3. 1. May- 2013 Micromechanics

7/28/2019 3. 1. May- 2013 Micromechanics

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1(L)

Figure 3.1b A unidirectionally fibre reinforced round bar

The in-plane elastic behaviour of a unidirectional lamina may be fully

described in terms of four basic lamina properties:

longitudinal modulus (E1=EL) transverse modulus (E2=ET) shear modulus (G12=GLT) the major Poissons ratio (12=LT)

Many failure theories for composite materials have been proposed up to

now. All theories can be expressed in terms of the basic strength parameters

referred to the principal material axes of the unidirectional lamina (Figure 3.2):

longitudinal tensile strength (fLt) longitudinal compressive strength (fLc) transverse tensile strength (fTt) transverse compressive strength (fTc) in-plane shear strength (fLTs)

The properties of a composite material depend on the properties of its

constituents and their distribution and physical and chemical interactions. These

properties can be determined by experimental measurements but one set of

experimental measurements determines the properties of a fibre-matrix system

produced by a single fabrication process. When any change in the system

variables occur, additional measurements are required. These experiments may

become time consuming and cost prohibitive, therefore a variety of methods

have been used to predict properties of composite materials (Agarwal et al, 2006

Daniel and Ishai, 2006). The mechanics of materials approach is based onmicromechanics.

In most of the composites literature micromechanics means the analysis

of the effective composite properties in terms of constituent material properties.

A complete definition of micromechanics (Lee 1989) is given below:

Micromechanics is a set of concepts, math-models and detailed studies

used to predict composite properties from constituent material properties,

stresses and strains, geometric configuration and fabrication process variables.

The unidirectional composite shows different properties in the material

axes directions. Thus, this type of composites are orthotropic with their axes1,2,3 as axes of symmetry (Figure 3.1a,b). These unidirectionally fibre

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reinforced composite elements have the strongest properties in the longitudinal

direction; material behaviour in the other two directions (2,3) is nearly identical

because of the random fibre distribution in the cross section. Therefore, a

unidirectional composite can be considered to be transverselyisotropic, that is,

it is isotropic in the 2-3 plane.

3.2 VOLUME AND MASS FRACTIONS

A key element in micromechanical analysis is the characterization of the

relative volume and/or weight content of the various constituent materials. The

mass fractions are easier to obtain during fabrication or using one of the

experimental methods after fabrication; the volume fractions are used in

micromechanics of composites. Therefore it is desirable to determine these

fractions and the relationships between the mass fractions and volume fractions.

Consider a volume vc of a composite material which consists of volume vf

of fibres and volume vm of the matrix material. The subscripts c,f and mrepresent the composite material, fibres, and the matrix material respectively.

74

Figure 3.2 Lamina loading schemes for basic strength parameters:

a) longitudinal tensile stress (fLt

);b) longitudinal compressive stress (fLc

);

c) transverse tensile stress (fTt

); d) transverse compressive stress (fTc

);

e) in-plane shear stress (fLTs

).

a. b.

1=

L

1=L

1=

L

1=L

c. d. e.

2=

T

2=

T

2=

T

2=

T

12

=LT

12

=LT

21

=

TL

21

=

TL


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Let us also considermc, mf and mm the corresponding masses of the composite,

fibres and the matrix material respectively. Let the volume fraction and the mass

fractions be denoted by V and M respectively. Assuming that no voids are

present in the composite the volume fractions and the mass fractions are defined

as follows:

mfc vvv += (3.1a)

c

f

fv

vV = and

c

mm

v

vV = (3.1b)

mfc mmm +=

(3.2a)

c

f

fm

mM = and

c

mmm

mM =

(3.2b)

The densitycof the composite can be obtained in terms of the densities

of the constituents (f andm) and their volume fractions or mass fractions. The

mass of a composite can be written as:

mmffcc vvv += (3.3)

Dividing both sides of Equation (3.3) by vc and using the definition for thevolume fractions, the following equation can be derived for the composite

material density:

mmffc VV += (3.4)

Equation (3.4) for a composite material with two constituents can be

generalized for a composite with n constituents:

==n

i

iic V1

(3.5)

The density of composite materials in terms of mass fractions can be

obtained as:

mmff

cMM

+

=1

(3.6)

which can also be generalized forn constituents:

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=

=n

i

ii

c

M1

1

(3.7)

Considering the definition of mass fractions and replacing the mass by the

product of density and volume, the conversion between the mass fractions andvolume fractions can be obtained:

m

c

mmf

c

f

f VMVM

== (3.8)

or, in general

i

c

ii VM

= (3.9)

The reverse relations are

m

m

cmf

f

cf MVMV

== (3.10)

and, in general

i

i

ci MV

= (3.11)

The composite density calculated theoretically from the mass fractionsmay not agree with the experimentally determined density. Assuming that the

theoretically calculated density isct and the experimentally determined density

isce the volume fraction of voids Vv is given by:

ct

cect

vV

= (3.12)

The void content may significantly influence some mechanical properties of a

composite material. A good composite must have less than 1% voids, whereas

a poorly made composite can have up to 5% void content (Agarwal and

Broutman, 1990). Higher void contents lead to increased scatter in strength

properties, to lower fatigue resistance and greater susceptibility to water

penetration.

For any number of constituent materials, n, the sum of the constituent

volume fractions must be unity:

=

=n

i

iV1

1 (3.13)

and, when the composite material consists of fibres, matrix and voids:

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1=++ gmf VVV (3.14)

The range of constituent volume fractions that may be expected in fibre

reinforced composites can be determined using representative area elements foridealized fibre-packing geometries such as the square and triangular arrays

shown in Figure 3.3. If it is assumed that the fibre spacing, s, and the fibre

diameter, d, do not change along the fibre length, then, the area fractions must

be equal to the volume fractions. The fibre volume fraction for the square array

is found by dividing the area of the fibre enclosed in the square by the total area

of square:

2

2

2

4

1

4

==s

d

s

dVf

(3.15)

The maximum theoretical fibre volume fraction occurs whens=d. In this case:

785.04

max ==

fV (3.16)

In case of a triangular array

2

32

=s

dVf

(3.17)

and, whens=d, the maximum fibre volume fraction is:

907.032

max ==

fV

(3.18)

These theoretical limits are not generally achievable in practice. In most

continuous fibre composites the fibre volume fractions range from 0.5 to 0.8.

77

Figure 3.3 Representative area elements for idealized fibre-

packing geometries: a) square; b) triangular

a. b.


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The four stages that can be identified during the composite loading are:

1-Both the fibres and the matrix deform in a linear elastic fashion.

2-The fibres continue to deform elastically but the matrix now deform

plastically.3-Both the fibres and the matrix deform plastically.

4-The fibres fracture followed by the composite failure.

Elementary mechanics of materials models have been adopted in the

elastic range, based on the following assumptions:

- A unidirectional composite may be modelled by assuming fibres to be

uniform in properties and diameter, continuous, and parallel throughout the

composite.

- It may be assumed that a perfect bonding exists at the interface, so that

no slip occurs between fibre and matrix materials.

- The fibre and matrix materials are assumed to be homogeneous and

linearly elastic. The matrix is assumed to be isotropic, but the fibre can be either

isotropic or orthotropic.

- Since it is assumed that the fibres remain parallel and that the

dimensions do not change along the length of the element, the area fractions

must equal the volume fractions.

Let us consider the model of the unidirectional composite shown in Figure 3.6.

Since no slippage occurs at the interface and the strains of fibre, matrix andcomposite are equal we can write:

1f = 1m = 1c (3.19)

in which subscriptsf, m and c refer to fibre, matrix and composite, respectively

and the second subscript refers to the direction.

For the model shown in Figure 3.6 the load (Pc=LAc) is shared between

the fibres (Pf=f1Af) and the matrix (Pm=m1Am).

79Figure 3.6 Model of FRP composite for predicting longitudinal behaviour

fibre

matrix

lc

L

L

cL

(2) T

(1) L


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Static equilibrium requires that the total force on the lamina cross section

must equal the sum of the forces acting on the fibre and matrix:

LAc=AAA

mffcc 111

+=m (3.20)

Since the area fractions are equal to the corresponding volume fractions,

Equation 3.20 can be rearranged to give an expression for the composite

longitudinal stress:

mmffcL VV +== 1 (3.21)

Equation (3.21) can be differentiated with respect to strain, which is the same

for the composite, fibres and matrix:

m

m

m

f

f

f

Lc

c Vd

dV

d

d

d

d

+

=

(3.22)

where (d/d) represents the slope of the corresponding stress-strain diagrams at

the given strain. If the stress-strain curves of the materials are linear, the slopes

(d/d) are constants and they can be replaced by the corresponding elastic

modulus in Equation (3.22). Thus:

mmffcLVEVEEE +==

11 (3.23)

Relationships (3.21) and (3.23) are known under the name rule of mixtures

indicating that the contributions of the fibres and the matrix to the composite

stress and elastic modulus respectively are proportional to their volume

fractions.

In Equation (3.23) it is assumed that the fibre can be anisotropic with

different properties in the longitudinal and transverse directions and that the

matrix is isotropic. For example aramid and carbon fibres are anisotropicwhereas glass is practically isotropic. Carbon and aramid fibres are orthotropic

and they have very different values of longitudinal modulus (E fl) and transverse

modulus, Eft. The ratio Efl/Eft =24 for Kevlar, 15.3 for high strength carbon and

65 for high modulus carbon (Gay et al 2007). The matrix modulus does not need

a second subscript. The rule of mixtures predictions for the longitudinal elastic

modulus is very close to the experimental results.

Equations (3.21) and (3.23) can be generalized forn constituents as:

( ) ==n

i

iic V1 (3.24)

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and

=

=n

i

iiL VEE1

(3.25)

Equation (3.23) can be also written as:

)1(1 fmffL

VEVEE += (3.23a)

and it predicts a linear variation of the longitudinal modulus EL with fibre

volume fraction as shown in Figure 3.7

3.3.2 Behaviour beyond initial deformation

The rule of mixtures accurately predicts the stress-strain behaviour of a

unidirectional composite subjected to longitudinal loads provided that Equation

(3.21) is used for the stress and Equation (3.22) for the slope of the stress-strain

curve. However, the replacement of slopes by the elastic moduli is possible only

when both constituents deform elastically. This may constitute only a small

portion of the stress-strain behaviour and is primarily applicable for glass fibre

reinforced thermosetting plastics. In general, the deformation of a fibre

reinforced composite may proceed in the four stages shown in Figure 3.4. Stage

2 usually occupies the largest portion of the composite stress-strain curve, and in

this period of development the matrix stress-strain curve is no longer linear, so

that the composite modulus must be predicted at each strain level by:

81

Figure 3.7 Variation of predicted EL

with

fibre volume fraction

EL

Em

Ef

0 0.25 0.5 0.75 1.0

Vf


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m

m

m

ffLV

d

dVEE

+=

1(3.26)

where (dm/dm) is the slope of stress-strain curve of the matrix at the

corresponding strain of the composite. The stress-strain diagrams forhypothetical composites with ductile and brittle fibres are shown in Figure 3.5. It

can be observed that the stress-strain curves of a composite fall between those of

the fibre and the matrix. However if the fibres are capable of deforming

plastically within the matrix, the fracture strain of fibres in the composite may

be larger than the fracture strain of the fibres working separately. Thus the

fracture strain of the composite may exceed that of the fibres, (Figure 3.5).

3.3.3 Longitudinal tensile strength

When a fibre reinforced composite is subjected to longitudinal tension the

constituent with the lower ultimate strain will fail first. Under assumption of

uniform strengths, two cases are distinguished depending on the relative

magnitudes of the ultimate strains of fibres and matrix. When the ultimate

tensile strain of the fibre is lower than that of the matrix (fu


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nonreinforced matrix, whereas Equation 3.27 predicts composite strength that

can be higher or lower than the matrix strength depending on the fibre volume

fraction. In Figure 3.9 the fibre volume fraction can be utmost 0.785 in case of

square fibre array and up to 0.907 for triangular fibre packing.

A critical fibre volume fraction, Vcrit, which must be exceeded forstrengthening can be defined as follows:

mtfmfftLt fVVff += )1( (3.30)

mft

mmt

critff

fVV

== (3.31a)

or

( )1( / )

/mt ft m fl cr

ft m fl

f f E EV

f E E

=

(3.31b)

The reinforcing fibres begin to act effectively only when Vf exceeds Vcrit,

a condition that is important to understand when composites are designed,

(Gerdeen and Rorrer, 2012).

In polymeric composites Vcrit and Vmin are very small because most polymers

exhibit only a limited amount of plastic flow and strain hardening. For example

in case of a glass-fibre reinforced epoxy composite Vmin would range between

0.25% and 1% a fibre volume fraction much lower than usual fibre content

(Agarwal et al, 2006).When the ultimate matrix tensile strain is lower than that of the fibre

(mu


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may break at a weak point, generating a nonuniform state of stress around the

fibre break (Figure 3.10). The effect of the fibre break on adjacent fibres is an

increase in fibre stress and in interfacial shear stress in the failure region . The

fibre stress increases from zero at the fibre end to the nominal far field value

after a characteristic value (l1) from the break. Thus the broken fibre becomesineffective on a length 2l1. Different failure mechanisms may develop,

depending on the properties of the constituents:

-transverse matrix cracking when composites have a brittle matrix and

strong interface (Figure 3.11a);

-fibre-matrix debonding in the case of weak interface and/or relatively

high ultimate fibre strain (Figure 3.11b);

-conical shear failures in matrix in the case of a relatively ductile matrix

and strong interface (Figure 3.11c).

As the longitudinal tension force increases, the fibre breaks increase in density,

these localized failure interact eventually leading to catastrophic fracture,

(Figure 3.12). In the case of brittle matrix composites, the ultimate strain of the

matrix is lower than that of the fibres and failure starts with the development of

multiple matrix cracks. These cracks produce local stress distributions and high

interfacial shear stresses, initiating fibre/matrix debonding and fibre breaks. The

failure mechanisms consist of transverse matrix cracks, fibre breaks and fibre

pull-out.

85

Figure 3.9 Variation of longitudinal tensile strength with

fibre volume fraction

Vf

Vcrit

Vmin

fmt

fLt

fmt

fft

fu

fft

)1( fmfftLt VVff +=

)1( fmtLt Vff =

Tensile

longitudinalstrength

Fibre volume fraction


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Figure 3.10 Stress distribution around a fibre break

Figure 3.11 Tensile failure mechanisms in a unidirectional composite

loaded in tension

3.3.4 Longitudinal compression

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When fibre reinforced composite materials are loaded in longitudinal

compression the models for tensile longitudinal strength cannot be used since

the failure of the composite is, in many cases, associated with microbuckling or

kinking of the fibre within the restraint of the matrix material.Accurate experimental values for the compressive strength are difficult to

be obtained and they are highly dependent on specimen geometry and the testing

method. There are three main longitudinal compression failure modes (Jones

1999):

-microbuckling of fibres in eitherextensionalorshearmode

-shear failure of fibres without buckling

-transverse tensile fracture due to Poisson strain

A fibre buckling in the matrix can be thought as a column on an elastic

foundation. For such a model the buckle wave length can be shown to be

proportional to the fibre diameter. Fibre buckling can also be caused by

shrinkage stresses developed during curing of the composite material. These

shrinkage stresses result from the matrix having a higher coefficient of thermal

expansion (CTE) than the fibres. Two modes of fibre buckling are possible in

the representative volume element, Figure 3.13. In the first mode the fibres can

buckle out of phase relative to one-another, to give the so calledextensional

buckling mode (Figure 3.13) in which the matrix is extended or compressed in

they direction.

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In the second mode, called the shear mode, the fibres buckle in phase with oneanother, Figure 3.13c, and the matrix shears in the x-y plane with all the shear

being relative to x axis. The fibres can be considered much stiffer than the

matrix (Gf>>Gm) and the fibre shear deformations may be neglected. Two-

dimensional models were used, with the fibres represented as plates separated by

matrix blocks.

The buckling load of a fibre surrounded by a supporting matrix material is

higher than if there is no matrix material around the fibre, since the lateral

support of the continuous matrix material has an effect similar to increasing the

number of discrete lateral supports for an Euler column.This effect leads to a buckling load:

EIl

mN

cr 2

22

= (3.33)

in whichEis the elastic modulus of the fibre material, Iis the fibre moment of

inertia, l is the fibre length, and m is the number of half sine waves (Figure

3.14). The buckling load depends on the number of lateral supports (m-1), and

the buckling load is much larger than ifm=1 (column without lateral support).

88

Figure 3.12 Failure evolution in unidirectional composite

under longitudinal tension

L

L

L

L

L

L

L

L


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To find the fibre buckling load in each buckling mode the energy method can be

utilised (Timoshenko and Gere 1961, Jones 1999). Using the energy method, the

work done by external forces (W) is equal to the corresponding change in strain

energy of fibres (Uf) plus the change in the strain energy of the matrix (

Um):

WUUmf=+ (3.34)

In the energy method the buckling loads are calculated with Equation (3.34)using deflection configurations approximated for the various buckle modes.

89

Figure 3.14 Buckling of an Euler column with discrete supports

a. b.

L

L

Ncr

Ncr

m=1 m=2 m=3

Figure 3.13 Modes of fibre buckling

a-representative volume element; b-extension mode; c-shear mode

a. b. c.

L

L

L

L

L

L

2c d

L

x

y


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Usually the unknown buckling transverse displacement can be represented by

the Fourier sine series. In case ofextension buckling mode a sinusoidal buckled

shape is assumed and the following formula can be developed for the fibre

critical stress:

)1(32

f

fmf

fcrV

EEV

= (3.35)

in which Vf is the fibre volume fraction.

The maximum compressive stress in the composite material is:

)1(3

2maxf

fmf

fLcc

V

EEVVf

== (3.36)

In Equation (3.35) it is assumed that the matrix is unstressed in the x-

direction. The strain at buckling in the x-direction can be calculated using the

expression ofcr from Equation (3.35):

f

m

f

f

fcrE

E

V

V

)1(32

= (3.37)

Assuming that the matrix has the same strain in the fibre direction as the fibre(m=fcr) the following equations can be written:

fcrmm E = (3.38)

fcrmffcrfcEVV )1(

max+= (3.39)

( ) fcrf

mffc

E

EVV

+= 1max (3.40)

( ))1(3

12max

f

fmf

f

m

ffcV

EEV

E

EVV

+= (3.41)

For high ratios Ef/Em the difference between Equations (3.36) and (3.41) are

insignificant. When the shear buckling mode occurs (Figure 3.13c) the fibre

displacements are equal and in phase with one another. The matrix material is

alternately sheared in one direction and then the other. It can be assumed that the

changes in deformations in the y direction are negligible and the shear strains

can be considered to be only a function of the fibre direction coordinate. The

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change in the strain energy can be expressed in terms of matrix shear properties

and the following formula for the fibre buckling stress is determined:

( )ffm

fcr

VV

G

=

1

(3.42)

and the maximum composite stress is:

f

mLcc

V

Gf

==1

max (3.43)

while the fibre strain at buckling is determined with:

= f

m

ff

crEG

VV )1(1 (3.44)

Equations (3.36) and (3.43)

( )ffmf

fLcV

EEVVf

=

132 and

f

mLc

V

Gf

=1

are plotted in Figure 3.15 for a glass-epoxy composite material. It can be noticed

that the shear mode has the minimum strength for the composite over a widerange of fibre volume fraction. It can also be observed that the extensional mode

governs the compressive strength for low fibre volume fractions (Vf=0.10.2)

and is not important for practical composites. The predicted strength should be

below the curve labelled elastic shear mode in Figure 3.15. The inelastic

shear mode curve is obtained by replacing the elastic matrix shear modulus in

Equation. (3.43) by a shear modulus that varies linearly from the elastic value at

1% strain to a zero value at 5% strain (Figure 3.16) but the predictions are still

too high. Reasonable predictions of compressive strength for graphite/epoxy

composites have been obtained on a model including the effects of materialnonlinearity and the effects of initial fibre curvature.

The predicted compressive strengths are now closer to the actual values

but they are still higher than the real ones. An explanation may be that the

analysis of the buckling problem has been performed two dimensionally instead

of the actual three-dimensional buckling problem. The influence of the matrix

shear modulus reduction due to inelastic deformation is illustrated in Figure

3.17. In this figure the composite material strain at buckling versus fibre volume

fraction is plotted.

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The results are calculated from Equation (3.44) for two values of the ratio

Ef/Gm and the matrix Poissons ratio 0.25. Again the shear mode governs the

material behaviour for the most range of the fibre volume fractions.

Experimental work carried out on boron/epoxy composite materials have shown

that theory agrees quite well with experimental data if the matrix moduli in

Equations (3.36) and (3.43) are multiplied by 0.63.

Another possible failure mode under longitudinal compression is the

failure of fibres in direct shear due to maximum shear stress. This occurs at an

angle =45o to the loading axis, Figure 3.18. At the highest values ofVf for wellaligned fibres pure compressive failure, which can be related to shear failure of

the fibres, may be encountered. In case of the shear mode governed by the shear

strength of the fibre, the predicted strength is:

92

Figure 3.16 Variation of shear

modulus with shear strain

1% 5%0

Gme

Matrix Gm

shear

modulus

Figure 3.15 Compressive strength of glass-epoxy composite materials

a-extension mode; b-shear mode

0.2 0.4 0.6 0.8 1.0

Lcf

L

L

a

b

elastic

inelastic

Vf


22/44

])1([2f

m

fffsLcE

EVVff += (3.45)

in which ffsis the shear strength of the fibres.

93

Figure 3.17 Compressive strain

at microbuckling for fibre

reinforced composite materials

0 0.2 0.4 0.6 0.8 1.0

1.0

0.5

0.1

0.05

50=m

f

G

E

100=m

f

G

E

extension

shear

fcr

Vf

Compressive

strain at

microbuckling

m=0.25


23/44

Good agreement with experimental data has been reported for graphite/epoxy

composite when the maximum shear stress is given by a rule of mixtures, so that

the compressive strength is:

)(2mLTsmfLTsfLc

VfVff +=

(3.46)

wherefLTsf andfLTsm are the shear strengths of fibre and matrix respectively in the

planeLT.

A model of failure under longitudinal compressive loading is based on

the transverse tensile fracture due to Poisson strains (Figure 3.19).

The longitudinal compressive stress L produces the longitudinal strain:

L

L

LE

= (3.47)

where EL is the longitudinal modulus of the composite in compression.

Under the compressive longitudinal stress, the transverse Poisson strain is:

==LLTT

L

L

LTE

(3.48)

Compressive failure of a unidirectional fibre reinforced composite loadedin the fibre direction may be caused by transverse splittingof the material.

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The transverse tensile strain resulting from the Poisson ratio effect can exceed

the ultimate strain capability of the composite resulting in longitudinal cracks at

the interface. At failure L is the ultimate compressive strength (fLc) such that

LTLequals the ultimate transverse tensile strain (Tu) of the composite:

Tu

L

L

LTE

= (3.49)

Tu

LT

LLc

Ef

= (3.50)

The ultimate transverse strain of the composite can be calculated from the

ultimate tensile strain (Agarwal and Broutman 2006) of the matrix (mu):

)1(3/1

fmuTuV= (3.51)

and the longitudinal compressive strength of the composite is:

)1(

)1)](1([ 3/1

fmff

muffmff

LcVV

VVEVEf

+

+=

(3.52)

Experimental results are in better agreement with predictions of Equation(3.52) than with the predictions based on microbuckling of fibres. The

95

Figure 3.18 Shear failure without fibre

buckling of a unidirectional compositeFigure 3.19 Transverse tensile rupture

due to Poisson strains

L

L

L

L


25/44

agreement is particularly good when mu=0.5% which is a reasonable value for

the epoxy resin.

A number of other factors have been shown to affect longitudinal

compressive strength, among them the fibre/matrix interfacial strength, in case

of transverse tensile rupture due to Poisson strain. Other experiments haveshown that the compressive strength of graphite/epoxy is strongly related to the

interfacial shear strength, which also depends on the fibre surface treatments.

3.4 TRANSVERSE STIFFNESS AND STRENGTH OF

UNIDIRECTIONAL COMPOSITES

3.4.1 Transverse modulus

The transverse modulus is a matrix-dominated property being sensitive to

the local state of stress. In the case of transverse loading the stateof stress in the

matrix surrounding the fibres is complex and more affected by interaction from

neighbouring fibres. Approaches leading to the mechanics of materials

predictions are based on simplified stress distributions and do not yield accurate

results. Let us consider a simple mathematical model shown in Figure 3.20. As

in the previous section the fibres are assumed to be uniform in properties and

diameter, continuous and parallel throughout the composite. The composite is

represented by a series model of matrix and fibre elements, and the main

assumption is that the stress is the same in the fibre and matrix. Both

constituents are assumed to be linear-elastic materials and the fibre-matrixbondis perfect. Considering the model made up of layers representing fibres and

matrix materials it is clear from Figure 3.20 that each layer has the same area on

which load acts, experiencing the same stress. The cumulative thicknesses of the

fibre and matrix layers are proportional to their respective volume fractions

because each layer is assumed to be uniform in thickness. In this case, the

composite transverse elongation ( cT ) is the sum of the fibre ( f ) and matrix (

m ) elongation respectively. The elongation of each constituent can be written

as the product of the strain and its cumulative thickness:

mmffccT

mmmTfffTccTcT

mTfTcT

lll

lll

+=

===

+=

;; (3.53)

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Since the dimensions of the representative volume element do not change

along the longitudinal direction, the length fractions must be equal to the volume

fractions:

tl

tlV

c

f

f =tl

tlV

c

m

m = (3.54)

Assuming the fibres and matrix to deform elastically and the stress is the same

in the fibre, matrix and composite, in the transverse direction, we can write:

( )m

m

mf

f

f

T

Tc VE

VEE

+= (3.55)

and:

=

TE

1

m

m

f

f

E

V

E

V+ (3.56)

which is the inverse rule of mixtures for the transverse modulus.

Equation (3.56) can be also written as:

mffm

mf

TVEVE

EEE

+= (3.57)

whereEf is the transverse modulus of the fibres.

For a composite made from n different constituents Equation (3.56) can be

generalised as :

97

fibre

Figure 3.20 Model of a unidirectional composite under transverse normal stress

matrix

lc

m

t

lf

lm

f

T

T


27/44

( )=

=n

i

ii

T

EV

E

1

1

(3.58)

The transverse modulus of a unidirectional composite as predicted byEquation (3.57) is plotted in Figure 3.21 as a function of fibre volume fraction.

The longitudinal modulus as predicted by the rule of mixtures it is also shown in

the same figure.

It can be noticed that the fibres do not contribute much to the transverse

modulus unless the fibre volume fraction is very high. This is in sharp contrast

to the effect of fibres on the longitudinal modulus. Theoretically, the transverse

modulus can be raised to seven times the matrix modulus by providing 90%

fibres, which is not practical.

The model utilised to determine the transverse modulus is not

mathematically rigorous. In a real composite the parallel fibres are dispersed in

the matrix material in a random fashion; generally both constituents will bepresent at any section perpendicular to the load, especially at the higher volume

fraction. Thus the load is shared between the fibres and the matrix and the

assumption that the stresses and the matrix are equal is inaccurate and the

mechanics of materials prediction underestimates the transverse modulus,(Kaw,

2006).

Halpin and Tsai developed (Halpin and Tsai 1967) semiempirical equations to

match the results of more exact micromechanics analyses.

98

m

TL

E

EorE

28

24

20

16

12

8

4

Em

Vf0 0.25 0.5 0.75 1.0

EL

ET

Ef=30Em

Figure 3.21 Variation of ELand E

Tas a function of fibre volume fraction


28/44

f

f

mTV

VEE

1

11

1

1

+=

(3.59)

where: (( )

1

1

1

+

=mf

mf

EE

EE(3.60)

and 1 is the reinforcing efficiency factor for transverse loading. Its valuedepends on the fibre geometry, packing geometry and loading conditions.

The prediction above tends to agree with experimental results for values of

1=1.0 to 2.0. For usual case of circular-section fibres, satisfactory results are

obtained by taking 1=2. Predictions of the Halpin-Tsai equations for transversemodulus of a unidirectional composite are shown as a function of fibre volumefraction in Figure 3.22 for different constituent modulus ratios.

The transverse modulus of a unidirectional composite is much smaller than its

longitudinal modulus. An increase in fibre volume fraction results in the

increase of transverse modulus similar to the longitudinal modulus, while anincrease in fibre modulus does not have a significant influence on the transverse

modulus. When =0, the Halpin-Tsai equation reduces to the inverse rule of

mixtures, whereas a value of= yields the rule of mixtures.

3.4.2 Transverse tensile strength

The transverse tensile loading is the most critical loading of a

unidirectional composite. Many factors influence the transverse tensile strength

and the most important are: the matrix strength, the fibre-matrix interfaceproperties, and defects in matrix such as microcraks and voids. In case of

99

12

8

4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

m

T

E

E

Vf

Figure 3.22 Predictions of Halpin-Tsai equation for transverse

modulus of a unidirectional composite

a

b

c

d

a.

b.

c.

d.


29/44

transverse loading, the high-modulus fibres act as effective constraints on the

deformation of the matrix, causing stress and strain concentrations in this

constituent and at the fibre-matrix interface, where the critical stresses and

strains usually occur (Gibson 1994). The maximum stress in the matrix for a

square fibre array is the axial stress at the interface along the loading direction,Figure 3.23.

The stress concentration factor (k) is defined as the ratio of the maximum

internal stress to the applied average stress. Its value depends on the relative

properties of the constituents and their volume fractions:

]/1[)/4(1

]/1[1

2/1fmf

fmf

EEV

EEV

k

= (3.61)

In Figure 3.24 the variation of the stress concentration factor for two

polymeric composites is illustrated. Knowing the value of k the composite

transverse strength,fTt, can be predicted dividing the tensile matrix strength, fmt,

to the stress concentration factor, k.

100

x

y

2=

T

2=

T

y= rx=

x=

Figure 3.23 Local stresses in transversely loaded

unidirectional composites


30/44

Another characteristic quantity for transversely loaded composites is the

strain concentration factor (k) (Daniel and Ishai 2006) defined as the ratio of

the maximum strain(max) to the average strain (2). A formula to calculate k isgiven below:

( ) ( )

( )m

mm

mE

Ekk

+

=

1

2112

2

max

(3.62)

where max and 2 are the maximum and average strains, respectively andm the matrix Poissons ratio. In the formula above it was assumed there is a

perfect bond between the constituents and the fibres are much stiffer than the

matrix. The strain concentration factor, also termed strain magnification factor

(Agarwal and Broutman, 2006) can also be determined with:

]/1[)/4(1

12/1

fmf EEVk

=

(3.63a)

A mechanics of materials approach based on the model shown in Figure 3.25

leads to another formula (Gibson 1994) for the strain concentration factor:

101

Figure 3.24 Stress concentration in matrix of unidirectional composites

with square fibre array under transverse tensile loading

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

3.5

3.0

2.5

2.0

1.5

1.0

0.5

Vf

Stressco

ncentration

factor,k

glass/epoxy

carbon/epoxy

2=

T

2=

T


31/44

11

1

+

=

f

m

E

E

s

dk

(3.63b)

where dis the fibre diameter ands the distance between the fibre centres.

In Figure 3.26 the variation of the strain concentration factor of a unidirectional

composite is illustrated. It can be seen that k increases sharply for fibre volume

fractions exceeding 0.5.

102

s

d

2

Figure 3.25 Mechanics of materials model for strain

concentration factor

2


32/44

The value of the transverse failure strain (cT) due to the strainconcentration in the matrix around the fibres is:

k

mTcT = (3.64)

where mT=m is the matrix tensile failure strain (matrix is assumedisotropic). If a linear behaviour to failure can be assumed, the corresponding

composite transverse strength can be determined with:

103

Figure 3.26 Variation of strain concentration factors (k

and k) of a

unidirectional composite with fibre volume fraction

s

d

2

2

0 0.2 0.4 0.6 0.8

20

16

12

8

4

Fibre volume fraction, Vf

Strainconcentrationcoefficient,k

ork

d


33/44

kE

fEEf

m

mtTTcTTt ==

(3.65)

An empirical approach (Nielsen 1974) for the prediction of transverse tensilestrength of fibrous composites is given below:

)1( 3/1fm

mtT

Tt VE

fEf =

(3.66)

The preceding equations above assume perfect adhesion between phases and

thus failure occurs by matrix fracture at or near the interface. A reduction

coefficient (Cv)to account for voids can be used to modify Equation (3.66). Cvcan be determined with:

)1(

41

f

vv

V

VC

=

(3.67)

where Vv is the void volume fraction.

Another empirical formula (Barbero 1999) can also account for voids:

+=

f

mffvmtTt

E

EVVCff 1)(1 (3.68)

The effect of voids is very detrimental to the transverse strength and this is

reflected by both empirical formulas. Although the results provided by these

formulas can be used for preliminary design, experimental data are usually

required if transverse strength is the controlling mode of failure of the

component. Failure of a transversely loaded composite is strongly influenced by

the residual stresses and strains caused by matrix curingor thermal stresses andstrains due to thermal expansion mismatch (Daniel and Ishai 2006). Assuming a

linear elastic behaviour of the matrix to failure, the maximum tensile stress or

strain failure criterion can be used to predict the tensile strength for a

unidirectional composite:

a) [ ]rmmtTt fk

f

=1

(3.70)

for the maximum tensile stress criterion and

104


34/44

b) )()21)(1(

1mrmmt

mm

m

Tt Efk

f

+

= (3.71)

for the maximum tensile strain criterion, where rm and rm are the maximum

residual stress and residual strain respectively.Failure of unidirectional composites subjected to transverse tensile loads occurs,

in most cases, because of matrix or interface tensile failure. In some cases they

may fail by fibre transverse tensile failure if the fibres are highly oriented and

weak in the transverse direction.

Therefore the unidirectional composite failure modes under transverse tensile

loads may be described as: matrix tensile failure, constituent debonding and

fibre splitting. Failure takes the form of isolated interfacial microcracks

increasing in number as load increases and finally coalescing into a catastrophic

macrocrack, Figure 3.27.

105

Figure 2.27 Failure stages of a unidirectional composite

under transverse tension

2

2

2

2

2

2

2

2


35/44

3.4.3 Transverse compression

A unidirectional fibre reinforced composite subjected to transverse

compressive load, Figure 3.28, may fail by shear failure of the matrix or matrixshear failure with constituent debonding and/or fibre crushing. The failure

modes are illustrated in Figure 3.29, where some portions of the failure surface

are created by constituent debonding.

The transverse compressive strength is lower than the longitudinal

compressive strength, but the comments made in the previous case are also

valid. For the most frequent failure mechanism the predicted transverse

compressive composite strength,fTc, is:

106

Figure 3.29 Failure of unidirectional

composite under compressive load

2

2

Figure 3.28 Unidirectional composite

under transverse compression

2

2


36/44

k

ff mcTc = (3.72)

where fmc is the compressive strength of the matrix and k is the stress

concentration factor discussed in the previous section. When the maximum

residual stresses at the interface, rm, are taken into account, Equation (3.72)becomes:

k

ff rmmcTc

+= (3.73)

Transverse compressive strength values are higher than tensile strength values

for both matrix and composite. Also the transverse compressive strength

increases with increase in the fibre volume fraction. This is explained by the

additional constraints placed on the matrix, preventing its deformation in thedirection perpendicular to the plane of load-fibre axes.

3.5 SHEAR STIFFNESS AND STRENGTH OF UNIDIRECTIONAL

COMPOSITES

3.5.1 In-plane shear modulus

The behaviour of unidirectional composites under in-plane shear loading

is dominated by the matrix properties and the local stress distributions. Themechanics of materials approach uses aseries model under uniform shear stress,

Figure 3.30 to determine the shear modulus. Using the notations shown in the

figure, the total shear deformation of the composite, c, is the sum of the shear

deformations of the fibre, f, and the matrix, m; each shear deformation can be

then expressed as the product of the corresponding shear strain (c, f, m) and the

cumulative widths of the material(lc, lf, lm):

107

Figure 3.30 a) Model of unidirectional composite for prediction of shear

modulus; b) shear deformations for constituents and for the model

t

lf

lmm

f

f

m

c

LT

TL

TL

LT

a. b.

L

T

lc

f

c

m


37/44

mfc += (3.74)mmffcc lll += (3.75)

Dividing both sides of Equation (3.75) by lc and recognising that the width

fraction is proportional to volume fractions, yields:

mmffc VV += (3.76)

Assuming linear shear stress-shear strain behaviour of fibres and matrix, the

shear strains can be replaced by the ratios of shear stress and the corresponding

shear modulus:

m

m

mf

f

f

c

LT

LT lG

lG

lG

+= (3.77)

where GLT is the in-plane shear modulus of the composite, Gf is the shear

modulus of fibres and Gm the shear modulus of matrix. But the shear stresses are

equal on composite, fibres and matrix and from Equation (3.77) we obtain:

m

m

fLT G

V

G

V

G

f +=1

(3.78)

or

mffm

mf

LTVGVG

GGG +=

(3.79)

Equation (3.79) can be rewritten as:

)/()1( fmff

mLT

GGVV

GG

+=

(3.79a)

and if the fibres are much stiffer than the matrix (Gf>>Gm) the in-plane shear

modulus can be approximated as:

f

mLT

V

GG

1 (3.79b)

As in the case of transverse modulus Equation (3.79) underestimates the values

of the in-plane shear modulus, and the Halpin-Tsai equations can be used to give

better predictions:

108


38/44

f

f

mLTV

VGG

2

22

1

1

+

= (3.80)

where:( )

( ) 22

1

+

=

mf

mf

GG

GG(3.81)

and 2 is the reinforcing efficiency factor for in-plane shear. The best agreement

with experimental results has been found for2=1. Shear modulus as predictedby Equation (3.80) is shown as a function of the constituent property ratios and

fibre volume fraction in Figure 3.31. Assuming 2=1, Equation (3.80) becomes:

)()(

)()(

mffmf

mffmf

mLTGGVGG

GGVGGGG

+

++=

(3.82)

Equation (3.82) is identical to the formula derived by the self-consistent field

model and to the lower bond of the variational approach. It may be underlined

that Gm has a significant influence on GLT, similar toEm onET.

In this section, the matrix and the fibres have been assumed to be isotropic; the

shear modulus of the constituents can be computed from the elastic modulus,E,

and Poissons ratio, using the following formula:

)1(2 +=

EG (3.83)

When the reinforcing fibres are anisotropic, the corresponding shear modulus

(G12) should be utilised.

109

Figure 3.31 Variation of shear modulus according to Halpin-Tsai equations

10

20

50100=m

f

G

G

Vf

m

LT

G

G 7

6

5

4

3

2

1

0 0.2 0.4 0.6 0.8


39/44

3.5.2 In-plane shear strength

Under in-plane shear, Figure 3.32, high shear stress concentration may

develop at the fibre-matrix interface. The failure could occur by matrix failure,

constituent debonding or a combination of the two. Shear failure may also occurwhen off-axis unidirectional composite elements are loaded in axial tension.

Based on matrix shear failure, the following formula can be used to determine

the in-plane shear strength,fLTs, of the composite:

k

ff msLTs = (3.84)

wherefms is the matrix shear strength and k is the shear concentration factor.

When the shear

strain

concentration factor, k, is utilised, fLTs can be estimated using the procedure

outlined in section 3.4.2 for transverse tensile strength. The representative

shaded rectangular element shown in Figure 3.33 loaded by in-plane shear stress

gives the following formula (Gibson 1994) to determine the shear strain

concentration factor, k:

11

1

+

=

f

m

G

G

s

dk

(3.85)

where dand s are shown in Figure 3.33. The shear strain concentration factor

causes the composite shear failure strain to be less than the matrix failure strain.

If a linear behaviour to failure can be assumed the corresponding shear strength

can be determined with:

110

12= LT

Figure 3.32 In-plane shear failure of unidirectional composite

12

=

LT

TL

=21

fms

fms

Failure surface

21

=TL

L

T


40/44

kG

fGf

m

msLTLTs = (3.86)

As in case of transverse tensile strength, the matrix shear strength can be used as

an upper bound on the composite shear strength. For a preliminary design, the

in-plane shear strength may be evaluated using a formula (Barbero 1999) similar

to (3.68) replacing the matrix tensile strength with the shear strength of the

matrix as follows:

+=

f

mffvmsLTs

G

GVVCff 1)(1 (3.87)

where Cv is the reduction coefficient (described in section 3.4.2) to account for

voids.

Again, in this section the matrix and the fibres have been assumed to be

isotropic; when the reinforcing fibres are anisotropic, the corresponding shear

modulus (G12) should be utilised.

3.6 PREDICTION OF POISSONS RATIO

111

d

s

21

= TL

21

= TL

12

= LT

12

= LT

d

matrix

fibre

Figure 3.33 Mechanics of materials model for strain concentration

factor under in plane shear loading


41/44

The Poisson coefficients (ij) are important engineering constants ofcomposite materials. The first subscript, i, refers to the direction of the applied

stress, and the second one, j, corresponds to the direction of the associated

transverse strain.

i

jij

= (3.88)

fori0, ij, and i,j=1,2. In case of a unidirectional composites ijji.Two Poisson ratios are considered for in-plane loading of a unidirectional

fibre reinforced unidirectional composite. Using the axis system shown in

Figure 3.34 the first Poisson ratio, LT, relates the longitudinal stress, L, to the

transverse strain, T, and is normally referred to as the major Poisson ratio:

L

T

LT

=

(3.89)

where L is the longitudinal strain and the loading scheme is: L0, T=0 and

LT=0. The second one called the minor Poisson ratio, TL, relates the transverse

stress, T, to the longitudinal strain, L:

T

LTL

= (3.90)

when T0, L=0 and LT=0.

A model similar to that used to predict ET can be used to determine LT;however, the load is applied parallel to the fibres, Figure 3.34. The deformationpattern illustrated in this figure, for cumulative thicknesses of layers is utilised

to express the transverse strains in the composite and constituents (fibre and

matrix) in terms of longitudinal strains and the Poisson ratio.

112

Figure 3.34 Model of unidirectional composite for

prediction of Poissons ratio

lf

lm

m

f

L

L lc

Deformed composite Undeformed composite

f

m


42/44

The total transverse deformation of the composite, c, is the sum of the

constituent transverse deformations, f andm. Transverse deformations can bewritten as the product of strain and cumulative thickness:

( ( ) ( ) cTccmTmmfTffmfc lll ===+= ;;; (3.91)

Transverse strains in the composite, (c)T, fibre (f)T and matrix, (m)T, can

be expressed in terms of the corresponding longitudinal strains, (c)L, (f)L, (m)Land the Poisson ratios as follows:

( ) ( ) ( ) ( )LffTfLmmTmLcLTTc

vvv === ;; (3.92)

Equation (3.91) now becomes:

( ) ( ) mLmmfffcLcLT lvlvlv = (3.93)

Assuming that no slippage occurs at the interface and the strains experienced by

the composite, fibre and matrix are equal: ( ) ( )LmLfLc == and that the widths

are proportional to the volume fractions the following formula is obtained for

the major Poisson ratio:

mmffLTVvVvv += (3.94)

Equation (3.94) is the rule of mixtures for the major Poisson ratio of a

unidirectional composite. The plot of LT with respect to the fibre volumefraction is similar to the plot of the longitudinal modulus (Taranu and Isopescu

1996), Figure 3.35.

The following functional relationship (presented in macromechanics of

composites) exists between engineering constants:

LTLTLT EE = (3.95)

Thus the minor Poisson ratio can be obtained from the already known

engineering constantsEL,ET and LT:

L

TLTTL

E

E = (3.96)

113


43/44

or in the extended form:

( )[ ]( )

fmff

fmffmf

fmffTLVEVE

VEVEEEVV

++

+=1

1/1

(3.97)

REFERENCES

1. Agarwal, B.D., Broutman, L.J., Chandrashekhara, K, (2006), Analysis and

performance of fiber composites. Third edition. Wiley-Interscience, New-York.

2. Barbero, E. J, (2011),Introduction to composite materials design. Second edition,

CRC Press, Taylor & Francis, Boca Raton.

3. Daniel I., Ishai O, (2006), Engineering mechanics of composite materials. Second

Edition, University Press, Oxford.4. Gay, D., Hoa, S.V, (2007), Composite materials. Design and applications. CRC Press,

Boca Raton.

5. Gerdeen, J.C., Rorrer, R.A.L, (2012),Engineering design with polymers and

composites. Second edition, CRC Press, Boca Raton

6. Gibson, R. F. (2012),Principles of composite material mechanics. Third edition, CRC

Press Boca Raton.

7. Jones, R. M. (1999), Mechanics of composite materials. Taylor & Francis,

Philadelphia.

8. Kaw, A.K., (2006), Mechanics of composite materials, Second edition, , CRC Press

Boca Raton.

9. Lee, S. M. (1989), Dictionary of composite materials technology. Technomic,Lancaster.

Vf

f

LT

m

0 0.2 0.4 0.6 0.8

114

Figure 3.35 Poisson ratio vLTas a function of

fibre volume fraction (m>f)


44/44

10. Nielsen, L. E. (1974), Mechanical properties of polymers and composites. Vol. 2,

Marcel Dekker, New York.

11. Taranu N., Isopescu D. (1996) Structures made of composite materials. Vesper, Iasi.

12. Timoshenko, S.P, Gere, J. M. (1961), Theory of elastic stability. McGraw Hill, New

York.

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