6. Rules of Mixture
The mechanical and physical properties of composite materials depend
in a complex way on the type, form, quantity and arrangement of the
constituents. Rules of mixture are equations which attempt to express
these dependencies in a predictable fashion. Several different
approaches have been used (summarised in Daniel and Ishai, 1994 and
discussed in more detail by Gibson, 1994), but they are all based on
various degrees of simplification, and may be semi-empirical. It
follows that the user should treat the predictions with caution,
especially if properties are to be used for anything more than
preliminary design, and should endeavour to understand the
assumptions underlying the formulae.
The rule of mixtures for composite density was derived in Chapter 5.
Here we concentrate on expressions for elastic and thermal properties.
Models of composite strength are discussed in Chapter 9.
6.1 Unidirectional Ply – longitudinal modulus
The sign and nomeclature convention used here is shown in Fig. 6.1.
The orthogonal axes 1, 2 and 3 relate to the fibre direction, in-plane
transverse and through-thickness transverse directions respectively.
3
3
1
1
2
Fig. 6.1: Orthogonal
directions in unidirectional
ply.
2
We note that the unidirectional ply has two different in-plane tensile
moduli (E1 and E2). To a first approximation, E3  E2. We also need
two values of Poisson’s ratio to describe the lateral contraction
resulting from in-plane tension, as shown in Fig. 6.2. The conventional
notation is that ij denotes the contraction in the j-direction when stress
is applied in the i-direction.
6-1
1
fibre
direction
(‘1’)
2 = - 12 1
2
fibre
direction
(‘1’)
Fig. 6.2: Definition of inplane Poisson’s ratios in
orthotropic material
1 = - 21 2
From Fig. 6.2, it can be deduced that the lateral strain (2) resulting
from a stress applied in the fibre direction (‘1’) is much larger than the
longitudinal strain (1) resulting from a transverse (‘2’) applied stress.
Hence, 12 > 21. In fact the Poisson’s ratios and moduli are related by:
12 E1

 21 E 2
(6.1)
In deriving the rule of mixtures, the following assumptions are made:



Fibres are uniform, parallel and continuous.
Perfect bonding exists between fibres and matrix.
A longitudinal load produces equal strain in fibre and matrix.
We apply a load F1 in the longitudinal direction – this is shared equally
between reinforcement and matrix so that F1 = Ff + Fm.
Writing loads in terms of stresses and areas we obtain:
1A  f Af  m Am
(6.2)
where A = Af + Am is the cross-sectional area of the ply.
Applying Hooke’s law we replace stress with the product of strain and
modulus:
E11A  Ef f Af  E mm Am
6-2
(6.3)
But we have assumed equal strain (1 = f = m), so:
E1A  E f A f  E m A m
or
E1  E f
Af
A
 Em m
A
A
(6.4)
(6.5)
The terms Af / A and Am / A are the ‘area fractions’ of fibre and matrix
respectively. In our unidirectional composite, these are clearly
equivalent to the constituent volume fractions, so we can write:
E1  Ef Vf  E m Vm  Ef Vf  E m 1  Vf 
(6.6)
A similar rule of mixtures is commonly used for Poisson’s ratio:
12  f Vf   m Vm
(6.7)
These two rules of mixture are generally accepted as corresponding
well with experimental data – Matthews and Rawlings (1999), for
example, state that agreement is within 5%.
To a reasonable approximation, if Ef >> Em, then E1 = Ef Vf.
We should, however, take note of the fact that reinforcing fibres may
not be isotropic. Carbon and aramid, for example, rely on a threedimensional oriented microstructure for their exceptional mechanical
properties, but quoted values almost always refer to the axial direction.
There seem to be few data available on transverse properties (partly
because they are obviously very difficult to measure). Shindo (2000)
quotes values for the ratio of longitudinal to transverse modulus from
about 16 for high strength carbon fibre, up to more than 80 for highly
oriented high modulus carbon fibre. Hence both Ef and possibly f in
Equations 6.6, 6.7 and the other rules of mixture presented here should
be interpreted with care – this applies to thermal as well as elastic
properties.
6.2 Unidirectional Ply – transverse modulus
For loading in the transverse (‘2’) direction, the state of stress in the
relatively flexible matrix is much more complex, and not surprisingly
the transverse modulus E2 turns out to be a matrix-dominated property.
Generally, the rules of mixture are based on simply assumptions of
stress distribution, and are much less reliable than those for longitudinal
properties.
The simplest approach is a series model of fibre and matrix in which
any Poisson’s contraction is ignored, and the stress in each of the
constituents is assumed to be the same. A detailed derivation can be
6-3
found, for example, in Matthews and Rawlings (1999) or Hull and
Clyne (1996). The result is:
V
V
1
 f  m
E2 Ef Em
or
E2 
Ef Em
Vf E m  Vm E f
(6.8)
Again, we note that Ef should properly be associated with the transverse
rather than longitudinal value of modulus.
If Ef >> Em and Vf  Vm then we may write
E2 
Em
1  Vf 
(6.9)
which is of course independent of the modulus of the reinforcement.
The literature abounds with alternative models for transverse modulus,
which seek to improve on the rather poor experimental agreement
observed with Equation 6.8. A simple modification (which allows for
fibres restricting the Poisson contraction) is to replace the matrix
modulus in Equation 6.8 with
Em 
Em
1   2m
(6.10)
where m is the Poisson’s ratio of the matrix.
A commonly-used alternative is the Halpin-Tsai model for transverse
modulus:
E2 
where  
E m 1  Vf 
1  Vf 
(6.11)
E f  E m 
E f  E m 
The parameter  is adjustable, but is usually close to unity. Equation
(6.10) is generally considered to more reliable than the simpler
alternative.
The various rules of mixture for longitudinal and transverse modulus in
UD glass/epoxy are plotted in Fig. 6.3. Fibre and matrix moduli were
taken as 70 GPa and 3 GPa respectively. Note that the simple constant
stress model for E2 is a lower bound.
6-4
60
modulus (GPa)
50
Fig. 6.3: Rules of
mixtures modulus for
UD glass/epoxy
composite. Matrix
Poisson’s ratio taken as
0.4.
E1 (equation 6.6)
40
30
E2 (equation 6.11)
20
E2 (equation 6.10)
10
E2 (equation 6.9)
0
0
0.2
0.4
0.6
0.8
fibre volume fraction
6.3 Unidirectional Ply – shear modulus
As the tensile modulus relates tensile stress and strain, so the shear
modulus is defined as the ratio of shear stress to shear strain:
G ij 
 ij
 ij
Note that the subscripts ‘ij’ indicate the plane in which the shear
modulus is defined.
The rule of mixtures for shear modulus is based on the same
assumptions as that for transverse tensile modulus, and hence should be
treated with similar caution:
V
V
1
 f  m
G12 G f G m
(6.12)
The Halpin-Tsai equations are applicable to shear modulus, giving the
preferred expression:
G 12 
with

G m 1  Vf 
1  Vf 
G f  G m 
G f  G m 
(6.13)
(6.14)
Again, the parameter  is approximately equal to 1.
6-5
Assuming transverse isotropy, we expect G13 = G12.
The third shear modulus is usually obtained from:
G 23 
E2
21   23 
(6.15)
6.4 Unidirectional Ply – Poisson’s ratio
If our unidirectional ply (Fig. 6.1) has transverse isotropy, we expect
13 = 12 (see Equation 6.7). Hull and Clyne (1996) give the following
expression for the other out-of-plane Poisson’s ratio in terms of the
bulk modulus (K):
 23  1   21 
where
1 Vf Vm


K Kf Km
with
Kf 
E2
3K
(6.16)
Ef
Em
and K m 
31  2 m 
31  2 f 
Three dimensional elastic constants are rarely required for routine
calculations, but they may well be needed in finite element or other
numerical analyses. For reference, Tables 6.1 and 6.2 give theoretical
values of all the elastic constants for UD glass and carbon/epoxy
composites over a limited range of fibre volume fractions. The other 3
Poisson’s ratios may be obtained from Equation 6.1. The constituent
properties used are listed in Table 6.3. There is a wide range of
published values, and these data should be used for illustrative purposes
only. Note that some of the properties in Table 6.3 have been obtained
by ‘back calculation’ – i.e. deduced from empirical composite data
rather than measured as primary data.
6-6
Table 6.1: Rules of mixtures values for elastic constants of UD E-glass
fibre/epoxy (units of tensile and shear moduli are GPa).
Vf
0.4
0.45
0.5
0.55
0.6
0.65
0.7
E1 E2 = E3 12 =
29.8
6.5
33.2
7.2
36.5
8.1
39.9
9.1
43.2
10.4
46.6
11.9
49.9
13.8
13
0.33
0.32
0.31
0.30
0.29
0.28
0.27
23 G12 = G13
0.65
4.3
0.64
4.8
0.63
5.3
0.62
6.0
0.60
6.7
0.59
7.6
0.57
8.6
G23
2.0
2.2
2.5
2.8
3.2
3.7
4.4
Table 6.2: Rules of mixtures values for elastic constants of UD high
strength carbon fibre/epoxy (units of tensile and shear moduli are
GPa).
Vf
0.4
0.45
0.5
0.55
0.6
0.65
0.7
E1 E2 = E3 12 = 13
89.8
6.8
0.32
100.7
7.7
0.31
111.5
8.7
0.30
122.4
9.9
0.29
133.2
11.4
0.28
144.1
13.3
0.27
154.9
15.8
0.26
23 G12 = G13
0.70
3.5
0.69
3.7
0.68
4.0
0.66
4.3
0.65
4.6
0.64
4.9
0.63
5.3
G23
2.0
2.3
2.6
3.0
3.5
4.1
4.9
Table 6.3: Constituent properties for Tables 6.1 and 6.2 (Gibson,
1994).
longitudinal tensile
modulus (GPa)
transverse tensile
modulus (GPa)
longitudinal shear
modulus (GPa)
transverse shear modulus
(GPa)
longitudinal Poisson’s
ratio
transverse Poisson’s ratio
6-7
E-glass HS (T300)
aramid
epoxy
fibre
carbon
fibre
resin
fibre
73
220
152
3
73
13.8
4.1
-
30.1
9.0
2.9
2.1
30.1
4.8
1.5
-
0.22
0.20
0.35
0.4
0.22
0.25
0.35
-
6.5 Multidirectional Ply – in-plane tensile modulus
The calculation of elastic properties for laminae or laminates is
normally undertaken with numerical laminate analysis (Chapter 8).
However, it is possible to modify Equation 6.6 to give an estimate of
tensile modulus for composites in which the fibres are neither
continuous or unidirectional.
A correction factor to allow for the loss of efficiency if fibres are not
perfectly aligned in the load direction was given by Krenchel (1964) as:
o  i cos 4 i
(6.18)
where a proportion i of the fibres has orientation i, and the
summation is carried out over the various angles present in the
reinforcement. For example, in a biaxial reinforcement with fibres
oriented at  45o to the load direction, we have
 


o  0.5 cos 4 45o  0.5 cos 4  45o  0.25
The orientation correction factor is then included in Equation 6.6,
giving the semi-empirical:
E  L o E f Vf  E m (1  Vf )
(6.19)
Here L is a length correction factor for ‘short’ reinforcing fibres. For
fibres greater than a critical length (see section 6.6), and for continuous
fibres, L = 1. Table 6.4 gives appropriate values of o for common
forms of reinforcement.
Table 6.4: Orientation correction factors for non-UD reinforcement
(Equations 6.18 and 6.19).
orientation
unidirectional
biaxial
 45o
random (in-plane)
random (3D)
o
1
0.5
0.25
0.375
0.2
Equation 6.19 applies strictly to in-plane reinforcement. In woven
fabrics, the fibres exhibit ‘waviness’ in the through-thickness direction.
This ‘crimp’ is greatest in plain weave styles, but less in twill and satin
forms (see Chapter 3). We can use Equation 6.18 to estimate an
additional orientation factor to allow for the loss in reinforcing
efficiency due to out-of-plane waviness.
6-8
Assuming that the path of a tow in a woven fabric is sinusoidal, the
orientation factor can be obtained numerically as a function of weave
crimp angle (Grove and Summerscales, 2000). The results are shown
in Fig. 6.4. In a woven reinforcement, both orientation factors should
be combined to allow for in-plane and out-of-plane deviation from the
load direction.
1
0.95
0.9
0.85
0.8
Fig. 6.4: Calculated
orientation distribution
factor for a plain weave
tow with varying crimp
angle.
0.75
0.7
0
5
10
15
20
25
30
Crimp angle (degrees)
6.6 Short Fibre Reinforcement
When considering continuous fibres, we can safely assume conditions
of uniform stress or strain in the longitudinal direction. This is not the
case with discontinuous fibres, due to the nature of the load transfer at
the fibre ends. When a stiff fibre is embedded in a relatively flexible
matrix, shear stress and strain are a maximum at the fibre ends (Fig.
6.5). The tensile stress in the fibre, on the other hand, is zero at the
fibre end and increases towards the centre (Fig. 6.6).
6-9
Fig. 6.5: Finite element
model of single glass
fibre (blue) embedded in
epoxy resin. The model
is axisymmetric about
the upper edge, and is
subjected to a uniaxial
tensile strain. Upper
picture shows
undeformed mesh – note
high shear strain at fibre
end.
Fig. 6.6: Contours of
tensile stress in the fibre
direction. Note low
stress at fibre end,
increasing to maximum
value about 7 radii
towards the centre.
Schematically, the shear stress at the fibre/matrix interface and the
tensile stress in the fibre are as shown in Fig. 6.7. The common
theoretical derivation of these stress distributions relies on the so-called
shear lag model, which describes the transfer of tensile stresses from
matrix to fibre via shear stress at the interface. This and other models
are described in detail in Hull and Clyne (1996).
For our purposes, a simplified model will suffice. We consider the
balance of the tensile force in the fibre (diameter D) with the shear
force at the interface:
D 2 DL c
f

4
2
(6.20)
Lc is the length of the fibre over which the interfacial shear forces act.
Referring to Fig. 6.7, it is clear that the average tensile stress in the
6-10
L
Lc / 2
Fig. 6.7: Tensile and
shear stres distribution
for a single fibre
embedded in a matrix
(Daniel & Ishai, 1994).
fibre is less than its maximum value of Ef m, where m is the strain in
the matrix. Moreover, if the fibre length L < Lc, then the tensile stress
nowhere reaches its maximum value. Lc is thus referred to as the
critical length – defined as the minimum length of fibre required for the
tensile stress to reach its failure value. At this point, from Equation
6.20 we have:
Lc 
f D
2
(6.21)
where f* is the fibre tensile strength. By inserting values of interfacial
or matrix shear strength for  in Equation 6.21, the critical aspect ratios
(Lc / D) for various fibre/matrix combinations can be estimated. Table
6.5 gives some typical values.
Table 6.5: Critical length and aspect ratio for some composite systems
(from Matthews and Rawlings, 1996).
matrix
aluminium
epoxy
epoxy
polyester
alumina
fibre
boron
boron
carbon
glass
SiC
Lc (mm) Lc / D
1.8
20
3.5
35
0.2
35
0.5
40
0.005
10
The average stress in a fibre reaches 90% of the value in a continuous
fibre when L  5Lc. For fibre lengths L > Lc, the average fibre stress is
given approximately by
6-11
 L 
 ave
  fmax 1  c 
f
 2L 
(6.22)
Although the length correction factor (L) in Equation 6.6 depends on
fibre length as expected, the theoretical models also include elastic
properties of fibre and matrix and the fibre distribution:
L  1 
where  
tanh L / 2
L / 2
(6.23)
8G m
E f D ln 2R D
2
Here 2R is the interfibre spacing and Gm the matrix shear modulus.
Calculations in Hull and Clyne (1996) for carbon and glass
reinforcement show that L = 0.2, 0.89 and 0.99 for fibre lengths of 0.1,
1.0 and 10.0 mm respectively. In Fig. 6.8, we have calculated the
length correction factor for glass fibres in epoxy resin, assuming a
mean interfibre spacing of 20 times the fibre diameter.
1
Fig. 6.8: Theoretical
length correction factor
for glass fibre/epoxy,
assuming inter-fibre
separation of 20 D.
length correction factor
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
fibre length (mm)
This would suggest that, so far as stiffness is concerned, the length
correction factor can be ignored for fibre lengths over about 1 mm. It
should be noted, however, that processes such as injection moulding of
thermoplastics can cause considerable fibre damage. Data from
Vetrotex examined glass fibre/nylon feedstock, containing 4 mm long
fibres. After moulding, almost all the fibres had lengths ranging from
0.1 to 0.7 mm – only 12% of all the fibres had a length greater than 0.8
mm. The lower stiffness of discontinuous fibre-reinforced composites
6-12
is of course also due to the fact that typical volume fractions are much
lower than with aligned, continuous reinforcements.
In short fibre-reinforced thermosetting polymer composites, it is
reasonable to assume that the fibres are always well above their critical
length, and that the elastic properties are determined primarily by
orientation effects. The following equations give reasonably accurate
estimates for the isotropic in-plane elastic constants:
E  83 E1  85 E 2
G  18 E1  14 E 2

E
1
2G
(6.24)
Here, E1 and E2 are the tensile moduli of a unidirectional ply of the
same fibre volume fraction, as given by Equations 6.6 and 6.11 (or its
alternatives). Results from Equation 6.23 are plotted in Fig. 6.9 for a
random glass / epoxy composite.
modulus (GPa) and
Poisson's ratio x 10
20
18
Fig. 6.9: Elastic
constants of 2D random
glass fibre-reinforced
epoxy. Poisson’s ratio is
multiplied by 10 for
clarity.
16
14
E
12
10
8
G
6
4
2
0
 x 10
0
0.1
0.2
0.3
0.4
0.5
fibre volume fraction
6.7 Thermal Expansion
Coefficients of thermal expansion are of considerable relevance, since
many polymer composites experience temperature changes as an
integral part of their processing cycle. Rules of mixture are discussed
in Chapter 10.
6-13
6.8 Exercises
1. Using an Excel spreadsheet, or otherwise, produce an equivalent set
of curves to those in Fig. 6.3 for high strength carbon fibre
reinforcement.
2. The following empirical data was published in an Owens Corning
Design Data Book. Compare these data with the appropriate
predictions from rules of mixtures.
3. An aligned short fibre composite is stressed to a mean interfacial
shear stress of 5 MPa. If the tensile strength of the reinforcement is
3400 MPa, what fibre length is required for the composite to have its
maximum load carrying capacity?
4. Using Fig. 6.8, plot an approximate graph of tensile modulus versus
fibre length for a short fibre glass/epoxy composite.
5. Using Equation 6.22, estimate the fibre length at which you would
expect a glass/polyester composite to have 95% of the strength of one
reinforced with continuous fibres.
6. Sketch and label the variation of tensile stress along the length of a
high strength carbon fibre of length L = Lc if it is aligned in the load
direction and the applied strain is 0.1%.
7. A test specimen containing a single HS carbon fibre is subjected to a
uniaxial tensile test. At maximum stress, the fibre breaks into pieces
each of length 0.625 mm. The fibre is 10 m in diameter, has
6-14
longitudinal tensile modulus 240 GPa and an ultimate tensile strength
of 2500 MPa. (a) What is the interfacial shear strength of the
composite? (b) If the composite has a longitudinal modulus of 80 GPa,
what was the applied stress at fibre failure? (Gibson, 1994).
6-15