5
5
9.9
13.9
6.8
3.3
272
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
FIGURE 4.23
The failure mode of a composite pressure vessel.
4.4 ALLOWABLE STRESSES FOR LAMINATES CONSISTING OF
UNIDIRECTIONAL PLIES
It follows from Section 4.3 (see also Section 2.4.2) that a unidirectional fibrous composite ply can
experience two modes of failure associated with
•
•
fiber failure, and
cracks in the matrix.
The first mode can be identified using the strength criterion in Eqs. (4.3), that is,
ðiÞ
ðiÞ
1
σ1 # σ1 if σ1 . 0
ðiÞ if σðiÞ
σ 1 # σ 2
1 ,0
1
(4.57)
2
in which σ1
1 and σ1 are the ultimate stresses of the ply under tension and compression along the
fibers, and i is the ply number. For the second failure mode, we have the strength criterion in
Eq. (4.18), that is,
σðiÞ
2
!2
2
1
1
ðσðiÞ
τ ðiÞ
2 Þ
12
2 2 1 1 21
51
σ1
σ2
σ2 σ2
τ 12
2
(4.58)
2
in which σ1
2 and σ2 are the ultimate stresses of the ply under tension and compression across the
fibers, and τ 12 is the ultimate in-plane shear stress.
Consider a laminate loaded with normal, σx and σy , and shear, τ xy , stresses as in Fig. 4.24.
Assume that the stresses are increased in proportion to some loading parameter p. Applying the
strength criteria in Eqs. (4.57) and (4.58), we can find two values of this parameter, that is, p 5 pf
which corresponds to fiber failure in at least one of the plies and p 5 pm for which the loading
causes a matrix failure in one or more plies. Since the parameters pf and pm usually do not coincide
with each other for modern composites, a question concerning the allowable level of stresses acting
in the laminate naturally arises. Obviously, the stresses under which the fibers fail must not be
4.4 ALLOWABLE STRESSES FOR LAMINATES CONSISTING
273
σy
τxy
σx
σx
τxy
σy
FIGURE 4.24
A laminate loaded with normal and shear stresses.
treated as allowable stresses. Moreover, the allowable value pa of the loading parameter must be
less than pf by a specified safety factor sf , that is,
pfa 5 pf =sf
(4.59)
However, for the matrix failure, the answer is not evident, and at least two different situations
may take place.
First, failure of the matrix can result in failure of the laminate. As an example, we can take a
6 φ angle-ply layer discussed in Section 2.5 whose moduli in the x and y directions are specified
by Eqs. (2.147), that is,
Ex 5 A11 2
A212
;
A22
Ey 5 A22 2
A212
A11
Ignoring the load-carrying capacity of the failed matrix, that is, taking E2 5 0, G12 5 0 and
ν 12 5 ν 21 5 0 in Eqs. (2.72) to get
A11 5 E1 cos4 φ;
A12 5 E1 sin2 φcos2 φ;
A22 5 E1 cos4 φ
we arrive at Ex 5 0 and Ey 5 0, which means that the layer under consideration cannot work without the matrix. For such a mode of failure, we should take the allowable loading parameter as
pm
a 5 pm =sm
(4.60)
where sm is the corresponding safety factor.
Secondly, matrix fracture does not result in laminate failure. As an example of such a structure,
we can take the pressure vessel considered in Section 4.3 (see Figs. 4.22 and 4.23). Now we have
another question as to whether the cracks in the matrix are allowed even if they do not affect the
structure’s strength. The answer to this question depends on the operational requirements for the
274
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
structure. For example, suppose that the pressure vessel in Fig. 4.21 is a model of a filamentwound solid propellant rocket motor case which works only once and for a short period of time.
Then, it is appropriate to ignore the cracks appearing in the matrix and take the allowable stresses
in accordance with Eq. (4.59). We can also suppose that the vessel may be a model of a pressurized
passenger cabin in a commercial airplane for which no cracks in the material are allowed in flight.
Then, in principle, we must take the allowable stresses in accordance with Eq. (4.60). However, it
follows from the examples considered in Sections 2.4.2 and 4.3 that for modern composites the
loading parameter pm is reached at the initial stage of the loading process. As a result, the allowable
loading parameter, pm
a in Eq. (4.60), is so small that modern composite materials cannot demonstrate their high strength governed by the fibers and cannot compete with metal alloys. A more
realistic approach allows the cracks in the matrix to occur but only if pm is higher than the operational loading parameter p0 . Using Eq. (4.60), we can assume that
p0 5 pm
a 5 pm =sm
The ultimate loading parameter p is associated with fiber failure, so that p 5 pf . Thus, the actual
safety factor for the structure becomes
s5
p
pf
5
sm
pm
p0
(4.61)
and depends on the ratio pf =pm .
To be specific, consider a four-layered ½03 =453 = 2 453 =903 quasi-isotropic carbon-epoxy laminate shown in Fig. 4.24 which is widely used in aircraft composite structures. The mechanical properties of quasi-isotropic laminates are discussed in Section 3.7. The constitutive equations for such
laminates have the form typical for isotropic materials, that is,
εx 5
1
ðσx 2 νσy Þ;
E
εy 5
1
ðσy 2 νσx Þ;
E
γ xy 5
τ xy
G
(4.62)
where E, ν, and G are given by Eqs. (3.110). For a typical carbon-epoxy fibrous composite,
Table 3.4 yields E 5 54:8 GPa, ν 5 0:31, and G 5 20:9 GPa. The strains in the plies’ principal coordinates (see Fig. 2.18) can be found using Eqs. (2.69) from which we have for the 03 layer,
ε1 5 εx ;
ε2 5 εy ;
γ 12 5 γ xy
for the 6 453 layer,
1
ε1 5 ðεx 1 εy 6 γ xy Þ;
2
1
ε2 5 ðεx 1 εy 7 γ xy Þ;
2
γ 12 5 6 ðεy 2 εx Þ
for the 903 layer,
ε1 5 εy ;
ε2 5 εx ;
γ 12 5 γ xy
The stresses in the plies are
σ1 5 E1 ðε1 1 ν 12 ε2 Þ;
σ2 5 E2 ðε2 1 ν 21 ε1 Þ;
τ 12 5 G12 γ 12
(4.63)
where E1;2 5 E1;2 =ð1 2 ν 12 ν 21 Þ, ν 12 E1 5 ν 21 E2 , and the elastic constants E1 , E2 , ν 21 , and G12 are
given in Table 1.5. For given combinations of the acting stresses σx , σy , and τ xy (see Fig. 4.24), the
strains εx , εy , and γ xy found with the aid of Eqs. (4.62) are transformed to the ply strains ε1 , ε2 , and
γ 12 , and then substituted into Eqs. (4.63) for the stresses σ1 , σ2 , and τ 12 . These stresses are
4.4 ALLOWABLE STRESSES FOR LAMINATES CONSISTING
275
substituted into the strength criteria in Eqs. (4.57) and (4.58), the first of which gives the combination of stresses σx , σy , and τ xy causing failure of the fibers, whereas the second one enables us to
determine the stresses inducing matrix failure.
Consider biaxial loading with stresses σx and σy as shown in Fig. 4.24. The corresponding failure envelopes are presented in Fig. 4.25. The solid lines determine the domain within which the
fibers do not fail, whereas within the area bound by the dashed lines no cracks in the matrix appear.
Consider, for example, the cylindrical pressure vessel discussed in Section 4.3 for which
σx 5
pR
;
2h
σy 5
pR
h
(4.64)
are the axial and circumferential stresses expressed in terms of the vessel radius and thickness, R
and h, and the applied internal pressure p. Let us take R=h 5 100. Then, σx 5 50p and σy 5 100p,
whereby σy =σx 5 2. This combination of stresses is shown with the line 0BA in Fig. 4.25. Point A
corresponds to failure of the fibers in the circumferential (903 ) layer and gives the ultimate loading
parameter (which is the pressure in this case) p 5 pf 5 8 MPa. Point B corresponds to matrix failure
in the axial (03 ) layer and to the loading parameter pm 5 2:67 MPa. Taking sm 5 1:5, which is a typical value for the safety factor preventing material damage under the operational pressure, we get
s 5 4:5 from Eq. (4.61). If such a high safety factor is not appropriate, then we should either allow
the cracks in the matrix to appear under the operational pressure, or to change the carbon-epoxy
composite to some other material. It should be noted that the significant difference between the
loading parameters pf and pm is typical mainly for tension. For compressive loads, the fibers usually fail before the matrix (see point C in Fig. 4.25).
σy, MPa
1200
A
800
σx = –200
σy = 600
400
–800
0
–400
–400
σx = –600
σy = –600
σx = 1000
σy = 1000
B
σy = 267
σx = 267
0
400
800
σx, MPa
1200
σx = 600
σy = –200
C
–800
FIGURE 4.25
Failure envelops for biaxial loading corresponding to the failure criteria in Eqs. (4.57) ( sss ) and
Eq. (4.58) ( ------- ).
276
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
τxy, MPa
1000
σx = –400
τxy = 900
σx = 670
τxy = 900
τxy = 900
800
600
σx = 670
400
σx = 0
τxy = 268
σx = –400
200
0
0
200
400
600
800
σx, MPa
σx = 268
FIGURE 4.26
Failure envelops for uniaxial tension and compression combined with shear corresponding to the failure criteria in
Eqs. (4.57) ( sss ) and Eq. (4.58) ( ------- ).
Similar results for uniaxial tension with stresses σx combined with shear stresses τ xy (see
Fig. 4.24) are presented in Fig. 4.26. As can be seen, shear can induce the same effect as tension.
The problem of matrix failure discussed earlier significantly reduces the application of modern
fibrous composites to structures subjected to long-term cyclic loading. It should be noted that
Figs. 4.25 and 4.26 correspond to static loading at room temperature. Temperature, moisture, and
fatigue can considerably reduce the areas bounded by the dashed lines in Figs. 4.25 and 4.26
(see Sections 5.1.2 and 5.3.3). Some methods developed to solve the problem of matrix failure are
discussed in Sections 2.4.3 and 2.4.4.
4.5 PROGRESSIVE FAILURE: GENERAL APPROACH TO MODELING
AND ANALYSIS
The development of a representative constitutive model for fiber reinforced composite materials
and structures should involve the consideration of their mechanical responses before the initiation
of damage, the prediction of damage initiation, and the modeling of postfailure behavior.
4.5 PROGRESSIVE FAILURE: GENERAL APPROACH TO MODELING
277
Continuum damage mechanics (CDM) provides a tractable framework for modeling damage initiation and development, as well as stiffness degradation. It is applied at the ply level and reflects on
the corresponding failure modes. A number of material models employing CDM have been reported
in the literature (see, e.g., Ladeveze & Le Dantec, 1992; Lapczyk & Hurtago, 2007; Maimi,
Camanho, Mayugo, & Dávila, 2007a, 2007b; Matzenmiller, Lubliner, & Taylor, 1995; Van Der
Meer & Sluys, 2009). Most of the CDM-based material approaches employ elastic-damage models
which are suitable for modeling the mechanical behavior of elasticbrittle composites that do not
exhibit noticeable nonlinearity or irreversible strains before the initiation of damage development.
However, these models may be insufficient in describing the nonlinear or plastic behavior that
some thermoset or thermoplastic composites might exhibit, particularly under transverse and/or
shear loading.
The results of experiments reported by Lafarie-Frenot and Touchard (1994), Van Paepegem, De
Baere, and Degrieck (2006), Vogler and Kyriakides (1999), and Wang, Callus, and Bannister
(2004) show that some composite materials experience significant nonlinearity or plasticity before
the collapse of the structures. In particular, such composites as carbon fiber reinforced epoxy T300/
914, T300/1034-C, and AS4/3501-6, and fiber reinforced PEEK composite AS4/PEEK might manifest this type of mechanical behavior under certain conditions.
In addition to plasticity effects, the deterioration of material properties under loading is another
significant feature of composite laminates. Defects such as fiber ruptures, matrix cracks, or fiber/
matrix debonding developing in a ply do not lead to the collapse of a laminate immediately as they
occur. These defects can accumulate gradually within the laminates. As a result, the material properties degrade progressively. Thus, the consideration of postfailure behavior is important for an
accurate prediction of failure loads.
Physically, the nonlinearity and/or irreversible deformations of fiber reinforced composites stem
from various mechanisms, such as the nonlinearity of each individual constituent, damage accumulation resulting from fiber or matrix cracking, and fiber/matrix interface debonding. Drucker (1975)
has proposed that such micromechanical phenomena can be described macroscopically within the
framework of plasticity theory. In combined plasticity and damage theories, the plastic strain represents all the irreversible deformations, including those caused by microcracks. This approach has
been employed by Morozov, Chen, and Shankar (2011) and Chen, Morozov, and Shankar (2012),
Chen, Morozov, and Shankar (2014a, 2014b) in the development of a combined elastoplastic damage model for progressive failure analysis that accounts for both the plasticity effects and material
properties degradation of composite materials under loading and which is presented in this section.
The plasticity effects are modeled using an equivalent form of Sun and Chen’s (1989) plastic
model. The prediction of the damage initiation and propagation in the laminated composites is performed taking into account various failure mechanisms employing Hashin’s (1980) failure criterion.
Once the damage development process initiates in a material, the local stresses are redistributed
in the undamaged area. As a result, the effective stresses in the undamaged area become higher
than the nominal stresses in the damaged material. Plasticity is assumed to be developed in the
undamaged area of the damaged material. Correspondingly, the effective stresses are used in the
plastic model. Since the nominal stresses in the postfailure branch of the stressstrain curves
decrease with an increase in strain, the use of these stresses in the failure criteria does not provide
a prediction for further damage growth. Thus, the effective stresses are also used in the failure criteria (Hashin, 1980). It is assumed that under transverse and shear loading, the irreversible
278
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
deformations are exhibited before the damage initiations; however, there is no stiffness degradation.
Beyond the damage initiation points, both irreversible deformations and stiffness degradations are
taken into account. Once the damage initiation is detected, both the damage development and plasticity evolve in the post-failure regime. In addition to the plastic strain accumulation at this stage,
the material’s stiffness degrades as well. The model accounts for both of these effects.
The plastic damage model can be implemented in Abaqus/Standard using a user-defined subroutine (UMAT). The strain-driven implicit integration procedure for the proposed model is developed
using equations of CDM, plasticity theory, and applying the return mapping algorithm. To ensure
the algorithmic efficiency of the Newton-Raphson method in the finite element analysis, a tangent
operator consistent with the developed integration algorithm is derived and implemented. The
efficiency of the proposed model is verified by performing progressive failure analysis of various
composite laminates subjected to in-plane tensile loads. The predicted results agree well with the
test data reported in the literature and provide accurate estimates of the failure loads.
4.5.1 CONSTITUTIVE EQUATIONS
Damage development affects the behavior of fiber-reinforced composite materials considerably.
Material properties, such as elastic moduli and Poisson’s ratio, degrade due to damage accumulation and growth. These effects are taken into account by introducing damage variables in the stiffness matrix using the CDM-based approach. According to this, the relationship between the
nominal stress and effective stress under uniaxial loading is given as
σ 5 ð1 2 dÞσ^
(4.65)
where σ 5 P=A0 is the Cauchy nominal stress (P is the normal internal force applied to the resisting
surface and A0 is the original area), σ^ 5 P=Aeff is the effective stress (Aeff is the effective resisting
area of the damaged surface), and d is the damage variable.
For composite materials exhibiting plasticity, the total strain tensor ε is presented as a sum of
the elastic and plastic strain parts εe and εp
ε 5 ε e 1 εp
(4.66)
where the plastic strain ε represents all the irreversible deformations, including those caused by
microcracks.
According to CDM theory, the stressstrain relationships for the damaged and undamaged composite materials are presented in the following forms:
p
σ 5 SðdÞ:εe ;
σ^ 5 S0 :εe
(4.67)
where bold-face symbols are used for variables of tensorial character and the symbol “:” denotes
the inner product of two tensors with double contraction, for example, ðSðdÞ:εe Þij 5 SðdÞijkl εekl , where
the summation convention is applied to the subscripts; σ and σ^ are the Cauchy nominal stress tensor and the effective stress tensor (both are of the second order); S0 is the fourth-order constitutive
tensor for linear-elastic undamaged unidirectional laminated composite; SðdÞ is the similar tensor
for the associated damaged material. The explicit form of S0 is determined by elasticity theory for
4.5 PROGRESSIVE FAILURE: GENERAL APPROACH TO MODELING
279
orthotropic materials (see, e.g., Eq. (2.195)). The Voigt form of SðdÞ adopted in the model under
consideration is similar to that presented by Matzenmiller et al. (1995)
2
ð1 2 d1 ÞE10
6
6
16
SðdÞ 5 6
ð1 2 d1 Þð1 2 d2 Þν 021 E20
D6
6
4
0
ð1 2 d1 Þð1 2 d2 Þν 012 E10
0
ð1 2 d2 ÞE20
0
0
Dð1 2 d3 ÞG012
3
7
7
7
7
7
7
5
(4.68)
where SðdÞ in italic is used to identify the Voigt form of SðdÞ; D 5 1 2 ð1 2 d1 Þð1 2 d2 Þν 012 ν 021 ; the
parameters d1 , d2 , and d3 denote damage developed in the fiber direction, transverse direction, and
under shear stress, respectively (they are scalar damage variables that remain constant throughout
the ply thickness); E10 , E20 , G012 and ν 012 , ν 021 are the elastic moduli and Poisson’s ratios of undamaged unidirectional composite ply, respectively (as earlier, E10 ν 012 5 E20 ν 021 Þ.
To differentiate between the effects of compression and tension on the failure modes, the damage variables are introduced as follows:
d1 5
d1t if σ^ 1 $ 0
d1c if σ^ 1 , 0
d2 5
d2t if σ^ 2 $ 0
d2c if σ^ 2 , 0
(4.69)
where d1t and d1c characterize the damage development caused by tension and compression in the
fiber direction, respectively; d2t and d2c reflect the damage development caused by tension and
compression in the transverse direction, respectively; and σ^ 1 and σ^ 2 are the effective stresses in the
fiber and transverse directions, respectively.
We assume that the shear stiffness reduction is caused by fiber and matrix cracking. To take
this effect into account, the corresponding damage variable d3 is expressed as:
d3 5 1 2 ð1 2 d6 Þð1 2 d1t Þ
(4.70)
where d6 represents the damage effects on shear stiffness caused by matrix cracking.
As previously mentioned, all the irreversible deformations are represented by the plastic strain
εp . These effects are allowed for by the plastic model which includes the yield criterion, plastic
flow rule, hardening rule, and the hardening law.
4.5.2 PLASTIC MODEL
In the damaged materials, the internal forces are resisted by the effective area. Thus, it is reasonable to assume that plastic deformation occurs in the undamaged area of the damaged composites.
According to this, the plastic flowp rule and hardening law are expressed in terms of effective stresB
B
^ equivalent plastic strain ε , and equivalent stress σ , which are based on the effective stress
ses σ,
space concept.
The plastic yield function is expressed in terms of effective stresses as follows:
Bp
Bp
^ ε Þ 5 F p ðσÞ
^ 2 κð ε Þ 5 0
Fðσ;
(4.71)
280
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
where F p is the plastic potential and κ is the hardening parameter,p which depends on the plastic
B
deformations and is expressed in terms of equivalent plastic strain ε .
Due to its simplicity and accuracy, an equivalent form of the one-parameter plastic potential for
the plane stress state proposed by Sun and Chen (1989) is adopted to describe the irreversible
strains exhibited by composites under transverse and/or shear loading:
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
3 2
B Bp
ðσ^ 1 2aσ^ 23 Þ 2 σ ð ε Þ 5 0
Fðσ;
^ ε Þ5
2 2
Bp
(4.72)
where a is a material parameter which describes the level of plastic deformation developed under
shearp loading compared to the transverse loading; σ^ 3 is the effective in-plane shear stress; and
B B
σ ð ε Þ represents the isotropic hardening law. Note that the use of the yield criterion in the form of
Eq. (4.72) in the model under consideration improves the efficiency and accuracy of the computational algorithm.
The equivalent stress is expressed in terms of σ^ 2 and σ^ 3 as follows (Sun & Chen, 1989):
1
B
σ 5
2
3 2
ðσ^ 2 12aσ^ 23 Þ
2
(4.73)
Assuming the associated plastic flow rule for composite materials, the plastic strain rate ε_ p is
expressed as:
ε_ p 5 λ_ @σ^ F
p
(4.74)
where λ_ $ 0 is a nonnegative plastic consistency parameter (hereafter in this section the following
notations are used for the derivatives: @x y 5 @y=@x, y_ 5 dy=dt).
Substituting Eq. (4.72) into Eq. (4.74), the following explicit form of plastic strain rate
is derived:
p
2
3
0
3
σ
^
2
6 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
7
62
7
2 p3
3 2
6
ε_ 1
2 7
6
ð
σ
^
1
2a
σ
^
Þ
p
p
3 7
4 ε_ p 5 5 λ_ @σ^ F 5 λ_ 6
7
2 2
2
6
7
p
3a
σ
^
6
7
3
ε_ 3
6 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 7
4
5
3 2
ðσ^ 1 2aσ^ 23 Þ
2 2
(4.75)
Also, the associated hardening rule is assumed for the equivalent plastic strain rate to have the
following form:
B
_p
ε 5 λ_ hp 5 2 λ_ @B F
σ
p
p
(4.76)
where hp defines the evolution direction of the equivalent plastic strain.
The equivalent plastic strain rate can be obtained from the equivalence of the rates of the plastic
work per unit volume W p presented as follows:
BB
p
_p
^ εp 5 σ ε
W_ 5 σ:_
(4.77)
Substituting Eqs. (4.75) and (4.73) into Eq. (4.77), the following relation is derived
B
_p
ε 5 λ_
p
(4.78)
4.5 PROGRESSIVE FAILURE: GENERAL APPROACH TO MODELING
281
It follows from the comparison of Eqs. (4.76) and (4.78) that the value of hp is unity. Note that
this does not hold if the original quadratic form of the Sun and Chen yield criterion is adopted (Sun
& Chen, 1989). As a result, the application of the original yield criterion leads to higher computational cost in the integration procedure. The current approach based on the use of Eq. (4.72) avoids
this inefficiency.
For the
sake of simplicity, an isotropic hardening law expressed in terms of equivalent plastic
Bp
strain ε is adopted in this study. The following formulation of this law proposed by Sun and Chen
(1989) is employed in this work to represent the equivalent stress versus equivalent plastic strain
hardening curve:
κð~εp Þ 5 σð~
~ εp Þ 5 βð~εp Þn
(4.79)
where β and n are coefficients that fit the experimental hardening curve. These parameters along
with the material parameter a are determined using an approach based on linear regression analyses
of the off-axis tensile tests performed on unidirectional composite specimens (Sun & Chen, 1989;
Winn & Sridharan, 2001).
4.5.3 DAMAGE MODEL
The prediction of the damage initiation and progression in the laminated composites under consideration is performed taking into account various modes of failure, including those caused by tensile
and compressive loading for the fibers and matrix, and also the combined action of transverse and
shear stresses.
4.5.3.1 Damage Initiation and Propagation Criteria
To predict the damage initiation and propagation of each intralaminar failure of the material and
evaluate the effective stress state, the damage initiation and propagation criteria fI are presented in
the following form:
fI ðφI ; rI Þ 5 φI 2 rI # 0
I 5 f1t; 1c; 2t; 2c; 6g
(4.80)
where φI are the loading functions for different failure mechanisms adopted in the form of
Hashin’s failure criterion (Hashin, 1980):
2
φ1t 5 σ^11
σ
1 2
φ1c 5 σ^21
σ
1 2 2
φ2t 5 σ^12 1 σ^ 3
τ 12
σ
2 2 2
φ2c 5 σ^22 1 σ^ 3
τ 12
σ2
ðσ^ 1 $ 0Þ ðtensile fiber damage modeÞ
ðσ^ 1 , 0Þ ðcompressive fiber damage modeÞ
(4.81)
ðσ^ 2 $ 0Þ ðtensile matrix damage modeÞ
ðσ^ 2 , 0Þ ðcompressive matrix damage modeÞ
2
1
where σ1
1 and σ1 are the tensile and compressive strengths in fiber direction, respectively; σ2 and
2
̅
σ2 are the transverse tensile and compressive strengths, respectively; and τ 12 is the shear strength.
Each damage threshold parameter rI controls the size of the expanding damage surface and depends
on the loading history. The damage development in the material initiates when the value of φI
282
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
exceeds the initial damage threshold rI;0 5 1. Further damage growth occurs when the value of φI
in the current stress state exceeds the value of rI in the previous loading history. As mentioned
previously, the damage variable d6 represents the damage effects on the shear stiffness due to matrix
fracture caused by the combined action of transverse and shear stresses. However, the compressive
transverse stress has beneficial effects on matrix cracking. Thus, it is reasonable to assume that the
damage effects are governed by the tensile matrix cracking only, that is, f6 5 f2t , r6 5 r2t .
Once the damage initiation criterion is satisfied, then further damage evolution could be developed. The way the damage would evolve depends on the level of the stress caused by the strain
increment. The damage propagation criteria, Eq. (4.80) define the growing damage surfaces when
the value of rI corresponding to the current stress state in Eq. (4.80) is greater than unity. The
modeling of the evolution of damage is discussed in the following section.
4.5.3.2 Damage Evolution
The evolution of damage is possible only when the stresses reach the level corresponding to the
damage surface, that is, when Eq. (4.80) is converted to equality. In the further process of damage
evolution, the following damage consistency condition should be satisfied: f_ðφI ; rI Þ 5 φ_ I 2 r_ I 5 0.
On the basis of Ðthis condition,
the values of the thresholds could be found by integrating equation
Ðt
t
r_ I 5 φ_ I , that is, 0 r_ I dt 5 0 φ_ I dt, and taking into account the initial condition φI;0 5 0. The following expressions for damage thresholds rI can be derived:
rI 5 maxf1; maxfφτI gg I 5 f1t; 1c; 2t; 2c; 6g
τA½0; t
(4.82)
Since damage is irreversible, the damage evolution rate should satisfy the following condition:
d_I $ 0. The exponential damage evolution law is adopted for each damage variable and expressed
in the following form (Faria, Oliver, & Cervera, 1998):
dI 5 1 2
1
expðAI ð1 2 rI ÞÞ
rI
I 5 f1t; 1c; 2t; 2c; 6g
(4.83)
where AI is the parameter that defines the exponential softening law. This parameter is determined
by regularizing the softening branch of the stressstrain curve to ensure that the dissipated energy
in the formation and opening of all microcracks per unit area of the fracture surface, computed
according to the numerical procedure proposed, is independent of mesh size. This approach helps
alleviate the mesh dependency of the finite element results. The regularization is based on Bazant’s
crack band theory (Bazant & Oh, 1983). According to this, the damage energy dissipated per unit
volume gI for a uniaxial or shear stress state is related to the critical strain energy release rate GI;c
along with the characteristic length of the finite element l as follows:
gI 5
GI;c
l
I 5 f1t; 1c; 2t; 2c; 6g
(4.84)
where the parameters G1t;c and G1c;c are the mode I fracture toughness parameters related to fiber
breakage under tension and compression. The critical strain energy release rates G2t;c and G2c;c are
referred to as the intralaminar mode I fracture toughness parameters under transverse tension and
compression. The parameter G6;c is the intralaminar mode II fracture toughness parameter. The
identification of these parameters and the characteristic length l is discussed in Maimi et al.
(2007b) and Pinho (2005).
4.5 PROGRESSIVE FAILURE: GENERAL APPROACH TO MODELING
283
The damage energy dissipated per unit volume for a uniaxial or shear stress state is obtained by
integration of the damage energy dissipation during the process of damage development:
gI ðAI Þ 5
ðN
YI d_I ðAI Þdt;
YI 5 2
0
@ψ
;
@dI
ψ5
1
σ:ε I 5 f1t; 1c; 2t; 2c; 6g
2
(4.85)
where YI is the damage energy release rate; d_I is the rate of damage development defined as
d_I 5 ddI =dt; ψ is the Helmholtz free energy. The parameters AI correspond to each uniaxial or
shear loading condition applied to the elementary ply, and these parameters should be determined
for a given material and a given finite element size to avoid mesh dependency in the finite element
analysis. Equating Eqs. (4.84) and (4.85), we can derive the following equation:
gI;c ðAI Þ 5
GI;c
l
I 5 f1t; 1c; 2t; 2c; 6g
(4.86)
The values of AI for I 5 f1t; 1c; 2t; 2c; 6g are found by solving Eq. (4.86) iteratively using a
root-finding algorithm described in Chen et al. (2012).
The loading/unloading stressstrain curves illustrating the combined elastoplastic damage
model under consideration are shown in Fig. 4.27. It is assumed that under loading in the fiber
direction the material exhibits linear elastic brittle behavior and the irreversible strain is not developed. Note that the model can be easily extended to the case where the material would exhibit
σ1
σ1
σ2
σ2
σ1
σ2
ε1
ε2
(A)
(B)
σ3
σ3
σ3
σ3
σ3
(C)
ε3
FIGURE 4.27
Loading/unloading stressstrain curves for in-plane loading along the fibers (A), in transverse direction (B), and
shear (C).
284
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
nonlinear inelastic behavior in the fiber direction, if required. After damage initiation, the elastic
modulus E1 is assumed to degrade gradually. It is also assumed that under transverse and shear
loading the irreversible deformations are exhibited before the damage initiations; however, there is
no stiffness degradation at that stage. Beyond the damage initiation points, both irreversible deformations and stiffness degradations are taken into account.
4.5.4 NUMERICAL IMPLEMENTATION
The combined elastoplastic damage material model discussed earlier has been implemented in the
Abaqus/Standard finite element software package using the user-defined subroutine UMAT. The
numerical integration algorithms updating the Cauchy nominal stresses and solution-dependent state
variables have been derived, as well as the tangent matrix that is consistent with the numerical integration algorithm ensuring the quadratic convergence rate of the Newton-Raphson method in the
finite element analysis (Chen et al., 2012; Morozov et al., 2011).
4.5.4.1 Integration Algorithm
The solution of the nonlinear inelastic problem under consideration is based on the incremental
approach and is regarded as strain-driven. The loading history is discretized into a sequence of time
steps ½tn ; tn11 ; nA0; 1; 2; 3; ::: where each step is referred to as the ðn 1 1Þth increment. Driven by
the strain increment Δεn11 , the discrete problem in the context of a backward Euler scheme for the
combined elastoplastic
damage model can be stated as follows: for a given variable set
Bp
B
fεn11 ; εpn ; ε n ; σ^ n ; σ
;
σ
;
r
n
n
I;n ; dI;n g at the beginning of the ðn 1 1Þth increment, find the updated variBp
B
able set fεpn11 , ε n11 , σ^ n11 , σ n11 , σ n11 , rI;n11 , dI;n11 g at the end of the ðn 1 1Þth increment. The
updated stresses and solution-dependent state variables are stored at the end of the ðn 1 1Þth increment and are passed on to the user subroutine UMAT at the beginning of the next increment.
The effective stressstrain relationship Eq. (4.67), yield criterion Eq. (4.72), associated plastic
flow rule Eq. (4.74), and hardening power law Eq. (4.79) represent the nonlinear plastic constitutive
material model. Using the backward Euler implicit integration procedure, the corresponding integration algorithm is formulated as follows:
εn11 5 εn 1 Δεn11
p
p
εn11 5 εpn 1 Δλpn11 @σ^ n11 Fn11
Bp
Bp
ε n11 5 ε n 1 Δλpn11
σ^ n11 5 S0 :ðεn11 2 εpn11 Þ
Bp
Fn11 5 Fðσ^ n11 ; ε n11 Þ # 0
(4.87)
p
where Δλpn11 5 λ_ n11 Δt is the increment of the plastic consistency parameter.
The closest point return
mapping algorithm is employed to solve this nonlinear coupled system.
Bp
The solutions fεpn11 , ε n11 , σ^ n11 g are the converged values at the end of the ðn 1 1Þth increment.
These ensure that upon yielding, the calculated stress state lies on the yield surface and they prevent drift from the yield surface due to the unconverged solutions obtained from the forward Euler
integration scheme.
The nonlinear system, Eq. (4.87), is linearized and solved iteratively using the Newton-Raphson
Bp;ðk11Þ
ðk11Þ p;ðk11Þ
scheme. The iterations are performed until the final set of state variables fσ^ n11
; εn11 ; ε n11
g
4.5 PROGRESSIVE FAILURE: GENERAL APPROACH TO MODELING
285
Bp;ðk11Þ
ðk11Þ
in the ðk 1 1Þth iteration fulfils the yield criterion Fðσ^ n11
; ε n11 Þ # TOL, where TOL is the error
26
tolerance which would typically be set to 1 3 10 in these type of analyses.
Substituting the effective stresses σ^ n11 into the damage model, the damage variables are
updated. According to Eq. (4.67), the Cauchy stresses are calculated as σ n11 5 Sðdn11 Þ:εen11 .
4.5.4.2 Consistent Tangent Stiffness Matrix
The consistent tangent matrix for the proposed constitutive model is derived in the following form:
dσ n11
5 ½Mn11 1 Sðdn11 Þ:C0 :Salg
n11
dεn11
(4.88)
in which the fourth-order tensor Mn11 can be presented in Voigt notation (in indicial form) as follows:
Mik @SðdÞij εej ^ q e @SðdÞij @dp @rξ @φξ @σ
5
5 εj
@dp @rξ @φξ @σ^ q @εek n11
@εek n11
n11
i; j; p; q; k 5 f1; 2; 3g; ξ 5 f1; 2g
(4.89)
where matrix Mn11 5 ½Mik n11 is asymmetric. This results in the asymmetry of the consistent tangent matrix of the combined elastoplastic damage model. The explicit expression for matrix M,
along with the associated derivatives, can be found in Morozov et al. (2011) and Chen et al.
(2012). In Eq. (4.88), C0 5 S21
0 is the fourth-order compliance tensor of the undamaged composite
material and Salg
is
the
consistent
tangent matrix for the discrete plastic problem Eq. (4.87). The
n11
latter is expressed as:
dσ^ n11
Salg
n11 5
dεn11
B
5 S n11 2
B
B
p
ðS n11 :@σ^ Fn11
Þ ðS n11 :@σ^ Fn11 Þ
B
p
@σ^ Fn11 :S n11 :@σ^ Fn11
2 @Bp Fn11
ε
(4.90)
p
where S~ n11 5 ðC0 1Δλpn11 @2σ^ σ^ Fn11
Þ21 , Δλpn11 denotes the increment of plastic consistency paramep
ter λ in the ðn 1 1Þth increment; @2xx y 5 @2 y=@x2 , and () denotes a tensor product. As the time
increment Δt approaches zero, the increment of the plastic consistency parameter Δλpn11 also
B
approaches zero. Thus, S n11 approaches S0 and Salg
n11 reduces to the elastoplastic tangent operator
when standard procedures of classical plasticity theory are applied.
4.5.4.3 Viscous Regularization
Numerical simulations based on the implicit procedures, such as Abaqus/Standard, and the use of
material constitutive models that are considering strain softening and material stiffness degradation
often abort prematurely due to convergence problems. To alleviate these computational difficulties
and improve convergence, a viscous regularization scheme has been implemented in the following
form (Lapczyk & Hurtado, 2007):
1
v
d_m 5 ðdm 2 dmv Þ;
η
m 5 1; 2; 3
(4.91)
where dm is the damage variable obtained as described previously, dmv is the regularized viscous
damage variable, and η is the viscosity coefficient.
286
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
The regularized damage variable in the ðn 1 1Þth increment is derived as
v
dm;n11
5
Δt
η
dm;n11 1
dv
η 1 Δt
η 1 Δt m;n
(4.92)
The corresponding regularized consistent tangent matrix is derived as
dσn11 v
Þ:C0 :Salg
5 ½Mvn11 1 Sðdn11
n11
dεn11 v
(4.93)
where the fourth-order tensor Mvn11 can be presented in Voigt notation (in indicial form) as
follows:
Δt
@Sðd v Þij εej @Sðd v Þij @dp @rξ @φξ @σ^ q e
Mvik n11 5
5
ε
U
j
@εek n11
@dpv @rξ @φξ @σ^ q @εek n11 η 1 Δt
i; j; p; q; k 5 f1; 2; 3g; ξ 5 f1; 2g
4.5.4.4 Computational Procedure
The user subroutine UMAT for the numerical algorithm based on the model discussed earlier has
been coded and is called by Abaqus/Standard at each material integration point of the elements
where user-defined materials are used to update the Cauchy nominal stresses and provide the consistent tangent matrix of the material for the calculation of the element tangent stiffness matrix.
This process includes the use of the elastic predictor, plastic corrector, damage corrector parts of
the algorithm, and calculation of the regularized consistent tangent matrix as follows:
1. Initial conditions:
Bp
B
v
εn11 ; Δεn11 ; εpn ; ε n ; σ^ n ; σn ; σ n ; rI;n ; dI;n ; dm;n ; dm;n
2. Elastic predictor:
a. Assign and compute the trial state variables:
Bp;trial
Bp
Btrial
B
p
εp;trial
ε n11 5 ε n ; σ n11 5 σ n ;
n11 5 εn ;
trial
trial
σ^ n11 5 σ^ n 1 S0 :Δεn11 ; dI;n11
5 dI;n
b. Check the yield criterion:
Btrial
p;trial
trial
Fn11
5 Fn11
2 σ n11 # 0
trial
If Fn11
# 0, then set σ^ n11 5 σ^ trial
n11 . Otherwise, plasticity evolves, go to step 3.
3. Plastic corrector:
The Newton-Raphson method is employed to solve the nonlinear problem
p Bp
^
Fn11 5 FðσðΔλ
Þ; ε ðΔλp ÞÞ 5 0 iteratively to compute the real plastic state variables
p
B
B
(εpn11 , ε n11 , σ n11 ):
a. Initialize:
Bp;ð0Þ
Bp
p;ð0Þ
p;ð0Þ
p
^ trial
k 5 0; σ^ ð0Þ
n11 ; εn11 5 εn ; ε n11 5 ε n ; Δλn11 5 0:
n11 5 σ
4.5 PROGRESSIVE FAILURE: GENERAL APPROACH TO MODELING
287
Bp
b. Calculate δðΔλp Þ, δεp , δ ε in the (k 1 1) iteration;
c. Update the plastic state variables in the (k 1 1) iteration:
p
Δλp;ðk11Þ
5 Δλp;ðkÞ
n11
n11 1 δðΔλ Þ;
p;ðkÞ
εp;ðk11Þ
5 εn11
1 δεp ;
n11
Bp;ðk11Þ
Bp;ðkÞ
ε n11 5 ε n11 1 δBε p
ðk11Þ
p;ðk11Þ
d. Calculate σ^ n11
5 S0 : εn11 2 εn11
ðk11Þ
ðk11Þ
p;ðk11Þ
ðk11Þ
e. Calculate Fn11
5 F σ^ n11
; Bε n11
, TOL, then σ^ n11 5 σ^ ðk11Þ
, If Fn11
n11 ,
Bp
Bp;ðk11Þ
p;ðk11Þ
εpn11 5 εn11
, ε n11 5 ε n11 , calculate Salg
n11 . Otherwise, set k 5 k 1 1, go to b.
4. Damage corrector (updating the nominal Cauchy stresses):
a. Calculate φI;n11 , updating rI;n11 5 maxf1; φI;n11 ; rI;n g
v
v
b. Calculate damage variables dI;n11 , dm;n11 , dm;n11
and Sðdm;n11
Þ
p
v
c. Update the nominal stresses: σ n11 5 Sðdm;n11 Þ:ðεn11 2 εn11 Þ
5. Calculation of the tensor Mvn11 and the regularized consistent tangent matrix:
i
dσ n11 h v
v
5 Mn11 1 Sðdm;n11
Þ :C0 :Salg
n11
dεn11
The flow chart of this subroutine is shown in Fig. 4.28.
4.5.5 NUMERICAL ANALYSES
To demonstrate the capability of the present model, it was applied to the progressive failure analysis of a double-notched AS4/3501-6 carbon/epoxy composite laminate. The experimental investigation of the laminate has been carried out by Pham, Sun, Tan, Chen, and Tay (2012). The layup of
the laminate is ½452 =902 =2452 =02 s . The laminate considered is subjected to tensile loading and
has the length, width, and thickness of 100, 40, and 2 mm, respectively. Each side of the specimen
has a V-notch with the same size. The geometry and dimension of the laminate is shown in
Fig. 4.29.
In addition, AS4/PEEK [ 6 45 ]2s composite laminate without cutouts under in-plane tensile
loading has been analyzed.
4.5.5.1 Analysis of AS4/3501-6 Double-Notched Laminate
The material properties of AS4/3501-6 composite materials reported by Wang et al. (2004) are
adopted in this work and the plastic model parameters (i.e., a,β,n) are obtained from Chen and Sun
(1987). The values of the fracture toughness parameters associated with fiber tensile and compressive failure G1t;c 5 G1c;c 5 128.0 N/mm which correspond to those of AS4/PEEK composite materials (Chen et al., 2012; Morozov & Chen, 2014, 2015) are adopted in this study. The Mode I
and Mode II intra-laminar fracture toughness parameters are G2t;c 5 0.238 N/mm and
G6;c 5 0.476 N/mm, respectively, as reported by Shahid and Chang (1995). The parameter G2c;c is
calculated according to the equation suggested by Maimi et al. (2007b), that is,
288
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
FIGURE 4.28
Flow chart of the user subroutine UMAT.
G2c;c G6;c =cos53 5 0.2865 N/mm. The material properties and model parameters are listed in
Table 4.2.
The failure load predicted for the laminate using the present model is compared with the experimental data and the predictions made using the MCDM and MChristensen methods reported in
Pham et al. (2012) (see Table 4.3). It follows from this table that the present model provides the
most accurate prediction compared with the MCDM and MChristensen methods. The predicted difference with respect to the test result using the present model is only 2.31%, while those predicted
using the MCDM and MChristensen methods are 12.92% and 9.64%, respectively. For the sake
of comparison, the experimental and predicted displacement vs applied load curves of the laminates
are shown in Fig. 4.30. It is clear from this figure that the present model provides closer predicted
4.5 PROGRESSIVE FAILURE: GENERAL APPROACH TO MODELING
289
FIGURE 4.29
Geometry and dimensions of the double-notched AS4/3501-6 laminate.
Table 4.2 Material Properties of AS4/3501-6 and Model Parameters
E10
E20
G012
ν 012
s1t
s1c
s2t
s2c
ss
142 GPa
10.3 GPa
7.2 GPa
0.27 GPa
2280 MPa
1440 MPa 57 MPa
228 MPa 71 MPa
G1c;c
G2t;c
G2c;c
G12;c
a
β
n
η
G1t;c
128.0 N/mm 128.0 N/mm 0.238 N/mm 0.2865 N/mm 0.476 N/mm 1.25
567.9092 0.272405 0.0002
Table 4.3 Experimental and Predicted Failure Load of the Double Notched Laminates
Failure Load (N)
Test (Pham et al., 2012)
MCDM (Pham et al., 2012)
MChristensen (Pham et al., 2012)
Present model
Specimen 1
Specimen 2
Specimen 3
Specimen 4
Average
Difference (%)
17,898.7
18,275.0
19,451.0
18,533.8
18,539.6
16,144
16,752
18,968.6
12.92
9.64
2.31
failure load of the laminate than the MCDM and MChristensen methods. The predicted patterns of
the damage variables d1 and d2 in the first 453 ply at the end of the analysis are shown in Fig. 4.31.
As can be seen, a significant amount of damage due to matrix cracking has been developed in the
cross section of the double-notches (see Fig. 4.31A), while much less damage due to fiber breakage
has been grown in the same area and propagates in the 2453 direction (see Fig. 4.31B).
FIGURE 4.30
Comparison of the experimental and predicted load vs displacement curves of the double-notched laminate.
FIGURE 4.31
The predicted patterns of damage variables d1 and d2 in the first 45 ply of the laminate at the end of the
analysis. (A) Damage variable d2 and (B) damage variable d1.
4.5 PROGRESSIVE FAILURE: GENERAL APPROACH TO MODELING
291
4.5.5.2 AS4/PEEK [ 6 45 ]2s Composite Laminate
To demonstrate the capability of the proposed model to represent the postfailure regime and to validate its ability to reflect on the plasticity behavior, a numerical analysis of the AS4/PEEK ½ 6 453 2s
composite laminate subjected to tensile loading has been performed. The length, width, and thickness of the coupon are 230, 20, and 1 mm, respectively. The experimental stressstrain curve for
this laminate was obtained by Lafarie-Frenot and Touchard (1994). The results of the simulation
and testing are shown in Fig. 4.32. The material properties of the unidirectional AS4/PEEK CFRP
material listed in Table 4.4 were adopted in the numerical analysis. The material elastic properties
and plastic model parameters were obtained from Sun and Yoon (1992). The compressive strengths
were taken from Sun and Rui (1990), whereas the tensile strength and shear strength were obtained
from Kawai, Masuko, Kawase, and Negishi (2001). The values of the critical strain energies G1t;c
and G2t;c were taken from the experimental data reported by Carlile, Leach, Moore, and Zahlan
(1989). The value of the parameter G6;c was taken from Donaldson (1985). Using this value, the
critical strain energy G2c;c was calculated approximately based on the equation proposed by Maimi
et al. (2007b).
It follows from Fig. 4.32 that the stressstrain curve predicted by the numerical simulation correlates well with that obtained from the experimental study.
FIGURE 4.32
Comparison of the experimental and predicted stress vs longitudinal strain curves of AS4/PEEK [ 6 45 ]2s
composite laminates.
292
CHAPTER 4 FAILURE CRITERIA AND STRENGTH OF LAMINATES
Table 4.4 Material Properties of AS4/PEEK and Model Parameters
E10
E20
G012
ν 021
σ 1t
σ 1c
σ 2t
σ 2c
σs
127.6 GPa
G1t;c
128.0 N/mm
10.3 GPa
G1c;c
128.0 N/mm
6.0 GPa
G2t;c
5.6 N/mm
0.32
G2c;c
9.31 N/mm
2023.0 MPa
G12;c
4.93 N/mm
1234.0 MPa
a
1.5
92.7 MPa
β
295.0274
176.0 MPa
n
0.142857
82.6 MPa
η
0.0002
The aforementioned results of numerical simulations show the capability of the model and
numerical procedure discussed in this section in simulating a progressive failure process for
composite laminates and capturing and reflecting the key features of the damage initiation
and progression.
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CHAPTER
ENVIRONMENTAL, SPECIAL
LOADING, AND MANUFACTURING
EFFECTS
5
The properties of composite materials, as well as those of all structural materials, are affected by
environmental and operational conditions. Moreover, for polymeric composites, this influence is
more pronounced than for conventional metal alloys because polymers are more sensitive to
temperature, moisture, and time than metals. There is also a specific feature of composites associated with the fact that they do not exist apart from composite structures and are formed while these
structures are fabricated. As a result, the material characteristics depend on the type and parameters
of the manufacturing process, for example, unidirectional composites made by pultrusion, hand layup, and filament winding can demonstrate different properties.
This section of the book is concerned with the effect of environmental, loading, and
manufacturing factors on the mechanical properties and behavior of composites.
5.1 TEMPERATURE EFFECTS
Temperature is the most important of environmental factors affecting the behavior of composite
materials. First of all, polymeric composites are rather sensitive to temperature and have relatively
low thermal conductivity. This combination of properties allows us, on one hand, to use these materials in structures subjected to short-term heating, and on the other hand, requires the analysis of
these structures to be performed with due regard to temperature effects. Secondly, there exist
composite materials, for example, carboncarbon and ceramic composites, that are specifically
developed for operation under intense heating, and materials such as mineral fiber composites that
are used to form heatproof layers and coatings. Thirdly, the fabrication of composite structures is
usually accompanied by more or less intensive heating (e.g., for curing or carbonization), and the
subsequent cooling induces thermal stresses and strains, to calculate which we need to utilize the
equations of thermal conductivity and thermoelasticity as discussed in the following.
5.1.1 THERMAL CONDUCTIVITY
Heat flow through a unit area of a surface with normal n is related to the temperature gradient in
the n direction according to Fourier’s law as
q52λ
@T
@n
Advanced Mechanics of Composite Materials and Structures. DOI: https://doi.org/10.1016/B978-0-08-102209-2.00005-0
© 2018 Elsevier Ltd. All rights reserved.
(5.1)
295
296
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
where λ is the thermal conductivity of the material. The temperature distribution along the n-axis is
governed by the following equation:
@
@T
@T
λ
5 cρ
@n
@n
@t
(5.2)
in which c and ρ are the specific heat and density of the material, and t is time. For a steady (timeindependent) temperature distribution, @T=@t 5 0, and Eq. (5.2) yields
ð
T 5 C1
dn
1 C2
λ
(5.3)
Consider a laminated structure referred to coordinates x, z as shown in Fig. 5.1. To determine
the temperature distribution along the x-axis only, we should take into account that λ does not
depend on x, and assume that T does not depend on z. Using conditions Tðx 5 0Þ 5 T0 and
Tðx 5 lÞ 5 Tl to find the constants C1 and C2 in Eq. (5.3), in which n 5 x, we get
x
T 5 T0 1 ðTl 2 T0 Þ
l
Introduce the apparent thermal conductivity of the laminate in the x direction, λx , and write
Eq. (5.1) for the laminate as
qx 5 2 λx
Tl 2 T0
l
The same equation can be written for the ith layer, that is,
qi 5 2 λi
Tl 2 T0
l
The total heat flow through the laminate in the x direction is
qx h 5
k
X
qi hi
i51
Combining the foregoing results, we arrive at
–
FIGURE 5.1
Temperature distribution in a laminate.
5.1 TEMPERATURE EFFECTS
λx 5
k
X
λi hi
297
(5.4)
i51
where hi 5 hi =h.
Consider the heat transfer in the z direction and introduce the apparent thermal conductivity λz
in accordance with the following form of Eq. (5.1)
qz 5 2 λz
Th 2 T0
h
(5.5)
Taking n 5 z and λ 5 λi for zi21 # z # zi in Eq. (5.3), using step-wise integration and the conditions Tðz 5 0Þ 5 T0 ; Tðz 5 hÞ 5 Th to find constants C1 , and C2 , we obtain for the ith layer
Th 2 T0
Ti 5 T0 1 Pk
i51 hi =λi
i21
X
z 2 zi21
hj
1
λi
λ
j51 j
!
(5.6)
The heat flow through the ith layer follows from Eqs. (5.1) and (5.6), that is,
qi 5 2 λi
@Ti
Th 2 T0
5 2 Pk
@z
i51 hi =λi
Obviously, qi 5 qz (see Fig. 5.1), and with due regard to Eq. (5.5)
k
X
1
hi
5
λz
λ
i51 i
(5.7)
where, as earlier, hi 5 hi =h.
The results obtained, Eqs. (5.4) and (5.7), can be used to determine the thermal conductivity of
a unidirectional composite ply. Indeed, comparing Fig. 5.1 with Fig. 1.34 showing the structure of
the first-order ply model, we can write the following equations specifying the thermal conductivity
of a unidirectional ply along and across the fibers:
λ1 5 λ1f vf 1 λm vm
1
vf
vm
5
1
λ2f
λm
λ2
(5.8)
Here, λ1f and λ2f are the thermal conductivities of the fiber in the longitudinal and transverse
directions (for some fibers these are different), λm is the corresponding characteristic of the matrix,
and vf and vm 5 1 2 vf are the fiber and matrix volume fractions, respectively. The conductivity
coefficients in Eqs. (5.8) are analogous to the elastic constants specified by Eqs. (1.76) and (1.78),
and the discussion presented in Section 1.3 is valid for Eqs. (5.8) as well. In particular, it should be
noted that application of higher order microstructural models has practically no effect on λ1 but
substantially improves λ2 , determined by Eqs. (5.8). Typical properties for unidirectional and fabric
composites are listed in Table 5.1.
Consider heat transfer in an orthotropic ply or layer in coordinate frame x, y whose axes x and y
make angle φ with the principal material coordinates x1 and x2 as in Fig. 5.2. Heat flows in coordinates x; y and x1 ; x2 are linked by the following equations:
qx 5 q1 cosφ 2 q2 sinφ;
Here, in accordance with Eq. (5.1)
qy 5 q1 sinφ 1 q2 cosφ
(5.9)
298
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
Table 5.1 Typical Thermal Conductivity and Expansion Coefficients of Composite Materials
Property
Longitudinal
conductivity, λ1 (W/mK)
Transverse conductivity,
λ2 (W/mK)
Longitudinal CTE, 106α1
(1/ C)
Transverse CTE, 106α2
(1/ C)
Glassepoxy
Carbonepoxy
Aramidepoxy
Boronepoxy
Glass
Fabricepoxy
Aramid Fabricepoxy
0.6
1
0.17
0.5
0.35
0.13
0.4
0.6
0.1
0.3
0.35
0.13
7.4
0.3
3.6
4.1
8
0.8
22.4
34
60
19.2
8
0.8
y
x2
x1
q1
– q2
qx
φ
x
q1
qy
q2
FIGURE 5.2
Heat flows in coordinates x; y and x1 ; x2 .
q1 5 2 λ1
@T
;
@x1
q2 5 2 λ2
@T
@x2
Changing variables x1 ; x2 to x, y with the aid of the following transformation relationships
x 5 x1 cosφ 2 x2 sinφ;
y 5 x1 sinφ 1 x2 cosφ
and substituting q1 and q2 into Eqs. (5.9), we arrive at
qx 5 2 λx
@T
@T
2 λxy
;
@x
@y
qy 5 2 λy
@T
@T
1 λxy
@y
@x
where
λx 5 λ1 cos2 φ 1 λ2 sin2 φ
λy 5 λ1 sin2 φ 1 λ2 cos2 φ
λxy 5 ðλ2 2 λ1 Þsinφcosφ
(5.10)
can be treated as the ply thermal conductivities in coordinates x, y. Since the ply is anisotropic in
these coordinates, the heat flow in for example, the x direction, induces a temperature gradient not
5.1 TEMPERATURE EFFECTS
299
only in the x direction, but in the y direction as well. Using Eq. (5.4), we can now determine the
in-plane thermal conductivities of the laminate as
Λx 5
k
X
λðiÞ
x hi ;
Λxy 5
i51
k
X
Λy 5
k
X
λðiÞ
y hi ;
i51
(5.11)
λðiÞ
xy hi
i51
ðiÞ
where λðiÞ
x;y are specified by Eqs. (5.10) in which λ1;2 5 λ1;2 and φ 5 φi . For 6 φ angle-ply laminates
that are orthotropic, Λxy 5 0.
As an example, consider the composite body of a space telescope, the section of which is shown
in Fig. 5.3. The cylinder having diameter D 5 1 m and total thickness h 5 13.52 mm consists of
four layers, that is,
•
6 φs angle-ply carbon-epoxy external skin with the following parameters:
φs 5 203 ; hes 5 3:5 mm; E1e 5 120 GPa;
E2e 5 11 GPa; Ge12 5 5:5 GPa; ν e21 5 0:27;
λe1 5 1 W=m K; λe2 5 0:6 W=m K;
αe1 5 2 0:3 3 1026 1=3 C; αe2 5 34 3 1026 1=3 C;
•
•
carbon-epoxy lattice layer (see Chapter 10) formed by a system of 6 φr helical ribs with angle
φr 5 263 , height and thickness of the rib cross section hr 5 9 mm and δr 5 4 mm, ribs spacing
ar 5 52 mm, rib modulus Er 5 80 GPa, λr 5 0:9 W=m K, αr 5 2 1 3 1026 1=3 C,
internal skin made of aramid fabric with
his 5 1 mm; Exi 5 Eyi 5 34 GPa; Gixy 5 5:6 GPa;
ν ixy 5 ν iyx 5 0:15; λix 5 λiy 5 0:13 W=m K;
FIGURE 5.3
A composite section of a space telescope.
300
•
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
αix 5 αiy 5 0:8 3 1026 1=3 C (x and y are the axial and the circumferential coordinates of the
cylinder),
internal layer of aluminum foil with
hf 5 0:02 mm; Ef 5 70 GPa; ν f 5 0:3;
λf 5 210 W=m K; αf 5 22:3 3 1026 1=3 C
The apparent thermal conductivity of the cylinder wall can be found with the aid of Eqs. (5.10),
(5.11) and the continuum model of the lattice layer described in Chapter 10 as
Λx 5
1
2
ðλe1 cos2 φe 1 λe2 sin2 φe Þhes 1 hr δr λr cos2 φr 1 λix his 1 λf hf
h
ar
Calculation yields Λx 5 0:64 W=m K. The thermal resistance of a unit length of this structure is
rx 5
1
5 36:8 K=W m
Λx πDh
5.1.2 THERMOELASTICITY
It is known that heating gives rise to thermal strains that, when restricted, induce thermal stresses.
Assume that the temperature distribution in a composite structure is known, and consider the problem of thermoelasticity.
Consider first the thermoelastic behavior of a unidirectional composite ply studied in
Section 1.3 and shown in Fig. 1.29. Total strains can be presented as
ε1T 5 ε1 1 εT1 ;
ε2T 5 ε2 1 εT2 ;
γ 12T 5 γ 12
(5.12)
Here and subsequently, the subscript “T” shows the strains that correspond to the problem of
thermoelasticity, whereas the superscript “T” indicates temperature terms. Elastic strains ε1 ; ε2 ,
and γ 12 in Eqs. (5.12) are related to stresses by Eqs. (1.58). Temperature strains, to a first approximation, can be taken as linear functions of the temperature change, that is,
εT1 5 α1 ΔT;
εT2 5 α2 ΔT
(5.13)
where α1 and α2 are the coefficients of thermal expansion (CTEs) along and across the fibers, and
ΔT 5 T 2 T0 is the difference between the current temperature T and some initial temperature T0 at
which thermal strains are zero. Taking into account Eqs. (5.12), we can generalize Hooke’s law,
Eqs. (1.59) as follows
σ1 5 E1 ðε1T 1 ν 12 ε2T Þ 2 E1 ðεT1 1 ν 12 εT2 Þ
σ2 5 E2 ðε2T 1 ν 21 ε1T Þ 2 E2 ðεT2 1 ν 21 εT1 Þ
τ 12 5 G12 γ 12T
(5.14)
where E1;2 5 E1;2 =ð1 2 ν 12 ν 21 Þ.
To describe the thermoelastic behavior of a ply, apply the first-order micromechanical model
shown in Fig. 1.34. Since the CTE (and elastic constants) of some fibers can be different in the longitudinal and transverse directions, generalize the first two equations of Eqs. (1.63) as
5.1 TEMPERATURE EFFECTS
1
ðσf ;m 2 ν f 2;m σf2;m Þ 1 αf 1;m ΔT
Ef 1;m 1
1
ðσf ;m 2 ν f 1;m σf1;m Þ 1 αf 2;m ΔT
εf2T;m 5
Ef 2;m 2
301
εf1T;m 5
(5.15)
Repeating the derivation of Eqs. (1.76)(1.79), we arrive at
E1 5 Ef 1 vf 1 Em vm ; ν 21 5 ν f 2 vf 1 ν m vm
1
vf
vm
ν2
5 ð1 2 ν f 1 ν f 2 Þ
1 ð1 2 ν 2m Þ
1 21
E2
Ef 2
Em
E1
1
α1 5
ðEf 1 αf 1 vf 1 Em αm vm Þ
E1
α2 5 ðαf 2 1 ν f 2 αf 1 Þvf 1 ð1 1 ν m Þαm vm 2 ν 21 α1
(5.16)
These equations generalize Eqs. (1.76)(1.79) for the case of anisotropic fibers and specify the
apparent CTE of a unidirectional ply.
As an example, consider the high-modulus carbon-epoxy composite tested by Rogers, Phillips,
and Kingston-Lee (1977). The microstructural parameters for this material are as follows:
Ef 1 5 411 GPa; Ef 2 5 6:6 GPa; ν f 1 5 0:06
ν f 2 5 0:35; αf 1 5 2 1:2 3 1026 1=3 C; αf 2 5 27:3 3 1026 1=3 C
Em 5 5:7 GPa; ν m 5 0:316; αm 5 45 3 1026 1=3 C; vf 5 vm 5 0:5
For these properties, Eqs. (5.16) yield
E1 5 208:3 GPa; E2 5 6:5 GPa; ν 21 5 0:33
α1 5 2 0:57 3 1026 1=3 C; α2 5 43:4 3 1026 1=3 C
whereas the experimental results were
E1 5 208:6 GPa; E2 5 6:3 GPa; ν 21 5 0:33
α1 5 2 0:5 3 1026 1=3 C; α2 5 29:3 3 1026 1=3 C
Thus, it can be concluded that the first-order microstructural model provides good results for
the longitudinal material characteristics, but fails to predict α2 with the required accuracy. The discussion and conclusions concerning this problem presented in Section 1.3 for elastic constants are
valid for thermal expansion coefficients as well. For practical applications, α1 and α2 are normally
determined by experimental methods. However, in contrast to the elasticity problem for which the
knowledge of experimental elastic constants and material strength excludes consideration of the
micromechanical models, for the thermoelasticity problems these models provide us with useful
information even if we know the experimental thermal expansion coefficients. Indeed, consider a
unidirectional ply that is subjected to uniform heating that induces only thermal strains, that is,
ε1T 5 εT1 , ε2T 5 εT2 , γ 12T 5 0. Then, Eqs. (5.14) yield σ1 5 0, σ2 5 0, τ 12 5 0. For homogeneous
materials, this means that no stresses occur under uniform heating. However, this is not the case for
a composite ply. Generalizing Eqs. (1.74) that specify longitudinal stresses in the fibers and in the
matrix, we obtain
σf1 5 Ef 1 ðα1 2 αf 1 ÞΔT;
σm
1 5 Em ðα1 2 αm ÞΔT
302
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
where α1 and α2 are specified by Eqs. (5.16). Thus, because the thermal expansion coefficients of
the fibers and the matrix are different from those of the material, there exist microstructural thermal
stresses in the composite structural elements. These stresses are self-balanced.
Indeed,
σ1 5 σf1 vf 1 σm
1 vm 5 0
Consider an orthotropic layer referred to coordinate axes x, y making angle φ with the principal
material coordinate axes (see Fig. 5.2). Using Eqs. (5.14) instead of Eqs. (2.56) and repeating the
derivation of Eqs. (2.71), we arrive at
σx 5 A11 εxT 1 A12 εyT 1 A14 γ xyT 2 AT11
σy 5 A21 εxT 1 A22 εyT 1 A24 γ xyT 2 AT22
(5.17)
τ xy 5 A41 εxT 1 A42 εyT 1 A44 γ xyT 2 AT12
where Amn are specified by Eqs. (2.72) and the thermal terms are
AT11 5 E1 εT12 cos2 φ 1 E2 εT21 sin2 φ
AT22 5 E1 εT12 sin2 φ 1 E2 εT21 cos2 φ
(5.18)
AT12 5 ðE1 εT12 2 E2 εT21 Þsinφcosφ
Here,
εT12 5 εT1 1 ν 12 εT2 ;
εT21 5 εT2 1 ν 21 εT1
and εT1 ; εT2 are determined by Eqs. (5.13). Expressing the total strains from Eqs. (5.17), we get
εxT 5 εx 1 εTx ;
εyT 5 εy 1 εTy ;
γ xyT 5 γ xy 1 γ Txy
(5.19)
where, εx ; εy , and γ xy are expressed in terms of stresses σx ; σy , and τ xy by Eqs. (2.75),
whereas the thermal strains are
εTx 5 εT1 cos2 φ 1 εT2 sin2 φ
εTy 5 εT1 sin2 φ 1 εT2 cos2 φ
γ Txy 5 ðεT1 2 εT2 Þsin2φ
Introducing thermal expansion coefficients in the xy coordinate frame with the following
equations
εTx 5 αx ΔT;
εTy 5 αy ΔT;
γ Txy 5 αxy ΔT
(5.20)
and using Eqs. (5.13), we obtain
αx 5 α1 cos2 φ 1 α2 sin2 φ
αy 5 α1 sin2 φ 1 α2 cos2 φ
αxy 5 ðα1 2 α2 Þsin2φ
(5.21)
It follows from Eqs. (5.19) that, in an anisotropic layer, uniform heating induces not only normal strains, but also a shear thermal strain. As can be seen in Fig. 5.4, Eqs. (5.21) provide fair
agreement with the experimental results of Barnes et al. (1989) for composites with carbon fibers
and thermoplastic matrix (dashed line and light circles).
5.1 TEMPERATURE EFFECTS
303
10 6 α x , 1/°C
30
20
10
φ°
0
15
45
60
75
90
–10
FIGURE 5.4
Calculated (lines) and experimental (circles) dependencies of thermal expansion coefficients on the ply
, ) and a 6 φ angle-ply layer
orientation angle for unidirectional thermoplastic carbon composite (
, K).
(
Consider a symmetric 6 φ angle-ply layer (see Section 2.5.1). This layer is orthotropic, and the
corresponding constitutive equations of thermoelasticity have the form of Eqs. (5.17) in which
A14 5 A41 5 0, A24 5 A42 5 0, and AT12 5 0. Expressing the total strains, we get
εxT 5 εx 1 εTx ;
εyT 5 εy 1 εTy ;
γ xyT 5 γ xy
where εx ; εy , and γ xy are expressed in terms of stresses by Eqs. (2.146), whereas the thermal
strains are
εTx 5
AT11 A22 2 AT22 A12
;
A11 A22 2 A212
εTy 5
AT22 A11 2 AT11 A12
A11 A22 2 A212
Using Eqs. (2.147), (5.13), (5.18), and (5.20), we arrive at the following expressions for apparent thermal expansion coefficients
αx 5
1 T
ða 2 ν yx aT22 Þ;
Ex 11
αy 5
1 T
ða 2 ν xy aT11 Þ
Ey 22
(5.22)
in which
aT11 5 E1 ðα1 1 ν 12 α2 Þcos2 φ 1 E2 ðα2 1 ν 21 α1 Þsin2 φ
aT22 5 E1 ðα1 1 ν 12 α2 Þsin2 φ 1 E2 ðα2 1 ν 21 α1 Þcos2 φ
Comparison of αx with the experimental results of Barnes et al. (1989) for a thermoplastic carbon composite is presented in Fig. 5.4 (solid line and dots). As can be seen in this figure, there
exists an interval ð0 # φ # 403 Þwithin which the coefficient αx of the angle-ply layer is negative.
304
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
10 6 α x , 1/°C
80
60
40
20
φ°
0
30
45
60
75
90
–20
FIGURE 5.5
Calculated (line) and experimental (circles) dependencies of thermal expansion coefficient on the ply orientation
angle for an aramid-epoxy 6 φ angle-ply layer.
The same type of behavior is demonstrated by aramid-epoxy angle-ply composites. A comparison
of calculated values based on Eqs. (5.22) with the experimental results of Strife and Prevo (1979)
is presented in Fig. 5.5. Looking at Figs. 5.4 and 5.5, we can hypothesize that by supplementing an
angle-ply laminate with plies having small thermal elongations in the x direction, we can synthesize
composite materials with zero thermal expansion in this direction. Such materials are important, for
example, for space telescopes (Fig. 5.3), antennas, measuring instruments, and other high precision,
thermally stable structures (Hamilton & Patterson, 1993).
Consider laminates with arbitrary structural parameters (see Chapter 3). Repeating the derivation of Eqs. (3.5) and using the thermoelasticity constitutive equations, Eqs. (5.17), instead of
Eqs. (2.71), we arrive at
T
Nx 5 B11 ε0xT 1 B12 ε0yT 1 B14 γ 0xyT 1 C11 κxT 1 C12 κyT 1 C14 κxyT 2 N11
T
Ny 5 B21 ε0xT 1 B22 ε0yT 1 B24 γ 0xyT 1 C21 κxT 1 C22 κyT 1 C24 κxyT 2 N22
T
Nxy 5 B41 ε0xT 1 B42 ε0yT 1 B44 γ 0xyT 1 C41 κxT 1 C42 κyT 1 C44 κxyT 2 N12
T
Mx 5 C11 ε0xT 1 C12 ε0yT 1 C14 γ 0xyT 1 D11 κxT 1 D12 κyT 1 D14 κxyT 2 M11
(5.23)
T
My 5 C21 ε0xT 1 C22 ε0yT 1 C24 γ 0xyT 1 D21 κxT 1 D22 κyT 1 D24 κxyT 2 M22
T
Mxy 5 C41 ε0xT 1 C42 ε0yT 1 C44 γ 0xyT 1 D41 κxT 1 D42 κyT 1 D44 κxyT 2 M12
These equations should be supplemented with Eqs. (3.15) for transverse shear forces, that is,
Vx 5 S55 γ xT 1 S56 γ yT ; Vy 5 S65 γ xT 1 S66 γ yT
(5.24)
5.1 TEMPERATURE EFFECTS
305
The temperature terms entering Eqs. (5.23) have the following form
ðs
T
Nmn
5
ðs
T
Mmn
5
ATmn dz;
2e
ATmn z dz
2e
where ATmn are specified by Eqs. (5.18). Performing the transformation that is used in Section 3.1 to
reduce Eqs. (3.6), (3.7), and (3.8) to (3.28) and (3.29), we get
T
ð0Þ
Nmn
5 Jmn
;
T
ð1Þ
ð0Þ
Mmn
5 Jmn
2 eJmn
(5.25)
Here (see Fig. 3.8),
ðrÞ
Jmn
5
ðh
ATmn tr dt;
(5.26)
0
where r 5 0; 1 and mn 5 11; 12; 22:
For a laminate, the temperature governed by Eq. (5.6) is linearly distributed over the layers’
thicknesses (see Fig. 5.1). The same law can obviously be assumed for the temperature coefficients
in Eqs. (5.18), that is, for the ith layer in Fig. 3.10
T
ATi
mn 5 ðAmn Þi21 1
1 T
ðAmn Þi 2 ðATmn Þi21 ðt 2 ti21 Þ
hi
where ðATmn Þi21 5 ATmn ðt 5 ti21 Þ and ðATmn Þi 5 ATmn ðt 5 ti Þ. Then, Eq. (5.26) takes the form
ðrÞ
Jmn
5
k
X
1 i51
hi
ðATmn Þi21 ti 2 ðATmn Þi ti21
r11
r12
tir11 2 ti21
tr12 2 ti21
1 ðATmn Þi 2 ðATmn Þi21 i
r11
r12
If the temperature variation over the thickness of the ith layer can be neglected, we can introduce some average value
T
ðAmn Þi 5
1 T
ðAmn Þi21 1 ðATmn Þi
2
and present Eq. (5.26) as
ðrÞ
Jmn
5
k
1 X
T
r11
ðA Þ ðtr11 2 ti21
Þ
r 1 1 i51 mn i i
(5.27)
The total (elastic and thermal) generalized strains εT ; γ T , and κT in Eqs. (5.23) and (5.24) can
be expressed in terms of the displacements and rotational angles of the laminate element with the
aid of Eqs. (3.3) and (3.14), that is,
@u
;
@x
ε0yT 5
@v
;
@y
γ 0xyT 5
@u @v
1
@y
@x
(5.28)
@θx
;
@x
κyT 5
@θy
;
@y
κxyT 5
@θx
@θy
1
@y
@x
(5.29)
@w
@y
(5.30)
ε0xT 5
κxT 5
γ xT 5 θx 1
@w
;
@x
γ yT 5 θy 1
It follows from Eqs. (5.23) that, in the general case, uniform heating of laminates induces, in contrast to homogeneous materials, not only in-plane strains but also changes to the laminate curvatures
306
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
and twist. Indeed, assume that the laminate is free from edge and surface loads so that the forces and
moments in the left-hand sides of Eqs. (5.23) are equal to zero. Since the CTE of the layers, in the
general case, are different, the thermal terms N T and M T in the right-hand sides of Eqs. (5.23) are
not equal to zero even for a uniform temperature field, and these equations enable us to find εT ; γ T ,
and κT specifying the laminate in-plane and out-of-plane deformation. Moreover, using the approach
described in Section 3.11, we can conclude that uniform heating of the laminate is accompanied, in
the general case, by stresses acting in the layers and between the layers.
As an example, consider the four-layered structure of the space telescope described in
Section 5.1.1.
First, we calculate the stiffness coefficients of the layers, that is,
•
•
for the internal layer of aluminum foil,
ð1Þ
Að1Þ
11 5 A22 5 E f 5 76:92 GPa;
Að1Þ
12 5 ν f Ef 5 23:08 GPa
ð2Þ
Að2Þ
11 5 A22 5 E x 5 34:87 GPa;
i
Að2Þ
12 5 ν xy E x 5 5:23 GPa
for the inner skin,
i
•
for the lattice layer,
Að3Þ
11 5 2Er
δr
cos4 φr 5 14:4 GPa
ar
Að3Þ
22 5 2Er
δr 4
sin φr 5 0:25 GPa
ar
Að3Þ
12 5 2Er
•
i
δr 2
sin φcos2 φ 5 1:91 GPa
ar
for the external skin,
4
4
e
e
2
2
Að4Þ
11 5 E 1 cos φe 1 E2 sin φe 1 2ðE1 ν 12 1 2G12 Þsin φe cos φe 5 99:05 GPa
e
e
e
4
4
e
e
2
2
Að4Þ
22 5 E 1 sin φe 1 E2 cos φe 1 2ðE1 ν 12 1 2G12 Þsin φe cos φe 5 13:39 GPa
e
e
e
e
e
e
2
2
Að4Þ
12 5 E 1 ν 12 1 ½E1 1 E2 2 2ðE1 ν 12 1 2G12 Þsin φe cos φe 5 13:96 GPa
e
e
e
e
Using Eqs. (5.18), we find the thermal coefficients of the layers (the temperature is uniformly
distributed over the laminate thickness)
ðAT11 Þ1 5 ðAT22 Þ1 5 Ef αf ΔT 5 1715 3 1026 ΔT GPa=3 C
ðAT11 Þ2 5 ðAT22 Þ2 5 Ex ð1 1 ν ixy Þαix ΔT 5 32:08 3 1026 ΔT GPa=3 C
i
δr
αr cos2 φr ΔT 5 4:46 3 1026 ΔT GPa=3 C
ar
δr
ðAT22 Þ3 5 2Er αr sin2 φr ΔT 5 1:06 3 1026 ΔT GPa=3 C
ar
e
e
ðAT11 Þ4 5 ½E1 ðαe1 1 ν e12 αe2 Þcos2 φ 1 E2 ðαe2 1 ν e21 αe1 Þsin2 φΔT
ðAT11 Þ3 5 2Er
5 132:43 3 1026 ΔT GPa=3 C
e
e
ðAT22 Þ4 5 ½E1 ðαe1 1 ν e12 αe2 Þsin2 φ 1 E2 ðαe2 1 ν e21 αe1 Þcos2 φΔT
5 317:61 3 1026 ΔT GPa=3 C
5.1 TEMPERATURE EFFECTS
307
Since the layers are orthotropic, AT12 5 0 for all of them. Specifying the coordinates of the layers
(see Fig. 3.10), that is,
t0 5 0 mm, t1 5 0:02 mm, t2 5 1:02 mm, t3 5 10:02 mm, t4 5 13:52 mmand applying
ðrÞ
Eq. (5.27), we calculate the parameters Jmn
for the laminate
ð0Þ
J11
5 ðAT11 Þ1 ðt1 2 t0 Þ 1 ðAT11 Þ2 ðt2 2 t1 Þ 1 ðAT11 Þ3 ðt3 2 t2 Þ
1 ðAT11 Þ4 ðt4 2 t3 Þ 5 570 3 1026 ΔT GPa mm=3 C
ð0Þ
5 1190 3 1026 ΔT GPa mm=3 C
J22
"
1
ð1Þ
ðAT11 Þ1 ðt12 2 t02 Þ 1 ðAT11 Þ2 ðt22 2 t12 Þ
5
J11
2
#
1 ðAT11 Þ3 ðt32 2 t22 Þ 1 ðAT11 Þ4 ðt42 2 t32 Þ 5 5690 3 1026 ΔT GPa mm=3 C
ð1Þ
J22
5 13150 3 1026 ΔT GPa mm=3 C
T
To determine Mmn
, we need to specify the reference surface of the laminate. Assume that this
surface coincides with the middle surface, that is, that e 5 h=2 5 6:76 mm. Then, Eqs. (5.25) yield
ð0Þ
T
N11
5 J11
5 570 3 1026 ΔT GPa mm=3 C
ð0Þ
T
N22
5 J22
5 1190 3 1026 ΔT GPa mm=3 C
ð1Þ
ð0Þ
T
5 J11
2 eJ11
5 1840 3 1026 ΔT GPa mm=3 C
M11
T
5 5100 3 1026 ΔT GPa mm=3 C
M22
Thus, the thermal terms in the constitutive equations of thermoelasticity, Eqs. (5.23), are specified. Using these results, we can determine the apparent CTEs for the space telescope section under
study (see Fig. 5.3). We can assume that, under uniform heating, the curvatures do not change in
the middle part of the cylinder so that κxT 5 0 and κyT 5 0. Since there are no external loads, the
free body diagram enables us to conclude that Nx 5 0 and Ny 5 0. As a result, the first two equations of Eqs. (5.23) for the structure under study become
T
B11 ε0xT 1 B12 ε0yT 5 N11
T
B21 ε0xT 1 B22 ε0yT 5 N22
Solving these equations for thermal strains and taking into account Eqs. (5.20), we get
1
T
T
2 B12 N22
Þ 5 αx ΔT
ðB22 N11
B
1
T
T
2 B12 N11
Þ 5 αy ΔT
ε0yT 5 ðB11 N22
B
ε0xT 5
where B 5 B11 B22 2 B212 . For the laminate under study, calculation yields
αx 5 2 0:94 3 1026 1=3 C;
αy 5 14:7 3 1026 1=3 C
Return to Eqs. (5.13) and (5.20) based on the assumption that the CTEs do not depend on temperature. For moderate temperatures, this is a reasonable approximation. This conclusion follows
from Fig. 5.6, in which the experimental results of Sukhanov, Lapotkin, Artemchuk, and Sobol’ (1990)
308
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
105ε xT
30
90°
20
± 10°
± 45°
10
100
0°
–100
–50
50
–10
± 45°
0°
T ,°C
± 10°
–20
–30
90°
FIGURE 5.6
Experimental dependencies of thermal strains on temperature (solid lines) for 6 φ angle-ply carbon-epoxy
composite and the corresponding linear approximations (dashed lines).
(shown by solid lines) are compared with Eqs. (5.20), in which ΔT 5 T 2 203 C (dashed lines) represent carbon-epoxy angle-ply laminates. However, for relatively high temperatures, some deviation
from linear behavior can be observed. In this case, Eqs. (5.13) and (5.20) for thermal strains can be
generalized as
ðT
ε 5
αðTÞ dT
T
T0
Temperature variations can also result in a change in material mechanical properties. As follows
from Fig. 5.7 in which the circles correspond to the experimental data of Ha and Springer (1987),
elevated temperatures result in either higher or lower reduction of material strength and stiffness
characteristics, depending on whether the corresponding material characteristic is controlled mainly
by the fibers or by the matrix. The curves presented in Fig. 5.7 correspond to a carbon-epoxy composite, but they are typical for polymeric unidirectional composites. The longitudinal modulus and
tensile strength, being controlled by the fibers, are less sensitive to temperature than longitudinal
compressive strength, and transverse and shear characteristics. Analogous results for a more temperature sensitive thermoplastic composite studied by Soutis and Turkmen (1993) are presented in
Fig. 5.8. Metal matrix composites demonstrate much higher thermal resistance, whereas ceramic
and carboncarbon composites have been specifically developed to withstand high temperatures.
For example, carboncarbon fabric composite under heating up to 2500 C demonstrates only a 7%
reduction in tensile strength and about 30% reduction in compressive strength without significant
change of stiffness.
5.1 TEMPERATURE EFFECTS
309
1
E1
σ 1+
0.8
σ 1–
E2
τ 12
G 12
0.6
σ 2+
0.4
0.2
0
0
50
100
150
200
T , °C
FIGURE 5.7
Experimental dependencies of normalized stiffness (solid lines) and strength (dashed lines) characteristics of
unidirectional carbon-epoxy composite on temperature.
1
0.8
E1
0.6
σ 1+
0.4
E2
0.2
σ 2+
σ 1–
T , °C
0
0
40
80
120
FIGURE 5.8
Experimental dependencies of normalized stiffness (solid lines) and strength (dashed lines) characteristics of
unidirectional glass-polypropylene composite on temperature.
Analysis of thermoelastic deformation for materials whose stiffness characteristics depend on
temperature presents substantial difficulties because total strains are caused not only by material
thermal expansion, but also by external forces. Consider, for example, a structural element under
temperature T0 loaded with some external force P0 and assume that the temperature is increased to
310
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
a value T1 . Then, the temperature change will cause a thermal strain associated with material
expansion, and the force P0 , being constant, also induces additional strain because the material
stiffness at temperature T1 is less than its stiffness at temperature T0 . To determine the final stress
and strain state of the structure, we should model the process of loading and heating using, for
example, the method of successive loading (and heating) presented in Section 2.1.2.
5.2 HYGROTHERMAL EFFECTS AND AGING
Effects that are similar to temperature variations, that is, expansion and degradation of properties,
can also be caused by moisture. Moisture absorption is governed by Fick’s law, which is analogous
to Fourier’s law, Eq. (5.1), for thermal conductivity, that is,
qW 5 2 D
@W
@n
(5.31)
in which qW is the diffusion flow through a unit area of surface with normal n, D is the diffusivity
of the material whose moisture absorption is being considered, and W is the relative mass moisture
concentration in the material, that is,
W5
Δm
m
(5.32)
where Δm is the increase in the mass of a unit volume material element due to moisture absorption
and m is the mass of the dry material element. The moisture distribution in the material is governed
by the following equation, similar to Eq. (5.2) for thermal conductivity
@
@W
@W
D
5
@n
@n
@t
(5.33)
Consider a laminated composite material shown in Fig. 5.9 for which n coincides with the z
axis. Despite the formal correspondence between Eq. (5.2) for thermal conductivity and Eq. (5.32)
for moisture diffusion, there is a difference in principle between these problems. This difference is
associated with the diffusivity coefficient D, which is much lower than the thermal conductivity λ
(A)
z
(B)
z
Wm
h
h
x
Wm
x
Wm
FIGURE 5.9
Composite material exposed to moisture on (A) both surfaces z 5 0 and z 5 h, and (B) on the surface z 5 0 only.
5.2 HYGROTHERMAL EFFECTS AND AGING
311
of the same material. As is known, there are materials, for example, metals, with relatively high λ
and practically zero D coefficients. Low D-value means that moisture diffusion is a rather slow process. As shown by Shen and Springer (1976), the temperature increase in time inside a surfaceheated composite material reaches a steady (equilibrium) state temperature about 106 times faster
than the moisture content approaching the corresponding stable state. This means that, in contrast
to Section 5.1.1 in which the steady (time-independent) temperature distribution is studied, we
must consider the time-dependent process of moisture diffusion. To simplify the problem, we can
neglect the possible variation of the mass diffusion coefficient D over the laminate thickness, taking
D 5 constant for polymeric composites. Then, Eq. (5.33) reduces to
D
@2 W
@W
5
@z2
@t
(5.34)
Consider the laminate in Fig. 5.9A. Introduce the maximum moisture content Wm that can exist
in the material under the preassigned environmental conditions. Clearly, Wm depends on the material nature and structure, temperature, relative humidity (RH) of the gas (e.g., humid air), or on the
nature of the liquid (distilled water, salt water, fuel, lubricating oil, etc.) to the action of which the
material is exposed. Introduce also the normalized moisture concentration as
w ðz; tÞ 5
Wðz; tÞ
Wm
(5.35)
Obviously, for t-N, we have w-1. Then, the function w(z, t) can be presented in the form
wðz; tÞ 5 1 2
N
X
wn ðzÞe2kn t
(5.36)
n51
Substitution into Eq. (5.34), taking into account Eq. (5.35), yields the following ordinary differential equation
00
wn 1 rn2 wn 5 0
in which rn2 5 kn =D and ð ::: Þ0 5 dð Þ=dz. The general solution is
wn 5 C1n sinrn z 1 C2n cosrn z
The integration constants can be found from the boundary conditions on the surfaces z 5 0 and
z 5 h (see Fig. 5.9A). Assume that on these surfaces W 5 Wm or w 5 1. Then, in accordance with
Eq. (5.36), we get
wn ð0; tÞ 5 0;
wn ðh; tÞ 5 0
(5.37)
The first of these conditions yields C2n 5 0, whereas from the second condition we have
sinrn h 5 0, which yields
rn h 5 ð2n 2 1Þπ
ðn 5 1; 2; 3; :::Þ
(5.38)
Thus, the solution in Eq. (5.36) takes the form
wðz; tÞ 5 1 2
N
X
n51
C1n sin
" #
2n 2 1
2n21 2 2
πzexp 2
π Dt
h
h
(5.39)
312
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
To determine C1n , we must use the initial condition, according to which
wð0 , z , h; t 5 0Þ 5 0
Using the following Fourier series:
15
N
4X
sinð2n 2 1Þzπ
π n51
2n 2 1
we get C1n 5 4=ð2n 2 1Þπ, and the solution in Eq. (5.39) can be written in its final form
" #
N
4X
sinð2n 2 1Þπz
2n21 2 2
exp 2
wðz; tÞ 5 1 2
π Dt
π n51
2n 2 1
h
(5.40)
where z 5 z=h.
For the structure in Fig. 5.9B, the surface z 5 h is not exposed to moisture, and hence
qW ðz 5 hÞ 5 0. So, in accordance with Eq. (5.31), the second boundary condition in Eqs. (5.37)
must be changed to w0 ðh; tÞ 5 0. Then, instead of Eq. (5.38), we must use
rn h 5
π
ð2n 2 1Þ
2
Comparing this result with Eq. (5.38), we can conclude that for the laminate in Fig. 5.9B, wðz; tÞ
is specified by the solution in Eq. (5.40) in which we must change h to 2h.
The mass increase of the material with thickness h is
ðh
ΔM 5 A Δm dz
0
where A is the surface area. Using Eqs. (5.32) and (5.35), we get
ðh
ΔM 5 AmWm w dz
0
Switching to a dimensionless variable z 5 z=h and taking the total moisture content as
C5
ΔM
Amh
(5.41)
we arrive at
ð1
C 5 Wm w dz
0
where w is specified by Eq. (5.40). Substitution of this equation and integration yields
" #
N
C
8 X
1
2n21 2 2
C5
512 2
exp 2
π Dt
Wm
π n51 ð2n21Þ2
h
For numerical analysis, consider a carbon-epoxy laminate for which D 5 1023 mm2 =h
(5.42)
5.2 HYGROTHERMAL EFFECTS AND AGING
313
(Tsai, 1987) and h 5 1 mm. The distributions of the moisture concentration over the laminate
thickness are shown in Fig. 5.10 for t 5 1; 10; 50; 100; 200; and 500 h. As can be seen, complete
impregnation of 1 mm-thick material takes about 500 h. The dependence of C on t found in accordance with Eq. (5.42) is presented in Fig. 5.11.
An
p
ffi interesting interpretation of the curve in Fig. 5.11 can be noted if we change the variable t
to t. The resulting dependence is shown in Fig. 5.12. As can be seen, the initial part of the curve
is close to a straight line whose slope can be used to determine the diffusion coefficient of the
material matching the theoretical dependence C(t) with the experimental one. Note that experimental methods usually result in rather approximate evaluation of the material diffusivity D with possible variations up to 100% (Tsai, 1987). The maximum value of the function C(t) to which it tends
to approach determines the maximum moisture content Cm 5 Wm .
Thus, the material behavior under the action of moisture is specified by two experimental parameters—D and Cm —which can depend on the ambient media, its moisture content, and temperature. The experimental dependencies of C in Eq. (5.41) on t for 0.6 mm thick carbon-epoxy
composite exposed to humid air with various RH levels are shown in Fig. 5.13. As can be seen, the
moisture content is approximately proportional to the air humidity. The gradients of the curves in
Fig. 5.13 depend on the laminate thickness (see Fig. 5.14).
Among polymeric composites, the highest capacity for moisture absorption under room temperature is demonstrated by aramid composites ð7 6 0:25% by weight) in which both the polymeric
matrix and fibers are susceptible to moisture. Glass and carbon polymeric composites are
w( z, t )
1
t = 500 h
0.8
200
0.6
100
0.4
50
0.2
10
1
0
0
0.2
0.4
0.6
0.8
1
z
FIGURE 5.10
Distribution of the normalized moisture concentration w over the thickness of 1 mm thick carbon-epoxy
composite for various exposure times t.
314
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
C (t )
1
0.8
0.6
0.4
0.2
0
t, h
0
100
200
300
400
500
FIGURE 5.11
Dependence of the normalized moisture concentration C on time t.
C( t )
1
0.8
0.6
0.4
0.2
t, h
0
0
5
10
FIGURE 5.12
Dependence of the normalized moisture concentration on
15
20
25
pffi
t.
characterized with moisture content 3:5 6 0:2% and 2 6 0:75%, respectively. In real aramid-epoxy
and carbon-epoxy composite structures, the moisture content is usually about 2% and 1%, respectively. The lowest susceptibility to moisture is demonstrated by boron composites. Metal matrix,
ceramic, and carboncarbon composites are not affected by moisture.
5.2 HYGROTHERMAL EFFECTS AND AGING
315
C (t ), %
1.6
3
1.2
2
0.8
1
0.4
t, h
0
0
10
20
30
40
50
FIGURE 5.13
Dependence of the moisture content on time for a carbon-epoxy composite exposed to air with 45% RH (1), 75%
RH (2), and 95% RH (3).
C (t ), %
1.6
3
1.2
0.8
2
0.4
1
t, h
0
0
10
20
30
40
50
FIGURE 5.14
Dependencies of the moisture content on time for a carbon-epoxy composite with thickness 3.6 mm (1), 1.2 mm
(2), and 0.6 mm (3) exposed to humid air with 75% RH.
The material diffusivity coefficient D depends on temperature in accordance with the Arrhenius
relationship
DðTa Þ 5
D0
ek=Ta
(Tsai, 1987) in which D0 and k are some material constants and Ta is the absolute temperature.
The experimental dependencies of the moisture content on time in a 1.2 mm thick carbon-epoxy
316
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
C (t ), %
1.2
3
0.8
2
1
0.4
t, h
0
0
10
20
30
40
50
FIGURE 5.15
Dependencies of the moisture content on time for 1.2 mm thick carbon-epoxy composite exposed to humid air
with 95% RH under temperatures 25 C (1), 50 C (2), and 80 C (3).
C (t ), %
16
75°C
12
60°C
40°C
8
T = 20°C
4
t,
0
0
10
20
30
40
h
50
FIGURE 5.16
Moisture content as a function of time and temperature for aramid-epoxy composites.
composite exposed to humid air with 95% RH at various temperatures are presented in Fig. 5.15.
The most pronounced effect of temperature is observed for aramid-epoxy composites. The corresponding experimental results of Milyutin, Bulmanis, Grakova, Popov, and Zakrzhevskii (1989) are
shown in Fig. 5.16.
When a material absorbs moisture it swells, demonstrating effects that are similar to thermal
effects, which can be modeled using the equations presented in Section 5.1.2, if we treat α1 ; α2 ,
and αx ; αy as coefficients of moisture expansion and change ΔT for C. Similar to temperature,
5.2 HYGROTHERMAL EFFECTS AND AGING
317
increase in moisture content reduces material strength and stiffness. For carbon-epoxy composites,
this reduction is about 12%, for aramid-epoxy composites, about 25%, and for glass-epoxy materials, about 35%. After drying out, the effect of moisture usually disappears.
The cyclic action of temperature, moisture, or sun radiation results in material aging, that is, in
degradation of the material properties during the process of material or structure storage. For some
polymeric composites, exposure to elevated temperature, which can reach 703 C, and radiation,
whose intensity can be as high as 1 kW=m2 , can cause more complete curing of the resin and some
increase of material strength in compression, shear, or bending. However, under long-term action
of the aforementioned factors, the material strength and stiffness decrease. To evaluate the effect of
aging, testing under transverse bending (see Fig. 1.66) is usually performed. The flexural strength
obtained
σf 5
3Pl
2bh2
allows for both fiber and matrix material degradation in the process of aging. Experimental results
showing the dependence of the normalized flexural strength on time for advanced composites are
presented in Fig. 5.17. For practical purposes, the effect of aging on composite material strength is
evaluated by testing of specimen after their exposure to temperature 70 C and moisture 96% RH
during n days, each of which simulates 1 year of operation under normal conditions. For carbonepoxy quasi-isotropic composites and n 5 30, the reduction of strength under tension is about 10%,
whereas it is about 15% for the strength under compression. Low temperature (255 C) decreases
the strength under tension by about 5% and practically does not affect the strength under compression. The most dramatic is the effect of aging on the ultimate transverse tensile deformation ε2 of
σf
1
1
2
0.8
3
4
0.6
0.4
0.2
t , year
0
0
1
2
3
4
5
FIGURE 5.17
Dependence of the normalized flexural strength on the time of aging for boron- (1), carbon- (2), aramid- (3), and
glass- (4) epoxy composites.
318
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
unidirectional composites, the low value of which results in cracking of the matrix as discussed in
Sections 2.4.2 and 4.4. After accelerated aging, that is, long-term moisture conditioning at the temperature 703 C, a 0.75% moisture content in carbon-epoxy composites results in about a 20% reduction of ε2 , whereas a 1.15% moisture content causes about a 45% reduction.
Environmental effects on composite materials are discussed in detail elsewhere (Tsai, 1987;
Springer, 1981, 1984, 1988).
5.3 TIME-DEPENDENT LOADING EFFECTS
Polymeric matrices are characterized by pronounced viscoelastic properties resulting in timedependent behavior in polymeric composites as discussed in the following.
5.3.1 VISCOELASTISITY
The viscoelastic properties of composite materials manifest themselves in creep, stress relaxation,
and dependence of the stressstrain diagram on the rate of loading. It should be emphasized that in
composite materials, viscoelastic deformation of the polymeric matrix is restricted by the fibers
that are usually linear elastic and do not demonstrate time-dependent behavior. The one exception
to existing fibers is represented by aramid fibers that are actually polymeric themselves by their
nature. The properties of metal matrix, ceramic, and carboncarbon composites under normal conditions do not depend on time. Rheological (time-dependent) characteristics of structural materials
are revealed in creep tests allowing us to plot the dependence of strain on time under constant
stress. Such diagrams are shown in Fig. 5.18 for the aramid-epoxy composite described by Skudra,
Bulavs, Gurvich, and Kruklinsh (1989). An important characteristic of the material can be established if we plot the so-called isochrone stressstrain diagrams shown in Fig. 5.19. Three curves in
this figure are plotted for t 5 0, t 5 100, and t 5 1000 days, and the points on these curves correspond to points 1, 2, and 3 in Fig. 5.18. As can be seen, the initial parts of the isochrone diagrams
ε1, %
σ 1 = 600 MPa
1.5
σ 1 = 450 MPa
3
1
2
0.5
0
ε1
0
0
σ 1 = 300 MPa
3
2
1
1
200
400
600
800
t , Days (24 h)
1000
FIGURE 5.18
Creep strain response of unidirectional aramid-epoxy composite under tension in longitudinal direction with three
constant stresses.
5.3 TIME-DEPENDENT LOADING EFFECTS
σ1 , MPa
t=0
319
t = 100
600
t = 1000
3
3
2
400
2
1
1
200
ε1 , %
0
0
0.5
1
1.5
FIGURE 5.19
Isochrone stressstrain diagrams corresponding to creep curves in Fig. 5.18.
are linear, which means that under moderate stress, the material under study can be classified as a
linear-viscoelastic material. To characterize such a material, we need to have only one creep diagram, whereby the other curves can be plotted, increasing strains in proportion to stress. For example, the creep curve corresponding to σ1 5 450 MPa in Fig. 5.18 can be obtained if we multiply
strains corresponding to σ1 5 300 MPa by 1.5.
Linear-viscoelastic material behavior is described with reasonable accuracy by the hereditary
theory according to which the dependence of strain on time is expressed as
2
3
ðt
14
σðtÞ 1 Cðt 2 τÞσðτÞ dτ 5
εðtÞ 5
E
(5.43)
0
Here, t is the current time, τ is some moment of time in the past ð0 # τ # tÞ at which stress σðτÞ
acts, and Cðt 2 τÞ is the creep compliance (or creep kernel) depending on time passing from the
moment τ to the moment t. The constitutive equation of hereditary theory, Eq. (5.43), is illustrated
in Fig. 5.20. As can be seen, the total strain εðtÞ is composed of the elastic strain εe governed by
the current stress σðtÞ and the viscous strain εv depending on the loading process as if the material
“remembers” this process. Within the framework of this interpretation, the creep compliance CðθÞ,
where θ 5 t 2 τ, can be treated as some “memory function” that should, obviously, be infinitely
high at θ 5 0 and tend to zero for θ-N as in Fig. 5.21.
The inverse form of Eq. (5.43) is
2
ðt
3
σðtÞ 5 E4εðtÞ 2 Rðt 2 τÞεðτÞ dτ 5
(5.44)
0
Here, Rðt 2 τÞ is the relaxation modulus or the relaxation kernel that can be expressed, as shown
below, in terms of Cðt 2 τÞ.
The creep compliance is determined using experimental creep diagrams. Transforming to a new
variable θ 5 t 2 τ, we can write Eq. (5.43) in the following form:
320
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
σ
σ (τ )
σ (τ )
t
dτ
τ
t
ε
εv
τ
t
σ (t )
εe =
t–τ
E
t
FIGURE 5.20
Geometric interpretation of the hereditary constitutive theory.
C (θ )
θ = t –τ
FIGURE 5.21
Typical form of the creep compliance function.
2
3
ðt
14
σðtÞ 1 CðθÞσðθ 2 tÞ dθ5
εðtÞ 5
E
(5.45)
0
For a creep test, the stress is constant, so σ 5 σ0 , and Eq. (5.45) yields
2
εðtÞ 5 ε0 41 1
ðt
0
3
CðθÞ dθ5
(5.46)
5.3 TIME-DEPENDENT LOADING EFFECTS
321
where ε0 5 σ0 =E 5 εðt 5 0Þ is the instantaneous elastic strain (see Fig. 5.18). Differentiating this
equation with respect to t, we get
CðtÞ 5
1 dεðtÞ
ε0 dt
This expression allows us to determine the creep compliance by differentiating the given experimental creep diagram or its analytical approximation. However, for practical analysis, CðθÞ is usually determined directly from Eq. (5.46) introducing some approximation for CðθÞ and matching
the function obtained εðtÞ with the experimental creep diagram. For this purpose, Eq. (5.46) is
written in the form
εðtÞ
511
ε0
ðt
CðθÞ dθ
(5.47)
0
Experimental creep diagrams for unidirectional glass-epoxy composite are presented in this
form in Fig. 5.22 (solid lines).
The simplest form is an exponential approximation of the type
CðθÞ 5
N
X
An e2αn θ
(5.48)
n51
Substituting Eq. (5.48) into Eq. (5.47), we obtain
N
X
εðtÞ
An
5
1
1
ð1 2 e2αn t Þ
0
ε
α
n51 n
For the curves presented in Fig. 5.22, calculation yields
•
Longitudinal tension: N 5 1, A1 5 0
ε 1 ε 2 γ 12
,
,
ε 10 ε 20 γ 120
2
γ 12 γ 120
1.8
1.6
ε 2 ε 20
1.4
1.2
ε 1 ε 10
t , Days (24 h)
1
0
50
100
150
FIGURE 5.22
Creep strain diagrams for unidirectional glass-epoxy composite (solid lines) under tension in longitudinal
direction ðε1 =ε01 Þ and the transverse direction ðε2 =ε02 Þ, and under in-plane shear ðγ 12 =γ 012 Þ and the corresponding
exponential approximations (dashed lines).
322
•
•
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
Transverse tension: N 5 1, A1 5 0:04, α1 5 0:06 day21
In-plane shear: N 5 2, A1 5 0:033, α1 5 0:04 day21 , A2 5 0:06, α2 5 0:4 day21
The corresponding approximations are shown in Fig. 5.22 with dashed lines. The main shortcoming of the exponential approximation in Eq. (5.48) is associated with the fact that, in contrast
to Fig. 5.21, CðθÞ has no singularity at θ 5 0 which means that it cannot properly describe material
behavior in the vicinity of t 5 0.
It should be emphasized that the one-term exponential approximation corresponds to a simple
rheological mechanical model shown in Fig. 5.23. The model consists of two linear springs simulating material elastic behavior in accordance with Hooke’s law
σ1 5 E1 ε1 ;
σ2 5 E2 ε2
(5.49)
and one dashpot simulating material viscous behavior obeying the Newton flow law
σv 5 η
dεv
dt
(5.50)
Equilibrium and compatibility conditions for the model in Fig. 5.23 are
σ 5 σ 2 1 σv ;
εv 5 ε2 ;
σ1 5 σ
ε1 1 ε2 5 ε
Using the first of these equations and Eqs. (5.49)(5.50), we get
σ 5 E2 ε2 1 η
dεv
dt
Taking into account that
ε2 5 εv 5 ε 2
σ
E
σ,ε
E1, σ1, ε1
E2, σ 2 , ε 2
η ,σv , ε v
σ,ε
FIGURE 5.23
Three-element mechanical model.
5.3 TIME-DEPENDENT LOADING EFFECTS
323
we finally arrive at the following constitutive equation relating the apparent stress σ to the apparent
strain ε
E2
η dσ
dε
1
σ 11
5 E2 ε 1 η
E1
E1 dt
dt
(5.51)
This equation allows us to introduce some useful material characteristics. Indeed, consider a
very fast loading, that is, such that stress σ and strain ε can be neglected in comparison with their
rates. Then, integration yields σ 5 Ei ε, where Ei 5 E1 is the instantaneous modulus of the material.
Now assume that the loading is so slow that stress and strain rates can be neglected. Then,
Eq. (5.51) yields σ 5 El ε, where
El 5
E1 E2
E1 1 E2
(5.52)
is the long-time modulus.
We can now apply the model under study to describe material creep. Taking σ 5 σ0 and integrating Eq. (5.51) with initial condition ε0 ð0Þ 5 σ0 =E, we get
ε5
σ0
E1
11
1 2 eðE2 =ηÞt
E1
E2
The corresponding creep diagram is shown in Fig. 5.24. As follows from this figure,
εðt-NÞ 5 σ0 =El , where El is specified by Eq. (5.52). This means that there exists some limit for
the creep strain, and materials that can be described with this model should possess the so-called
limited creep.
Now assume that the model is loaded in such a way that the apparent strain is constant, that is,
that ε 5 ε0 . Then, the solution of Eq. (5.51) which satisfies the condition σð0Þ 5 E1 ε0 is
σ5
E1 ε0
E2 1 E1 e2t=tr ;
E1 1 E2
tr 5
η
E1 1 E2
The corresponding dependence is presented in Fig. 5.25 and illustrates the process of stress
relaxation. The parameter tr is called the time of relaxation. During this time, the stress decreases
by the factor of e.
ε
σ0
Ei
σ0
El
t
FIGURE 5.24
Creep diagram corresponding to the mechanical model in Fig. 5.23.
324
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
σ
Eiε 0
σ
e
El ε 0
t
tr
FIGURE 5.25
Relaxation diagram corresponding to the mechanical model in Fig. 5.23.
Consider again Eq. (5.51) and express E1 ; E2 , and η in terms of Ei ; El , and tr . The resulting
equation is as follows:
dσ
dε
5 El ε 1 Ei tr
dt
dt
σ 1 tr
(5.53)
This first-order differential equation can be solved for ε in the general case. Omitting rather
cumbersome transformations, we arrive at the following solution
2
3
ðt
14
1
El
εðtÞ 5
σðtÞ 1
12
e2ðEl =Ei tr Þðt2τÞ σðτÞ dτ 5
Ei
tr
Ei
0
This result corresponds to Eq. (5.45) of the hereditary theory with one-term exponential approximation of the creep compliance in Eq. (5.48), in which N 5 1. Taking more terms in Eq. (5.48),
we get more flexibility in the approximation of experimental results with exponential functions.
However, the main features of material behavior are, in principle, the same as that for the one-term
approximation (see Figs. 5.23 and 5.24). In particular, there exists the long-time modulus that follows from Eq. (5.46) if we examine the limit for t-N, that is,
εðtÞ-
σ0
;
El
E
El 5
N
Ð
11
CðθÞ dθ
0
For the exponential approximation in Eq. (5.48),
N
ð
I5
CðθÞ dθ 5
0
N
X
An
n51
αn
Since the integral I has a finite value, the exponential approximation of the creep compliance
can be used only for materials with limited creep. There exist more complicated singular approximations, for example,
CðθÞ 5
A
;
θα
CðθÞ 5
A 2βθ
e
θα
5.3 TIME-DEPENDENT LOADING EFFECTS
325
for which I-N and El 5 0. This means that for such materials, the creep strain can be
infinitely high.
A useful interpretation of the hereditary theory constitutive equations can be constructed with
the aid of the integral Laplace transformation, according to which a function f ðtÞ is associated with
its Laplace transform f ðpÞ as
N
ð
f ðpÞ 5
f ðtÞe2pt dt
0
For some functions that we need to use for the examples presented in the following, we have
f ðtÞ 5 1;
f ðpÞ 5
f ðtÞ 5 e2αt ;
1
p
1
α1p
f ðpÞ 5
(5.54)
The importance of the Laplace transformation for the hereditary theory is associated with the
existence of the so-called convolution theorem, according to which
2t
3
ð
4 f1 ðθÞf2 ðθ2tÞ dθ5 5 f ðpÞf ðpÞ
1
2
0
Using this theorem and applying Laplace transformation to Eq. (5.45), we get
ε ðpÞ 5
1 ½σ ðpÞ 1 C ðpÞσ ðpÞ
E
This result can be presented in a form similar to Hookes’s law, that is,
σ ðpÞ 5 E ðpÞε ðpÞ
(5.55)
where
E 5
E
1 1 C ðpÞ
Applying Laplace transformation to Eq. (5.44), we arrive at Eq. (5.55) in which
E 5 E½1 2 R ðpÞ
(5.56)
Comparing Eqs. (5.55) and (5.56), we can relate Laplace transforms of the creep compliance to
the relaxation modulus, that is,
1
5 1 2 R ðpÞ
1 1 C ðpÞ
Taking into account Eq. (5.55), we can formulate the elasticviscoelastic analogy or the correspondence principle, according to which the solution of the linear viscoelasticity problem can be
obtained in terms of the corresponding Laplace transforms from the solution of the linear elasticity
problem if E is replaced with E and all the stresses, strains, displacements, and external loads are
replaced with their Laplace transforms.
326
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
For an orthotropic material in a plane stress state, for example, for a unidirectional composite
ply or layer referred to the principal material axes, Eqs. (2.55) and (5.43) can be generalized as
2
3
ðt
1 4
ε1 ðtÞ 5
σ1 ðtÞ 1 C11 ðt 2 τÞσ1 ðτÞ dτ 5
E1
0
2
3
ðt
ν 12 4
2
σ2 ðtÞ 1 C12 ðt 2 τÞσ2 ðτÞ dτ 5
E2
0
2
3
1 4
σ2 ðtÞ 1 C22 ðt 2 τÞσ2 ðτÞ dτ 5
ε2 ðtÞ 5
E2
ðt
0
2
3
ðt
ν 21 4
2
σ1 ðtÞ 1 C21 ðt 2 τÞσ1 ðτÞ dτ 5
E1
0
2
3
1 4
γ 12 ðtÞ 5
τ 12 ðtÞ 1 K12 ðt 2 τÞτ 12 ðτÞ dτ 5
G12
ðt
0
Applying Laplace transformation to these equations, we can reduce them to a form similar to
Hooke’s law, Eqs. (2.55), that is,
ε1 ðpÞ 5
σ1 ðpÞ
ν ðpÞ 2 12
σ ðpÞ
E1 ðpÞ
E2 ðpÞ 2
ε2 ðpÞ 5
σ2 ðpÞ
ν 21 ðpÞ 2
σ ðpÞ
E2 ðpÞ
E1 ðpÞ 1
γ 12 ðpÞ 5
(5.57)
τ 12 ðpÞ
G12 ðpÞ
where
E1 ðpÞ 5
E1
ðpÞ ;
1 1 C11
E2 ðpÞ 5
1 1 C12
ðpÞ
ν 12 ðpÞ 5
ðpÞ ν 12 ;
1 1 C22
E2
ðpÞ ;
1 1 C22
G12 ðpÞ 5
1 1 C21
ðpÞ
ν 21 ðpÞ 5
ðpÞ ν 21
1 1 C11
G12
ðpÞ
1 1 K12
(5.58)
For a unidirectional composite ply, the foregoing equations can be simplified by neglecting
material creep in the longitudinal direction ðC11 5 0Þ and assuming that Poisson’s effect is linear
elastic and symmetric, that is, that
ν 12
ν 12
5
;
E2
E2
ν 21
ν 21
5
E1
E1
5.3 TIME-DEPENDENT LOADING EFFECTS
327
Then, Eqs. (5.57) take the form
σ1 ðpÞ ν 12 2
σ ðpÞ
E1
E2 2
σ ðpÞ ν 21 σ ðpÞ
ε2 ðpÞ 5 2 2
E2
E1 1
τ ðpÞ
γ 12 ðpÞ 5 12
G12 ðpÞ
ε1 ðpÞ 5
(5.59)
Supplementing constitutive equations, Eqs. (5.57) or (5.59), with straindisplacement and equilibrium equations written in terms of Laplace transforms of stresses, strains, displacements, and
external loads and solving the problem of elasticity, we can find Laplace transforms for all the variables. To represent the solution obtained in this way in terms of time t, we need to take the inverse
Laplace transformation, and this is the most difficult stage of the problem solution. There exist
exact and approximate analytical and numerical methods for performing inverse Laplace transformation discussed, for example, by Schapery (1974). The most commonly used approach is based
on approximation of the solution written in terms of the transformation parameter p with some
functions for which the inverse Laplace transformation is known.
As an example, consider the problem of torsion for an orthotropic cylindrical shell similar to
that shown in Fig. 4.20. The elastic solution of the problem is presented in Section 4.3. Using the
elasticviscoelastic analogy, we can write the shear strain induced by torque T as
γ xy ðpÞ 5
T ðpÞ
2πR2 B44 ðpÞ
(5.60)
Here, B44 ðpÞ 5 A44 ðpÞh, where h is the shell thickness.
Let the shell be made of glass-epoxy composite whose mechanical properties are listed in
Table 1.5 and creep diagrams are shown in Fig. 5.22. To simplify the analysis, we assume that for
the unidirectional composite under study E2 =E1 5 0:22, G12 =E1 5 0:06, ν 12 5 ν 21 5 0 and introduce
the normalized shear strain
γ 5 γ xy
21
T
R2 hE1
Consider a 6 453 angle-ply material discussed in Section 2.5 for which, with due regard to
Eqs. (2.72), and (5.58), we can write
1
1
E2
E1 1
A44 ðpÞ 5 ðE1 1 E2 Þ 5
ðpÞ
4
4
1 1 C22
Exponential approximation, Eq. (5.48), of the corresponding creep curve in Fig. 5.22 (the lower
dashed line) is
C22 5 A1 e2α1 θ
where A1 5 0:04 and α1 5 0:06 day21 . Using Eqs. (5.54), we arrive at the following Laplace transforms of the creep compliance and the torque which is constant
C22
ðpÞ 5
A1
;
α1 1 p
T ðpÞ 5
T
p
328
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
The final expression for the Laplace transform of the normalized shear strain is
γ ðpÞ 5
2Eðα1 1 A1 1 pÞ
πpðα1 1 A1 E 1 pÞ
(5.61)
where E 5 E1 =ðE1 1 E2 Þ
To use Eqs. (5.54) for the inverse Laplace transformation, we should decompose the right-hand
part of Eq. (5.61) as
γ ðpÞ 5
2E
α1 1 A1
A1 ð1 2 EÞ
2
πðα1 1 A1 EÞ
α1 1 A1 E 1 p
p
Applying Eqs. (5.54), we get
γðtÞ 5
2E
α1 1 A1 2 A1 ð1 2 EÞe2ðα1 1A1 EÞt
πðα1 1 A1 EÞ
This result is demonstrated in Fig. 5.26. As can be seen, there is practically no creep because
the cylinder’s deformation is controlled mainly by the fibers.
Quite different behavior is demonstrated by the cylinder made of 03 =903 cross-ply composite
material discussed in Section 2.4. In accordance with Eqs. (2.114) and (5.58), we have
A44 ðpÞ 5 G12 ðpÞ 5
G12
ðpÞ
1 1 K12
Exponential approximation, Eq. (5.48), of the shear curve in Fig. 5.22 (the upper dashed line)
results in the following equation for the creep compliance
K12 5 A1 e2α1 θ 1 A2 e2α2 θ
in which A1 5 0:033, α1 5 0:04 day21 , A2 5 0:06, α2 5 0:4 day21 . Omitting simple rearrangements,
we finally get
γ5
E1
A1
A2
11
ð1 2 e2α1 t Þ 1
ð1 2 e2α2 t Þ
2πG12
α1
α2
γ
5
0 ° 90 °
4
3
2
± 45°
1
t , Days ( 24 h)
0
0
50
100
150
FIGURE 5.26
Dependencies of the normalized shear strain on time for 03 =903 cross-ply and 6 453 angle-ply glass-epoxy
composite cylinders under torsion.
5.3 TIME-DEPENDENT LOADING EFFECTS
329
The corresponding creep diagram is shown in Fig. 5.26.
Under relatively high stresses, polymeric composites demonstrate nonlinear viscoelastic
behavior. The simplest approach to study nonlinear creep problems is based on experimental
isochrone stressstrain diagrams of the type shown in Fig. 5.19. Using the curves corresponding to time moments t1 , t2 , t3 , etc., we can solve a sequence of nonlinear elasticity
problems for these time moments and thus determine the change of strains and stresses with
time. This approach, sometimes referred to as the aging theory, is approximate and can be
used to study structures loaded with forces that do not change with time, or that change very
slowly.
There exist also several variants of nonlinear hereditary theory described, for example, by
Rabotnov (1980). According to the most common versions, Eq. (5.43) is generalized as
2
ðt
14
σðtÞ 1 C1 ðt 2 τÞσðτÞ dτ
εðtÞ 5
E
ðt ðt
1
0
3
C2 ðt 2 τ 1 ; t 2 τ 2 Þσðτ 1 Þσðτ 2 Þ dτ 1 dτ 2 1 :::5
00
or
εðtÞ 5
X
2
ðt
Ak 4σðtÞ1 Cðt2τÞσðτÞ dτ 5
k
or
3k
0
2
3
ðt
14
f ½εðtÞ 5
σðtÞ 1 Cðt 2 τÞσðτÞ dτ 5
E
0
or
ðt
εðtÞ 5 φ½σðtÞ 1
Cðt 2 τÞψ½σðτÞ dτ
0
In conclusion, it should be noted that correctly designed composite structures (see the next
chapter) in which the material behavior is controlled by the fibers usually do not exhibit pronounced time-dependent behavior. For example, consider the filament-wound glass-epoxy pressure vessel studied in Section 4.3 (see Fig. 4.21 and the second row in Table 4.1 for
parameters of the vessel). The vessel consists of 6 363 helical plies and circumferential plies,
and has structural parameters that are close to optimal. The experimental dependence of circumferential strain on time for step-wise loading with internal pressure p presented in
Fig. 5.27 does not indicate any significant creep deformation. It should be noted that this conclusion is valid for composite materials reinforced with glass or carbon fibers and working
under normal conditions (room temperature) whose creep is mainly associated with the polymeric matrix.
Composite materials reinforced with aramid and other polymeric fibers can demonstrate significant creep associated with creep of the fibers, and under relatively high stress the behavior of these
materials is not linear.
330
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
εy, %
1
P = 6 MPa
0.8
P = 4 MPa
0.6
0.4
P = 2 MPa
0.2
t , Days ( 24 h)
0
0
5
10
15
20
FIGURE 5.27
Dependence of the circumferential strain on time for a glass-epoxy cylindrical pressure vessel loaded in steps
with internal pressure p.
5.3.2 DURABILITY
Composite materials, to be applied to structures with long service life, need to be guaranteed for
the corresponding period of time from failure, which usually occurs as a result of an evolutionary
process of material degradation in the service environment. To provide adequate durability of the
material, we need, in turn, to study its long-term behavior under load and its endurance limits.
The most widely used durability criteria establishing the dependence of material strength on the
time of loading are based on the concept of the accumulation of material damage induced by the
acting stresses and intensified by the degrading influence of service conditions such as temperature and moisture. Particular criteria depend on the accepted models simulating the material damage accumulation. Although there exist microstructural approaches to the durability evaluation of
composite materials (see, e.g., Skudra et al. (1989)), for practical purposes, the experimental
dependencies of the ultimate stresses on the time of their action are usually evaluated. In particular, these experiments allow us to conclude that fibers, which are the major load-carrying elements of composite materials, possess some residual strength σN 5 σðt-NÞ, which is about
30% for glass fibers, 50% for aramid fibers, and 60% for carbon fibers (for t about 15 years)
with respect to the corresponding static strength σ0 5 σðt 5 0Þ. Typical dependencies of the longterm strength of composite materials on time are presented in Fig. 5.28. As can be seen, the time
of loading dramatically affects material strength. However, being unloaded at any moment of
time t, composite materials demonstrate practically the same static strength that they had before
long-term loading.
5.3 TIME-DEPENDENT LOADING EFFECTS
331
σ 1 (t ) / σ 1 (0)
1
0.8
0.6
1
0.4
2
0.2
t , Days (24 h)
0
0
100
200
300
400
500
FIGURE 5.28
Normalized long-term longitudinal strength of aramid-epoxy (1) and glass-epoxy (2) unidirectional composites.
Approximation of the curves shown in Fig. 5.28 can be performed using exponential functions
as follows:
σðtÞ 5 σN 1
X
An e2λn t
(5.62)
n
in which σN ; An and λn are coefficients providing the appropriate approximation. The initial static
strength is
σð0Þ 5 σ0 5 σN 1
X
An
n
The simplest is a one-term approximation
σðtÞ 5 σN 1 ðσ0 2 σN Þe2λt
(5.63)
To approximate the initial part of the curve, we can put σN 5 0 and arrive at the following
equation
σðtÞ 5 σ0 e2λt
(5.64)
Now assume that we can solve Eqs. (5.62), (5.63), or (5.64) for t and find the material durability
td ðσÞ, that is, the time during which the material can withstand stress σ. Consider the process of
loading as a system of k stages such that the duration of each stage is ti and the stress acting at this
stage is σi ði 5 1; 2; 3; ::: ; kÞ. Then, the whole period of time during which the material can
withstand such step-wise loading can be calculated with the aid of the following equation
k
X
ti
51
t
ðσ
Þ
i51 d i
in which td ðσi Þ is the material durability corresponding to stress σi .
332
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
The strength criteria discussed in Chapter 4 can be generalized for the case of long-term loading
if we change the static ultimate stresses in these criteria for the corresponding long-term strength
characteristics.
Experimental study of material durability requires long-term testing that takes considerable
time. This time can be substantially reduced by applying the Zhurkov equation (Zhurkov &
Narzullaev, 1953), which links the material lifetime t with the stress level σ and absolute temperature T as
t 5 t0 Ueð1=RTÞðU0 2γσÞ
(5.65)
in which, originally, t0 is the period of atoms’ thermal oscillations, U0 is the energy activating
material fracture, γ is the coefficient indicating the influence of stress on the fracture energy, and R
is Boltzmann’s constant. However, irrespective of the physical meaning of the forgoing constants,
we can treat them as some coefficients that can be found from direct experiments. For this purpose,
rewrite Eq. (5.65) as
t 5 t0 Ueð1=TÞða2bσÞ
(5.66)
in which t is the time to failure (lifetime), T is the absolute temperature, σ is the stress expressed as
a percentage of the static ultimate stress, and t0 , a, and b are experimental constants. Thus, to determine these constants (t0 , a, and b), we need to perform three tests under different temperatures and
loading levels.
Suppose that we have the results of these tests, that is,
σ 5 σð1Þ ;
T 5 T 1;
t 5 t1
σ 5 σð2Þ ;
T 5 T 2;
t 5 t2
σ 5 σð3Þ ;
T 5 T 3;
t 5 t3
Taking the logarithm of Eq. (5.66) and writing the result for the three test cases, we get
lnti 5 lnt0 1
1
a 2 bσðiÞ
Ti
(5.67)
where i 5 1, 2, 3. Solving Eqs. (5.67) for a, b, and t0 , we find
a5
1 b σð2Þ T3 2 σð3Þ T2 2 T2 T3 ðlnt3 2 lnt2 Þ
T3 2 T2
b 5 T1
lnt1 2 lnt2
σð2Þ 2 σð1Þ
ð2Þ
t0 5 t2 e2ð1=T2 Þða2bσ Þ
Application of Eq. (5.65) to an organic fiber Kevlar 49 unidirectional composite material under
tension along the fibers has been undertaken by Chiao, Sherry, and Hetherington (1977). The composite material with static tensile strength σ 1 5 2280MPa has been tested under various temperatures and the loading levels σ. The results are presented in Fig. 5.29, which demonstrates dramatic
effects of the temperature and stress on the lifetime t.
Using the test results shown in Fig. 5.29, Chiao et al. have arrived at the following form of
Eq. (5.66) (Chiao et al., 1977):
5.3 TIME-DEPENDENT LOADING EFFECTS
333
t, h
(A)
0.5
0 .5
0.4
0.3
0 . 246
0 . 166
0.2
0.1
T °, K
0
373
383
(B)
393
(C)
t, h
t . 10 – 4, h
5
4
5
4
4 . 055
3.69
3
3
1 . 894
2
2
1.93
1 . 267
1
1
T °, K
0
373
383
393
0
373
0.79
T °, K
383
393
FIGURE 5.29
Dependencies of the material lifetime on temperature for loading levels of 80.4% (A), 73.2% (B), and 67.2% (C).
Table 5.2 Predicted (tp) and Experimental (te) Lifetimes under Room Temperature
σ, %
tp , h
te , h
70
8725
8800
75
1108
1150
80
141
150
t 5 7:9 3 10211 eð1=TÞð18;2462123σÞ
85
17.9
7.6
90
2.3
0.8
(5.68)
in which t is measured in hours and σ in percentage. Comparison of the results calculated using
Eq. (5.68) and those obtained from the experiment is presented in Table 5.2 (Chiao et al., 1977).
As can be seen, the approach under consideration provides a fair prediction of the material
lifetime.
334
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
5.3.3 CYCLIC LOADING
Consider the behavior of composite materials under the action of loads periodically changing with
time. For qualitative analysis, first consider a material that can be simulated with the simple
mechanical model shown in Fig. 5.23. Applying stress acting according to the following form:
σðtÞ 5 σ0 sin ωt
(5.69)
where σ0 is the amplitude of stress and ω is the frequency, we can solve Eq. (5.53) that describes
the model under study for strain εðtÞ. The result is
where
εðtÞ 5 ε0 sinðωt 1 θÞ
(5.70)
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 1 tr2 ω2
ε0 5 σ0
2
El 1 Ei2 tr2 ω2
(5.71)
tr ωðEi 2 El Þ
tanθ 5 2
El 1 Ei tr2 ω2
It follows from these equations that a viscoelastic material is characterized by a phase shift of
strain with respect to stress. Eliminating the time variable from Eqs. (5.69) and (5.70), we arrive at
the following relationship between stress and strain:
2 2
σ
ε
σε
1
2 2cosθ
5 sin2 θ
σ0
ε0
σ0 ε0
This is the equation of an ellipse shown in Fig. 5.30A. The absolute value of the area A inside
this ellipse (its sign depends on the direction of integration along the contour) determines the
energy dissipation per single cycle of vibration, that is,
ΔW 5 jAj 5 πσ0 ε0 jsinθj
σ
(A)
(5.72)
σ
(B)
σ0
σ0
ε
ε0
FIGURE 5.30
Stressstrain diagrams for viscoelastic (A) and elastic (B) materials.
ε
ε0
5.3 TIME-DEPENDENT LOADING EFFECTS
335
Following Zinoviev and Ermakov (1994), we can introduce the dissipation factor as the ratio of
energy loss in a loading cycle, ΔW, to the value of the elastic potential energy in a cycle, W, as
ψ5
ΔW
W
where, in accordance with Fig. 5.30B, W 5 ð1=2Þσ0 ε0 . Transforming Eq. (5.72) with the aid of
Eqs. (5.71), we arrive at
ψ5
2πtr ω
El
1
2
1 1 tr2 ω2
Ei
As follows from this equation, ψ depends on the number of oscillations accomplished during
the period of time equal to the material relaxation time, tr , and reaches a maximum value for
tr ω 5 1.
As shown by Zinoviev and Ermakov (1994), for anisotropic materials, the dissipation factor
also depends on the direction of loading. Particularly, for a unidirectional composite ply, referred
to axes x and y and making angle φ with the principal material axes 1 and 2 as in Fig. 2.18, the dissipation factors are
ψ1
ψ
cos2 φ 2 2 sin2 φ cos2φ 1 ψ45 μ12 sin2 φcos2 φ
E1
E2
ψ2
ψ1 2
2
2
2
cos φ 2 sin φ cos2φ 1 ψ45 μ12 sin φcos φ
ψy 5 Ey
E2
E1
2ψ1
2ψ
ψ
1 2 2 ψ45 μ12 sin2 φcos2 φ 1 12 cos2 2φ
ψxy 5 Gxy
E1
E2
G12
ψx 5 Ex
where
μ12 5
1 2 ν 12
1 2 ν 21
1
1
1
G12
E1
E2
Ex ; Ey , and Gxy are specified by Eqs. (2.76) and ψ1 ; ψ2 ; ψ12 ; and ψ45 are the ply dissipation
factors corresponding to loading along the fibers, across the fibers, under in-plane shear and at 453
with respect to principal material axes 1 and 2. As follows from Fig. 5.31, calculations based on
the foregoing equations provide fair agreement with experimental results of Ni and Adams (1984).
Energy dissipation in conjunction with the relatively low heat conductivity of composite materials induces their self-heating during cyclic loading. The dependence of an aramid-epoxy composite
material’s temperature on the number of cycles under tensile and compressive loading with frequency 103 cycles per minute is shown in Fig. 5.32 (Tamuzh & Protasov, 1986).
Under cyclic loading, structural materials experience a fatigue fracture caused by material damage accumulation. As already noted in Section 1.2.4, the heterogeneous structure of composite
materials provides relatively high resistance of these materials to crack propagation resulting in
their specific behavior under cyclic loading. It follows from Fig. 5.33 that stress concentration in
aluminum specimens, which has practically no effect on the material’s static strength due to the
plasticity of aluminum, dramatically reduces its fatigue strength. Conversely, the static strength of
carbon-epoxy composites, which are brittle materials, is reduced by stress concentration that has
low effect on the slope of the fatigue curve. On average, the residual strength of carbon composites
336
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
ψ,%
8
6
4
2
0
0
15
30
45
60
75
90
φ°
FIGURE 5.31
Calculated (lines) and experimental (circles) dependencies of dissipation factor on the ply orientation for glass) and carbon-epoxy (
) unidirectional composites.
epoxy (
T °C
100
75
1
50
25
2
N
0
1 . 105
2 . 105
FIGURE 5.32
Temperature of an aramid-epoxy composite as a function of the number of cycles under tension (1) and
compression (2).
5.3 TIME-DEPENDENT LOADING EFFECTS
337
σ f /σ 0
1
1
1
0.8
2
0.6
2
0.4
0.2
0
0
1
2
3
4
5
6
7
log N
FIGURE 5.33
Typical fatigue diagrams for carbon-epoxy composite (solid lines) and aluminum alloy (dashed lines) specimens
without (1) and with (2) stress concentration (fatigue strength is normalized to static strength of specimens
without stress concentration).
(without stress concentration) after 106 loading cycles is 70%80% of the material’s static
strength, in comparison with 30%40% for aluminum alloys. Moreover, as follows from existing
experiments, cyclic longitudinal compression of unidirectional carbon-epoxy composites with the
stresses that are less than 55% of the static strength does not result in failure for practically unlimited number of cycles. Qualitatively, this comparative evaluation is true for all fibrous composites
that are widely used in structural elements subjected to intensive vibrations, for example, helicopter
rotor blades, airplane propellers, drive shafts, and automobile leaf springs.
The forgoing discussion concerns the fatigue strength of unidirectional composites loaded along
the fibers. However, composites are anisotropic materials having different strengths in different
directions and, naturally, different responses under cyclic loading. As shown in Fig. 5.34 presenting
the approximations of the experimental results given by Tsai (1987), the degradation of material
strength under tension across the fibers (line 2) is much higher than under tension along the fibers
(line 1). Recall that the stress σ2 induces the cracks in the matrix as discussed in Sections 2.4.2
and 4.4.
A typical composite materials fatigue diagram, constructed from the experimental results of
Apinis, Mikelsons, and Khonichev (1991), is shown in Fig. 5.35. Standard fatigue diagrams usually
determine the material strength for 103 # N # 106 and are approximated as
σR 5 a 2 blogN
(5.73)
Here, N is the number of cycles to failure under stress σR , a and b are experimental constants
depending on frequency of cyclic loading, temperature and other environmental factors, and on the
stress ratio R 5 σmin =σmax , where σmax and σmin are the maximum and the minimum stresses. It
338
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
σ 1 / σ 10 , σ 2 / σ 20
1
1
0.8
0.6
2
0.4
0.2
log N
0
0
1
2
3
4
5
6
7
FIGURE 5.34
Normalized fatigue strength of carbon-epoxy composites loaded along (1) and across (2) the fibers.
σ –1 / σ
1
0.8
0.6
0.4
0.2
log N
0
0
1
2
3
4
5
6
7
8
FIGURE 5.35
Normalized fatigue diagram for fabric carboncarbon composite material (σ-static strength): (
experimental part of the diagram loading frequency 6 Hz () and 330 Hz (3); (s s s) extrapolation.
)
5.3 TIME-DEPENDENT LOADING EFFECTS
339
should be taken into account that the results for fatigue tests are characterized, as a rule, with high
scatter.
The factor R specifies the cycle type. The most common bending fatigue test provides a symmetric cycle for which σmin 5 2 σ, σmax 5 σ, and R 5 2 1. A tensile load cycle
ðσmin 5 0; σmax 5 σÞ has R 5 0, whereas a compressive cycle ðσmin 5 2 σ; σmax 5 0Þ has
R- 2N. Cyclic tension with σmax . σmin . 0 corresponds to 0 , R , 1, whereas cyclic compression with 0 . σmax . σmin corresponds to 1 , R , N. Fatigue diagrams for the unidirectional
aramid-epoxy composite studied by Limonov and Anderson (1991) corresponding to various Rvalues are presented in Fig. 5.36. Similar results (Anderson, Mikelsons, Tamuzh, & Tarashuch,
1991) for carbon-epoxy composites are shown in Fig. 5.37.
Since only σ21 is usually available from standard tests under cyclic bending, fatigue strengths
for other load cycles are approximated as
σ21
σR 5 σ21 1 σm 1 2
σt
where σm 5 ðσmin 1 σmax Þ=2 is the mean stress of the load cycle and σt is the material long-term
strength (see Section 5.3.2) for the period of time equal to that of the cyclic loading.
Fabric composites are more susceptible to cyclic loading than materials reinforced with straight
fibers. This fact is illustrated in Fig. 5.38, showing the experimental results of Schulte, Reese, and
Chou (1987).
The foregoing discussion deals with high-cycle fatigue. The initial interval
1 # N # 103 corresponding to so-called low-cycle fatigue is usually studied separately, because the
slope of the approximation in Eq. (5.73) can be different for high stresses. A typical fatigue diagram for this case is shown in Fig. 5.39 (Tamuzh & Protasov, 1986).
Fatigue has also some effect on the stiffness of composite materials. This can be seen in
Fig. 5.40, demonstrating a reduction in the elastic modulus for a glass-fabric-epoxy-phenolic
σ 1R , MPa
1600
1200
R = 0 .1
800
0
400
–∞
–1
log N
0
3
4
5
6
FIGURE 5.36
Fatigue diagrams for unidirectional aramid-epoxy composite loaded along the fibers with various stress ratios.
340
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
σ1R , MPa
1200
R = 0 .1
10
800
–1
400
0
log N
3
4
5
6
FIGURE 5.37
Fatigue diagrams for a unidirectional carbon-epoxy composite loaded along the fibers with various stress ratios.
σ R , MPa
800
1
600
2
400
200
log N
0
3
4
5
6
FIGURE 5.38
Tensile fatigue diagrams for a cross-ply (1) and fabric (2) carbon-epoxy composites.
composite under low-cycle loading (Tamuzh & Protasov, 1986). This effect should be accounted
for in the application of composites to the design of structural members such as automobile leaf
springs that, being subjected to cyclic loading, are designed under stiffness constraints. Stiffness
degradation can be used as an indication of material damage to predict fatigue failure. The most
sensitive characteristic of the stiffness change is the tangent modulus Et 5 dσ=dε: The dependence
of Et on the number of cycles, N, normalized to the number of cycles that cause material fatigue
fracture under the preassigned stress, is presented in Fig. 5.41, corresponding to a 6 453 angle-ply
carbon-epoxy laminate studied by Murakami, Kanagawa, Ishida, and Tsushima (1991).
5.3 TIME-DEPENDENT LOADING EFFECTS
341
σ1R , MPa
2000
1600
1200
800
400
log N
0
0
1
2
3
FIGURE 5.39
Low-cycle fatigue diagram for unidirectional aramid-epoxy composite loaded along the fibers with R 5 0.1.
E, GPa
30
20
10
log N
0
0
1
2
3
FIGURE 5.40
Dependence of elastic modulus of glass-fabric-epoxy-phenolic composite on the number of cycles at stress
σ 5 0:5σ ðσ is the static ultimate stress).
5.3.4 IMPACT LOADING
Thin-walled composite laminates possessing high in-plane strength and stiffness are rather susceptible to damage initiated by transverse impact loads that can cause fiber breakage, cracks in the
matrix, delamination, and even material penetration by the impactor. Depending on the impact
energy determined by the impactor mass and velocity and the properties of laminate, impact
342
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
Et
1
0.8
0.6
0.4
0.2
0
N
0
0.2
0.4
0.6
0.8
1
FIGURE 5.41
Dependence of the tangent modulus normalized to its initial value on the number of cycles related to the ultimate
number corresponding to fatigue failure under stress σmax 5 120 MPa and R 5 2 1 for 6 453 angle-ply carbonepoxy laminate.
loading can result in considerable reduction in material strength under tension, compression, and
shear. One of the most dangerous consequences of an impact loading is an internal delamination in
laminates (so-called barely visible impact damage (BVID)), which can sometimes be hardly noticed
by visual examination.
Impact loading causing delamination of composite materials is critical for design of the laminated skin for modern composite airframe structures with load-carrying skin supported by stringers and rings as shown in Fig. 5.42. To study the impact loading, a special tool shown in
Fig. 5.43 is recommended by ASTM Standard D7136. To model the impact loading, a rectangular 100 3 150 mm composite plate is fixed on a metal plate with a rectangular 75 3 125 mm cutout made in the center. The central part of the composite panel situated above the cutout is
impacted by the steel drop-weight cylinder having a hemispherical nose of 8 mm in diameter.
The mass of the cylinder is mW 5 5:45 kg. The cylinder moves inside the pipe as shown in
Fig. 5.43.
The intensity of impact is specified by the given kinetic energy E0 , which is measured in
joules (1 J 5 1 Nm) and is expressed in terms of the mass m0 and velocity V0 of the actual flying
objects as
1
E0 5 m0 V02
2
(5.74)
The velocity of the drop-weight falling from the height H and the corresponding kinetic
energy are
VW 5
pffiffiffiffiffiffiffiffiffi
2gH ;
EW 5
1
2
mW VW
2
(5.75)
5.3 TIME-DEPENDENT LOADING EFFECTS
343
FIGURE 5.42
Composite airframe structure with load-carrying skin.
The real impact is simulated under condition EW 5 E0 . Then, the velocity and the height of the
drop-weight following from Eqs. (5.74) and (5.75) are calculated as
VW 5 V0
rffiffiffiffiffiffiffi
m0
;
mW
H5
2
VW
2g
After the impact, the compressive strength of the composite plate is found using the test fixture
shown in Fig. 5.44. The fixed metal parts, as well as the upper part that compresses the plate, have
5 mm deep grooves whose width is equal to the plate thickness. The plate is inserted into the
grooves which fix the plate edges.
To perform the impact tests, quasi-isotropic (03 =903 = 6 453 ) symmetric 8 mm thick carbonepoxy plates have been fabricated by filament winding on the flat mandrel and tested as shown in
Figs. 5.43 and 5.44. Relatively high impact energy causes complete delamination of the plate.
The delamination area (black) for the plate after impact with energy E 5 90 J is shown in
Fig. 5.45A, whereas Fig. 5.45B demonstrates the interlaminar cracks in this plate. Reduction of
the impact energy decreases the delamination area. This is illustrated in Fig. 5.46 where the contour of the delamination area is marked by the white line. This delamination was induced by an
impact with energy E 5 35 J and represents the BVID. The depth of the crater on the plate surface
is only 0.3 mm, and this crater can hardly be detected by visual examination of the surface.
Nevertheless, the diameter of the delamination area is about 70 mm, and further testing using the
test rig shown in Fig. 5.44 results in a reduction in the ultimate compressive force by a factor of
four.
The dependence of the diameter of delamination area on the impact area plotted in accordance
with experimental results of Hwang, Kwon, Lee, and Hwang (2000) for 2.3 mm thick carbon-epoxy
plates is shown in Fig. 5.47 (solid line). Extrapolation of the experimental curve to the axis d 5 0
344
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
FIGURE 5.43
The test rig simulating the impact loading of a composite panel.
5.3 TIME-DEPENDENT LOADING EFFECTS
FIGURE 5.44
Composite plate in a clamping fixture for compression test.
(A)
(B)
FIGURE 5.45
Delamination area (black) (A) and interlaminar cracks (B) in the plate after impact with energy E 5 90 J.
345
346
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
FIGURE 5.46
Delamination area of the plate after impact with energy E 5 35 J.
d , mm
50
40
30
20
10
E, J
0
0
5
10
15
20
25
FIGURE 5.47
Dependence of diameter of delamination area on the impact energy for quasi-isotropic 2.3 mm-thick carbon); extrapolation (
).
epoxy plates: experiment (
gives the minimum energy inducing material delamination: Em 5 3 J. This is a very low energy
level, corresponding to a 1 kg mass falling from a 0.3 m height.
It follows from the foregoing results and discussion that laminated composite materials have
rather poor resistance to impact. This conclusion is supported by the experiments with relatively
thin skin, the results of which are presented in Fig. 5.48. The skin with a thickness of 1.6 mm has
been made of carbon fabric by winding. As can be seen in Fig. 5.48, the skin is penetrated by the
steel impactor with an energy exceeding just 15 J.
Thus, the obvious conclusion that composite material should not be used in load-carrying structures subjected to impact suggests itself. However, this pessimistic conclusion relates more to the
5.3 TIME-DEPENDENT LOADING EFFECTS
347
FIGURE 5.48
Damage of a thin fabric skin caused by impact loading with energies 15 and 35 J.
methodology of the impact test (see Fig. 5.43) rather than to the real ability of composite materials
to resist impact loading. Indeed, the experiments with ice hail accelerated in a gas gun show that
the damage demonstrated in Fig. 5.48 is caused by hail ice with energy exceeding 400 J which is
by an order of magnitude higher than the energy of the steel impactor. The same is true for the
delamination under impact loading. The dependence of the energy causing material delamination
on the thickness of a quasi-isotropic carbon-epoxy laminate plotted in accordance with experimental results of Kim and Kedward (1999a, 1999b) who have studied the hail ice impact is presented
in Fig. 5.49.
The difference between the steel impactor and hail ice is in the different masses and velocities
of the impacting objects. The velocity of the steel impactor Vm can be found from Eqs. (5.75),
whereas the velocity of the hail ice is specified by the following equation (Davis & Sakata, 1981):
pffiffiffiffi
Vh 5 17:1 D
in which Vh is measured in m/s and the hail ice pellet diameter D is substituted in mm. The parameters of the hail ice with density 850 kg/m3 are listed in Table 5.3.
Compare now Fig. 5.47 and Fig. 5.49, both showing the energy of delamination under impact.
For a 2.3 mm thick plate (see Fig. 5.47), the energy of the steel impactor with mass mW 5 4:8 kg
which causes plate delamination is EW 5 3 J. For a plate of the same thickness and hail ice
with diameter D 5 42:7 mm and mh 5 0:035 kg, it follows from Fig. 5.49 that Eh 5 240 J, which is
higher than EW by the factor of 80. Thus, the higher the impactor mass, the lower the delamination
energy.
348
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
E, J
300
200
100
0
h , mm
1
2
3
FIGURE 5.49
Dependencies of the delamination energy on the thickness of composite plates impacted by the hail ice with
diameter D 5 47.2 mm.
Table 5.3 Mass, Velocity, and Kinetic Energy of Hail Ice with Various Diameters
Hail Ice Diameter, D (mm)
Mass, mh (g)
Velocity, Vh (m/s)
Kinetic energy, E (J)
10
20
30
40
50
0.44
3.56
12.01
28.47
55.60
54.1
76.5
93.7
108.1
120.9
0.64
10.41
52.7
166.3
406.3
To explain this effect, consider Fig. 5.50, which shows the impactor whose mass and initial
velocity are m and V0 , respectively. Assume that V0 is much less than the velocity of the shock
wave propagation in the plate. Then, after contact between the impactor and the plate, the contact
velocity reduces to V1 , V0 , whereas the velocity of the plate points becomes V 5 V1 f ðr; zÞ in which
f ðr 5 0; z 5 hÞ 5 1 and f ðr-N; zÞ 5 0. The balance of momentum is
0
mV0 5 V1 @m 1 2πρ
N
ð
1
ðh
dr f ðr; zÞr dr A
0
(5.76)
0
where ρ is the density of plate material and h is the plate thickness. The balance equation,
Eq. (5.76), allows us to determine V1 as
V1 5
V0
;
1 1 m1 =m
N
ð
m1 5 2πρ
0
ðh
dr f ðr; zÞr dr
0
(5.77)
5.3 TIME-DEPENDENT LOADING EFFECTS
349
z
m
V0
m
V1
h
r
FIGURE 5.50
Interaction between the impactor and the plate.
in which m1 is the reduced mass of the plate. The function f ðr; zÞ is not known; however, for qualitative analysis, it is not necessary to know it. Let m1 =m , , 1, which means that the impactor
mass is much higher than the reduced mass of the plate. In this case, which is typical for the test
shown in Fig. 5.43, V1 V0 and the plate practically does not reduce the kinetic energy of the
impactor E0 . If m1 =m . . 1, Eq. (5.77) gives V1 , , V0 and the actual kinetic energy of the
impactor, E1 , can be considerably lower than E0 . The energy dissipation ΔE 5 E0 2 E1 is absorbed
by the impactor, which slows down and experiences deformation and sometimes fracture (this is
typical for hail ice).
Thus, the conventional impact test (Fig. 5.43) based on the condition EW 5 E0 cannot provide
adequate evaluation of the material impact resistance if the drop-weight mass in experiment is considerably higher than the mass of the impactor in real conditions.
The other comment concerns the conventional compression test (see Fig. 5.44). The results of
this test are usually presented in the form shown in Fig. 5.51 which demonstrates the dependencies
of the normalized ultimate compression stress on the impact energy referred to the plate thickness
(Verpoest, Li, & Doxsee, 1989). As can be seen, the ultimate compression stress dramatically
reduces with increase in the impact energy. However, simple considerations allow us to conclude
that the plate strength cannot depend on delamination induced by impact. Indeed, consider two
plates clamped at the edges, the first of which is monolithic and the second one has interlaminar
cracks parallel to the plate surfaces (see Fig. 5.45). Under in-plane compression, the plate does not
bend and the cross-sectional area of both plates is the same, as well as the material strength. So,
interlaminar cracks cannot reduce the plate’s compressive strength. The observed reduction of the
ultimate compressive force for the delaminated plate is associated with local buckling of the plate
surface layers (see Fig. 5.52). However, the buckling load, in contrast to strength, is not a material
characteristic. The critical force depends on the relative thickness of the delaminated part of the
plate hd =h (see Fig. 5.52), delamination size, a, and the boundary conditions (type of plate support).
350
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
σ i− / σ 0−
1
0.8
1
0.6
2
0.4
3
0.2
Ei , J/mm
0
0
5
10
15
20
FIGURE 5.51
Dependence of compression strength after impact normalized to the initial compressive strength on the impact
energy related to the plate thickness for glass-fabric-epoxy (1), unidirectional glass-epoxy (2), and carbon-epoxy
composite plates (3).
F
F
hd
h
a
l
FIGURE 5.52
Local buckling of the surface layers under compression.
The dependence of the load F (see Fig. 5.52) on the axial deformation ε calculated applying the
finite-element method by Liang and Yuan (1993) for a plate with hd =h 5 0:2 and a=l 5 0:5 is shown
in Fig. 5.53. The diagram consists of two parts corresponding to prebuckling and postbuckling
deformation (as opposed to a compression diagram, which is usually linear up to the failure). Thus,
the conventional dependences of the effective ultimate compression stress on the impact energy,
like that shown in Fig. 5.51, are of a qualitative nature and cannot completely characterize the
material behavior after impact.
5.3 TIME-DEPENDENT LOADING EFFECTS
351
F , kN
6
5
4
3
2
1
0
ε ,%
0
0.1
0.2
0.3
0.4
FIGURE 5.53
Load vs plate shortening under compression for the plate with a crack.
h
i
1
k
z
P(t)
zi–1
hi
zi
FIGURE 5.54
Laminate under impact load.
Delamination as discussed earlier is caused mainly by interlaminar shear stresses. Under highvelocity impact, delamination can be also caused by normal stresses induced by superposition of
shock waves propagating through the thickness.
To study the mechanism of material delamination caused by transverse normal stresses, consider
the problem of wave propagation through the thickness of the laminate shown in Fig. 5.54. The
motion equation has the following well-known form
352
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
@
@uz
@2 uz
Ez
5ρ 2
@z
@z
@t
(5.78)
Here, uz is the displacement in the z direction, Ez is material modulus in the same direction
depending, in the general case on z, and ρ is the material density. For the laminate in Fig. 5.54, the
solution of Eq. (5.70) should satisfy the following boundary and initial conditions
σz ðz 5 0; tÞ 5 2 pðtÞ
σz ðz 5 h; tÞ 5 0
uz ðz; t 5 0Þ 5 0;
@uz
ðz . 0; t 5 0Þ 5 0
@t
(5.79)
(5.80)
in which
σz 5 Ez
@uz
@z
(5.81)
is the interlaminar normal stress.
Consider first a homogeneous layer such that Ez and ρ do not depend on z. Then, Eq. (5.78)
takes the form
c2
@2 uz
@2 uz
5 2
2
@z
@t
where c2 5 Ez =ρ. Transform this equation introducing new variables, that is, x1 5 z 1 ct and
x2 5 z 2 ct. Performing conventional transformation and rearrangement, we arrive at
@2 uz
50
@x1 @x2
The solution for this equation can be readily found and presented as
uz 5 φ1 ðx1 Þ 1 φ2 ðx2 Þ 5 φ1 ðz 1 ctÞ 1 φ2 ðz 2 ctÞ
Here, φ1 and φ2 are some arbitrary functions. Using Eq. (5.81), we get
σz 5 Ez ½f1 ðx 1 ctÞ 1 f2 ðx 2 ctÞ
where
f1 5
@φ1
;
@z
f2 5
@φ2
@z
Applying the boundary and initial conditions, Eqs. (5.79) and (5.80), we arrive at the following
final result:
σz 5 E½ f ðx 1 ctÞ 2 f ðx 2 ctÞ
(5.82)
in which the form of function f is governed by the shape of the applied pulse. As can be seen, the
stress wave is composed of two components having opposite signs and moving in opposite directions with one and the same speed c, which is the speed of sound in the material. The first term in
Eq. (5.82) corresponds to the applied pulse that propagates to the free surface z 5 h (see Fig. 5.55,
demonstrating the propagation of a rectangular pulse), whereas the second term corresponds to the
pulse reflected from the free surface z 5 h. It is important that for a compressive direct pulse (which
5.3 TIME-DEPENDENT LOADING EFFECTS
353
σ
FIGURE 5.55
Propagation of direct and reflected pulses through the layer thickness.
is usually the case), the reflected pulse is tensile and can cause material delamination since the
strength of laminated composites under tension across the layers is rather low.
Note that the speed of sound in a homogeneous material, that is,
sffiffiffiffiffi
Ez
c5
ρ
(5.83)
is the same for the tensile and compressive waves in Fig. 5.55. This means that the elastic modulus
in Eq. (5.83) must be the same for both tension and compression. For composite materials, tensile
and compressive tests sometimes produce modulus values that are slightly different. Usually, the
reason for such a difference is that the different specimens and experimental techniques are used
for tensile and compression tests. Testing of fiberglass fabric coupons (for which the difference in
the experimental values of tensile and compressive moduli is sometimes observed) involving continuous loading from compression to tension through zero load does not show any “kink” in the
stressstrain diagram at zero stress (Tarnopol’skii & Kincis, 1985). Naturally, for heterogeneous
materials, the apparent (effective) stiffness can be different for tension and compression as, for
example, in materials with cracks which open under tension and close under compression.
Sometimes stressstrain diagrams with a “kink” at the origin are used to approximate nonlinear
experimental diagrams that actually do not have a “kink” at the zero stress level at all.
For laminates, such as that shown in Fig. 5.54, the boundary conditions, Eqs. (5.79), should be
ði21Þ
ði21Þ
supplemented with the interlaminar conditions uðiÞ
and σðiÞ
. Omitting the rather
z 5 uz
z 5 σz
cumbersome solution that can be found elsewhere (Vasiliev & Sibiryakov, 1985), we present some
numerical results.
Consider a two-layered structure: the first layer has thickness 15 mm and is made of aramid-epoxy
composite material with Ezð1Þ 5 4:2 GPa, ρ1 5 1:4 g=cm3 , and the second layer is made of boron-epoxy
composite material and has Ezð2Þ 5 4:55 GPa, ρ2 5 2 g=cm3 , and h2 5 12 mm. The duration of a rectangular pulse of external pressure p acting on the surface of the first layer is tp 5 5 3 1026 s. The dependence of the interlaminar (z 5 15 mm) stress on time is shown in Fig. 5.56. As can be seen, at t 3tp
the tensile interface stress exceeds the intensity of the pulse of pressure by a factor of 1.27. This stress
is a result of interaction of the direct stress wave with the waves reflected from the laminate’s inner,
outer, and interface surfaces. Thus, in a laminate, each interface surface generates elastic waves.
354
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
σz P
1.5
1
0.5
10 6 t, s
0
4
12
16
20
–0.5
–1
–1.5
FIGURE 5.56
Dependence of the interlaminar stress referred to the acting pressure on time.
(A)
(B)
P
(C)
P
P
h1
h2
a
a
h3
a
a
a
a
FIGURE 5.57
Structures of the original laminate (A) and the laminates with one (B) and two (C) additional aluminum layers in
the foam core.
For laminates consisting of more than two layers, the wave interaction becomes more complicated and, what is more important, can be controlled by the appropriate stacking sequence of layers.
As an example, consider the sandwich structure shown in Fig. 5.57A. The first (loaded) layer is
made of aluminum and has h1 5 1 mm, Ezð1Þ 5 72 GPa,ρ1 5 2:7 g=cm3 ; the second layer is a foam
core with h2 5 10 mm, Ezð2Þ 5 0:28 GPa, ρ2 5 0:25 g=cm3 ; and the third (load-carrying) composite
layer has h3 5 12 mm, Ezð3Þ 5 10 GPa, ρ3 5 1:4 g=cm3 . The duration of a rectangular pulse of external pressure is 1026 s. The maximum tensile stress occurs in the middle plane of the load-carrying
layer (plane aa in Fig. 5.57). The normal stress induced in this plane is presented in Fig. 5.58A.
5.3 TIME-DEPENDENT LOADING EFFECTS
355
10 2 σ z / p
(A)
1
10 5 t, s
0
1.1 1.2 1.3
1.4 1.5 1.6
1.8
–1
–2
(B)
10 5 t, s
0
3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9
–2
–4
–6
(C)
10 2 σ z / p
10 5 t, s
0
2.7
2.8
2.9
3
3.1
3.2
3.3
–0.5
–1
–1.5
10 2 σ z / p
FIGURE 5.58
Normal stress related to external pressure acting in section aa of the laminates in Fig. 5.57(A)(C),
respectively.
As can be seen, at the moment of time t equal to about 1:75 3 1025 s, this stress is tensile and can
cause delamination of the structure.
Now introduce an additional aluminum layer in the foam core as shown in Fig. 5.57B. As follows from Fig. 5.58B, this layer suppresses the tensile stress in section aa. Two intermediate aluminum layers (Fig. 5.57C) working as generators of compressive stress waves eliminate the
appearance of tensile stress in this section. Naturally, the effect under discussion can be achieved
for a limited period of time. However, in reality, the impact-generated tensile stress is dangerous
soon after the application of the pulse. The damping capacity of real structural materials (which is
not taken into account in the foregoing analysis) dramatically reduces the stress amplitude in time.
A flying projectile with relatively high kinetic energy can penetrate through the laminate. As is
known, composite materials, particularly high-strength aramid fabrics, are widely used for protection against flying objects. To demonstrate the mechanism of this protection, consider a square
composite plate clamped in the steel frame shown in Fig. 5.59 and subjected to impact by a
356
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
(A)
(B)
FIGURE 5.59
Plate No. 2 (see Table 5.4) after the impact test: (A) front view; (B) back view.
rectangular plane projectile (see Fig. 5.59) simulating the blade of the turbojet engine compressor.
The plate consists of layers of thin aramid fabric impregnated with epoxy resin at a distance from
the window in the frame (see Fig. 5.59) and cocured together as shown in Fig. 5.60. The central
part of the plate is composed of unbonded layers. The front (loaded) surface of the plate has a
1 mm thick cover sheet made of glass-fabric-epoxy composite. The results of ballistic tests are presented in Table 5.4. Front and back views of plate No. 2 are shown in Fig. 5.59, and the back view
of plate No. 3 can be seen in Fig. 5.60. Since the mechanical properties of the aramid fabric used
to make the plates are different in the warp and fill directions, the plates consist of couples of
mutually orthogonal layers of fabric that are subsequently referred to as 03 =903 layers. All the
plates listed in Table 5.4 have n 5 32 of such couples.
To calculate the projectile velocity below which it fails to perforate the plate (the so-called ballistic limit), we use the energy conservation law, according to which
5.3 TIME-DEPENDENT LOADING EFFECTS
357
FIGURE 5.60
Back view of plate No. 3 (see Table 5.4) after the impact test.
Table 5.4 Ballistic Test of Plates Made of Aramid Fabric
Plate No.
Projectile Velocity (m/s)
Test Results
1
2
3
315
320
325
No penetration
The projectile is caught by the containment
Penetration
1
mp ðVs2 2 Vr2 Þ 5 nðW 1 TÞ
2
(5.84)
where Vs is the projectile striking velocity, Vr is its residual velocity, mp 5 0:25 kg is the projectile
mass, n 5 32 is the number of the 03 =903 layers, W is the fracture work for the 03 =903 layers, and T
is the kinetic energy of the layer. All other factors and the fiberglass cover of the plate are
neglected.
The fracture work can be evaluated using the quasi-static test shown in Fig. 5.61. A couple of
mutually orthogonal fabric layers are fixed along the plate contour and loaded by the projectile.
The area under the forcedeflection curve (solid line in Fig. 5.61) can be treated as the work of
fracture which, for the fabric under study, has been found to be W 5 120 Nm.
To calculate T, the deformed shape of the fabric membrane has been measured. Assuming that
the velocities of the membrane points are proportional to deflections f and that dfm =dt 5 Vs , the
kinetic energy of the fabric under study (the density of the layer unit surface is 0.2 kg/m2) turns out
to be Tc 5 0:0006 Vs2 .
To find the ballistic limit, we should take Vr 5 0 in Eq. (5.84). Substituting the foregoing results
into this equation, we get Vb 5 190:5 m=s, which is much lower than the experimental result
ðVb 5 320 m=s) following from Table 5.4.
358
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
P
P, kN
7
fm
6
5
4
3
2
1
0
5
10
15
20
25
30
35
40
45
f m , mm
FIGURE 5.61
) couple of layers with orthogonal
Forcedeflection diagrams for square aramid fabric membranes: (
) superposition of the diagrams for individually tested layers.
orientations; (
Let us change the model of the process and assume that the fabric layers fail one after
the other rather than all at once, as is assumed in Eq. (5.84). The result is expected to be
different because the problem under study is not linear, and the principle of superposition
is not applicable. Bearing this in mind, we write Eq. (5.84) in the following incremental
form
1
2
mp ðVk21
2 Vk2 Þ 5 W 1 Tk21
2
(5.85)
Here, Vk21 and Vk are the projectile velocities before and after the failure of the kth couple of
fabric layers, W is, as earlier, the fracture work consumed by the kth couple of layers,
2
Tk21 5 0:0006 Vk21
, and the last term in the right-hand side of Eq. (5.85) means that we account
for the kinetic energy of only those fabric layers that have been already penetrated by the projectile.
Solving Eq. (5.85) for Vk , we arrive at
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 2 2 W
Vk 5 ½1 2 0:0048ðk 2 1ÞVk21
mp
(5.86)
5.3 TIME-DEPENDENT LOADING EFFECTS
359
Vk , m / s
350
300
250
200
150
100
50
k
0
0
8
16
24
32
FIGURE 5.62
Dependence of the residual velocity of the projectile on the number of penetrated layers.
For k 5 1, we take V0 5 320 m=s, in accordance with the experimental ballistic limit, and have
V1 5 318:5 m=s from Eq. (5.86). Taking k 5 2, we repeat the calculation and find that, after the failure of the second couple of fabric layers, V2 5 316:2 m=s. This process is repeated until Vk 5 0, and
the number k thus determined gives an estimate of the minimum number of 03 =903 layers that can
stop a projectile with striking velocity Vs 5 320 m=s. The result of the calculation is presented in
Fig. 5.62, from which it follows that k 5 32. This is exactly the same number of layers that have
been used to construct the experimental plates.
Thus, it can be concluded that the high impact resistance of aramid fabrics is determined by
two main factors. The first factor is the relatively high work of fracture, which is governed not
only by their high strength, but also by the interaction of the fabric layers. The dashed line in
Fig. 5.61 shows the fracture process constructed as a result of the superposition of experimental
diagrams for individual 03 and 903 layers. The solid line corresponds, as noted, to 03 and 903
layers tested together (the ratio of the fabric strength under tension in the warp and the fill direction is 1.3). As can be seen, the area under the solid line is much larger that under the dashed
one, which indicates the high contribution of the layers’ interaction to the work of fracture. If
this conclusion is true, we can expect that for layers with higher anisotropy and for laminates in
which the principal material axes of the adjacent layers are not orthogonal, the fracture work
would be higher than for the orthotropic laminate under study. The second factor increasing the
impact resistance of aramid fabrics is associated with the specific process of the failure, during
which the fabric layers fail one after the other, but not all at once. Plates of the same number of
layers, but consisting of resin-impregnated and cocured layers that fail at once, demonstrate
much lower impact resistance.
360
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
5.4 MANUFACTURING EFFECTS
As has been already noted, composite materials are formed in the process of the fabrication of a
composite structure, and their properties are strongly dependent on the type and parameters of the
processing technology. This means that material specimens that are used to determine the mechanical properties should be fabricated using the same manufacturing method that is expected to be
applied to fabricate the structure under study.
5.4.1 POROSITY EFFECT
Industrial manufacturing processes usually result in composite materials characterized with a certain level of porosity which can occur in three main ways. First, the resin which is used to impregnate the fibers can contain air voids left after the mixing of the resin components. To reduce this
type of porosity, a certain time exposure or evacuation should be applied before the mixed resin is
used. Second, voids can appear in the impregnated tows because the high viscosity of the resin prevents full impregnation. To reduce this type of porosity, the resin is usually heated in the process
of impregnation. Third, the voids can appear in materials made by winding or lay-up methods
because air is trapped between the layers. All three types of porosity can be reduced by applying
pressure in the process of curing. For properly organized manufacturing processes, the volume fraction of material porosity can vary from 0.25 to 3.5%. Even such a relatively low level of porosity
can dramatically reduce a material’s strength. The most significant effect is observed on material
strength in compression and interlaminar shear. Shmergelskii and Avrasin (1987) established the
following experimental dependence of interlaminar shear strength on porosity: τ 5 71 2 3p (MPa),
where 0:25% # p # 3%: As can be seen, 3% of porosity reduces the shear strength by 12.6%.
Porosity practically does not affect the tensile strength of unidirectional composites but significantly reduces the compressive strength because thin fibers can buckle when passing through air
voids.
5.4.2 TAPE OVERLAP EFFECT
Typical fiber placement technology results in a continuous composite layer consisting of a system
of tapes placed one beside the other. To prevent gaps between the tapes that can occur because of
insufficient accuracy of fiber placement machines, some guaranteed tape overlaps are normally preassigned. To discuss the effect of tape overlaps on material properties, consider the process of circumferential winding on a cylindrical surface as in Fig. 5.63 in which the tapes are wound with
some overlap w0 shown in Fig. 5.64A. Introducing the dimensionless parameter
λ5
w0
w
(5.87)
we can conclude that for the case of complete overlap (Fig. 5.64B) we have λ 5 1. The initial position of the tape placed with overlap w0 as in Fig. 5.64A is shown in this figure with a dashed line,
whereas the final position of the tapes is shown with solid lines. Assume that after the winding and
curing are completed, the resulting structure is a unidirectionally reinforced ring that is removed
from the mandrel and loaded with internal pressure, so that the ring radius, being R before the
5.4 MANUFACTURING EFFECTS
361
FIGURE 5.63
Winding of a circumferential layer.
w
(A)
(B)
w0
δ
δ
C
B
D
R
w
w
FIGURE 5.64
Circumferential winding with (A) partial overlap w0 , w and (B) complete overlap w0 5 w.
loading, becomes R1 . Decompose the resultant force acting in the ring cross-section into two components, that is,
F 5 F 0 1 Fv
(5.88)
and introduce the apparent stress acting along the fibers of the ring as
σ1 5
F
A
(5.89)
where A 5 2wδ is the cross-sectional area of the ring made from two tapes as shown in Fig. 5.64.
The force F 0 corresponds to part BC of the ring (Fig. 5.64A) and can be found as
F 0 5 A0 E1
R1 2 R
R
362
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
where A0 5 ðw 1 w0 Þδ is the cross-sectional area of this part of the ring and E1 is the modulus of
elasticity of the cured unidirectional composite. To calculate the force Fv that corresponds to part
CD of the ring (Fig. 5.64A), we should take into account that the fibers start to take the load only
when this part of the tape reaches the position indicated with dashed lines, that is,
Fv 5 AvE1
R1 2 ðR 1 δÞ
R
where Av 5 ðw 2 w0 Þδ. Taking into accont Eqs. (5.87), (5.88), and (5.89), we can write the result of
the foregoing analysis in the following form:
δ
σ1 5 E1 ε1 2 ð1 2 λÞ
2R
(5.90)
Here, ε1 5 ðR1 2 RÞ=R is the apparent strain in the fiber direction. For complete overlap in
Fig. 5.64B, λ 5 1 and σ1 5 E1 ε1 . It should be noted that there exists also the so-called tape-to-tape
winding for which λ 5 0. This case cannot be described by Eq. (5.90) because of assumptions introduced in the derivation, and the resulting equation for this case is σ1 5 E1 ε1 .
It follows from Eq. (5.89), which is valid for winding without tension, that overlap of the tape
results in a reduction of material stiffness. Since the levels of loading for the fibers in the BC and CD
parts of the ring (Fig. 5.64A) are different, a reduction in material strength can also be expected.
Filament winding is usually performed with some initial tension of the tape. This tension
improves the material properties because it straightens the fibers and compacts the material.
However, high tension may result in fiber damage and reduction in material strength. For glass and
carbon fibers, the preliminary tension usually does not exceed 5% of the tape strength, whereas for
aramid fibers, which are less susceptible to damage, the level of initial tension can reach 20% of
the tape strength. Preliminary tension reduces the effect of the tape overlap discussed earlier and
described by Eq. (5.90). However, this effect can show itself in a reduction in material strength
because the initial stresses, which are induced by preliminary tension in the fibers, can be different
and some fibers can be overloaded or underloaded by the external forces acting on the structure in
operational conditions. Strength reduction of aramid-epoxy unidirectional composites with tape
overlap has been observed in the experiments of Rach and Ivanovskii (1986) for winding on a
200 mm diameter mandrel, as demonstrated in Fig. 5.65.
The absence of tape preliminary tension or low tension can cause ply waviness as shown in
Fig. 5.66, which can occur in filament-wound laminates as a result of the pressure exerted by the
overwrapped plies on the underwrapped plies or in flat laminates due to material shrinkage in the
process of curing.
The simplest model for analysis is a regular waviness as presented in Fig. 5.66A. To determine
the apparent modulus in the x direction, we can use an expression similar to the one presented in
Eqs. (2.76), that is,
1
cos4 α sin4 α
1
2ν 31
sin2 αcos2 α
5
1
1
2
Ex
E1
E3
G13
E1
(5.91)
Then, since the structure is periodic,
ðl
1 dx
5
l Ex
ExðrÞ
1
0
(5.92)
σ 1λ / σ 10
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
λ
FIGURE 5.65
Dependence of the normalized longitudinal strength of unidirectional aramid-epoxy composite on the tape
overlap.
(A)
z
δ
α
a
x
l
z
(B)
h
x
l
z
(C)
h
x
l10
lW
l20
FIGURE 5.66
Regular (A), through-the-thickness (B), and local (C) ply waviness.
364
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
Approximating the ply wave as
z 5 asin
πx
l
where a is the amplitude, we get
tanα 5
dz
πx
5 f cos
dx
l
where f 5 πa=l. Substitution into Eqs. (5.91) and (5.92) and integration yields
1
5
ðrÞ
Ex
1 21f2
1
1
2ν 31 2
f
1
ð2λ 2 2 2 3f 2 Þ 1
2
E1
2λ E1
E3
G13
(Tarnopol’skii & Roze, 1969), where λ 5 ð11f 2 Þ3=2 . Simplifying this result using the assumption that f 2 , , 1, we arrive at
ExðrÞ 5
E1
1 1 ðE1 f 2 =2G13 Þ
(5.93)
For glass-, carbon-, and aramid-epoxy composites with the properties listed in Table 1.5, the
dependencies corresponding to Eq. (5.93) are presented in comparison with the experimental results
of Tarnopol’skii and Roze (1969) in Fig. 5.67.
If the ply waviness varies over the laminate thickness, as in Fig. 5.66B, Eq. (5.93) can be generalized as
Ex( r )/E1
1
1
0.8
2
0.6
3
0.4
0.2
f
0
0
0.04
0.08
0.12
0.16
0.2
FIGURE 5.67
Reduction of the normalized modulus with the ply waviness parameter, f, for (1) glass-, (2) carbon-, and (3)
) Eqs. (5.93) and (x) experiment for glass-epoxy composite.
aramid-epoxy composites. (
5.4 MANUFACTURING EFFECTS
ExðtÞ 5
ðh
E1
dz
h 1 1 ðE1 =2G13 Þf 2 ðzÞ
365
(5.94)
0
Finally, for only local waviness (see Fig. 5.66C), we obtain
0
0
1
5
ðlÞ
Ex
lw
l
l1
1 ðtÞ 1 2
E
E1
1
Ex
where
l0
0
l1;2 5 0 1;2 0 ;
l1 1 lw 1 l2
lw
lw 5 0
l1 1 lw 1 l02
and ExðtÞ is specified by Eq. (5.94).
Even moderate ply waviness dramatically reduces the material strength under compression
along the fibers, as can be seen in Fig. 5.68 for a unidirectional carbon-epoxy composite. The other
strength characteristics of unidirectional composites are only slightly affected by the ply waviness.
Ply waviness can be a result of not only a low tension, but of a high tension as well. Winding
of relatively thick structures with high tension can result in buckling of the initially placed plies
under the pressure induced by the high tension of the subsequent plies. To prevent this type of waviness, winding with tension that is continuously reduced in the process of winding or step-by-step
curing of the material are utilized.
σ 1W / σ 1−
1
0.8
0.6
0.4
0.2
a
0
0
0.1
0.2
0.3
δ
FIGURE 5.68
Experimental dependence of carbon-epoxy composite longitudinal compression strength related to the
corresponding strength of material without ply waviness on the ratio of the waviness amplitude to the ply
thickness.
366
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
5.4.3 WARPING AND BENDING OF LAMINATES IN FABRICATION PROCESS
There exist also some manufacturing operations that are specific for composites that cause stresses
and strains appearing in composite structural elements in the process of their fabrication.
As an example, consider the problem of bending and warping of unsymmetric laminates after
fabrication. Assume that a laminated polymeric composite panel is cured at temperature Tc and
cooled to room temperature T0 . Under slow cooling, the temperature change, ΔT 5 T0 2 Tc , is the
same for all the layers. Since the thus-fabricated panel is free of loading (i.e., no loads are applied
to its edges or surfaces), the forces and moments in the left-hand sides of Eqs. (5.23) and (5.24) are
zero, and these equations form a linear algebraic system for generalized strains εT ; γ T , and κT .
Integration of the straindisplacement equations, Eqs. (5.28), allows us to determine the shape of
the fabricated panel.
Analysis of Eqs. (5.25) and (5.26), similar to that performed in Section 3.4, shows that for symT
metric laminates Mmn
5 0. Since Cmn 5 0 for such laminates, the last three equations of Eqs. (5.23)
in which Mx 5 My 5 Mxy 5 0 form a set of homogeneous equations whose solution is
κxT 5 κyT 5 κxyT 5 0. This means that a flat symmetric panel does not acquire curvature in the process of cooling. Naturally, the in-plane dimensions of the panel become different from those that
the panel had before cooling. The corresponding thermal strains ε0xT ; ε0yT , and γ 0xyT can be found
T
T
T
from the first three equations of Eqs. (5.23) in which Nx 5 Ny 5 Nxy 5 0, but N11
; N22
, and N12
are
not zero.
T
However, for unsymmetric laminates, in general, Mmn
6¼ 0, and these laminates experience
bending and warping in the process of cooling. To demonstrate this, consider the two antisymmetric
laminates studied in Section 3.8.
The first one is a two-layered orthotropic cross-ply laminate shown in Fig. 3.28. Using the stiffness coefficients calculated in Section 3.8, taking into account that for a cross-ply laminate
T
T
N12
5 M12
5 0, and applying Eqs. (5.23) for Nxy and Mxy , we get γ 0xyT 5 0 and κxyT 5 0. Thus, cooling of such a cross-ply laminated panel does not induce in-plane shear or twisting in it. The other
four parts of Eqs. (5.23) take the form
axx ε0xT 1 axy ε0yT 2 cxx κxT 5 nx
ayx ε0xT 1 ayy ε0yT 1 cyy κyT 5 ny
2cxx ε0xT 1 bxx κxT 1 bxy κyT 5 mx
cyy ε0yT 1 bxy κxT 1 byy κyT 5 my
where
axx 5 ayy 5 hE; axy 5 ayx 5 E1 ν 12 h
h2
h3 E
cxx 5 cyy 5 ðE1 2 E2 Þ; bxx 5 byy 5
12
8
h3
1
bxy 5 byx 5
E1 ν 12 ; E 5 ðE1 1 E2 Þ
12
2
h
nx 5 ny 5 E1 ðα1 1 ν 12 α2 Þ 1 E2 ðα2 1 ν 21 α1 Þ ΔT
2
h2 E1 ðα1 1 ν 12 α2 Þ 2 E2 ðα2 1 ν 21 α1 Þ ΔT
2 mx 5 my 5
8
(5.95)
5.4 MANUFACTURING EFFECTS
367
The solutions to Eqs. (5.95) can be written as
ε0xT 5
nx
cxx
1 2
ðaxx κxT 1 axy κyT Þ
axx 1 axy
axx 2 a2xy
ε0yT 5
nx
cxx
2 2
ðaxx κyT 1 axy κxT Þ
axx 1 axy
axx 2 a2xy
(5.96)
where
κxT 5 2 κyT 5
mx ðaxx 1 axy Þ 1 nx cxx
ðaxx 1 axy Þðbxx 2 bxy Þ 2 c2xx
(5.97)
As follows from Eqs. (5.96) and (5.97), ε and κ do not depend on x and y.
To find the in-plane displacements, we should integrate Eqs. (5.28) which have the form
@u
5 ε0xT ;
@x
@v
5 ε0yT ;
@y
@u @v
1
50
@y
@x
Referring the panel to coordinates x and y shown in Fig. 5.69 and assuming that
uðx 5 0; y 5 0Þ 5 0 and vðx 5 0; y 5 0Þ 5 0, we get
u 5 ε0xT x;
v 5 ε0yT y
(5.98)
Now consider Eqs. (5.24) in which Vx 5 Vy 5 0. Thus, γ xT 5 γ yT 5 0, and Eqs. (5.30) yield
θx 5 2 @w=@x; θy 5 2 @w=@y. The plate deflection can be found from Eqs. (5.29), which reduce to
@2 w
5 2 κxT ;
@x2
@2 w
5 2 κyT ;
@y2
@2 w
50
@x@y
Assuming that wðx 5 0; y 5 0Þ 5 0; θx ðx 5 0; y 5 0Þ 5 0; θy ðx 5 0; y 5 0Þ 5 0, we can write
the result of the integration as
1
w 5 2 ðκxT x2 1 κyT y2 Þ
2
(5.99)
To present this solution in an explicit form, consider, for the sake of brevity, material with zero
Poisson’s ratios ðν 12 5 ν 21 5 0Þ. Then, Eqs. (5.96)(5.99) yield
x
y
FIGURE 5.69
Deformed shape of a cross-ply antisymmetric panel.
368
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
ΔTx
E1 E2 ðα2 2 α1 Þ
E1 α1 1 E2 α2 1 6ðE1 2 E2 Þ 2
E1 1 E2
E1 1 14E1 E2 1 E22
ΔTy
E1 E2 ðα2 2 α1 Þ
E1 α1 1 E2 α2 1 6ðE1 2 E2 Þ 2
v5
E1 1 E2
E1 1 14E1 E2 1 E22
12ΔT E1 E2 ðα2 2 α1 Þ
w52
ðx2 2 y2 Þ
h E12 1 14E1 E2 1 E22
u5
(5.100)
The deformed shape of the panel is shown in Fig. 5.69. Note that displacements u and v correspond to the panel reference plane, which is the contact plane of the 00 and 903 layers (see
Fig. 3.28).
Compare the obtained results with experimental data of Papadopoulos and Bowles (1990). An
experimental 00 =903 cross-ply rectangular plate with sides a 5 204 mm and b 5 76:5 mm and thickness h 5 0:762 mm is made of polyimide PMR-15 resin reinforced with Celion 6000 fibers. The
characteristics of
the unidirectional
composite are E1 5 132 GPa, E2 5 9:6 GPa,
α1 5 2 1:03 3 1026 1= C, α2 5 27:9 3 1026 1= C. The glass transition temperature is 330 C, and
room temperature is taken as 23 C, so that ΔT 5 307 C. The maximum deflections in the x 5 0
and y 5 0 plane calculated in accordance with Eq. (5.100) are
a
Wx 5 wðx 5 0; y 5 0Þ 2 w x 5 ; y 5 0 5 52:25 mm
2
b
Wy 5 wðx 5 0; y 5 0Þ 2 w x 5 0; y 5
5 7:37 mm
2
The corresponding experimental results (Papadopoulos & Bowles, 1990) are
Wx 5 38:1 mm, Wy 5 4:5 mm
The higher theoretical values can be explained by the high value of the transverse modulus E2 ,
which corresponds to the room temperature value and is considerably lower at elevated temperature
(heating up to 300 C can reduce the stiffness of PMR-15 resin by 40%). To calculate the deflection W appearing under cooling, the process can be divided into a system of steps with different
parameters, and the following equation can be used:
W5
X
ðiÞ
Wi ΔTi ; E1ðiÞ ; E2ðiÞ ; αðiÞ
1 ; α2
(5.101)
i
in which “i” is the number of the cooling step, ΔTi 5 Ti11 2 Ti , EðiÞ and αðiÞ are the material characteristics corresponding to the temperature Ti . Assume that the function E2 ðTÞ looks like that
shown in Fig. 5.70, whereas the other parameters correspond to room temperature. The process of
cooling from 325 C to 25 C is divided into seven steps for which
E2ð2Þ 5 3; E2ð2Þ 5 5:5; E2ð3Þ 5 6:6; E2ð4Þ 5 7:6; E2ð5Þ 5 8:6; E2ð6Þ 5 9:2; E2ð7Þ 5 9:5 ðGPaÞ;
ΔT1 5 ΔT2 5 ΔT3 5 25 C; ΔT4 5 ΔT5 5 ΔT6 5 50 C; ΔT7 5 75 C
Calculations using Eq. (5.101) yield
Wx 5 40:1mm, Wy 5 5:6mm
As can be seen, these results are fairly close to the experimental data.
Another typical antisymmetric structure is the two-layered angle-ply laminate shown in
Fig. 3.29. Using the stiffness coefficients for this laminate calculated in Section 3.8 and Eqs. (5.25)
and (5.27), we can write Eqs. (5.23) in the following form:
5.4 MANUFACTURING EFFECTS
E z , GPa
10
T8
T7
8
T6
T5
T4
T3
6
T2
4
T1
2
T °C
0
0
100
200
300
400
FIGURE 5.70
Dependence of the transverse modulus Ez on temperature.
h
A11 ε0xT 1 A12 ε0yT 2 A14 κxyT 5 AT11
4
h
A12 ε0xT 1 A22 ε0yT 2 A24 κxyT 5 AT22
4
h
A44 γ 0xyT 2 ðA14 κxT 1 A24 κyT Þ 5 0
4
h
0
2 A14 γ xyT 1 ðA11 κxT 1 A12 κyT Þ 5 0
3
h
0
2 A24 γ xyT 1 ðA12 κxT 1 A22 κyT Þ 5 0
3
h
0
A14 εxT 1 A24 ε0yT 2 A44 κxyT 5 AT12
3
where
AT11 5 ½E1 ðα1 1 ν 12 α2 Þcos2 φ 1 E2 ðα2 1 ν 21 α1 Þsin2 φΔT
AT22 5 ½E1 ðα1 1 ν 12 α2 Þsin2 φ 1 E2 ðα2 1 ν 21 α1 Þcos2 φΔT
AT12 5 ½E1 ðα1 1 ν 12 α2 Þ 2 E2 ðα2 1 ν 21 α1 ΔTsinφcosφ
The solution is
1 T
h
A11 A22 2 AT22 A12 1 ðA14 A22 2 A24 A12 ÞκxyT
A
4
1 T
h
T
0
εyT 5
A22 A11 2 A11 A12 1 ðA24 A11 2 A14 A22 ÞκxyT
A
4
γ 0xyT 5 0; κxT 5 0; κyT 5 0
ε0xT 5
κxyT 5
A14 ðAT11 A22 2 AT22 A12 Þ 1 A24 ðAT22 A11 2 AT11 A12 Þ 2 AT12
h½ðA=3ÞA44 1 ð1=4Þð2A14 A24 A12 2 A214 A22 2 A224 A11 Þ
369
370
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
x
y
FIGURE 5.71
Deformed shape of an angle-ply antisymmetric panel.
y ,1
x ,2
FIGURE 5.72
A unidirectional circumferential layer on a cylindrical mandrel.
where A 5 A11 A22 2 A212 .
Thus, the panel under study experiences only in-plane deformation and twisting. Displacements
u and v can be determined by Eqs. (5.28), whereas the following equations should be used
to find w
@2 w
5 0;
@x2
@2 w
5 0;
@y2
@2 w
5 2 κxyT
@x@y
The result is
w 5 2 κxyT xy
The deformed shape of the panel is shown in Fig. 5.71.
Depending on the laminates’ structures and dimensions there exists a whole class of stable and
unstable laminate configurations as studied by Hyer (1989).
5.4.4 SHRINKAGE EFFECTS AND RESIDUAL STRAINS
Deformation and warping of laminates appearing after the manufacturing process is completed can
occur not only due to cooling of the cured composite but also as a result of material shrinkage due
to the release of tension in the fibers after the composite part is removed from the mandrel or to
chemical setting of the polymeric matrix.
To demonstrate these effects, consider a thin unidirectional layer formed from circumferential
plies wound on a metallic cylindrical mandrel (see Fig. 5.72) under tension. Since the stiffness of
5.4 MANUFACTURING EFFECTS
371
the mandrel is much higher than that of the layer, we can assume that, on cooling from the curing
temperature Tc to room temperature T0 , the strains in the principal material coordinates of the layer
are governed by the mandrel with which the cured layer is bonded, that is,
εT1 5 εT2 5 α0 ΔT
(5.102)
where α0 is the CTE of the mandrel material and Δ T 5 T0 2 Tc . On the other hand, if the layer is
cooled after being removed from the mandrel, its strains can be calculated as
ε1 5 α1 ΔT 1 ε01 ;
ε2 5 α2 ΔT 1 ε02
(5.103)
The first terms in the right-hand sides of these equations are the free temperature strains along
and across the fibers (see Fig. 5.72), whereas ε01 and ε02 correspond to the possible layer shrinkage
in these directions.
Using Eqs. (5.102) and (5.103), we can determine the strains that appear in the layer when it is
removed from the mandrel, that is,
ε1 5 ε1 2 εT1 5 ε01 1 ðα1 2 α0 Þ ΔT
ε2 5 ε2 2 εT2 5 ε02 1 ðα2 2 α0 Þ ΔT
(5.104)
These strains can be readily found if we measure the layer diameter and length before and after
it is removed from the mandrel. Then, the shrinkage strains can be determined as
ε01 5 ε1 2 ðα1 2 α0 ÞΔT
ε02 5 ε2 2 ðα2 2 α0 ÞΔT
For a glass-epoxy composite with the following thermomechanical properties:
E1 5 37:24 GPa; E2 5 2:37 GPa; G12 5 1:2 GPa
v12 5 0:26; α1 5 3:1 3 1026 1= C; α2 5 25 3 1026 1= C
the measurements of Morozov and Popkova (1987) gave ε01 5 2 93.6 3 1025, ε02 5 2 64 3 1025.
Further experiments performed for different winding tensions and mandrel materials have shown
that, although the strain ε01 strongly depends on these parameters, the strain ε02 practically has no
variation. This supports the assumption that the strain ε02 is caused by chemical shrinkage of the
resin and depends only on the resin’s characteristics and properties.
For a cylinder in which the fibers make an angle φ with the x-axis in Fig. 5.72, the strains
induced by removal of the mandrel can be found from Eqs. (2.70), that is,
εx 5 ε1 cos2 φ 1 ε2 sin2 φ
εy 5 ε1 sin2 φ 1 ε2 cos2 φ
γ xy 5 ðε1 2 ε2 Þsin2φ
(5.105)
where ε1 and ε2 are specified by Eqs. (5.104). The dependencies of εx , εy , and γ xy on φ, plotted
with the aid of Eqs. (5.105), are shown in Fig. 5.73 together with the experimental data of
Morozov and Popkova (1987). As can be seen, the composite
cylinder experiences,
in the general
case, not only a change in its length ðεx Þ and diameter εy , but also twist γ xy .
To study the 6 φ angle-ply layer, we should utilize the thermoelasticity constitutive equations,
Eqs. (5.23). Neglecting the bending and coupling stiffness coefficients, we can write for the case
under study
372
CHAPTER 5 ENVIRONMENTAL, SPECIAL LOADING
(ε x , ε y , γ xy ) ⋅ 105
120
80
40
φ°
0
45
60
75
90
–40
–80
FIGURE 5.73
Dependence of residual strains in a glass-epoxy filament-wound cylinder on the winding angle: (
calculation; (x) experiment.
Nx 5 B11 εx 1 B12 εy 2 N1T
Ny 5 B21 εx 1 B22 εy 2 N2T
)
(5.106)
Applying these equations to an angle-ply composite cylinder removed from its mandrel, we
should put Nx 5 0, Ny 5 0 because the cylinder is free of loads, and take εT1 5 ε1 , εT2 5 ε2 in
Eqs. (5.18), (5.25), and (5.26) that specify N1T and N2T . Then, Eqs. (5.106) yield the following
expressions for the strains that appear in the angle-ply cylinder after it is removed from the
mandrel
1
ðN1T B22 2 N2T B12 Þ
B
1
εy 5 ðN2T B11 2 N1T B12 Þ
B
εx 5
where B 5 B11 B22 2 B212 ,
N1T 5 h E1 ðε1 1 ν 12 ε2 Þcos2 φ 1 E2 ðε2 1 ν 21 ε1 Þsin2 φ
N2T 5 h E1 ðε1 1 ν 12 ε2 Þsin2 φ 1 E2 ðε2 1 ν 21 ε1 Þcos2 φ :
Here, ε1 and ε2 are given by Eqs. (5.104), Bmn 5 Amn h, where Amn is specified by Eqs. (2.72)
and h is the thickness of cylinder. The results of calculations for the experimental cylinder studied
by Morozov and Popkova (1987) are presented in Fig. 5.74.
As follows from Figs. 5.60 and 5.61, the approach described earlier, based on the constitutive
equations for laminates, Eqs. (5.23), with the shrinkage characteristics of a unidirectional ply or an
elementary layer determined experimentally, provides fair agreement between the predicted results
and the experimental data.
REFERENCES
373
(ε x , ε y ) ⋅ 10 5
80
40
φ°
0
45
60
75
90
–40
–80
FIGURE 5.74
Residual strains in the 6 φ angle-ply filament-wound glass-epoxy cylinder: (
experiment.
) calculation; (x)
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CHAPTER
LAMINATED COMPOSITE
BEAMS AND COLUMNS
6
High modulus carbon fiber reinforced composites are successfully used to fabricate laminated composite beams loaded with axial and transverse forces and to reinforce traditional metal and concrete
beam elements to increase their stiffness. Various approaches to the analysis and design of composite beams have been discussed by Vasiliev (1993), Gay, Hoa, and Tsai (2003), Kollar and Springer
(2003), Dekker (2004), and Vinson and Sierakowski (2004). This chapter is concerned with the
analysis of static, stability, and dynamic problems using elastic beam theory, allowing for transverse shear deformation.
6.1 BASIC EQUATIONS
Consider a beam as shown in Fig. 6.1 and loaded with a transverse distributed load (p, q) and/or
concentrated forces (R, F), and end forces and moment (Vl , Nl , Ml ). The analysis of such a beam is
based on a specific feature of the structure, namely, the height h is much smaller than its length l.
This feature allows us to introduce some assumptions concerning the distribution of beam displacements through the beam height which, in turn, enables us to reduce the problem of the analysis of
beams to a solution of ordinary differential equations. We also assume that the beam is loaded only
in the xz-plane (see Fig. 6.1), so that the structure is in a state of plane stress.
Single out the shaded element of the beam as shown in Fig. 6.2 and assume that the stresses acting on this element are uniformly distributed over the element width b. The equilibrium equations
for this element have the following form:
b
@σx
@
1 ðbτ xz Þ 5 0;
@x
@z
@
@τ xz
50
ðbσz Þ 1 b
@x
@z
(6.1)
in which bðzÞ is the width of the beam cross section (see Fig. 6.1). The stresses σz and τ xz must satisfy the boundary conditions on the beam surfaces. Note that in accordance with Fig. 3.1, we
measure the z-coordinate from some reference plane which is located at distances e and s from the
beam bottom and top surfaces (see Fig. 6.1). Then, the boundary conditions can be presented as
σz ðx; z 5 2 eÞ 5 2 p;
σz ðx; z 5 sÞ 5 2 q;
τ xz ðx; z 5 2 eÞ 5 0
τ xz ðx; z 5 sÞ 5 0
(6.2)
Since the beam height h is relatively small, we assume that the beam deflection, uz , does not
depend on coordinate z (see Fig. 6.1), whereas the axial displacement, ux , is a linear function of z.
Advanced Mechanics of Composite Materials and Structures. DOI: https://doi.org/10.1016/B978-0-08-102209-2.00006-2
© 2018 Elsevier Ltd. All rights reserved.
377
378
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
uz z
q
Fm
q
bk
s
V
dz
z
dx
N
t
e
M
l
p
x
ux
Vl
Ml
Nl
dz
b(z)
b1
Rm
xm
h
t
p
FIGURE 6.1
A beam loaded with surface and end forces and moment.
σ zb + b
∂
(σ z b)
∂z
σ xb
dz
τ xz b
b(z )
dx
τ xzb
τ xzb + b
∂τ xz
∂x
σ xb + b
∂σ x
∂x
σ xb
FIGURE 6.2
Stress state of a beam element.
These assumptions are the same as those introduced in Section 3.1 for thin laminates. Thus, in
accordance with Eqs. (3.1) and (3.2), we have
uz 5 wðxÞ; ux 5 uðxÞ 1 zθðxÞ
(6.3)
Here, w is the beam deflection, whereas u and θ are the axial displacement and the angle of
rotation of the beam cross section, respectively. The axial strain is specified by the first equation of
Eq. (3.3), that is,
εx 5 u0 ðxÞ 1 zθ0
0
in which ð. . .Þ 5 dð. . .Þ=dx. Using Hooke’s law, we get
0
0
σx 5 Eεx 5 E u 1 zθ
(6.4)
for the axial stress, where E is the axial modulus of the beam material. It follows from this equation
0
that σx depends on two functions, u0 ðxÞ and θ ðxÞ, that can be replaced by the corresponding stress
resultants, that is, with the internal axial force N acting in the reference plane and directed along
6.1 BASIC EQUATIONS
379
the beam axis, and the internal bending moment M exerted on the reference plane and acting in the
xz-plane, that is,
ðs
N5
ðs
σx bdz;
M5
2e
σx bzdz
(6.5)
2e
(see Fig. 6.1). Substitution of Eq. (6.4) for σx into Eq. (6.5) yields
0
0
0
N 5 B11 u 1 C11 θ0 ;
M 5 C11 u 1 D11 θ
(6.6)
where the coefficients
ðs
B11 5
ðs
Ebdz;
C11 5
2e
ðs
D11 5
Ebzdz;
2e
Ebz2 dz
(6.7)
2e
are similar to the stiffness coefficients specified by Eqs. (3.6)(3.8).
The constitutive equations, Eq. (6.6) can be simplified by a suitable choice of reference plane.
Introduce a new coordinate for the shaded beam element
t5z1e
(6.8)
(see Fig. 6.1) which changes from t 5 0 (z 5 2 eÞ to t 5 h (z 5 sÞ. Then, the stiffness coefficients in
Eq. (6.7) become
B11 5 I0 ;
C11 5 I1 2 eI0 ;
D11 5 I2 2 2eI1 1 e2 I0
(6.9)
in which
ðh
In 5
Ebtn dt; n 5 0; 1; 2
(6.10)
0
Now assume that the coordinate of the reference plane e (see Fig. 6.1) determines the location
of the beam neutral axis, so that the internal axial force N acting along this axis induces only an
axial displacement u, whereas the bending moment M causes only a rotation of the cross section.
Then, it follows from Eq. (6.6) that C11 5 0, and the second equation of Eq. (6.9) yields
e5
I1
I0
(6.11)
Under this condition, Eq. (6.6) reduce to
0
0
N 5 Bu ; M 5 Dθ
(6.12)
I2
B 5 I0 ; D 5 I2 2 1
I0
(6.13)
where
are the axial and bending stiffnesses of the beam. Substituting Eq. (6.12) into Eq. (6.4) and using
Eq. (6.8), we arrive at the following final expression for the axial stress:
σx 5 E
N
M
1 ðt 2 eÞ
B
D
(6.14)
380
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
Consider now the shear stress τ xz . According to the first assumption in Eq. (6.3), the beam
height does not change with deformation of the beam. So, the distribution of the shear stress over
the beam height does not affect the beam behavior which is governed by the resultant transverse
shear force V only. In accordance with Eq. (3.12),
ðs
ðh
V5
τ xz bdz 5
τ xz bdt
2e
(6.15)
0
The second part of this equation is written with the aid of Eq. (6.8). The transverse shear force
V (see Fig. 6.1) is linked with the average shear strain γ (see Eq. 3.14) by the constitutive equation
V 5 Sγ
(6.16)
in which
0
γ5θ1w
(6.17)
The shear stiffness coefficient S is defined and discussed in Section 3.5. To determine this
coefficient, we average the shear strain γ xz as
ðh
ðh
1
1 τ xz
dt
γ5
γ dt 5
h xz
h G
0
(6.18)
0
where G 5 Gxz is the shear modulus of the beam material in the xz-plane. Since the actual distribution of shear stress over the beam height, as mentioned earlier, does not affect the beam deformation, we can introduce the averaged shear stress τ a 5 V=bh which has the same resultant force V as
the actual stress τ xz . Taking τ xz 5 τ a in Eq. (6.18), we arrive at Eq. (6.16) in which
S5 Ð
h2
h dt
0 bG
(6.19)
For a homogeneous rectangular cross section with b 5 constant and G 5 constant, we get
S 5 Gbh
(6.20)
The internal axial force N, bending moment M and transverse shear force V must satisfy the
equilibrium equations following from Eq. (6.1) and boundary conditions in Eq. (6.2). To derive
these equations, consider the first equation of Eq. (6.1) and using the transformation defined by
Eq. (6.8) present it in the form
@
@σx
ðbτ xz Þ 5 2 b
@t
@x
Integrating with respect to t from t 5 0 and taking into account the boundary conditions,
Eq. (6.2), according to which τ xz ðt 5 0Þ 5 0, we get
τ xz 5 2
ðt
1 @σx
bdt
b @x
0
6.1 BASIC EQUATIONS
381
Substituting the axial stress specified by Eq. (6.14) into this equation, we have
2 t
3
ð
ðt
1 4N 0
M0
τ xz 5 2
Ebdt 1
Eðt 2 eÞbdt5
b B
D
0
(6.21)
0
Now, take t 5 h and again use the boundary conditions, Eq. (6.2) (according to which
τ xz ðt 5 hÞ 5 0), to get
N0
M0
I0 1
ðI1 2 eI0 Þ 5 0
B
D
where the integrals In ðn 5 0; 1; 2Þ are specified by Eq. (6.10). Applying Eq. (6.11), we can prove
that the second term is zero, so that
0
N 50
(6.22)
This is the first equilibrium equation for the beam under study showing that if the beam is not
loaded with any distributed axial forces, which is the case, the internal axial force does not change
along the beam axis. Thus, Eq. (6.21) reduces to
ðt
M0
τ xz 5 2
Eðt 2 eÞbdt
bD
(6.23)
0
Substitute this result into Eq. (6.15) for the transverse shear force to get
V52
ðh ðt
M0
dt Eðt 2 eÞbdt
D
0
(6.24)
0
Calculating the integral by parts and using Eqs. (6.10), (6.11), and (6.23), we get in several
steps
ðh
ðt
ðh
ðh
dt Eðt 2 eÞbdt 5 h Ebtdt 2
0
0
0
0
ðh
ðh
0
0
Ebt2 dt 2 e@h Ebdt 2
0
1
EbtdtA
(6.25)
5 hI1 2 I2 2 eðhI0 2 I1 Þ 5 2 I2 1 eI1 1 hðI1 2 eI0 Þ 5 2 D
Thus, Eq. (6.22) yields
V 5M
0
(6.26)
This is the moment equilibrium equation for the beam element. Using Eq. (6.26), we can present the following final form of Eq. (6.23) for the shear stress:
τ xz 5 2
ðt
V
Eðt 2 eÞbdt
bD
0
in which e is specified by Eq. (6.11).
(6.27)
382
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
Consider the so-called shear correction factor discussed, for example, by Birman and Bert
(2002). In its application to homogeneous beams with rectangular cross section, Eq. (6.20) for the
beam shear stiffness is modified as
S 5 kGbh
(6.28)
in which k is the shear correction factor. The concept of the shear correction factor (k 5 2/3 and
k 5 0.889) was introduced by Timoshenko (1921, 1922). Reissner (1945) actually used k 5 5/6 as
proposed by Goens (1931), whereas Mindlin (1951) proposed k 5 π2 =12. This problem is further
discussed in Section 6.7.4.
Recall that Eq. (6.19) for S is derived under the condition that the shear stress τ xz is averaged
through the beam height. Now we have Eq. (6.27), which specifies the actual distribution for
the shear stress. Substituting Eq. (6.27) into the second part of Eq. (6.18), we arrive at Eq. (6.16)
in which
S52
Dh
ð
Ð h dt t
Eðt 2 eÞbdt
0 bG
0
(6.29)
For a rectangular homogeneous beam, D 5 Ebh3 =12, e 5 h/2 and Eq. (8.29) yields S 5 Gbh.
Hence, k 5 1 in Eq. (6.28).
Thus, we have two different equations for the beam shear stiffness, that is, Eqs. (6.19) and
(6.29). This contradiction is natural because the analysis is based on approximate equations,
Eq. (6.3). For a homogeneous beam, both equations give the same result S 5 Gbh For laminated
beams a question arises as to what equation should be used for practical analysis. The difference
between Eqs. (6.19) and Eq. (6.29) is most pronounced for sandwich beams with a lightweight
core. Consider the cross section shown in Fig. 6.3 for which b 5 h0 5 25 mm and δ 5 2:5 mm. The
relevant parameters for aluminum facings are E 5 70 GPa, ν 5 0:3, and G 5 26.9 GPa, whereas
for the core we take E0 5 140 MPa, ν 0 5 0, and G0 5 70 MPa. Thus, E=E0 5 500, G=G0 5 384,
and h0 =δ 5 10. For the beam shown in Fig. 6.3, Eq. (6.19) gives S 5 63 kN, whereas Eq. (6.29)
yields S 5 58:2 kN, that is, the difference is 8.2%. However, Eq. (6.19) is much simpler than
Eq. (6.29) and for this reason it is recommended for practical analysis.
b
δ
h0
δ
FIGURE 6.3
Cross section of a sandwich beam.
6.1 BASIC EQUATIONS
383
Return to the equilibrium equations and consider the second equation of Eq. (6.1) which can be
presented as
@
@τ xz
ðbσz Þ 5 2 b
@x
@t
Substituting τ xz from Eq. (6.24), integrating with respect to t from t 5 0, and taking into account
the boundary conditions in Eq. (6.2) according to which σz ðt 5 0Þ 5 2 p, we arrive at
2 t t
3
ð ð
1 4V 0
σz 5
dt Eðt 2 eÞbdt 2 pb1 5
b D
0
(6.30)
0
in which b1 5 bðt 5 0Þ (see Fig. 6.1). Taking t 5 h and using the boundary conditions in Eq. (6.2)
according to which σz ðt 5 hÞ 5 2 q, we get
ðh ðt
V0
dt Eðt 2 eÞbdt 5 pb1 2 qbk
b
0
0
where bk 5 bðt 5 hÞ (see Fig. 6.1). Using Eq. (6.25) for the integral in the left-hand side of this
equation, we arrive at the last equilibrium equation for the beam element, that is,
0
V 1 p 5 0;
p 5 pb1 2 qbk
(6.31)
This is the equilibrium equation for the beam element, providing the equilibrium in the z-direction (see Fig. 6.1). Now, Eq. (6.30) allows us to determine the transverse normal stress as
2 t t
3
ð ð
1 4p
σz 5 2
dt Eðt 2 eÞbdt 1 pb1 5
b D
0
(6.32)
0
Thus, we have constructed the theory of composite beams which includes the equilibrium equations, Eqs. (6.22), (6.26), and (6.31), that allow us to determine the stress resultants N, V, and M,
and the constitutive equations, Eqs. (6.12) and (6.16), (6.17) that can be used to find the displacements u, w, and the angle of rotation θ. These equations can be divided into two independent
groups.
The first group describes the axial loading and includes Eqs. (6.22) and (6.12), that is,
0
N 5 0;
N 5 Bu
0
(6.33)
This set of equations is of the second order and requires two boundary conditions (one for each
end). At the fixed end u 5 0, whereas at the free end N 5 0. For the end x 5 l loaded with an axial
force Nl , N 5 Nl . Note that the force must be applied to the neutral axis of the beam. Otherwise,
the force must be moved to the neutral axis and the additional bending moment must be applied in
the end cross section.
The second group of equations, Eqs. (6.26) and (6.31) and Eqs. (6.12), (6.16), and (6.17),
that is,
0
M 5 V;
0
M 5 Dθ ;
0
V 1p 50
0
V 5 Sðθ 1 w Þ
(6.34)
384
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
describes the beam bending. This set is of the fourth order and requires four boundary conditions
(two for each end of the beam). At a clamped end w 5 0 and θ 5 0, at a hinged end w 5 0 and
M 5 0, whereas for a free end V 5 0 and M 5 0. For the end x 5 l loaded with the transverse force
Vl and the bending moment Ml , we have V 5 Vl and M 5 Ml .
6.2 STIFFNESS COEFFICIENTS
The stiffness coefficients B, D, and S depend on the beam structure. For homogeneous or quasihomogeneous (consisting of identical layers) beams with rectangular cross sections (see Fig. 6.4),
E 5 constant, G 5 constant, and b 5 constant and Eqs. (6.10), (6.11), and (6.20) yield
I0 5 Ebh;
h
e5 ;
2
1
I1 5 Ebh2 ;
2
B 5 Ebh;
D5
1
I2 5 Ebh3
3
1
Ebh3 ;
12
S 5 Gbh
The axial and shear stresses specified by Eqs. (6.14) and (6.27) are
σx 5
1
12M
h
N1 2 t2
;
bh
h
2
τ xz 5
6V
tðh 2 tÞ
bh3
For laminated beams (see Fig. 6.5), it follows from Eqs. (3.42) and (3.46) that
I0 5
k
X
Ei bi hi ;
I1 5
i51
k
1X
Ei bi hi ðti21 1 ti Þ
2 i51
k
2
1X
Ei bi hi ti21
1 ti21 ti 1 ti2
3 i51
I2
h2
B 5 I0 ;
D 5 I2 2 1 ;
S5 P
hi
I0
k
i51
Gi bi
I2 5
e5
I1
;
I0
t
h
b
FIGURE 6.4
Rectangular cross section.
(6.35)
6.2 STIFFNESS COEFFICIENTS
385
bk
k
i
tk = h
t i−1
ti
2
t1
1
b1
FIGURE 6.5
Laminated cross section.
where hi 5 ti 2 ti21 is the thickness of the i-th layer. The axial stress in the i-th ply is
σðiÞ
x 5 Ei
N
M
1 ðzi 2 eÞ
B
D
(6.36)
where ti21 2 e # zi # ti 2 e.
The shear stress acting between the i-th and the (i 1 1)-th layers is
ði;i11Þ
τ xz
52
i
X
V
Ej bj hj tj21 1 tj 2 2e
2Dbi;i11 j51
(6.37)
where bi;i11 is the lower value of bi and bi11 .
Composite tows can be used to reinforce metal profiles as shown in Fig. 6.6. In this case,
Eq. (6.10) for coefficients In is generalized as
ðh
In 5
Ebtn dt 1
m
X
Ej Aj tjn ðn 5 0; 1; 2Þ
j51
0
Here m is the number of composite tows, Ej and Aj are the modulus and the cross-sectional area
of the j-th tow and tj is the tow coordinate (see Fig. 6.6). The axial stress in the composite tow is
σðjÞ
x 5 Ej
N
M
1
tj 2 e
B
D
Finally, consider a laminated beam whose layers are parallel to the plane of bending as shown
in Fig. 6.7. The structure is symmetric with respect to the plane of bending (xz-plane). For such a
beam,
h
e5 ;
2
B5h
k
X
Ei bi ;
D5
i51
The axial stress is
σðiÞ
x 5 Ei
k
h3 X
Ei bi ;
12 i51
N
M
1 z
B
D
S5h
k
X
i51
Gi bi
386
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
tj
FIGURE 6.6
Metal profiles reinforced with composite tows.
z
h
h/ 2
bi
b1 b 2
bk
FIGURE 6.7
Laminated beam with layers parallel to the plane of bending.
Repeating the derivation of Eq. (6.27) for the shear stress, we arrive at
2
h
2
2z
2D 4
V
τ ðiÞ
xz 5 Ei
6.3 BENDING OF LAMINATED BEAMS
Consider the problem of bending described by Eq. (6.34). Integration of these equations with
respect to x from x 5 0 yields
6.3 BENDING OF LAMINATED BEAMS
V 5 V0 2 Vp 2 VR
M 5 M0 1 V0 x 2 Mp 2 MR
M0
V0 2
x1
x 2 θp 2 θR
θ 5 θ0 1
D
2D
1
M0 2 V0 3
x 2
x 1 wp 1 wR
w 5 w 0 1 V 0 x 2 Mp 2 MR 2 θ 0 x 2
S
2D
6D
387
(6.38)
where V0 , M0 , θ0 , and w0 are the initial values of V, M, θ, and w corresponding to the cross section
x 5 0. These values can be found from the boundary conditions at the beam ends x 5 0 and x 5 l
(see Fig. 6.1). The load terms with subscript “p” correspond to the distributed loads and have the
following form:
ðx
Vp 5
ðx
Mp 5
pdx;
0
θp 5
Vp dx;
ðx
1
Mp dx;
D
0
ðx
wp 5
θp dx
(6.39)
1
p x4
24D 0
(6.40)
0
0
For uniform pressure p 5 p 0 ,
Vp 5 p 0 x;
Mp 5
1
p x2 ;
2 0
θp 5
1
p x3 ;
6D 0
wp 5
The load terms with subscript “R” in Eq. (6.38) correspond to the concentrated forces Rm and
Fm shown in Fig. 6.1. These terms can be written with the aid of Eq. (6.39) if we present them in
the form p 5 R m δðx 2 xm Þ where R m 5 Rm 2 Fm and δ is the delta function. Using the rules of integration of this function, we get from Eq. (6.39)
VR 5
n
X
VRðmÞ ;
MR 5
m51
n
X
MRðmÞ ;
θR 5
m51
n
X
θðmÞ
R ;
wR 5
m51
n
X
wðmÞ
R
(6.41)
m51
where n is the number of the beam cross sections in which the forces act and for xm , x
VRðmÞ 5 0;
MRðmÞ 5 0;
θRðmÞ 5 0;
wðmÞ
R 50
(6.42)
Rm
ðx2xm Þ2 ;
2D
(6.43)
and for x $ xm
VRðmÞ 5 R m ;
MRðmÞ 5 R m ðx 2 xm Þ;
wRðmÞ 5
θðRmÞ 5
Rm
ðx2xm Þ3
6D
The solution given by Eq. (6.38) is universal and allows us to study both statically determinate
and redundant beams using one and the same procedure. This solution can be applied also to
multisupported beams. Introducing forces Fm as the support reactions at the support cross sections
x 5 xm , we can find Fm using the conditions wðx 5 xm Þ 5 0.
The second term with S in Eq. (6.38) for w accounts for the transverse shear deformation. As
can be seen, the allowance for this deformation practically does not hinder the analysis of the
beam. If the shear deformation is neglected, we must take S-N in Eq. (6.38) for w. As a result,
we arrive at the solution corresponding to classical beam theory.
To demonstrate the application of the general solution provided by Eq. (6.38), consider a beam
similar to that supporting the passenger floor of an airplane fuselage shown in Fig. 6.8. Since the
cross section x 5 0 is clamped, we must take w0 5 0 and θ0 5 0 in Eq. (6.38). The beam consists of
388
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
R
R
Q
x
(A)
x=c
0.5
x=l
V
(B)
M
0.05
(C)
w
0.001
(D)
FIGURE 6.8
Distributions of the normalized shear force V 5 V=Ql (B), bending moment M 5 M=Ql2 (C), and deflection
w 5 Dw=Ql4 (D) along the x-axis, of the clamped beam (A).
two parts corresponding to 0 # x , c and c # x # l (see Fig. 6.8). For the first part, p 5 2 qbk (see
Fig. 6.1), so we take p 5 2 Q, where Q 5 qbk . Then, Eqs. (6.39) and (6.42) yield
Vpð1Þ 5 2 Qx;
Mpð1Þ 5 2
1 2
Qx ;
2
θpð1Þ 5 2
Q 3
x ;
6D
wð1Þ
p 52
Q 4
x
24D
VR 5 0; MR 5 0; θR 5 0; wR 5 0
and the solution in Eq. (6.38) can be written as
V1 5 V0 1 Qx
1
M1 5 M0 1 V0 x 1 Qx2
2
M0
V0 2
Q 3
x
x1
x 1
θ1 5
6D
D
2D
1
1
M0 2 V0 3
Q 4
x 2
x 2
w1 5
V0 x 1 Qx2 2
x
2D
6D
S
2
24D
Consider the second part for which p 5 0. Then, the load terms in Eq. (6.39) become
Vpð2Þ 5
ðx
ðc
pdx 5 2
0
Qdx 5 2 Qc
0
(6.44)
6.3 BENDING OF LAMINATED BEAMS
Mpð2Þ 5
ðx
ðc
Vp dx 5
0
Vpð1Þ dx 1
ðx
389
1
Vpð2Þ dx 5 2 Qcð2x 2 cÞ
2
c
0
0c
1
ð
ðx
1
1@
Qc 2
ð2Þ
ð1Þ
ð2Þ A
ðc 1 3x2 2 3cxÞ
θp 5
Mp dx 5
Mp dx 1 Mp dx 5 2
D
D
6D
ðx
0
wð2Þ
p 5
0
ðx
ðc
θp dx 5
0
θð1Þ
p dx 1
0
0
ðx
θpð2Þ dx 5 2
Qc 3
4x 2 c3 1 4xc2 2 6x2 c
24D
c
The support reaction R (see Fig. 6.8) is treated as the unknown concentrated force. Then,
Eq. (6.43) yield
R
R
ðx2cÞ2 ; wR 5
ðx2cÞ3
VR 5 R; MR 5 Rðx 2 cÞ; θR 5
2D
6D
Finally, we get for the second part of the beam
V2 5 V0 1 Qc 2 R
1
M2 5 M0 1 V0 x 1 Qcð2x 2 cÞ 2 Rðx 2 cÞ
2
M0
V0 2 Qc 2
R
x1
x 1
c 1 3x2 2 3cx 2
ðx2cÞ2
θ2 5
D
2D
6D
2D
1
1
M0 2 V0 3
x 2
x
w2 5
V0 x 1 Qcð2x 2 cÞ 2 Rðx 2 cÞ 2
2D
6D
S
2
Qc 3
R
4x 2 c3 1 4xc2 2 6x2 c 1
ðx2cÞ3
2
24D
6D
(6.45)
The solution obtained, Eqs. (6.44) and (6.45), includes three unknown parameters: V0 , M0 , and
R, which can be found from two symmetry conditions, that is, V2 ðx 5 lÞ 5 0, θ2 ðx 5 lÞ 5 0, and the
condition for the reaction force w1 ðx 5 cÞ 5 w2 ðx 5 cÞ 5 0. The result is as follows
V0 5 R 2 Qc;
1
1
M0 5 Qc2 ð1 2 4ks Þ 2 Rcð1 2 6ks Þ
4
3
1
3lð1 1 4ks Þ 2 2c
R 5 Qc
2
4lð1 1 3ks 2 3c
(6.46)
where ks 5 D=Sc2 .
For numerical analysis, neglect the shear deformation by taking ks 5 0. Then, the solution in
Eq. (6.46) reduces to
V0 5 2
Qcð5l 2 4cÞ
;
2ð4l 2 3cÞ
M0 5
Qc2 ð6l 2 5cÞ
;
12ð4l 2 3cÞ
R5
Qcð3l 2 2cÞ
2ð4l 2 3cÞ
The dependencies of the normalized shear force, the bending moment, and the beam deflection
on the axial coordinate are presented in Fig. 6.8.
Beam deflections are often used as approximation functions in the solutions of plate bending
problems (see Chapter 7: Laminated Composite Plates). Solutions of typical beam problems are
presented in Table 6.1.
As an example, consider a simply supported I-beam loaded with uniform pressure q (see
Fig. 6.9). The beam is made of an aluminum alloy for which E 5 70 GPa and G 5 26:9 GPa. At
390
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
Table 6.1 Solutions for the Beams Loaded With Uniform Pressure for Typical Boundary
Conditions
Case
Beam Type
Solution
q
1
bk
w
x
l
q
2
bk
w
x
l
q
3
bk
w
x
l
V 5 2 qbk ðl 2 xÞ
1
M 5 qbk ðl2xÞ2
2
qbk 2
ð3l 2 3lx 1 x2 Þx
θ5
6D
qbk 3
x 2 4lx2 1 6l2 x
w52
24D
D
1 12 ð2l 2 xÞx
S
1
V 5 qbk ð2x 2 lÞ
2
1
M 5 2 qbk ðl 2 xÞx
2
qbk 3
ðl 2 6lx2 1 4x3 Þ
θ5
24D
qbk 3
l 2 2lx2 1 x3
w52
24D
D
1 12 ðl 2 xÞx
S
V 5 qbk ðx 2 2l Þ
1
qbk ð6x3 2 6lx 1 l2 Þx
M5
12
qbk
ð2x2 2 3lx 1 l2 Þx
θ5
12D qbk
D
xðl 2 xÞ 1 12 ðl 2 xÞx
w52
S
24D
q
q
δ
δ
h0
w
l
x
FIGURE 6.9
A simply supported I-beam.
b
δ
δ
6.3 BENDING OF LAMINATED BEAMS
391
b4 = b
4
b3 = δ
3
t3
e
2
1
t0 = 0
t1
t4 = h
t2
b1 = b2 = b
FIGURE 6.10
Coordinates of the layers.
the bottom, where the maximum tensile stress acts, the beam is reinforced with a unidirectional
carbon-epoxy layer with modulus of elasticity Ec 5 140 GPa and shear modulus Gc 5 3:5 GPa.
The beam dimensions are
l 5 1250 mm; δ 5 2:5 mm; h0 5 100 mm; b 5 50 mm
(6.47)
The layer coordinates are shown in Fig. 6.10. The beam is composed of four layers with the following parameters:
b1 5 50 mm; t0 5 0; t1 5 2:5 mm; h1 5 2:5 mm; E1 5 140 GPa; G1 5 3:5 GPa ðlayer1Þ
b2 5 50 mm; t2 5 5 mm; h2 5 2:5 mm; E2 5 70 GPa; G2 5 26:9 GPa ðlayer2Þ
b3 5 2:5 mm; t3 5 105 mm; h3 5 100 mm; E3 5 70 GPa; G3 5 26:9 GPa ðlayer3Þ
b4 5 50 mm; t4 5 h 5 107:5 mm; h4 5 2:5 mm; E4 5 70 GPa; G4 5 26:9 GPa ðlayer4Þ
(6.48)
The beam strength is analyzed in accordance with the following procedure.
1. Determine the maximum shear force, bending moment, and deflection. The beam under study
corresponds to Case 2 in Table 6.1 from which it follows that
1
Vm 5 V ðx 5 0Þ 5 2 qbl
2
l
1
5 2 qbl2
Mm 5 M x 5
2
8
l
5qbl4
48D
52
ð1 1 αÞ; α 5
Wm 5 w x 5
2
5Sl2
384D
(6.49)
2. Determine the stiffness coefficients. First, calculate the I-coefficients specified by Eq. (6.35),
that is,
I0 5
4
X
Ei bi hi 5 5:25 104 GPa mm2
i51
I1 5
4
1X
2 i51
Ei bi hi ðti21 1 ti Þ 5 1:95 106 GPa mm3
392
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
I2 5
4
1X
2
Ei bi hi ðti21
1 ti21 ti 1 ti2 Þ 5 1:66 108 GPa mm4
3 i51
The coordinate of the neutral axis can be found from Eq. (6.11), which yields
e5
I1
5 37:14 mm
I0
The bending stiffness of the beam is calculated according to Eq. (6.13) as
I2
D 5 I2 2 1 5 0:936 108 GPa mm4
I0
For a beam without the composite layer D 5 0:573 108 GPa mm4 , that is, the composite
layer increases the beam bending stiffness by 63%.
The transverse shear stiffness of the beam is specified by Eq. (6.35), which give
S5 P
h2
4
i51
hi
Gi bi
5 7:68 103 GPa mm2
3. Calculate the axial stress using Eqs. (6.36) and (6.49), according to which
σðiÞ
x 5 Ei
Mm
qbl2 h
ðt 2 eÞ
ðt 2 eÞ 5 2 Ei
8D
D
e
where t 5 t=h, e 5 ; and ti21 # t # ti . For the beam with dimensions as per Eqs. (6.47) and
h
(6.48), the maximum tensile stress in the composite layer corresponds to t 5 0 and is equal to
ð2Þ
σð1Þ
x 5 542q. The maximum tensile stress in the metal part is σ x ðt 5 t1 Þ 5 253q. The maximum
compressive stress in the metal part is σðx4Þ ðt 5 hÞ 5 2 514q.
4. Calculate the shear stress. The most dangerous is the shear stress which acts between composite
layer 1 and metal layer 2 and can cause delamination. This stress is specified by Eq. (6.37)
which yields
ð1;2Þ
τ xz
52
Vm
ql
E1 bh1 ðh1 2 2eÞ
E1 b1 h1 ðt1 2 2eÞ 5
4D
2Db
ð1;2Þ
For the beam under study, τ xz
5 2 4:2q.
5. Calculate the maximum deflection. The deflection is specified by the third equation of
Eq. (6.49) in which α 5 0:075. Thus, the allowance for transverse shear deformation increases
the maximum deflection by 7.5%.
6.4 NONLINEAR BENDING
To demonstrate the concept of nonlinear bending, consider Fig. 6.11. A conventional simply supported beam is shown in Fig. 6.11A. As can be seen, the beam deflection (dashed line) causes the
displacements of the beam supports toward each other. The deflection is small and the curve slope
is also so small that it can be neglected. In this case, the beam equations can be written for the
6.4 NONLINEAR BENDING
393
q
(A)
z
x=−
q
l
2
x=
l
2 x
(B)
FIGURE 6.11
Bending (A) and bending-stretching (B) of the beam.
N0
dx
dw
N0
ds
α2
α1
FIGURE 6.12
A curved element of the beam axis.
initial rectilinear shape of the beam resulting in Eqs. (6.33) and (6.34). For the beam in Fig. 6.11B,
in contrast to the beam in Fig. 6.11A, the ends do not move under bending. The deflection of this
beam is also small (it is even smaller than the deflection of the beam shown in Fig. 6.11A), however, the obvious difference between the length of the dashed curve and the initial length of the
beam allows us to conclude that the beam axis experiences tension under bending.
The slope of the beam axis shown in Fig. 6.11B is small, so that the first equation of
Eq. (6.33) is still valid and shows that the internal axial force N which appears as a result of the
beam bending does not depend on the axial coordinate, so that N 5 N0 5 constant. To determine
this force, consider a curved element of the beam axis shown in Fig. 6.12 from which it
follows that
ds 5
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2
1
dx2 1 dw2 5 1 1 ðw0 Þ dxD 1 1 ðw0 Þ2 dx
2
0
The last part of this equation can be obtained if we take into account that w 5 dw=dx is much
smaller than unity. So, the elongation of the beam axis due to bending is
ε5
ds 2 dx 1 0 2
5 ðw Þ
dx
2
394
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
pn
dx
FIGURE 6.13
Imaginary pressure simulating the action of forces N0 .
This elongation should be added to u0 in Eq. (6.33), and the final expression for the axial
force becomes
1
0
N0 5 B u 1 ðw0 Þ2
2
(6.50)
0
Solving this equation for u and integrating the resulting equation with respect to x from
x 5 2 l=2 (see Fig. 6.11B), the following expression for uðxÞ is derived:
1
1
u 5 N0 x 2
B
2
ðx
ðw0 Þ2 dx 1 c
2l=2
where c is the constant of integration. Since uðx 5 2 l=2Þ 5 0 (see Fig. 6.11B), c 5 0. However,
uðx 5 l=2Þ is also zero, and we arrive at the following equation for N0 :
B
N0 5
2
l=2
ð
ðw0 Þ2 dx
(6.51)
2l=2
As can be seen, the dependence of N0 on w is not linear. The axial force N0 also affects the
equilibrium equation, particularly the second equation of Eq. (6.34) presenting the projection of the
forces on the axis normal to the beam axis. The effect of N0 can be simulated if we introduce some
imaginary pressure shown in Fig. 6.13. Static equivalence of the forces in Figs. 6.12 and 6.13 gives
Pn dx 5 2 N0 α1 1 N0 α2
0
Here, α1 and α2 are small angles, such that α1 w0 and α2 w 1 w00 dx. Then, Pn 5 N0 w00 ,
and the second equation of Eq. (6.34) can be generalized as
0
00
V 1 N0 w 1 p 5 0
(6.52)
where p is the acting transverse load. The rest of Eq. (6.34) remain the same, that is,
0
0
0
M 5 V; M 5 Dθ ; V 5 S θ 1 w
(6.53)
Reduce the set of equations, Eqs. (6.52) and (6.53), to one equation for the beam deflection. For
this purpose, first, substitute V from the first equation of Eq. (6.53) into Eq. (6.52) to get
00
00
M 1 N0 w 1 p 5 0
Second, eliminate θ from the last two equations of Eq. (6.53), that is,
M5D
0
V
2 w00
S
(6.54)
6.4 NONLINEAR BENDING
395
Third, substitute V 0 in this equation with its expression from Eq. (6.52) and determine
00
M 5 2 Dð1 1 λÞw 2
D
p
S
(6.55)
where λ 5 N0 =S. Substituting Eq. (6.55) into Eq. (6.54), we finally arrive at
00
wIV 2 k2 w 5
1
D
N0
p 2 p 00 ; k2 5
Dð1 1 λÞ
S
Dð1 1 λÞ
(6.56)
Consider the solution of this equation for the beam shown in Fig. 6.11B for which, as earlier,
p 5 2 qbk 5 2 Q (see Fig. 6.1). The solution of Eq. (6.56) is symmetric with respect to x 5 0 and
has the following form:
w 5 C1 1 C2 coshkx 1
Q 2
x
2N0
The bending moment is specified by Eq. (6.55), which yields
Q
M 5 2 D k2 C1 ð1 1 λÞcoshkx 1
N0
The constants of integration C1 and C2 can be found from the boundary conditions
wðx 5 6 l=2Þ 5 0 and Mðx 5 6 l=2Þ 5 0. The resulting solution is
w5
Q D
coshkx
x2
l2
QD coshkx
1
21
2
; M5
12
N0 N0
coshη
N0 coshη
2
8
(6.57)
in which η 5 kl=2.
To find N0 , apply Eq. (6.51). Substitution of the first equation of Eq. (6.57) into Eq. (6.51) gives
the following transcendental equation:
"
#
N0 l
Q2 l3
2D
l
1
D2 k 2
1
l
2
cosh
η
2
sinhη
1
sinh
η
cosh
η
2
50
2
2B
k
2
2N02 24 N0 cosh η 2
2N02 ðcoshηÞ2 k
(6.58)
Recall that in accordance with the foregoing derivation
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lu
N0
u
η5 u N0
t
2
D 11
S
To simplify the solution, neglect the shear deformation taking S-N. Then, Eq. (6.58) can be
transformed into the following dimensionless form:
q2 1
1
1
1
1
2
tanh
η
1
sinhη
cosh
η
2
1
51
1
2
3
η
4N ðcosh ηÞ2 η
N 24 N
Ql4
in which q 5
D
rffiffiffiffi
B
;
D
N5
N0 l2
;
D
η5
1 pffiffiffiffi
N
2
The function N ðq) is presented in graphical form by Reismann (1988).
396
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
For the beam shown in Fig. 6.11B and having square ðb 5 hÞ uniform cross section, the dependencies of the normalized maximum deflection wm 5 wðx 5 0Þ on the loading parameter q
are shown in Fig. 6.14. As can be seen, for relatively low loads (q # 60Þ the nonlinear solution
(curve 2) practically coincides with the linear one. The normalized maximum stress
σm 5
σm l2
;
Eh2
σm 5 E
N0
Mm h
6
B
D 2
is presented in Fig. 6.15. As can be seen, the loading parameter q, for which the linear solution is
valid, is much smaller for the maximum compressive stress than for the maximum tensile stress. It
should be taken into account that the ultimate value of the loading parameter is limited by material
wm /h
1.0
1
0.8
2
0.6
0.4
0.2
q
0
200
400
600
800
1000
FIGURE 6.14
Dependencies of the normalized maximum deflection on the loading parameter corresponding to the linear (1)
and nonlinear (2) solutions.
σm
6
1
4
2
2
q
0
200
–2
–4
1
400
600
800
1000
2
FIGURE 6.15
Dependencies of the normalized maximum stress on the loading parameter corresponding to the linear (1) and
nonlinear (2) solutions.
6.4 NONLINEAR BENDING
397
z,w
l /2
l/ 2
P
h
A
c
P
A
x
e
FIGURE 6.16
Eccentric tension of a laminated beam.
strength. For example, for an aluminum beam with l/h 5 10, the material yields at q 5 10 for
which the linear solution is valid. However, for composite beams, the situation can be different.
For example, for a unidirectional glass-epoxy composite that has relatively low modulus and high
strength (see Table 1.5), the ultimate value of q can reach 250, which is far beyond the limit for
the linear solution.
In conclusion, consider another nonlinear problem. Suppose that the laminated beam in
Fig. 6.16 is loaded with two tensile forces P applied at the beam ends x 5 6 l=2. If the forces are
applied to the beam neutral axis, that is, at points A with coordinate e, then no bending is observed.
A natural question arises as to what happens if the forces are applied at some points with coordinate c which is not equal to e.
The problem of a beam subjected to eccentric tension (for a homogeneous rectangular beam
e 5 h=2 and a force which does not induce any beam bending passes through the center of the cross
section) can be studied using Eq. (6.56) in which p 5 0 and N0 5 P, that is,
00
wIV 2 k2 w 5 0;
k2 5
P
;
Dð1 1 λÞ
λ5
P
S
(5.59)
The solution of this equation which satisfies the boundary conditions wðx 5 6 l=2Þ 5 0 is
w 5 2 Cðcoshη 2 coshkxÞ
(6.60)
where η 5 kl=2. The rotation angle and the bending moment can be found with the aid of
Eqs. (6.53) and (6.55), that is,
θ 5 2 kC ð1 1 λÞsinh kx; M 5 2 Dk2 Cð1 1 λÞcoshkx
(6.61)
Assume that the beam in Fig. 6.16 is clamped, so that θðx 5 6 l=2Þ 5 0. Then, the first equation
of Eq. (6.61) yields C 5 0 and w 5 0. Thus, bending of a clamped beam does not occur irrespective
of the location of the force. Now assume that the beam ends are hinged and can rotate around
points A (see Fig. 6.16) under a bending moment M0 5 Pðc 2 eÞ. Taking Mðx 5 6 l=2Þ 5 M0 in the
second equation of Eq. (6.61), we can present Eq. (6.60) as
coshkx
w 5 ðC 2 eÞ 1 2
coshη
(6.62)
As an example, consider a two-layer beam composed of a unidirectional glass-epoxy and
carbon-epoxy layers with the following parameters: h1 5 1 mm, E1 5 60 GPa, h2 5 1 mm,
and E2 5 140 GPa. The beam length and width are l 5 250 mm; b 5 25 mm. Calculate the
I-coefficients in Eq. (6.35), that is,
I0 5 5 103 GPa mm2 ;
I1 5 6 103 GPa mm3 ;
I2 5 8:67 103 GPa mm4
398
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
− w/h
0.3
α =1
0.2
0 .1
0.1
0 .01
x/l
0
0
0.1
0.2
0.3
0.4
0.5
FIGURE 6.17
Deflection of the beam under eccentric tension.
Then, the beam stiffness in Eq. (6.13), and the coordinate of the neutral axis in Eq. (6.11)
become
B 5 5 103 GPa mm2 ;
D 5 4:5 103 GPa mm4 ;
e 5 1:6 mm
The axial strain of the beam under tension is specified by Eq. (6.12), that is,
0
ε5u 5
P
B
(6.63)
The maximum strains of the carbon-epoxy and glass-epoxy composites are (see Table 1.5)
εc 5 1:43% and εg 5 3%. To find the ultimate force P, we should substitute ε5εc into Eq. (6.62) to
get P 5 71:5 kN.
Introduce
the ratio α 5 P=P and dimensionless coordinate x 5 x=l (see Fig. 6.16).
pffiffiffi
ffi
Then, η 5 15:71 α and in accordance with Eq. (6.62)
w5
pffiffiffiffi w
cosh31:42 αx
5 ðc 2 e Þ 1 2
h
cosh15:71
Let the force P be applied at the layers’ interface, so that c 5 h1 , c 5 0:5, and e 5 e=h 5 0:8.
The functions wðxÞ are presented in Fig. 6.17 for various values of the loading parameter. As can
be seen, under a force equal to the ultimate value ðα 5 1Þ, bending is concentrated in the vicinity of
the beam ends.
6.5 BUCKLING OF COMPOSITE COLUMNS
Consider a column compressed by an axial force T as shown in Fig. 6.18. For a particular value of
the force T, which is called the critical value, the initial rectilinear shape becomes unstable, and the
column axis becomes curved as shown in Fig. 6.18 with dashed lines. According to the static buckling criterion, the critical force Tc is the minimum force under which two equilibrium states of the
column are possible, that is, the initial rectilinear shape and the slightly curved shape. To apply this
criterion, we need to induce some deflection (see the dashed lines in Fig. 6.18) and to determine
6.5 BUCKLING OF COMPOSITE COLUMNS
T
399
T
T
l
w
x
(A)
(B)
(C)
FIGURE 6.18
Axial compression and buckling modes of a column with simply supported ends (A), with clamped end and free
end (B), and with clamped ends (C).
whether this deflection can be supported by the acting force T. Note that this small deflection is
artificially induced because the column is analyzed assuming that it is “perfect” (i.e., it is perfectly
straight, no transverse forces acting, etc.). In real structures, the tendency to bending always exists.
For example, a small eccentricity of the load application (see Fig. 6.16) causes bending from the
very beginning of the loading process (see Fig. 6.17).
The equilibrium of a compressed column with some small deflection w can be described by
Eq. (6.59) in which P 5 2 T, that is,
00
wIV 1 k2 w 5 0;
k2 5
T
;
Dð1 2 λÞ
λ5
T
S
(6.64)
Note that this equation has a trivial solution w 5 0 which corresponds to the initial rectilinear
shape of the column. We need to find out whether the critical force Tc exists, for which Eq. (6.64)
has a nonzero solution. The general solution of Eq. (6.64) is
w 5 C1 x 1 C2 1 C3 sin kx 1 C4 cos kx
(6.65)
Applying Eqs. (6.53) and (6.55), we can also find
θ 5 2 kð1 2 λÞðC3 cos kx 2 C4 sin kxÞ 2 C1
M 5 Dk2 ð1 2 λÞðC3 sin kx 1 C4 cos kxÞ
(6.66)
V 5 Dk3 ð1 2 λÞðC3 cos kx 2 C4 sin kxÞ
Consider the simply supported column shown in Fig. 6.18A for which
wðx 5 0Þ 5 wðx 5 lÞ 5 0;
M ðx 5 0Þ 5 M ðx 5 lÞ 5 0
(6.67)
For these boundary conditions, Eqs. (6.65) and (6.66) give the following algebraic equations
written in terms of the constants of integration Ci ði 5 1; . . . ; 4Þ:
C2 1 C4 5 0;
C4 5 0;
C1 l 1 C2 1 C3 sinkl 1 C4 coskl 5 0
C3 sinkl 1 C4 coskl 5 0
400
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
The last two equations yield
C3 sinkl 5 0
Obviously, if C3 5 0, then all the constants are zero and w 5 0, which corresponds to the initial
shape of the column. So, C3 6¼ 0 and
sinkl 5 0
(6.68)
As can be readily seen, C3 is the only one nonzero constant and Eq. (6.65) reduces to
w 5 C3 sinkx
(6.69)
Note that within the approach under discussion the constant C3 cannot be determined. Recall
that Eq. (6.64) is valid for small deflections only. So, for the unknown C3 , we can formally assume
that it is small enough so that the deflection, Eq. (6.69), is also small. The nonzero deflection given
by Eq. (6.69) exists if the parameter k satisfies Eq. (6.68) from which kl 5 nπ ðn 5 1; 2; 3; . . .Þ.
Using the corresponding Eq. (6.64) for k, we get the values of forces
Tn 5
π2 n2 D
ðn 5 1; 2; 3; . . .Þ
π2 n2 D
l2 1 1
Sl2
(6.70)
for which the nonzero deflection of the column can take place. Only one of these values, that is,
the smallest one, has a physical meaning and represents the critical force. If we neglect the
shear deformation taking S-N, the result is evident: the minimum value of Tn takes place when
n 5 1, that is,
TE 5
π2 D
l2
(6.71)
This is a well known Euler formula for the critical load. In the general case, the situation looks
more complicated. However, it can be proved that n 5 1 in Eq. (6.70) as well. Indeed, using
Eq. (6.71), that equation can be transformed as follows
Tn 5
n2 TE
TE
1 1 n2
S
Consider the inequality Tn . T1 for n . 1. This inequality reduces to n . 1 which is true. So, in
the general case, n 5 1 and the critical force is determined by
Tc 5
TE
TE
11
S
(6.72)
For n 5 1, Eq. (6.69) specifies the mode of buckling
w 5 C3 sin
πx
l
shown in Fig. 6.18A with the dashed line.
In conclusion, consider the case λ 5 1, or T 5 S which is sometimes associated with the so00
called shear mode of buckling. In this case, Eq. (6.64) reduces to w 5 0 whose solution is
6.6 FREE VIBRATIONS OF COMPOSITE BEAMS
401
w 5 C1 x 1 C2 . Applying the boundary conditions in Eq. (6.67), we get C1 5 C2 5 0 and w 5 0.
Thus, the case λ 5 1 corresponds to the straight column and T 5 S is not a critical load. Anyway,
the force T 5 S is always higher than Tc given by Eq. (6.72). Indeed, the inequality T 5S . Tc is
equivalent to the inequality S . 0.
The result obtained for the critical force, that is, Eq. (6.72), can be used for columns with various boundary conditions substituting the corresponding expression for TE instead of that given by
Eq. (6.71). Particularly for the cantilever (see Fig. 6.18B) and clamped (see Fig. 6.18C) columns,
we have, respectively,
TE 5
π2 D
;
4l2
TE 5
4π2 D
l2
It should be noted that a buckling analysis must be accompanied by a study of stresses because
composite beams can fail before buckling if the stresses in the layers reach the material ultimate
compressive stresses. The stresses in the layers are specified by Eq. (6.36) in which N 5 2 T and
M 5 0, that is,
σðiÞ
x 5 2 Ei
T
B
Buckling can take place if these stresses calculated for T 5Tc are less than material strength.
6.6 FREE VIBRATIONS OF COMPOSITE BEAMS
Composite beams are good models allowing us to discuss some specific features of the dynamic
behavior of laminated composite structures. The equations of motion of the beam element shown in
Fig. 6.2 can be written adding the inertia terms to the equilibrium equations, Eq. (6.1), that is,
b
@σx
@
@2 ux
1 ðbτ xz Þ 2 ρb 2 5 0
@z
@x
@t
b
@τ xz
@
@2 uz
1 ðbσz Þ 2 ρb 2 5 0
@x
@t
@z
in which t is the time and ρ is the material density. Applying the assumptions introduced in
Section 6.1, that is, using Eq. (6.3) for the displacements and Eqs. (6.5) and (6.15) for forces and
moment, we can arrive at the following equations for composite beams [these equations are generalizations of the equilibrium equations, Eqs. (6.22), (6.26), and (6.31)]:
@N
@2 u
@2 θ
2 Bρ 2 2 Cρ 2 5 0
@x
@t
@t
@M
@2 u
@2 θ
2 V 2 Cρ 2 2 Dρ 2 5 0
@x
@t
@t
@V
@2 w
2 Bρ 2 2 pðx; tÞ 5 0
@x
@t
(6.73)
402
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
where
Bρ 5 J0 ;
Cρ 5 J0 eρ 2 e ;
Dρ 5 J2 2 2eJ1 1 e2 J0
ðh
Jn 5 bρtn dt
ðn 5 0; 1; 2Þ
(6.74)
0
are the inertia coefficients which are analogous to the stiffness coefficients in Eq. (6.9). For
laminated beams,
Jn 5
k
1 X
n11
bi ρi tin11 2 ti21
n 1 1 i51
(6.75)
(see Fig. 6.5). The parameter eρ in Eq. (6.74)
eρ 5
J1
J0
(6.76)
is similar to the coordinate of the neutral axis e defined by Eq. (6.11). In particular, Bρ is the inertia
term corresponding to translational displacement of the beam element in Fig. 6.2, Dρ corresponds
to the element rotation, and Cρ is the coupling term linking transverse and rotational motions of the
beam element. In the general case, eρ does not coincide with e because the integrals In [see
Eq. (6.10)] depend on the stiffness distribution, whereas the coefficients Jn determined by
Eq. (6.75) are governed by the density distribution through the height of the beam cross section.
However, there are some particular cases for which eρ 5 e. For homogeneous beams
(E 5 constant and ρ 5 constant), we have
Ðh
eρ 5 e 5 0h
Ð
btdt
;
C50
bdt
0
For beams with a symmetric laminated structure,
h
eρ 5 e 5 ;
2
C50
(6.77)
The forces and moments in Eq. (6.73) are linked with the corresponding deformations by the
constitutive equations, Eqs. (6.12), (6.16), and (6.17), that is,
0
N 5 Bu ;
0
M 5 Dθ ;
0
V 5S θ1w
(6.78)
Substituting Eq. (6.78) into Eq. (6.73), we can derive the motion equations in terms of displacements. Taking p 5 0 for the case of free vibrations, we get
@2 u
@2 u
@2 θ
2 Bρ 2 2 Cρ 2 5 0
2
@x
@t
@t
@2 θ
@w
@2 u
@2 θ
D 2 2S θ1
2 Cρ 2 2 Dρ 2 5 0
@x
@x
@t
@t
2
2
@θ @ w
@ w
S
1 2 2 Bρ 2 5 0
@x
@x
@t
B
(6.79)
(6.80)
(6.81)
6.6 FREE VIBRATIONS OF COMPOSITE BEAMS
403
In the general case,eρ 6¼ e and Cρ 6¼ 0, so that Eq. (6.79) which describes the axial vibrations
includes the rotation angle θ and is coupled through this term with Eqs. (6.80) and (6.81) which
describe the flexural vibrations of the beam.
Consider here the flexural vibrations. For this purpose, assume that the beam has a symmetric
structure, so that Eq. (6.77) hold. Then, Eqs. (6.80) and (6.81) can be transformed to the following
equation written in terms of the beam deflection:
D
Bρ D @4 w
Bρ Dρ @4 w
@4 w
@2 w
2
D
1
1
1 Bρ 2 5 0
ρ
4
2
2
4
@x
@x @t
@t
S @t
S
(6.82)
In this equation, the terms including S allow us to take into account the shear deformation,
whereas the terms with coefficient Dρ allow for the rotation inertia of the beam cross section.
Following Uflyand (1948), Eq. (6.82) can be presented in the form
4
2
@4 w
1
1
@ w
1 @4 w
2@ w
2
1
1
1
C
50
@x4
@t2
C12
C22 @x2 @t2
C12 C22 @t4
or
@2
1 @2
2 2 2
2
@x
C1 @t
where
2
@ w
1 @2 w
@2 w
1 C2 2 5 0
2
2
2
2
@x
@t
C2 @t
sffiffiffiffiffi
S
C2 5
;
Bρ
sffiffiffiffiffiffi
D
;
C1 5
Dρ
rffiffiffiffiffi
Bρ
C5
B
To clarify the physical meaning of the coefficients C1 and C2 , consider a beam with rectangular
cross section (see Fig. 6.19) for which
D5
1
1
Ebh3 ; S 5 Gbh; Bρ 5 bhρ; Dρ 5 ρbh3
12
12
Then,
sffiffiffiffi
E
;
C1 5
ρ
(6.83)
sffiffiffiffi
G
C2 5
ρ
are the velocities of normal stress (C1 ) and shear stress (C2 ) wave propagation along the beam axis.
Neglecting rotatory inertia ðDρ 5 0Þ or shear deformation ðS-NÞ in Eq. (6.82), we actually
assume that C1 or C2 become infinitely high.
w
x
h
l
b
FIGURE 6.19
Free vibrations of a simply supported beam.
404
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
For qualitative analysis, consider flexural vibrations of a simply supported beam shown in
Fig. 6.19 in which case the beam deflection can be presented as
wðx; tÞ 5 Wm sin
πmx
sinωm t
l
(6.84)
where m is the number of the vibration mode (the mode corresponding to m 5 1 is shown in
Fig. 6.19 with a dashed line), Wm is the amplitude, and ωm is the frequency of the m-th vibration
mode. Recall that the number of the beam vibrations per 1 second (Hz) can be found as
νm 5
ωm
2π
Substituting the deflection, Eq. (6.84), into Eq. (6.82) results in the following equation for
the frequency:
Bρ Dρ 4
DBρ
πm 2
πm 2
Dρ 1
50
ωm 2 Bρ 1
ω2m 1 D
l
l
S
S
(6.85)
Neglecting both transverse deformation ðS-NÞ and rotatory inertia ðDρ 5 0Þ, we arrive at a
result corresponding to classical beam theory, that is,
πm
l
ω0m 5
sffiffiffiffiffi
D
Bρ
(6.86)
If we neglect the rotatory inertia ðDρ 5 0Þ and take into account the shear deformation,
Eq. (6.85) yields
ω0m
ωm 5 pffiffiffiffiffiffiffiffiffiffiffiffiffi
1 1 αS
where ω0m is specified by Eq. (6.86) and
αS 5
π2 m2 D
l2 S
allows for shear deformation. For a homogeneous beam with a rectangular cross section, using
Eq. (6.83), we get
αS 5
π2 m2 Eh2
12Gl2
Thus, the shear deformation reduces the frequency and the effect depends on the ratios E/G, h/l,
and the mode number m.
Consider now the effect of rotatory inertia. Taking S-N in Eq. (6.85), we have
ω0m
ωm 5 pffiffiffiffiffiffiffiffiffiffiffiffiffi
1 1 αr
where
αr 5
π2 m2 Dρ
l2 Bρ
6.6 FREE VIBRATIONS OF COMPOSITE BEAMS
405
Again, for a homogeneous beam with a rectangular cross section
αr 5
π2 m2 h2
12l2
(see Fig. 6.19). As can be seen, the effect depends on the beam relative thickness and
mode number.
Consider coupled longitudinal-transverse vibrations of the beams (for which Cρ 6¼ 0Þ described
by Eqs. (6.79)(6.81). Taking, for the sake of brevity, Dρ 5 0 and S-N, we can transform these
equations into the following form:
BD
@6 w
@6 w
@6 w
@4 w
@4 w
2 Bρ D 4 2 2 Cρ2 2 4 1 BBρ 2 2 2 B2ρ 4 5 0
6
@x
@x @t
@x @t
@x @t
@t
Substituting the deflection from Eq. (6.84), we arrive at the following equation for the
frequency:
Bλ2m 2 Bρ ω2m Dλ4m 2 Bρ ω2m 2 Cρ2 λ2m ω4m 5 0
in which λm 5 πm=l. Taking into account Eq. (6.74) for Cρ , we can present this equation in
the form
h
2 i
ω4m B2ρ 1 2 λ2m eρ 2e 2 Bρ λ2m ω2m Dλ2m 1 B 1 BDλ6m 5 0
It follows from this equation that the effect of vibration coupling can be evaluated by comparing
the parameter
αc 5
2
π2 m2 e2eρ
2
l
with unity. If αc {1, the coupling effect can be ignored.
Finally, consider flexural vibrations of beams loaded with axial compressive force T (see
Fig. 6.18). For this purpose, we need to modify Eq. (6.81), adding to it the term with the axial force
as in Eq. (6.52), that is, present it as
@θ @2 w
@2 w
@2 w
1 2 2 T 2 2 Bρ 2 5 0
S
@x
@x
@x
@t
(6.87)
Taking, for the sake of brevity, Dρ 5 0 and Cρ 5 0 in Eq. (6.80), we get the second equation
D
@2 θ
@w
50
2
S
θ
1
@x2
@x
(6.88)
These two equations, Eqs. (6.87) and (6.88) can be reduced to the following equation:
T @4 w DBρ @4 w
@2 w
@2 w
D 12
2
1 T 2 1 Bρ 2 5 0
S @x4
@x
@t
S @x2 @t2
For a simply supported beam (see Fig. 6.18A), substituting the deflection in accordance with
Eq. (6.84), we finally get
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
u1 2 T 1 1 αm
u
αm
S
u
ωm 5 ω0m t
αm
11
S
(6.89)
406
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
where αm 5 π2 m2 D=l2 . As can be seen, a compressive force reduces the frequency of flexural
vibrations. For the first mode (m 5 1), we have α1 5 TE , where TE is the Euler critical force specified by Eq. (6.71), and Eq. (6.89) becomes
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
T
u
u1 2 TT 1 1 E
E
u
S
ω1 5 ω01 u
t
TE
11
S
For the critical force T 5 Tc in Eq. (6.72), ω1 5 0, that is, compressive loading up to the critical
load results in zero frequency of flexural vibrations. A tensile force, naturally, increases the
frequency.
6.7 REFINED THEORIES OF BEAMS AND PLATES
The composite beam, being a simple but representative model, allows us to consider the so-called
“higher-order” theories of beams, plates, and shells intensively discussed in the literature. Note that
although the equations presented in the text following are valid only for beams, the main references
are made to plates, for which such theories have been originally constructed.
Consider two typical problems of beam theory, that is, the cantilever beam shown in
Fig. 6.20 and the simply supported beam shown in Fig. 6.21. Both beams are homogeneous and
have rectangular cross sections.
z,w
h/2
F
M
h
x, u
h/2
V
l
dx
b
FIGURE 6.20
A cantilever beam.
z, w
q
x, u
l
FIGURE 6.21
A simply supported beam.
6.7 REFINED THEORIES OF BEAMS AND PLATES
407
6.7.1 FIRST-ORDER THEORY
The theory described in the foregoing sections is sometimes referred to as the “first-order” theory
and is based on the approximations given by Eq. (6.3) for the displacements which, for the case of
bending, yield
ux 5 zθðxÞ;
uz 5 wðxÞ
(6.90)
The corresponding static variables are the internal bending moment and the transverse shear
force specified by Eqs. (6.5) and (6.15), that is,
ð2
ð2
h
h
M5
σx bzdz;
V5
2h2
τ xz bdz
(6.91)
2h2
This approximation reduces the theory to Eq. (6.34), that is,
0
0
M 5 V;
V 1p 50
0
V 5 Sγ;
γ5 θ1w
0
M 5 Dθ ;
(6.92)
(6.93)
For a beam with a rectangular cross section
D5
1
Ebh3 ;
12
S 5 Gbh
(6.94)
(see Fig. 6.20). The normal and shear stresses are specified by Eqs. (6.12) and (6.27) which, for the
stiffnesses in Eq. (6.94), give
σx 5
12M
z;
bh3
τ xz 5
6V h2
2
2
z
bh3 4
(6.95)
For the cantilever beam (see Fig. 6.20), the solution of Eqs. (6.92) and (6.93) is
F 5 V;
6F
M 5 2 F ðl 2 xÞ;
θ52
ð2l 2 xÞx
3
Ebh
2
2F
Eh
x
w5
3lx 2 x2 1
Ebh3
2G
(6.96)
For a simply supported beam (see Fig. 6.21), the solutions are given in Table 6.1 (Case 2) and
have the following form:
1
1
M 5 2 qbðl 2 xÞx
V 5 2 qbðl 2 2xÞ;
2
2
q 3
q
Eh2
2
3
3
2
3
θ5
;
w
5
2
ðl
2
xÞ
x
l
2
6lx
1
4x
l
2
2lx
1
x
1
2Eh3
2Eh3
G
(6.97)
Traditionally, the main shortcoming of the “first-order” theory is associated with the shear stress
distribution following from Eq. (6.90). Indeed, the exact constitutive equation for the shear stress is
@ux
@uz
1
τ xz 5 Gγ xz 5 G
@z
@x
(6.98)
408
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
Substituting Eq. (6.90), we get
0
τ xz 5 G θ 1 w
(6.99)
which does not depend on z. Thus, the stress τ xz in Eq. (6.99) does not satisfy the boundary conditions (see Fig. 6.20) according to which
h
50
τ xz z 5 6
2
(6.100)
However, as discussed in Section 3.1, Eq. (6.99) is not correct. The displacements in Eq. (6.90)
approximate the distribution of the actual displacements through the beam height. It is known that
approximate functions must not be differentiated, as has been done to derive Eq. (6.99). The second
equation in Eq. (6.90) states that uz does not depend on z, which means that the beam element
shaded in Fig. 6.20 is absolutely rigid in the z direction. As is known, the distribution of forces
over the surface of an absolutely rigid body does not affect the body motion. Only the forces resultant, that is, the transverse shear force V for the beam, affects the beam deflection. Integration in
the second equation of Eq. (6.91) allows us to avoid differentiation of ux in Eq. (6.98) with respect
to z. Indeed, using Eqs. (6.90), (6.91), and (6.98), we get in several steps
ð2
h
V5
2h2
ð2 @ux
@uz
1
dz
τ xz bdz 5 bG
@z
@x
h
2h2
h
h
0
0
2 ux 2
1 hw 5 bGhðθ 1 w Þ
5 bG ux
2
2
Thus in the theory under consideration, the constitutive equation exists for the shear force V
only and does not exist for the shear stress τ xz . The shear stress following from the equilibrium
equation is specified by the second equation of Eq. (6.95) and, naturally, satisfies the boundary conditions as per Eqs. (6.100). The same situation occurs in classical beam theory in which Eq. (6.98)
simply does not exist, so there is no temptation to use it and the shear stress is found from the equilibrium equation.
6.7.2 HIGHER-ORDER THEORIES
Consider the “higher-order” theories which are widely discussed in the literature. These theories
are based on the displacement approximations which generalize Eq. (6.90) as
ux 5 zu1 ðxÞ 1 z2 u2 ðxÞ 1 z3 u3 ðxÞ 1 ? 1 zk uk ðxÞ
(6.101)
uz 5 wðxÞ
(6.102)
To be specific, we shall discuss the so-called “third-order” theory. To derive the equations of
this theory, take k 5 3 in Eq. (6.101), that is,
ux 5 zu1 ðxÞ 1 z2 u2 ðxÞ 1 z3 u3 ðxÞ; uz 5 wðxÞ
(6.103)
6.7 REFINED THEORIES OF BEAMS AND PLATES
409
Following the conventional procedure, substitute Eqs. (6.103) into Eq. (6.98) to get
0
τ xz 5 Gðu1 1 w 1 2u2 z 1 3u3 z2 Þ
Apply the boundary conditions in Eq. (6.100) and determine the functions u2 and u3 . The resulting displacement field becomes
ux 5 zu1 2
4z3 0
u1 1 w ;
2
3h
uz 5 wðxÞ
(6.104)
The displacements in Eq. (6.104) have been originally proposed by Vlasov (1957) and
Ambartsumyan (1958) and are used in numerous papers and books to construct refined plate and
shell theories (e.g., Reddy, 2004).
It seems, at first glance, that the displacements in Eq. (6.103) are more accurate than the linear
approximation given by Eq. (6.90). However, this is not actually the case. In general, introduction
of force constraints into the kinematic field is not the best way to improve it. To show this, apply
the constitutive equations and find the stresses corresponding to the displacements presented by
Eq. (6.104), that is,
@ux
4z3 0
0
00
5 E zu1 2 2 ðu1 1 w Þ
σx 5 Eεx 5 E
@x
3h
@ux
@uz
4z2
0
τ xz 5 Gγ xz 5 G
1
5 Gðu1 1 w Þ 1 2 2
@z
@x
h
(6.105)
Following Ambartsumyan (1987), introduce the bending moment and the shear force in
Eq. (6.91). Substituting the stresses given by Eq. (6.105), we get
M5E
bh3 0
1 00
u1 2 w ;
4
15
2
0
V 5 Gbh u1 1 w
3
(6.106)
Consider first the cantilever beam (see Fig. 6.20) for which the equilibrium equations,
Eq. (6.92) for p 5 0, have the solution given by Eq. (6.96). Integration of Eq. (6.106) yields
3F
12F
x2
2 3 lx 2
2 C1
10Ghb Eh b
2
6Fx
6F
x3
1 3 lx2 2
1 C1 x 1 C2
w5
5Ghb Eh b
3
u1 5
(6.107)
where C1 and C2 are the constants of integration which should be determined from the boundary
conditions at x 5 0 (see Fig. 6.20). According to these conditions, we should put wðx 5 0Þ 5 0 and,
taking into account that u1 is actually the angle of the cross section rotation, put u1 ðx 5 0Þ 5 0:
Then, Eq. (6.107) reduce to
u1 5 2
6F
ð2l 2 xÞx;
Eh3 b
w5
2F
3Eh2
2
3lx
2
x
x
1
Eh3 b
4G
(6.108)
Comparing this result with Eq. (6.96), we can conclude that u1 coincides with θ, whereas the
deflection has a different coefficient in the term corresponding to the shear deformation (0.75
instead of 0.5). However, the main difference between the solutions is more significant. Indeed, for
the clamped edge of the beam, Eq. (6.96) of the “first-order” theory give θ 5 0 and w 5 0, and
Eq. (6.90) for the displacements yield ux ðx 5 0Þ 5 0 and uz ðx 5 0Þ 5 0. Thus, the displacements are
410
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
identically zero at the clamped edge of the beam. But for the “third-order” theory, Eq. (6.104) and
(6.108) give
ux ðx 5 0Þ 5 2
2Fz3
;
Gh3 b
uz ðx 5 0Þ 5 0
Thus the boundary conditions for the clamped end ðux 5 0Þ cannot be satisfied within the framework of the “third-order” theory.
A similar conclusion can be derived for a simply supported beam shown in Fig. 6.21.
Substitution of the bending moment M and the shear force V (neither of which depend on the shear
deformation) from Eq. (6.97) into Eq. (6.106) and integration yield the following solution:
q
3Eh2
3
2
3
ðl
2
2xÞ
l
2
6lx
1
4x
2
2Eh3
10G
q
6Eh2
3
2
3
w52
ðl 2 xÞ x
l 2 2lx 1 x 2
2Eh3
5G
u1 5
(6.109)
in which the constants of integration are found from the boundary conditions
wðx 5 0Þ 5 wðx 5 lÞ 5 0 (see Fig. 6.21). Comparing this solution with the corresponding equation in
Eq. (6.97), we can conclude that the results coincide within the accuracy of the terms including
the shear modulus G. The bending moment M specified by the first equation of Eq. (6.106)
and Eq. (6.109) satisfies the boundary conditions (see Fig. 6.21), according to which
Mðx 5 0Þ 5 M ðx 5 lÞ 5 0. However, the axial stress is not zero at the beam ends. Substituting
Eq. (6.109) into the first equation of Eq. (6.105), we arrive at
σx 5 2
6q
qE 3h2
2
2
2z
z
ð
l
2
x
Þxz
1
h3
Gh3 10
(6.110)
At the beam ends (x 5 0 and x 5 l), the first term vanishes, but the second term gives nonzero
stress, which cannot exist at the ends of a simply supported beam.
Usually, simply supported beams and plates are studied applying a trigonometric series, each
term of which satisfies the boundary conditions wð0Þ 5 wðlÞ 5 0 and Mð0Þ 5 M ðlÞ 5 0, that is,
u1 5
X
Um cosλm x;
w5
m
X
Wm sinλm x
(6.111)
m
in which λm 5 πm=l (see Fig. 6.21). Substituting expansions (6.111) into Eq. (6.106), we get
bh3 X 1
λm Wm 2 Um sinλm x
15 m 4
X
2
ðλm Wm 1 Um Þcosλm x
V 5 Gbh
3
m
M5E
(6.112)
Decomposing the load p 5 2 qb into a similar series
p 5 2 qb 5 2 b
X
m
qm sinλm x
(6.113)
6.7 REFINED THEORIES OF BEAMS AND PLATES
411
and substituting Eqs. (6.112) and (6.113) into the equilibrium equations, Eq. (6.92), we arrive at the
following set of algebraic equations derived in terms of Wm and Um :
E
h2 2 1
λm λm Wm 2 Um 5 2Gðλm Wm 1 Um Þ
4
5
2
Ghλm ðλm Wm 1 Um Þ 5 2 qm
3
Solving these equations for Wm and Um , we have
Wm 5 2
12qm
Eh2 2
λ
;
1
1
10G m
Eh3 λ4m
Um 5
12qm
Eh2 2
λ
1
2
40G m
Eh3 λ3m
(6.114)
Substitution of Eqs. (6.111) and (6.114) into Eq. (6.105) yields the following equation for σx :
σx 5 2
X
12z X qm
Ez 3h2
2
2
2z
sinλ
x
1
qm sinλm x
m
h3 m λ2m
Gh3 10
m
Retaining a finite number of terms in the series of this equation, we can arrive at the conclusion
that σx ðx 5 0Þ 5 σx ðx 5 lÞ 5 0. However, for an infinite number of terms, the second series converges, and in accordance with Eq. (6.113) becomes equal to the second term in Eq. (6.110), which
means that σx is not zero at the beam ends.
Thus, the “third-order” theory does not allow one to satisfy the boundary conditions which are
satisfied within the framework of the “first-order” theory. It is important that, following
Ambartsumyan (1987), we used the displacements in Eq. (6.104) in conjunction with the equilibrium equations, Eq. (6.92), which follow from the free-body diagrams for the shaded element of
the beam shown in Fig. 6.20. In more recent versions of the “third-order” theory (e.g., Reddy,
2004), the governing equations are derived with the aid of the variational principle. To discuss
this approach, apply the minimum condition for the total potential energy of the beam, according
to which
h=2
ðl
ð
ðl
δT 5 b dx
σx δεx 1 τ xz δγ xz dz 2 b pδwdx 5 0
2h=2
0
(6.115)
0
Consider the “first-order” theory. Using Hooke’s law and displacements as per Eq. (8.90), we
get the following equation for the axial stress:
σx 5 Eεx 5 E
@ux
0
5 Eθ z
@x
(6.116)
As has been noted earlier, the constitutive equation for the shear stress in the “first-order” theory exists only in integral form, Eq. (6.93). To apply this form, we need to use Eq. (6.91) for V and
replace γ xz with the average shear strain γ 5 θ 1 w0 . Taking into account Eq. (6.116) and Eq. (6.91)
for the bending moment, we can present Eq. (6.115) in the following form:
ðl
0
0
Mδθ 1 V ðδθ 1 δw0 Þ 2 pδw dx 5 0
(6.117)
412
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
Integration by parts yields the following variational equations and natural boundary conditions:
0
M 2 V 5 0;
x5l
½Mδθx50
5 0;
0
V 1p 50
(6.118)
½Vδwx5l
x50 5 0
(6.119)
As can be seen, Eq. (6.118) coincide with equilibrium equations, Eq. (6.92), whereas the boundary conditions, Eq. (6.119) give M 5 0 and V 5 0 which are equivalent to σx 5 0 and τ xz 5 0 for
the free end [see Eq. (6.95)], and for the clamped end give θ 5 0 and w 5 0 which are equivalent to
ux 5 0 and uz 5 0 [see Eqs. (6.90)]. Thus, the variational formulation of the “first-order” theory is
equivalent to the direct formulation based on the equilibrium equations and the corresponding
boundary conditions.
The situation is totally different for the “third-order” theory for which we have constitutive
equations, Eq. (6.105), for both normal and shear stresses. Substituting Eq. (6.105) into Eq. (6.115),
and integrating with respect to z, we get
ðl
0
0
Mδu1 2 H δu1 1 δw00 1 V ðδu1 1 δw0 Þ 2 Qðδu1 1 δw0 Þ 2 pδw dx 5 0
0
where, M and V are specified by Eq. (6.91) and
H5
4b
3h2
h=2
ð
σx z3 dz;
Q5
2h=2
4b
h2
h=2
ð
τ xz z2 dz
(6.120)
2h=2
The corresponding variational equations and the natural boundary conditions are
0
0
M 2 H 2 V 1 Q 5 0;
½ðM2HÞδu1 x5l
x50 5 0;
0
0
V 2 Q 1 H 00 1 p 5 0
½Hδw0 x5l
x50 5 0;
½ðV 2QÞδwx5l
x50 5 0
(6.121)
(6.122)
in which, in accordance with Eqs. (6.91), (6.105) and (6.120)
Ebh3 0
00 4u1 2 w ;
60
2
0
V 5 Gbh u1 1 w ;
3
M5
Ebh3 0
00 16u1 2 5w
1260
2
Q 5 Gbhðu1 1 wÞ
15
H5
(6.123)
The equations obtained, Eq. (6.121) in conjunction with Eq. (6.123), are of the sixth order
which corresponds to three natural boundary conditions in Eq. (6.122). For a cantilever beam
shown in Fig. 6.20, at the clamped end u1 5 0, w0 5 0; and w 5 0. Correspondingly, it follows from
Eq. (6.104) that ux 5 0 and uz 5 0 at x 5 0. Thus, the boundary conditions at the clamped end can
be satisfied. However, it follows from Eq. (6.105) that, under these conditions, τ xz ðx 5 0Þ 5 0 and
the force F acting at x 5 l (see Fig. 6.20) is not balanced. This result is expected because the variational equations, Eq. (6.121), do not coincide with equilibrium equations, Eq. (6.92), and hence, the
equilibrium equations of a beam element as a solid are violated. The reason for this is associated
with the formulation of the displacement field. In contrast to Eq. (6.90) of the “first-order” theory
which allow for two mutually independent degrees of freedom for the shaded beam element shown
in Fig. 6.20 (i.e., displacement w and rotation angle θ), in Eq. (6.104) of the “third-order” theory
6.7 REFINED THEORIES OF BEAMS AND PLATES
413
the displacement and the rotation are not mutually independent, and the variational equations,
Eq. (6.121), cannot provide the equilibrium of the beam element as a solid.
6.7.3 CONSISTENCY CONDITIONS
Energy consistency conditions for the refined beam, plate, and shell theories have been proposed
by Vasiliev and Lurie (1990a, b, and 1992). According to these conditions, equilibrium equations
following from the free body diagram for the structure element and the boundary conditions corresponding to the assumed displacement field must coincide with the variational equations and the
natural boundary conditions providing the minimum of the structure’s total potential energy. As follows from Section 6.7.2, the “first-order” theory is energy consistent. In case the energy consistency conditions are not satisfied, there can exist, in principle, two versions of the refined theories.
The theories of the first type can be referred to as energy inconsistent, but statically consistent.
As an example of such a theory, consider classical beam theory. Because transverse shear deformation is ignored in classical beam theory, we must put the shear stiffness S-N in Eq. (6.93). Then,
as follows from these equations, γ 5 0, θ 5 w0 , and
M 5 2 Dwv
Consider now Eq. (6.117) for the variation of the total potential energy of the beam. Since
θ 1 w0 5 0 the second term in this equation is zero and it reduces to
ðl
ðl
ðM δθ0 2 pδwÞdx 5 2 ðDδwv 1 pδwÞdx 5 0
0
(6.124)
0
It is clear that classical beam theory cannot be energy consistent because the beam element
must have two mutually independent degrees of freedom, that is, displacement w and rotation θ
whereas the functional in Eq. (6.124) includes only deflection w. Integrating by parts, we arrive at
the following variational equation:
Mv 1 P 5 0
(6.125)
Compare this equation with the equilibrium equations, Eq. (6.34), which follow from the free
body diagram for the beam element. As can be seen, the direct equilibrium approach yields two
equilibrium equations, whereas the energy approach gives only one equation. Thus the theory is not
energy consistent. However, it is statically consistent because Eq. (6.125) follows from Eq. (6.34).
The theories of the second type can be referred to as energy inconsistent and statically inconsistent. The example of such a theory is the “third order” theory described by Eq. (6.121) derived by
the energy method. These equations are not equilibrium equations because they do not provide
equilibrium of a beam element as a solid.
6.7.4 THEORY OF ELASTICITY SOLUTION
Return to the “third-order” theory. Irrespective of the energy inconsistency of Eq. (6.121) and
(6.123), the corresponding governing equations have the specific structure which shows us a way to
414
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
improve the “third-order” theory. Integrating Eq. (6.121), substituting Eq. (6.123), and performing
some transformations, we arrive at the following two equations with regard to u1 and w:
G 30C1 x2 1 C2 x 1 C3 2 1260C1
2
Eh
0
w 5 4u1 2 30C1 x2 2 C2 x 2 C3
00
u1 2 k2 u1 5 2 672
(6.126)
in which k2 5 840G=Eh2 and C1 , C2 and C3 are the constants of integration. As can be seen,
Eq. (6.126), in addition to the traditional polynomial solution, contain a solution which includes the
exponential function Expð2 kxÞ which rapidly vanishes at a distance from the beam end and corresponds to the so-called boundary-layer solution. Goldenveizer (1958) was the first who stated that
the nonlinear approximation of the displacement distribution through the plate thickness [as in
Eq. (6.104)] must be supplemented with the boundary-layer solutions.
To do this, consider substitution of Eq. (6.108) for u1 and w corresponding to the cantilever
beam shown in Fig. 6.20 into Eq. (6.104) for ux and uz to get the displacements
6F
E 3
z
ð
2b
2
x
Þxz
1
Eh3 b
3G
2F
3Eh2
x
u0z 5 3 3lx 2 x2 1
Eh b
4G
u0x 5 2
(6.127)
which are treated now as the basic approximation. Following Vasiliev and Lurie (1972), and
Vasiliev (2010), approximate the total displacements ux and uz supplementing the displacements in
Eq. (6.127) with a set of boundary-layer solutions which vanish at a distance from the clamped end
of the beam (see Fig. 6.20), that is,
ux 5 u0x 1 u x ; uz 5 u0z 1 u z ðxÞ
where
u x 5 C0 f0 ðzÞ 1
N
X
e2sk x fk ðzÞ
(6.128)
(6.129)
k51
Here, C0 is a constant coefficient, f0 ðzÞ 5 z, and fk ðzÞ (k 5 0; 1; 2 . . .Þ is a complete system of
functions which should be found (note that without f0 the functions fk do not form a complete system). Substituting Eqs. (6.128) and (6.129) into the equations of Hooke’s law, find the stresses
X
@ux
12F
5 2 3 ðl 2 xÞz 2 E
sk e2sk x fk ðzÞ
bh
@x
k
"
#
X
@ux
@uz
6F h2
du z
2
2sk x dfk
1
5 G C0 1
2z 1
1
e
τ xz 5 G
@z
@x
dz
dx
Gbh3 4
k
σx 5 E
(6.130)
Apply the equilibrium equation, that is, the first equation of Eq. (6.1) in which b 5 constant
@σx
@τ xz
1
50
@x
@z
(6.131)
Substituting for the stresses their expressions, Eq. (6.130), we arrive at the following equation
for fk ðzÞ ðk 5 1; 2; 3; . . .Þ:
d 2 fk
1 α2k fk 5 0
dz2
6.7 REFINED THEORIES OF BEAMS AND PLATES
415
The antisymmetric solution with respect to z of this equation is
fk ðzÞ 5 Ck sinαk z;
α2k 5 s2k
E
G
(6.132)
Now, Eq. (6.130) for τ xz becomes
"
τ xz 5 G C0 1
#
X
6F h2
du z
2
2sk x
2
z
1
C
α
e
cosα
z
1
k k
k
Gbh3 4
dx
k
(6.133)
According to the boundary conditions, Eq. (6.102), τ xy ðz 5 6 h=2Þ 5 0. Thus
X
du z
5 2 C0 2
Ck αk e2sk x cosλk ;
dx
k
1
λk 5 αk h
2
(6.134)
Integrating this equation and taking into account that for the clamped end (see Fig. 6.20)
uz ðx 5 0Þ 5 0, we get
uz 5 2 C0 x 1
X
Ck
k
αk 2sk x
ðe
2 1Þ cosλk
sk
(6.135)
Substituting Eq. (6.134) into Eq. (6.133), we arrive at the following expression for the shear
stress
"
#
X
6F h2
2
2sk x
2z 2
τ xz 5 G
Ck αk e ϕk ðzÞ
Gbh3 4
k
(6.136)
ϕk ðzÞ 5 cosλk 2 cosαk z
(6.137)
in which
Apply the equilibrium equations, Eqs. (6.91) and (6.92) according to which
h=2
ð
V 5b
τ xz dz 5 F
(6.138)
2h=2
for the cantilever beam (see Fig. 6.20) under consideration. Substituting Eq. (6.136) into
Eq. (6.138), we derive the following equation for the parameter λk :
tanλk 5 λk
(6.139)
As can be readily checked, the functions ϕk(z) in Eq. (6.137) and their derivatives are orthogonal on the interval [2h/2, h/2], that is,
8
>
<h 2
sin λk ; n 5 k
ϕk ϕn dz 5 2
;
>
:
0;
n
¼
6
k
2h=2
h=2
ð
(
h=2
ð
0
0
ϕk ϕn dz 5
2h=2
2 2 2
λ sin λk ; n 5 k
h k
0; n 6¼ k
(6.140)
Moreover, the following integral conditions are also valid:
h=2
ð
h=2
ð
ϕk dz 5 0;
2h=2
0
h=2
ð
ϕk dz 5 0;
2h=2
0
zϕk dz 5 0
2h=2
(6.141)
416
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
0
It follows from Eq. (6.141) that the system of functions fk 5 Ck =αk ϕk is complete if it is
supplemented with function f0 5 z [see Eq. (6.129)] in which this property of functions fk
is used.
To determine the constants C0 and Ck , apply the boundary condition for the clamped end
of the beam, that is, ux ðx 5 0Þ 5 0. Using Eqs. (6.127) and (6.128), and (6.132) and (6.137),
we have
ux ðx 5 0Þ 5 2
X Ck 0
2F 3
z 1 C0 z 1
ϕ ð zÞ 5 0
3
Gbh
αk k
k
(6.142)
Multiplying this equation by z, integrating from h/2 to h/2 and using Eq. (6.141), we get
0
C0 5 3F=ð10GbhÞ.
Multiplying by ϕk ðzÞ, integrating, and using Eq. (6.140), we find
2
Ck 5 F= Gbλk sinλk .
Thus, we have constructed the mathematically exact solution of the problem. To study the solution, present the explicit expressions for the displacements and the stresses. The final expression
for the axial displacement follows from Eqs. (6.127) and (6.142), that is,
ux 5 2
X sin αk z
6F
F
3z
2z3
2 3 1
ð2l 2 xÞxz 1
e2sk x
2
3
Eh b
Gb 10h
h
k λk sin λk
!
(6.143)
This displacement satisfies exactly the boundary conditions. We have two ways to simplify it.
First, we can neglect the second term and arrive at the “first-order” theory, Eqs. (6.90) and (6.96).
This approach looks attractive because all the boundary conditions are satisfied. Second, we can
neglect the boundary-layer part of the solution, that is, omit the sum in Eq. (6.143). In this case, we
arrive at the displacement that does not satisfy the boundary condition as per Eq. (6.142) and does
not coincide with Eq. (6.127) for u0x which corresponds to the “third-order” theory. Thus, the
“third-order” theory displacement must be supplemented with the boundary-layer solution.
The beam deflection is specified by Eqs. (6.127), (6.128) and (6.135) which give
rffiffiffiffi
2F
3Eh2
F
E X cosλk
2
ð1 2 e2sk x Þ
x2
uz 5 3 3lx 2 x 1
Eh b
Gb G k λ2k sin λk
5G
(6.144)
Direct calculation shows that the second term in this equation corresponding to the boundarylayer solution is negligible. Returning to the shear correction factor in Eq. (6.28), we can conclude
that for the exact solution (without the boundary-layer part) k 5 5=6, for the “first-order” theory
[see Eq. (6.96)], k 5 1, and for the “third-order” theory [see Eq. (6.108)], k 5 2=3:
The stresses corresponding to the foregoing exact solution can be found from Eqs. (6.130),
(6.136) and presented in the following final form:
rffiffiffiffi
12F
2F E X sinαk z 2sk x
ð
l
2
x
Þz
2
e
bh3
bh G k λk sinλk
6F h2
2F X cosλk 2 cosαk z 2sk x
2 z2 2
e
τ xz 5 3
bh
bh k
λk sinλk
4
σx 5 2
(6.145)
(6.146)
It follows from these equations that the solution contains a polynomial part, which is referred to
as a penetrating solution because it does not vanish at a distance from the beam clamped end, and a
boundary-layer part. It is important that as follows from Eq. (6.141), the boundary-layer solution is
6.7 REFINED THEORIES OF BEAMS AND PLATES
417
self-balanced, that is, it does not contribute to the bending moment and the shear force. Introduce
the normalized normal stress and dimensionless coordinates as follows
σ 5 σx
bh2
x
z
; x5 ; z5
l
h
6F
and present Eq. (6.145) as
h
σ 5 2 2ð1 2 x Þz 2
3
rffiffiffiffi
E X sin 2λk z 2s k x
e
G k λk sin λk
where
2l
sk 5
h
rffiffiffiffi
E
λk
G
and λk are the roots of Eq. (6.139) listed in Table 6.2. The higher roots corresponding to kc10 can
be approximately found as λk ð2k 1 1Þ=ð2πÞ. For a beam with parameters h=l 5 0:1, E=G 5 10,
we have the following stress distribution:
σ 5 2 ð1 2 x Þσ ð0Þ ðz Þ 1 0:1054 σ ð1Þ ðz Þe2s 1 x 1 σ ð2Þ ðz Þe2s 2 x 1 σ ð3Þ ðz Þe2s 3 x 1 . . .
(6.147)
ðiÞ
The functions σ ðz Þ for i 5 0; 1; 2; 3 and the corresponding values of the parameter s i are presented in Fig. 6.22. As can be seen, the higher the gradient of the function σðiÞ ðz Þ, the higher the
rate with which the corresponding boundary-layer solution vanishes at a distance from the clamped
end x 5 0. For example, the contribution of the first boundary-layer solution σð1Þ ðz Þ to the
maximum stress σðz 5 0:5Þ at a distance from the clamped end x 5 0:01 (which is 0.1 of the beam
thickness) does not exceed 0.2 %.
Thus, the improvement of the penetrating solution provided by the boundary-layer solutions is
more of a formal nature. Moreover, for practical analysis, the penetrating solution looks more realistic than the derived exact solution. Indeed, at the corner of the clamped edge (x 5 0; z 5 h=2 in
Fig. 6.20) the series in Eq. (6.145) does not converge, giving rise to infinitely high normal stresses
at these corners. This problem is discussed further in Section 6.7.5.
Consider the shear stress acting in the clamped cross section of the beam. Taking x 5 0 in
Eq. (6.146), we get
τ xz ðx 5 0Þ 5
6F h2
2F X ϕk ðzÞ
2
2
z
2
bh3 4
bh k λk sinλk
(6.148)
Table 6.2 Nonzero Roots of the Equation tanλk 5 λk ðk 5 1; 2; 3. . .10Þ
k
1
2
3
4
5
λk
4.4934
7.7253
10.9041
14.0661
17.2208
k
6
7
8
9
10
λk
20.3713
23.5194
26.6661
29.8116
32.9563
418
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
z
z
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
σ
0
s1 = 28 .42
–0.2
–0.4
–0.6
–0.8 –1.0
10 ⋅ σ (1)
(0)
0.3
0.2
0.1
0
z
0.5
0.4
0.4
0.3
0.3
s 2 = 48 .86
s 3 = 68 .96
0.2
0.1
0.1
–0.3
z
0.5
0.2
0.2
–0.1 –0.2
0.1
10 ⋅ σ ( 2 )
0
–0.1 –0.2
0.2
0.1
10 ⋅ σ ( 3 )
0
–0.1
–0.2
FIGURE 6.22
Functions σðiÞ ðzÞði 5 0; 1; 2; 3 Þand the corresponding values of parameter si .
where ϕk ðzÞ is given by Eq. (6.137). Since ϕk z 5 6 h=2 5 0, we have τ xz z 5 6 h=2 5 0, so
the boundary conditions are satisfied on the beam top and bottom surfaces. Now, use the orthogonality conditions in Eq. (6.141) and decompose the quadratic function in the first term in
Eq. (6.148) into a series with respect to the functions ϕk ðzÞ. Taking into account the fact that in
accordance with Eq. (6.141) the system of functions ϕk ðzÞ is complete only in conjunction with
unity, we have
X
h2
2 z2 5 C 1
Bk ϕk ðzÞ
4
k
where C and Bk are some constant coefficients. Multiplying this equation first by unity and second
by ϕn ðzÞ, integrating from h/2 to h/2, and using Eqs. (6.140) and (6.141), we can determine C
and Bk and finally arrive at
h2
h2
2 z2 5
4
6
X ϕ ðzÞ
k
112
sin
λk
λ
k
k
!
6.7 REFINED THEORIES OF BEAMS AND PLATES
419
Substituting this result into Eq. (6.148), we get
τ xz ðx 5 0Þ 5
F
bh
(6.149)
As can be seen, the shear stress at the clamped end of the beam does not depend on z. This
result is expected, because it follows directly from the constitutive equation, Eq. (6.98) for the
shear stress. Indeed, at the clamped end of the beam [see Fig. (6.20)] ux 5 0 and, hence,
ð@ux Þ=@z 5 0, whereas uz does not depend on z. So, at x 5 0 the shear stress does not change
through the beam height and is specified by Eq. (6.149). However, this equation is valid only for
the open interval 2 h=2 , z , h=2. At the interval ends z 5 6 h=2, the shear stress must be zero
due to the symmetry of the stress tensor (τ xz 5 0 at z 5 6 h=2 according to the boundary conditions). Thus, the distribution of the shear stress over the beam height for a clamped edge of the
beam is not continuous, that is, τ xz is constant for 2 h=2 , z , h=2 and zero for z 5 6 h=2.
Naturally, the derivative @τ xz =@z is infinitely high at the corner points (x 5 0, z 5 6 h=2), and
as follows from the equilibrium equation, Eq. (6.131), the normal stress σx is singular at the
corner points.
6.7.5 NONCLASSICAL ELASTICITY SOLUTION
It follows from the previous section that the solution corresponding to classical theory of elasticity
is singular, that is, the shear stress distribution over the clamped edge of the beam shown in
Fig. 6.20 is not continuous, whereas the normal stresses in the corners of this edge are infinitely
high. This result cannot be treated as correct both mathematically and physically. First, it is obvious
that the derivatives of the stresses which enter the equilibrium equation, Eq. (6.131), do not exist at
the corner points with coordinates x 5 0, z 5 6 h=2 (see Fig. 6.20) and hence, this equation is not
valid at these points. Second, cantilever beams similar to those shown in Fig. 6.20 are widely used
in engineering and are analyzed with the aid of the equations of strength of materials (mechanics of
materials) which do not provide singular solutions. The foregoing results according to which the
maximum stress in the beam is infinitely high actually mean that the cantilever beams cannot work
which is obviously not the case.
Singularities, being quite natural for mathematics, cannot exist in reality. Singular solutions of
physical problems mean that the mathematical model of the problem is not consistent and does not
adequately correspond to the physical problem. Vasiliev and Lurie (2014) have shown experimentally by testing glass beams that the stress singularity does not reduce the beam strength and, hence,
does not have physical nature.
The classical theory of elasticity is based on the so-called phenomenological model of a solid
according to which an infinitesimal element of a solid has the same properties as the whole solid.
This means that the equilibrium equation, Eq. (6.131), and the rest of equations of the theory are
derived for an infinitely small element of a solid. Apply a more general model of a solid (Vasiliev
& Lurie, 2015a, b) assuming that it consists of the elements with small but finite dimensions, a as
shown in Fig. 6.23. Consider an arbitrary point A with coordinates x and z and introduce local coordinates α and β in the vicinity of this point such that 2 a=2 # α # a=2 and 2 a=2 # β # a=2. Let
some continuous field function f (stress, strain, or displacement) be specified in the domain
420
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
z
Tz
4
a 2
Txz
3
β
A
( x, z )
Txz
α
a 2
1
a 2
x
Tx
a 2
2
FIGURE 6.23
Element of a solid with finite dimensions.
½2a=2; a=2 with the derivatives existent up to the third order. Decompose this function into a
Taylor series as
@f
@f
1 @2 f 2
@2 f
@2 f 2
αβ
1
f ðx; z;α; β Þ 5 f ðx; zÞ 1 α 1 β 1
α
1
2
β
@x
@z
2! @x2
@x@z
@z2
3
3
3
3
1 @ f 3
@ f 2
@ f
@ f 3
2
1
α
α
1
3
β
1
3
αβ
1
β
3! @x3
@x2 @z
@x@z2
@z3
(6.150)
Assume that the function f represents the stress σx , σz , or τ xz and construct the equilibrium
equation for the element shown in Fig. 6.23. The resultants of the stresses σx and τ xz acting on the
sides of this element in the x direction are
a
ð2
R223;124 ðσx Þ 5
a
σx x; z; α 5 6 ; β dβ 5
2
a
2
a @σx
a2 @2 σx
1 @2 σx
a3 @3 σx
@3 σ x
1
6
ðx; yÞ
1
1
5 a σx 6
2 @x
3 @z2
8 @x2
48 @x3
@x@z2
2
a
ð2
R324;122 ðτ xz Þ 5
a
dα 5
τ xz x; z; α; β 5 6
2
a
2
a @τ xz
a2 1 @2 τ xz
@2 τ xz
a3 @3 τ xz
@3 τ xz
6
ðx; yÞ
1
1
1
5 a τ xz 6
8 3 @x2
@z2
48 @x2 @z
@z3
2 @z
2
(6.151)
(6.152)
Here, symbol (x, y) means that changing x to y and y to x we can write similar equations for the
y direction. The equilibrium equations can be presented as
R223 ðσx Þ 2 R124 ðσx Þ 1 R324 ðτ xz Þ 2 R124 ðτ xz Þ 5 0 ðx; yÞ
6.7 REFINED THEORIES OF BEAMS AND PLATES
421
Substituting the resultants with their expressions given by Eqs. (6.151) and (6.152), we get
Lx ðT Þ 5
@Tx
@Txz
1
5 0 ðx; yÞ
@x
@z
in which
T ðTx ; Tz ; Txz Þ 5 σðσx ; σz ; τ xz Þ 1
(6.153)
a2 @2
@2
σðσx ; σz ; τ xz Þ
1
24 @x2
@z2
(6.154)
As can be seen, Eq. (6.153) has the same form as the traditional equilibrium equation,
Eq. (6.131), but it includes the generalized stresses T instead of traditional stresses σ.
Using the traditional definition of strain as the ratio of the absolute deformation of the element
in Fig. 6.23 to the corresponding initial dimensions, we get for the normal and shear strains
1
Ex 5 u223
2 u124
ðx; zÞ
x
x
a
1
2 u122
1 u223
2 u124
Exz 5 u324
x
z
z
a x
(6.155)
Here
u223;124
5
x
1
a
a
ð2
a
ux x; z; α 5 6 ; β dβ
2
a
2
2
a
ð2 1
a
5
x;
z;
α;
β
5
6
dα
u324;122
u
x
x
a
2
a
2
2
ðx; zÞ
(6.156)
ðx; zÞ
are the average displacements of the edges of the element in Fig. 6.23 expressed in terms of the
actual displacements ux and uz . Decomposing these displacements with the aid of Eq. (6.150) (i.e.,
taking f 5 ux and f 5 uz , and substituting the resulting expressions into Eqs. (6.154) and (6.155),
we finally get
Ex 5
@Ux
;
@x
Ez 5
@Uz
;
@z
@Ux
@Uz
1
@z
@x
(6.157)
a2 @2
@2
uðux ; uz Þ
1
24 @x2
@z2
(6.158)
Exz 5
where
U ðUx ; Uz Þ 5 uðux ; uz Þ 1
are the generalized displacements similar to the generalized stresses in Eq. (6.154). The generalized
displacements have a simple physical meaning: they correspond to the displacements of the element
in Fig. 6.23 averaged over its area. Using Eq. (6.150), we have
1
ux 5 2
a
ZZa=2
ux ðx; z; α; β Þdxdβ 5 Ux
2a=2
ðx; zÞ
(6.159)
422
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
Now assume that the generalized stresses and strains are linked by Hooke’s law, that is,
Tx 5
E
ðEx 1 νEz Þ
1 2 ν2
ðx; zÞ;
Txz 5
E
Exz
2ð1 1 νÞ
(6.160)
For a homogeneous stress-strain state (i.e., not depending on coordinates), T 5 σ, U 5 u and
Eq. (6.160) reduce to the traditional Hooke’s law.
To complete the theory, apply the principle of virtual displacements to the element in
Fig. 6.23 to get
ZZ
½Lx ðT ÞδUx 1 Lz ðTÞδUz dxdz 5 0
(6.161)
Here, L are specified by Eq. (6.153) and δU are the variations of the average displacements of
the element in Eq. (6.159). Performing variation in Eq. (6.161), we arrive at the following natural
boundary conditions:
Tx δUx 5 0;
Txz δUz 5 0
(6.162)
Thus, the proposed generalized theory reduces to equilibrium equations, Eq. (6.153), straindisplacement relationships, Eq. (6.157), constitutive equations, Eq. (6.160), and the boundary
conditions in Eq. (6.162). As can be seen, these equations completely coincide with the corresponding equations of the classical theory of elasticity, but include generalized stresses, strains and
displacements given by Eqs. (6.154), (6.157), and (6.158). Consequently, the generalized solution
of the problem coincides with the traditional solution of the classical theory of elasticity. Actual
stresses and displacements can be found from Eqs. (6.154) and (6.158).
In conclusion, a main principal point needs to be made. As can be seen, Eqs. (6.154) and
(6.158) include some structural parameter a which is not known and must be determined experimentally. However, the constructed theory is of a phenomenological nature, that is, it does not
allow for the actual microstructure of a solid. Consequently, the structural parameter a, in principle,
cannot be determined using this theory and possible macrostructural
experiments. To avoid this dispffiffiffi
crepancy, we put (see Vasiliev & Lurie, 2015a, b) a 5 2i 6r, where i2 5 2 1 and r is the real
parameter that can be found experimentally. As a result, Eqs. (6.154) and (6.158) reduce to
2
@ σ
@2 σ
1
@x2
@z2
2
@ u @2 u
1
U 5 u 2 r2
@x2
@z2
T 5 σ 2 r2
(6.163)
(6.164)
Now consider bending of the cantilever beam shown in Fig. 6.20. To find the axial displacement
ux , we should use Eq. (6.164) in which U is specified by the corresponding solution of the classical
theory of elasticity, that is, by Eq. (6.143). The resulting equation is presented as
@2 ux
@2 ux
1
6F
1 2 2 2 ux 5 2
ð2l 2 xÞxz
2
@x
@z
r
Ebh3 r2
F
2
Gbr 2
X sinαk z
3z
2z3
e2sk x
2 3 1
2
h
10h
sinλ
λ
k
k
k
!
(6.165)
6.7 REFINED THEORIES OF BEAMS AND PLATES
423
The general solution of this equation can be given by
6F 2
4F 3h2
z2
2
2
2
3r
z
x
1
2ðr
2
lxÞ
z
1
2
Ebh3
Gbh3 40
2
F X
sinα z
k
e2sk x
2
Gb k λ2k sinλk r 2 s2k 2 α2k 2 1
X
X
1 A z3 1 3rxz e2x=r 1
Ak1 e2ηk x sinαk z 1
Ak2 e2ξk x sinβ k z
ux 5
k
(6.166)
k
where
η2k 5 α2k 1
1
;
r2
ξ 2k 5 β 2k 1
1
;
r2
βk 5
πð2k 2 1Þ
h
The terms including the force F in Eq. (6.166) are the particular solutions of Eq. (6.165),
whereas the other terms represent the solutions of the homogeneous equation corresponding to
Eq. (6.165). These last terms allow us to satisfy the boundary condition ux ðx 5 0; zÞ 5 0 at
the clamped end of the beam (see Fig. 6.20). Using this condition and Eqs. (6.139) and (6.140),
we get
F
Gbλ2k sinλk r 2 s2k 2 α2k 2 1
48F
E h2
2k 2 1
2
2
2
r
sin
r
1
Ak2 5
G 40
2
Ebh2 π2 ð2k21Þ2
A5
2F
;
Gbh
Ak1 5
(6.167)
Thus, the axial displacement is specified by Eqs. (6.166) and (6.167). The final expression is
rather cumbersome and is omitted.
The beam transverse displacement uz satisfies the equation following from Eqs. (6.164) and
(6.144), that is,
d 2 uz
1
2F
3Eh2
2
2
u
5
2
3lx
2
x
1
z
r2
Ebh3 r 2
dx2
5G
rffiffiffiffi
X
F
E
cosλk
ð1 2 e2sk x Þ
1
Gbr 2 G k λ2k sinλk
The solution of this equation satisfying the boundary condition uz ðx 5 0Þ 5 0 at the clamped end
of the beam (see Fig. 6.20) is
6F
x
6F
2
2
ð
Þ
1
2r
x
x
l
2
l
2
x
1
Ebh3
3
5Gbh
r
ffiffiffi
ffi
12Flr2 2x=r
F
EX 1
e2sk x 2 r 2 s2k e2x=r
2
e
2
11
Ebh3
Gb G k λ3k
r 2 s2k 2 1
uz 5
(6.168)
Determine the stresses. The shear stress τ xz can be found from Eq. (6.163) in which Txz is specified by Eq. (6.146), that is,
@2 τ xz
@2 τ xz
1
6F h2
2
2
z
1
2
τ
5
2
xz
@x2
@z2
r2
bh3 r2 4
2F X cosλk 2 cosαk z 2sk z
1
e
2bhr 2 k
λk sinλk
(6.169)
424
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
The general solution of this equation is
(
)
6F h2
2F X
1
cosαk z
2
2
2
2 z 2 2r 1
τ xz 5 2 3
e2sk x
bh
bh k
4
λk sinλk r 2 s2k 2 α2k 2 1
λ2k r 2 s2k 2 1
X
z
1Bcosh 1
Bk e2sk x cosγ k z
r
k
(6.170)
where γ 2k 5 s2k 2 r12 .
As earlier, the terms with F in Eq. (6.170) are the particular solutions of Eq. (6.169), whereas
the other terms are the solutions of the homogeneous
to Eq. (6.169)
equation corresponding
which allow us to satisfy the boundary conditions τ xz x; z 5 6 h=2 5 0 on the longitudinal top
and bottom surfaces of the beam (see Fig. 6.22). Using these conditions and Eqs. (6.139) and
(6.140), we get
B5
12Fr 2
bh3 cosh h=2r
;
Bk 5
bhλ2k
2Fr 2 α2k
2
2
2
2
r sk 2 1 r sk 2 α2k 2 1 cos γ k h=2
The final expression for the shear stress is
τ xz 5
"
#)
cosh z=r
h2
2 z2 2 2r2 1 2
4
cosh h=2r
(
)
2F X
1
cosαk z
r 2 α2k cosγ k z
1
2
e2sk x
1
bh k
λk sinλk r2 s2k 2 α2k 2 1
λ2k r 2 s2k 2 1
λ2k r2 s2k 2 1 r2 s2k 2 α2k 2 1 cos γ k h=2
6F
bh3
(
(6.171)
The series entering this solution converge and, in contrast to Eq. (6.146), it specifies the continuous distribution of τ xz over the beam height.
Finally, determine the normal stress σx which can be found from Eq. (6.163) in which Tx is
specified by Eq. (6.145), that is,
@2 σx
@2 σx
1
12F
1 2 2 2 σx 5 3 2 ðl 2 xÞz
2
r
bh r
@x
@z
rffiffiffiffi
4F
E X sinαk z 2sk x
1
e
bhr 2 G k λk sinλk
(6.172)
Since there are no boundary conditions imposed on σx , we can take only the particular solution
of Eq. (6.172), which has the form
σx 5 2
12F
2F
ðl 2 xÞz 2
bh3
bh
rffiffiffiffi
EX
e2sk x sinαk y
G k λk sinλk 1 2 r 2 s2k 2 α2k
(6.173)
The series in this solution converges and, as opposed to the classical case given by Eq. (6.145),
this solution is not singular.
The solutions obtained in this study include three experimental constants: E, G, and r. To determine these constants, the experimental beam shown in Fig. 6.24 has been studied. The beam is
made of hybrid carbon-glass-epoxy plies of composite material made by angle-ply
circumferential winding (see Section 2.6). The beam has the following geometric parameters
6.7 REFINED THEORIES OF BEAMS AND PLATES
425
FIGURE 6.24
Experimental composite beam.
(Fig. 6.20): l 5 300 mm, b 5 89.7 mm, h 5 53.8 mm. The axial modulus Ex 5 105:84 GPa has
been measured under axial compression of the beam. To find the shear modulus G, a bending test
has been used. For a beam with the length l, which is much larger than the beam height h, we can
neglect the boundary-layer parts of the solution and present the beam deflection in Eq. (6.168) as
uz ðxÞ 5
i
6F h 2
x
6F
x l 2 1 2r 2 ðl 2 xÞ 1
x
3
Ebh
3
5Gbh
(6.174)
The deflection of the beam end is
uz ðlÞ 5
4Fl3
6Fl
1
5Gbh
Ebh3
Measuring uz ðlÞ corresponding to the applied force F, we get G 5 5.926 GPa.
To determine the parameter r, calculate the shear strain εxz . Using Eqs. (6.166)(6.168) and
neglecting the boundary-layer parts of the solution, we have for a relatively long beam
εxz 5
@ux
@uz
12Fr 2
6F h2
2
2
1
5
2
z
1
2
2r
@z
@x
Ebh3
Gbh3 4
(6.175)
The maximum value of this strain corresponds to z 5 0, that is,
εm
xz 5
3Fr 2
6F h2
2
2
2r
1
Gbh3 4
Ebh3
Solving this equation for r 2 , we get
h2
2Gbh m
12
εxz
r2 5 3F
8 1 2 E=G
(6.176)
426
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
FIGURE 6.25
Bending of the experimental beam.
Thus, to determine r, we need to measure εm
xz . The idea of the experiment is demonstrated in
Fig. 6.25. A laser 1 whose beam is focused on the screen at point A is placed on the top surface of
a composite beam. A laser 2 whose beam is focused at point B of the screen is fixed at the level of
the middle plane of the beam cross section. The lasers fixed to the beam are shown in Fig. 6.24.
Under the action of the load F applied to the beam end, the points A and B move to their new locations at A1 and B1 , respectively. The angles of the laser beam rotations are α1 5 ðAA1 Þ=L and
α2 5 ðBB1 Þ=L where L is the distance between the end cross section of the composite beam and the
screen (see Fig. 6.25). Then, the shear strain can be calculated as εm
xz 5 α1 2 α2 . For a load
F 5 24.5 kN, the experiment yields εm
5
0:001271.
Then,
Eq.
(6.176)
gives r 5 0.0381h 5
xz
2.05 mm. Note that the experimental beam has been assembled of the layers having a thickness of
about 1.5 mm.
To demonstrate a specific feature of the theory, consider the shear stresses at a distance
from the clamped end of the beam. Neglecting the boundary-layer part of the solution in
Eq. (6.171), we get
6F
τ xz 5 3
bh
(
"
#)
cosh z=r
h2
2
2
2 z 2 2r 1 2
4
cosh h=2r
Comparing this stress with the shear strain in Eq. (6.175), we can conclude that Hooke’s law
for shear, that is, τ xz 5 Gεxz , is satisfied only if r 5 0. For r 6¼ 0, Hooke’s law for shear is violated.
The reason is associated with the nonzero gradients of the functions τ xz ðzÞ and εxz ðzÞ, because of
which τ xz and εxz do not coincide with the corresponding generalized stress and strain Txz and Exz .
The fact that Hooke’s law is not valid for the stress and strain fields with gradients has been
demonstrated experimentally by Andreev (1976).
Consider the results of the numerical analysis for the beam with the foregoing parameters. The
dependence of the normalized beam deflection uz 5 uz =um , where um 5 4Fl3 =Ebh3 is the maximum
deflection corresponding to the conventional solution of strength of materials, on coordinate x 5 x=l
is shown in Fig. 6.26. The solid line corresponds to the strength of materials solution (i.e., the classical beam theory), whereas the dashed line denotes the solution given by Eq. (6.168). Note that
6.7 REFINED THEORIES OF BEAMS AND PLATES
427
FIGURE 6.26
Dependencies of the normalized deflection on the axial coordinate corresponding to strength of materials (ss)
and the solution in Eq. (6.168) (s s s).
FIGURE 6.27
Distributions of the normalized normal stress acting in the fixed cross section of the beam over the beam
thickness corresponding to strength of materials (ss), Eqs. (6.145) (s s s) and (6.173) ( ).
the results of the calculation for the first-order theory, Eq. (6.96), practically coincide with the
dashed line.
The distribution of the normalized stress σx 5 σx =σm , where σm 5 2 6Fl=bh2 is the maximum
stress corresponding to strength of materials solution, over the height of the fixed cross section of
the beam (x 5 0; y 5 y=hÞ is presented in Fig. 6.27. The solid line demonstrates the strength of
materials (and the first-order theory) solution, the dashed line corresponds to the singular solution
of the classical theory of elasticity given by Eq. (6.145), whereas the dotted line shows a refined
solution obtained as per Eq. (6.173). As can be seen, the refined solution provides a maximum
stress which is 12% higher than that calculated using the strength of materials formula. The distribution of the normalized shear stress τ xz 5 τ xz =τ m , where τ m 5 3F=2bh is the maximum stress corresponding to the strength of materials theory, over the height of the beam clamped cross section
(x 5 0; y 5 y=hÞ is shown in Fig. 6.28. The solid line corresponds to the strength of materials
428
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
FIGURE 6.28
Distributions of the normalized shear stress acting in the fixed cross section of the beam over the beam thickness
corresponding to strength of materials (ss), Eqs. (6.146) (s s s), and (6.171) ( ).
solution, the dashed line shows the discontinuous solution of the classical theory of elasticity as per
Eq. (6.146), whereas the dotted line presents the solution given by Eq. (6.171). As can be seen, the
shear stress, specified by the new solution is continuous.
6.8 LAMINATED BEAMS
Consider laminated beams. The first solution for a laminated cantilever beam was obtained by S.G.
Lekhnitskii in 1935 (Lekhnitskii, 1935) and reviewed by Carrera (2003). The biharmonic equation
of elasticity theory for a laminated beam was solved with the aid of a polynomial representation of
the stress function for each layer. Thus, S.G. Lekhnitskii has constructed a penetrating solution
ignoring boundary-layer solutions. Equations describing the combination of a penetrating and
boundary-layer solutions have been derived by Bolotin (1963). The laminate consists of a system
of load-carrying layers alternating with core layers which take only transverse stresses (see
Fig. 6.29). The axial displacements of the load-carrying layers correspond to classical plate theory,
according to which
0
0
uðiÞ
x 5 ui 2 zi wi ;
uði11Þ
5 ui11 2 zi11 wi11
x
(6.177)
0
(Bolotin & Novichkov, 1980), where ð. . .Þ 5 dð. . .Þ=dx, zi is counted from the middle surface of
the i-th layer (2hi =2 # zi # hi =2Þ, and wi is the deflection of the i-th layer. The axial stress in the
i-th layer is
00 ðiÞ ðiÞ
ðiÞ 0
σðiÞ
x 5 A11 εx 5 A11 ui 2 zi wi
For the core layers, the axial stress is zero and the shear strain does not depend on z, that is,
ði;i11Þ
γ xz
5
1
hi;i11
ui11 2 ui 1
1 0
1 0
0
0
wi hi 1 wi11 hi11 1 wi11 1 wi
2
2
6.8 LAMINATED BEAMS
429
z i +1
hi +1
u i +1
h i , i +1
zi
hi
ui
FIGURE 6.29
Distribution of the axial displacement over the height of a laminated beam.
The transverse normal strain of the core layers is
εði;i11Þ
5
z
1
ðwi11 2 wi Þ
hi;i11
The governing equations for the functions ui and wi and the corresponding natural boundary
conditions are derived with the aid of the variational principle. The total order of these equations
depends on the number of layers. The solutions of numerous problems have been obtained by
Bolotin and Novichkov (1980). They include the penetrating solutions and the system of boundarylayer solutions which describe local bending of load-carrying layers, as well as shear and normal
strains of the core layers rapidly vanishing from the beam edge.
A theory that is more realistic for composite laminates (that do not have core layers) has been
proposed by Grigolyuk and Chulkov (1964). In this theory, the axial stiffness of the core layers is
not ignored. A similar theory with membrane load-carrying layers have been developed by
Elpatievskii and Vasiliev (1965).
The main shortcoming of the theories under consideration is the high order of the governing
equations (which depends on the number of layers) in which the penetrating solution is mixed with
boundary-layer solutions. In modern versions of the theory (Reddy, 2004), the penetrating solution
(the corresponding displacement is shown in Fig. 6.29 with a dashed line) is singled out and the
shear deformation of the layers is taken into account. The resulting expressions for the displacements become
P
ux 5 uðxÞ 1 zθðxÞ 1 i ui ðxÞϕi ðzÞ
P
uz 5 w 1 i wi ψi ðzÞ
(6.178)
where z is the global normal coordinate of the laminate and ϕi ðzÞ, ψi ðzÞ are some functions that provide continuity of the displacements, the derivatives of which with respect to z are not continuous
at the layer interfaces. Different forms of such functions have been presented by Carrera (2003).
Since three functions for the whole laminate (u; θ; and w) are introduced in Eq. (6.178), the number
of functions ui and wi can be respectively reduced. For example, we can take for the first and the
last (k-th) layers u1 5 0, w1 5 0, uk 5 0 (Reddy, 2004).
430
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
z2
h
2
x
F
z1
1
h0
h
l
b
FIGURE 6.30
A cantilever sandwich beam.
In the simplest version of the theory, the deflection is assumed to be the same for all the
layers, that is, uz 5 wðxÞ, whereas the functions ϕi ðzÞ in Eq. (6.178) are some linear functions
providing the continuity of the axial displacement ux with respect to z (Carrera, 2003). Applying
the continuity conditions for the shear stresses at the layer interface surfaces, we can express
ui ðxÞ in terms of two functions θðxÞ and wðxÞ (Carrera, 2003). As a result, the order of the governing equations for this theory does not depend on the number of layers. The theory constructed in
this way is similar to the “higher-order” theory for homogeneous beams discussed earlier.
Return to the general theory based on Eqs. (6.177) or (6.178). The most pronounced difference
between this theory and the theory based on a linear distribution of the axial displacement through
the beam height is manifested in sandwich beams with stiff load-carrying layers and a shear
deformable lightweight core.
Consider as an example the cantilever sandwich beam shown in Fig. 6.30. The axial displacements of the first and the second layers (see Fig. 6.30) following from Eqs. (6.177) and (6.178) are
H
0
2 z1 θðxÞ 2 z1 w ðxÞ
u1 5 uðxÞ 2
2
H
0
2 z2 θðxÞ 2 z2 w ðxÞ
u2 5 uðxÞ 2
2
(6.179)
Here, uðxÞ, θðxÞ, and wðxÞ are the axial displacement, rotational angle, and the deflection
(the same for both layers), H 5 h0 1 h, 2h=2 # zi # h=2, ði 5 1; 2Þ. The shear strain in the
core is
γ5
1
h
h
H
0
0
2 u1 z1 5
1 w 5 ðθ 1 w Þ
u2 z2 5 2
h0
2
2
h0
Then, the normal stresses in the layers and the shear stress in the core become
H
0
0
σð1Þ 5 E u 2
2 z1 θ 2 z1 w00
2
H
0
0
ð2Þ
00
σ 5E u 2
2 z2 θ 2 z2 w
2
H
0
τ 5 G ðθ 1 w Þ
h0
(6.180)
6.8 LAMINATED BEAMS
431
where E is the modulus of the layer and G is the shear modulus of the core material. The forces
and the moments in the layers are
h=2
ð
N1 5 b
2h=2
h=2
ð
N2 5 b
2h=2
h=2
ð
M1 5 b
H
0
σð1Þ dz1 5 Bl u 2 θ0
2
H
0
σð2Þ dz2 5 Bl u 1 θ0
2
(6.181)
0
σð1Þ z1 dz1 5 2 Dl θ 1 w00
2h=2
h=2
ð
M2 5 b
0
σð2Þ z2 dz2 5 Dl θ 2 w00
2h=2
where
Bl 5 Ebh; Dl 5
1
Ebh3
12
(6.182)
are the membrane and bending stiffnesses of the layers. The equilibrium equations for the layers
(see Fig. 6.31A) can be written as
1
0
M1 2 V1 1 τbh 5 0;
2
0
M2 2 V2 1
1
τbh 5 0
2
(6.183)
From Eq. (6.180) for τ, (6.181) for M, and Eq. (6.183) we can find the transverse forces in the
layers, that is,
00
1
1
τbh 5 2 Dl θ 1 w000 1 GbhH ðθ 1 w0 Þ
2
2
00
1
1
0
000
V2 5 M2 1 τbh 5 Dl θ 2 w 1 GbhH ðθ 1 w0 Þ
2
2
0
V1 5 M1 1
Then, the total shear force
V 5 V1 1 V2 1 bh0 τ
becomes
0
000
V 5 2 2Dl w 1 S w 1 θ
(6.184)
S 5 GbH
(6.185)
where
is the shear stiffness of the beam. Finally, the total axial force and bending moment can be
presented as
0
N 5 N1 1 N2 5 2Bl u ;
00
M 5 M1 1 M2 5 2 2Dl w 1
1
0
Bl H 2 θ
2
(6.186)
432
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
V1+ V1′dx
(A)
N1 + N1′dx
N1
M 1 V1
N2
M2
V2
τ
M2 + M2′ dx
τ
V 2 + V2′dx
N2 + N2′ dx
M1+ M1′dx
dx
(B)
V + V ′dx
N
N + N ′dx
M + M ′dx
M
V
dx
FIGURE 6.31
Free-body diagrams for a sandwich beam: equilibrium of the layers (A), and equilibrium of the beam
element (B).
The equilibrium equations of the beam element follow from Fig. 6.31B and have the form
0
N 5 0;
0
M 5 V;
0
V 50
(6.187)
The boundary conditions for the clamped end of the beam (see Fig. 6.30), that is,
uðx 5 0Þ 5 0;
θðx 5 0Þ 5 0;
0
w ðx 5 0Þ 5 0
(6.188)
provide with regard to Eq. (6.179) u1 ðx 5 0Þ 5 0 and u2 ðx 5 0Þ 5 0. For the end x 5 l (see
Fig. 6.30), we have
N 5 0;
V 5 F;
M50
(6.189)
The first equations in Eqs. (6.186)(6.189) yield N 5 0 and u 5 0. Eq. (6.187) for V and the
corresponding boundary condition in Eq. (6.189) give V 5 F. Applying Eq. (6.184), we get
0
22Dl ww 1 S w 1 θ 5 F
(6.190)
0
Now, according to the second equilibrium equation in Eq. (6.187), M 5 F and M 5 Fx 1 C1 .
Since Mðx 5 lÞ 5 0, we arrive at M 5 2 Fðl 2 xÞ. Then, the second equation of Eq. (6.186) yields
1
0
22Dl wv 1 Bl H 2 θ 5 2 Fðl 2 xÞ
2
6.8 LAMINATED BEAMS
Integration gives
0
2Dl w 1
433
1
x2
Bl H 2 θ 5 F lx 2
1 C2
2
2
Using the boundary conditions in Eq. (6.188), we have C2 5 0 and
0
2Dl w 1
1
F
Bl H 2 θ 5
2lx 2 x2
2
2
(6.191)
Thus, the problem of a beam in bending is reduced to Eqs. (6.190) and (6.191). Eliminating θ
from these equations and integrating the resulting equation, we arrive at
00
w 2 k2 w 5 2
FS
x3
Bl H 2
2
1
x
1 C3
lx
2
2Dl Bl H 2
3
S
(6.192)
where C3 is the constant of integration and
k2 5
Dt S
1
; Dt 5 2Dl 1 Bl H 2
Dl Bl H 2
2
(6.193)
Here, Dt is the total bending stiffness of the beam. The solution of Eq. (6.192) consists of two
parts. The first part is the polynomial (i.e., penetrating solution) and the second part is the
boundary-layer solution. Retaining the boundary-layer solution in the vicinity of the clamped end
only, we get
w 5 C4 e2kx 1
FBl H 2
2Dl
Fl 2
F 3
FSl
Dl Bl H 2
12
x1
x 2
x 1
2
C
3
2Dt
6Dt
2Dl Bl H 2
2Dt S
Dt
Dt S
The constants of integration C3 and C4 are found from the boundary conditions wð0Þ 5 0 and
θð0Þ 5 0. Then, the final expressions for w and θ become
w5
rffiffiffiffiffiffiffiffiffi
FBl H 2
2Dl
Bl Dl 2kx
Fl 2
F 3
12
e 21 1x 1
H
x 2
x
2Dt S
Dt
Dt S
2Dt
6Dt
F
2Dl 2Dl Bl H 2 2kx
2
1
2
e
θ5
1
2
2
2lx
1
x
Bl H 2
Dt
Dt S
(6.194)
These expressions include the parameter d 5 Dl =Dt which is the ratio of the bending stiffness of
the layer to the total bending stiffness of the beam. Using Eqs. (6.182) and (6.193) for the stiffness
coefficients, we get
d5
h2
h2 1 6H 2
;
H 5 h0 1 h
For practical sandwich beams, this parameter is normally rather small. For example,
for h0 5 15mm, h 5 1mm (see Fig. 6.30), d 5 0:00065. Thus, we can simplify Eq. (6.194) taking
d 5 0. Then,
w5
F
6Dt
x;
3lx 2 x2 1
6Dt
S
θ52
F
ð2l 2 xÞx
2Dt
where Dt 5 Bl H 2 =2. These equations coincide exactly with Eq. (6.96) which correspond to a linear
distribution of the axial displacement through the beam height.
434
CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS
Table 6.3 Normalized Maximum Deflections of Sandwich Beams
E, GPa
G, MPa
l, mm
h, mm
h0 , mm
wð1Þ
m
ð2Þ
wm
wð3Þ
m
70
70
70
70
70
55.7
915
915
38.5
38.5
280
480
280
280
280
0.7
0.7
0.7
1.0
2.4
3.7
4.8
5.1
17.0
18.8
0.990
0.969
0.856
0.329
0.171
0.992
0.983
0.986
0.923
0.921
1.027
1.020
1.033
0.985
1.053
Taking into account the transverse normal strain of the core εz , we arrive at an additional
boundary-layer solution rapidly vanishing at a distance from the clamped end.
For simply supported sandwich beams (see Fig. 6.21) loaded with pressure q 5 q0 sin πx=l ,
the comparison of the maximum deflection wm corresponding to various beam models with the
experimental results wem obtained by Alexandrov, Bryukker, Kurshin, and Prusakov (1960) is presented in Table 6.3. In this table, wm 5 wm =wem , wð1Þ
m corresponds to the classical beam theory
for which S-N, wð2Þ
specifies the results corresponding to the displacements given by
m
Eq. (6.179), whereas deflections wð3Þ
m show the solution based on the linear approximation of the
ð3Þ
axial displacement. It follows from Table 6.3 that wð2Þ
m and wm are characterized by a comparable
accuracy for all the beams. Thus, the theory constructed in Section 6.1 allows us to describe the
beam behavior with reasonable accuracy and is generalized in Chapter 7, Laminated Composite
Plates for composite plates.
REFERENCES
Alexandrov, A. Ya, Bryukker, L. E., Kurshin, L. M., & Prusakov, A. P. (1960). Analysis of sandwich plates.
Moscow: Oborongiz, in Russian.
Ambartsumyan, S. A. (1958). On a theory of bending of anisotropic plates. Proceedimgs of USSR Academy of
Sciences (Izvestia AN SSSR, Otdelenie Technicheskih Nauk), 5, 6977, in Russian.
Ambartsumyan, S. A. (1987). Theory of anisotropic plates. Moscow: Nauka, in Russian.
Andreev, A. V. (1976). Experimental study of stress concentration in machine parts. Moscow: Masinostroenie,
in Russian.
Birman, V., & Bert, C. W. (2002). On the choice of shear correction factor in sandwich structures. Journal of
Sandwich Structures and Materials, 4, 8395.
Bolotin, V. V. (1963). To the theory of laminated plates. Proceedings of Russian Academy of Sciences
(Izvestia AN SSSR, Otdelenie Technicheskih Nauk, Mechanika and Mashinostroenie), 3, 6572, in
Russian.
Bolotin, V. V., & Novichkov, Yu. N. (1980). Mechanics of multilayered structures. Moscow: Machinostroenie,
in Russian.
Carrera, E. (2003). Historical review of zig-zag theories for multilayered plates and shells. Applied Mechanics
Reviews, 56(3), 287308.
Dekker, M. (2004). Structural analysis of polymeric composite materials. New York: Marcell Dekker, Inc.
Elpatievskii, A. N., & Vasiliev, V. V. (1965). Analysis of the stress state of filament-wound fiberglass
cylindrical shell. Sov. Engineering Journal, 5(1), 129142, in Russian.
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Gay, D., Hoa, S. V., & Tsai, S. W. (2003). Composite materials design and applications. Boca Raton: CRC
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Goens, E. (1931). Uber die Bestimmung des Elastizitatsmodule von Staben mit Hilfe von
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Goldenveizer, A. L. (1958). On Reissner’s theory of plate bending. Proceedings of USSR Academy of Sciences
(Izvestia AN SSSR, Otdelenie Technicheskih Nauk), 4, 102109, in Russian.
Grigolyuk, E. I., & Chulkov, P. P. (1964). Theory of viscoelastic multilayered shells with load-carrying
core layers. Applied Mechanics and Technical Physics (Prikladnaya Mechanika i Technicheskaya
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Kollar, L. P., & Springer, G. S. (2003). Mechanics of composite structures. Cambridge University Press:
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Lekhnitskii, S. G. (1935). Strength calculation of composite beams. Vestnik inzhenerov i tekhnikov(9), in
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Mindlin, R. D. (1951). Influence of rotator inertia and shear on flexural motions of isotropic, elastic plates.
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Reddy, J. N. (2004). Mechanics of laminated composite plates and shells: Theory and Analysis (2nd ed). Boca
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Reismann, H. (1988). Elastic plates: theory and application. New York, USA: John Wiley and Sons.
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Timoshenko, S. P. (1921). On the correction for shear of the differential equation for transverse vibrations of
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Timoshenko, S. P. (1922). On the transverse vibrations of bars of uniform cross-section. Philosophical
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Uflyand, Ya. S. (1948). Propagation of waves under transverse vibrations of beams and plates. Applied
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Vasiliev, V. V. (1993). Mechanics of composite structures. Washington, USA: Taylor & Francis.
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Vasiliev, V. V., & Lurie, S. A. (1972). A version of refined beam theory for laminated plastics. Mechanics of
Polymers (Mechanika Polymerov), Riga, Zinatne(4), , 674681, in Russian.
Vasiliev, V. V., & Lurie, S. A. (1990a). To the problem of development of nonclassical plate theories.
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Vasiliev, V. V., & Lurie, S. A. (1990b). To the problem of refinement of the shallow shell theory.
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Vasiliev, V. V., & Lurie, S. A. (1992). On refined theories of beams, plates and shells. Journal of Composite
Materials, 26(4), 546557.
Vasiliev, V. V., & Lurie, S. A. (2014). On singular solution in plane problem of theory of elasticity for a
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CHAPTER
LAMINATED COMPOSITE PLATES
7
Laminated composite plates possessing high strength and stiffness under in-plane loading and bending are widely used nowadays in composite aircraft and marine structures and have been discussed
by many authors, particularly by Lekhnitskii (1957), Ambartsumian (1967, 1987), Ashton and
Whitney (1970), Whitney (1987), Jones (1975, 1999), Pikul (1977, 1985), Vinson and Sierakowski
(1986, 2004), Vasiliev (1993), Reddy (1997, 2004), Andreev and Nemirovskii (2001), Decolon
(2002), Gay, Hoa, and Tsai (2003), Ye (2003), Kollar and Springer (2003), Vinson (2005), and
Dekker (2009). This chapter is concerned with traditional and advanced specific problems and
applications of the theory of anisotropic plates.
7.1 EQUATIONS OF THE THEORY OF ANISOTROPIC LAMINATED PLATES
Consider an anisotropic laminated plate referred to coordinates x, y, z as shown in Fig. 7.1. The reference plane z 5 0 is located at distances e and s from the bottom and top surfaces (see Chapter 3:
Mechanics of Laminates). The displacement distributions over the plate thickness are specified by
Eqs. (3.1) and (3.2) according to which
ux 5 uðx; yÞ 1 zθx ðx; yÞ;
uy 5 vðx; yÞ 1 zθy ðx; yÞ;
uz 5 wðx; yÞ
(7.1)
where u, v, and w are the displacements of the normal element cd (see Fig. 7.1) in the directions
of the coordinate axes x, y, and z, respectively, whereas θx and θy are the rotation angles of the
element in the xz and yz planes, respectively.
In plate theory, the displacements given by Eq. (7.1) were originally introduced by Hencky
(1947) and Bolle (1947) and later used by Uflyand (1948) and Mindlin (1951), whereas the first
version of the theory based on a stress formulation has been constructed by Reissner (1944). Brief
historical reviews of the problem have been presented by Reissner (1985), Vasiliev (1992, 1998,
2000), and Jemielita (1993).
For the displacements presented by Eq. (7.1), the in-plane stresses σx , σy , and τ xy can be found
from Eqs. (3.4), i.e.,
σx 5 A11 ε0x 1 A12 ε0y 1 A14 γ 0xy 1 z A11 κx 1 A12 κy 1 A14 κxy
σy 5 A21 ε0x 1 A22 ε0y 1 A24 γ 0xy 1 z A21 κx 1 A22 κy 1 A24 κxy
τ xy 5 A41 ε0x 1 A42 ε0y 1 A44 γ 0xy 1 z A41 κx 1 A42 κy 1 A44 κxy
(7.2)
where Amn 5 Anm are the stiffness coefficients of the material, which, in general, depend on the
z-coordinate and
Advanced Mechanics of Composite Materials and Structures. DOI: https://doi.org/10.1016/B978-0-08-102209-2.00007-4
© 2018 Elsevier Ltd. All rights reserved.
437
438
CHAPTER 7 LAMINATED COMPOSITE PLATES
z
b
s
a
d
c
x
e
q
h
y
p
FIGURE 7.1
A laminated plate.
Vx
dx
dy
Vy
z
0
Nx
Nxy
Mxy
My
Mx
Mxy
Ny
FIGURE 7.2
Stress resultants and couples acting on the reference plane of a plate.
@u
;
@x
@θx
κx 5
;
@x
ε0x 5
@v
;
@y
@θy
κy 5
;
@y
ε0y 5
@u @v
1
@y
@x
@θx
@θy
κxy 5
1
@y
@x
γ 0xy 5
(7.3)
are the in-plane and bending deformations of the reference surface z 5 0. The stresses in Eq. (7.2)
are reduced to stress resultants and couples as follows:
ðs
Nx 5
ðs
σx dz;
2e
σx zdz;
2e
σy dz;
2e
ðs
ðs
Mx 5
ðs
Ny 5
My 5
σy zdz;
2e
Nxy 5
τ xy dz
2e
ðs
Mxy 5
(7.4)
τ xy zdz
2e
which are applied to the reference surface as shown in Fig. 7.2. Substitution of Eq. (7.2) into
Eq. (7.4) yields the constitutive equations similar to Eq. (3.5), i.e.,
Nx 5 B11 ε0x 1 B12 ε0y 1 B14 γ 0xy 1 C11 κx 1 C12 κy 1 C14 κxy
Ny 5 B21 ε0x 1 B22 ε0y 1 B24 γ 0xy 1 C21 κx 1 C22 κy 1 C24 κxy
Nxy 5 B41 ε0x 1 B42 ε0y 1 B44 γ 0xy 1 C41 κx 1 C42 κy 1 C44 κxy
Mx 5 C11 ε0x 1 C12 ε0y 1 C14 γ 0xy 1 D11 κx 1 D12 κy 1 D14 κxy
My 5 C21 ε0x 1 C22 ε0y 1 C24 γ 0xy 1 D21 κx 1 D22 κy 1 D24 κxy
Mxy 5 C41 ε0x 1 C42 ε0y 1 C44 γ 0xy 1 D41 κx 1 D42 κy 1 D44 κxy
(7.5)
7.1 EQUATIONS OF THE THEORY OF ANISOTROPIC LAMINATED PLATES
439
in which
ðs
Bmn 5
ðs
Amn dz;
Cmn 5
2e
ðs
Dmn 5
Amn zdz;
2e
Amn z2 dz
(7.6)
2e
are membrane, coupling, and bending stiffness coefficients, respectively.
As discussed in Sections 3.1 and 6.7, for the displacements given by Eq. (7.1), the distribution
of the transverse shear stresses τ xz and τ yz over the plate thickness does not affect the plate behavior which is governed by the resultant shear forces
ðs
ðs
Vx 5
τ xz dz;
2e
Vy 5
τ yz dz
(7.7)
2e
shown in Fig. 7.2. The constitutive equations for the shear forces, Eq. (3.15), become
Vx 5 S55 γ x 1 S56 γ y ;
Vy 5 S66 γ y 1 S56 γ x
(7.8)
in which, in accordance with Eq. (3.14)
γ x 5 θx 1
@w
;
@x
γ y 5 θy 1
@w
@y
(7.9)
are the shear strains averaged over the plate thickness. In accordance with Eq. (3.19), the stiffness
coefficients in Eq. (7.8) are calculated as follows:
Ðs
h2 2e A mn dz
Smn 5 Ð s
Ð s
Ð s
2
2e A 55 dz
2e A 66 dz 2 2e A 56 dz
(7.10)
where mn 5 55; 56; 66 and
A mn 5
Amn
A55 A66 2 A256
(7.11)
in which Amn are the transverse shear stiffness coefficients specified by Eq. (2.72). Since the stiffness coefficients Amn in Eqs. (7.6) and (7.10) include the coordinates of the reference surface e and
s as the integrals limits, we can simplify the expressions for these coefficients by introducing the
new coordinate t 5 z 1 e (see Fig. 3.8) which changes from t 5 0 to t 5 h; where h is the plate
thickness (see Fig. 7.1). Then, Eq. (7.6) reduce to the form specified by Eqs. (3.28) and (3.29), i.e.,
0
Bmn 5 Imn
;
ð1Þ
ð0Þ
Cmn 5 Imn
2 eImn
;
ð2Þ
ð1Þ
ð0Þ
Dmn 5 Imn
2 eImn
1 e2 Imn
(7.12)
r 5 0; 1; 2
(7.13)
where mn 5 11; 12; 22; 14; 24; 44 and
ðrÞ
Imn
5
ðh
Amn tr dt;
0
Eq. (7.10) takes the form
Smn 5
where mn 5 55; 56; 66 and
h2 Jmn
2
J55 J66 2 J56
(7.14)
440
CHAPTER 7 LAMINATED COMPOSITE PLATES
ðh
Jmn 5
A mn dt
(7.15)
0
in which A mn is given by Eq. (7.11). Typical composite material laminates have been discussed in
Sections 3.23.9.
For an orthotropic plate A14 5 A24 5 A56 5 0 and Eqs. (7.5) and (7.8) are simplified as
Nx 5 B11 ε0x 1 B12 ε0y 1 C11 κx 1 C12 κy
Ny 5 B21 ε0x 1 B22 ε0y 1 C21 κx 1 C22 κy
Nxy 5 B44 γ 0xy 1 C44 κxy
Mx 5 C11 ε0x 1 C12 ε0y 1 D11 κx 1 D12 κy
(7.16)
My 5 C21 ε0x 1 C22 ε0y 1 D21 κx 1 D22 κy
Mxy 5 C44 γ 0xy 1 D44 κxy
Vx 5 Sx γ x ; Vy 5 Sy γ y
where Bmn , Cmn , and Dmn are specified by Eq. (7.12), whereas
Sx 5 Ð h 0
h2
dt=Gxz
;
Sy 5 Ð h 0
h2
dt=Gyz
(7.17)
and Gxz and Gyz are the material shear moduli which, in the general case, vary through the
plate thickness.
Consider the equilibrium equations for a solid element referred to Cartesian coordinates, i.e.,
@σx
@τ xy
@τ xz
1
1
50
@x
@y
@z
ðx; yÞ
@σz
@τ xz
@τ yz
1
1
50
@z
@x
@y
(7.18)
(7.19)
in which the notation (x, y) means that changing x to y and y to x we can obtain one more equation
from the presented one. The boundary conditions on the plate surfaces are (see Fig. 7.1)
σz ðz 5 2 eÞ 5 2 p; τ xz ðz 5 2 eÞ 5 τ yz ðz 5 2 eÞ 5 0
:
σz ðz 5 sÞ 5 2 q; τ xz ðz 5 sÞ 5 τ yz ðz 5 sÞ 5 0
(7.20)
We can reduce Eqs. (7.18) and (7.19), which include three independent variables x, y, and z, to the
two-dimensional equations of plate theory. Integrating Eq. (7.18) from t 5 0 to t and applying the
boundary conditions, we get
τ xz 5 2
ðz @σx
@τ xy
1
dz ðx; yÞ
@x
@y
(7.21)
2e
Putting z 5 s, taking into account that τ xz ðz 5 sÞ 5 0, and using Eq. (7.4) for the stress resultants,
we have
@Nx
@Nxy
1
5 0 ðx; yÞ
@x
@y
(7.22)
7.1 EQUATIONS OF THE THEORY OF ANISOTROPIC LAMINATED PLATES
441
These two equations correspond to the projections of the forces acting on the plate element
shown in Fig. 7.2 on axes x and y, respectively. Integrating Eq. (7.19) and using the boundary conditions given by Eq. (7.20) for σz , we obtain
ðz @τ xz
@τ yz
1
dz 2 p
@x
@y
σz 5 2
(7.23)
2e
Taking z 5 s and applying the boundary condition for σz given in Eq. (7.20) in conjunction
with Eq. (7.7) for the shear forces, we arrive at the third equilibrium equation which
corresponds to the projection of the forces acting on the plate element shown in Fig. 7.2 on the
z-axis
@Vx
@Vy
1
1 p 5 0; p 5 p 2 q
@x
@y
(7.24)
Finally, substitute Eq. (7.19) for the shear stresses into Eq. (7.7) for the shear force, i.e.,
ðs
Vx 5
dz
2e
ðz @σx
@τ xy
1
dz ðx; yÞ
@x
@y
(7.25)
2e
Integration by parts, taking into account Eq. (7.4), yields
ðs
ðz
dz
2e
2e
ðz
ðs
dz
2e
2e
@σx
dz 5 s
@x
@τ xy
dz 5 s
@y
and Eq. (7.23) becomes
Vx 5 s
ðs
2e
ðs
2e
@σx
dz 2
@x
@τ xy
dz 2
@y
ðs
z
2e
ðs
z
2e
@σx
@Nx
@Mx
dz 5 s
2
@x
@x
@x
@τ xy
@Nxy
@Mxy
dz 5 s
2
@y
@y
@y
@Nx
@Nxy
@Mx
@Mxy
1
2
2
@x
@y
@x
@y
Applying Eq. (7.22), we finally arrive at the following two moment equations for the plate element shown in Fig. 7.2:
@Mx
@Mxy
1
2 Vx 5 0 ðx; yÞ
@x
@y
(7.26)
Thus, the plate theory under consideration reduces to five equilibrium equations,
Eqs. (7.22), (7.24), and (7.26), eight constitutive equations, Eqs. (7.16), and eight
straindisplacement equations, Eqs. (7.3) and (7.9). This set consists of 21 equations in total
which include the same number of unknown functions, i.e., eight forces and moments, Nx , Ny ,
Nxy , Vx , Vy , Mx , My , and Mxy , eight strains, ε0x , ε0y , γ 0xy , κx , κy , κxy , γ x , and γ y , and five displacements and rotation angles, u, v, w, θx , and θy . The equilibrium equations can be obtained as variational equations using the minimum conditions for the total potential energy functional in the
form
442
CHAPTER 7 LAMINATED COMPOSITE PLATES
ða ðb
δT 5
ðNx δε0x 1 Ny δε0y 1 Nxy δγ 0xy 1 Mx δκx 1 My δκy
(7.27)
00
1Mxy δκxy 1 Vx δγ x 1 Vy δγ y 2 pδwÞdxdy 5 0
The derived governing set of equations consists of two groups of equations describing the plane
stress state of the plate and plate bending. Assume that for some of the reasons discussed, e.g., in
Section 3.10, the coupling stiffness coefficients Cmn in Eq. (7.16) are zero. Then, these equations
link the in-plane forces with the corresponding strains, i.e., for an orthotropic plate,
Nx 5 B11 ε0x 1 B12 ε0y ;
Ny 5 B21 ε0x 1 B22 ε0y ;
Nxy 5 B44 γ 0xy
(7.28)
The forces satisfy the equilibrium equations, Eq. (7.22)
@Nx
@Nxy
1
5 0;
@x
@y
@Ny
@Nxy
1
50
@y
@x
(7.29)
and the strains are expressed in terms of the in-plane displacements by Eq. (7.3) according to
which
ε0x 5
@u
;
@x
ε0y 5
@v
;
@y
γ 0xy 5
@u @v
1
@y
@x
(7.30)
The set of equations obtained consisting of eight equations is of the overall fourth order and
requires two boundary conditions for the plate edge. For the edges x 5 constant and y 5 constant,
respectively, the natural boundary conditions following from the energy condition given by
Eq. (7.27) are
Nx δu 5 0;
Ny δv 5 0;
Nxy δv 5 0 for x 5 constant
Nxy δu 5 0 for y 5 constant
(7.31)
According to Eq. (7.31), force, displacement, or mixed boundary conditions can be formulated
at the plate edges.
For the problem of plate bending, the constitutive equations follow from Eq. (7.16), i.e.,
Mx 5 D11 κx 1 D12 κy ; My 5 D21 κx 1 D22 κy ;
Vx 5 Sx γ x ; Vy 5 Sy γ y
Mxy 5 D44 κxy
(7.32)
whereas the corresponding equilibrium equations are specified by Eqs. (7.24) and (7.26), i.e.,
@Vx
@Vy
1
1 p 5 0;
@x
@y
@Mx
@Mxy
1
2 Vx 5 0;
@x
@y
@My
@Mxy
1
2 Vy 5 0
@y
@x
(7.33)
The straindisplacement equations, Eqs. (7.3) and (7.9), become
κx 5
@θx
;
@x
κy 5
@θy
;
@y
κxy 5
@θx
@θy
1
;
@y
@x
γ x 5 θx 1
@w
;
@x
γ y 5 θy 1
@w
@y
(7.34)
The set of 13 equations obtained is of the overall sixth order and requires the following three
boundary conditions for the plate edge:
Mx δθx 5 0; Mxy δθy 5 0; Vx δw 5 0 for x 5 constant
My δθy 5 0; Mxy δθx 5 0; Vy δw 5 0 for y 5 constant
(7.35)
7.2 EQUATIONS FOR THE ORTHOTROPIC PLATES
443
7.2 EQUATIONS FOR THE ORTHOTROPIC PLATES WITH
SYMMETRIC STRUCTURE
It follows from the constitutive equations Eq. (7.5) that in the general case, composite plates demonstrate a rather complicated behavior which involves various types of coupling effects, i.e., shear
is coupled with tension and bending is coupled with torsion due to material anisotropy, whereas
bending is coupled with in-plane deformation due to, in general, the nonsymmetrical laminated
structure of the plate. Fortunately, plates with such a general structure are rarely used as composite
structural elements.
First of all, it should be noted that the majority of real composite plates are orthotropic.
Anisotropic composite plates are usually designed for special applications, e.g., for the so-called
adaptive and/or passive control composite structures, whose bending must be accompanied by controlled torsion. A unique example of such application is the composite panel of a forward swept airplane wing (Weisshaar, 1980, 1981). The anisotropic properties of such panels allow us to provide
the appropriate aeroelastic tailoring of a wing’s structure. In the vast majority of applications, plates
designed as load-carrying structural elements are usually orthotropic.
Second, as shown in Section 3.4, the maximum bending stiffness of the plate composed of a
given number and given properties of layers is reached if the plate structure is symmetric with
respect to the plate middle plane. Note that the thickness of the elementary composite ply is normally rather small (0.10.2 mm) and that actual composite plates are usually composed of a large
number of plies. Thus, it is practically always possible to arrange these plies symmetrically with
respect to the plate middle plane. Note that homogeneous plates whose properties do not vary
across the thickness represent one of the most widespread particular cases of plates with a balanced
and symmetric structure. Typical examples of real nonsymmetrically laminated plates include sandwich panels with different facing layers and also stiffened panels.
Formally, the difference between symmetrically and nonsymmetrically laminated plates is associated with coupling stiffness coefficients Cmn in Eq. (7.5) which are zero for plates with a symmetric structure. As shown in Section 3.10, the condition Cmn 5 0 can be approximately satisfied using
the reduced bending stiffness method. Thus, formally, even nonsymmetric plates can sometimes be
analyzed using the equations presented in this section.
For the foregoing reasons, we consider further mainly balanced and symmetrically laminated
orthotropic plates. Plates with more complicated structures are discussed in the following sections
of this chapter.
Thus, assume that the coupling stiffness coefficients Cmn 5 0 and, hence, the problem of plate
bending can be separated from the in-plane problem and described by Eqs. (7.32)(7.34) derived
in the previous section. Substitute the generalized strains, Eq. (7.34) into the constitutive equations,
Eq. (7.32), and the resulting expressions for the moments and forces substitute into the equilibrium
equations, Eq. (7.33). As a result, we arrive at the following three equilibrium equations written in
terms of the deflection, w, and rotation angles, θx and θy :
L1x ðθx Þ 1 L1y θy 1 L1w ðwÞ 5 0
L2x ðθx Þ 1 L2y θy 1 L2w ðwÞ 5 0
L3x ðθx Þ 1 L3y θy 1 L3w ðwÞ 5 2 p
(7.36)
444
CHAPTER 7 LAMINATED COMPOSITE PLATES
in which the differential operators L are
1
@2 θx
@2 θx
D11 2 1 D44 2 2 θx
Sx
@x
@y
D @2 θy
@w
; L1w ðwÞ 5 2
L1y θy 5
Sx @x@y
@x
2
D @ θx
L2x ðθx Þ 5
Sy @x@y
1
@2 θy
@2 θy
@w
L2y θy 5
D22 2 1 D44 2 2 θy ; L2w ðwÞ 5 2
Sy
@y
@y
@x
@θx
@θy
L3x ðθx Þ 5 Sx
; L3y θy 5 Sy
@x
@y
@2 w
@2 w
L3w ðwÞ 5 Sx 2 1 Sy 2
@x
@y
L1x ðθx Þ 5
(7.37)
where D 5 D12 1 D44 .
We can reduce Eq. (7.36) to one governing equation using the operational method
(Vasiliev, 1992). According to this method, consider formally the determinant constructed of
the operators given by Eq. (7.37) and decompose it using the minors corresponding to the third
row. Then, the functions θx , θy , and w can be expressed in terms of some resolving function
Wðx; yÞ as
L1y L1w ðW Þ 5 L1y L2w 2 L1w L2y ðWÞ
θx 5 L2y L2w
L1x L1w ðW Þ 5 ðL1w L2x 2 L1x L2w ÞðW Þ
θy 5 2 L2x L2w L1x L1y ðW Þ 5 L1x L2y 2 L1y L2x ðWÞ
w 5 L2x L2y (7.38)
where we use the formal multiplication operation for the differential operators with constant coefficients, e.g.,
L1y L2w ðW Þ 5 L2w L1y ðW Þ 5 2
D @3 W
Sx @x@y2
As can be directly verified, substitution of Eq. (7.38) into the first two equations of Eq. (7.36)
satisfies identically these equations, whereas the third equation takes the following operational
form:
½L3x L1y L2w 2 L1w L2y 2 L3y ðL1x L2w 2 L1w L2x Þ
1L3w L1x L2y 2 L1y L2x ðW Þ 5 2 p
(7.39)
The left-hand part of this equation is actually the determinant composed of the operators given
by Eq. (7.37). The explicit form of Eq. (7.39) is
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
D11
@4 W
@4 W
@4 W
D11 D44 @6 W
D22 D44 @6 W
1 2ðD12 1 2D44 Þ 2 2 1 D22 4 2
2
Sy @x6
Sx @y6
@x4
@x @y
@y
6
D11 D44
1
@ W
2
1 ½D11 D22 2 D12 ðD12 1 2D44 Þ
Sy
@x4 @y2
Sx
6
D22 D44
1
@ W
2
1 ½D11 D22 2 D12 ðD12 1 2D44 Þ
5p
Sx
@x2 @y4
Sy
445
(7.40)
The rotation angles and the deflection are expressed in terms of the resolving function W as
@
D44 @2 W
D22
D @2 W
2W 1
1
2
@x
Sx @y2
Sy @x2
Sy
@
D44 @2 W
D11
D @2 W
2W 1
1
2
θy 5
@y
Sy @x2
Sx @y2
Sx
θx 5
D11
D44 @2 W
D22
D44 @2 W
1
2
1
w5W 2
Sx
Sy @x2
Sy
Sx @y2
1
@4 W
@4 W
@4 W
1
D11 D44 4 1 ½D11 D22 2 D12 ðD12 1 2D44 Þ 2 2 1 D22 D44 4
Sx Sy
@x
@x @y
@y
(7.41)
(7.42)
The equations obtained, Eqs. (7.41) and (7.42), have specific structures that are studied in the
next section.
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY
ISOTROPIC PLATES
To analyze the structure of the governing equations of plate theory, Eqs. (7.41) and (7.42), consider
a transversely isotropic plate composed of identical isotropic layers. For such plates,
D11 5 D22 5 D;
D12 5 νD;
1
D44 5 ð1 2 ν ÞD;
2
Sx 5 Sy 5 S
(7.43)
where D and S are the plate bending and transverse shear stiffnesses, and ν is the in-plane
Poisson’s ratio. For the stiffnesses presented by Eq. (7.43), Eqs. (7.40)(7.42) are markedly simplified and take the form
DΔΔLðW Þ 5 p
@
@
θx 5 2 LðW Þ; θy 5 2 LðW Þ
@x
@y
D
w 5 LðW Þ 2 ΔLðWÞ
S
(7.44)
in which
LðW Þ 5 W 2
Dð1 2 νÞ
ΔW;
2S
Δð. . .Þ 5
@2 ð. . .Þ @2 ð. . .Þ
1
@x2
@y2
(7.45)
446
CHAPTER 7 LAMINATED COMPOSITE PLATES
7.3.1 CLASSICAL PLATE THEORY
Consider first classical plate theory which ignores the transverse shear deformation. Taking S-N
in Eqs. (7.44) and (7.45), we get LðW Þ 5 w and arrive at the following well-known equations in
classical plate theory:
DΔΔw 5 p
θx 5 2
@w
;
@x
(7.46)
θy 5 2
@w
@y
(7.47)
in which w is the plate deflection. The bending moments are expressed in terms of the curvature
changes as
Mx 5 D κx 1 νκy ;
My 5 D κy 1 νκx ;
D
ð1 2 ν Þκxy
2
(7.48)
@2 w
@x@y
(7.49)
Mxy 5
where, in accordance with Eq. (7.47),
κx 5 2
@2 w
;
@x2
κy 5 2
@2 w
;
@y2
κxy 5 2 2
The shear forces follow from the last two equilibrium equations in Eq. (7.33), i.e.,
Vx 5
@Mx
@Mxy
1
;
@x
@y
Vy 5
@My
@Mxy
1
@y
@x
(7.50)
Substitution of these forces into the first equilibrium equation of Eq. (7.33) yields
@ 2 Mx
@2 Mxy
@ 2 My
1
12
1p 50
2
@x
@x@y
@y2
(7.51)
If we express the moments in Eq. (7.48) in terms of deflection with the aid of Eq. (7.49) and
substitute the resulting equations into Eq. (7.51), we arrive at the biharmonic equation, Eq. (7.46)
for the plate deflection.
Specific features of classical plate theory follow from the variational formulation of the problem. For the classical theory, Eq. (7.27) can be presented as
ða ðb
δT 5
ðMx δκx 1 My δκy 1 Mxy δκxy 2 pδwÞdxdy 5 0
0 0
Substituting for κx , κy , and κxy their expressions from Eq. (7.49), we get
ða ðb
Mx δ
0 0
2 2 2 @ w
@ w
@ w
1
M
1
2M
δ
δ
1 pδw dxdy 5 0
y
xy
@x2
@y2
@x@y
Integration by parts for the first and the second terms of Eq. (7.52) yields in several steps
ða ðb
0 0
2b
3
2 ð
ða ðb
@ w
@w
@Mx @w
4 Mx δ
5 2
dy
dxdy
δ
dxdy
5
Mx δ
@x2
@x
@x
@x
0
x
0 0
(7.52)
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
447
2b
3
2b
3
ð
ð
ða ðb 2
@w
@M
@ Mx
x
4
5
4
5
dy 2
δwdy 1
5
Mx δ
δwdxdy ðx; yÞ
@x
@x
@x2
0
0
x
0 0
x
The notations ½ x and ½ y mean that the term ½ corresponds to either the plate edge
x 5 constant or y 5 constant. The third term in Eq. (7.52) can be transformed in two ways, i.e.,
ða ðb
0 0
2b
3
2 ð
ða ðb
@ w
@w
@Mxy @w
dxdy 5 4 Mxy δ
dy5 2
dxdy
δ
Mxy δ
@x@y
@y
@y
@x
0
0 0
x
2b
3
2a
3
ð
ð
ða ðb 2
@w
@M
@ Mxy
xy
4
5
4
5
dy 2
δwdy 1
δwdxdy
5
Mxy δ
@y
@x
@x@y
0
0
x
y
0 0
and
ða ðb
0 0
2a
3
2 ð
ða ðb
@ w
@w
@Mxy @w
dxdy 5 4 Mxy δ
dx5 2
dxdy
Mxy δ
δ
@x@y
@x
@x
@y
0
0 0
y
2b
3
2a
3
ð
ð
ða ðb 2
@w
@Mxy
@ Mxy
4
5
4
5
δwdy 1
δwdxdy
5
Mxy δ
dx 2
@y
@x@y
@x
0
y
0
x
0 0
Thus, using Eq. (7.50) for the transverse forces, we can present the variational equation,
Eq. (7.52), in the following form:
ða ðb 2
@ Mx
@2 Mxy
@2 My
δwdxdy
1
2
1
@x2
@x@y
@y2
00
2b
3 2a
3
2b
3 2a
3
ð
ð
ð
ð
@w
@w
4
5
4
5
4
5
4
dy 1 My δ
dx 2 Vx δwdy 2 Vy δwdx5
1 Mx δ
@x
@y
0
x
0
y
0
x
2b
3
2a
3
ð
ð
1 4 Mxy δ @w dy5 1 4 Mxy δ @w dx5 5 0
@y
@x
0
0
x
0
(7.53)
y
y
To demonstrate some special features of classical plate theory, consider a simply supported
plate loaded with a pressure
q 5 q0 sin
πx
πy
sin
a
b
(7.54)
as shown in Fig. 7.3. The boundary conditions for this plate
½wx 5 ½wy 5 0;
½Mx x 5 0;
My y 5 0
(7.55)
448
CHAPTER 7 LAMINATED COMPOSITE PLATES
x
a
b
q
y
FIGURE 7.3
A simply supported plate loaded with sine pressure.
can be satisfied if the plate deflection has the form
w 5 w0 sin
πx
πy
sin
a
b
(7.56)
Substituting Eqs. (7.54) and (7.56) into the governing equation, Eq. (7.46), we arrive at the
exact solution of the problem
w52
q0 a4 b4
π4 Dða2 1b2 Þ2
sin
πx
πy
sin
a
b
(7.57)
The transverse shear forces can be determined using Eqs. (7.48)(7.50) and Eq. (7.57), to give
@
q0 ab2
πx
πy
cos sin
ðΔwÞ 5 2
πða2 1 b2 Þ
@x
a
b
2
@
q0 a b
πx
πy
sin cos
Vy 5 2 D ðΔwÞ 5 2
@y
πða2 1 b2 Þ
a
b
Vx 5 2 D
(7.58)
Remove the plate shown in Fig. 7.3 from its supports and find the resultant of the edge
reactions, i.e.,
ðb
ða
½Vx ðx 5 aÞ 2 Vx ðx 5 0Þdy 1
R5
0
Vy ðy 5 bÞ 2 Vy ðy 5 0Þ dx
(7.59)
0
Substitution of Eq. (7.58) yields
R5
4q0 ab
π2
(7.60)
The applied load in Eq. (7.54), gives, as expected, the same resultant.
Now return to the variational equation, Eq. (7.53). The first integral in this equation gives
Eq. (7.51). The next four integrals including bending moments and transverse forces are zero
according to the boundary conditions, Eq. (7.55). Moreover, the derivatives of the deflection given
by Eq. (7.56) are zero along the plate edges and the last two integrals in Eq. (7.53) with the twisting moment are also zero. Thus, the variational equation, Eq. (7.53), is satisfied and the equilibrium
equations are also satisfied along with the corresponding boundary conditions.
Note that actually the deflection determined by Eq. (7.57), being the solution of the fourth-order
differential equation, Eq. (7.46), satisfies three boundary conditions at the plate edges. For example,
for the edge x 5 0 we have
w 5 0;
Mx 5 0;
θy 5 0
(7.61)
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
449
The analysis for clamped plates is similar.
Now, suppose that instead of θy we specify the twisting moment Mxy at the plate edge, then the
last two integrals in Eq. (7.53) are not zero and the variational equation is not satisfied. The reason
for this is associated with Eq. (7.47) for the rotational angles, according to which the plate in-plane
displacements are
ux 5 zθx 5 2 z
@w
;
@x
uy 5 zθy 5 2 z
@w
@y
(7.62)
Torsion of the plate induces rotation of the plate element. The angle of in-plane rotation is
specified by the following equation:
ωz 5
1 @uy
@ux
2
2 @x
@y
(7.63)
Substitution of Eq. (7.62) yields ωz 5 0, so that no rotation can take place. The problem of
plate torsion is considered in Section 3.5. The solution given by Eq. (3.93) depends on parameter
λ, Eq. (3.94), which includes the transverse shear stiffness of the plate Sx . However, in classical
plate theory Sx -N so this solution does not exist. Thus, classical plate theory cannot be
applied to the problems that involve plate torsion. This problem is further discussed in
Section 7.3.2.3.
Nevertheless, some transformations traditionally have been introduced (Timoshenko &
Woinowsky-Krieger, 1959) into classical plate theory that formally allows us to study the problem.
Consider the last two integrals in Eq. (7.53) and apply integration by parts which gives
2b
3
2b
3
ð
ð
4 Mxy δ @w dy5 5 Mxy δw 2 4 @Mxy δwdy5 ðx; yÞ
x;y
@y
@y
0
x
0
(7.64)
x
The last term can be added to the integral including Qx in Eq. (7.53) resulting in the so-called
Kirchhoff’s shear force
Kx 5 Vx 1
@Mxy
@ @2 w
@2 w
52D
1 ð2 2 νÞ 2
2
@x @x
@y
@y
ðx; yÞ
(7.65)
whereby the first term in the right-hand part of Eq. (7.63) is treated as the force applied at the plate
corner. Since the plate has four edges, the force R1 5 2 2Mxy is formally acting at each corner in
the direction which is opposite to the direction of the forces Kx and Ky given by Eq. (7.65). As a
result, the reactive forces of the plate shown in Fig. 7.3 become as those shown in Fig. 7.4. For the
plate shown in Fig. 7.3, Eq. (7.65) yield
Kx 5 Vx 1 1
ð1 2 νÞa2
;
a2 1 b2
Ky 5 Vy 1 1
ð1 2 νÞb2
a2 1 b2
The distributions of the normalized forces V x 5 Vx =Vxm and K x 5 Kx =Vxm (in which Vxm is the
maximum shear force acting at the center) along the edge x 5 0 for a square (a 5 b) plate with
ν 5 0:3 are presented in Fig. 7.5. As can be seen, Kx (the dashed line) is about 35% higher than Vx
(Alfutov, 1992). The resultant force can be found from Eq. (7.59) if we replace V with K.
The result
450
CHAPTER 7 LAMINATED COMPOSITE PLATES
2 Mxy
Ky
2 Mxy
Kx
2 Mxy
2 Mxy
FIGURE 7.4
Support reactions corresponding to classical plate theory.
Vx , K x
1.6
Kx
1 .35
1.2
Vx
1 .0
0.8
0.4
y
0
0
0.2
0.4
0.6
0.8
1.0 b
FIGURE 7.5
Distribution of the normalized transverse shear forces over the plate width coordinate y for x 5 a.
R2 5
4q0 ab 8ð1 2 νÞq0 a3 b3
1
π2
π2 ða2 1b2 Þ2
shows that R2 does not balance the applied pressure. The resulting force coincides with the actual
force given by Eq. (7.60) if we add four concentrated forces
R1 5 2
2ð1 2 νÞq0 a3 b3
π2 ða2 1b2 Þ2
applied at the corners. However, for the plate shown in Fig. 7.3, the exact solution of the theory of
elasticity equations is known (Galerkin, 1931). For an isotropic plate (see Fig. 7.1) whose surfaces
z 5 6 h=2 are free of shear stresses and for which the pressure is specified by Eq. (7.54), the shear,
τ xz and τ yz , and normal, σz , transverse stresses are given by
τ xz 5
@
@2 F
ð1 2 ν ÞΔF 2 2 ;
@x
@z
τ yz 5
@
@2 F
ð2 2 ν ÞΔF 2 2 ;
@z
@z
ΔðÞ 5
σz 5
@
@2 F
ð1 2 ν ÞΔF 2 2
@y
@z
@2 ðÞ @2 ðÞ @2 ðÞ
1 2 1 2
@x2
@y
@z
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
451
where F is the stress function which has the following form:
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
πx
πy
1
1
F 5 ðC1 sinhλz 1 C2 coshλz 1 C3 z coshλz 1 C4 z sinhλzÞsin sin ; λ 5 π 2 1 2
a
b
a
b
The constants C1 C4 can be found from the boundary conditions for the stresses σz and τ xz
(or τ yz ) at the plate surfaces z 5 6 h=2. Omitting the determination of these constants, we can conclude
anyway that the foregoing equations for τ xz and τ yz have no singularities at the plate corners and,
hence, no corner forces exist at the plate corners. So, it seems that the transformation in Eq. (7.64),
and the result given by Eq. (7.65) following from this transformation, are not correct for the plate
under study.
Having derived the basic equations, consider the history of the problem. The boundary problem
in classical plate theory was originally formulated by Poisson (1829), who arrived at the governing
equation of the fourth order, Eq. (7.46), accompanied by three boundary conditions at the plate
edge. No comments were given concerning the contradiction between the equation order and the
number of boundary conditions, because Poisson considered the problems of bending of simply
supported and clamped plates, for which three boundary conditions automatically reduce to two,
and the problem of axisymmetric bending of circular plates, for which only two boundary conditions existed. The inconsistency of Poisson’s theory was criticized by Kirchhoff (1850), who
applied the variational approach and arrived at two natural boundary conditions at the plate edge.
For the free edge x 5 constant, Kirchhoff found that
@2 w
@2 w
@ @2 w
@2 w
50
1
ν
5
0;
1
ð2
2
νÞ
@x2
@y2
@x @x2
@y2
(7.66)
It seems as though the first of these conditions corresponds to the bending moment Mx , whereas
the second, in accordance with Eq. (7.65), provides the absence of Kirchhoff shear forces at the
free plate edges. However, Kirchhoff derived the conditions in Eq. (7.66) from a variational equation similar to Eq. (7.52) which corresponded to the plate vibration problem. The resulting equation
and the natural boundary conditions were not associated with any forces and moments. Kirchhoff
forces, Eq. (7.65), were introduced later by Thomson and Tait (1883), who reduced the twisting
moment acting at the plate edge to transverse shear forces [as per Eq. (7.65)], applying a static
transformation which is valid, in general, for an absolutely rigid body. For an elastic plate, a
moment cannot be reduced to forces (Zhilin, 1992, 1995). Thus, the transformation in Eq. (7.64)
formally allows us to construct the theory for the plate with free edges or with force boundary conditions. However, this transformation is not valid for the plates whose edges are fixed with respect
to the deflection.
To show this, return to the functional in Eq. (7.53) and to the plate shown in Fig. 7.3 considered
as an example. As can be readily seen, the plate deflection w, being zero at the plate edges, provides zero derivatives along the edges. This means that the last two integrals in Eq. (7.53) are zero,
and their transformation in Eq. (7.64) has no physical meaning. Consequently, Kirchhoff and
ThomsonTait transformations are not valid for a simply supported plate for which the transverse
shear forces are specified by Eq. (7.58) and no generalized and corner forces following from these
transformations can exist. The same is true for a clamped plate and for any plate whose deflection
is zero at the boundary. For the force boundary conditions, the situation is different and the sixthorder plate theory discussed in the next section must be applied to study such plates.
452
CHAPTER 7 LAMINATED COMPOSITE PLATES
7.3.2 THEORY OF SHEAR DEFORMABLE PLATES
7.3.2.1 Theory equations
Any plate theory which allows us to study the plates with force boundary conditions which are not
consistent with the ThomsonTait transformation must result in governing equations of the sixth
order. Such a theory has been proposed by Reissner (1944), who used a rather cumbersome stress
formulation based on the through-the-thickness distribution of normal and shear stresses corresponding to classical plate theory and the variational principle of minimum complementary energy.
In the displacement formulation proposed originally by Hencky (1947) and based on Eq. (7.1), i.e.,
ux 5 uðx; yÞ 1 zθx ðx; yÞ; uy 5 vðx; yÞ 1 zθy ðx; yÞ; uz 5 wðx; yÞ
(7.67)
this theory reduces to Eq. (7.44) which are studied in this section.
Assume that LðWÞ 6¼ 0 in which case
LðW Þ 5 W 2
Dð1 2 νÞ
ΔW;
2S
ΔW 5
@2 W
@2 W
1
@x2
@y2
(7.68)
and introduce the new function ϕðx; yÞ such that LðW Þ 5 ϕ. Then, Eq. (7.44) are transformed into
the form
DΔΔϕ 5 p
(7.69)
@ϕ
@ϕ
D
; θy 5 2
; w 5 ϕ 2 Δϕ
θx 5 2
@x
@y
S
(7.70)
It follows from these equations that ϕðx; yÞ is a potential function for the rotation angles,
whereas zϕðx; yÞ is a potential function for the in-plane displacements in Eq. (7.62). As is known,
the potential displacement field cannot represent the continuum rotation. Indeed, substituting
Eq. (7.70) into Eq. (7.63), we get
ωz 5
1 @uy
@ux
2
50
2 @x
@y
(7.71)
The potential function determines the so-called penetrating solution that does not vanish at a
distance from the plate edges. Note that Eq. (7.69) for ϕ and Eq. (7.70) for θx and θy are similar to
Eqs. (7.46) and (7.47) in classical plate theory and coincide with them if ϕ 5 w. However, in the
theory under discussion, w is expressed by the last equation of Eq. (7.70) in which the second term
allows for the influence of the transverse shear deformation on the plate deflection.
We can expect that the theory described by Eq. (7.69) and (7.70) is not complete and these
equations must be supplemented with an additional equation for the following reasons:
1. The initial set of equations, Eq. (7.44), is of the sixth order, whereas Eq. (7.69) and (7.70) are
only of the fourth order.
2. The equations obtained, Eq. (7.69) and (7.70), cannot describe the plate torsion.
3. The initial equations, Eq. (7.44), are reduced to Eqs. (7.69) and (7.70) under the condition that
LðWÞ 6¼ 0. So, a natural question arises as to what happens if LðWÞ in Eq. (7.68) is zero.
To derive the desired equation, return to the initial equations, Eq. (7.36) and write them for a
transversely isotropic plate. Note that because the pressure p is allowed for by the penetrating
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
453
potential ϕ in accordance with Eq. (7.69), the equation under derivation must be homogeneous, i.e.,
correspond to p 5 0. So, taking p 5 0 and using Eq. (7.43) for the stiffness coefficients, we can
transform Eq. (7.36) into the following form:
D @2 θx
1 2 ν @2 θx
Dð1 1 νÞ @2 θy
@w
50
1
2 θx 2
1
2
2
S @x
2 @y
2S @x@y
@x
D @2 θy
1 2 ν @2 θx
Dð1 1 νÞ @2 θx
@w
50
1
2 θy 2
1
S @y2
2 @x2
2S @x@y
@y
(7.72)
@θx
@θy
1
1 Δw 5 0
@x
@y
(7.73)
As can be seen, Eq. (7.73) is satisfied if we take
θx 5
@ψ
;
@y
θy 5 2
@ψ
; w50
@x
(7.74)
Then, Eq. (7.72) become
@
LðψÞ 5 0;
@y
@
LðψÞ 5 0
@x
(7.75)
in which operator L is specified by Eq. (7.68). It follows from Eq. (7.75) that LðψÞ 5 c. The solution of this equation is ψ 5 ψ0 1 c where ψ0 is the solution for the corresponding homogeneous
equation LðψÞ 5 0 and c is a constant. Since θx and θy are the derivatives of ψ, this constant is not
important and we can take c 5 0. Thus, the final equation is
Δψ 2
2S
ψ50
Dð1 2 νÞ
(7.76)
The function ψ can be referred to as the rotational potential. Indeed, substituting Eq. (7.75) into
Eq. (7.67) for ux and uy and then the resulting expressions into Eq. (7.71) for ωz , we get
ωz 5 2
z
Δψ
2
For a homogeneous plate, S 5 Gh 5 Eh=2ð1 1 νÞ and D 5 Eh3 =12ð1 2 ν 2 Þ: Introducing the normalized coordinates x 5 x=a and y 5 y=b, we can transform Eq. (7.76) into the form
@2 ψ a2 @2 ψ
12a2
1 2 2 2 k2 ψ 5 0; k2 5 2
2
b @y
h
@x
(7.77)
For relatively thin plates, i.e., for high ratios of a/h, the value of the parameter k is also high
and the foregoing equation specifies the boundary-layer solution associated with plate torsion in the
vicinity of the edges. Assume, for example, that ψ 5 ψðxÞ. Then, Eq. (7.77) has the solution
pffiffi
ψðx Þ 5 ce2ð2 3a=hÞx
(7.78)
rapidly changing in the vicinity of the plate edge.
Historically, an equation analogous to Eqs. (7.76) and (7.77) was obtained by M. Lévy, M.J.
Boussinesq, and W. Thomson and P. Tait at the end of the 19th century (Todhunter & Pearson,
1960), who formulated a theory of elasticity problem for a plate with particular boundary conditions. According to these conditions, the plate is loaded with edge forces that are statically
454
CHAPTER 7 LAMINATED COMPOSITE PLATES
equivalent to the transverse force Vx and twisting moment Mxy such that the Kirchhoff shear force
Kx in Eq. (7.65) is zero. As a result, a second-order three-dimensional equation specifying the
boundary-layer solution has been obtained. The two-dimensional version of this equation has been
derived for plates by Reissner (1944).
Thus, the rotation angles of the plate element are composed of two functions specified by
Eqs. (7.70) and (7.74), i.e.,
θx 5 2
@ϕ @ψ
1
;
@x
@y
θy 5 2
@ϕ @ψ
2
@y
@x
(7.79)
in which ϕ is the penetrating potential which satisfies the biharmonic equation
DΔΔϕ 5 p
(7.80)
and ψ is the rotational potential that can be found as the boundary-layer solution from
Eq. (7.76), i.e.,
Δψ 2 s2 ψ 5 0;
s2 5
2S
Dð1 2 νÞ
(7.81)
The plate deflection depends on the penetrating potential only, i.e.,
w5ϕ2
D
Δϕ
S
(7.82)
The following equations for the bending and twisting moments are derived using Eq. (7.48),
(7.34), and (7.79):
@2 ϕ
@2 ϕ
@2 ψ
1 ν 2 2 ð1 2 νÞ
2
@x
@y
@x@y
@2 ϕ
@2 ϕ
@2 ψ
My 5 2 D
1
ν
2
ð1
2
νÞ
@y2
@x2
@x@y
2
2
@ ϕ
1 @ ψ @2 ψ
2
1
Mxy 5 2 Dð1 2 νÞ
@x@y 2 @x2
@y2
Mx 5 2 D
(7.83)
Finally, the equations for the transverse shear forces follow from Eqs. (7.32), (7.34), and
(7.82), i.e.,
@w
@ψ
@
Vx 5 S θx 1
5S
2 D Δϕ;
@x
@y
@x
@w
@ψ
@
Vy 5 S θy 1
5S
2 D Δϕ
@y
@x
@y
(7.84)
The variational equation
ða ðb
Mx δ
0 0
@θx
@θy
@θx
@θy
@w
@w
1 My δ
1 Mxy δ
1
1 Vx δ θx 1
1 Vy δ θy 1
2 pδw dxdy 5 0
@x
@y
@y
@x
@x
@y
(7.85)
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
455
provides the equilibrium equations, Eq. (7.33), and the following natural boundary conditions:
2b
3
ð
4 Mx δθx 1Mxy δθy 1Vx δw dy5 5 0
0
2a
3x
ð
4 My δθy 1Mxy δθx 1Vy δw dx5 5 0
0
(7.86)
y
Consider typical boundary conditions for the plate edge x 5 constant. First, consider the simply
supported plates for which Eq. (7.86) provides two possible types of boundary conditions. The
classical version of the boundary conditions is
w 5 0; Mx 5 0; θy 5 0
(7.87)
The physical interpretation (see Fig. 7.6A) is that the plate is supported by the walls that are
absolutely rigid in the wall plane providing the conditions w 5 0 and θy 5 0 but have zero out-ofplane bending stiffness, so that Mx 5 0. Substituting Eqs. (7.79), (7.82), and (7.83) into Eq. (7.87),
we get
ϕ2
D
Δϕ 5 0;
S
@2 ϕ
@2 ϕ
@2 ψ
5 0;
1 ν 2 2 ð1 2 ν Þ
2
@x
@y
@x@y
@ϕ
@ψ
52
@y
@x
(7.88)
Using the third of these equations to eliminate ψ from the second equation, we get Δϕ 5 0.
Thus, it follows from the first equation of Eq. (7.88) that ϕ 5 0, and the boundary conditions given
by Eq. (7.88) reduce to
ϕ 5 0;
@2 ϕ
5 0;
@x2
@ψ
50
@x
(7.89)
For the edge y 5 constant, Eq. (7.89) can be supplemented with
ϕ 5 0;
@2 ϕ
5 0;
@y2
@ψ
50
@y
(7.90)
Consider the last of the boundary conditions in Eqs. (7.89) and (7.90). The rotational potential
ψ satisfies the homogeneous equation, Eq. (7.81), and the zero boundary conditions for the function
(B)
(A)
x
x
y
y
FIGURE 7.6
Two versions of a simply supported plate edge corresponding to Eq. (7.87) (A) and (7.92) (B).
456
CHAPTER 7 LAMINATED COMPOSITE PLATES
normal derivatives at the plate contour. In this case, Eq. (7.81) has only a trivial solution, so that
ψ 5 0. Then, the boundary conditions (7.89) and (7.90) become
at x 5 constant
ϕ 5 0;
at y 5 constant
ϕ 5 0;
@2 ϕ
50
@x2
@2 ϕ
50
@y2
(7.91)
Thus, for the conventional (classical) simply supported edge, the boundary layer solution does
not exist (Pelekh, 1970; Vasiliev, 1992, 2000), and the plate bending is governed by the penetrating
solution which is specified by the fourth-order equation, Eq. (7.80), with two boundary conditions
in Eq. (7.91) at the plate edge. The second term in Eq. (7.82) for the plate deflection allows for the
transverse shear deformation. Clearly, no Kirchhoff or corner forces exist in this case.
It follows from the natural boundary condition given by Eq. (7.86) that the other version of a
simple support can exist if we replace the condition θy 5 0 with Mxy 5 0, i.e.,
w 5 0; Mx 5 0; Mxy 5 0
(7.92)
The physical interpretation of these boundary conditions is presented in Fig. 7.6B. The plate is
supported by absolutely rigid columns (w 5 0) and zero bending stiffnesses in the edge plane and
in the direction normal to the edge (Mxy 5 0, Mx 5 0). Note that the boundary conditions in
Eqs. (7.92) cannot exist in classical plate theory based on Eqs. (7.62). Indeed, if w 5 0 at the edge
x 5 constant, then θy 5 0 at this edge and Mxy cannot be zero. However, for the sixth-order theory,
the boundary conditions given by Eqs. (7.92) can formally exist. Using Eqs. (7.82) and (7.83),
we get
ϕ2
D
Δϕ 5 0
S
(7.93)
@2 ϕ
@2 ϕ
@2 ψ
50
1
ν
2
ð
1
2
ν
Þ
@x2
@y2
@x@y
(7.94)
@2 ϕ
@2 ψ @2 ψ
1 2 2 2 50
@x@y
@x
@y
(7.95)
2
These boundary conditions cannot be reduced to two equations and the set of the sixth-order
equations, Eqs. (7.80) and (7.81), must be solved with these conditions. The problems of plate
bending with the boundary conditions in Eq. (7.92) have been solved by Schäfer (1952) for a rectangular plate under sine pressure (see Fig. 7.3), by Kromm (1955) for a square plate loaded with
uniform pressure, and by Sheremetiev, Pelekh, and Dyachina (1968) for a square plate under the
action of a central concentrated force.
7.3.2.2 Bending problem
Consider a square plate loaded with cosine pressure as shown in Fig. 7.7. Placing the center of the
coordinate frame at the plate center, we assume that
p 5 2 q 5 q0 cosλ1 x cosλ1 y
(7.96)
where λ1 5 π=a. The boundary conditions in Eqs. (7.93)(7.95) must be satisfied at x 5 6 a=2.
The penetrating potential ϕ satisfies Eq. (7.80) and can be decomposed as
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
p
457
a/2
x
a/2
a/2
a/2
y
FIGURE 7.7
A square plate loaded with cosine pressure.
ϕ 5 ϕh 1 ϕq
(7.97)
where
ϕq 5 ϕ0 cosλ1 x cosλ1 y;
ϕ0 5 2
q0 a4
4π4 D
(7.98)
is the particular solution, whereas ϕh is the solution of the corresponding homogeneous equation
ΔΔϕh 5 0
(7.99)
Taking into account the symmetry of the problem (see Fig. 7.7), the solution of this equation
can be represented as
ϕh 5
X
ϕm ðyÞ cosλm x 1 ϕm ðxÞ sinλm y
(7.100)
m
where λm 5 ð2m 2 1Þπ=a and ϕm is one and the same function of coordinates x or y.
Substituting the solution, Eq. (7. 100), into Eq. (7.99), we arrive at the following equation for,
e.g., ϕm ðyÞ:
00
2
4
ϕIV
m 2 2λm ϕm 1 λm ϕm 5 0
The symmetric solution of this equation is
ϕm 5 C1m coshλm y 1 C2m ysinhλm y
Thus, the penetrating solution becomes
ϕ 5 ϕ0 cosλ1 x cosλ1 y 1
X C1m coshλm y 1 C2m ysinhλm y cosλm x 1 C1m coshλm x 1 C2m xsinhλm x cosλm y
m
(7.101)
The rotational potential ψ can be found as the combination of antisymmetric functions, i.e.,
ψ5
X
ψm ðyÞsinλm x 2 ψm ðxÞsinλm y
m
Substituting this solution into Eq. (7.81), we get, e.g., for ψm ðyÞ,
00
ψm 1 λ2m 1 s2 ψm 5 0
458
CHAPTER 7 LAMINATED COMPOSITE PLATES
Thus, ψm 5 Bm sinhkm y and the rotational potential becomes
ψ5
X
Bm ðsinhkm y sinλm x 2 sinhkm x sinλm yÞ
(7.102)
m
where
2S
Dð1 2 ν Þ
km2 5 λ2m 1 s2 5 λ2m 1
(7.103)
To determine the constants C1m , C2m , and Bm in Eqs. (7.101) and (7.102), we must apply the
boundary conditions, Eqs. (7.93)(7.95) for the edge x 5 a=2. Substituting the solutions,
Eqs. (7.101) and (7.102), into the boundary conditions, Eqs. (7.93) and (7.94), we arrive at the
following two algebraic equations:
D m
C2 5 0
λm C1m 1 λ m tanhλ m 1 2λ2m
S
ð1 2 ν Þ coshλ m C1m 1 λm 2 coshλ m 1 ð1 2 ν Þλ m sinhλ m C2m
1
1
1 ð1 2 ν Þλm km coshk m Bm 5 0; λ m 5 λm a; k m 5 km a
2
2
(7.104)
which allows us to express the constants C1m and C2m in terms of Bm . The third boundary condition,
Eq. (7.95), yields the following equation:
2ð1 2 ν Þϕ0 λ21 sinλ1 y 2 2ð1 2 ν Þ
P
2
m ½λm ðsinλ m sinhλm y
1sinhλ m sinλm yÞC1m 1 λm ðsinλ m sinhλm y 1 λm ysinλ m coshλm y
2
P
1sinhλ m sinλm y 1 λ m coshλ m sinλm yÞC2m (7.105)
2
2
m ðλm 1 km Þðsinλ m sinhkm y 1 sinhk m sinλm yÞBm 5 0
Multiplying this equation by sinλn y, integrating with respect to y from a/2 to a/2 and using
the following integrals:
a=2
ð
sin2 λn dy 5 a;
2a=2
a=2
ð
sinhλm y sinλn ydy 5
2a=2
a=2
ð
y coshλm y sinλn ydy 5
2a=2
2
ðλm coshλ m sinλ n 2 λn sinhλ m cosλ n Þ;
λ2m 1 λ2n
2
λ2m 1 λ2n
ðλ m sinhλ m sinλ n 2
λ2m 2 λ2n
coshλ m sinλ n Þ;
λ2m 1 λ2n
we arrive at an infinite set of algebraic equations for the constants Bm . Only the first of these equations includes ϕ0 , whereas the rest of the equations are homogeneous. For a one-term approximation of the acting pressure given by Eq. (7.96), we can retain in the foregoing series only the first
term taking m 5 1, λm 5 λ1 5 π=a. Then, Eq. (7.105) yields
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
2
2C1 λ21 coshλ 1 ð1 1 2λ 1 tanhλ 1 Þ 1 2C2 λ1 coshλ 1 1 1 3λ 1 tanhλ 1 1 2λ 1 tanh2 λ 1
1B1 λ21 1 k12 ðcoshλ 1 1 2λ 1 sinhk 1 Þ 5 4ϕ0 λ21 λ 1
459
(7.106)
in which λ 1 5 λ1 a=2 5 π=2, k 1 5 k1 a=2.
The equations obtained, Eqs. (7.104) and (7.106), allow us to determine the constants C1 , C2 ,
and B1 . The final result looks rather cumbersome; however, for relatively thin plates, it can be significantly simplified. Neglecting the effect of shear deformation on the penetrating potential (i.e.,
taking ϕ 5 w) and taking into account that k1 cλ1 , we get
C1 5
ð1 2 ν Þλ1 λ 1 tanhλ 1
ϕ0 ;
k1 coshλ 1
B1 5
C2 5 2
ð1 2 ν Þλ21 ϕ0
k1 coshλ 1
2λ21 ϕ0
k12 sinhk 1
The final approximate expression for the transverse shear force at the edge x 5 a=2 (see
Fig. 7.7) becomes
V
ð1 2 ν Þ
V x 5 xm 5 cosλ1 y 2
π
Vx
pffiffiffi
3a coshk1 y π
2 cosλ1 y
2
h sinhk 1
(7.107)
where the notation ( ) shows that Vx corresponds to the theory of shear deformable plates and Vxm
is the maximum shear force at the edge corresponding to classical plate theory. The function
V x ðyÞ is shown in Fig. 7.8. At the central part of the edge, the shear force specified by Eq. (7.107)
coincides with the generalized Kirchhoff shear force Kx in Eq. (7.65) as per classical plate theory
(this force is shown in Fig. 7.5). As discussed in Section 7.3.1, the force Kx is accompanied by a
concentrated corner force which is also shown in Fig. 7.8. In the vicinity of the corner, the force
Vx changes its sign (see the dashed line in Fig. 7.8). At the plate corner (y 5 a=2), Eq. (7.107)
yields
Vx
a
2
52
pffiffiffi
ð1 2 ν Þ 3a
πh
If the relative thickness of the plate, h 5 h=a becomes infinitely small, V x ða=2Þ becomes infinitely high and can be treated as a concentrated force. Thus, for the particular boundary conditions
in Eq. (7.92), the Kirchhoff shear force is close to the actual reaction of the plate support and the
corner force following from Kirchhoff and ThomsonTait transformations corresponds to the
asymptotic value of this reaction for an infinitely thin plate.
Thus, for a plate with the boundary conditions given by Eq. (7.92), the solution corresponding
to the theory of shear deformable plates demonstrates plate behavior which is asymptotically close
to the traditional solution corresponding to classical theory which is shown in Fig. 7.4. However, it
should be emphasized that this solution exists only for special boundary conditions, Eq. (7.92), and
the existing recommendation (Timoshenko & Woinowsky-Krieger, 1959) to use it for simply supported plates with classical boundary conditions, Eq. (7.87), is not consistent.
Consider a clamped plate, for which
w 5 0;
θx 5 0;
θy 5 0
460
CHAPTER 7 LAMINATED COMPOSITE PLATES
2
V x* , K x
1 .35
1
Kx
y
b
0
0.1
0.3
0.2
0.4
–1
V x*
–2
–3
3 .85
–4
FIGURE 7.8
Distribution of the normalized transverse shear forces over the edge x 5 a/2.
at the edge x 5 constant. Applying Eqs. (7.79) and (7.82), we arrive at
ϕ2
D
Δϕ 5 0;
S
@ϕ @ψ
5
;
@x
@y
@ϕ
@ψ
52
@y
@x
(7.108)
In contrast to a simply supported plate with classical boundary conditions, Eq. (7.87), these
three conditions cannot be exactly reduced to two. However, as demonstrated in the foregoing
example, the boundary-layer effect hinders the solution considerably, and an attempt can be
made to solve this problem using only the penetrating potential. So, take ψ 5 0 and reduce
Eq. (7.108) to
ϕ2
D
Δϕ 5 0;
S
@ϕ
5 0;
@x
@ϕ
50
@y
(7.109)
The third of these conditions can be integrated along the edge x 5 constant which yields
ϕ 5 C, where C is some constant. The same result can be obtained for the clamped edge
y 5 constant. Thus, we can replace ϕ with ϕ 1 C. Since C is some arbitrary constant, it can
be taken equal to zero. As a result, the boundary conditions for the penetrating solution
become
@ϕ
5 0;
@x
ϕ2
D
Δϕ 5 0
S
(7.110)
and are consistent with the fourth order of the equation for the penetrating solution, Eq. (7.80).
Thus, for clamped plates, the theory of shear deformable plates can be approximately reduced to
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
461
the fourth-order equation, Eq. (7.80), for the penetrating potential and two boundary conditions at
the plate edge.
To support the results obtained, consider the cylindrical bending of a plate. Taking ϕ 5 ϕðxÞ,
we arrive at the following equation and boundary conditions:
0
DϕIV 5 p;
ϕ x 5 ϕ2
D 00
ϕ 50
S
x
0
where ðÞ 5 dðÞ=dx. As can be seen, four constants are required for the solution for constant
pressure, i.e.,
ϕ5
px4
1
1
1 C1 x3 1 C2 x2 1 C3 x 1 C4
3
2
24
which can be found from the four boundary conditions at x 5 0 and x 5 a. We can formally add the
constant C discussed above to ϕ, but it can be included into the unknown constant C4 .
For the free edge, as well as for any edge on which the twisting moment Mxy is preassigned,
three boundary conditions cannot be reduced to two and the sixth-order equations of the theory of
shear deformable plates should be used to study the problem.
As an example, consider a semiinfinite plate with simply supported longitudinal edges (see
Fig. 7.9), for which the exact solutions can be found. Let the plate be loaded with pressure
p 5 p 5 p0 sinλy
(7.111)
where λ 5 π=b. For the longitudinal edges y 5 0 and y 5 b, the boundary conditions are
w 5 0;
My 5 0;
θx 5 0
or exactly
ϕ2
D
Δϕ 5 0;
S
@2 ϕ
@2 ϕ
@2 ψ
5 0;
1 ν 2 2 ð1 2 ν Þ
2
@y
@x
@x@y
@ϕ @ψ
5
@x
@y
These boundary conditions can be satisfied if we take
ϕ 5 ΦðxÞ sinλy; ψ 5 ΨðxÞ cosλy
(7.112)
Substitution of Eq. (7.112) into the governing equations, Eqs. (7.80) and (7.81), yields
00
D ΦIV 2 2Φ 1 Φ 5 p0
00
Ψ 2 s2 Ψ 5 0; s2 5
(7.113)
2S
Dð1 2 ν Þ
(7.114)
y
b
x
FIGURE 7.9
A semiinfinite simply supported plate.
462
CHAPTER 7 LAMINATED COMPOSITE PLATES
The solutions of these equations in conjunction with Eq. (7.112) for the plate shown in
Fig. 7.9 give
p0
sinλy
Dλ4
(7.115)
k2 5 λ2 1 s2
(7.116)
ϕ 5 ðC1 1 C2 xÞe2λx 1
ψ 5 Be2kx cosλy;
The following three cases can be discussed:
1. Consider first a plate simply supported at the edge x 5 0. The classical boundary conditions
w 5 0;
Mx 5 0;
θy 5 0
(7.117)
can be represented as
ϕ2
D
Δϕ 5 0;
S
@2 ϕ
@2 ϕ
@2 ψ
1
ν
2
ð
1
2
ν
Þ
5 0;
@x2
@y2
@x@y
@ϕ
@ψ
52
@y
@x
Substitution of Eqs. (7.115) and (7.116) yields
C1 5 2
p0
;
Dλ4
C2 5 2
p0
;
2Dλ3
B50
Thus, ψ 5 0 and the solution is governed by the penetrating potential ϕ only.
For an isotropic plate with parameters
h
5 0:05;
b
ν 5 0:3;
E
5 2:6;
G
λ5
π
;
b
k5
69:4
b
(7.118)
the final solution specifies the following expressions for the plate deflection, moments and
transverse force:
p0
x
2λx
1
2
e
1
1
1:56
sinλy
b
Dλ4
p0
x
sinλy
Mxð1Þ 5 0:3 2 1 2 e2λx 1 2 3:66
b
λ
p0
x
ð1Þ
5 2 0:35 2 e2λx 1 1 3:14 cosλy
Mxy
b
λ
p0
Vxð1Þ 5 e2λx sinλy
λ
wð1Þ 5 1:0059
(7.119)
Compare this solution with the results corresponding to classical plate theory which gives
wð1Þ
c 5
p0 h
x i
1 2 e2λx 1 1 1:57
sinλy
4
b
Dλ
This deflection is close to that determined by the first equation in Eq. (7.119), whereas the
moments and the force are exactly the same as those given by Eq. (7.119).
2. Consider the second version of the simple support, according to which at the edge x 5 0 we
have
w 5 0; Mx 5 0; Mxy 5 0
(7.120)
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
463
or
ϕ2
D
Δϕ 5 0;
S
@2 ϕ
@2 ϕ
@2 ψ
5 0;
1 ν 2 2 ð1 2 ν Þ
2
@x
@y
@x@y
2
@2 ϕ
@2 ψ @2 ψ
1 2 2 2 50
@x@y
@x
@y
The final solution is
p0 h
x i
2λx
sinλy
1
2
e
1
1
1:49
b
Dλ4
i
p0 h
x
Mxð2Þ 5 0:3 2 1 2 0:89e2λx 1 2 4:08 2 0:11e2kx sinλy
b
λ
i
p0 h 2λx x
ð2Þ
1 1 2:95 2 e2kx cosλy
Mxy 5 2 0:36 2 e
b
λ
p0 2λx
ð2Þ
e
1 0:38e2kx sinλy
Vx 5 0:96
λ
wð2Þ 5 1:0057
As can be seen, wð2Þ is rather close to wð1Þ . The results of these calculations are presented in
Figs. 7.107.12. The normalized bending moments M x 5 Mx λ2 =p0 acting at y 5 b=2 (see
Fig. 7.9) are shown in Fig. 7.10 as functions of x. It follows from the figure that the moment
ð2Þ
M x corresponding to the boundary conditions given by Eq. (7.120) is quite close to the
ð1Þ
moment M x corresponding to the classical boundary conditions, Eq. (7.117). The difference in
ð1Þ
ð2Þ
the twisting moments M xy and M xy is significant only in the vicinity of the plate edge, where
ð2Þ
ð1Þ
M xy 5 0 in accordance with Eq. (7.120), whereas M xy is a reactive moment. The distributions
ð1Þ
ð2Þ
of the shear forces are presented in Fig. 7.12, where V x and V x correspond to the boundary
conditions in Eqs. (7.117) and (7.120), respectively. For comparison, the Kirchhoff shear force
M x(1) , M x( 2 )
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
x
b
1.0
FIGURE 7.10
ð1Þ
ð2Þ
Distribution of the normalized bending moments M x ðy 5 b=2Þ ( sss ) and M x ðy 5 b=2Þ ( 3
the x-axis.
3
3 ) along
464
CHAPTER 7 LAMINATED COMPOSITE PLATES
M xy(1) ,
M xy( 2 )
0.4
0.3
0.2
0.1
x
0
0
0.2
0.4
0.6
1.0 b
0.8
FIGURE 7.11
ð1Þ
ð2Þ
Distribution of the normalized twisting moments M xy ðy 5 0Þ ( sss ) and M xy ðy 5 0Þ (22 22 2 ) along the
x-axis.
Kxð1Þ 5 Vxð1Þ 1
ð1Þ
@Mxy
p0
x
5 e2λx 1:35 1 1:10 sinλy
@y
λ
b
corresponding to the classical boundary conditions in Eq. (7.117) is presented in Fig. 7.12.
As can be seen, this force, being close to the force Vxð2Þ following from the sixth-order theory at
the plate edge x 5 0, does not represent the actual force Vxð2Þ at a distance from the plate edge,
where Vxð2Þ is close to Vxð1Þ . Thus, the force Kxð1Þ is a formal static variable allowing us to
satisfy the boundary conditions and not reflecting the actual shear at a distance from the edge.
3. Consider a plate with the clamped edge x 5 0. The boundary conditions
w 5 0; θx 5 0; θy 5 0
can be presented as
ϕ2
D
@ϕ @ψ @ϕ
@ψ
Δϕ 5 0;
5
;
52
S
@x
@y @y
@x
whereupon the solution becomes
p0
x
2λx
sinλy
1
2
e
1
1
3:12
Dλ4
b
p0
x
Mxð3Þ 5 2 0:3 2 1 2 4:307e2λx 1 2 1:69
1 0:0135e2kx sinλy
b
λ
p0 2λx
Vxð3Þ 5 1:998
e
2 0:0225e2kx sinλy
λ
wð3Þ 5 1:0059
If we neglect the boundary-layer solution and satisfy two boundary conditions in
Eq. (7.110), i.e.,
(7.121)
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
465
V x(1) , V x( 2 ) , K x(1)
1.5
1.0
K x(1)
V x(1)
0.5
V x( 2 )
0
0
0.2
0.4
0.6
0.8
x
1.0 b
FIGURE 7.12
ð1Þ
ð2Þ
ð1Þ
Distribution of the normalized shear forces V x ðy 5 b=2Þ ( sss ), V x ðy 5 b=2Þ (22 22 2 ) and K x ðy 5 b=2Þ
(UUUUUUUUU) along the x-axis.
ϕ2
D
Δϕ 5 0;
S
@ϕ
50
@x
the solution reduces to
p0 h
x i
1 2 e2λx 1 1 3:14
sinλy
4
b
Dλ
i
p0 h
x
Mxð4Þ 5 0:3 2 1 2 4:333e2λx 1 2 1:69 1 0:0135e2kx sinλy
b
λ
p
0
Vxð4Þ 5 2 e2λx sinλy
λ
wð4Þ 5
These equations closely coincide with the corresponding Eq. (7.121). Thus, clamped plates
can be studied retaining the penetrating potential only.
This conclusion is valid for relatively thin plates for which the parameter k characterizing the
rate of the boundary layer solution is much higher than the corresponding parameter of the
466
CHAPTER 7 LAMINATED COMPOSITE PLATES
penetrating solution λ. Consider again the plate shown in Fig. 7.9 with the clamped edge x 5 0 and
having the following parameters:
h
5 1;
b
ν 5 0;
E
5 100;
G
λ5
π
;
b
k5
5:82
b
(7.122)
In contrast to Eq. (7.118), the plate with parameters given by Eq. (7.122) is relatively thick and
the ratio E/G is high. As a result, the parameter k is comparable with λ. Compare the following
three solutions:
1. Exact solution
2. Penetrating solution
3. Solution corresponding to the classical plate theory
The exact solutions for the plate deflection and the bending moment are
i
p0 h
x
1 2 1 1 1:352 e2λx sinλy
4
b
Dλ
i
p0 h
x 2λx
ðeÞ
2 0:451e2kx sinλy
Mx 5 2 1:057 2 1 2 2:375 e
b
λ
wðeÞ 5 1:823
The dependencies of the normalized deflection and moment, i.e., w ðeÞ 5 Dλ4 wðeÞ =p0 and
corresponding to y 5 b=2 (see Fig. 7.9) on the longitudinal coordinate, are
represented in Fig. 7.13 by solid lines.
The penetrating solution gives
ðeÞ
M x 5 λ2 MxðeÞ =p0
i
p0 h
x
1 2 1 1 1:187 e2λx sinλy
4
b
Dλ
p
x
0
MxðpÞ 5 2 0:689 2 1 2 3:14 e2λx sinλy
b
λ
wðpÞ 5 1:823
As can be seen, the deflection corresponding to the penetration solution (shown with circles in
Fig. 7.13A) practically coincides with the exact solution, whereas the bending moment (circles in
Fig. 7.13B) is different. The maximum bending moment for the penetrating solution is 1.43 times
higher than the exact moment.
The solution corresponding to classical plate theory
p0 h
x 2λx i
e
1
2
1
1
3:14
sinλy
b
Dλ4
p0
x
MxðcÞ 5 2 2 1 2 3:14 e2λx sinλy
b
λ
wðcÞ 5
is shown by the dashed lines and is, as expected, rather far from the exact solution.
In conclusion, we can formulate some practical recommendations. Note that they follow
from experience and are not formally justified. It follows from the foregoing discussion
that the effect of transverse shear deformation provides two basic contributions to plate
theory.
Consider the first contribution. For plates fixed at the edges (note that for the majority of the
real boundary conditions, w 5 0 at the plate edges), the shear deformation, first, affects the penetrating solution, i.e., the plate deflection specified by Eq. (7.82). This effect is the same as that discussed in Chapter 6, Laminated Composite Beams and Columns for beams and, as for the beams, it
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
467
w
(A) 2.0
1.6
1.2
0.8
0.4
0
(B) 0.2
0
0
0.4
0.8
1.2
1.6
2.0
x
b
2.0
x
b
Mx
0.4
0.8
1.2
1.6
–0.2
–0.4
0.59
–0.6
0.842
–0.8
1.0
–1.0
FIGURE 7.13
Dependencies of the normalized deflection (A) and bending moment (B) on the longitudinal coordinate: exact
solution ( sss ); penetrating solution ( 3 3 3 ); classical plate theory (22 22 2 ).
does not increase the fourth order of the governing equation, Eq. (7.80). Second, the allowance for
transverse shear deformation gives rise to an additional second-order differential equation,
Eq. (7.81), which does not exist in beam theory and describes the boundary layer solution corresponding to plate torsion.
468
CHAPTER 7 LAMINATED COMPOSITE PLATES
The effect of the transverse shear deformation can be evaluated by comparing the parameter
r2 5
π2 Dm2
Sl2
(7.123)
with unity. Here, m specifies the rate of the penetrating solution (e.g., the half-wave number) and l
is the minimum in-plane dimension of the plate. If r 2 is of the order of unity, both shear effects
must be taken into account (this is the case for the last example, the results of which are shown in
Fig. 7.13). If the order of r 2 is 0.1, then the boundary-layer solution can be neglected. Finally, if
the order of r 2 is 0.01, both shear effects can be neglected and classical plate theory can be used to
study the plate.
The second basic contribution of the shear deformation is associated with the problems for
which the sixth-order plate theory must be used irrespective of the order of the parameter r 2 given
by Eq. (7.123). Such problems involve, particularly, plates loaded with twisting moments applied at
the plate edges which are discussed below.
7.3.2.3 Torsion problem
The problem of plate torsion has been discussed in the literature since the traditional solution of
this problem was obtained by G. Lamb, W. Thomson, and P. Tait (Todhunter & Pearson, 1960;
Timoshenko, 1983) and experimentally confirmed by Nadai (1925). This problem is of a principal
nature: its solution is used to support the existence of Kirchhoff’s shear forces [see Eq. (7.65)] and
provides the background for the experimental determination of the shear moduli of composite materials (Tarnopol’skii & Kincis, 1985).
Consider the classical problem of pure torsion of a square plate with twisting moments M as
shown in Fig. 7.14A. Within the framework of classical plate theory, the plate deflection should
satisfy the biharmonic equation, Eq. (7.46), in which the load p 5 0, and the following boundary
conditions at the plate edges
Mx 5 My 5 0;
Vx 5 Vy 5 0;
Mxy 5 M
(7.124)
in which the moments and the shear forces are specified by Eqs. (7.48)(7.50). The classical solution which satisfies Eq. (7.46) and boundary conditions in Eq. (7.124) is
FIGURE 7.14
Plate torsion by classically applied twisting moments (A), corner forces (B) and uniformly distributed twisting
moments (C).
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
w5 2
Mxy
Dð1 2 νÞ
469
(7.125)
Since Eq. (7.46) is of the fourth order, only two of three boundary conditions in Eq. (7.124) can
be satisfied. As discussed in Section 7.3.1, Kirchhoff’s transformation allows us to reduce the last
two boundary conditions in Eq. (7.124) to one condition Kx 5 Ky 5 0; where K are the forces specified by Eq. (7.65). As can be seen, the solution in Eq. (7.125) satisfies these boundary conditions.
However, as follows from Section 7.3.1, reduction of three boundary conditions to two conditions
is accompanied with concentrated forces P 5 2Mxy applied at the plate corners. For the plate in
Fig. 7.14A, we have P 5 2M and the problem of the plate torsion with moments is formally reduced
to the problem of plate torsion with corner forces (Fig. 7.14B). Expressing M in terms of P, in
Eq. (7.125), we get
w5
Pxy
2Dð1 2 νÞ
(7.126)
Thus, the problems shown in Fig. 7.14A and B in classical plate theory are identical, naturally,
if Kirchhoff’s transformation is correct. Formally, this is the case, but in reality, these problems are
different. To show this, single out the plate element 0-1-2-3 (Fig. 7.14B). For both plates in
Fig. 7.14A and B, the shear forces are zero. But whereas the element in Fig. 7.14A is loaded only
with moments and is in equilibrium, the element in Fig. 7.14B is loaded with the corner force
which is not balanced. Thus, the application of Kirchhoff’s transformation to the problem of plate
torsion is questionable.
Apply the equations of shear deformable plate theory to study the problem. First, consider the
plate shown in Fig. 7.14A. The solution of Eq. (7.80) (in which p 5 0) and (7.81) that satisfies the
boundary conditions in Eq. (7.124) is
ϕ5
Mxy
;
Dð1 2 νÞ
ψ50
and Eq. (7.82) yields w 5 ϕ which coincides with Eq. (7.126). Thus, for a plate loaded with edge
moments, solutions corresponding to the classical and shear deformable plate theories are the same.
Consider a plate loaded with corner forces (Fig. 7.14B). Introduce normalized coordinates
x 5 x=a; y 5 y=a, where 2a is the length of the plate side (Fig. 7.14), so that 21 # x; y # 1, and
present the corner forces as the surface load expanded into the trigonometric series
p5
4P X X
sinλm sinλn sinλm x sinλn y;
a2 m n
λm 5
2m 2 1
π;
2
λn 5
2n 2 1
π
2
Substituting p into Eq. (7.80), determine the particular solution of this equation
ϕq 5
4Pa2 X X sin λm sin λn
2
2 sin λm x sin λn y
D m n
λ 1λ2
m
(7.127)
n
Taking into account symmetry conditions, present the general solutions of homogeneous equation corresponding to Eqs. (7.80) and (7.81) as
ϕ0 5
P m
m C1 ðsinλm x sinhλm y 1 sinhλm x sinλm yÞ
P
1 m C2m ðxcoshλm x sinλm y 1 ycoshy sinλm xÞ
(7.128)
470
CHAPTER 7 LAMINATED COMPOSITE PLATES
ψ0 5
X
C3m cosλm x coshηm y 2 coshηm x cosλm y ;
η2m 5 λ2m 1 λ2 5 λ2m 1
m
12a2
h2
(7.129)
Then, the general solutions of Eqs. (7.80) and (7.81) take the form ϕ 5 ϕ0 1 ϕq and ψ 5 ψ0 .
Substituting these solutions into Eqs. (7.82)(7.84), we get the following expressions for the plate
deflection, moments, and shear forces:
4Pa2
w5
D
1
X
(
X
m
C 1 ðsinλm xsinhλm y 1 sinhλm xsinλm yÞ
m
m
C 2 xcoshxsinλm y 1 ycoshλm ysinλm x 2
m
1
X X sinλm sinλn
11
λm h2
ðsinhλm xsinλm y 1 sinλm xsinhλm yÞ
3ð1 2 νÞa2
h2 ðλ2m 1 λ2n Þ
sinλm xsinλn y
6ð1 2 νÞa2
)
ðλ2m 1λ2n Þ2
P m
Mx 5 2 4Pð1 2 νÞf " m C 1 λ2m ðsinhλm xsinλm y 2 sinλm xsinhλm yÞ
P m
2
ðsinhλm xsinλm y 1 sinλm xsinhλm yÞ 2 ð2sinhλm y 1 λm ycoshλm yÞsinλm x
1 m C2 λm
12ν
#
P m
1 λm xcoshλm xsinλm y 1 m C 3 λm ηm ðsinλm xsinhηm y 2 sinhηm xsinλm yÞg
m
n
(7.130)
(7.131)
P P λ2m 1 νλ2n
sinλm sinλn sinλm xsinλn y
m
n
ðλ2m 1λ2n Þ2
(
P m
Mxy 5 2 2Pð1 2 νÞ 2 m C1 λ2m ðcosλm xcoshλm y 1 coshλm xcosλm yÞ
1 4P
P m
1 2 m C2 λm ½ðcoshλm y 1 λm ysinhλm yÞcosλm x 1 ðcoshλm x 1 λm xsinhλm xÞcosλm y
P m
2 m C3 ðλ2m 1 η2m Þðcoshηm xcosλm y 1 cosλm xcoshηm yÞ
)
P P
λm λn
sinλm sinλn cosλm xcosλn y
12 m n 2
ðλm 1λ2n Þ2
"
Pa X m
Vx 5 24 2
C 3 ðηm cosλm xsinhηm y 1 λm coshηm xsinλm yÞ
h
m
#
P m λ2m h2
2 m C2
ðcoshλm xsinλm y 1 cosλm xsinhλm yÞ
3ð1 2 νÞa2
4P X X sinλm sinλn
λm 2
cosλm xsinλn y
1
a m n
λm 1 λ2n
where
m
Ci 5 Cim
D
4Pa2
ði 5 1; 2; 3Þ
(7.132)
(7.133)
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
m
m
471
m
The coefficients C 1 , C2 and C 3 can be found from the boundary conditions which have the following form for the edge x 5 1
Vx ðyÞ 5 0;
Mxy ðyÞ 5 0;
Mx ðyÞ 5 0
(7.134)
Using the first and the second conditions, we have
m
C3 5
λm h2 coshλm
m
C ;
3ð1 2 νÞa2 coshηm 2
m
m
C 1 5 2 cm C 2 ;
cm 5
1
ðλ2 1 η2m Þh2
1 tanhλm 2 m
λm
6λm ð1 2 νÞa2
Substituting these coefficients into the third boundary condition in Eq. (7.134), multiplying it by
sin λi y, and integrating with respect to y from 0 up to 1, we arrive at the following linked system
m
of linear algebraic equations for coefficients C 2 :
i λi coshλi
C2
λi ηi h2
2tanhλi 1 λi ð1 2 νÞð1 2 ci tanhλi Þ 2
tanhηi
sinλi
3a2
"
!
#
P m 2λ2m sinλm coshλm
η2m h2 λ2m 1 λ2i
λ2m 2 λ2i
1 2ν 1
ð1 2 νÞ cm λm 2 λm tanhλm 1 2
1 m C2
λ2m 1 λ2i
λm 1 λ2i
3a2 η2m 1 λ2i
P λ2 1 νλ2i
5 m m2
ðλm 1λ2i Þ2
Nadai (1925) presented the results of an experimental study of a steel plate with the following
parameters: a 5 81.32 mm, h 5 14.96 mm (h/(2a)) 5 0.092), E 5 198.44 GPa, ν 5 0:3, and
P 5 4.96 kN. The plate deflection was measured with high accuracy (up to 0.001 mm) at the points
located onpthe
ffiffiffi diagonal 02 of the plate shown in Fig. 7.14B. The coordinates of these points are
x 5 y 5 s= 2; where s 5 s=a and s is the diagonal coordinate (see Fig. 7.14B). For the first 10
terms retained in the series, results of calculation are presented in Fig. 7.15 demonstrating the
dependence of the normalized deflection w 5 wD=ðPa2 Þ on the diagonal coordinate s. Circles correspond to the experimental data. The solid line shows the solution given by Eq. (7.130) corresponding to the shear deformable theory, whereas the dashed line shows the classical theory solution as
per Eq. (7.126). For the last experimental point in Fig. 7.15, the difference between the deformable
shear theory solution and the experiment does not exceed 1%, whereas the difference between the
classical theory solution and the experiment is 4.6%. For the plates with thicknesses 10 mm (h/
(2a) 5 0.0615) and 5 mm, the differences between the classical theory solutions and the experiment
are 1.7% and 0.42%, respectively. The dotted line in Fig. 7.15 corresponds to the classical solution
for which the conditions in Eqs. (7.134) are changed to Ky ðyÞ 5 0 and Mx ðyÞ 5 0, where Ky is specified by Eq. (7.65). Thus, the classical plate theory allows us to predict the deflection of relatively
thin plates loaded with corner forces with reasonable accuracy.
However, this conclusion is true for the deflection only. Fig. 7.16 demonstrates the distribution of
the normalized twisting moment M xy 5 Mxy =P in the cross-section x 5 0 of the plate. The solid line corresponds to the solution in Eq. (7.132) of the shear deformable theory. As can be seen, the moment is
zero at the plate edge which agrees with the second boundary condition in Eq. (7.134). The dashed line
corresponds to the classical theory solution which yields M xy 5 0:5 and the boundary condition for the
twisting moment is not satisfied. Similar results are shown in Fig. 7.17 for the normalized transverse
shear force V x 5 Vx a=P acting in the plate cross-section x 5 0. The solid line corresponds to the solution of the shear deformable plate theory in Eq. (7.133) and results in nonzero force acting in the crosssection which is orthogonal to the plate edge. This force balances the force acting at the corner point of
the plate. The classical theory solution (dashed line) gives Vx 5 0 and the corner force is not balanced.
472
CHAPTER 7 LAMINATED COMPOSITE PLATES
FIGURE 7.15
Dependence of the plate normalized deflection on the diagonal coordinate corresponding to the shear deformable
(solid line) and classical (dashed line) plate theories.
Thus, the classical plate theory results in a reasonable solution for the deflection of the plate
loaded with corner forces and fails to predict the moment and the shear force. The Kirchhoff transformation in classical plate theory actually allows us to change the edge twisting moment to statically equivalent shear force. To demonstrate what happens under this transformation, consider a
plate loaded with an edge force and moment. Particularly, for the edge x 5 1 assume that
Vx 5 V1 sinλ1 y;
Mxy 5 M1 cosλ1 y
(7.135)
where λ1 5 π=2 and coefficients V1 and M1 satisfy the following condition:
λ1 M1 5 aV1
If these conditions are valid, the Kirchhoff stress resultant in Eq. (7.65) is zero and the applied
moment and force in Eq. (7.135) are self-balanced. Note that Mxy 5 0 at the plate corners and
Kirchhoff’s transformation does not result in corner forces in this case. For qualitative analysis,
take m 5 1 in series, Eqs. (7.128) and (7.129). Omitting rather cumbersome transformations, we
arrive at the following expression for the plate deflection:
w5
V1 ah2
Fðx; yÞ
6Dð1 2 νÞ
(7.136)
where F is some complicated function whose form is not important for further analysis. Assume
that V1a 5 P in which P is the force in Eq. (7.130) for the deflection of the plate loaded with corner
forces. Then, comparing Eqs. (7.130) and (7.136), we can conclude that the deflection in
7.3 ANALYSIS OF PLATE THEORY FOR TRANSVERSELY ISOTROPIC PLATES
473
FIGURE 7.16
Dependencies of the twisting moment corresponding to shear deformable (solid line) and classical (dashed line)
plate theories.
Eq. (7.136) is (h/a)2 times smaller than the deflection in Eq. (7.130). Thus, the Kirchhoff transformation results in a plate deflection which is very small for thin plates. This is the reason for which
the classical solution in Eq. (7.126) provides rather accurate results for the deflection of thin plates.
In conclusion, consider a plate loaded with twisting moments which do not change their direction as in Fig. 7.14C. In contrast to the boundary conditions in Eq. (7.124) for the plate in
Fig. 7.14A, the boundary conditions in this case become
Mx 5 0;
Vx 5 0;
Mxy 5 2 M;
for x 5 a
(7.137)
My 5 0;
Vy 5 0;
M xy 5 M;
for y 5 a
(7.138)
To satisfy these boundary conditions, we change signs in Eqs. (7.128) and (7.129) for ϕ and ψ,
and write them in the form
ϕ5
P
m
m ½C 1 ðsinλm xsinhλm y 2 sinhλm xsinλm yÞ
m
1C2 ðycoshλm ysinλm x 2 xcoshλm xsinλm yÞ
P
ψ 5 m C3m ðcosλm xcoshηm y 1 coshηm xcosλm yÞ
x
x5 ;
a
y
y5 ;
a
λm 5
2m 2 1
π;
2
η2m 5 λ2m 1
(7.139)
12a2
h2
To solve the problem for a square plate, we need to satisfy the conditions in Eq. (7.137) only
because the conditions in Eq. (7.138) are automatically satisfied in this case. Present the twisting
moment acting on the plate edge as
474
CHAPTER 7 LAMINATED COMPOSITE PLATES
FIGURE 7.17
Dependencies of the shear force corresponding to shear deformable (solid line) and classical (dashed line)
plate theories.
M5
X
m
Mm cosλm y;
Mm 5
2M
sinλm
λm
(7.140)
Then, the third and the second boundary conditions in Eq. (7.137) yield
λm h2 coshλm
2Ma2 sinλm
m
m
m
C 2 ; C1 5 2 dm C2 2
2
3ð1 2 νÞa coshηm
Dð1 2 νÞλ3m coshλm
2
2 2
1
ðλ 1 ηm Þh
dm 5
1 tanhλm 1 m
λm
6λm ð1 2 νÞa2
m
C3 5
The first boundary condition in Eq. (7.137) results in the following linked system of algebraic
equations for the coefficients C2m :
i λi coshλi
C2
λi ηi h2
λi ð1 2 νÞð1 1 di tanhλi Þ 2 2tanhλi 1
sinλi
3a2
"
#
!
2
P m 2λm sinλm coshλm
2ð2 2 νÞ λ2m 2 λ2i
η2m h2 ðλ2m 1 λ2i Þ
1
ð1 2 νÞ dm λm 1 λm tanhλm 1
2 2
1 m C2
12ν
λ2m 1 λ2i
λm 1 λ2i
3a2 ðη2m 1 λ2i Þ
X
2
1
5 2 tanhλi 2 4
2
2
λi
m λm 1 λi
m
where C j 5 Ma2 Cjm =D ðj 5 1; 2; 3Þ. The solution for the plate deflection is
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
475
FIGURE 7.18
Deflections of the plates loaded with twisting moments in accordance with Fig. 7.14A (solid line) and Fig. 7.14C
(dashed line).
Ma2 X m
w5
C
D m 2
(
λm h2
ðsinhλm xsinλm y 2 sinλm xsinhλm yÞ
3ð1 2 νÞa2
2Ma2 X sinλm
1 ycoshλm y 2 xcoshλm xsinλm y 2
ðsinλm xsinhλm y 2 sinhλm xsinλm yÞ
Dð1 2 νÞ m λ3m coshλm
dm 2
The dependence of the normalized deflection w 5 Dw=ðMa2 Þ on the coordinate x 5 x=a for the
edge y 5 a (Fig. 7.14C) is shown in Fig. 7.18 with the solid line. For comparison, the dashed line
corresponds to Eq. (7.125).
For classical plate theory, we should take ϕ 5 w and ψ 5 0. Since the edge twisting moment
does not depend on coordinates, the Kirchhoff shear forces Kx and Ky in Eq. (7.65) are equal to Vx
and Vy, respectively. As a result, we arrive at the first two boundary conditions in Eqs. (7.137) and
m
m
m
m
(7.138) for coefficients C1 and C2 from which it follows that C1 5 C 2 5 0 and hence, w 5 0.
Thus, classical plate theory does not allow us to study plates loaded in accordance with Fig. 7.14C.
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
The general set of equations obtained in Section 7.1 describes a plate with arbitrary structure.
However, in the majority of applications, composite plates are orthotropic and the material structure
is symmetric with respect to the plate middle plane. Such plates are considered in this section.
476
CHAPTER 7 LAMINATED COMPOSITE PLATES
7.4.1 EXACT SOLUTIONS OF CLASSICAL PLATE THEORY
The problem of bending of orthotropic symmetric plates is described by Eqs. (7.40)(7.42). As follows from the foregoing discussion, thin plates with a moderate ratio of E/G can be studied using
classical plate theory. Note that for the majority of real composite laminated plates, classical plate
theory provides rather accurate results, assuming that the plate edges are fixed with respect to plate
deflection. Naturally (see Section 7.3), no generalized transverse shear forces and corner forces
should be introduced into the analysis.
Neglecting transverse shear deformation, take Sx -N and Sy -N in Eqs. (7.40)(7.42). Then,
W 5 w and Eq. (7.40) reduces to the well-known equation for plate deflection, i.e.,
@4 w
@4 w
@4 w
1 2D3 2 2 1 D2 4 5 p
4
@x
@x @y
@y
(7.141)
D1 5 D11 ; D2 5 D22 ; D3 5 D12 1 2D44
(7.142)
D1
where p 5 p 2 q (see Fig. 7.1) and
The moments and the shear forces are expressed in terms of w as
@2 w
@2 w
@2 w
@2 w
@2 w
Mx 5 2 D11 2 1 D12 2 ; My 5 2 D12 2 1 D22 2 ; Mxy 5 2 2D44
@x
@y
@x
@y
@x@y
2
2
2
2
@
@ w
@ w
@
@ w
@ w
D11 2 1 D3 2 ; Vy 5 2
D22 2 1 D3 2
Vx 5 2
@x
@x
@y
@y
@y
@x
(7.143)
For simply supported edges
w 5 0;
w 5 0;
@2 w
5 0; for x 5 constant
@x2
2
@ w
5 0; for y 5 constant
@y2
(7.144)
@w
5 0; for x 5 constant
@x
@w
5 0; for y 5 constant
@y
(7.145)
whereas for clamped edges
w 5 0;
w 5 0;
Consider the exact solutions obtained for plates with various boundary conditions.
7.4.1.1 Simply supported plates
Plates supported at the contour by walls that are absolutely rigid in-plane and flexible out-of-plane
(see Fig. 7.19) are usually treated as simply supported plates with boundary conditions specified by
Eq. (7.144). The method of exact solution for such plates proposed in 1823 by L.M. Navier is based
on the following trigonometric series for w:
wðx; yÞ 5
XX
m
wmn sinλm x sinλn y
(7.146)
n
in which wmn are constant coefficients, λm 5 πm=a, λn 5 πn=b and a, b are the plate in-plane
dimensions (see Fig. 7.19). Each term of the series in Eq. (7.146) satisfies the boundary conditions
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
477
z
x
a
y
b
FIGURE 7.19
A simply supported plate.
given by Eqs. (7.144). In addition, the pressure p in Eq. (7.141) is also expanded into a similar
series, i.e.,
p ðx; yÞ 5
XX
m
pmn sinλm xsinλn y
(7.147)
n
in which
pmn 5
4
ab
ða ðb
pðx; yÞsinλm x sinλn y
0 0
For uniform pressure p 5 p0 ,
8
16p0
>
<
ðm; n 5 1; 3; 5; . . .Þ
2 mn
π
pmn 5
>
:
0ðm; n 5 2; 4; 6; . . .Þ
(7.148)
Substituting Eqs. (7.146) and (7.147) into Eq. (7.144), we can express wmn in terms of pmn and
obtain finally the following solution:
wðx; yÞ 5
XX
m
n
pmn sinλm x sinλn y
D1 λ4m 1 2D3 λ2m λ2n 1 D2 λ4n
(7.149)
The moments and forces acting in the plate can be found using Eq. (7.145).
Convergence of the series in Eq. (7.149) depends on the plate in-plane aspect ratio a/b (see
Fig. 7.20). For square and close to square plates (see Fig. 7.20A)
1#
a
# 1:5;
b
(7.150)
using only the first term in Eq. (7.149) provides reasonable accuracy for the solution. Convergence
of the corresponding series for the moment and the forces in Eq. (7.143) which require differentiation of Eq. (7.149) is much slower than the convergence of Eq. (7.149), and to calculate the
moments and the forces, we need to retain up to five terms in the series (Vasiliev, 1993).
478
CHAPTER 7 LAMINATED COMPOSITE PLATES
a
(A)
x
b
y
a
(B)
x
b
y
FIGURE 7.20
Close to square (A) and long (B) rectangular plates.
wm .102
1.5
1.3
1.0
0.5
0.34
0
2
4
6
8
a
b
FIGURE 7.21
Dependence of the normalized deflection on the aspect ratio a/b.
For long plates (see Fig. 7.20B), with an aspect ratio
rffiffiffiffiffiffi
a
D1
. 3:5
;
b
D2
(7.151)
the convergence of the series is getting rather slow. It follows from Fig. 7.20B that in the middle
part of the plate the deflection is practically independent of x and is close to constant. To predict
this deflection, we must retain a large number of terms in Eq. (7.149).
For a 6 45 composite plate with D1 5 D2 5 D and D3 5 1:4D, loaded with uniform pressure
p0 , the dependence of the normalized maximum deflection w m 5 wm D=p0 b4 on the plate aspect
ratio a/b is shown in Fig. 7.21. As can be seen, for those ratios a/b which satisfy the condition
given by Eq. (7.151), the plate can be considered as infinitely long. Taking a =b-N, we get from
Eq. (7.149) in which pmn is specified by Eq. (7.148)
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
479
FIGURE 7.22
A rectangular plate with two simply supported opposite edges.
w5
X1
16p0 b4 X 1
p0
sinλm x
sinλn y 5
ðy4 2 2by3 1 b3 yÞ
6
n5
π D2 m m
24D2
n
This result corresponds to the deflection of a beam of length b simply supported at its ends.
7.4.1.2 Plates simply supported at opposite edges
A more general method whose convergence does not depend on the aspect, a/b ratio was proposed
by M. Levy in 1899. Consider a plate with simply supported edges y 5 0 and y 5 b (see Fig. 7.22)
and assume that a $ b, i.e., that the plate is simply supported along the long edges, whereas the
boundary conditions for the short edges x 5 6 a=2 can be arbitrary. The solution of Eq. (7.141) is
presented in the form
wðx; yÞ 5
X
Wn ðxÞ sinλn y
(7.152)
n
where, as earlier, λn 5 πn=b. The deflection given by Eq. (7.152) satisfies the boundary conditions,
Eq. (7.144) for the edges y 5 0 and y 5 b. The pressure, p, in Eq. (7.144) is expanded into a similar
series, i.e.,
p ðx; yÞ 5
X
pn ðxÞ sinλn y
(7.153)
n
where
pn ðxÞ 5
ðb
2
pðx; yÞ sinλn ydy
b
0
For uniform pressure p 5 p0 ,
8
>
< 4p0 ðn 5 1; 3; 5; . . .Þ
pn 5 bλn
>
:
0ðn 5 2; 4; 6; . . .Þ
(7.154)
480
CHAPTER 7 LAMINATED COMPOSITE PLATES
Substituting Eqs. (7.152) and (7.154) into Eq. (7.141), we arrive at the following ordinary differential equation for the function Wn ðxÞ:
00
2
4
WnIV 2 2k1n
Wn 1 k2n
Wn 5 kpn
(7.155)
in which
2
k1n
5
D3 2
λ ;
D1 n
4
k2n
5
D2 4
pn
λ ; kpn 5
D1 n
D1
(7.156)
The solution of Eq. (7.155) is
Wn 5
4
X
Ci Φi ðxÞ 1 Wpn
(7.157)
i51
where Ci are the constants of integration and Wpn is the particular solution. For a uniform pressure
with pn specified by Eq. (7.154),
Wpn 5
pn
4
D1 k2n
(7.158)
The functions Φi ðxÞ have a form which depends on the relationship between the coefficients in
Eq. (7.156). Possible forms for these functions are presented in Table 7.1.
For orthotropic plates, it is usually the case that k22 . k12 . In this case,
Φ1 5 coshrxcostx; Φ2 5 sinhrxsintx
Φ3 5 coshrxsintx; Φ4 5 sinhrxcostx
(7.159)
00
Table 7.1 Solutions of the Differential Equation W IV 2 2k12 W 1 k24 W 5 0
Relations Between
Coefficients
Roots of the
Characteristic Equation
k22 . k12
k22 5 k12
k22 , k12
Φ1
Φ2
Φ3
Φ4
x1 5 2 x2 5 r 1 it
x3 5 2 x4 5 r 2 it
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r 5 1=2 k22 1 k12
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
t 5 1=2 k22 2 k12
coshrx costx
sinhrx sintx
coshrx sintx
sinhrx costx
e2rx costx
e2rx sintx
erx costx
erx sintx
x1 5 x2 5 r
x3 5 x4 5 2 r
coshrx
xsinhrx
sinhrx
xcoshrx
r 5 k1
x1 5 2 x2 5 r
x3 5 2 x4 5 t
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r 5 k12 2 k14 2 k24
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
t 5 k12 1 k14 2 k24
e2rx
coshrx
xe2rx
coshtx
erx
sinhrx
xerx
sinhtx
e2rx
e2tx
erx
etx
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
(A)
a/2
(B)
a/2
481
a /2
a /2
x
x
b
b
y
y
FIGURE 7.23
Plates with simply supported longitudinal edges and simply supported (A) and clamped (B) transverse edges.
The derivatives of these functions are expressed in terms of the initial functions as
0
0
Φ1 5 rΦ4 2 tΦ3 ; Φ2 5 rΦ3 1 tΦ4 ;
00
Φ1 5 r 2 2 t2 Φ1 2 2rtΦ2 ;
00
Φ3 5 r 2 2 t2 Φ3 1 2rtΦ4 ;
Φv10 5 r r 2 2 3t2 Φ4 1 t t2 2 3r 2 Φ3 ;
Φv30 5 r r 2 2 3t2 Φ2 2 t t2 2 3r 2 Φ1 ;
0
0
Φ3 5 rΦ2 1 tΦ1 ; Φ4 5 rΦ1 2 tΦ2
00
Φ2 5 r 2 2 t2 Φ2 1 2rtΦ1
00
Φ4 5 r 2 2 t2 Φ4 2 2rtΦ3
Φv20 5 r r 2 2 3t2 Φ3 1 t t2 2 3r 2 Φ4
Φv40 5 r r 2 2 3t2 Φ1 1 t t2 2 3r 2 Φ2
(7.160)
Consider the action of a uniform pressure p0 on the plates whose boundary conditions are
symmetric with respect to the x-coordinate (see Fig. 7.23). In this case, C3 5 C4 5 0 and
Eq. (7.157) becomes
Wn 5 C1 Φ1 ðxÞ 1 C2 Φ2 ðxÞ 1 Wp
(7.161)
in which Wp and Φ1 , Φ2 are specified by Eqs. (7.158) and (7.159).
If the transverse edges x 5 6 a=2 are simply supported (see Fig. 7.23A) and the boundary conditions correspond to Eq. (7.144), the solution is
Wn 5 Wp ½1 2 F1 ðxÞ
where
F1 ðxÞ 5
(7.162)
00 00 Φ1 ðxÞΦ2 a=2 2 Φ2 ðxÞΦ1 a=2
00 00 Φ1 a=2 Φ2 a=2 2 Φ2 a=2 Φ1 a=2
and the derivatives of the functions Φ are specified by Eq. (7.160). In contrast to the Navier
solution given by Eq. (7.149), the series following from Eq. (7.152), (7.161), and (7.162) rapidly
converge for any aspect ratio b/a.
For a plate with clamped edges x 5 6 a=2 (see Fig. 7.23B), the boundary conditions correspond
to Eq. (7.145) and
Wn ðxÞ 5 Wp ½1 2 F2 ðxÞ
where
0 0 Φ1 ðxÞΦ2 a=2 2 Φ2 ðxÞΦ1 a=2
0
0
F2 ðxÞ 5 Φ1 a=2 Φ2 a=2 2 Φ2 a=2 Φ1 a=2
(7.163)
482
CHAPTER 7 LAMINATED COMPOSITE PLATES
(A)
(B)
b/ 2
a /2
1
x
b/2
a/2
1
1
x
1
y
y
FIGURE 7.24
Clamped plate referred to coordinates x, y (A) and x, y(B).
7.4.1.3 Clamped plates
The exact solution for a clamped plate has been found by Lurie (1982) with the aid of the method
of homogeneous solutions developed by Shiff (1883), Dougall (1904), Brahtz (1935), Tolcke
(1938), Fadle (1941), and Papkovich (1941). The mathematical justification of the method and
the exact solutions of biharmonic problems in the theory of elasticity are presented by Lurie and
Vasiliev (1995).
Consider a plate clamped at its edges (see Fig. 7.24A) and introduce the normalized coordinates
x5
2x
;
a
y5
2y
b
(7.164)
The initial rectangular plate (see Fig. 7.24A), being referred to the coordinates given in
Eq. (7.148), reduces to a square plate (see Fig. 7.24B) for which 2 1 # x # 1 and 2 1 # y # 1.
Then, the governing equation, Eq. (7.141), is transformed as follows:
D1 b4 @4 w D3 b2 @4 w
@4 w
pb4
1
1 4 5
4
2
2
4
2
D2 a @x
D2 a @x @y
16D2
@y
(7.165)
The solution of Eq. (7.165) corresponding to a uniform pressure p 5 p0 has the form
w 5 w0 1 wp
(7.166)
2
p0 b4 12y 2
384D2
(7.167)
in which
wp 5
satisfies
the
boundary
condition,
Eq.
(7.145)
for
clamped
edges,
i.e.,
0
wp ðy 5 6 1Þ 5 wp ðy 5 6 1Þ 5 0, whereas w0 satisfies the homogeneous equation corresponding to
Eq. (7.165). The function w0 , which is symmetric with respect to the x-coordinate, is given by
w0 ðx; y Þ 5
X
Cm coshðλm xÞFm ðy Þ
(7.168)
m
where Cm and λm are constants and Fm ðy Þ are the so-called homogeneous solutions which can
be found if we substitute Eq. (7.168) into Eq. (7.165) where p 5 0. The resulting equation for
Fm ðy Þ is
00
FmIV 1 2α2 Fm 1 β 4 Fm 5 0
(7.169)
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
483
0
where ðÞ 5 dðÞ=dy and
α2 5
D3 b2
;
D2 a2
β4 5
D1 b4
D2 a4
The solution of Eq. (7.169), symmetric with respect to the y coordinate, is
Fm ðy Þ 5 C1m cosðr1 λm yÞ 1 C2m cosðr2 λm y Þ
where
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b
4
2
4
r1;2 5 α 6 α 2 β 5 pffiffiffiffiffiffi D3 6 D23 2 D1 D2
a D2
(7.170)
(7.171)
and r1 , r2 are the roots of the characteristic equation.
pffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffi
Note that the solution given by Eq. (7.170) is valid if D3 . D1 D2 . If D3 , D1 D2 , then the
roots r1;2 are complex numbers, i.e.,
0sr
1
sr
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffiffiffiffiffiffi
ffiffiffiffiffiffi
b @
D1 D3
D1 D3 A
7i
2
1
r1;2 5 pffiffiffi
D2 D2
D2 D2
a 2
(7.172)
where i is the imaginary unit (i2 5 2 1). For this case
F m ðy Þ 5 C1m coshðr1 λm yÞ cosðr2 λm yÞ 1 C2m sinhðr1 λm yÞ sinðr2 λm y Þ
Further derivation is performed for the homogeneous solution, Eq. (7.170), because it is easier
to differentiate this equation, although the roots r1 and r2 can be either real, Eq. (7.171), or
complex, Eq. (7.172), numbers.
To determine the parameter λm in Eq. (7.170), apply the boundary conditions, Eq. (7.145) to the
edges y 5 6 1, i.e.,
0
Fm ðy 5 6 1Þ 5 0; Fm ðy 5 6 1Þ 5 0
(7.173)
Substituting Eq. (7.170) into Eq. (7.173), we arrive at the following two equations for the constants C1m and C2m :
C1m cosðr1 λm Þ 1 C2m cosðr2 λm Þ 5 0
C1m r1 sinðr1 λm Þ 1 C2m r2 sinðr2 λm Þ 5 0
The first of these equations yields
C2m 5 2 C1m
cosðr1 λm Þ
cosðr2 λm Þ
(7.174)
whereas substitution of Eq.(7.174) into the second equation results in the following equation for the
parameter λm :
r1 tanðr1 λm Þ 5 r2 tanðr2 λm Þ
(7.175)
Using Eq. (7.174), we can write the homogeneous solutions given by Eq. (7.170) as
Fm ðy Þ 5 C1m ½cosðr2 λm Þ cosðr1 λm yÞ 2 cosðr1 λm Þ cosðr2 λm y Þ
(7.176)
484
CHAPTER 7 LAMINATED COMPOSITE PLATES
Finally, substituting Eq. (7.176) into Eq. (7.168) and including the constant C1m into the constant Cm , we arrive at
w0 ðx; y Þ 5
X
Cm coshðλm xÞF m ðy Þ
(7.177)
m
where
F m ðy Þ 5 cosðr2 λm Þcosðr1 λm yÞ 2 cosðr1 λm Þcosðr2 λm y Þ
(7.178)
The constants Cm in Eq. (7.177) can be found from the boundary conditions given by
Eq. (7.145) for the plate edges x 5 6 1, i.e.,
wðx 5 6 1Þ 5
X
Cm coshλm F m ðy Þ 1 wp ðy Þ 5 0
m
X
@w
ðx 5 6 1Þ 5
Cm λm sinhλm F m ðy Þ 5 0
@x
m
(7.179)
It follows from these equations that the two mutually independent functions (wp and 0) must be
expanded into two series with respect to the homogeneous solutions with one and the same system
of constants C1m . If the functions Fm ðy Þ were orthogonal and complete, this problem could not be
solved. However, the functions Fm ðy Þ specified by Eq. (7.176) are of a more complicated nature,
i.e., they are biorthogonal and bicomplete which, in principle, allows us to solve the problem (see
Lurie & Vasiliev, 1995).
To derive the biorthogonality condition, consider Eq. (7.169) written for the functions Fm
and Fn such that m 6¼ n. Multiplying the equation for Fm by Fn and the equation for Fn by Fm and
integrating the resulting equations from 1 to 11, we get
ð1
ð1
00
FmIV Fn dy 1 2a2 Fm Fn dy 1 b4 Fm Fn 5 0
21
ð1
21
ð1
FnIV Fm dy 1 2a2
21
(7.180)
00
Fn Fm dy 1 b Fn Fm 5 0
4
21
Integrating by parts, with allowance for the boundary conditions, Eq. (7.173) yields for the
first equation
ð1
ð1
FmIV Fn dy 5 2
21
ð1
000
21
00
ð1
Fm Fn dy 5 2
21
ð1
0
Fm Fn dy 5
00
00
Fm Fn dy
21
0
0
Fm Fn dy
21
Similar relations are valid for the second equation of Eq. (7.180). Now, subtract the second
equation from the first one and take into account that λm 6¼ λn . Then,
ð1
00
00
ðFm Fn 2 b4 λ2m λ2n Fm Fn Þdy 5 0; λm 6¼ λn
21
(7.181)
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
485
The biorthogonality condition in Eq. (7.181) allows us to determine directly the constants in
Eq. (7.177) only for a simply supported plate. Indeed, the boundary conditions in Eq. (7.144) for a
simply supported plate take the form
wðx 5 6 1Þ 5
X
Cm coshλm F m ðy Þ 1 wp ðy Þ 5 0
m
X
@2 w
ðx 5 6 1Þ 5
Cm λ2m coshλm F m ðy Þ 5 0
2
@x
m
(7.182)
00
Differentiate twice the first of these conditions, multiply it by Fn , subtract the second condition
multiplied by b4 λ2n Fn , and integrate the result from 1 to 1 to get
X
01
1
ð
00
00
Cm coshλm @ ðF m F n 2 b4 λ2m λ2n F m F n Þdy A 5 0
m
21
Applying the biorthogonality condition as per Eq. (7.181) for functions F m and F n , we can find Cn as
Ð1
Cn 5 2
21
00
wp ðyÞF n dy
Ð1 0 0 2
2
coshλn
F n 2 b4 F n dy
21
Return now to the clamped plate. Application of the biorthogonality condition, Eq. (7.181), to
the boundary conditions given by Eq. (7.179) requires some additional transformations. First,
rewrite Eq. (7.178) as
F m ðy Þ 5 um ðy Þ 1 vm ðy Þ
(7.183)
um ðy Þ 5 C2m cosðr1 λm yÞ; vm ðy Þ 5 C1m cosðr2 λm yÞ
(7.184)
where
in which
C1m 5 2 cosðr1 λm Þ; C2m 5 cosðr2 λm Þ
Second, substitute Eqs. (7.183) and (7.184) into Eq. (7.181). Taking into account that
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r12 1 r22 5 2a2 ; r12 2 r22 5 2 a4 2 b4 ; r12 r22 5 b4
where a2 and b4 are the coefficients of Eq. (7.169) and r1 and r2 are the roots of the corresponding
characteristic equation, we can write the biorthogonality condition, Eq. (7.181) in the following
form:
ð1
2
r1 um un 2 r22 vm vn dy 5 0
(7.185)
21
in which m 6¼ n.
Finally, rewrite the boundary conditions in Eq. (7.179) in terms of the functions um and vm , i.e.,
X
m
Cm coshλm ðum 1 vm Þ 5 2 wp ;
X
m
Cm λm sinhλm ðum 1 vm Þ 5 0
486
CHAPTER 7 LAMINATED COMPOSITE PLATES
Differentiate these equations twice, taking into account that in accordance with Eq. (7.184)
00
00
um 5 2 r12 λ2m um ; vm 5 2 r22 λ2m vm
Thus,
X
00
Cm λ2m coshλm ðr12 um 1 r22 vm Þ 5 wp
(7.186)
m
X
Cm λ3m sinhλm ðr12 um 1 r22 vm Þ 5 0
(7.187)
m
Decompose Eq. (7.186) as
X
Cm λ2m r12 coshλm um ðyÞ 5 ϕ1 ðyÞ
(7.188)
Cm λ2m r22 coshλm vm ðyÞ 5 ϕ2 ðyÞ
(7.189)
m
X
m
where ϕ1 ðyÞ and ϕ2 ðyÞ are some unknown functions such that
00
ϕ1 ðy Þ 1 ϕ2 ðy Þ 5 wp
(7.190)
Note that if we knew the functions ϕ1 ðyÞ and ϕ2 ðyÞ, we could readily find the constants Cm .
Indeed, multiply Eqs. (7.188) and (7.189) by un and vn (n 6¼ m), respectively, subtract the second
equation from the first one and integrate from 1 to 1 to get
X
ð1
ð1
Cm λ2m coshλm
m
ðr12 um un 2 r22 vm vn Þdy 5
21
ðϕ1 un 2 ϕ2 vn Þdy
21
Applying Eq. (7.185), we arrive at
Ð1
21
Cn 5
ðϕ1 un 2 ϕ2 vn Þdy
λ2n coshλn
Ð1
21
(7.191)
ðr12 u2n 2 r22 v2n Þdy
However, we do not know the functions ϕ1 ðyÞ and ϕ2 ðyÞ in Eqs. (7.188) and (7.189). To determine these functions, decompose Eq. (7.187) in a similar way, i.e.,
X
Cm λ3m r12 sinhλm um ðyÞ 5 ψ1 ðyÞ
(7.192)
Cm λ3m r22 sinhλm vm ðyÞ 5 ψ2 ðyÞ
(7.193)
m
X
m
where
ψ1 ðy Þ 1 ψ2 ðy Þ 5 0
(7.194)
Consider Eqs. (7.188) and (7.192). Expand the symmetric functions coshλm and λm sinhλm into
the following power series:
coshλm 5
N
X
i51
ai λ2i
m;
λm sinhλm 5
N
X
i51
bi λ2m
i
(7.195)
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
487
where ai and bi are some coefficients, the values of which are not important for further derivation.
Then, Eqs. (7.188) and (7.192) can be presented as
ϕ1 ð y Þ 5
X
Cm r12 λ2m um ðy Þ
m
ψ1 ðy Þ 5
X
Cm r12 λ2m um ðy Þ
X
i
X
m
2
ai λ2i
m 5 r1
2
bi λ2i
m 5 r1
i
X
i
X
ai
X
Cm λ2ði11Þ
um ðy Þ
m
m
bi
i
X
Cm λ2ði11Þ
um ðy Þ
m
(7.196)
m
Summation with respect to m and i is interchanged in the second parts of Eq. (7.196). Now,
differentiate the functions um 2n times. Taking into account Eq. (7.184) for um , we get
2n
uðm2nÞ ðy Þ 5 ð21Þn r 2n
1 λm um ðy Þ
Thus, Eq. (7.196) after differentiation become
ϕ1ð2nÞ ðy Þ 5 ð21Þn r 2ðn11Þ
1
X
ai
X
Cm λ2ði1n11Þ
um ðy Þ
m
m
i
ψ1ð2nÞ ðy Þ 5 ð21Þn r 2ðn11Þ
1
X
bi
i
X
2ði1n11Þ
Cm λm
um ðy Þ
m
Multiply these equations by
bn
ð21Þn 2ðn11Þ
r1
and
an
ð21Þn 2ðn11Þ ;
r1
respectively, and sum them up with respect to n, i.e.,
X
X
XX
bn
2ði1n11Þ
ð21Þn 2ðn11Þ ϕ1ð2nÞ ðy Þ 5
ai bn
Cm λm
um ðy Þ
r1
n
n
m
i
X
XX
X
an
ð21Þn 2ðn11Þ ψ1ð2nÞ ðy Þ 5
an bi
Cm λ2ði1n11Þ
um ðy Þ
m
r1
n
n
m
i
The right-hand parts of these equations are the same and, hence, so are the left-hand
parts. Thus,
X
n
X
bn
an
ð21Þn 2ðn11Þ ϕ1ð2nÞ ðy Þ 5
ð21Þn 2ðn11Þ ψ1ð2nÞ ðy Þ
r1
r
n
1
(7.197)
Repeating the foregoing procedure for the functions ϕ2 ðy Þ and ψ2 ðy Þ in Eqs. (7.189) and
(7.193), we can write the following equations similar to Eq. (7.197):
X
n
X
bn
an
ð21Þn 2ðn11Þ ϕ2ð2nÞ ðy Þ 5
ð21Þn 2ðn11Þ ψ2ð2nÞ ðy Þ
r2
r2
n
(7.198)
Thus, we have four equations, Eqs. (7.190), (7.194), (7.197), and (7.198), for the functions
ϕ1 ðy Þ, ϕ2 ðy Þ, ψ1 ðy Þ, and ψ2 ðy Þ. However, Eqs. (7.197) and (7.198) are differential equations of
infinitely high order. In principle, these equations can be solved using the method based on the
algebra of pseudodifferential operators (Lurie & Vasiliev, 1995). Since the description of these
techniques is rather cumbersome, we use an alternative approach allowing us to reduce the differential operators in Eqs. (7.197) and (7.198) to polynomials. The particular solution wp specified
488
CHAPTER 7 LAMINATED COMPOSITE PLATES
by Eq. (7.167), being an even function which is zero at y 5 6 1, can be expanded in the following Fourier series:
w p ðy Þ 5
N
X
wj cosðαj yÞ; αj 5
j51
In a similar way
ϕ1;2 ðy Þ 5
X
π
p0 b4
ð3 2 α2j Þ sinðαj yÞ
ð2j 2 1Þ; wj 5
2
24D2 α5j
ϕ1j;2j cosðαj yÞ;
ψ1;2 ðy Þ 5
j
X
ψ1j;2j cosðαj yÞ
(7.199)
(7.200)
j
where, in accordance with Eqs. (7.190) and (7.194)
ϕ1j 1 ϕ2j 5 2 α2j wj ;
ψ1j 1 ψ2j 5 0
(7.201)
Substituting Eqs. (7.200) into Eqs. (7.197) and (7.198), we arrive at
2n
X αj 2n
αj
5 ψ1j
an
r1
r1
n
n
X
X
αj 2n
αj 2n
ϕ2j
bn
5 ψ2j
an
r2
r2
n
n
ϕ1j
X
bn
Applying the expressions for the power series in Eq. (7.195), we get
ϕ1j
αj
αj
αj
5 ψ1j cosh
;
sinh
r1
r1
r1
ϕ2j
αj
αj
αj
5 ψ2j cosh
sinh
r2
r2
r2
Solving these equations in conjunction with Eq. (7.201), we find
ϕ1j 5 2 wj β j r1 sinh
αj
;
r2
ϕ2j 5 wj β j r2 sinh
αj
r1
(7.202)
where
βj 5
α2
j
r1 sinh αj =r2 2 r2 sinh αj =r1
Using Eqs. (7.200) and (7.202) and applying Eq. (7.191), we arrive at the following expression
for the constants in Eq. (7.168):
λ2n coshλn
Ð1
21
ð1
αj
wj β j ½r1 sinh
un cosðαj yÞdy
r2
j51
N
X
1
Cn 5 2
ðr12 u2n 2r22 v2n Þdy
21
ð1
αj
1 r2 sinh
vn cosðαj yÞdy
r1
21
in which λn are the roots of Eq. (7.175), whereas un ðyÞ and vn ðyÞ are specified by Eq. (7.184).
Calculating the integrals, we finally get
N
P
C1n 5
j51
2wj αj β j cosðr1 λn Þcosðr2 λn Þsinαj
h
i
r1 sinh αj =r2 = r12 λ2n 2 α2j
2 r2 sinh αj =r1 = r22 λ2n 2 α2j
λn coshλn fr1 cos2 ðr2 λn Þ½r1 λn 1 sinðr1 λn Þcosðr1 λn Þ 2 r2 cos2 ðr1 λn Þ½r2 λn 1 sinðr2 λn Þcosðr2 λn Þg
(7.203)
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
489
pffiffiffiffiffiffiffiffiffiffiffi
The foregoing solution corresponds to the case
D3 . D1 D2 for which r1 and r2 are real numpffiffiffiffiffiffiffiffiffiffiffi
bers specified by Eq. (7.171). For the case D3 , D1 D2 , r1 and r2 are complex numbers, given by
Eq. (7.172), according to which
r1;2 5 t1 7 it2 ;
b
t1;2 5 pffiffiffi
a 2
sr
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffiffiffiffiffiffi
D1 D3
7
D2 D2
In this case, the foregoing solution should be transformed using the following relations:
sinðt1 7 it2 Þ 5 sint1 cosht2 7 icost1 sinht2
cosðt1 7 it2 Þ 5 cost1 cosht2 7 isint1 sinht2
sinhðt1 7 it2 Þ 5 sinht1 cost2 7 icosht1 sint2
coshðt1 7 it2 Þ 5 cosht1 cost2 7 isinht1 sint2
pffiffiffiffiffiffiffiffiffiffiffi
For the case D3 5 D1 D2 , i.e., for an isotropic plate, the characteristic equation corresponding
to Eq. (7.169) has double-reiterated roots r1;2 5 c, where c 5 b=a. To study this case, we should put
in the foregoing equations r1 5 c, r2 5 c 1 ε and go to the limit taking ε-0. For example,
Eq. (7.175) for an isotropic plate reduces to
sin2cλm 1 2cλm 5 0
(7.204)
This equation has only complex roots
2cλn 5 6 rn 6 itn
(7.205)
The real, rn , and imaginary, tn , parts of the first five roots are presented in Table 7.2. Consider,
for example, a square (a 5 b) isotropic plate loaded with uniform pressure p0 . The normalized
deflection at the plate center w 5 wðx 5 0; y 5 0ÞD2 =ð16p0 a4 Þ calculated with the aid of the foregoing equations is w 5 0:00128 (Lurie & Vasiliev, 1995). The relatively accurate approximate solution presented by Timoshenko and Woinowsky-Krieger (1959) is w 5 0:00126.
7.4.2 APPROXIMATE SOLUTIONS FOR CLASSICAL PLATE THEORY
The exact solutions presented in the previous section, particularly the solution for the clamped plate
given by Eqs. (7.168), (7.176), and (7.203), are rather cumbersome for practical applications. For
this purpose, efficient approximate solutions can be constructed using applied methods of analysis
(Vasiliev, 1993).
Consider a plate for which a . b (see Fig. 7.22) and represent its deflection as
wðx; yÞ 5 W ðxÞ f ðyÞ
(7.206)
Table 7.2 Roots [Eq. (7.205)] of Eq. (7.204)
n
1
2
3
4
5
rn
tn
4.211
2.250
10.713
3.103
17.073
3.550
23.398
3.859
29.700
4.093
490
CHAPTER 7 LAMINATED COMPOSITE PLATES
(A)
a /2
(B)
x
a /2
x
a /2
a/2
b
b
y
y
FIGURE 7.25
Plates with clamped longitudinal edges and simply supported (A) and clamped (B) transverse edges.
where f ðyÞ is some function that satisfies the boundary conditions at the plate long edges y 5 0
and y 5 b (see Fig. 7.22) and approximates the plate deflection in the transverse direction. The
function f ðyÞ can be taken as the deflection of the beam of unit width with the corresponding
boundary conditions. Such functions are presented in Table 6.1. Neglecting the shear deformation,
i.e., taking S -N, and changing x to y and l to b, we get
f ðyÞ 5 y4 2 2by3 1 b3 y
(7.207)
for a beam with simply supported edges y 5 0 and y 5 b (see Fig. 7.23) and
f ðyÞ 5 y2 ðb2yÞ2
(7.208)
for a plate with clamped edges y 5 0 and y 5 b (see Fig. 7.25).
Since the functions in Eqs. (7.207) and (7.208) satisfy the boundary conditions, we can apply
the BubnovGalerkin method (Grigolyuk, 1996) to obtain the approximate solutions. Consider the
governing equation, Eq. (7.141). Substitute the deflections, Eq. (7.206), into Eq. (7.141), multiply
it by f ðyÞ and integrate with respect to y from 0 to b to get the following ordinary differential
equation for W(x):
ðb
ðb
ðb
ðb
00
00
D1 W IV f 2 dy 1 2D3 W f fdy 1 D2 W f IV fdy 5 pfdy
0
0
0
(7.209)
0
Using the boundary conditions for the functions given by Eqs. (7.207) and (7.208) and integrating by parts, we have
ðb
00
ðb
f fdy 5 2
0
ðb
0 2
ðf Þ dy;
ðb
f fdy 5
IV
0
0
ðf 00 Þ2 dy
0
Thus, Eq. (7.209) reduces to
00
W IV 2 2k12 W 1 k24 W 5 kp
(7.210)
where
k12 5
D 3 c2
;
D 1 c1
k24 5
D 2 c3
;
D 1 c1
kp 5
cp
D 1 c1
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
491
in which
ðb
c1 5
ðb
c2 5
2
f dy;
0
ðb
0 2
ðf Þ dy;
c3 5
0
00 2
ðf Þ dy;
ðb
cp 5
0
pfdy
(7.211)
0
For a plate loaded with uniform pressure p 5 p0 ,
ðb
cp 5 p0 c;
c5
fdy
(7.212)
0
The solution of Eq. (7.210) is specified by Eq. (7.157), i.e.,
W5
4
X
ci Φi ðxÞ 1 Wp
i51
The functions Φi are given in Table 7.1. For uniform pressure p 5 p0 and the same boundary
conditions at the plate transverse edges x 5 6 a=2 (see Figs. 7.23 and 7.25)
W 5 c1 Φ1 ðxÞ 1 c2 Φ2 ðxÞ 1 Wp
(7.213)
in which
Wp 5
p0 c
D 1 c1
The constants c1 and c2 can be found from the boundary conditions at x 5 6 a=2.
If the plate longitudinal edges (y 5 0; y 5 b) are simply supported (Fig. 7.18), the function f ðyÞ
is specified by Eq. (7.207) and the coefficients given by Eqs. (7.211) and (7.212) are
c1 5 0:04921b9 ;
c2 5 0:48571b7 ;
c3 5 4:8b5 ;
c 5 0:2b5
If the plate transverse edges (x 5 6 a=2) are also simply supported (see Fig. 7.23A), the solution in Eqs. (7.206) and (7.213) can be written as
wðx; yÞ 5 Wp ½1 2 F1 ðxÞ y4 2 2by3 1 b3 y
(7.214)
where F1 ðxÞ is specified by Eq. (7.162).
For a plate with simply supported longitudinal edges and clamped transverse edges (see
Fig. 7.23B)
wðx; yÞ 5 Wp ½1 2 F2 ðxÞ y4 2 2by3 1 b3 y
(7.215)
where F2 ðxÞ is given by Eq. (7.163).
For a plate with clamped longitudinal edges (see Fig. 7.25), the function f ðyÞ corresponds to
Eq. (7.208), whereas the coefficients in Eqs. (7.211) and (7.212) become
c1 5 0:001587b9 ;
c2 5 0:01905b7 ;
c3 5 0:8b5 ;
c 5 0:03333b5
If the plate transverse edges are simply supported (see Fig. 7.25A)
wðx; yÞ 5 Wp ½1 2 F1 ðxÞy2 b2 2 y2 ;
(7.216)
492
CHAPTER 7 LAMINATED COMPOSITE PLATES
where F1 ðxÞ is given by Eq. (7.162).
Finally, for a plate with both transverse and longitudinal clamped edges (see Fig. 7.25B)
wðx; yÞ 5 Wp ½1 2 F2 ðxÞy2 b2 2 y2 ;
(7.217)
in which F2 ðxÞ is specified by Eq. (7.163).
The simple approximate solutions obtained are typically fairly accurate. In particular, the deflection closely coincides with the corresponding exact solution, whereas the accuracy of the moment
is within about 5% (Vasiliev, 1993). Approximate solutions for orthotropic plates with various
boundary conditions have been presented by Whitney (1971), Dalaei and Kerr (1995), and Bhaskar
and Kaushik (2004).
7.4.3 SHEAR DEFORMABLE ORTHOTROPIC SYMMETRIC PLATES
The behavior of shear deformable orthotropic symmetric plates under transverse bending is
described by Eqs. (7.40)(7.42). The exact solution of these equations can be found for simply
supported plates (see Fig. 7.19) with the classical boundary conditions
w 5 0; Mx 5 0; θy 5 0 for the edges x 5 constant
w 5 0; My 5 0; θx 5 0 for the edges y 5 constant
For the simply supported plate shown in Fig. 7.19 with the foregoing boundary conditions, the
function W(x,y) in Eq. (7.40) can be taken in the form of the following series:
W ðx; yÞ 5
XX
m
Wmn sinλm x sinλn y
(7.218)
n
where λm 5 πm=a; λn 5 πn=b: The pressure pðx; yÞ should be expanded into a similar series specified by Eq. (7.147). Then, substitution into Eq. (7.40) yields
Wmn 5
pmn
Rmn
(7.219)
where
λ2n
λ2
Rmn 5 Dmn λ2m 1 Dnm λ2n 1 2Dλ2m λ2n 1 Dmn Dnm 2 Dλ2m λ2n
1 m
Sx
Sy
in which
Dmn 5 D11 λ2m 1 D44 λ2n ; Dnm 5 D22 λ2n 1 D44 λ2m ; D 5 D12 1 D44
The rotation angles and the deflection can be found from Eqs. (7.41) and (7.42) as
θx 5
XX
m
θy 5
w5
θmn
x cosλm x sinλn y
n
XX
m
n
m
n
XX
θmn
y sinλm x cosλn y
wmn sinλm x sinλn y
(7.220)
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
where
Dnm
D 2
θmn
5
2
λ
W
1
1
2
λ
m mn
x
Sx n
Sy
Dmn
D 2
θmn
5
2
λ
W
1
1
2
λ
n
mn
y
Sy m
Sx
wmn 5 cmn Wmn
493
(7.221)
in which
cmn 5 1 1
Dmn
Dnm
1 1
1
Dmn Dnm 2 Dλ2m λ2n
Sx Sy
Sx
Sy
The bending moments and transverse shear forces can be found with the aid of Eqs. (7.32)
and (7.34).
For plates with other boundary conditions, approximate analytical solutions are usually derived.
Note that the governing equations of the theory of shear deformable plates have a special structure
that allows us to simplify the problem. To study this structure, consider a semiinfinite plate with
simply supported longitudinal edges as shown in Fig. 7.9. The pressure is specified by Eq. (7.111),
according to which
p 5 p0 sinλy; λ 5
π
b
(7.222)
The solution of Eq. (7.40) can be presented in the form
W ðx; yÞ 5 FðxÞ sinλy
Substitution into Eq. (7.40) results in the following ordinary differential equation:
D1 D44 VI
D1 D44
1
2
2
F 1 D1 1
1 ðD1 D2 2 D12 D3 Þ λ F IV
Sy Sy
Sx
D1 D44
1
D44 2 4
00
2
2 2D3 λ 1
1 ðD1 D2 2 D12 D3 Þ λ4 F 1 D2 1
λ λ F 5 p0
Sx
Sy
Sx
(7.223)
where D1 , D2 , and D3 are specified by Eq. (7.142). Recall that for transversally isotropic plates
(see Section 7.3.2), Eq. (7.223) degenerates into a set of two separate equations, Eqs. (7.113) and
(7.114). For orthotropic plates, we have a unique equation of the sixth order which cannot be
reduced exactly to two independent equations. However, applying the asymptotic method
(Reissner, 1989, 1991), we can approximately reduce the sixth-order equation to a set of a fourthorder and second-order equations.
It follows from Section 7.3.2 that Eq. (7.223) has two types of solution, i.e., a penetrating solution and a boundary layer solution. The penetrating (or internal) part of the general solution Fi relatively slowly changes over the plate surface, which means that the derivatives of this solution FiðnÞ
are of the same order of magnitude as the function Fi . However, Eq. (7.223) contains the terms
2
D=S which have the order of magnitude h , where h 5 h=b, h being the plate thickness and b the
in-plane dimensions. To determine the penetration solution for relatively thin ðh{1Þ plates, these
small terms can be neglected and Eq. (7.223) can be reduced to
00
D1 FiIV 2 2D3 λ2 Fi 1 D2 λ4 Fi 5 p0
This equation is similar to Eq. (7.113).
(7.224)
494
CHAPTER 7 LAMINATED COMPOSITE PLATES
The boundary-layer (or edge) solution Fe is a rapidly changing function whose derivatives can
ðnÞ
2
be evaluated as FeðnÞ 5 Fe =h . Taking into account that D=S is proportional to h , we can conclude that only two terms of Eq. (7.223) (namely, the first one and the second) have the order
4
ð1=h Þ. Neglecting the rest of the terms, we arrive at the following equation for the boundary-layer
solution:
2
D1 D44 VI
Fe 1 D1 FeIV 5 0
Sy
Reducing the order of this equation, we finally transform it into the following form:
00
Fe 2
Sy
Fe 5 0
D44
(7.225)
which is similar to Eq. (7.114). Thus, the solution of Eq. (7.223) becomes
F 5 Fi 1 Fe
(7.226)
where Fi and Fe are the solutions of Eqs. (7.224) and (7.225), respectively. The total solution given
by Eq. (7.226) allows us to satisfy three boundary conditions at the plate edge.
Returning to the general case, i.e., to Eq. (7.40), we can present its approximate solution in a
form analogous to Eq. (7.226), i.e.,
W 5 Wi 1 We
(7.227)
where Wi can be found from the following equation:
D1
@4 W
@4 W
@4 W
1
2D
1
D
5p
3
2
@x4
@x2 @y2
@y4
(7.228)
In the vicinity of the edge x 5 constant, the boundary-layer solution satisfies the following equation similar to Eq. (7.225):
@2 We
Sy
2 s2y We 5 0; s2y 5
@x2
D44
(7.229)
whereas in the vicinity of the edge y 5 constant
@2 We
Sx
2 s2x We 5 0; s2x 5
@y2
D44
(7.230)
Thus, the solution should be constructed separately for the edges x 5 constant and
y 5 constant.
It follows from Section 7.3.2 that for relatively thin plates whose edges are fixed with respect to
deflection, the boundary-layer solution can be neglected in comparison with the penetrating solution. Naturally, in this case, only two boundary conditions can be formulated at the plate edge.
However, for the plates under consideration, three boundary conditions can be reduced to two as
shown in Section 7.3.2. The penetrating solution can be found from Eq. (7.228), whereas the
expressions for the rotation angles and deflection in Eqs. (7.41) and (7.42) are simplified to
θx 5 2
@Wi
;
@x
θy 5 2
@Wi
;
@y
w 5 Wi 2
D1 @2 Wi
D 2 @ 2 Wi
2
2
Sx @x
Sy @y2
(7.231)
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
495
The moments and the forces are
@2 Wi
@2 Wi
Mx 5 2 D1 2 1 D12 2 ;
@x
@y
@2 Wi
@2 Wi
My 5 2 D2 2 1 D12 2
@y
@x
@ 2 Wi
Mxy 5 2 2D44
@x@y 2
@
@ Wi
@2 Wi
@
@2 Wi
@2 Wi
D1 2 1 D3 2 ; Vy 5 2
D2 2 1 D3 2
Vx 5 2
@x
@y
@y
@x
@x
@y
Note that the foregoing set of equations for the penetrating solution in the theory of shear
deformable plates has been obtained for isotropic plates by Donnell (1959, 1976) as a generalization of the corresponding Timoshenko (1921) beam theory. In this theory, the first term in
Eq. (7.231) is the part of the plate deflection caused by bending, whereas the last two terms take
into account the part of the deflection induced by shear deformation. To obtain the penetrating
solution in the theory of shear deformable plates the exact or approximate methods discussed in
Sections 7.4.1 and 7.4.2 as applied in classical plate theory can be used. It is only necessary to take
into account that in accordance with Donnell’s interpretation Wi is the deflection w corresponding
to classical plate theory. Thus, the approximate solutions of classical plate theory specified by
Eqs. (7.214)(7.217) are valid for the shear deformable plate theory under consideration if we
change w(x,y) to Wi ðx; yÞ in the left-hand parts of these equations and apply Eq. (7.231) to determine the plate deflection.
However, as shown in Section 7.3.2, the boundary-layer solution can be neglected if the
parameters sx and sy in Eqs. (7.229) and (7.230) are much larger than the parameter λ for the penetrating solution in Eq. (7.222). If these parameters are of the same order of magnitude, the
boundary-layer solution cannot be neglected in comparison with the penetrating solution. This
situation usually occurs for thick sandwich plates with a shear deformable core.
Consider two particular cases for which the problem can be readily solved, i.e., square and
close-to-square plates (see Fig. 7.20A) and long plates (see Fig. 7.20B).
To study square and close-to-square plates, apply the initial equilibrium equations, i.e.,
Eq. (7.36) for θx , θy , and w which describe both the penetrating and the boundary-layer states
of the plate. For the plates under consideration, introduce the following one-term
approximations:
θx ðx; yÞ 5 Cx θxx ðxÞ θxy ðyÞ
θy ðx; yÞ 5 Cy θyx ðxÞθyy ðyÞ
wðx; yÞ 5 W wx ðxÞ wy ðyÞ
(7.232)
in which Cx , Cy , and W are some constants and θxx ; θxy , θyx , θyy , wx , and wy are preassigned coordinate functions that satisfy the boundary conditions. Applying the BubnovGalerkin method, multiply Eq. (7.36) by θxx θxy ; θyx θyy ; and wx wy , respectively, and integrate the resulting equations over
the plate surface (see Fig. 7.1). As a result, Eq. (7.36) reduce to the following algebraic equations
for the coefficients Cx , Cy , and W:
A11 Cx 1 A12 Cy 1 A13 W 5 0
A21 Cx 1 A22 Cy 1 A23 W 5 0
A31 Cx 1 A32 Cy 1 A33 W 1 Ap 5 0
(7.233)
496
CHAPTER 7 LAMINATED COMPOSITE PLATES
where
A11 5 D11 a1 b1 1 D44 a2 b2 2 Sx a2 b1
A12 5 Da3 b3 ; A13 5 2 Sx a4 b4 ; A21 5 Da5 b5
A22 5 D22 a6 b6 1 D44 a7 b7 2 Sy a6 b7
A23 5 2 Sy a8 b8 ; A31 5 Sx a9 b9 ; A32 5 Sy a10 b10
ða ðb
A33 5 Sx a11 b11 1 Sy a12 b12 ; Ap 5 pwx wy dxdy
00
in which
ða
a1 5
00
θxx θxx dx;
ða
ða
a2 5
θ2xx dx;
a3 5
0
0
0
ða
ða
ða
a5 5
0
θxx θyx dx;
a6 5
0
ða
a9 5
θ2yx dx;
a7 5
0
θxx wx dx;
a10 5
0
a4 5
ða
00
θyx θyx dx;
a8 5
wx θyx dx
0
0
ða
θyx wx dx;
0
wx θxx dx
0
ða
0
ða
0
θyx θxx dx;
a11 5
00
wx wx dx;
0
ða
a12 5
0
w2x dx
0
and
ðb
b1 5
ðb
θ2xy dy;
0
ðb
b5 5
b2 5
θxy θxy dy;
0
0
θyy θyy dy;
00
b6 5
θyy θyy dy;
b4 5
wy θxy dx
b10 5
0
ðb
b7 5
ðb
θ2yy dy;
b8 5
0
ðb
0
θyy θxy dy;
0
ðb
θxy wy dy;
b3 5
ðb
0
0
ðb
0
b9 5
ðb
00
0
θyy wy dy;
0
ðb
b11 5
wy θyy dx
0
ðb
w2y dy;
0
b12 5
00
wy wy dx
0
The solution of Eq. (7.233) is
Ap
ðA12 A23 2 A22 A13 Þ
A
Ap
θy 5
ðA21 A13 2 A11 A23 Þ
A
Ap
W5
ðA11 A22 2 A12 A21 Þ
A
θx 5
(7.234)
where
A 5 ½ðA32 A23 2 A22 A33 ÞA11 1 ðA22 A31 2 A32 A21 ÞA13 1 ðA21 A33 2 A31 A23 ÞA12 We take the solutions of the theory of shear deformable beams (see Table 6.1) presented in
Table 7.3 as the coordinate functions in Eq. (7.232).
7.4 BENDING OF ORTHOTROPIC SYMMETRIC PLATES
497
Table 7.3 Coordinate Functions for the Plates Loaded with Uniform Pressure
Boundary Conditions
a
Coordinate functions
wx ðxÞ 5 θyx ðxÞ 5 x4 2 2ax3 2 12D1 =Sx x2 1 a3 1 12D1 a=Sx x
3
2
3
θxx ðxÞ 5 4x 2 6ax 1 a
wy ðyÞ 5 θxy ðyÞ 5 y4 2 2by3 2 12D2 =Sy y2 1 b3 1 12D2 b=Sy y
3
2
3
θyy ðyÞ 5 4y 2 6by 1 b
x
b
y
wx ðxÞ 5 θyx ðxÞ 5 2x4 1 2ax3 1 12D1 =Sx 2a2 x 2 12D1 ax=Sx
θxx ðxÞ 5 2x3 2 3ax2 1 a2 x
wy ðyÞ 5 θxy ðyÞ 5 2y4 1 2by3 1 12D2 =Sy 2b2 y 2 12D2 by=Sy
θyy ðyÞ 5 2y3 2 3by2 1 b2 y
x
a
b
y
a
wx ðxÞ 5 θyx ðxÞ 5 2x4 1 2ax3 1 12D1 =Sx 2a2 x 2 12D1 ax=Sx
θxx ðxÞ 5 2x3 2 3ax2 1 a2 x wy ðyÞ 5 θxy ðyÞ 5 y4 2 2by3 2 12D2 =Sy y2 1 b3 1 12D2 b=Sy y
θyy ðyÞ 5 4y3 2 6by2 1 b3
x
b
y
Table 7.4 Equations for Infinitely Long Orthotropic Plates under Uniform Pressure p0
Boundary Conditions
Equations
x
1
b
y
1
x
b
y
3
2
w 5 p0 =24D
1 y3 1 12 D
2 =Sy ðb 2 yÞ y
2 b 2
2by
3
2
3
θy 5 2
p0=24D2 b 2 6by 1 4y
My 5 1=2 p0 ðb 2 yÞy
Vy 5 1=2 p0 ðb 2 2yÞ
w 5 p0 =24D
2 yðb2
yÞ21 12 D2 =S2y ðb 2 yÞy
θy 5 2 p0 =12D2 2y 2 3by 1 b y
2
2
My 5 2 p0 =12
6y 2 6by 1 b
Vy 5 2 p0 =2 ð2y 2 bÞ
498
CHAPTER 7 LAMINATED COMPOSITE PLATES
wm . 3
10
b
5
4
3
2
1
1
2
3
4
a
b
5
FIGURE 7.26
Dependence of the normalized maximum deflection of a simply supported rectangular plate on the aspect ratio a/b:
exact solution( sss ), solution for a square plate (22 22 2 ), solution for an infinitely long plate (UUUUUUUUUU),
numerical solution (3).
Consider the second particular case, i.e., long plates as depicted in Fig. 7.20B. For such
plates, we can assume that in the middle part of the plate the stress-strain state does not
depend on the x-coordinate. Then, the problem reduces to the bending of a beam of unit
width, the solution for which is obtained in Chapter 6, Laminated Composite Beams and
Columns (see Table 6.1). The basic equations for infinitely long plates are presented in
Table 7.4.
As an example, consider a simply supported rectangular plate with thickness h 5 15 mm, width
b 5 250 mm, and various aspect ratios a/b. The plate has a sandwich structure with carbonepoxy
facing layers and an aluminum honeycomb core. The plate parameters are D1 5 D2 5 D,
D12 =D 5 D44 =D 5 0:25, D=b2 Sx 5 0:04, and D=b2 Sy 5 0:08. The normalized maximum deflection
of the plate wm =b as a function of the aspect ratio a/b is shown in Fig. 7.26. The solid line corresponds to the exact solution, Eq. (7.219), and the dashed line shows the solution given by
Eq. (7.234) corresponding to square and close-to-square plates, whereas the dotted line demonstrates the solution for an infinitely long plate (see Table 7.4). As can be seen, for 1 # a=b # 1:75
the plate can be treated as close to square and for a=b . 3 the solution for an infinitely long plate
can be applied.
In the general case (or for 1:75 # a=b # 3), the problem cannot be considerably simplified and
the general equations, Eq. (7.36) must be solved. Apply the approximate method described in
Section 7.4.2 and represent the unknown functions in the following form:
θx ðx; yÞ 5 Fx ðxÞ θxy ðyÞ;
θy ðx; yÞ 5 Fy ðxÞθyy ðyÞ;
wðx; yÞ 5 WðxÞ wy ðyÞ
(7.235)
where Fx , Fy , and W are the unknown functions of the coordinate x, whereas the functions θxy ,
θyy , and wy are presented in Table 7.3 for various boundary conditions. Multiply Eq. (7.36)
by θxy , θyy , and wy , respectively, and integrate them with respect to y from y 5 0 to y 5 b
7.5 BUCKLING OF ORTHOTROPIC SYMMETRIC PLATES
499
(see Figs. 7.23 and 7.25). As a result, we arrive at the following set of ordinary differential
equations for the functions Fx ðxÞ, Fy ðxÞ, and WðxÞ:
L1x ðFx Þ 1 L1y Fy 1 L1w ðW Þ 5 0
L2x ðFx Þ 1 L2y Fy 1 L2w ðW Þ 5 0
L3x ðFx Þ 1 L3y Fy 1 L3w ðW Þ 1 Lp 5 0
(7.236)
where
L1x ðFx Þ 5
1
00
D1 a11 Fx 1 D33 b11 Fx 2 a11 Fx
Sx
1
0
L1y Fy 5 Da21 Fy ; L1w ðW Þ 5 2 a13 W 0
Sx
1
00
L2y Fy 5
D2 a22 Fy 1 D33 b22 Fy 2 b22 Fy
Sy
1
0
Da21 Fx ; L2w ðW Þ 5 2 a23 W
Sy
0
L3x ðFx Þ 5 Sx a31 Fx ; L3y Fy 5 Sy a32 Fy
L2x ðFx Þ 5
ðb
00
L3w ðW Þ 5 Sx a33 W 1 Sy Wb33 ; Lp 5
pwy dy
0
in which
ðb
a11 5
ðb
θ2xy dy;
b11 5
0
ðb
a21 5
ðb
00
θxy θxy dy;
0
0
θxy θyy dy;
a12 5
ðb
0
θyy θxy dy;
a13 5
wy θxy dy
0
ðb
a22 5
00
θyy θyy dy;
0
ðb
b22 5
ðb
θ2yy dy;
a23 5
0
0
0
0
ðb
ðb
ðb
ðb
a31 5
θxy wy dy;
0
a32 5
0
θyy wy dy;
0
a33 5
w2y dy;
0
wy θyy dy
b33 5
0
00
wy wy dy
0
In principle, Eq. (7.236) can be solved analytically. They can be reduced to sixth-order ordinary
differential equation and further to third-order algebraic characteristic equation, the roots of which
specify the form of the general solution. This solution includes six constants which can be found
from the boundary conditions for the plate edges x 5 6 a=2 (see Fig. 7.23 or 7.25). However, it is
easier to apply numerical integration of Eq. (7.236). For the aspect ratio a/b 5 2, the result of such
integration is shown in Fig. 7.26 with a circle.
7.5 BUCKLING OF ORTHOTROPIC SYMMETRIC PLATES
It is also possible for plates subjected to in-plane compression and shear to buckle after the load
reaches some critical value. The problem of buckling is studied in this section.
500
CHAPTER 7 LAMINATED COMPOSITE PLATES
Ty
Txy
Tx
x
Txy
Txy
Tx
Txy
y
Ty
FIGURE 7.27
A plate loaded with in-plane loads.
x
a
Tx
Tx
b
y
FIGURE 7.28
Buckling mode of a plate under uniaxial compression.
7.5.1 CLASSICAL PLATE THEORY
Consider a rectangular plate simply supported at the edges and loaded with external in-plane line
loads Tx , Ty , and Txy (see Fig. 7.27). Under the action of in-plane compression or shear, the initial
plane shape of the plate can become unstable and the plate takes a curved shape shown, for example, in Fig. 7.28. The corresponding forces that are referred to as critical loads can be found using
the Euler buckling criterion according to which the critical load is the minimum load which provides, in addition to the initial plane state of the plate, some curved shape of the plate close to
this state which is balanced by the applied in-plane forces. Thus, to determine the critical load, we
must induce a small deflection of the plate (see the dashed lines in Fig. 7.28) and find the minimum in-plane load under which the plate is in equilibrium with this load. The simplest way to
derive the buckling equation which allows us to determine the critical load is to apply the method
described in Section 6.4 with application to beams. Consider an element of the plate reference surface and induce small curvatures κx , κy , and twist κxy deforming the element as shown in
Fig. 7.29, which is similar to Fig. 6.12 for a beam. It follows from Fig. 7.29 that the forces Tx , Ty ,
and Txy now acquire a nonzero projection on the z-axis, which can be simulated by an imaginary
pressure pn which is shown in Fig. 7.30; compare Fig. 6.13 for the beam. This imaginary pressure
can be found as
pn 5 Tx κx 1 Ty κy 1 Txy κxy
(7.237)
7.5 BUCKLING OF ORTHOTROPIC SYMMETRIC PLATES
501
z
Ty
Txy
Txy
Tx
Tx
Txy
Txy
Ty
FIGURE 7.29
A deformed element of the plate reference surface.
pn
FIGURE 7.30
Imaginary pressure simulating the action of forces Tx , Ty , and Txy .
in which κx , κy , and κxy are specified by Eq. (7.49). Using these equations and taking p 5 pn in the
equilibrium equation, Eq. (7.141) of classical plate theory, we arrive at the following buckling
equation for orthotropic plates:
D1
@4 w
@4 w
@4 w
@2 w
@2 w
@2 w
50
1 2D3 2 2 1 D2 4 1 Tx 2 1 Ty 2 1 2Txy
4
@x
@x @y
@y
@x
@y
@x@y
(7.238)
where, as earlier, D1 5 D11 , D2 5 D22 , D3 5 D12 1 2D44 . For plates that are simply supported at the
edges x 5 0, x 5 a and y 5 0, y 5 b (see Fig. 7.27), the deflection can be presented in the form of
double trigonometric series, i.e.,
wðx; yÞ 5
XX
m
wmn sinλm x sinλn y;
n
λm 5
πm
;
a
λn 5
πn
b
(7.239)
each term of which satisfies the boundary conditions, Eq. (7.144).
If the plate is subjected to uniaxial compression (see Fig. 7.28), Ty 5 Txy 5 0 and Eq. (7.238)
reduces to
LðwÞ 1 Tx
@2 w
5 0;
@x2
LðwÞ 5 D1
@4 w
@4 w
@4 w
1 2D3 2 2 1 D2 4
4
@x
@x @y
@y
(7.240)
502
CHAPTER 7 LAMINATED COMPOSITE PLATES
Lm /D1
16
m 1
m 2
12
m 3
m 4
8
a
0
1
2
3
4 b
FIGURE 7.31
Dependence of Lm =D1 on the plate aspect ratio a/b.
Substitution of Eq. (7.239) into Eq. (7.240) allows us to determine the force Tx which provides
equilibrium of the plate with a nonzero deflection in Eq. (7.239), i.e.,
Tx 5
π2
Lmn
b2
(7.241)
where
2
Lmn 5 D1 λ 1 2D3 n2 1 D2
n4
λ
(7.242)
in which λ 5 mb=a . Thus, for each combination of the half-wave numbers m and n in
Eq. (7.239) for the deflection, there exists a force Tx specified by Eq. (7.241). For example, the
curved shape of the plate corresponding to m 5 2 and n 5 1 is shown in Fig. 7.28 with dashed lines.
The critical force Txc corresponds to the minimum value of the coefficient Lmn with respect to
m 5 1; 2; 3; . . . and n 5 1; 2; 3; . . .. It follows from Eq. (7.242) that Lmn increases with an increase in
n. Thus, the minimum possible value n 5 1 must be taken in Eq. (7.242) and
Lmn 5 Lm 5 D1 λ 1 2D3 1
D2
λ
(7.243)
The dependence of Lm =D1 on the plate aspect ratio a/b for various m-values is shown in
Fig. 7.31 for a carbonepoxy angle-ply 6 45 rectangular plate. The solid parts of the lines correspond to the minimum values of coefficient Lm . For example, for a=b , 1, we have the curve corresponding to m 5 1. The points of intersection of the curves in Fig. 7.31 corresponding to (m 1),
m, and (m 1 1) can be found from the following conditions: Lm ðm 2 1Þ 5 Lm ðmÞ and
Lm ðmÞ 5 Lm ðm 1 1Þ. These conditions allow us to determine the value of m which provides the minimum value of Lm for a plate with given stiffness and geometric parameters. The corresponding
value of m satisfies the following inequality:
mðm 2 1Þ ,
a2
b2
rffiffiffiffiffiffi
D2
, mðm 1 1Þ
D1
For example, let D2 5 D1 and a/b 5 1.75. Then, the foregoing condition, i.e.,
mðm 2 1Þ , 3:06mðm 1 1Þ
7.5 BUCKLING OF ORTHOTROPIC SYMMETRIC PLATES
503
is satisfied from m 5 2 which corresponds to the plate shown in Fig. 7.28.
For a=b $ 2, i.e., for relatively long plates, the approximate minimum condition dLm =dλ 5 0
can be used for Lm in Eq. (7.243). As a result, we get
λ2 5
D2
;
D1
pffiffiffiffiffiffiffiffiffiffiffi
Lm 5 2 D1 D2 1 D3
(7.244)
For the example illustrated by Fig. 7.31, Lm =D1 5 7:22. This result is depicted in Fig. 7.31 by
the dotted line. Substitution of Eq. (7.244) into Eq. (7.241) yields the following final expression for
the critical load:
Txc 5
2π2 pffiffiffiffiffiffiffiffiffiffiffi
D3
p
ffiffiffiffiffiffiffiffiffiffiffi
D
D
1
1
1 2
b2
D1 D2
For a homogeneous plate,
π2 h3
Txc 5
2
6b
(7.245)
qffiffiffiffiffiffiffiffiffiffiffi
E x E y 1 ν xy E x 1 2Gxy
where
E x;y 5
Ex;y
;
1 2 ν xy ν yx
Ex ν xy 5 Ey ν yx
Consider biaxial loading with forces Tx and Ty (see Fig. 7.27). Taking Txy 5 0 in Eq. (7.238),
we get
LðwÞ 1 Tx
@2 w
@2 w
1 Ty 2 5 0
2
@x
@y
where LðwÞ is obtained from Eq. (7.240). Substitution of the deflection into Eq. (7.239) yields
Tx λ2m 1 Ty λ2n 5 D1 λ4m 1 2D3 λ2m λ2n 1 D2 λ4n
Introducing the loading parameter p 5 Ty =Tx , we get Eq. (7.241) for Tx in which
Lmn 5
n2 λ
D2
D1 λ 1 2D3 1
;
λ1p
λ
λ5
2
mb
na
(7.246)
For the given loading parameter p, the minimum value of Lmn can be found by sorting
m; n 5 1; 2; 3; . . .. Various combinations of loads Tx and Ty have been studied by Jones
(2006).
For composite plates, a biaxial stress state can be induced by uniaxial loading with forces Tx
only. Consider the plane state of the plate loaded in the general case with forces Tx and Ty which
cause the in-plane strains ε0x and ε0y linked with the forces Tx and Ty by the constitutive equations,
Eq. (7.5), i.e.,
Tx 5 B11 ε0x 1 B12 ε0y ;
Ty 5 B21 ε0x 1 B22 ε0y
(7.247)
Assume that under compression in the x-direction with forces Tx the plate longitudinal
edges y 5 0 and y 5 b (see Fig. 7.28) can move in the y-direction. Then, Ty 5 0 and Eqs. (7.247)
yield
504
CHAPTER 7 LAMINATED COMPOSITE PLATES
ε0y 5 2
B21 0
ε ;
B22 x
B12 B21 0
Tx 5 B11 2
εx
B22
The critical force for such plates is specified by Eq. (7.245). Now assume that the plate longitudinal edges are fixed, so that ε0y 5 0. Then, it follows from Eqs. (7.247) that a reactive force
Ty 5
B21
Tx 5 ν xy Tx
B11
appears in the plate due to Poisson’s effect. For composite plates, Poisson’s ratio ν xy depends on
the material structure. For example, for 0 =90 cross-ply carbonepoxy composite ν xy 5 0:03,
whereas for angle-ply 6 45 composite ν xy 5 0:75 and the effect of reactive biaxial loading can be
significant. The corresponding critical force Txc can be found from Eqs. (7.241) and (7.246) if we
take p 5 ν xy , i.e.,
Tx 5
π2 n2 λ
D2
D
λ
1
2D
1
1
3
ðλ 1 ν xy Þb2
λ
(7.248)
and minimize Tx with respect to the parameter λ. For an angle-ply 6 ϕ carbonepoxy square plate,
the dependence of the normalized critical load T 5 Txc ðφÞ=Txc ðφ 5 0Þ on the angle φ is shown in
Fig. 7.32. Curve 1 corresponds to Eq. (7.245), whereas curve 2 is plotted in accordance with
Eq. (7.248). As can be seen, the reactive transverse force Ty can significantly reduce the critical
load. Comparison of the calculated and experimental critical loads for carbonepoxy square plates
Txc
2.0
1
1.6
1.2
0.8
2
0.4
0
0
15
30
45
60
75
90
FIGURE 7.32
Dependencies of the normalized critical load for angle-ply 6 φ square carbonepoxy plates on the ply angle φ as
per Eq. (7.245) (curve 1) and Eq. (7.248) (curve 2).
7.5 BUCKLING OF ORTHOTROPIC SYMMETRIC PLATES
505
Table 7.5 Calculated and Experimental Critical Loads for Square Composite Plates
Plate
h/a
ν xy
Critical Load, kN
Calculated
Experimental
1
2
3
4
5
6
0.006
0.009
0.009
0.009
0.015
0.02
0.32
0.03
0.75
0.22
0.03
0.15
1.07
1.95
3.28
2.45
8.48
23.32
1.06
1.86
3.04
2.45
8.33
22.93
with various ν xy coefficients is presented in Table 7.5. Plate 1 consists of 7 plies with angles 0 and
6 45 . Plates 24 are composed of 11 plies with angles 0 and 90 (plate 2), 6 45 (plate 3),
and 0 , 90 , and 6 45 (plate 4). Plate 5 is a cross-ply (0 , 90 ) laminate consisting of 15 plies and
plate 6 has 9 plies with angles 0 and 90 and two aluminum face sheets with thickness 0.5 mm.
For the plates with ν xy # 0:3 (Nos 1, 2, and 46) the critical load is calculated using Eq. (7.245),
whereas for plate 3, Eq. (7.248) is applied because Eq. (7.245) gives a load which is about 80%
higher than the corresponding experimental result. It follows from Table 7.5 that Eq. (7.245) or
(7.248) predict the critical load with reasonable accuracy.
Consider the action of combined compression and shear. Taking Ty 5 0 (see Fig. 7.27), generalize Eq. (7.240) as
LðwÞ 1 Tx
@2 w
@2 w
50
1 2Txy
@x2
@x@y
(7.249)
For a simply supported plate, apply Eq. (7.239) for the deflection and substitute it into
Eq. (7.249). Using the BubnovGalerkin method, multiply the resulting equation by sinλi x sinλj y
and integrate it over the plate surface. Taking into account that
(a
ða
sinλm x sinλi x dx 5
0
ða
8
>
<
2
0
0
2ai
cosλm x sinλi x dx 5
2
2
>
: πði 2 m Þ
0
0
for m 5 i
for m 6¼ i
for m 5 i
for m 1 i 5 2; 4; 6 . . .
for m 1 i 5 1; 3; 5 . . .
and the similar integrals for the y-coordinate, we arrive at the following set of coupled algebraic
equations for the coefficients wmn :
π2 pffiffiffiffiffiffiffiffiffiffiffi 4
n4
2 2
D
α
1
2m
n
β
1
2 m2 Tx
D
m
1 2
b2
α
amn X X
wij ij
50
132Txy 2
π b i j ðm2 2 i2 Þðn2 2 j2 Þ
wmn
(7.250)
506
CHAPTER 7 LAMINATED COMPOSITE PLATES
K
30
β
3 .6
20
3 .2
2 .8
2 .4
2 .0
1 .6
1 .2
10
0 .8
0 .4
α
0
0.2
0.4
0.6
0.8
1.0
FIGURE 7.33
Dependence of the shear buckling coefficient on the parameters α and β.
where
b2
α5 2
a
rffiffiffiffiffiffi
D1
D3
; β 5 pffiffiffiffiffiffiffiffiffiffiffi
D2
D1 D2
(7.251)
and m 6 i 5 1; 3; 5; . . ., n 6 i 5 1; 3; 5; . . .. Usually the loads Tx and Txy are increased in proportion
to some loading parameter p. To find the critical value of this parameter, we must use the zero condition for the determinant of Eq. (7.250) and minimize the root p of the resulting algebraic
equation.
Consider the case of pure shear, taking Tx 5 0 in Eq. (7.250). The critical shear force can be
presented as (Azikov, Vasiliev, & Paterekas, 1990)
c
Txy
5K
π2 pffiffiffiffiffiffiffiffiffiffiffi
D1 D2 :
ab
(7.252)
The buckling coefficient calculated with the aid of Eq. (7.250) [22 terms in Eq. (7.239) have
been retained] is presented in Fig. 7.33 as the function of the parameters α and β in Eq. (7.251).
7.5 BUCKLING OF ORTHOTROPIC SYMMETRIC PLATES
507
FIGURE 7.34
Buckling and failure modes of 6 45 angle-ply carbonepoxy shear web.
As an example, consider a 6 45 angle-ply carbonepoxy plate (see Fig. 7.34) with the following
parameters: a 5 196 mm, b 5 96 mm, D1 5 D2 5 0:685 N=m, and D3 5 1:718 N=m. For this plate,
Eq. (7.251) yield α 5 0:24, β 5 2:5 and interpolation of the curves in Fig. 7.33 gives K 5 19.8. The
corresponding experimental result is K 5 19.7.
Note that carbonepoxy plates loaded with shear forces are used as composite shear webs in
modern aircraft structures. In the corresponding aluminum structures buckling of the shear webs is
usually allowed at the limit loads, because aluminum webs can buckle many times without failure.
However, this is not the case for composite webs made of relatively brittle carbonepoxy materials. After several cycles of loading, a crack which is parallel to the buckling lobe direction appears
in the composite plate (see Fig. 7.34).
Consider the case of combined loading with compressive forces Tx and shear forces Txy
(see Fig. 7.27). The critical combination of the forces can be found from the following
condition:
Tx
Txy
1
c
Txc
Txy
!2
51
(7.253)
c
in which Txc and Txy
are specified by Eq. (7.245) or (7.246) and Eq. (7.252). The dependencies
of the shear critical parameter K in Eq. (7.252) on the ratio Tx =Txc for carbonepoxy composite
plates with various values of α and β parameters are shown in Fig. 7.35. The solid lines correspond to Eq. (7.253), whereas the circles correspond to the exact solutions based on
Eq. (7.250).
In conclusion, consider anisotropic plates which demonstrate specific behavior under shear. For
anisotropic plates, Eq. (7.143) for the moments become [see Eqs. (7.5)]
508
CHAPTER 7 LAMINATED COMPOSITE PLATES
K
20
3
16
12
2
8
1
4
T x / T xc
0
0.2
0.4
0.6
0.8
1.0
FIGURE 7.35
Dependencies of the shear buckling coefficient K on the normalized axial compressive force Tx =Txc as
per Eq. (7.253) ( sss ) and according to the exact solution (33333); 1 (α 5 1, β 5 0:2); 2 (α 5 1, β 5 1); 3
(α 5 1, β 5 3)
@2 w
@2 w
@2 w
Mx 5 2 D11 2 1 D12 2 1 2D14
@x
@y
@x@y
2
2
@ w
@ w
@2 w
My 5 2 D21 2 1 D22 2 1 2D24
@x
@y
@x@y
@2 w
@2 w
@2 w
Mxy 5 2 D41 2 1 D42 2 1 2D44
@x
@y
@x@y
(7.254)
and the buckling equation, Eq. (7.240), is generalized as
D1
@4 w
@4 w
@4 w
@4 w
@4 w
1 2D3 2 2 1 D2 4 1 4D14 3 1 4D24
4
@x
@x @y
@y
@x @y
@x@y3
2
2
2
@ w
@ w
@ w
1 Ty 2 5 0
1Tx 2 1 2Txy
@x
@x@y
@y
For simply supported plates, the deflection in Eq. (7.239) provides the absence of deflection at
the plate edges. However, it follows from Eqs. (7.239) and (7.254) that the bending moments are
not zero at x 5 0, x 5 a and y 5 0, y 5 b (see Fig. 7.28). Nevertheless, Eq. (7.239) can be used to
approximate the deflection within the framework of the Ritz method which allows us to violate the
force boundary conditions (Langhaar, 1962). To use this method, consider the total potential energy
of an anisotropic plate (Whitney, 1987), i.e.,
7.5 BUCKLING OF ORTHOTROPIC SYMMETRIC PLATES
Txy
Tx
y
x
+φ
Txy
Tx Tx
−φ
Txy
x
Txy
Tx Tx
Tx
−φ
Txy
y
(A)
+φ
x
(B)
y
509
Txy
(C)
FIGURE 7.36
Angle-ply 6 φ (A) and unidirectional 1φ (B) and 2φ (C) composite plates under compression and shear.
ða ðb
1
@2 w
@2 w
@2 w
½Mx 2 1 My 2 1 2Mxy
2
@x
@y
@x@y
:
0
0
2
2
@w
@w
dxdy
2Tx @w 2 Ty @w 2 2Txy
@x
@y
@x @y
T5
(7.255)
Substituting the deflection, Eq. (7.239), into Eq. (7.254) for the moments and the expressions
obtained for the moments into Eq. (7.255), we can express T in terms of wmn . Applying the minimum conditions @T=@wmn 5 0, we arrive at the infinite set of coupled algebraic equations whose
determinant, being equal to zero, allows us, in principle, to determine the critical combination of
forces Tx , Ty ; and Txy .
Consider three composite plates as shown in Fig. 7.36. The first plate (see Fig. 7.36A) has 6 φ
orthotropic angle-ply structure; the second and the third plates (see Fig. 7.36B and C) have unidirectional structure and are reinforced at angles 1 φ and φ to the x-axis, respectively. Under
compression, the second and the third plates demonstrate, as expected, similar behavior. However
under shear, their behavior is different because in the 1 φ plate the force Txy induces tension of the
fibers, whereas in the φ plate, the fibers are compressed. Recall that the stiffness of unidirectional
composites in the fiber direction is much higher than the stiffness in the transverse direction. So,
we can expect that the critical shear force is different for the second and third plates, though the
material structure is similar.
The results of the buckling analysis are presented in Figs. 7.37 and 7.38. The critical load corresponding to axial compression is considerably higher for the angle-ply plate (curve 6 φ in
Fig. 7.37), whereas for the second and the third plates it is the same (curves 1 φ and φ in
Fig. 7.37). The critical force corresponding to shear is the highest for the third plate (curve φ
in Fig. 7.38) in which the fibers are compressed and lowest for the second plate (curve 1 φ in
Fig. 7.38) in which compression is directed across the fibers.
The foregoing buckling analysis has been undertaken for simply supported plates. For other
boundary conditions, approximate solutions can be obtained by approximating the plate deflection
with some suitable functions that satisfy the boundary conditions. Krylov beam functions and
Legendre polynomials were used for this purpose by Khaziev and Boshers (2011). Various types of
loading and boundary conditions were considered by Veres and Kollar (2001), Shan and Qiao
(2008), and Weaver and Nemeth (2008).
510
CHAPTER 7 LAMINATED COMPOSITE PLATES
Txc
2.0
1.6
±φ
1.2
+φ
−φ
0.8
0.4
0
φ
0
15
30
45
60
75
90
FIGURE 7.37
Dependencies of the normalized compressive force on the angle φ for 6 φ angle-ply and 1φ or 2φ
unidirectionally reinforced plates.
Txyc
4
3
−φ
2
±φ
1
+φ
0
0
15
30
45
60
75
90
φ
FIGURE 7.38
Dependencies of the normalized shear force on the angle φ for 6 φ angle-ply and 1φ or 2φ unidirectionally
reinforced plates.
7.5 BUCKLING OF ORTHOTROPIC SYMMETRIC PLATES
511
7.5.2 THEORY OF SHEAR DEFORMABLE PLATES
The foregoing results (Section 7.5.1) correspond to classical plate theory which ignores transverse
shear deformation. In practical problems, this deformation is generally negligible. However, there
are at least two cases in which it should be taken into account. The first case corresponds to relatively thick plates such that the ratio of the plate thickness to the smallest in-plane dimension
exceeds 0.02. Those plates with relatively low transverse shear stiffness (e.g., sandwich plates with
lightweight core) belong to the second case.
First, consider transversely isotropic plates as discussed in Section 7.3. For these plates
D11 5 D22 5 D;
D12 5 νD;
D44 5
12ν
D;
2
Sx 5 Sy 5 S
and the governing equation, Eq. (7.40) for orthotropic plates, reduces to
D
Dð1 2 νÞ
Δ 2 1 ΔΔW 1 p 5 0
2S
(7.256)
For the case of bending considered in Section 7.3.2, Eq. (7.256) can be decomposed into two
mutually independent equations, Eqs. (7.69) and (7.76). For the buckling problem, the pressure p
should be changed to the imaginary pressure pn specified by Eq. (7.237) according to which
@2 w
@2 w
@2 w
pn 5 2 Tx 2 1 Ty 2 1 2Txy
@x
@y
@x@y
(7.257)
For the case of uniaxial compression with forces Tx (see Fig. 7.28), the buckling equation, following from Eqs. (7.256) and (7.257), is
D
Dð1 2 νÞ
@2 w
Δ 2 1 ΔΔW 2 Tx 2 5 0
2S
@x
(7.258)
in which the deflection w is specified by Eq. (7.42) which for transversely isotropic plates takes
the form
w5W 2
Dð3 2 νÞ
D2 ð1 2 νÞ
ΔΔW
ΔW 1
2S
2S2
(7.259)
For a simply supported plate, we can present the functions W as follows:
W5
XX
m
Wmn sinλm x sinλn y
(7.260)
n
where λm 5 πm=a and λn 5 πn=b (see Fig. 7.28). Substituting the deflection w, Eq. (7.259) into the
buckling equation, Eq. (7.258), and using Eq. (7.260) for W, we arrive at the following expression for Tx :
Tx 5
2
D λ2m 1λ2n
λ2m 1 1 D=S λ2m 1 λ2n
(7.261)
Note that the same result can be obtained if we neglect the boundary-layer solution and consider
Eqs. (7.80) and (7.82) which specify only the penetrating solution. The buckling equation, following from Eqs. (7.80), (7.82), and (7.257) can be represented by
DΔΔϕ 2 Tx
@2
D
ϕ
2
Δϕ
50
@x2
S
512
CHAPTER 7 LAMINATED COMPOSITE PLATES
Expanding ϕ into a series similar to Eq. (7.260), we arrive at Eq. (7.261) for Tx .
Consider orthotropic plates. The solution of the buckling problem can be obtained using
Eq. (7.36) in which the pressure p is changed to the imaginary pressure pn specified by Eq. (7.257).
Then, the buckling equations become
L1x ðθx Þ 1 L1y θy 1 L1w ðwÞ 5 0
L2x ðθx Þ 1 L2y θy 1 L2w ðwÞ 5 0
@2 w
@2 w
@2 w
L3x ðθx Þ 1 L3y θy 1 L3w ðwÞ 5 Tx 2 1 Ty 2 1 2Txy
@x
@y
@x@y
(7.262)
For simply supported plates, the angles θx and θy , and the deflection w can be taken in the form
of double trigonometric series, Eq. (7.220). Substitution of these series into Eq. (7.262) yields three
mn
homogeneous algebraic equations for the series coefficients θmn
x , θy , and wmn : The critical combination of loads Tx , Ty , and Txy is the first combination of loads (corresponding to the minimum
loading parameter according to which these loads increase) for which the determinant of the algebraic equations is zero.
Particularly, for unidirectional compression with forces Tx (see Fig. 7.28), we get
2
Dmn λ2m 1 Dnm λ2n 1 2Dλ2m λ2n 1 Dmn Dnm 2 D λ2m λ2n
λ2n =Sx 1 λ2m =Sy
h
i
Tx 5
2
λ2m 1 1 Dmn =Sx 1 Dnm =Sy 1 1=Sx Sy Dmn Dnm 2 D λ2m λ2n
(7.263)
where, as earlier,
Dmn 5 D11 λ2m 1 D44 λ2n ;
Dnm 5 D22 λ2n 1 D44 λ2m ;
πm
πn
; λn 5
λm 5
a
b
D 5 D12 1 D44
Ignoring the transverse shear deformation, i.e., taking Sx -N and Sy -N, we arrive at
Eq. (7.241) corresponding to classical plate theory. The critical load Txc can be found if we minimize Tx in Eq. (7.263) with respect to half-wave numbers m; n 5 1; 2; 3; . . .. The allowance for
transverse shear deformation usually does not affect the critical values of half-wave numbers and
they can be evaluated using the equations derived in Section 7.5.1.
The solution obtained, Eq. (7.263), can be significantly simplified if we take into account the
results obtained previously for transversely isotropic plates and neglect the boundary-layer solution.
Thus, determine the critical load using Eqs. (7.228) and (7.231) which specify the penetrating
solution. Then, the buckling equation can be written as
D1
@4 Wi
@4 Wi
@4 Wi
@2 w
@2 w
@2 w
1 2D3 2 2 1 D2 4 1Tx 2 1 2Txy
1 Ty 2 5 0
4
@x
@x @y
@y
@x
@x@y
@y
D1 @2 Wi
D2 @2 Wi
2
w 5 Wi 2
Sx @x2
Sy @y2
For simply supported plates, we can take Wi in accordance with Eq. (7.260) and for the case of
unidirectional compression (Ty 5 0; Txy 5 0Þ, derive the following expression for Tx which generalizes Eq. (7.261):
Tx 5
π2
b2
D1 λ 1 2D3 n2 1 D2 n4 =λ
;
2
π2 1 1 2 D1 =Sx λ 1 D2 =Sy n
b
λ5
2
mb
a
(7.264)
7.5 BUCKLING OF ORTHOTROPIC SYMMETRIC PLATES
513
Tc
1.0
0.8
0.6
0.4
0.2
η
0
0.2
0.4
0.6
0.8
1.0
FIGURE 7.39
Dependence of the normalized critical forces on parameter η for transversely isotropic square plates.
It can be readily proved that @Tx =@n2 . 0, i.e., that function Tx ðn2 Þ increases with increase in n2 .
Thus, the minimum possible value n 5 1 must be taken, and Eq. (7.264) reduces to
Tx 5
π2
b2
D1 λ 1 2D3 1 D2 1=λ
π 2 1 1 2 D1 =Sx λ 1 D2 =Sy
b
(7.265)
Using the minimum condition @Tx =@λ 5 0, we can arrive at a quadratic equation for λ, but the
corresponding solution looks rather cumbersome. For practical calculation, it is easier to apply a
sorting procedure to find the parameter m 5 1; 2; 3; . . .. Consider for example a square (a 5 b) transversely isotropic plate for which m 5 1, n 5 1 and Eq. (7.263) or (7.264) yield
Tcð1Þ 5
4π2 D
a2 ð1 1 2ηÞ
;
η5
π2 D
a2 S
For classical plate theory, η 5 0 and
Tcð2Þ 5
4π2 D
a2
The dependence of the normalized critical force T c 5 Tcð1Þ =Tcð2Þ on the parameter η is presented
in Fig. 7.39. Note that for homogeneous and sandwich plates, the parameter η is evaluated as
ηh 5
π2 Eh2
;
12ð1 2 ν 2 ÞGa2
ηs 5
π2 Ef δh
2Gc a2
in which Ef and δ are the modulus and the thickness of the facing layers, and Gc and h are the
shear modulus and the thickness of the core of a sandwich plate. For composite plates, ηh can reach
10h2 =a2 and for h=a # 0:05 we have ηh # 0:025, so that the shear deformation is negligible. For
sandwich plates ηs can reach 1000h2 =a2 and even for thin plates ðh=a 5 0:01Þ, ηs 5 0:1 and the
shear deformation must be taken into account.
514
CHAPTER 7 LAMINATED COMPOSITE PLATES
7.6 POSTBUCKLING BEHAVIOR OF ORTHOTROPIC SYMMETRIC PLATES
UNDER AXIAL COMPRESSION
It is known that rectangular plates fixed at the edges with respect to deflection can take considerable load after the onset of buckling. Consider a simply supported plate loaded with forces Tx
uniformly distributed over the plate transverse edges x 5 0 and x 5 a (see Fig. 7.40). After
buckling, the plate has a curved shape characterized by some deflection w(x, y) and should be
described by nonlinear equations (Vasiliev, 1993). We assume that the deflection is small so that
the in-plane equilibrium equation can be taken in accordance with Eq. (7.29) corresponding to a
flat plate, i.e.,
@Nx
@Nxy
1
5 0;
@x
@y
@Ny
@Nxy
1
50
@y
@x
(7.266)
where the stress resultants Nx , Ny , and Nxy are linked to the corresponding strains ε0x , ε0y , and γ 0xy by
constitutive equations, Eq. (7.28), according to which
Nx 5 B11 ε0x 1 B12 ε0y ;
Ny 5 B21 ε0x 1 B22 ε0y ;
Nxy 5 B44 γ 0xy
(7.267)
Here, in contrast to Eq. (7.30), the strains allow for the plate deflection and can be written as
(Vasiliev, 1993)
ε0x 5
@u 1 @w 2
1
;
@x
2 @x
ε0y 5
@v 1 @w 2
1
;
@y 2 @y
γ 0xy 5
@u @v @w @w
1
1
:
@y
@x
@x @y
(7.268)
Substituting Eq. (7.268) into Eq. (7.267) and the expressions obtained for the stress resultants
into the equilibrium equations, Eq. (7.266), we get
@2 u
@2 u
@2 v
@w @2 w
@w @2 w
@w @2 w
5
2
B
1
B
1
B
1
B
1
B
44
11
44
@x2
@y2
@x@y
@x @x2
@y @x@y
@x @y2
2
2
2
2
2
@ v
@ v
@ u
@w @ w
@w @ w
@w @2 w
5 2 B22
1
B
B22 2 1 B44 2 1 B
1
B
44
@y
@x
@x@y
@y @y2
@x @x@y
@y @x2
B11
(7.269)
where B 5 B12 1 B44 . Assume that the plate deflection corresponds to a mode of plate buckling
with some unknown coefficient wm and half-wave number m (see Fig. 7.28), i.e.,
w 5 wm sinλm xsinλ1 y
(7.270)
x
y
Tx
b
a
FIGURE 7.40
A simply supported plate after buckling.
Tx
7.6 POSTBUCKLING BEHAVIOR OF ORTHOTROPIC SYMMETRIC PLATES
515
where λm 5 πm=a and λ1 5 π=b. Substituting the deflection, Eq. (7.270), into Eq. (7.269), we can
obtain the following solution of these equations (Azikov & Vasiliev, 1986):
x
w2
B12 2
B22 Tx 2 λ2m w2m 2 m λ2m ð1 2 cos2λ1 yÞ 2
λ sin2λm x
B
16λm
B11 1
y
w2
B12 2
B12 Tx 2 λ21 w2m 2 m λ21 ð1 2 cos2λm xÞ 2
v5
λ sin2λ1 y
B
16λ1
B22 m
u52
(7.271)
where B 5 B11 B22 2 B212 . To determine wm , apply the principle of minimum total potential
energy of the plate
ða ðb
P5
Nx ε0x 1 Ny ε0y 1 Nxy γ 0xy 1 Mx κx 1 My κy 1 Mxy κxy 1 Tx
0 0
!
@u
dxdy
@x
(7.272)
where the moments M are expressed in terms of curvatures κ as
Mx 5 D11 κx 1 D12 κy ;
My 5 D12 κx 1 D22 κy ;
Mxy 5 D44 κxy
(7.273)
@2 w
@x@y
(7.274)
where within the framework of classical plate theory
κx 5 2
@2 w
;
@x2
κy 5 2
@2 w
;
@y2
κxy 5 2 2
Using Eqs. (7.267), (7.268), (7.273), and (7.274), we can write the energy P, Eq. (7.272), in
terms of displacements u, v, and w. Substituting Eqs. (7.270) and (7.271) for the displacements and
integrating, we can further obtain P as the following function of Tx , wm , and λm . Applying the minimum condition with respect to wm , i.e., taking @P=@wm 5 0, we obtain the equation for wm whose
solution is
16B11 B22 λ2m
ðTx 2 Tm Þ
w2m 5 B B11 λ4m 1 B22 λ41
(7.275)
1 D1 λ4m 1 2D3 λ2m λ21 1 D2 λ41
2
λm
(7.276)
in which
Tm 5
It follows from Eq. (7.275) that the deflection exists if Tx . Tm , where Tm in Eq. (7.276)
coincides with Tx given by Eqs. (7.241) and (7.243). Minimization of Tm with respect to m gives
the critical load Txc 5 Tm1 and the corresponding half-wave number m1 specified by Eq. (7.244).
Thus, if Tx # Tm , wm 5 0 and the plate remains flat. If Tx . Txc , wm is determined by Eq. (7.275)
in which m 5 m1 . Under further loading, the so-called mode jumping can occur and the half-wave number m might increase. To find the half-wave number, substitute the displacements,
Eqs. (7.271) and (7.275), into Eq. (7.272) for the total potential energy. After integration, we
get
P5
"
#
ab 2
2B11 λ2m λ21
Tm 2
Tx B22 1 1
12
2B
Tx
B11 λ4m 1 B22 λ41
(7.277)
516
CHAPTER 7 LAMINATED COMPOSITE PLATES
0.04
p̂1
m=5
m=4
0.03
m=4
m=3
0.02
m=3
m=2
0.01
1.4 T23
1.0
T4
T3
T2
1.8 T34 2.2
Tx
T5
2.6 T45 3.0
3.4
FIGURE 7.41
Dimensionless loads T x corresponding to the eigenvalues ( ) and to the mode jumping loads (3).
The half-wave number m corresponding to the acting force Tx is found from the minimum condition for the total potential energy P. First, single out the second term in the brackets of
Eq. (7.277) which depends on m and present it as follows
P1 5
1
B22 ab 2
λ2 ðTx 2Tm Þ2
T
5 m 4
P
2
x
2
2B
B11 λm 1 B22 λ41
abλ1
(7.278)
where Tx is the acting load and Tm is the eigenvalue specified by Eq. (7.276). To demonstrate
the procedure, consider an isotropic plate for which
D11 5 D22 5 D12 5 D44 5 D;
B11 5 B22 5 Eh;
E5
E
ð1 2 ν 2 Þ
For this isotropic plate, Eqs. (7.278) and (7.276) become
P1 5
m 2 c2
ðTx 2Tm Þ2 ;
m4 1 c4
Tm 5
2
π2 Dc2 m2
11
b2 m2 c2
where c 5 b/a. Introduce the dimensionless functions of m
2
P1 b4
m 2 c2 5
T x 2T m
π4 D2 c4
m 4 1 c4
2
Tx b2
Tm b2
1 m2
Tx 5 2 2 ; Tm 5 2 2 5 2
11
m c2
π Dc
π Dc
p^ 1 ðmÞ 5
(7.279)
(7.280)
For the plate considered as an example in Section 7.5.1, c 5 1.75 and m1 5 2. The corresponding
dimensionless critical force following from Eq. (7.280) is T 2 5 1:33. The other eigenvalues are
T 3 5 1:723, T 4 5 2:421, and T 5 5 3:36. The eigenvalues T m ðm 5 2; 3; 4; 5Þ are shown on the horizontal axis in Fig. 7.41 with dots. The curves in Fig. 7.41 show the dependencies of p^ 1 ðmÞ specified
by Eq. (7.279) on the dimensionless load T x for various m-values. As can be seen, for T 2 # T x # T 23 ,
we have m 5 2. For T 23 # T x # T 3 , the minimum values of p^ 1 correspond to m 5 3, for T x $ T 34 ,
m 5 3 changes to m 5 4 and for T x $ T 45 , we have m 5 5. The forces Tm;m11 at which mode jumping
7.6 POSTBUCKLING BEHAVIOR OF ORTHOTROPIC SYMMETRIC PLATES
140
517
Tx , kN/m
120
Txu
100
80
60
40
Txc
20
0
0
15
30
45
60
75
90
FIGURE 7.42
Dependencies of the ultimate load Txu and the critical load Txc on angle φ.
takes place can be found as the T x -coordinate of the point of intersection of the curves p^ 1 ðmÞ corresponding to m and m 11 (see Fig. 7.41), i.e., from the condition P ðmÞ 5 Pðm 1 1Þ. Using
Eq. (7.277), we finally get
Tx 2Tm
Tx 2Tm11
2
λ2m11 B11 λ4m 1 B22 λ41
5 2
λm B11 λ4m11 1 B22 λ41
(7.281)
For the example under consideration, this equation yields T 23 5 1:51, T 34 5 2:03, T 45 5 2:84.
The half-wave number m found from Eq. (7.281) and the corresponding value of the acting load
Tx are substituted into Eq. (7.275) for wm and this value of wm is further substituted into
Eqs. (7.270) and (7.271) which specify the plate displacements. Then, using Eq. (7.268), we can
find the strains and, following the procedure described in Section 3.11, determine the strains and
the stresses in the ply principle coordinates. Applying the strength criteria discussed in Chapter 4,
Failure Criteria and Strength of Laminates, we can evaluate the load-carrying capacity of the plate
after buckling. As an example, consider a carbonepoxy 6 φ angle-ply square plate with
a 5 b 5 0.4 m and h 5 1 mm. The dependence of the ultimate force Txu (solid line) along with the
critical force Txc (dashed line) on the angle φ is shown in Fig. 7.42. As can be seen, the optimal
angle which provides the maximum load-carrying capacity (close to 0 ) does not coincide with the
518
CHAPTER 7 LAMINATED COMPOSITE PLATES
σ x , MPa
20
15
2
10
1
5
y/b
0
0.2
0.4
0.6
0.8
1.0
FIGURE 7.43
Theoretical ( sss ) and experimental (33333) distributions of the normal stress over the plate width after buckling
for Tx =Txc 5 2(1) and Tx =Txc 5 2:53(2).
optimal angle for buckling (45 Þ: In contrast to metal plates for which the ultimate load is much
higher than the critical load, the relation between these loads for composite plates depends on the
material structure.
The foregoing approach provides reasonably accurate evaluation of the stresses and displacements of composite plates after buckling. Experimental results for 2.5-mm-thick composite plates
have been presented by Banks and Harvey (1979). The predicted
σx 5
Nx
Tx
Bλ2m 2
52
1
wm cos2λ1 y
h
h
8B22 h
and experimental distributions of σx along the y-coordinate of the plate (see Fig. 7.40) after buckling is shown in Fig. 7.43 for two values of the acting load. The dependence of the plate shortening
Δu 5 uðx 5 aÞ 2 uðx 5 0Þ under the action of compressive forces Tx (see Fig. 7.40) is shown in
Fig. 7.44. The knee on the curve corresponds to the plate buckling.
Numerical analysis of the postbuckling behavior of composite plates has been presented by
Ovesy and Assaee (2005).
7.7 GENERALLY LAMINATED PLATES
The foregoing sections of this chapter are concerned with symmetrically laminated plates for which
bending-stretching coupling stiffness coefficients Cmn in the constitutive equations, Eq. (7.5) are
zero. For generally laminated plates, these coefficients are not zero, and the plate analysis becomes
much more complicated. As shown in Section 3.4, a symmetric arrangement of the layers in laminated plates provides the maximum bending stiffness for the plates with a given number of layers.
It should be also taken into account that the symmetric structure of the plate substantially simplifies
7.7 GENERALLY LAMINATED PLATES
519
Tx / Txc
6
4
2
Δu
Δu (T = Txc )
2
0
4
6
10
8
FIGURE 7.44
Theoretical ( sss ) and experimental (33333) dependencies of the plate shortening on the normalized
compression load.
2
1
a
(A)
h
t
b
2
1
(B)
FIGURE 7.45
Reduction of a stiffened plate to a laminated plate.
the plate behavior, i.e., the plate bending and in-plane loading are described by independent sets of
equations and the plane remains flat before buckling under in-plane loading. In unsymmetrically
laminated plates in-plane stressstrain state and bending are coupled and the plate behavior is
described by a coupled set of equations which include all three displacements u, v, and w. Thus,
without special reasons it does not make much sense to design unsymmetrically laminated plates.
Note that the thickness of the composite ply is relatively small (0.10.2 mm) and tens of such plies
with various orientations are usually placed to fabricate real composite plates. Thus, the possibility
to make a symmetrically laminated plate practically always exists.
However, there are special cases in which the plates cannot be designed as symmetric laminates.
The first class of such plates comprises plates stiffened with ribs which are usually located on one
side of the plate. Being smeared over the plate surface, the ribs can be simulated by continuous
layers resulting in an unsymmetrically laminated structure. For example, consider a stiffened plate
shown in Fig. 7.45. The ribs in Fig. 7.45A can be approximately reduced to the layer 1 shown in
Fig. 7.45B whose stiffness coefficients in the coordinates 1,2 are
b
ð1Þ
ð1Þ
A11
5 Er; Að1Þ
12 5 A22 5 0
a
where Er is the modulus of the rib material. In conjunction with the skin (layer 2 in Fig. 7.45B),
the resulting structure is unsymmetric. The second group of unsymmetric structures corresponds to
plates consisting of two layers, one of which is a load-carrying layer, whereas the second one protects this layer against, for example, temperature or impact loading.
520
CHAPTER 7 LAMINATED COMPOSITE PLATES
7.7.1 BENDING OF UNSYMMETRIC PLATES
Consider a rectangular orthotropic plate loaded with transverse pressure as shown in Fig. 7.1. The
plate behavior is described by equilibrium equations, Eqs. (7.22), (7.24), and (7.26), constitutive
equations, Eq. (7.16), and straindisplacement equations, Eq. (7.3), in which the curvatures corresponding to classical plate theory are given by Eq. (7.49). Within the framework of classical plate
theory, the constitutive equations become
@u
@v
@2 w
@2 w
1 B12 2 C11 2 2 C12 2
@x
@y
@x
@y
@u
@v
@2 w
@2 w
1 B22 2 C12
Ny 5 B12
2 C22 2
@x @y @x2
@y
@u @v
@2 w
1
2 2C44
Nxy 5 B44
@y
@x
@x@y
@u
@v
@2 w
@2 w
1 C12 2 D11 2 2 D12 2
Mx 5 C11
@x
@y
@x
@y
@u
@v
@2 w
@2 w
1 C22 2 D12
My 5 C12
2 D22 2
@x @y @x2
@y
@u @v
@2 w
1
2 2D44
Mxy 5 C44
@y
@x
@x@y
Nx 5 B11
(7.282)
In classical plate theory, the equilibrium equations reduce to two in-plane equations, Eq. (7.22),
and Eq. (7.51) describing the plate bending. Substituting stress resultants and couples specified
by Eq. (7.242) into these three equations, we arrive at the following set of equations for
u, v, and w:
L1 ðu; v; wÞ 5 0;
L2 ðu; v; wÞ 5 0;
L3 ðu; v; wÞ 5 p
(7.283)
in which
@2 u
@2 u
@2 v
@3 w
@3 w
1 B44 2 1 ðB12 1 B44 Þ
2 C11 3 2 ðC12 1 2C44 Þ
@x2
@y
@x@y
@x
@x@y2
@2 v
@2 v
@2 u
@3 w
@3 w
2 C22 3 2 ðC12 1 2C44 Þ 2
L2 ðu; v; wÞ 5 B22 2 1 B44 2 1 ðB12 1 B44 Þ
@y
@x
@x@y
@y
@x @y
3
@4 w
@4 w
@4 w
@3 u
@ u
@3 v
@3 v
L3 ðu; v; wÞ 5 D1 4 1 2D3 2 2 1 D2 4 C11 3 2 ðC12 1 2C44 Þ
2 C22 3
2
2
2
@x
@x @y
@y
@x
@x@y
@x @y
@y
L1 ðu; v; wÞ 5 B11
(7.284)
where, as earlier, D1 5 D11 , D2 5 D22 , and D3 5 D12 1 2D44 .
Consider a simply supported plate (see Fig. 7.3) and assume that the plate edges can move in
the directions normal to the edges, so that the corresponding forces are zero. Then, the boundary
conditions on the plate edges are at x 5 0 and x 5 a,
w 5 0;
Mx 5 0;
Nx 5 0;
v50
w 5 0;
My 5 0;
Ny 5 0;
u50
(7.285)
and at y 5 0 and y 5 b,
It can be proved with the aid of Eq. (7.282), that these boundary conditions can be satisfied if
we take the displacements in the form
7.7 GENERALLY LAMINATED PLATES
u5
v5
w5
XX
m
n
X
X
m
n
m
n
XX
521
umn cosλm x sinλn y
vmn sinλm x cosλn y
(7.286)
wmn sinλm x sinλn y
where λm 5 πm=a and λn 5 πn=b (see Fig. 7.1). Expanding the pressure p into a similar series
specified by Eq. (7.147) and substituting the displacements, Eq. (7.286), into the equilibrium equations, Eq. (7.283), we arrive at the following three algebraic equations for the coefficients of
Eq. (7.285):
a11 umn 1 a12 vmn 1 a13 wmn 5 0
a21 umn 1 a22 vmn 1 a23 wmn 5 0
a31 umn 1 a32 vmn 1 a33 wmn 5 pmn
(7.287)
in which
a11 5 B11 λ2m 1 B44 λ2n ; a12 5 a21 5 ðB12 1 B44 Þλm λn
a22 5 B44 λ2m 1 B22 λ2n ; a13 5 a31 5 C11 λ3m 1 ðC12 1 2C44 Þλm λ2n
a23 5 a32 5 C22 λ3n 1 ðC12 1 2C44 Þλ2m λn
(7.288)
a33 5 D1 λ4m 1 2D3 λ2m λ2n 1 D2 λ4n
The solution of Eq. (7.287) is
umn 5
wmn
wmn
ða13 a22 2 a12 a23 Þ; vmn 5
ða11 a23 2 a12 a13 Þ
A1
A1
A1
wmn 5 pmn ; A1 5 a11 a22 2 a212
A2
A2 5 a33 A1 2 a11 a223 2 a22 a213 1 2a12 a13 a23
(7.289)
It is possible to demonstrate a remarkable property of the solution obtained. Recall that in
reducing the three-dimensional equations of the theory of elasticity to the two-dimensional equations of plate theory, we actually reduce the plate to its reference surface located at some distance e
from the lower surface of the plate. Hence, the forces Nx , Ny , and Nxy in plate theory are applied to
the reference surface and the moments Mx , My , and Mxy are calculated with respect to this surface.
Only the transverse forces Vx and Vy and the plate deflection w (see Fig. 7.46) correspond to any
x
Vx (w)
M xy (θ y )
Vy (w)
M y (θ y )
y
Mx (θ x )
Nx (u )
M xy (θ x )
e
Ny (v)
Nxy (v)
Nxy (u )
FIGURE 7.46
Forces and moments applied to the plate reference surface and the corresponding displacements u, v, w and
rotation angles θx , θy .
522
CHAPTER 7 LAMINATED COMPOSITE PLATES
FIGURE 7.47
Sandwich plate with ramps.
surface parallel to the reference surface. Thus, when formulating the boundary conditions for
in-plane forces Nx , Ny , and Nxy or in-plane displacements u and v, we must apply the loads to the
reference surface or fix the reference surface on the plate edges. In some cases, the location of the
reference surface can be preassigned. For example, if the skin of the stiffened plate shown in
Fig. 7.45 is fixed, the reference surface should be located within the skin thickness. For sandwich
plates with ramps (see Fig. 7.47), the reference surface should pass through the fixed boundary contour. However, in the majority of cases the boundary conditions do not help us to identify the location of the reference surface and the coordinate e (see Fig. 7.46) can be chosen arbitrarily. Clearly,
in such cases the feasible solutions for real plates must not depend on e, and this is the case for the
solution given by Eq. (7.289). Indeed, the stiffness coefficients, Eqs. (3.28) and (3.29), are
ð0Þ
Bmn 5 Imn
;
ð1Þ
ð0Þ
Cmn 5 Imn
2 eImn
ð2Þ
ð1Þ
ð0Þ
Dmn 5 Imn
2 2eImn
1 e2 Imn
;
ðh
ðrÞ
Imn
5 Amn tr dt
ðr 5 0; 1; 2Þ
(7.290)
0
where mn 5 11, 12, 22, and 44. Substituting Eq. (7.290) into Eq. (7.289) for A1 and A2 , we
arrive at
2
ð0Þ 2
ð0Þ 2
ð0Þ 2
ð0Þ 2
ð0Þ
ð0Þ
A1 5 I11
λm 1 I44
λn I22
λm 1 I44
λn 2 I12
1I44
λ2m λ2n
h
i
ð2Þ 4
ð2Þ
ð2Þ
ð2Þ 4
λ2m λ2n 1 I22
A2 5 A1 I11
λm 1 2 I12
12I44
λn
h
i
ð0Þ 2
ð0Þ 2
ð1Þ 3
ð1Þ
ð1Þ
2 I11 λm 1 I44 λn I22 λn 1 I12 12I44 λ2m λn
h
i
ð0Þ 2
ð0Þ 2
ð1Þ 3
ð1Þ
ð1Þ
2 I44
λm λ2n
λm 1 I22
λn I11
λm 1 I12
12I44
h
ih
i
ð0Þ
ð0Þ
ð1Þ 2
ð1Þ
ð1Þ
ð1Þ 3
ð1Þ
ð1Þ
12 I12
1 I44
λm 1 I12
12I44
λn 1 I12
12I44
I11
λm λ2n I22
λ2m λn λm λn
ðrÞ
As can be seen, the terms explicitly including e disappear, whereas the integrals Imn
do not
depend on e. Thus, the deflection coefficient wmn in Eq. (7.289) does not depend on e either. This
result is expected, since the boundary conditions given by Eq. (7.285) actually correspond to the
plate resting on thin walls which have infinitely high in-plane stiffness and are absolutely flexible
with respect to the out-of-plane displacements. The conditions N 5 0 and M 5 0 provide zero stresses which are normal to the plate edges, whereas the conditions imposed on the displacements
provide zero displacements in the plane of the edges. Obviously, the deflection of the plate with
such boundary conditions must not depend on the choice of the reference surface.
To demonstrate the opposite situation, change the first two series in Eq. (7.286) to the
following ones:
u5
XX
m
n
umn sinλm x sinλn y;
v5
XX
m
n
vmn sinλm x sinλn y
(7.291)
7.7 GENERALLY LAMINATED PLATES
523
a
90
0
x
a
h /2
h /2
y
FIGURE 7.48
Two-layered cross-ply square plate.
In conjunction with Eq. (7.286) for w, these expansions correspond to the following boundary
conditions. The boundary conditions on the plate edges are, at x 5 0 and x 5 a,
w 5 0; Mx 5 0;
u 5 0; v 5 0
w 5 0; My 5 0;
v 5 0; u 5 0
and at y 5 0 and y 5 b
Thus, the in-plane displacements are zero at the plate edges. Substitution of the series for u, v,
and w into Eq. (7.283) yields
XX
m
XX
m
ða11 umn sinλm x sinλn y 2 a12 vmn cosλm x cosλn y
n
2a13 wmn cosλm x sinλn yÞ 5 0
ða21 umn cosλm x cosλn y 2 a22 vmn sinλm x sinλn y
n
(7.292)
1a23 wmn sinλm x cosλn yÞ 5 0
XX
ða31 umn cosλm x sinλn y 1 a32 vmn sinλm x cosλn y
m
n
1ða33 2 pmn Þwmn sinλm x sinλn yÞ 5 0
where the coefficients aij are specified by Eq. (7.288). These equations cannot be directly
reduced to algebraic equations similar to Eq. (7.287). To arrive at an algebraic set of equations
for umn , vmn , and wmn , we can use the BubnovGalerkin method, i.e., multiply Eq. (7.292) by
sinλi x sinλj y and integrate them over the plate surface area. As a result, we get an infinite set of
algebraic equations for umn , vmn , and wmn .
As an example, consider a square (a 5 b) cross-ply carbonepoxy plate (see Fig. 7.48) composed of 0 and 90 layers. The plate dimensions are a 5 100 mm, h 5 2 mm. For uniform pressure
p 5 p0 , the maximum plate deflection corresponding to the solution given by Eq. (7.289) is
w 5 5:24 106 p0 =E1 in which E1 is the longitudinal modulus of the unidirectional ply. As has been
noted, this deflection does not depend on the location of the reference surface. Now assume that
524
CHAPTER 7 LAMINATED COMPOSITE PLATES
the reference surface coincides with the contact surface of the layers (e 5 h/2). The solution of the
equation corresponding to Eq. (7.292) in which 10 terms are retained is w 5 4:3 106 p0 =E1 . This
deflection is about 20% lower than that calculated for a simply supported plate. However, this solution depends on the coordinate of the reference surface and can hardly be used for real plates for
which the location of the reference is not known.
We can expect that in addition to the solution given by Eq. (7.289) for simply supported plates,
that the deflection does not depend on the location of the reference surface if the plate edges are
clamped and the boundary conditions are, at x 5 0 and x 5 a,
w 5 0;
θx 5
@w
5 0;
@x
u 5 0;
v50
w 5 0;
θy 5
@w
5 0;
@y
u 5 0;
v50
(7.293)
and at y 5 0 and y 5 b,
These boundary conditions can be satisfied if in addition to Eq. (7.291) for u and v, the plate
deflection is presented as
w5
XX
m
wmn ð1 2 cos2λm xÞð1 2 cos2λn yÞ
(7.294)
n
where, as earlier, λm 5 πm=a, λn 5 πn=b (see Fig. 7.1). Substitution of the displacements,
Eqs. (7.291) and (7.294), into the equilibrium equations, Eq. (7.283), results in equations similar to
Eq. (7.292) which can be solved with the aid of BubnovGalerkin method.
7.7.2 IN-PLANE LOADING
Consider unsymmetrically laminated plates loaded by compressive forces Tx as shown in Fig. 7.28.
As opposed to symmetric plates, unsymmetrically laminated plates can experience bending under
compression. To allow for the in-plane loading of the plate, change the pressure p to the imaginary
pressure pn specified by Eq. (7.237). Then, the equilibrium equations, Eq. (7.283) become
L1 ðu; v; wÞ 5 0;
L2 ðu; v; wÞ 5 0;
L3 ðu; v; wÞ 1 Tx
@2 w
50
@x2
(7.295)
where the operators L1 , L2 , and L3 are specified by Eq. (7.284).
Consider first the conditions under which the plate remains flat under in-plane loading (Leissa,
1986). Assume that
w 5 0;
Nx 5 2 Tx ;
Ny 5 0; Nxy 5 0
Then, the first three equations of Eq. (7.282) yield
B11
@u
@v
1 B12 5 2 Tx ;
@x
@y
B12
@u
@v
1 B22 5 0;
@x
@y
@u @v
1
50
@y
@x
(7.296)
7.7 GENERALLY LAMINATED PLATES
525
The solution of these equations is
u52
Tx B22
x 1 C1 y 1 C2 ;
B
v5
B12 Tx
y 2 C1 x 1 C3
B
(7.297)
where C1 , C2 , and C3 are the constants corresponding to the plate in-plane displacements and rotation as a solid. Substituting Eq. (7.297) into the last three equations of Eq. (7.282), we get
Mx 5
Tx
ðB11 C12 2 B22 C11 Þ;
B
My 5
Tx
ðB12 C22 2 B22 C12 Þ; Mxy 5 0
B
(7.298)
Since Mx and My do not depend on x and y, the moment equilibrium equations, Eq. (7.26) show
that Vx 5 0 and Vy 5 0. Thus, the plate is in a state of a pure bending by the moments specified by
Eq. (7.298) and applied at the plate edges. These moments are of a reactive nature. An unsymmetrically laminated plate, being loaded with in-plane forces Tx (see Fig. 7.28), tends to bend, but the
edge moments, Eq. (7.298), prevent this bending. Such moments can appear if the plate edges are
clamped with respect to bending, i.e., if the first two boundary conditions in Eq. (7.293) are valid
at the plate edges.
Thus, unsymmetrically laminated plates clamped at the edges (clearly, this is a sliding clamp
allowing nonzero in-plane displacements of the plate) remain flat under in-plane loading by uniformly distributed edge forces. Such plates can buckle and conventional methods to determine the
critical loads can be used (Qatu & Leissa, 1993; Majeed & Hyer, 2005).
Consider simply supported plates. First of all, note that the equilibrium equations,
Eq. (7.295), are homogeneous and, in conjunction with the homogeneous boundary conditions,
Eq. (7.285), allow us to formulate the boundary-value problem and to find the corresponding
eigenvalues for the load Tx . For symmetrically laminated plates, the lowest of these values is
associated with the critical load which causes plate buckling. Consider this problem for unsymmetrically laminated plates. Substituting the displacements, Eq. (7.286), into Eq. (7.295), we
arrive at the homogeneous set of algebraic equations, Eq. (7.287), in which pmn 5 Tx λ2m wmn . The
determinant of this set must be zero, so that we arrive at the following system of eigenvalues
for the load Tx :
Tmn 5
A2
λ2m A1
(7.299)
in which A1 and A2 are specified by Eq. (7.289). Recall that the load Tx must be applied to the
reference surface with coordinate e (see Fig. 7.46) which is not specified. However, it follows
from the foregoing analysis of the solution given by Eq. (7.289) that the coefficients A1 and A2
do not depend on e and hence, the eigenvalues Tmn do not depend on e either. For the symmetrically laminated plate problem, the so-called bifurcation of the plate equilibrium takes place
when Tx reaches the first eigenvalue Tx0 . This means that for the load Tx 5 Txc two equilibrium
states of the plate are possible: a flat shape (w 5 0) which is not stable for Tx . Txc and a
curved shape to which the plate is transformed as a result of buckling. However, an unsymmetrically laminated plate starts bending from the very beginning of the loading process, i.e.,
w 6¼ 0 for Tx # Txc , which means that there is no bifurcation and, hence, buckling cannot take
place at the first eigenvalue specified by Eq. (7.299) (Azikov & Vasiliev, 1992; Qatu &
Leissa, 1993).
526
CHAPTER 7 LAMINATED COMPOSITE PLATES
Consider the initial phase of the plate bending under compression assuming that the plate
deflection is small in comparison with the plate thickness. In accordance with Eq. (7.286), the
deflection of a simply supported plate can be taken in the form
w5
XX
m
wmn sinλm x sinλn y
(7.300)
n
Substitute this expansion into the first two equilibrium equations, Eq. (7.295), and determine u
and v from these equations. Taking into account Eq. (7.297) in which we can eliminate the plate
displacements as a solid body putting C1 5 C2 5 C3 5 0, we finally arrive at
XX
Tx B22
x1
wmn λm rmn cosλm x sinλn y
B
m
n
XX
Tx B12
y1
wmn λm tmn sinλm x cosλn y
v5
B
m
n
u52
where
! #
λ2n
2 C22 2 1 C4 B
λm
"
!
!
! #
1
λ2m
λ2n
λ2m
C22 1 C 2
B11 1 B44 2 2 C11 2 1 C4 B
tmn 5
B1
λn
λm
λn
1
rmn 5
B1
"
(7.301)
λ2
C11 1 C 2n
λm
in which
B1 5 B 1 B44
!
λ2
B22 1 B44 m2
λn
!
!
λ2n
λ2m
B22 2 1 B11 2 2 2B12 ; B 5 B11 B22 2 B212
λm
λn
C 5 C12 1 2C44 ; B 5 B12 1 B44
To determine the deflection coefficients wmn , apply the principle of minimum total potential
energy of the plate
1
P5
2
ða ðb (
00
"
#)
@u
@v
@u @v
@2 w
@2 w
@2 w
@u 1 @w 2
1 Ny 1 Nxy
1
2 Mx 2 2 My 2 2 2Mxy
1Tx
2
dxdy
Nx
@x
@y
@y
@x
@x
@y
@x@y
@x
2 @x
(7.302)
Note that in contrast to Eq. (7.272) which corresponds to the nonlinear problem, Eq. (7.302)
corresponds to the linearized problem and we must additionally take into account the work
performed by the force Tx on the plate shortening caused by deflection. Substituting the stress
resultants and couples from Eq. (7.282) into Eq. (7.302), using Eqs. (7.300) and (7.301) for displacements, and integrating, we arrive at the total potential energy P as a function of wmn : Applying
the minimum condition @P=@wmn 5 0, we finally get
wmn 5
Tx kmn
Tx 2 Tmn
(7.303)
7.7 GENERALLY LAMINATED PLATES
527
in which
kmn 5
1
λm
λn
; m; n 5 1; 3; 5; . . .
ðB12 C12 2 B22 C11 Þ
1 ðB12 C11 2 B22 C12 Þ
λn
λm
B1
and Tmn is specified by Eq. (7.299). Let the minimum value of Tmn ðm; nÞ be Te . Then, it follows
from Eq. (7.303) that the plate deflection increases infinitely while Tx approaches Te . Consider, for
example, a carbonepoxy three-layered plate with dimensions a 5 0.4 m and b 5 0.2 m composed
of layers with angles φ1 5 0 , φ2 5 6 45 , and φ3 5 6 20 and equal thicknesses
h1 5 h2 5 h3 5 0:26 mm. The dependencies of the plate shortening Δu 5 uðx 5 aÞ 2 uðx 5 0Þ on the
normalized load Tx =Te are shown in Fig. 7.49. Curve 1 corresponds to the foregoing linearized
solution, Eq. (7.303). As can be seen, Δu becomes infinitely high at Tx =Te 5 1, i.e., when the acting load becomes equal to the minimum eigenvalue. Clearly, this result is quite formal and does
not have any physical meaning for a load that is close to Te . Indeed, the solution, Eq. (7.303), is
obtained under the condition that the deflection is small in comparison to the plate thickness and is
not valid for large (moreover, infinite) deflection.
To describe large deflections, we must use the nonlinear equations presented in Section 7.6. For
unsymmetrically laminated plates, Eq. (7.269) are generalized as follows:
@w @2 w
@w @2 w
@w @2 w
1 B44
1B
50
2
@x @x
@y @x@y
@x @y2
2
2
2
@w @ w
@w @ w
@w @ w
1 B44
L2 ðu; v; wÞ 1 B22
1B
50
@y @y2
@x @x@y
@y @x2
L1 ðu; v; wÞ 1 B11
(7.304)
where the operators L1 and L2 are specified by Eq. (7.284).
Tx /Te
3.2
m=2
n =1
2.4
m =1
n =1
1.6
2
0.8
1
0
1
2
3
4
Δu ⋅10 −5 , m
FIGURE 7.49
Dependence of the plate shortening on the normalized in-plane load: linearized (1) and nonlinear (2) solutions.
528
CHAPTER 7 LAMINATED COMPOSITE PLATES
For square or close to square plates ( a=b # 1:75), the initial deflection can be approximated by
Eq. (7.270), according to which
w 5 wm sinλm x sinλ1 y
(7.305)
where, as earlier, λm 5 πm=a and λ1 5 π=b. For the initial loading (at least for Tx =Te , 1 in
Fig. 7.49), we take m 5 1. The problem we are going to consider is to identify the dependence
ΔuðTx =Te Þ which can be different from curve 1 shown in Fig. 7.49 and to study the stability of the
initial plate bending, i.e., the possibility of mode-jumping to the half-wave number m which is
higher than the initial value m 5 1 (Azikov & Vasiliev, 1992).
Substituting the deflection, Eq. (7.305), into Eq. (7.304), we can derive the following expressions for the in-plane displacements which generalize Eq. (7.271) for unsymmetrically laminated
plates:
x
1
B22 Tx 1 w2m λ2m 1 wm λm rm cosλm x cosλ1 y
B
8
2
w
1
B12 2
2 m
λ2m 2
λ1 2 λm cos2λ1 y sin2λm x
16 λm
B11
y
1 2 2
B12 Tx 2 wm λ1 1 wm λ1 tm sinλm x cosλ1 y
v5
B
8
w2m 1
B12 2
2
λ21 2
λm 2 λ1 cos2λm x sin2λ1 y
16 λ1
B22
u52
(7.306)
where rm 5 rmn ðm; n 5 1Þ and tm 5 tmn ðm; n 5 1Þ are given in notations to Eq. (7.301).
Substituting the displacements, Eqs. (7.305) and (7.306), into Eq. (7.302) for the total potential
energy, and integrating, we arrive at
P5
ab 2
1
1
1
ab
λm k1 wm 1 k2 w2m 1 k3 w3m 1 k4 w4m 2
B22 Tx2
4
2
3
4
2B
where
k1 5
k3 5
1
ab
(
12
16Tx
km ;
λ2m ab
!
B12 λ21
B11 λ2m
!
k2 5 Tm 2 Tx
ðC11 2 B11 rm Þ
λm
λ1
1 ðC12 2 B12 tm Þ
λ1
λm
λ21
B12
λ1
λm
2
ðC22 2 B22 tm Þ
1 ðC12 2 B12 rm Þ
B11
λm
λ1
λ2m
" #
λ1 λ2m
B12
λ21
B12
B12
B12
C11 2 C12
2 C22
1 2C12
1 2 C22 2 C12
2 C11
1
λm λ21
B22
B11
B11
B22
λm
!)
λm rm
λ31 tm
1
2B
B22 λ1
B11 λ3m
!
Bλ21
λ2m
λ21
1
k4 5
16 B22 λ21
B11 λ2m
1
(7.307)
7.7 GENERALLY LAMINATED PLATES
529
where km 5 kmn ðm; n 5 1Þ is given in notation to Eq. (7.303). The minimum condition,
@P=@wm 5 0, yields the following equation for wm :
k4 w3m 1 k3 w2m 1 k2 wm 1 k1 5 0
This equation has one real root which specifies the plate deflection.
The half-wave number m depends on the acting load Tx and can be found by comparing the
levels of the total potential energy P in Eq. (7.307) as demonstrated in Section 7.6.
For the carbonepoxy plate considered previously as an example, the nonlinear solution is
shown in Fig. 7.49 with line 2. As can be seen, the initial plate bending with m 5 1 is not affected
by the eigenvalue of the load Te , which, as noted, does not have any physical meaning. At the load
Tx 5 2:32Te the bending mode corresponding to m 5 1 changes to the mode corresponding to
m 5 2.
It should be taken into account that the behavior of in-plane loaded unsymmetrically laminated
plates depends on the coordinate of the reference surface e (see Fig. 7.46) to which the load Tx is
applied. Changing e, we actually move the load through the plate thickness and change the plate
bending conditions. Since the location of the reference surface is usually not known, it seems reasonable to take it in accordance with the condition C11 5 0, under which the bending-stretching
coupling disappears in the direction of plate loading. It follows from Eq. (7.290) that the coordinate
of the reference surface can be calculated in this case as
I ð1Þ
e 5 11
ð0Þ
I11
The foregoing procedure allows us to determine the plate displacements. Then, applying
Eq. (7.268), we can find the strains and, using the procedure described in Section 3.11, obtain the
strains and the stresses in the principal coordinates of the plies. Applying the strength criteria discussed in Chapter 4, Failure Criteria and Strength of Laminates, we can evaluate the load-carrying
capacity of the plate.
As a numerical example, consider a square (a 5 b 5 0.2 m) carbonepoxy plate consisting of
four layers with the following parameters: φ1 5 0 , φ2 5 6 30 , φ3 5 6 60 , φ4 5 90 , and
h1 5 h2 5 h3 5 h4 5 0:26 mm. For this plate, the first eigenvalue of the load is Te 5 1:91 kN=m,
the maximum deflection is w 5 5.82 mm, and the ultimate load corresponding to the quadratic
failure criterion, Eq. (4.11), is Txu 5 75:9 kN=m. For a symmetrically laminated plate with the same
structure and thickness, the critical load is Txc 5 5:21 kN=m, the ultimate load is Txu 5 81:7 kN=m,
and the deflection at the ultimate load is w 5 6 5:66 mm. Thus, there is no significant difference
between the ultimate loads and the maximum deflections of unsymmetrically and symmetrically
laminated plates composed of the same layers.
7.7.3 SHEAR DEFORMABLE UNSYMMETRICALLY LAMINATED PLATES
Shear deformable unsymmetrically laminated plates are described by the most complicated equations of linear plate theory which allow for both bending-stretching coupling and transverse shear
530
CHAPTER 7 LAMINATED COMPOSITE PLATES
deformation. Such equations are required to analyze sandwich plates with different facing layers or
sandwich plates with edge ramps (see Fig. 7.47).
To derive the corresponding governing equations, we need to use the complete set of the
equations of plate theory presented in Section 7.1, i.e., equilibrium equations
@Nx
@Nxy
1
5 0;
@x
@y
@Mx
@Mxy
1
2 Vx 5 0;
@x
@y
@Ny
@Nxy
1
50
@y
@x
(7.308)
@My
@Mxy
1
2 Vy 5 0
@y
@x
(7.309)
@Vx
@Vy
1
1p 50
@x
@y
(7.310)
constitutive equations for orthotropic plates
Nx 5 B11 ε0x 1 B12 ε0y 1 C11 κx 1 C12 κy
Ny 5 B21 ε0x 1 B22 ε0y 1 C12 κx 1 C22 κy
(7.311)
Nxy 5 B44 γ 0xy 1 C44 κxy
Mx 5 C11 ε0x 1 C12 ε0y 1 D11 κx 1 D12 κy
My 5 C21 ε0x 1 C22 ε0y 1 D12 κx 1 D22 κy
Mxy 5 C44 γ 0xy 1 D44 κxy
Vx 5 Sx γ x ;
Vy 5 Sy γ y
(7.312)
(7.313)
and straindisplacements equations
@u
;
@x
ε0y 5
@θx
;
@x
κy 5
ε0x 5
κx 5
γ x 5 θx 1
@v
;
@y
γ 0xy 5
@u @v
1
@y
@x
(7.314)
@θy
;
@y
κxy 5
@θx
@θy
1
@y
@x
(7.315)
@w
@y
(7.316)
@w
;
@x
γ y 5 θy 1
To arrive at the minimum number of the governing equations, we apply the mixed formulation
of the problem. First, introduce the stress function Fðx; yÞ
Nx 5
@2 F
;
@y2
Ny 5
@2 F
;
@x2
Nxy 5 2
@2 F
@x@y
(7.317)
which allows us to satisfy identically the first two equilibrium equations, Eq. (7.308). To find the
stress function, we need to use the compatibility equation which follows from Eq. (7.314), i.e.,
@2 ε0y
@2 γ 0xy
@2 ε0x
1 2 2
50
2
@y
@x
@x@y
(7.318)
The strains that enter this equation are expressed using Eq. (7.311). Using Eqs. (7.317) and
(7.315), we finally get
7.7 GENERALLY LAMINATED PLATES
@2 F
@2 F
@θx
@θy
2 Bxy
2 bxy 2 2 Bx
@y2
@x
@x
@y
2
2
@ F
@ F
@θy
@θx
2 Byx
ε0y 5 by 2 2 byx 2 2 By
@x
@y
@y
@x
2
@
F
@θ
@θ
x
y
2c
1
γ 0xy 5 2 b
@x@y
@y
@x
531
ε0x 5 bx
(7.319)
where
bx 5
B22
;
B
by 5
B11
;
B
b5
1
;
Byy
1
ðB22 C11 2 B12 C12 Þ;
B
1
Bxy 5 ðB22 C12 2 B12 C22 Þ;
B
Bx 5
B12
; B 5 B11 B22 2 B212
B
Cyy
c5
B44
1
By 5 ðB11 C22 2 B12 C12 Þ
B
1
Byx 5 ðB11 C12 2 B12 C11 Þ
B
bxy 5 byx 5
Substituting the strains, Eq. (7.319), into the compatibility equation, Eq. (7.318), we arrive at
the first governing equation
@4 F
@4 F @4 F
@3 θx
1 b 2 2bxy
1 bx 4 2 Byx 3
4
2
2
@x
@x @y
@y
@x
@3 θx
@3 θx
@3 θy
2 Bxy 3 5 0
2 ðBx 2 cÞ
2 By 2 c
@x@y2
@x2 @y
@y
by
(7.320)
To proceed, substitute Eqs. (7.319) and (7.315) into the constitutive equations for the moments,
Eq. (7.312), to get
@θx
@θy
@2 F
@2 F
1 Dxy
1 cx 2 1 cxy 2
@y
@x
@x
@y
@θy
@θx
@2 F
@2 F
1 Dyx
1 cy 2 1 cyx 2
My 5 D y
@x
@y
@y
@x
@2 F
@θx
@θy
1D
1
Mxy 5 2 c
@x@y
@y
@x
Mx 5 D x
(7.321)
where
Dx 5 D11 2 Bx C11 2 Byx C12 ; Dy 5 D22 2 By C22 2 Bxy C12
Dxy 5 D12 2 Bxy C11 2 By C12 ; Dyx 5 D12 2 Byx C22 2 Bx C12
1 2
D5
B44 D44 2 C44
B44
byx
bxy
cx 5 bx C11 2
C12 ; cy 5 by C22 2
C12
bx
by
bxy
byx
C11 ; cyx 5 bx C12 2
C22
cxy 5 by C12 2
by
bx
The remaining three equations of the governing set follow from the three equilibrium equations,
Eqs. (7.309) and (7.310), if we substitute the moments from Eq. (7.321) and the transverse shear
forces from Eqs. (7.313) and (7.316), i.e.,
532
CHAPTER 7 LAMINATED COMPOSITE PLATES
Dx
@2 θy
@2 θx @2 θx
@3 F
1 D 2 1 cxy 3
1 D 1 Dxy
@x
@x2
@x@y
@y
3
@ F
@w
50
1 ðcx 2 cÞ
2 Sx θx 1
@x@y2
@x
@2 θx
@2 θy @2 θy
@3 F
1 D 1 Dyx
1 D 2 1 cyx 3
2
@y
@y
@x@y @x @3 F
@w
50
2 Sy θy 1
1 cy 2 c
@x2 @y
@y
@θx
@2 w
@θy
@2 w
Sx
1 2 1 Sy
1 2 1p 50
@x
@y
@x
@y
(7.322)
Dy
(7.323)
(7.324)
Thus, the theory reduces to four equations, Eqs. (7.320) and (7.322)(7.324), which include
four unknown functions: F, θx , θy , and w. These equations have a total tenth order with respect to x
and y variables.
For a simply supported plate with the boundary conditions at x 5 0 and x 5 a (see Fig. 7.1),
w 5 0;
Mx 5 0;
Nx 5 0;
v 5 0;
θy 5 0;
and at y 5 0 and y 5 b
w 5 0;
My 5 0;
Ny 5 0;
u 5 0 θx 5 0
the solution of the equations obtained can be presented in the form
F5
XX
m
θx 5
n
XX
m
XX
Fmn sinλm x sinλn y; w 5
m
θmn
x cosλm x sinλn y; θy 5
n
wmn sinλm x sinλn y
n
XX
m
θmn
y sinλm x cosλn y
(7.325)
n
where, as earlier, λm 5 πm=a and λn 5 πn=b. Expanding the pressure p in a similar series specified
by Eq. (7.147) and substituting Eq. (7.325) into Eqs. (7.320) and (7.322)(7.324), we arrive at the
following set of algebraic equations for the coefficients in Eq. (7.325):
mn
a11 Fmn 1 a12 θmn
x 1 a13 θy 1 a14 wmn 5 0
mn
a21 Fmn 1 a22 θx 1 a23 θmn
y 1 a24 wmn 5 0
mn
a31 Fmn 1 a32 θx 1 a33 θmn
y 1 a34 wmn 5 0
mn
a41 Fmn 1 a42 θmn
x 1 a43 θy 1 a44 wmn 1 pmn 5 0
where
a11 5 by λ4m 1 b 2 2bxy λ2m λ2n 1 bx λ4n ; a12 5 2 Byx λ3m 2 ðc 1 Bx Þλm λ2n
a13 5 2 Bxy λ3n 2 c 1 By λ2m λn ; a14 5 0
a21 5 2 cxy λ3m 2 ðcx 2 cÞλm λ2n ; a22 5 2 Dx λ2m 2 Dλ2n 2 Sx
a23 5 2 D 1 Dxy λm λn ; a24 5 a42 5 2 Sx λm
a31 5 2 cyx λ3n 2 cy 2 c λ2m λn ; a32 5 2 D 1 Dyx λm λn
a33 5 2 Dy λ2n 2 Dλ2m 2 Sy ;
a34 5 a43 5 2 Sy λn
a41 5 0; a44 5 2 Sx λ2m 2 Sy λ2n
(7.326)
7.8 OPTIMAL DESIGN OF LAMINATED COMPOSITE PLATES
533
The solution of Eq. (7.326) is
Fmn 5 2
pmn
½a12 ða23 a34 2 a24 a33 Þ 2 a13 ða22 a34 2 a24 a32 Þ
A1
pmn
½a11 ða23 a34 2 a24 a33 Þ 2 a13 ða21 a34 2 a24 a31 Þ
A1
pmn
½a11 ða22 a34 2 a24 a32 Þ 2 a12 ða21 a34 2 a24 a31 Þ
θmn
y 52
A1
A1
wmn 5 pmn
A2
θmn
x 5
where
A1 5 ½a11 ða22 a33 2 a23 a32 Þ 2 a12 ða21 a33 2 a23 a31 Þ 1 a13 ða21 a32 2 a22 a31 Þ
A2 5 a24 ½a11 ða32 a43 2 a33 a42 Þ 2 a31 ða13 a42 2 a12 a43 Þ
2a34 ½a11 ða22 a43 2 a23 a42 Þ 2 a21 ða13 a42 2 a12 a43 Þ
1a44 ½a11 ða22 a33 2 a23 a32 Þ 2 a12 ða21 a33 2 a31 a23 Þ 1 a13 ða21 a32 2 a22 a31 Þ
To study the in-plane loading under axial compression by forces Tx (see Fig. 7.28), we should
change the pressure p in Eq. (7.324) to the imaginary pressure
pn 5 2 Tx
@2 w
@x2
and apply the method described in Section 7.7.2.
7.8 OPTIMAL DESIGN OF LAMINATED COMPOSITE PLATES
Advanced composite materials are characterized by their high specific strength and stiffness and, in
combination with automatic manufacturing processes, make it possible to fabricate composite structures with high levels of weight and cost efficiency. The replacement of metal alloys by composite
materials, in general, reduces a structure’s mass by 20%30%. However, in some special cases,
the number of which progressively increases, the combination of material directional properties
with design concept utilizing these properties, being supported by the advantages of modern composite technology, provides a major improvement in the structural performance. Such efficiency is
demonstrated by composite structures of uniform strength in which the load is taken by uniformly
stressed fibers.
7.8.1 OPTIMAL FIBROUS STRUCTURES
To introduce composite structures of uniform strength, consider a laminated panel as shown in
Fig. 7.50 and loaded with in-plane forces Nx, Ny, and Nxy uniformly distributed along the panel
edges. Let the laminate consist of k unidirectional composite layers with thicknesses hi and fiber
orientation angles φi ði 5 1; 2; 3; :::; kÞ. For a plane stress state, the stacking-sequence of the
layers is not important.
534
CHAPTER 7 LAMINATED COMPOSITE PLATES
hi
φi
h
1
2
Nxy
y
x
Nx
Nxy
Nx
FIGURE 7.50
A laminated plate in a plane state of stress.
To derive the optimality criterion specifying the best structure for the plate in Fig. 7.50, we
first use the simplest monotropic model of a unidirectional composite (see Section 1.3) assuming
that the forces Nx, Ny, and Nxy are taken by the fibers only. For a design analysis, this is a reasonable approach because the transverse and shear strengths of a unidirectional composite ply (stresses σ2 and τ 12 ) are much lower than the ply strength in the longitudinal direction (stress σ1 ).
Using Eq. (2.68) in which we put σ2 5 0 and τ 12 5 0, we can write the following equilibrium
equations relating the applied forces to the stresses σð1Þ
1 in the direction of the fibers of the ith
layer:
Nx 5
k
k
X
X
2
σðiÞ
σðiÞ
x hi 5
1 hi cos φi ;
i51
i51
i51
k
X
i51
k
X
i51
i51
k
k
X
X
σðiÞ
h
5
σ1ðiÞ hi sin2 φi ;
Ny 5
i
y
Nxy 5
τ ðiÞ
xy hi 5
(7.327)
σ1ðiÞ hi sinφi cosφi
The strain εðiÞ
1 in the fiber direction in the ith layer can be expressed in terms of strains in coordinates x, y with the aid of the first equation of Eq. (2.69). Using the constitutive equations for the
monotropic model of a ply, Eq. (1.61), we arrive at
ðiÞ
2
2
σðiÞ
1 5 E1 ε1 5 E1 ðεx cos φi 1 εy sin φi 1 γ xy sinφi cosφi Þ
(7.328)
where it is assumed that the layers are made of one and the same material.
Consider the design problem and stipulate, for example, that the best structure for the laminate
is that which provides the minimum total thickness
h5
k
X
hi
(7.329)
i51
for the given combination of loads. Thus, we should minimize the laminate thickness in
Eq. (7.329) subject to the constraints imposed by Eqs. (7.327) and (7.328). To solve this problem,
7.8 OPTIMAL DESIGN OF LAMINATED COMPOSITE PLATES
535
we can use the method of Lagrange multipliers, according to which we should introduce multipliers
λ and minimize the following augmented function:
!
!
k
k
k
X
X
X
ðiÞ
ðiÞ
2
2
L5
hi 1 λx Nx 2
σ1 hi cos φi 1 λy Ny 2
σ1 hi sin φi
i51
i51
i51
!
k
X
σðiÞ
1 λxy Nxy 2
1 hi sinφi cosφi
i51
k
X
λi ½σi 2 E1 ðεx cos2 φi 1 εy sin2 φi 1 γ xy sinφi cosφi Þ
1
i51
with respect to the design variables hi ; φi and multipliers λ, i.e.,
@L
5 0;
@hi
@L
5 0;
@φi
@L
@L
@L
@L
5
5
5
50
@λx
@λy
@λxy
@λi
(7.330)
Minimization with respect to λ gives, obviously, the constraints in Eqs. (7.327) and (7.328),
whereas the first two of Eq. (7.330) yield
2
2
σðiÞ
1 ðλx cos φi 1 λy sin φi 1 λxy sinφi cosφi Þ 5 1
(7.331)
hi σðiÞ
1 ½ðλy 2 λx Þsin2φi 1 λxy cos2φi 5 E1 λi ½ðεy 2 εx Þsin2φi 1 εxy cos2φi (7.332)
The solution of Eq. (7.332) is
λi
λx 5 E1 εx
hi σðiÞ
1
;
λy 5 E1 εy
λi
hi σðiÞ
1
;
λxy 5 E1 γ xy
λi
hi σðiÞ
1
;
These equations allow us to conclude that
λi
hi σðiÞ
1
5
λx
λy
λxy
1
5
5
5 2
c
E1 εx
E1 εy
E1 γ xy
where c is a constant. Substituting λx ; λy , and λxy from these equations into Eq. (7.331) and taking
into account Eq. (7.328), we have
2
2
ðσðiÞ
1 Þ 5c
(7.333)
This equation has two solutions: σðiÞ
1 5 6 c.
Consider the first case, i.e., σðiÞ
1 5 c. Adding the first two equations of Eq. (7.327) and taking
into account Eq. (7.329), we have
1
h 5 ðNx 1 Ny Þ
c
Obviously, the minimum value of h corresponds to c 5 σ1 , where σ1 is the ultimate stress.
Thus, the total thickness of the optimal plate is
h5
1
ðNx 1 Ny Þ
σ1
(7.334)
536
CHAPTER 7 LAMINATED COMPOSITE PLATES
Taking now σðiÞ
1 5 σ 1 in Eq. (7.327) and eliminating σ 1 with the aid of Eq. (7.334), we arrive at
the following two optimality conditions in terms of the design variables and acting forces:
k
X
hi ðNx sin2 φi 2 Ny cos2 φi Þ 5 0
(7.335)
hi ½ðNx 1 Ny Þsinφi cosφi 2 Nxy 5 0
(7.336)
i51
k
X
i51
Thus, 2k design variables, i.e., k values of hi and k values of φi, should satisfy these three equations, Eqs. (7.334)(7.336). All possible optimal laminates have the same total thickness in
Eq. (7.334). As follows from Eq. (7.328), the condition σðiÞ
1 5 σ1 is valid, in the general case, if
εx 5 εy 5 ε and γ xy 5 0. Applying Eq. (2.69) to determine the strains in the principal material coordinates of the layers, we arrive at the following result: ε1 5 ε2 5 ε and γ 12 5 0. This means that the
optimal laminate is the structure of uniform stress and strain in which the fibers in each layer coincide with the directions of principal strains. An important feature of the optimal laminate follows
from the last equation of Eq. (2.168) which yields φ0i 5 φi . Thus, the optimal angles do not change
under loading.
Introducing the new variables
hi 5
hi
;
h
ny 5
Ny
;
Nx
nxy 5
Nxy
;
Nx
λ5
1
1 1 ny
and taking into account that
k
X
hi 5 1
(7.337)
i51
we can transform Eqs. (7.334)(7.336) that specify the structural parameters of the optimal laminate to the following final form:
h5
k
X
hi cos2 φi 5 λ;
i51
Nx
λσ1
k
X
(7.338)
hi sin2 φi 5 λny
(7.339)
i51
k
X
hi sinφi cosφi 5 λnxy
(7.340)
i51
For uniaxial tension in the x-direction, we have ny 5 nxy 5 0; λ 5 1. Then, Eq. (7.339) yield
φi 5 0 (i 5 1, 2, 3,. . ., k) and Eq. (7.338) gives the obvious result h 5 Nx =σ.
To describe tension in two orthogonal directions x and y, we should put nxy 5 0. It follows from
Eq. (7.340) that the laminate structure in this case should be in-plane symmetric, i.e., each ply with
angle 1φi should be accompanied by a ply of the same thickness but with angle 2φi .
Consider, for example, uniform biaxial tension such that Nx 5 Ny 5 N , Nxy 5 0, ny 5 1, nxy 5 0,
and λ 5 0:5. For this case, Eqs. (7.338) and (7.339) yield
7.8 OPTIMAL DESIGN OF LAMINATED COMPOSITE PLATES
(A)
537
(B)
FIGURE 7.51
Cross-ply (A) and 6 45 angle-ply (B) optimal structures for uniform tension.
h5
2N
;
σ1
k
X
hi cos2φi 5 0
(7.341)
i51
The natural structure for this case corresponds to the cross-ply laminate for which k 5 2,
φ1 5 0 , and φ2 5 90 (Fig. 7.51A). Then, the second equation of Eq. (7.341) gives the evident
result h1 5 h2 .
Consider the first equation in Eq. (7.341), from which it follows that the total thickness
of the optimal laminate is twice the thickness of a metal plate of the same strength under the
same loading conditions. This result is expected because, in contrast to isotropic materials,
the monotropic layer can work in only one direction, i.e., along the fibers. So, we need to have
the 0 layer to take Nx 5 N and the same, but 90 layer to take Ny 5 N. From this we can
conclude that the directional character of a composite ply’s stiffness and strength is actually a
material shortcoming rather than an advantage. The real advantages of composite materials are
associated with their high specific strength provided by thin fibers (see Section 1.2.1), and if we
had isotropic materials with such high specific strength, no composites would be developed
and implemented.
Return now to the second equation of Eq. (7.341) which shows that, in addition to a cross-ply
laminate, there exists an infinite number of optimal structures. For example, this equation is satisfied for a symmetric 6 45 angle-ply laminate (Fig. 7.51B). Moreover, all the quasi-isotropic laminates discussed in Section 3.7 and listed in Table 3.3 satisfy the optimality conditions for uniform
tension.
Laminates reinforced with uniformly stressed fibers can exist under some restrictions imposed
on the acting forces Nx, Ny, and Nxy. Such restrictions follow from Eqs. (7.339) and (7.340) under
the conditions that hi $ 0, 0 # sin2 φi , cos2 φi # 1 and have the form
0 # λ # 1;
2
1
1
# λnxy #
2
2
In particular, Eqs. (7.339) and (7.340) do not describe the case of pure shear for which only the
shear stress resultant, Nxy, is not zero. This is quite reasonable because the strength condition
σðiÞ
1 5 σ 1 under which Eqs. (7.338)(7.340) are derived is not valid for shear inducing tension and
compression in angle-ply layers.
To study in-plane shear of the laminate, we should use both solutions of Eq. (7.333) and
assume that for some layers, e.g., with i 5 1, 2, 3, . . ., n 1, σðiÞ
1 5 σ1 , whereas for the
538
CHAPTER 7 LAMINATED COMPOSITE PLATES
other layers (i 5 n, n 1 1, n 1 2, . . ., k), σðiÞ
1 5 2 σ1 . Then, Eq. (7.327) can be reduced to the
following form:
Nx 1 Ny 5 σ1 ðh1 2 h2 Þ
n21
X
Nx 2 Ny 5 σ1
h1
i cos2φi 2
k
X
!
h2
i cos2φi
i5n
i51
n21
k
X
X
1
Nxy 5 σ1
h1
h2
i sin2φi 2
i sin2φi
2
i5n
i51
(7.342)
(7.343)
!
(7.344)
where
h1 5
n21
X
h1
i ;
h2 5
k
X
h2
i
i5n
i51
are the total thicknesses of the plies with tensile and compressive stresses in the fibers,
respectively.
For the case of pure shear ðNx 5 Ny 5 0Þ, Eqs. (7.342) and (7.343) yield h1 5 h2 and
φi 5 6 45 . Then, assuming that φi 51 45 for the layers with hi 5 h1
i , whereas φi 5 2 45 for the
layers with hi 5 h2
i , we get from Eq. (7.344)
h 5 h1 1 h2 5
2Nxy
σ1
The optimal laminate, as follows from the foregoing derivation, corresponds to a 6 45 angleply structure as shown in Fig. 7.51B.
7.8.2 COMPOSITE LAMINATES OF UNIFORM STRENGTH
Consider again the plate in Fig. 7.50 and assume that the unidirectional plies or fabric layers that
form the panel are orthotropic, i.e., in contrast to the previous section, we do not now neglect stresses σ2 and τ 12 in comparison with σ1 (see Fig. 1.29). Then, the constitutive equations for the panel
in a plane stress state are specified by the first three equations in Eq. (3.35), i.e.,
Nx 5 B11 εx 1 B12 εy 1 B14 γ xy
Ny 5 B21 εx 1 B22 εy 1 B24 γ xy
Nxy 5 B41 εx 1 B42 εy 1 B44 γ xy
where, in accordance with Eqs. (2.72), (3.28), and (3.42)
(7.345)
7.8 OPTIMAL DESIGN OF LAMINATED COMPOSITE PLATES
B11 5
539
k
X
ðiÞ
ðiÞ
ðiÞ
hi ðE1 cos4 φi 1 E2 sin4 φi 1 2E12
sin2 φi cos2 φi Þ
i51
B12 5 B21 5
k
X
ðiÞ
ðiÞ
ðiÞ
ðiÞ
2
2
hi ½E1 ν ðiÞ
12 1 ðE1 1 E2 2 2E12 Þsin φi cos φi i51
k
X
ðiÞ
ðiÞ
ðiÞ
hi ðE1 sin4 φi 1 E2 cos4 φi 1 2E12
sin2 φi cos2 φi Þ
B22 5
i51
k
X
ðiÞ
ðiÞ
ðiÞ
hi ðE1 cos2 φi 2 E2 sin2 φi 2 E12
cos2φi Þsinφi cosφi
B14 5 B41 5
(7.346)
i51
k
X
ðiÞ
ðiÞ
ðiÞ
B24 5 B42 5
hi ðE1 sin2 φi 2 E2 cos2 φi 1 E12
cos2φi Þsinφi cosφi
i51
k
X
ðiÞ
ðiÞ
ðiÞ
ðiÞ
2
2
2
B44 5
hi ½ðE1 1 E2 2 2E1 ν ðiÞ
12 Þsin φi cos φi 1 G12 cos 2φi i51
ðiÞ
E1;2
ðiÞ
ðiÞ
ðiÞ
ðiÞ
and E1;2 5
; E12
5 E1 ν ðiÞ
12 1 2G12 :
1 2 ν ðiÞ ν ðiÞ
12 21
In the general case, the plate can consist of layers made of different composite materials.
Using the optimality criterion developed in the previous section for fibrous structures, we
consider that the fibers in each layer are directed along the lines of principal strains, or
ðiÞ
ðiÞ
principal stresses, because τ ðiÞ
12 5 G12 γ 12 for an orthotropic layer and the condition γ 12 5 0 is
ðiÞ
equivalent to the condition τ 12 5 0. Using the third equation in Eq. (2.69), we can write these
conditions as
2ðεy 2 εx Þsinφi cosφi 1 γ xy cos2φi 5 0
(7.347)
This equation can be satisfied for all the layers if we take
εx 5 εy 5 ε;
γ xy 5 0
(7.348)
Then, Eq. (7.345) yield
Nx 5 ðB11 1 B12 Þε; Ny 5 ðB21 1 B22 Þε;
Nxy 5 ðB41 1 B42 Þε
These equations allow us to find the strain, i.e.,
ε5
Nx 1 Ny
B11 1 2B12 1 B22
and to write two relationships specifying the optimal structural parameters of the laminate
ðB11 1 B12 ÞNy 2 ðB21 1 B22 ÞNx 5 0
ðB41 1 B42 ÞðNx 1 Ny Þ 2 ðB11 1 2B12 1 B22 ÞNxy 5 0
(7.349)
540
CHAPTER 7 LAMINATED COMPOSITE PLATES
Substitution of Bmn from Eq. (7.346) results in the following explicit form of these conditions:
k
X
ðiÞ
2
2
hi ½E1 ð1 1 ν ðiÞ
12 ÞðNx sin φi 2 Ny cos φi Þ
i51
ðiÞ
2
2
1 E2 ð1 1 ν ðiÞ
21 ÞðNx cos φi 2 Ny sin φi Þ 5 0
k
X
ðiÞ
ðiÞ
hi fðNx 1 Ny ÞðE1 2 E2 Þsinφi cosφi
i51
ðiÞ
ðiÞ
ðiÞ
2 Nxy ½E1 ð1 1 ν ðiÞ
12 Þ 1 E 2 ð1 1 ν 21 Þg 5 0
(7.350)
To determine the stresses that act in the optimal laminate, we use Eqs. (2.69) and (7.348) that
specify the strains in the principal material coordinates of the layers as ε1 5 ε2 5 ε; γ 12 5 0.
Applying the constitutive equations, Eq. (2.56), substituting ε from Eq. (7.349), and writing the
result in the explicit form with the aid of Eq. (7.346), we arrive at
ðiÞ
E1
ðiÞ
ð1 1 ν 12
ÞðNx 1 Ny Þ
Si
ðiÞ
E2
ðiÞ
ð1 1 ν 21
ÞðNx 1 Ny Þ
σðiÞ
2 5
Si
ðiÞ
τ 12 5 0
σðiÞ
1 5
(7.351)
where
Si 5
k
X
ðiÞ
ðiÞ
ðiÞ
hi ½E1 ð1 1 ν ðiÞ
12 Þ 1 E2 ð1 1 ν 21 Þ
i51
is the laminate stiffness coefficient.
If all the layers are made from the same material, Eqs. (7.350) and (7.351) are simplified as
k
X
hi ½Nx sin2 φi 2 Ny cos2 φi 1 nðNx cos2 φi 2 Ny sin2 φi Þ 5 0
i51
(7.352)
k
X
hi ½mðNx 1 Ny Þsinφi cosφi 2 ð1 1 nÞNxy 5 0
i51
σðiÞ
1 5 σ1 5
Nx 1 Ny
;
hð1 1 nÞ
σðiÞ
2 5 σ2 5
nðNx 1 Ny Þ
;
hð1 1 nÞ
ðiÞ
τ 12
50
(7.353)
in which
n5
E2 ð1 1 ν 21 Þ
;
E1 ð1 1 ν 12 Þ
m5
E1 2 E2
;
E1 ð1 1 ν 12 Þ
h5
k
X
hi
i51
Laminates of uniform strength exist under the following restrictions:
n
Nx
1
#
;
#
1 1 n Nx 1 Ny
11n
Nxy 12n
N 1 N # 2ð1 1 nÞ
x
y
For the monotropic model of a unidirectional ply considered in the previous section, n 5 0,
m 5 1, and Eq. (7.352) reduce to Eqs. (7.335) and (7.336).
7.8 OPTIMAL DESIGN OF LAMINATED COMPOSITE PLATES
541
Table 7.6 Parameters of Typical Advanced Composites
FabricEpoxy Composites
UnidirectionalEpoxy Composites
Parameter
Glass
Carbon
Aramid
Glass
Carbon
Aramid
Boron
BoronAl
σ2 =σ1
n
0.99
0.85
0.99
1.0
0.83
1.0
0.022
0.28
0.025
0.1
0.012
0.072
0.054
0.11
0.108
0.7
To determine the thickness of the optimal laminate, we should use Eq. (7.353) in conjunction
with one of the strength criteria discussed in Chapter 4, Failure Criteria and Strength of Laminates.
For the simplest case, using the maximum stress criterion in Eq. (4.2), the thickness of the laminate
can be found from the following conditions: σ1 5 σ1 or σ2 5 σ2 , so that
h1 5
Nx 1 Ny
;
ð1 1 nÞσ1
h2 5
nðNx 1 Ny Þ
ð1 1 nÞσ2
(7.354)
Obviously, for the optimal structure, we would like to have h1 5 h2 . However, this can happen
only if material characteristics meet the following condition
σ2
E2 ð1 1 ν 21 Þ
5n5
E1 ð1 1 ν 12 Þ
σ1
(7.355)
The results of calculations for typical materials whose properties are listed in Tables 1.5 and 2.6
are presented in Table 7.6. As can be seen, Eq. (7.355) is approximately valid for fabric composites
with their stiffness and strength in the warp and fill directions (see Section 2.7) controlled by fibers
of the same type. However, for unidirectional polymeric and metal matrix composites, whose longitudinal stiffness and strength are governed by the fibers and transverse characteristics are determined by the matrix properties, σ2 =σ1 {n. In accordance with Eqs. (7.354), this means that
h1 {h2 , and the ratio h2/h1 varies from 12.7 for glass-epoxy to 2.04 for boron-epoxy composites.
Now, return to the discussion presented in Section 2.4.2 from which it follows that in laminated
composites transverse stresses σ2 reaching their ultimate value, σ2 , cause cracks in the matrix,
which do not result in failure of the laminate whose strength is controlled by the fibers. To describe
such a laminate with cracks in the matrix (provided the cracks are allowable for the structure under
design), we can use the monotropic model of the ply and, hence, the results of the optimization
presented in Section 7.8.1.
Consider again the optimality condition Eq. (7.347). As can be seen, this equation can be satisfied not only by strains in Eq. (7.348), but also if we take
tan2φi 5
γ xy
εx 2 εy
(7.356)
Since the left-hand side of this equation is a periodic function with period π, Eq. (7.356) determines two angles, i.e.,
γ xy
1
φ1 5 φ 5 tan21
;
2
εx 2 εy
φ2 5
π
1φ
2
(7.357)
542
CHAPTER 7 LAMINATED COMPOSITE PLATES
Table 7.7 Effective Elastic Constants of an Optimal Laminate
Property
Elastic modulus,
E (GPa)
Poisson’s ratio, ν
Glass
Epoxy
Carbon
Epoxy
Aramid
Epoxy
Boron
Epoxy
BoronAl
Carbon
Carbon
Al2O3Al
36.9
75.9
50.3
114.8
201.1
95.2
205.4
0.053
0.039
0.035
0.21
0.06
0.035
0.176
Thus, the optimal laminate consists of two layers, and the fibers in both layers are directed
along the lines of principal stresses. Suppose that the layers are made of the same composite material and have the same thickness, i.e., h1 5 h2 5 h=2, where h is the thickness of the laminate.
Then, using Eqs. (7.346) and (7.357), we can show that B11 5 B22 and B24 5 2 B14 for this laminate. After some transformation involving elimination of γ 0xy from the first two equations of
Eq. (7.345) with the aid of Eq. (7.356) and similar transformation of the third equation from which
ε0x and ε0y are eliminated, using again Eq. (7.356), we get
Nx 5 ðB11 1 B14 tan2φÞε0x 1 ðB12 2 B14 tan2φÞε0y
Ny 5 ðB12 2 B14 tan2φÞε0x 1 ðB11 1 B14 tan2φÞε0y
Nxy 5 ðB44 1 B14 cot2φÞγ 0xy
Upon substitution of coefficients Bmn from Eq. (7.346) we arrive at
i
hh
ðE1 1 E2 Þε0x 1 ðE1 ν 12 1 E2 ν 21 Þε0y
2
i
hh
Ny 5 ðE1 ν 12 1 E2 ν 21 Þε0x 1 ðE1 1 E2 Þε0y
2
h
Nxy 5 E1 ð1 2 ν 12 Þ 1 E2 ð1 2 ν 21 Þ γ 0xy
4
Nx 5
Introducing average stresses σx 5 Nx =h, σy 5 Ny =h, and τ xy 5 Nxy =h and solving these equations
for strains, we have
ε0x 5
1
ðσx 2 νσy Þ;
E
ε0y 5
1
ðσy 2 νσx Þ;
E
γ 0xy 5
τ xy
G
(7.358)
where
E5
1
E2 ð1 2 ν 212 Þ 1 E22 ð1 2 ν 221 Þ
2E1 E2 1 1
2ðE1 1 E2 Þ
1 2 ν 12 ν 21
ν5
E1 ν 12 1 E2 ν 21
;
E1 1 E2
G5
E
2ð1 1 νÞ
(7.359)
Changing strains for stresses in Eq. (7.357), we can write the expression for the optimal orientation angle as
1
2 τ xy
φ 5 tan21
σx 2 σy
2
(7.360)
It follows from Eq. (7.358) that a laminate consisting of two layers reinforced along the directions of principal stresses behaves like an isotropic layer, and Eq. (7.359) specify the elastic constants of the corresponding isotropic material. For typical advanced composites, these constants are
7.8 OPTIMAL DESIGN OF LAMINATED COMPOSITE PLATES
543
listed in Table 7.7 (the properties of the unidirectional plies are taken from Table 1.5). Comparing
the elastic moduli of the optimal laminates with those for quasi-isotropic materials (see Table 3.4),
we can see that for polymeric composites the characteristics of the first group of materials are about
40% higher than those for the second group. However, it should be emphasized that whereas the
properties of quasi-isotropic laminates are universal material constants, the optimal laminates demonstrate characteristics shown in Table 7.7 only if the orientation angles of the fibers are found
from Eq. (7.357) or (7.360) and correspond to a particular distribution of stresses σx ; σy ; and τ xy .
It can be noted from Table 7.7 that the modulus of a carbonepoxy laminate is close to the
modulus of aluminum, whereas the density of the composite material is lower by a factor of 1.7.
This is the theoretical weight saving factor that can be expected if we change from aluminum to
carbonepoxy composite in a thin-walled structure. Since the stiffness of both materials is approximately the same, to find the optimal orientation angles of the structure elements, we can substitute
into Eq. (7.360) the stresses acting in the aluminum prototype structure. A composite structure
designed in this way will have approximately the same stiffness as the prototype structure and, as a
rule, higher strength because carbon composites are stronger than aluminum alloys.
To evaluate the strength of the optimal laminate, we should substitute strains from Eq. (7.358)
into Eq. (2.69) and thence, these strains in the principal material coordinates of the layers, into constitutive equations, Eq. (4.56), that specify the stresses σ1 and σ2 ðτ 12 5 0Þ acting in the layers.
Applying the appropriate failure criterion (see Chapter 4: Failure Criteria and Strength of
Laminates), we can evaluate the laminate strength.
Using Table 7.7, we can see that boron-epoxy optimal laminates have approximately the same
stiffness as titanium (but are lighter by a factor of about 2). Boronaluminum can be used to
replace steel with a weight saving factor of about 3.
For preliminary evaluation, we can use a monotropic model of unidirectional plies neglecting
the stiffness and load-carrying capacity of the matrix. Then, Eq. (7.359) take the following
simple form:
E5
E1
;
2
ν 5 0;
G5
E1
4
(7.361)
As an example, consider an aluminum shear web with thickness h 5 2 mm, elastic constants
Ea 5 72 GPa and ν a 5 0:3, and density ρa 5 2:7 g=cm3 . This panel is loaded with a shear stress τ.
Its shear stiffness is Ba44 5 57:6 GPa mm and the mass of a unit surface is ma 5 5:4kg/m2. For the
composite panel, taking σx 5 σy 5 0 in Eq. (7.360) we have φ 5 453 . Thus, the composite panel
consists of 1453 and 2453 unidirectional layers of the same thickness. The total thickness of the
laminate is h 5 2 mm, i.e., the same as for an aluminum panel. Substituting E1 5 140 GPa and taking into account that ρ 5 1:55 g=cm3 for a carbonepoxy composite, which is chosen to substitute
for aluminum, we get Bc44 5 70 GPa mm and mc 5 3:1 kg/m3. The stresses acting in the fiber directions of the composite plies are σc1 5 6 2τ. Thus, the composite panel has a 21.5% higher stiffness
and its mass is only 57.4% of the mass of a metal panel. The composite panel also has higher
strength because the longitudinal strength of unidirectional carbonepoxy composite under tension
and compression is more than twice the shear strength of aluminum.
The potential performance of the composite structure under discussion can be enhanced if we
use different materials in the layers with angles φ1 and φ2 specified by Eq. (7.357). According to
the derivation of Obraztsov and Vasiliev (1989), the ratio of the layers’ thicknesses is
544
CHAPTER 7 LAMINATED COMPOSITE PLATES
ð1Þ
ð1Þ
1
2
h2
E 2 E2
5 1ð2Þ
ð2Þ
h1
E 2E
and the elastic constants in Eq. (7.359) are generalized as
ð1Þ ð2Þ
E5
ð1Þ ð2Þ
E
E E 2 E2 E 2
5 ð1Þ 1 1ð2Þ
ð1Þ
ð2Þ
1 2 ν2
E 1E 2E 2E
1
1
2
2
ð1Þ ð2Þ
ð1Þ ð2Þ
ð2Þ
E E ðν ð1Þ 1 ν ð2Þ Þ 2 E2 E2 ðν ð1Þ
21 1 ν 21 Þ
ν 5 1 1 12 ð1Þ 12ð2Þ
ð1Þ ð2Þ
E1 E1 2 E2 E 2
Superscripts 1 and 2 correspond to layers with orientation angles φ1 and φ2 , respectively.
7.8.3 OPTIMAL DESIGN OF LAMINATED PLATES
The foregoing sections of this chapter are concerned with composite plates of uniform strength
which can exist under special loading conditions only. In the general case, the optimal structure can be found using numerical methods. The optimization of composite materials and
structures using analytical and numerical methods has been discussed by Obraztsov, Vasiliev,
and Bunakov (1977), Obraztsov and Vasiliev (1989), Teters, Rikards, and Narusberg (1978),
Tsai (1987), Banichuk, Kobelev, and Rikards (1988), Narusberg and Teters (1988),
Kanibolotsky and Urzhumtsev (1989), Vasiliev and Gurdal (1999), and Gurdal, Haftka, and
Hajela (1999).
In their application to laminated composite structures, the numerical methods of optimization,
which have been widely developed and used, encounter serious problems. To demonstrate the
nature of these problems, consider the laminate in the general plane stress state shown in
Fig. 7.50. The laminate consists of k layers with unknown thicknesses hi and orientation angles φi
(i 5 1; 2; 3; :::; k). Thus, we have 2k unknown structural parameters and, moreover, the number
of layers k is not known either. As follows from Section 7.8.1, the optimal structure of a laminated composite is not unique. For example, two equations, Eqs. (7.335) and (7.336), link 2k
unknown structural parameters hi and φi and, being satisfied, provide an infinite number of optimal laminates with one and the same total thickness h. Applying a numerical method of optimization, we use an iterative procedure starting from an initial approximation of the design item and
improving it by the process of optimization. For a composite laminate with multiple optimal solutions, this process will converge to an occasional optimal design dependent on the initial approximation. The complete set of optimal solutions can be found by direct sorting of initial
approximations; however sorting of the design variables, i.e., thicknesses hi and angles φi, does
not require any specific optimization procedure, because the optimal parameters can be found by
direct matching of the laminate thickness h corresponding to the particular values of hi and φi.
Such an optimization process, proposed by Vasiliev and Khaziev (2009), is applied further to
optimize composite laminates.
For orthotropic laminates with symmetric structures, the constitutive equations, Eq. (7.345), can
be presented in the following form:
Nx 5 h B11 εx 1 B12 εy ;
Ny 5 h B12 εx 1 B22 εy ;
Nxy 5 hB44 γ xy
(7.362)
7.8 OPTIMAL DESIGN OF LAMINATED COMPOSITE PLATES
545
where
Bmn 5
k
X
hi AðiÞ
mn ;
i51
hi 5
hi
h
2
2
AðiÞ
AðiÞ
11 5 E2 1 R11 ci 1 R12 ci ;
22 5 E1 2 R21 ci 1 R12 ci
ðiÞ
ðiÞ
A12 5 E1 ν 12 1 R12 ci ð1 2 ci Þ; A44 5 G12 1 R12 ci ð1 2 ci Þ; ci 5 cos2 φi
R11 5 2ðE1 ν 12 2 E2 1 2G12 Þ; R12 5 ð1 2 2ν 12 ÞE1 1 E2 2 4G12 ;
R21 5 2½ð1 2 ν 12 ÞE1 2 2G12 Solving Eq. (7.362) for strains, we have
1
ðB22 Nx 2 B12 Ny Þ;
Bh
2
B 5 B11 B22 2 B12
εx 5
εy 5
1
ðB11 Ny 2 B12 Nx Þ;
Bh
γ xy 5
Nxy
B33
(7.363)
The strains in the principal coordinates of the ply 1, 2 (see Fig. 2.18) for plies with angles 6 φi
with respect to axis x (see Fig. 7.50) are specified by Eq. (2.69) which yield
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ci ð1 2 ci Þ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ε2i6 5 εx ð1 2 ci Þ 1 εy ci 7 γ xy ci ð1 2 ci Þ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
6
γ 12i
5 γ xy ð2ci 2 1Þ 6 2ðεy 2 εx Þ ci ð1 2 ci Þ
ε1i6 5 εx ci 1 εy ð1 2 ci Þ 6 γ xy
(7.364)
The stresses in plies with angles 6 φi follow from Hooke’s law, i.e.,
σ1i6 5 E1 ðε1i6 1 ν 12 ε2i6 Þ;
σ2i6 5 E2 ðε2i6 1 ν 21 ε1i6 Þ;
6
6
τ 12i
5 G12 γ 12i
(7.365)
Substituting the strains from Eq. (7.364) into Eq. (7.365), we arrive at the following expressions
for the stresses:
σ1i6 5
E1
ððν 12 B22 2 B12 ÞNx 1 ðB11 2 ν 12 B12 ÞNy
Bh
1 ðB22 1 B12 ÞNx 2 ðB11 1 B12 ÞNy ð1 2 ν 12 Þci 6 Nxy
σ2i6 5
E2
ððB22 2 ν 21 B12 ÞNx 1 ðν 21 B11 2 B12 ÞNy
Bh
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
B
ð1 2 ν 12 Þ ci ð1 2 ci ÞÞ
B33
(7.366)
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
B
2 ðB22 1 B12 ÞNx 2 ðB11 1 B12 ÞNy ð1 2 ν 21 Þci 7 Nxy
ð1 2 ν 21 Þ ci ð1 2 ci ÞÞ
B33
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
G12
B
6
Nxy
ð2ci 2 1Þ 7 2 ðB22 1 B12 ÞNx 2 ðB11 1 B12 ÞNy ci ð1 2 ci Þ
τ 12i 5
Bh
B44
Thus, the strains in the laminate loaded in accordance with Fig. 7.50 and the stresses in the plies
are specified, in the general case, by Eqs. (7.364) and (7.366).
Furthermore, we consider the optimization of symmetric laminates consisting of one, three and
five symmetrically arranged layers. Since the structure is symmetric, we have, respectively, k 5 1,
k 5 2, and k 5 3 (see Fig. 7.52). Each layer has an angle-ply 6 φi structure and is orthotropic.
7.8.3.1 Optimization under strength constraints
Since the signs of the stresses are not known, the ply strength is evaluated with the aid of the polynomial strength criterion valid both for tension and compression and given by Eq. (4.15), i.e.,
546
CHAPTER 7 LAMINATED COMPOSITE PLATES
h1
k =1
h2 / 2
h3 /2
h2 /2
h1
h1
h2 / 2
h2 /2
h3 /2
k =2
k =3
FIGURE 7.52
Laminates composed of one ðk 5 1Þ, three ðk 5 2Þ; and five ðk 5 3Þ layers.
σi1
i 2
1
1
1
1
ðσi1 Þ2
ðσi2 Þ2
τ 12
i
#1
1 2 2 1 σ2
1 2 2 1 1 2 1 1 2 1
σ1
σ1
σ2
σ2
σ1 σ1
σ2 σ2
τ 12
(7.367)
Substituting the stresses from Eq. (7.366) into Eq. (7.367), we arrive at the following strength
constraints:
ri
si
1 #1
h2
h
where
ði 5 1; 2; 3 ::: kÞ
2 (
1
ðν 12 B22 2B12 ÞNx 1ðB11 2ν 12 B12 ÞNy
ri 5 1 2 E 1
σ 1 σ1 B
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 ðB221B12 ÞNx2ðB111B12 ÞNy ð12ν 12 Þci 6Nxy B ð12ν 12 Þ ci ð12ci Þ
B44
2 (
1
ðB22 2ν 21 B12 ÞNx 1ðν 21 B11 2B12 ÞNy
1 1 2 E2
σ2 σ2 B
(7.368)
)2
)2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 ðB221B12 ÞNx2ðB111B12 ÞNy ð12ν 21 Þci 7Nxy B ð12ν 21 Þ ci ð12ci Þ
B44
2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2
1 G12
Nxy B ð2ci21Þ72 ðB221B12 ÞNx2ðB111B12 ÞNy ci ð12ci Þ
τ 12 B
B44 (
E1
1
1
ðν 12 B22 2B12 ÞNx 1ðB11 2ν 12 B12 ÞNy
si 5
2
σ1
σ2
B
1
1
)
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
B
1 ðB22 1B12 ÞNx 2ðB11 1B12 ÞNy ð12ν 12 Þci 6Nxy
ð12ν 12 Þ ci ð12ci Þ
B44
(
E2
1
1
ðB22 2ν 21 B12 ÞNx 1ðν 21 B11 2B12 ÞNy
2
1
σ1
σ2
B
2
2
)
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
B
2 ðB22 1B12 ÞNx 2ðB11 1B12 ÞNy ð12ν 21 Þci 7Nxy
ð12ν 21 Þ ci ð12ci Þ
B44
(7.369)
7.8 OPTIMAL DESIGN OF LAMINATED COMPOSITE PLATES
547
Thus, we need to determine hi , φi and the minimum value of the laminate thickness satisfying
the strength constraints given by Eq. (7.368) for all the layers. Note that the normalized layer thickness must satisfy the following condition:
k
X
hi 5 1
(7.370)
i51
To be specific, consider a carbonepoxy composite material with the following properties:
E1 5 143 GPa, E2 5 10 GPa, G12 5 6 GPa, ν 21 5 0:3 , σ1
σ2
1 5 2172 MPa,
1 5 1558 MPa,
1
2
σ2 5 54 MPa, σ2 5 186 MPa, and τ 12 5 87 MPa. To describe various loading cases, normalize the
acting loads as
Nx 5 NUN x ;
where
Ny 5 NUN y ;
Nxy 5 NUN xy
N x 1 N y 1 N xy 5 1
and N is an arbitrary factor which is preassigned here as the load causing failure of a 1mm thick
unidirectional composite layer under tension, i.e., N 5 2:172U106 N=m. The results of the calculations are presented in Table 7.8.
Consider first the laminate consisting of one layer (k 5 1 in Fig. 7.8). In accordance with
Eq. (7.370), h1 5 1 and we have only one design variable φ1. The strength constraint for i 5 1
becomes an equality from which it follows that the thickness that should be minimized with respect
to φ1 is
hð1Þ 5 h 5
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
s1 1 s21 1 4r1
2
(7.371)
Consider the case of uniaxial tension (Case 1 in Table 7.8). In contrast to the obvious solution
(φ1 5 0, h 5 1 mm) following from the maximum stress criterion (σ1 # σ1 ), we have φ1 5 6:49
and h 5 0:97 mm. The difference in results is attributed to the form of the strength criterion in
Eq. (7.367) which allows for the stress interaction. Indeed, if we plot the function hðφ1 Þ shown in
Fig. 7.53 for small angles φ1, we can conclude that the optimal angle is not zero.
Consider the case k 5 2, i.e., the sandwich laminate shown in Fig. 7.52. Taking into account
that in accordance with Eq. (7.370), h2 5 1 2 h1 and applying Eq. (7.368) for i 5 1 and i 5 2 as
equalities, we can find two values of the laminate thickness, i.e.,
hð1Þ 5
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
s1 1 s21 1 4r1 ;
2
hð2Þ 5
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
s2 1 s22 1 4r2
2
(7.372)
in which si ðh1 ; c1 ; c2 Þ and ri ðh1 ; c1 ; c2 Þ are specified by Eq. (7.369). Of the two thickness
values hð1Þ and hð2Þ we must select the largest and minimize it with respect to h1 ; c1 ; and c2 . The
results are presented in Table 7.8 (k 5 2). It follows from this table that for the loading cases 1, 3,
6, 7, and 9, calculation gives h1 5 1, h2 5 0 and the sandwich laminate reduces to the singlelayered structure. For the cases 4 and 5 for which an analytical solution can be found (see
Section 7.8.2), a number of optimal solutions can exist for k 5 2 and the laminate thickness is the
same as that for k 5 1. There are only four loading cases (2, 8, 10, and 11) for which the solution
corresponding to k 5 2 delivers a smaller thickness than that given by the solution for k 5 1.
Table 7.8 Optimal Structure of CarbonEpoxy Composite Panels Consisting of One k 5 1, Three k 5 2, and Five k 5 3 Layers
under Various Loading Conditions
Optimal Structural Parameters
k51
k52
k53
Loading Case N x , N y , N xy
φ
h; mm
h i ; φi
h; mm
h i ; φi
h; mm
1
6.49
0.97
0.97
0.97
0
1.39
φ1 5 0; φ2 5 90
1.36
1.36
45
5.25
5.25
5.25
45
3.29
h 1 5 0:85; h 2 5 0:15
3.29
1.31
1, 0, 0
h 1 5 0:95; h 2 5 0:05
2
2 1, 0, 0
3
0, 0, 1
4
φ1 5 40; φ2 5 85
φ1 5 50; φ2 5 5
0.5, 0.5, 0
h 1 5 h 2 5 0:5
φ1 5 φ; φ2 5 90 2 φ
45
1.31
3.29
3.29
h 1 5 0:85; h 2 5 0:15
5
2 0.5, 20.5, 0
φ1 5 40; φ2 5 85
φ1 5 50; φ2 5 0
1.31
h 1 5 0:65; h 2 5 0:35
φ1 5 30; φ2 5 80
φ1 5 35; φ2 5 65
φ1 5 55; φ2 5 25
φ1 5 60; φ2 5 10
1.31
h 1 5 0:55; h 2 5 0:45
φ1 5 20; φ2 5 80
φ1 5 70; φ2 5 10
1.31
h 1 5 h 2 5 0:5
φ1 5 φ; φ2 5 90 2 φ
1.31
Table 7.8 Optimal Structure of CarbonEpoxy Composite Panels Consisting of One k 5 1, Three k 5 2, and Five k 5 3 Layers
under Various Loading Conditions Continued
Optimal Structural Parameters
k51
k52
k53
Loading Case N x , N y , N xy
φ
h; mm
h i ; φi
h; mm
h i ; φi
h; mm
6
45
3.48
3.48
3.48
45
1.48
1.48
1.48
90
5.81
h 1 5 0:64; h 2 5 0:36
2.4
2.4
4.01
0.33, 0.33, 0.33
7
20.33, 20.33, 0.33
8
φ1 5 90; φ2 5 0
2 0.5, 0.5, 0
9
26.76
4.01
4.01
24.53
5.43
h 1 5 0:71; h 2 5 0:29
3.82
h 1 5 0:28; h 2 5 0:03
0.5, 0, 0.5
10
φ1 5 53:2; φ2 5 0
h 3 5 0:69
φ1 5 0; φ2 5 63:3
φ3 5 52:6
2 0.5, 0, 0.5
h 1 5 0:76; h 2 5 0:24
71.26
11
2 0.33, 0.33, 0.33
5.48
φ1 5 64:1; φ2 5 0
3.88
h 1 5 0:4; h 2 5 0:2
h 3 5 0:4
φ1 5 76:8; φ2 5 0
φ3 5 50:1
3.82
3.88
550
CHAPTER 7 LAMINATED COMPOSITE PLATES
h, mm
1.0
0.99
0.98
0.97
0.96
φ1
0.95
0
2
4
6
8
10
FIGURE 7.53
Dependence of the thickness on the layer angle.
For higher values of k, the solution by direct sorting of the structural parameters encounters
considerable computational problems. However, calculations show that by increasing the number
of layers we do not improve the structure. It follows from Table 7.8 that only two loading
cases (10 and 11) for which the solutions for k 5 3 do not degenerate to the solutions corresponding
to k 5 2 have been found. In both cases, the optimal thickness of laminate is the same for k 5 2
and k 5 3.
Thus, for the majority of applications, the optimal laminate structure can be found within the
class of symmetric sandwich laminates.
7.8.3.2 Optimization under strength and buckling constraints
Consider a laminate subjected to uniaxial compression with force Nx 5 2 Tx (Ny 5 Nxy 5 0) as in
Figs. 7.27 and 7.50. Applying the strength constraint only, we arrive at the result presented in
Table 7.8 (Case 1), i.e., to a single-layered structure with angle φ1 5 6:49 . However, the plate can
buckle under the critical load specified by Eq. (7.245) which can be written as
Txc 5 Kc
h3
b2
where
pffiffiffiffiffiffiffiffiffiffiffi
D3
Kc 5 2π D1 D2 1 1 pffiffiffiffiffiffiffiffiffiffiffi
D1 D2
(7.373)
!
2
where, as previously, D1 5 D11 , D2 5 D22 , D3 5 D12 1 2D44 and the normalized bending stiffness for symmetric laminates in Fig. 3.11 (see Section 3.4) are
7.8 OPTIMAL DESIGN OF LAMINATED COMPOSITE PLATES
551
Table 7.9 Optimal Structural Parameters of the Laminated Plates Subjected to Axial
Compression Under Strength and Buckling Constraints
a=b
Structural Parameters
k51
k52
1.0
1.25
1.5
1.75
2.0
24.2
24.8
24.9
22.9
24.8
φ1
6.2
3.6
6.4
12.7
6.5
12.5
6.4
3.4
6.4
3.6
φ2
45.7
50.0
44.4
40.9
45.7
h1
0.24
0.27
0.32
0.27
0.24
h2
h; mm
0.76
0.73
0.68
0.73
0.76
φ1
5.8
1.3
5.7
2.4
5.7
2.5
5.8
0.9
5.8
1.4
φ2
10.3
52.5
79.3
10.8
10.3
φ3
49.3
48.8
45.0
44.5
45.5
h1
0.08
0.21
0.14
0.08
0.08
h2
0.18
0.43
0.13
0.18
0.18
h3
h; mm
0.74
0.36
0.73
0.74
0.74
5.8
5.7
5.7
5.8
5.8
φ1
h; mm
k53
ð1Þ
k 5 1 Dmn 5
1
Amn
12
1 ð2Þ
2
2
A h2 ð3h1 1 3h1 h2 1 h2 Þ
12 mn
h
i
1
2
ð3Þ
ð2Þ
2
k 5 3 Dmn 5 Dmn 1 Að3Þ
mn h3 3ðh1 1h2 Þ 1 3ðh1 1 h2 Þh3 1 h3 Þ
12
ð2Þ
ð1Þ
k 5 2 Dmn 5 Dmn 1
If we apply only the buckling constraint for a single-layered plate, we get, in accordance
with Fig. 7.32 (curve 1) φ1 5 45 .
To apply both the strength and the buckling constraints we need to supplement the strength
constraints given by Eqs. (7.371) and (7.372) with the buckling constraints following from
Eq. (7.373), i.e.,
sffiffiffiffiffiffiffiffiffi
2
3 b Tx
;
h 5
Kc
ðcÞ
and select the maximum value of the laminate thickness (of hðiÞ and hðcÞ ) and minimize it with
respect to the structural parameters. As previously, minimization is performed by direct sorting of
hi and φi . For the carbonepoxy composite laminates considered in Section 7.8.3.1 and a plate
with width b 5 0:15 m and various aspect ratios a/b, the results of the optimization are presented in
Table 7.9. As can be seen, allowance for both constraints increases the angle of the single-layered
(k 5 1) structure in comparison with the design under strength constraints only, and reduces this
angle compared to the case where only the buckling constraint is imposed. In sandwich laminates
552
CHAPTER 7 LAMINATED COMPOSITE PLATES
r
dr
dθ
q
Vr
Mr
qr
Nr
Mθ
Nθ
p
FIGURE 7.54
Element of a circular plate.
(k 5 2), both constraints can be met by two layers and the laminate thickness can be reduced. If we
further increase the number of layers (k 5 3), the laminate thickness remains practically the same.
Thus, the results obtained allow us to conclude that the optimal laminate designed for compression
is close to the sandwich symmetric structure in which the middle layer provides the strength and its
reinforcement angle is close to 0, whereas the outer layers ensure buckling resistance and have an
angle for the reinforcement orientation close to 45 .
7.9 CIRCULAR COMPOSITE PLATES AND DISKS
7.9.1 GOVERNING EQUATIONS
Circular plates experience, as a rule, axisymmetric loading which is studied below. Equilibrium
equations providing the balance of the plate element shown in Fig. 7.54 have the following form (it
is assumed that forces, moments and loads depend on the radial coordinate only):
0
ðrNr Þ 2 Nθ 1 qr 5 0
0
ðrVr Þ 2 rp 5 0
0
ðrMr Þ 2 Mθ 2 rVr 5 0
(7.374)
(7.375)
(7.376)
0
in which ð. . .Þ 5 dð. . .Þ=dr and Nr , Mr , and Nθ , Mθ are the radial and the circumferential stress
resultants and bending moments, whereas Vr is the transverse shear force, and p 5 p 2 q (see
Fig. 7.54).
Let the material be cylindrically orthotropic. Then, the constitutive equations, Eq. (3.44),
become
Nr 5 B11 εr 1 B12 Eθ 1 C11 κr 1 C12 κθ ;
Nθ 5 B21 εr 1 B22 Eθ 1 C21 κr 1 C22 κθ
(7.377)
Mr 5 C11 εr 1 C12 Eθ 1 D11 κr 1 D12 κθ ;
Mθ 5 C21 εr 1 C22 Eθ 1 D21 κr 1 D22 κθ
(7.378)
Vr 5 S11 γ r
(7.379)
The stiffness coefficients in these equations are specified by Eqs. (3.45), (3.46), and (3.42). The
generalized strains are given by the straindisplacement equations which for the axisymmetric
case are
7.9 CIRCULAR COMPOSITE PLATES AND DISKS
553
R2
R1
p1
p2
FIGURE 7.55
Annular plate loaded with internal and external pressure.
0
εr 5 u ; εθ 5
0
κr 5 θr ; κθ 5
θr
;
r
u
r
(7.380)
0
θr 5 γ r 2 w
(7.381)
where u is the radial displacement, w is the normal deflection, θr is the rotation of the plate element in the radial direction and γ r is the transverse shear deformation. The set of 12 equations,
Eqs. (7.374)(7.381), includes 12 unknown functions of coordinate r, i.e., three forces Nr , Nθ , and
Vr ; two moments Mr and Mθ ; five generalized strains εr , εθ , κr , κθ , and γ r ; and two displacements
u and w. If these functions are found, we can determine the strains and the stresses in the ith layer
in accordance with equations presented in Section 3.11. For symmetric laminates, Cmn 5 0
in Eqs. (7.377) and (7.378) and we arrive at two mutually independent sets of equations,
i.e., Eqs. (7.374), (7.377), and (7.380) which describe in-plane deformation of the plate and
Eqs. (7.375), (7.376), (7.378), (7.379), and (7.381) which correspond to plate bending.
7.9.2 PLANE STRESS STATE OF SYMMETRICALLY LAMINATED PLATES
If the coupling stiffness coefficients Cmn are zero, the equations that describe the plate deformation
in the plate plane, Eqs. (7.374), (7.377), and (7.380) can be reduced to the following set:
0
ðrNr Þ 2 Nθ 1 qr 5 0
u
u
0
0
Nr 5 B11 u 1 B12 ; Nθ 5 B21 u 1 B22
r
r
(7.382)
(7.383)
Substituting Eq. (7.383) into Eq. (7.382), we arrive at the following equation for u:
00
0
ru 1 u 2
n2p
qr
B22
u1
5 0; n2p 5
r
B11
B11
(7.384)
The general solution of the equation is
u 5 C1 r np 1 C2 r2np 1 up
where up is a particular solution depending on the function qr ðrÞ.
(7.385)
554
CHAPTER 7 LAMINATED COMPOSITE PLATES
p
np < 1
np = 1
np >1
FIGURE 7.56
Stress distribution over the radius of a solid orthotropic plate.
Consider an annular plate loaded by internal and external pressure, so that Nr ðr 5 R1 Þ 5 2 p1 h
and Nr ðr 5 R2 Þ 5 2 p2 h (see Fig. 7.55). Determining C1 and C2 in Eq. (7.385) from these boundary
conditions and taking up 5 0, we get from Eq. (7.383)
Nr 5 F1 ðr Þ; Nθ 5 np F2 ðr Þ
where
(7.386)
"
#
n 11
r np 21 n 21
n 11 R2 np 11
h
p
p
p
F1;2 ðr Þ 5
p1 c
2 p2
6 p2 c
2 p1 c
1 2 c2np
R2
r
where c 5 R1 =R2 and h is the plate thickness. For a plate without a central hole, R1 5 0 and
c 5 0. Taking R2 5 R and p2 5 p, we can derive the following expressions for the stresses using
Eq. (7.386):
σr 5
r np 21
r np 21
Nr
Nθ
52p
5 2 pnp
; σθ 5
R
R
h
h
(7.387)
Qualitatively, the stress distribution over the plate radius is shown in Fig. 7.56. For an isotropic
plate, we have np 5 1 and σr 5 σθ 5 2 p. If np . 1, which means that the radial stiffness coefficient
B11 is less than the circumferential coefficient B22 , we have σr 5 σθ 5 0 at the plate center.
However, if B11 . B22 and np , 1, the stresses become infinitely high at the plate center. In the limiting case, assume that B22 5 0. Then, np 5 0 and
σr 5 2
pR
; σθ 5 0
r
(7.388)
Refer now to Section 6.7.5 in which it is stated that singular solutions do not have physical
sense and are the result of shortcomings in the mathematical models. Particularly, for the problem
under consideration, the equilibrium equation, Eq. (7.374) is derived under the condition that the
plate element in Fig. 7.54 is infinitely small. However, this equation is definitely not valid at
the plate center r 5 0 because the stress in Eq. (7.388) has no derivative at this point. Applying the
approach described in Section 6.7.5, assume that the element shown in Fig. 7.54 has small but finite
radial dimension as in Fig. 7.57. To shorten the derivation introduce a radial force fr ðr Þ 5 σr hr.
Then, the equilibrium equation, Eq. (7.382) can be presented as
7.9 CIRCULAR COMPOSITE PLATES AND DISKS
f
dθ
a /2
−
r
555
a /2
α
f r+
h
FIGURE 7.57
A plate element with finite radial dimension.
0
fr 5 0
(7.389)
We take qr 5 0 and σθ 5 0 in accordance with Eq. (7.388). Introduce a local coordinate α such
that 2 a=2 # α # a=2 (see Fig. 7.57) and expand fr ðrÞ into the Taylor series as
0
fr ðr; αÞ 5 fr ðr Þ 1 αfr ðr Þ 1
α2 0 0
α3 0 0 0
fr ðr Þ 1
f ðr Þ
2!
3! r
(7.390)
The equilibrium equation in the radial direction becomes fr1 2 fr2 5 0, where fr1 5 fr ðr; α 5 a=2)
and fr2 5 fr ðr; α 5 2 a=2). Using Eq. (7.390), we arrive at
0
Fr 5 0
(7.391)
a2 0 0
00
f 5 fr 2 s2 fr
24 r
(7.392)
where
Fr 5 fr 1
pffiffiffi
The second part of this equation is derived under the condition a 5 2i 6s discussed in
Section 6.7.5 (the parameter r in this condition is changed to s). Thus, the generalized equilibrium
equation, Eq. (7.397), is the same as the conventional equation, Eq. (7.389), however it includes the
generalized force F r instead of fr . Then the solution of Eq. (7.391) is the same as the solution of
Eq. (7.389), i.e., F r 5constant. Since the solution of Eq. (7.389) is f r 5 2 pRh, we have F r 5 2 pRh
(see Section 6.7.5). Then the actual force can be found from Eq. (7.392) which takes the form
00
fr 2 s2 fr 5 2 pRh
The general solution of this equation is
r
r
fr 5 C1 sinh 1 C2 cosh 2 pRh
s
s
and the radial stress can be presented as
σr 5
fr
1
r
r
R
5
C1 sinh 1 C2 cosh 2 p
rh rh
s
s
r
556
CHAPTER 7 LAMINATED COMPOSITE PLATES
To eliminate the stress singularity at r 5 0, we must take C2 5 pRh to get
σr 5
1
r
pR r
C1 sinh 1
cosh 2 1
rh
s
r
s
The constant C1 can be found from the boundary condition σr ðr 5 RÞ 5 p. Finally, we arrive at
σr 5
"
#
pR
r
coshðR=sÞ
r
sinh
cosh 2 1 2
r
s
s
sinh R=s
The stress at the plate center r 5 0
(7.393)
pR cosh R=s
σ0r 5 2
s sinh r=s
is finite. The parameter s can be found if we test a real plate, measure σr at some point and use
Eq. (7.393).
Composite materials have high specific (with respect to density) strength and are used in rotating disks in application to inertial accumulators of mechanical energy: flywheels (Portnov &
Tarnopolskii, 1982). For a disk rotating with an angular velocity ω, qr 5 ρhω2 r 2 , where ρ is the
material density and h is the disk thickness.
Usually composite disks have stiffness coefficients and thicknesses which depend on the radial
coordinate r. In such cases, Eqs. (7.382) and (7.383) can only be solved numerically. Taking
0
0
qr 5 ρhω2 r 2 in Eq. (7.382) and solving this equation and Eq. (7.383) for the derivatives Nr and u ,
we can derive the following equations:
0
Nr 5
1 B12
1
B2
2 1 Nr 1 2 B22 2 12 u 2 ρhω2 r
B11
r B11
r
0
u 5
1
B12
Nr 2
u
rB11
B11
which are suitable for numerical integration. Typical boundary conditions for rotating disks are
uðr 5 r0 Þ 5 0; Nr ðr 5 RÞ 5 0
(7.394)
where r0 is the radius of the central shaft and R is the external radius of the disk. The efficiency of
a numerical solution can be enhanced by a preliminary transformation of Eqs. (7.382) and (7.383)
(Vasiliev, 1982). Assume that the function Nr ðr Þ is known and integrate the first equation in
Eq. (7.383). Taking into account the first boundary condition in Eq. (7.394), we get
2f ðrÞ
ðr
u5e
r0
Nr f ðrÞ
e dr; f ðr Þ 5
B11
ðr
r0
B12 dr
rB11
(7.395)
The second equation in Eq. (7.383), transformed with the aid of the first equation, yields
Nθ 5
B12
B2 u
Nr 1 B22 2 12
B11
B11 r
(7.396)
Substituting Eqs. (7.395) and (7.396) into Eq. (7.382) and integrating it, we can find Nr as
1
C
Nr 5 F ðr; Nr Þ 1
r
r
(7.397)
7.9 CIRCULAR COMPOSITE PLATES AND DISKS
where
ðr
F ðr; Nr Þ 5
r0
B12
Nr dr 1
B11
557
0r
1
ðr ð
ðr
B212 1 2f ðrÞ @ Nr f ðrÞ A
2
e
B22 2
e dr dr 2 ω ρhr 2 dr
B11 r
B11
r0
r0
r0
The constant C can be found from the second boundary condition in Eq. (7.394), i.e.,
C 5 Fðr 5 R; Nr Þ
The integral equation, Eq. (7.397) obtained above can be solved numerically by the method of
successive approximations. Taking for an initial approximation Nr 5 0 in the right-hand part of
Eq. (7.397), we get for the first approximation
ω
Nrð1Þ 5
2
r
0R
1
ð
ðr
2
2
@ ρhr dr 2 ρhr drA
r0
r0
Substituting Nrð1Þ into the right-hand part of Eq. (7.397), we obtain Nrð2Þ and so on. The iterative
procedure converges rather rapidly. The displacement u and the stress resultant Nθ can be found
from Eqs. (7.393) and (7.396).
7.9.3 OPTIMAL FIBROUS STRUCTURE OF A SPINNING DISK
Consider the problem of the optimization of a spinning composite disk. Let the disk rotate around
its axis with an angular velocity ω be reinforced with fibers making angles 1 φ and 2 φ with the
radius as in Fig. 7.58 and find the optimal trajectories of the fibers. The radial Nr, and circumferential, Nβ, stress resultants are related to the stresses σ1 acting in the composite material along the
fibers by Eq. (2.68). Using the monotropic material model and putting σ2 5 0 and τ 12 5 0 in these
equations, we get
Nr 5 hσ1 cos2 φ;
Nβ 5 hσ1 sin2 φ
(7.398)
The equilibrium equation, Eq. (7.382) for the rotating disk is
ðrNr Þ0 2 Nθ 1 ρhω2 r 2 5 0
+φ
R
r0
ω
FIGURE 7.58
Fibers’ trajectories in the spinning disk.
r
−φ
(7.399)
558
CHAPTER 7 LAMINATED COMPOSITE PLATES
y
φ
w
w/cos φ
r
θ
x
FIGURE 7.59
To the calculation of the disk thickness.
The disk thickness can be found taking into account that the fibrous tapes that form the disk are
continuous. Assume that the number of tapes crossing a circle with radius r is equal to k, whereas
the tape width and thickness are w0 and δ. Then, the disk thickness can be found from the equation
following from Fig. 7.59
πrhðrÞ 5 kδ
w
cosφ
which yields
hðrÞ 5
kw0 δ
πrcosφ
(7.400)
For a disk of uniform strength, we take σ1 5 σ1 in which σ1 is the ultimate stress for the unidirectional composite under tension along the fibers. Correspondingly, we take ω 5 ω, where ω is the
ultimate angular velocity of the disk. Then, substituting Eq. (7.400) into Eqs. (7.398) and the calculated stress resultants Nr and Nβ into Eq. (7.399), we arrive at the following equation for the fiber
angle:
rφ0 sinφcosφ 1 sin2 φ 5
1
ρω2 r2
σ1
(7.401)
The solution of this equation must satisfy the boundary conditions. For a disk with radius R and
with a central opening of radius r0 as in Fig. 7.58, we must have Nr 5 0 at r 5 r0 and r 5 R. Taking
into account the first expression in Eq. (7.398), we arrive at the following boundary conditions for
Eq. (7.401):
φðr 5 r0 Þ 5
π
;
2
φðr 5 RÞ 5
π
2
(7.402)
Since Eq. (7.401) is of the first order, its solution can, in general, satisfy only one of these conditions. Using the second condition in Eq. (7.402), we get
R
sinφ 5
r
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
λ
r4
12
12 4
R
2
(7.403)
where
λ5
ρω2 R2
σ1
(7.404)
7.9 CIRCULAR COMPOSITE PLATES AND DISKS
559
Applying the first condition in Eq. (7.402), we arrive at the following equation specifying the
parameter λ:
λ5
2
2
1 1 r0 =R
(7.405)
In conjunction with Eq. (7.404), this result enables us to determine the maximum value of the
disk angular velocity, i.e.,
ω2 5
2σ1
ρR2 ð1 1 r 20 Þ
(7.406)
where r 0 5 r0 =R. It follows from Eq. (7.406) that the maximum value for the ultimate angular
velocity corresponds to r 0 5 0, i.e., to a disk without a central opening, for which Eq. (7.406)
reduces to
ω2m 5
2σ1
ρR2
(7.407)
Note that relatively small central openings have practically no effect on the ultimate angular
velocity. For example, for r 0 5 0:1, Eq. (7.406) gives ω which is only 0.005% less than the maximum value ωm following from Eq. (7.407).
For further analysis, we take r 0 5 0 and consider disks without a central opening. Then,
Eq. (7.405) yields λ 5 2, and Eq. (7.403) for the fiber angle becomes
sinφ 5
r
R
(7.408)
To find the fiber trajectory, consider Fig. 7.60 showing the tape element in Cartesian x, y and
polar r, β coordinate frames. As follows from the figure,
x 5 rsinβ;
y 5 rcosβ;
tanφ 5
rdβ
dr
(7.409)
Applying the last equation of Eq. (7.409) and using Eq. (7.408) for φ, we arrive at the following
differential equation:
dβ
1
5 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
dr
R 2 r2
y
x
r
β
dβ
rdβ
φ
dr
y
x
FIGURE 7.60
A tape element in Cartesian and polar coordinate frames.
560
CHAPTER 7 LAMINATED COMPOSITE PLATES
y
x
FIGURE 7.61
Fiber patterns in the spinning optimal composite disk.
whose general solution is
sinðβ 2 β 0 Þ 5
r
R
(7.410)
in which β 0 is the constant of integration. Changing β and r to x and y with the aid of Eq. (7.409),
we can write Eq. (7.410) in Cartesian coordinates as
2 2
R
R
R2
x2 cosβ 0 1 y2 sinβ 0 5
2
2
4
(7.411)
For each value of β 0 , this equation specifies a circle with radius R=2 with its center located on
the circumference r 5 R=2. Changing β 0 , we get the system of circles shown in Fig. 7.61.
As has been noted, composite disks can be efficiently used as inertial accumulators of mechanical energy—flywheels. A disk with the fiber trajectories specified by Eq. (7.411) is shown in
Fig. 7.62. Note that the disk in Fig. 7.62 can be made using the technology described in
Section 2.5.2. The disk composite structure is made by winding onto an inflated elastic mandrel
having the shape of a shallow surface of revolution. After the shell, with the winding patterns calculated with aid of equations presented in Section 2.5.2 is fabricated, the pressure is continuously
reduced and the shell is compressed in the axial direction between two plates. Once the shell is
transformed into a disk, the resin in the composite material is cured.
The maximum kinetic energy that can be stored and the mass of the disk are
KE 5 πω2 ρ
ðR
hr 3 dr;
0
M 5 2πρ
ðR
hrdr
0
Substituting h in accordance with Eq. (4.400) and using Eq. (7.408) for the winding patterns,
we get
7.9 CIRCULAR COMPOSITE PLATES AND DISKS
561
FIGURE 7.62
Carbonepoxy flywheel with optimal structure
Table 7.10 Maximum Values of the Circumferential Velocities for Fibrous Composite Disks of
Uniform Strength
Composite Material
GlassEpoxy
CarbonEpoxy
AramidEpoxy
vR ; m/s
1309
1606
1946
KE 5
π 3 2
R ρω kw0 δ;
8
M 5 πRρ kw0 δ
(7.412)
where k, w0 , and δ are specified in notations to Eq. (7.400). Transforming Eq. (7.412) for KE with
the aid of the corresponding equation for M, we have
1
KE 5 MR2 ω2
8
Substituting ω from Eq. (7.407), we finally arrive at
KE 5
Mσ1
4ρ
Introducing the linear circumferential velocity at the outer circumference r 5 R of the disk as
vR 5 ωR and using Eq. (7.407) for ω, we obtain the following result
sffiffiffiffiffiffiffiffi
2σ1
vR 5
ρ
(7.413)
which means that vR depends only on the material’s longitudinal specific strength ðσ1 =ρÞ. The
results of calculations for the composites whose properties are listed in Table 1.5 are presented in
Table 7.10. Note that in order to use Eq. (7.413) for vR we must substitute σ1 in N=m2 , ρ in kg=m3
and take into account that 1 N 5 1 kg m =s2 .
562
CHAPTER 7 LAMINATED COMPOSITE PLATES
FIGURE 7.63
Composite disks with external rings.
Spinning disks of uniform strength were originally studied by Kyser (1965) and later by Bokov,
Vasiliev, and Portnov (1982). The effect of filament winding mosaic pattern on the stress state of
filament wound composite flywheel disk was studied by Uddin, Morozov, and Shankar (2014)
using finite element modeling and analysis.
Fig. 7.62 demonstrates the structure of the composite disk with a free external boundary. To
increase the energy capacity of the composite, they are designed with massive outer rings. Disks
with such structures are presented in Fig. 7.63.
7.9.4 BENDING OF PLATES WITH SYMMETRICALLY LAMINATED STRUCTURE
Consider the problem of plate bending under pressures p and q as shown in Fig. 7.54. For symmetrically laminated plates, the coupling stiffness coefficients Cmn 5 0 in Eqs. (7.377) and (7.378), and
the problem is reduced to the following set of equations:
0
0
ðrVr Þ 1 rp 5 0; ðrMr Þ 2 Mθ 2 rVr 5 0
0
Mr 5 D11 θr 1 D12
θr
0
θr
; Mθ 5 D21 θr 1 D22
r
r
(4.414)
(7.415)
7.9 CIRCULAR COMPOSITE PLATES AND DISKS
0
Vr 5 S11 θr 1 w
563
(7.416)
Integrating the first equation in Eq. (7.414), we get
Vr 5
ð
1
C1 2 prdr
r
(7.417)
where C1 is the constant of integration. Substituting the moments from Eq. (7.415) and the shear
force from Eq. (7.417) into the second equilibrium equation in Eq. (7.414), we arrive at
00
0
rθr 1 θr 2
ð
1 2
1
nb θr 5
C1 2 prdr
r
D11
(7.418)
where
n2b 5
D22
D11
(7.419)
The general solution of Eq. (7.418) is
θr 5 C2 r nb 1 C3 r2nb 1 θp
(7.420)
where θp is the particular solution. For constant pressure p 5 p0 , Eq. (7.417) yields
Vr 5
C1
1
2 p0 r 2
2
r
(7.421)
and θp has the form
θp 5
p0 r3
C r
2
1
D11 n2b 2 1
2D11 n2b 2 9
As can be seen, this solution is not valid if nb 5 3 and nb 5 1. In these cases, the functions
θ 5 r 3 and θ 5 r coincide with the fundamental solutions of the homogeneous equation corresponding to Eq. (7.418). The case nb 5 1 corresponds, as follows from Eq. (7.419), to isotropic material
for which D11 5 D22 5 0 and the particular solution is
θp 5
C1 r
p0 r 3
lnr 2
2D
16D
θp 5 2
C1 r
p0 r 3
2
lnr
2D11
8D11
For nb 5 3, we get
The plate deflection w can be found from Eq. (7.416), i.e.,
0
w 5
ð
1
C1 2 prdr 2 θr
rS11
Substituting θr from Eq. (7.420) and integrating, we have
w5
ð ð
ð
C1
1 dr
C2 11nb
C3 12nb
lnr 2
prdr 2
r
2
r
2 θp dr 1 C4
S11
1 1 nb
1 2 nb
S11 r
(7.422)
The four integration constants in the solution in Eqs. (7.420)(7.422) can be found from the
boundary conditions. For a simply supported edge, we should put w 5 0 and Mr 5 0; for a clamped
564
CHAPTER 7 LAMINATED COMPOSITE PLATES
edge: w 5 0 and θr 5 0; and for a free edge: Vr 5 0, Mr 5 0. For a solid plate without a central
hole, the solution must exist at the plate center r 5 0 and, as follows from Eqs. (7.150) and (7.151)
we should take C1 5 0 and C3 5 0.
Conventional manufacturing processes (e.g., laying-up and winding) usually result in circular
plates whose stiffness coefficients and thicknesses depend on the radial coordinate r. To analyze
such plates, we can use the approximate method described in Section 7.9.2.
First, present Eq. (7.417) in the form
0
1
ðr
1@
V0 2 prdrA
Vr 5
r
(7.423)
r0
where r0 is the internal radius of the plate and V0 5 Vr ðr 5 r0 Þ: Then, assuming that we know the
moment Mr ðrÞ integrate the first equation in Eq. (7.415), i.e.,
0r
1
ðr
ð
M
D12
r ϕðrÞ
2ϕðrÞ @
A
θr 5 e
e dr 1 θ0 ; ϕðr Þ 5
dr
D11
rD11
r0
(7.424)
r0
0
where θ0 5 θr ðr 5 r0 Þ. Substituting θr from the first equation in Eq. (7.415) into the second of
these equations, we express Mθ as
Mθ 5
D12
D2 θr
Mr 1 D22 2 12
D11
D11 r
(7.425)
Substituting Eqs. (7.424) and (7.425) into Eq. (7.414) and integrating, we get
1
r0 M0
Mr 5 Φðr; Mr Þ 1 1 2
r
r
where M0 5 M ðr 5 r0 Þ and
ðr
D12
Φðr; Mr Þ 5
Mr dr 1
D11
r0
ðr 1θ0
r0
ðr r0
(7.426)
0r
1
ð
D212 1 2ϕðrÞ @ Mr ϕðrÞ A
e
D22 2
e dr dr
D11 r
D11
r0
ðr ðr
dr
e2ϕðrÞ 1 ðr 2 r0 ÞV0 2 dr prdr
D22 2
D11
r
D212
r0
r0
The plate deflection can be found from Eqs. (7.416), (7.423), and (7.424), i.e.,
ðr
w 5 V0
r0
0r
1
0r
1
ð
ð
ðr
ðr
dr
1 @
M
r
2
prdr Adr 2 e2ϕðrÞ @
eϕðrÞ drAdr
D11
rS11
rS11
r0
r0
ðr
r0
2ϕðr Þ
2θ0 e
r0
(7.427)
dr 1 w0
r0
where w0 5 wðr 5 r0 Þ.
To demonstrate the successive approximation procedure, consider a plate without a central hole
loaded with uniform pressure p 5 p0 for which r0 5 0, V0 5 0, and θ0 5 0. Take Mr0 5 0 as the initial approximation. Then, for the first approximation, Eq. (7.426) yields
7.9 CIRCULAR COMPOSITE PLATES AND DISKS
F
565
r
h
p(r )
Vr
Mr
R
FIGURE 7.64
A circular plate loaded through a body of revolution.
1
Mrð1Þ 5 M0 2 p0 r 2
6
Substituting this result into Eq. (7.426), we can determine Mrð2Þ and so on. Once Mr is found, we
can obtain θr from Eq. (7.424), Mθ from Eq. (7.425) and w from Eq. (7.427). The solution obtained,
Eqs. (7.423)(7.427), includes four constants of integration, i.e., V0 , θ0 , M0 , and w0 , that can be
found from the boundary conditions at the plate edges r 5 r0 and r 5 R.
To demonstrate the importance of the shear deformable plate theory, consider the contact problem for circular plates. Assume that the plate is loaded by a force F through an absolutely rigid
body of revolution as shown in Fig. 7.64 and determine the contact pressure pðrÞ acting between
the body and the plate. Consider first classical plate theory. Taking S11 -N in Eq. (7.416), we get
θr 5 2 w0 and Eq. (7.415) become
0
Mr 5 2ðD11 w00 1 D12
0
w
w
Þ; Mθ 5 2ðD22 1 D12 w00 Þ
r
r
(7.428)
Substitute Vr from the second equation in Eq. (7.14) into the first of these equations to get
00
0
ðrMr Þ 2 Mθ 1 rp 5 0
(7.429)
and p 5 p, q 5 0. Using Eq. (7.428), we can write Eq. (7.429) in terms of the plate deflection as
00
pr
000
0
r 3 wIV 1 2r 2 w 2 n2b rw 2 w 1
50
D11
(7.430)
where nb is specified by Eq. (7.419).
Consider the plate shown in Fig. 7.64 and assume that the body surface which is in contact
with the plate has a parabolic shape. Then, the plate deflection in the contact area 0 # r # R can be
presented as
w 5 w0 2
w0 2 wR 2
r
R2
(7.431)
in which w0 5 wðr 5 0Þ and wR 5 wðr 5 RÞ. Substituting Eq. (7.431) into Eq. (7.430), we get p 5 0.
Thus, according to classical plate theory, there is no pressure between the plate and the body.
For the contact area 0 # r # R, the bending moments can be found from Eqs. (7.428) and (7.431)
which yield
566
CHAPTER 7 LAMINATED COMPOSITE PLATES
Mr 5
2
ðD11 1 D12 Þðw0 2 wR Þ;
R2
Mθ 5
2
ðD22 1 D12 Þðw0 2 wR Þ
R2
Then, the equilibrium equations, Eq. (7.414), give (for p 5 0)
2
ðD11 2 D12 Þðw0 2 wR Þ
rR2
Vr 5
As can be seen, Vr becomes infinitely high at the plate center r 5 0. This result is evidently not
correct because, in accordance with the symmetry conditions, Vr must be zero at r 5 0.
Thus, classical plate theory does not provide the correct solution of a contact problem for a
circular orthotropic plate.
Now, apply the shear-deformable plate theory. Substituting the moments and the shear force
from Eqs. (7.415) and (7.416) into the equilibrium equations, Eq. (7.414), we arrive at
0 0
θr
S11 0
rθr 2 n2b 2
r θ1w 50
r
D11
0 0
0
rp
50
ðrθÞ 1 rw 1
D11
(7.432)
(7.433)
Integrating Eq. (7.433), we have
ðr
1
C
0
prdr 2 w 1
rS11
r
θ52
0
where C is the constant of integration which is obviously equal to zero. Substituting θ into
Eq. (7.432), we get
11
D11 0
w0
000
00
rp 5 D11 rw 1 w 2 n2b
S11
r
ðr
D11 2
nb 2 1
2
r S11
prdr 2
0
Using Eq. (7.431) for w, we finally obtain
11
D11 2
nb 2 1
2
r S11
ðr
prdr 2
D11 0
2D11 rp 5 2 2 1 2 n2b ðw0 2 wR Þ
S11
R
(7.434)
0
This integral-differential equation allows us to determine the contact pressure pðrÞ. To simplify
the further analysis, consider a transversely isotropic plate for which
D11 5 D22 5 D 5
Eh3
Eh
; nb 5 1; S11 5 S 5 Gh 5
2
12ð1 2 ν Þ
2ð1 1 νÞ
and Eq. (7.434) reduces to
ðr
prdr 2
D 0
rp 5 0
S
0
Differentiating this equation, we get
00
0
rp 1 p 2
S
pr 5 0
D
Introduce the following parameters (see Fig. 7.64):
(7.435)
7.9 CIRCULAR COMPOSITE PLATES AND DISKS
r5
567
r
SR2
6ð1 2 νÞR2
; λ2 5
5
R
D
h2
Then, Eq. (7.435) can be reduced to the Bessel equation
r2
d2 p
dp
2 r 2 λ2 p 5 0
1r
dr
dr 2
which has the following solution:
p 5 C1 I0 ðλr Þ 1 C2 K0 ðλrÞ
Since for r-0 the Macdonald function K0 -N, we must take C2 5 0. The constant of integration C1 can be found from the equilibrium condition for the body (see Fig. 7.64), i.e.,
ðR
F 5 2π prdr
0
Finally,
pðr Þ 5
FS
f ðrÞ
2πD
where f ðr Þ 5 I0 ðλrÞ=I1 ðλÞ. The function f ðr Þ is shown in Fig. 7.65 for R=h 5 5 and R=h 5 10. As
can be seen, the contact pressure is concentrated in the vicinity of the boundary of the contact area
r 5 1ðr 5 R; see Fig. 7.64).
7.9.5 PLATES WITH NONSYMMETRICALLY LAMINATED STRUCTURE
For plates with nonsymmetrically laminated structure (e.g., for sandwich plates with different facing layers), the coupling stiffness coefficients Cmn in Eqs. (7.377) and (7.378) are not zero and the
problems of the plate in-plane deformation and bending cannot be considered separately. The
behavior of such plates is described by the general equations, Eqs. (7.374)(7.381). To reduce this
set to one governing equation, integrate Eq. (7.375) and derive Eq. (7.417) for Vr . Then, substitute
Vr into the equilibrium equations, Eqs. (7.374) and (7.376), express the stress resultants and couples
with the aid of the constitutive equations, Eqs. (7.377) and (7.378) and substitute the strains from
the straindisplacement equations, Eqs. (7.380) and (7.381). As a result, we arrive at the following
two equilibrium equations which include u and θr , i.e.,
00
00
u
0 θr
0
B11 ru 1 u 2 B22 1 C11 rθr 1 θr 2 C22 1 qr 5 0
r
r
ð
00
00
0 θr
u
0
D11 rθr 1 θr 2 D22 1 C11 ru 1 u 2 C22 5 C1 2 pdr
r
r
(7.436)
Consider Eq. (7.12) for the stiffness coefficients. It follows from the second equation that we
can select the coordinate of the reference surface e in such a way that one of the coupling coeffið1Þ ð0Þ
cients Cmn becomes zero. According to this, take e 5 I11
=I11 , so that C11 5 0. Then, Eq. (7.436) are
simplified as
00
1
0
B11 ru 1 u 2 ðB22 u 1 C22 θr Þ 1 qr 5 0
r
(7.437)
568
CHAPTER 7 LAMINATED COMPOSITE PLATES
f (r )
1.2
1.0
0.8
0.6
0.4
R
=5
h
0.2
10
r
0
0.2
0.4
0.6
0.8
1.0
FIGURE 7.65
Function f ðrÞ for various ratios R/h.
00
0 1
D11 rθr 1 θr 2 ðD22 θr 1 C22 uÞ 5 C1 2
r
ð
pdr
(7.438)
Express θr from Eq. (7.437), i.e.,
θr 5
00
1
0
rB11 ru 1 u 2 B22 u 1 rqr
C22
(7.439)
and substitute it into Eq. (7.438) to get
00 0
000
r 3 uIV 1 6r2 u 1 7 2 n2p 2 n2b ru 1 1 2 n2p 2 n2b u
ð
u
00
0
1
1
1 n2p n2b 2 n4c
C1 2 prdr 2 r ðrqr Þ 2 ðrqr Þ 1 n2b qr 5 0
5
r
B11 D11
Here, np and nb are specified by Eqs. (7.384) and (7.419), whereas
(7.440)
7.9 CIRCULAR COMPOSITE PLATES AND DISKS
n4c 5
569
2
C22
B11 D11
(7.441)
Ai r ni 1 u0
(7.442)
The general solution of Eq. (7.440) is
u5
4
X
i51
where u0 is a particular solution, and ni are the roots of the following characteristic equation:
n4 2 n2p 1 n2b n2 1 n2p n2b 2 n4c 5 0
and have the form
1
n1;2;3;4 5 6
2
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2
np 1 nb 6
n2p 2n2b 1 4n4c
The rotation angle can now be found from Eq. (7.439), i.e.,
"
#
4
B11 X
1
0
ni
2
2
0
θr 5
Ai r ni 2 np 1 rðru0 Þ 2
ðB22 u0 2 rqr Þ
C22 i51
C22
(7.443)
and the plate deflection is found from Eqs. (7.416), (7.417), and (7.422), i.e.,
ð ð
1
dr
C1 lnr 2
prdr
S11
r
"
#
4
ð
X
B11
Ai 11ni 2
0
2
0
r
ni 2 np 2 rðru0 Þ dr
2
C22 i51 1 1 ni
ð
ð
B22
1
u0 dr 2
rqr dr 1 C2
2
C22
C22
w5
(7.444)
The solution obtained, Eqs. (7.442)(7.444), includes six constants of integration A1 A4 and
C1 , C2 which allows us to preassign u or Nr , w or Vr , and θr or Mr at the plate edge.
(A)
(B)
y
R
x
FIGURE 7.66
Circular plate orthotropic in cylindrical (A) and Cartesian (B) coordinates.
570
CHAPTER 7 LAMINATED COMPOSITE PLATES
7.9.6 CIRCULAR PLATES ORTHOTROPIC IN CARTESIAN COORDINATES
The previous sections are concerned with circular plates in which the material principal orthotropy
directions coincide with the axes of the cylindrical coordinates (see Fig. 7.66A). However, circular
plates can also have a structure which is orthotropic in Cartesian coordinates as shown in
Fig. 7.66B. The material structure of such plates is not axisymmetric, and the analysis is much
more complicated than that described in the previous sections. Nevertheless, for one particular
case, the solution is very simple (Lekhnitskii, 1957). Let the plate shown in Fig. 7.66B be clamped
at the edge r 5 R and loaded with a uniform pressure p0 . Within the framework of classical plate
theory, the plate deflection w satisfies Eq. (7.141) in which p 5 p0 . For the plate in Fig. 7.66B, take
2
x2 1y2
w 5 w0 12
R2
(7.445)
As can be readily proved, the deflection and its derivatives with respect to x and y are zero at
the plate contour whose equation is x2 1 y2 5 R2 . This means that the plate is clamped at the contour. The constant w0 in Eq. (7.437) can be found if we substitute Eq. (7.445) into Eq. (7.141). The
result is
w0 5
p0 R4
24D
in which D 5 D1 1 2=3 D3 1 D2 and D1 , D2 , and D3 are given by Eq. (7.142). The bending
moments and transverse shear forces in the plate can be found with the aid of Eq. (7.143).
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