7. Anisotropy – direction-dependence of elastic
properties
We have seen in Chapter 6 that the elastic properties of a fibrereinforced composite may depend on the loading direction. Anisotropy
is a general term which simply means ‘not isotropic’. Most of the
composites of commercial interest are actually orthotropic – they have
three mutually perpendicular planes of symmetry, with different
properties in each of the three directions. In the unidirectional
composite in Fig. 6.1, these so-called principal directions were
assigned the labels ‘1’, ‘2’, and ‘3’, and appropriate subscripts were
added to the elastic properties to identify them. Other forms of
reinforcement will use the same nomenclature, with ‘1’ and ‘2’ being
always ‘in-plane, and ‘3’ through the thickness of the layer. The UD
composite is referred to as ‘transversely isotropic’ because it is
assumed that E2 = E3. A random reinforcement such as chopped strand
mat will result in the in-plane properties being isotropic (E1 = E2),
although due to the way in which composites are laminated, it may not
be appropriate to assume E1 = E2 = E3.
Anisotropy should not be confused with homogeneity. A homogeneous
material has uniform properties throughout, but this description clearly
depends on the scale we are considering. The micromechanical
equations presented in Chapter 6 describe an inhomogeneous or
heterogeneous solid, as the individual properties of fibre and matrix are
taken into account. On a larger scale (say greater than a few mm), we
are interested in the properties of composites in engineering
applications, so we assume the composite to be macroscopically
homogeneous, even though we know it is heterogeneous on the fibre
scale.
The purpose of this chapter is to establish how the orthotropic
composite responds to applied loads which may not be aligned with the
principal directions of the material. The concepts serve as a basis for
calculating the effective engineering properties of composite laminates,
which comprise many individual layers, each with different
orientations.
The mathematics in Chapters 7 and 8 has deliberately been kept to a
minimum. The detailed derivations of laminate theory are to be found
in all the existing text books, and only the essential results are included
here. What is important, however, is that the engineer understands the
assumptions and approximations which lie behind the mathematical
theory, so that the results can be interpreted appropriately. For a
detailed discussion of laminate analysis, the reader is recommended to
consult Powell (1994).
7-1
7.1 Hooke’s law in an orthotropic material
The situation considered here is one of plane stress, so we only
consider stresses and strains in the ‘1-2’ plane. This is appropriate as
long as the thickness of our composite is negligible compared to its
lateral dimensions.
For an isotropic material, Hooke’s law is
 1

 1   E
  
 2   
   E
 12 
 0


E
1
E

0
 1 
 
0   2 

 
1  12 

G
0
(7.1)
where the applied stresses 1, 2 and 12 result in strains 1, 2 and 12.
Generalising Equation (7.1) to an orthotropic material, we can write:
 1

 1   E1
    12
 2   
    E1
 12  
 0

 12
E1
1
E2
0

0 
 1 
 
0   2 
 
1  12 
G12 
(7.2)
Using [S] to denote the 3 x 3 compliance matrix in Equation (7.2)
gives:
  S
(7.3)
We can invert any of the above equations to express the stresses in
terms of the strains. In our abbreviated notation:
  Q
(7.4)
where [Q] is the stiffness matrix. For an orthotropic material:
7-2
Q  S1
with
 E1

 J
 E
  12 2
 J
 0


2
J  1  12 21  1  12
12 E 2
J
E2
J
0

0 

0 

G12 


(7.5)
E2
E1
The form of the compliance and stiffness matrices in Equations 7.2 and
7.5 confirms that the application of direct stresses alone (i.e. 12 = 0)
gives rise only to direct strains - there are no shear strains, and the
deformation is independent of G12. Also, an applied shear stress
produces only shear strain. We say that there is no coupling between
tensile and shear stress/strain. This is always true for an isotropic
material, and also applies to an orthotropic materials providing that the
loading is aligned with the material axes (‘1-2’).
7.2 Off-axis loading
In general, loads may be applied at any angle to the principal axes of
the composites. We have already seen that E1  E2 in the UD lamina.
But how does the effective tensile modulus vary with the direction of
the applied load?
In order to progress, we need to be able to transform stresses and strains
from one coordinate system to another. We identify ‘x-y’ as the
loading axes, and ‘1-2’ as the principal material axes, as in Fig. 7.1.
y
2
1
Fig. 7.1: Definition of
loading axes (‘x-y’) and
material axes (‘1-2’)
x

The transformation laws for stress and strain are completely standard,
and their derivation may be found in any solid mechanics text book.
For two-dimensional stress we have
7-3
 x 
 1 
 
 
  2   T   y 
 
 
 12 
 xy 
(7.6)
and for strain:
 x 
 1 




  2   T   y 
1  
1  
 2 12 
 2 xy 
(7.7)
The transformation matrix is given by
 c2 s2
2sc 
 2

2
T   s c  2sc 
  sc sc c 2  s 2 


(7.8)
where c = cos  and s = sin .
Thus, the application of a uniaxial stress x results in stresses in the
material axes given by:
1   x cos 2 
 2   x sin 2 
(7.9)
12   x sin  cos 
The final stage is to apply Hooke’s law (in its generalised form),
combined with stress and strain transformation. Formally, what we are
doing is, starting with applied strains in ‘x-y’:
(i) Calculate the strains in the material axes (Equation 7.7)
(ii) Apply Hooke’s law to obtain stresses in the material axes
(Equation 7.4)
(iii) Transform the stresses back to the loading axes.
The 3rd stage involves the use of Equation 7.6, but with  replaced by
–.
These operations can be combined mathematically to give:
 x 
 x 
 
 
(7.10)
 y   Q  y 
 
 
 xy 
 xy 

where the reduced stiffness matrix is given by
7-4
Q  T
1
QT
Similarly, we can start with applied stresses in the loading axes, and
derive Hooke’s law in the form:
 x 
 x 
 
 
 y   S  y 
 
 
 xy 
 xy 

(7.11)
where we now have a reduced compliance matrix given by
S  T
1
ST 
Comparing Equation 7.11 with the isotropic version (Equation 7.1)
enables us to associate the terms in the reduced compliance matrix with
the effective elastic properties of the off-axis composite. Matrix
multiplication results in:
S11 
2 
1
cos 4  sin 4   1


 
 12  sin 2  cos 2 
Ex
E1
E2
E1 
 G 12
S 22 
2 
1
sin 4  cos 4   1


 
 12  sin 2  cos 2 
Ey
E1
E2
E1 
 G 12
S 66 
 1 2 12
1
1  2
1
 sin  cos 2  
 4


cos 2   sin 2 
G xy
E1
E2 
G 12
 E1


2


 1
1
1  2
 sin  cos 2 
 E x S12   xy  E x  12 sin 4   cos 4   


 E 1 E 2 G 12 
 E1



(7.12)
The above equations may appear daunting, but they are easily displayed
with the aid of a spreadsheet. Off-axis elastic properties are plotted in
Fig. 7.2 and 7.3 for UD carbon/epoxy, using the principal properties
given in Table 6.2 for a fibre volume fraction of 0.6.
7-5
normalised tensile modulus
1
Fig. 7.2: Normalised
elastic moduli for off-axis
UD carbon/epoxy
composite.
0.9
Ex/E1
0.8
Ey/E1
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
20
40
60
80
angle (degrees)
2.5
Gxy / G12
2
1.5
1
 xy /  12
0.5
Fig. 7.3: Normalised inpane shear modulus and
Poisson’s ratio for UD
carbon/epoxy composite.
0
0
20
40
60
80
angle (degrees)
7.3 Interpretation of off-axis properties
Fig. 7.2 shows that the elastic modulus of a unidirectional composite
falls sharply as the angle between the fibres and the loading direction
increases – at only 10o off axis, over 40% of the longitudinal modulus
has been lost. It is this sensitivity to orientation that makes
unidirectionally-reinforced composites rather rare in practical
applications.
The effective in-plane shear modulus is a maximum when the fibres are
at 45o to the loading axis. This is reasonable if we consider an applied
7-6
shear stress to be equivalent to tension and compression at 45o. What is
less intuitive is the increase in Poisson’s ratio. It is common for the
off-axis Poisson’s ratio in a UD reinforced composite to be greater than
0.5, which is, of course, not permitted in an isotropic material.
We an see that the result of uniaxial loading of an off-axis composite
may be a combination of direct, transverse and shear strains. The
possibilities are shown in Fig. 7.4 and 7.5.
Fig. 7.4: Stretching,
lateral contraction and
shear in an off-axis ply.
Fibres are at 30o to the
applied load.
Fig. 7.5: Stretching,
lateral contraction and
shear in an off-axis ply.
Fibres are at 30o to the
applied load.
We say that direct stress and shear strain are coupled in the off-axis
composite. This arises not because the material is orthotropic, but
because the principal axes (‘1-2’) are not aligned with the loading axes
(‘x-y’).
We also note that in an isotropic material, the direction of the shear
stress is not important. In our off-axis composite, however, the
direction does matter, as shown in Fig. 7.6. Positive shear is equivalent
7-7
to a tensile stress in the fibre direction and a compressive stress in the
transverse direction; negative shear stress gives the opposite state of
stress in the material axes.

fibre
direction
positive shear stress

Fig. 7.6: Positive and
negative shear stresses
expressed as equivalent
direct stresses in the
material directions.
negative shear stress
7.4 Principal Stresses in Isotropic Materials
The concept of principal stresses is used widely in the failure analysis
of isotropic materials, particularly ductile metals. The procedure of
two-dimensional stress transformation already outlined gives the direct
stress () and shear stress () on any chosen plane inclined at angle 
to the reference (loading) axes, as a result of applied stresses. The
equations are:
    x cos 2    y sin 2    xy sin 2
   ( x   y ) sin  cos    xy cos 2
(7.13)
We define principal stresses with respect to the plane on which the
shear stress is zero. The angle of this plane is obtained by putting  =
0 in Equation 7.13. The mutually perpendicular principal stresses are
then obtained at this angle:
7-8
1 
1
2
2 
1
2


x
  y   12
x
 y  
1
2


x
  y   4 2xy
x
  y   4
2
2
(7.14)
2
xy
and the maximum shear stress, given by  = ½(1-2), is then on the
plane at 45o to that of the principal stresses (Fig. 7.7).

1
max
x
Fig. 7.7: Principal
stresses and maximum
shear stress in an isotropic
material.
2
xy
y
In assessing failure, it is sometimes convenient to calculate the socalled ‘von Mises’ stress. This combines the principal stresses to give:
 VM  12  1 2   22
(7.15)
which is then compared with the uniaxial yield stress of the material.
Thus the orientation of the principal stresses in an isotropic material, as
well as their magnitude, depend only on the applied stresses (x, y and
xy). In fact, as far as failure is concerned, the orientation is not of
particular interest.
This is quite different from the situation in orthotropic materials. Here
we are interested in stresses on planes which correspond to the axes in
which the material is aligned. So where we refer to ‘principal stresses’
we really mean the material principal stresses. Unfortunately, we use
the same nomenclature (1, 2, etc.), but these stresses must not be
confused with the isotropic principal stresses defined in Equation 7.14.
Note that, unlike in isotropic materials, the shear stress is not generally
zero on the 1-2 plane. The concept of von Mises stress also has no
relevance in orthotropic materials, although we shall manipulate
7-9
material principal stresses in a similar way when considering failure
criteria in Chapter 9.
7.5 Stresses, strains and curvatures in bending
It is convenient to start by considering an isotropic plate in bending,
then generalise the equations for an orthotropic material. Here we
simply state the standard results from bending theory, which are based
on the assumptions of small deflections and a linear variation of strain
through the thickness of the plate. The stresses due to bending are:
Ez
 x   y 
1  2
Ez
 y  x 
y 
1  2
x 




(7.16)
where z is the distance from the mid-plane and x and y are the
curvatures in the x and y directions. Curvature has the dimensions of
length-1, since it is equal to the reciprocal of radius of curvature ( =
1/R).
Consideration of equilibrium in a plate of thickness h results in
relationships between the external applied moments per unit length (M)
and the curvatures:
M x  D x   y 
M y  D y  x 
(7.17)
Eh 3
where D 
is the flexural rigidity of the plate.
12 1   2


Rearranging Equation 7.17, we obtain expressions for the plate
curvatures in terms of the applied bending moments:
x 
y 
M x  M y


D 1 
M y  M x
2

D 1  2


12
M x  M y 
Eh 3
12
M y  M x 

Eh 3
(7.19)
The plate may also experience twisting if an appropriate moment
(labelled Mxy) is applied. Twisting is a little more difficult to visualise
than ordinary bending. Formally, it is defined in terms of the rate of
change of the slope of the plate. This is illustrated in Fig. 7.8.
7-10
Mxy
Fig. 7.8: Twisting
curvature resulting from
applied twisting moment
Mxy (after Powell, 1994).
Mxy
The centre of the plate is at (0, 0). At any point (x, y), the vertical
displacement is given by
w
d2w
xy   xy xy
dxdy
(7.20)
The shear strain in the plate at a distance z from the neutral plane is
related to the curvature by
 xy  z xy
(7.21)
and the application of equilibrium gives
M xy
Gh 3
D

 xy  1    xy
12
2
(7.22)
which follows from G = E / [2(1+)] for an isotropic material.
We can now write the relationship between moments and curvatures for
isotropic thin plates in matrix form:
 M x   D D


 M y    D D
M  
0
 xy   0
7-11

 x 
  x 



0
 y   [D]  y 


D
 
1     xy 
 xy 
2

0
(7.23)
The matrix [D] is called the bending stiffness matrix of the plate.
Equation 7.23 would appear to be an unnecessarily complicated way of
describing the plate’s response, but the form of this equation will be
useful when mode complex laminated composites are considered later.
For our orthotropic composite ply, which may, in general, be loaded off
axis, Equation 7.23 becomes more complex. As before, we only state
the result here, which is
 Mx 
 x 
 x 





h3 
 M y   [D]  y   Q    y 
12 
M 
 

 xy 
 xy 
  xy 
(7.24)
This now includes the reduced stiffness matrix, as already used in
Equation 7.10. This term allows for the elastic properties of the
composite and the necessary transformations between loading and
material axes. By inversion, we can write curvatures in terms of
applied moments:
 x 
 Mx 
 Mx 






1
  y   [D]  M y   [d] M y 
 
M 
M 
 xy 
 xy 
 xy 
(7.25)
Both [D] and its inverse [d] are fully populated (i.e. there are no zero
terms) – this means that the application of, for example a uniaxial
bending moment may result in both bending and twisting. This is
another form of coupling which does not occur in homogeneous,
isotropic materials, and is illustrated schematically in Fig. 7.9.
Fig. 7.9: Coupling
between bending and
twisting in a single UD ply
with fibres oriented at 30o
to the x-axis.
y
x
7-12