The Design of Composite Materials and Structures
What is a composite material?
A composite material is a material in which two or more distinct materials are combined together but remain uniquely identifiable in the mixture. The most common example is, perhaps, fibreglass, in which glass fibres are mixed with a polymeric resin. If one were to cut the fibreglass and, after suitable preparation of the surface, look at the material, the glass fibres and polymer resin would be easy to distinguish. This is not the same as making an alloy by mixing two distinct materials together where the individual components become indistinguishable. An example of an alloy that most people are familiar with is brass, which is made from a mixture of copper and zinc. After making the brass by melting the copper and zinc together and solidifying the resultant mixture, it is impossible to distinguish either between or where the atoms of copper and zinc are. There are many composite materials and while we may be aware of some, such a fibreglass and carbon epoxy, there are many others ranging from the mundane, reinforced concrete ( a mixture of steel rod and concrete (itself a composite of rock particles and cement), pneumatic tyres ( steel wires in vulcanised rubber), many cheap plastic moldings (polyurethane resin filled with ceramic particles such as chalk and talc) to the exotic metal matrix composites used in the space program (metallic titanium alloys reinforced with SiC ceramic fibres), and your automobile, such as engine pistons (aluminium alloys filled with fibrous alumina) and brake discs (aluminum alloys loaded with wear resistant SiC particles). Regardless of the actual composite, the two [or more] constituent materials that make up the composite are always readily distinguished when the material is sectioned or broken.
Is it possible to design a composite material?
Obviously the answer to that question is "Yes"! First, we must identify the numerous materials related variables that contribute to the mechanical and physical properties of the composite material. Secondly, the appropriate physical and mathematical models that describe how the properties of the individual components of the composite are combined to produce the properties of the composite material itself must be derived. So, "Yes", it is possible to design a composite material such that it has the attributes desired for a specific application. Those attributes might be as simple has having a specified stiffness and strength, a desired thermal conductivity, or have a minimum specified stiffness at the cheapest possible cost per unit volume. Whatever the specifications it should be possible to design a suitable composite material. As in all design processes, it may not be possible to meet all the specifications exactly and compromise and trade offs will be required, but by understanding the physical origin of the required properties and developing an appropriate mathematical description, a suitable composite can be designed. We should also keep in mind that there may be an exisitng conventional material that is more suitable for the application than a composite. So the composite must offer a specific advantage in terms of cost or performance than conventional alternatives. It is one of the goals of this resource to show you the logical steps needed to implement the design process.
How do we get started?
Perhaps the easiest way to demonstrate how the design process required to develop a composite material is implemented is to start with a familiar composite material and examine just what factors control its properties. So I will start by asking a simple question, "How strong is a piece of fibreglass?” As you should be aware, there is no single answer to that question and one might be tempted to reply, "How strong do you want it to be?”
The amount of load that it takes to break a piece of fibre glass depends on the size of the piece of fibreglass, its thickness, width and length, whether we are simply pulling it in tension, compressing it, or bending it. It also depends on what the fibreglass is made of. There are many types of glass and many different polymeric resins that are used to make fibreglass. There are also many different ways in which the glass can be combined into the resin, for example, are the fibres all aligned in the same direction, are the fibres woven into a cloth, what type of cloth, are the fibres aligned at random, and are the fibres long or short? Then, if the fibres are oriented, at what angle relative to the fibres, is the fibreglass being loaded? Finally, just what is the ratio of fibres to resin and is that by volume or by weight?
By looking at the range of fibreglass products available and by seeking clarification on the structure and composition of the fibreglass we have begun to identify the micro structural variables that will control the properties of the composite. These may be summarised as

The properties of the fibre reinforcement

The properties of the matrix in which the reinforcement is placed

The amount of reinforcement in the matrix.

The orientation of the reinforcement

The size and shape of the reinforcement.
In order to get started, it is tempting to rephrase the initial question "How strong is fibreglass?" to "What is the tensile strength of fibreglass ?" thus eliminating the size and loading mode variables, or better still, "What is the tensile strength of fibreglass when all the fibres are aligned in the same direction?" Now we only need consider the mechanical properties of the glass fibres, the polymeric resin used to bind them and the relative proportions of the two. It would be relatively simple, having selected a resin and a fibre, to manufacture a number of flat plates of the composite with various ratios of fibre to resin, test them and produce a graph of tensile strength vs. volume fraction from which we could select a volume fraction of fibre that gives a composite with the required strength. However, if strength outside the range of measured strengths was required or other factors dictated a change in resin or fibre then the whole process would have to be repeated. While this approach does work, it rapidly becomes very time consuming and costly.
If we were to look at the various test materials that were made in the first trial and error experiments and observe the stressstrain behaviour up to the point of fracture we could infer that failure resulted from either a critical strain in the matrix or fibre being exceeded or a critical stress in either component being exceeded. We would also observe that for the most part, the composite behaved elastically almost to the point of failure, primarily because the glass fibres and the polymeric resin were both linear elastic solids with a brittle fracture mode, i.e., no plastic deformation. We would also note from the mechanical tests that the elastic modulus of the composite also varied with the amount of fibre added to the resin. Since we are already familiar with HookeÕs Law that defines the elastic modulus as the ratio of stress to strain, then to start answering the question "How strong is fibreglass?" we will first examine how the elastic modulus of the composite, measured parallel to the aligned fibres, varies as a function of the volume fraction of fibres.
Aligned Continuous Fibres
If the composite material is to stay in equilibrium then the force we apply to the composite as a whole, F, must be balanced by an equal and opposite force in the fibre, F_{f} and the matrix F_{m}.
When considering 'Strength of Materials' problems we usually work in terms of stress () (force per unit area) rather than force itself. So the force on the fibres is simply the stress on the fibres, _{f}, multiplied by the crosssectional area of the fibres lying perpendicular to the stress. The cross sectional area of the composite occupied by the fibres is just f, the volume fraction of the fibres multiplied by the crosssectional area of the composite itself  we'll call that "A"  i.e. f.A. Similarly the force on the matrix is just the stress in the matrix multiplied the crosssectional area of the matrix in the composite, i.e. (1f).A . Since the crosssectional area of the composite itself, A, is in each term on both sides of the equation we can cancel it out. So the stress in the composite is just the sum of the stresses in the fibre and the matrix multiplied by their relative crosssectional areas.
The stress in the fibre and the stress in the matrix are not the same. Now the tricky bit!
We can now use Hooke's Law, which states that the stress (or Force) experienced by a material is proportional to the strain (or deflection). This applies as long as the stresses are low (below the elastic limit  we'll come to that soon) and the material in question is linear elastic  which is true for metals, ceramics, graphite and many polymers but not so for elastomers (rubbers).
Where E is the elastic modulus; the bigger these number the stiffer the material. For compatibility, the strain, , must be the same in both the fibres and the matrix otherwise holes would appear in the ends of the composite as we stretched it. This is known as the ISOSTRAIN RULE.
Since the fibre and matrix often have quite different elastic moduli then the stress in each must be different  in fact the stress is higher in the material with the higher elastic modulus (usually the fibre). In fibreglass, the elastic modulus of the glass (~75GPa) is much greater than that of the polyester matrix (~5GPa) so as the volume fraction of fibres is increased, the elastic modulus of the composite (measured parallel to the fibres) increases linearly.
Try selecting different types of polymer matrices or different types of fibres and see how the different elastic properties change as you increase the volume fraction of fibres. The greyed areas to the right of the graph represent fibre contents which are either difficult to achieve in practice (light grey) or just plain impossible (dark grey).
In practice it is very difficult to get more than 60% by volume of fibres which puts a practical limit on the maximum stiffness of the composite of 0.6xE_{f}.
While the rule of mixtures has proved adequate for tensile modulus (E) in the axial direction, the isostrain rule of mixtures does not work for either the shear (G) or bulk (k) moduli. Instead, these are dependent on the phase morphology. An example of shear modulus (G) and bulk modulus (k) dependencies for an assemblage of cylindrical fibres is shown below.
What about the stiffness perpendicular to the fibres?
If we were to look down on the top of the composite or along the axis of the fibres and apply a load perpendicular to the fibre axis then the composite would respond in a very different way.
In a fibrous composite with the applied stress aligned perpendicular to the fibres, the stress is transferred to the fibres through the fibre matrix interface and both the fibre and the matrix experience the same stress If the matrix and fibre have different elastic properties then each will experience a different strain and the strain in the composite will be the volume average of the strain in each material. Since the stress is the same in each phase this is known as the ISOSTRESS rule of mixtures.
If a force is applied perpendicular to the fibres then the fibres and matrix will stretch in the same direction. The total deflection (d) is just the sum of the deflections in the fibre (d_{f}) and the matrix (d_{m}).
Again, we can use Hooke's law to introduce the elastic modulus and since the stress is the same in both the matrix and fibre we can get the elastic modulus perpendicular to the fibres
Note that the stiffness of the composite, measured perpendicular to the fibres increases much more slowly than stiffness measured parallel to the fibres as the volume fraction of fibres is increased. Since the properties of the composite are different in different directions, the composite is anisotropic. Back to Calculator.
See also Calculation of Shear modulus and Poissons ratio in aligned fibre composites using the halpiTsai equations.
Woven Fibres
Fibre Packing
In all systems the equations which predict the properties of a composite breakdown at high volume fractions of reinforcement because of geometric packing limitations and the necessity for the reinforcing phase to be surrounded by the matrix in order that load can be transferred to it. There are two simple packing models which we can use to establish an upper bound for the volume fraction, a square array and a hexagonal array with circular section reinforcement.
From the two figures it is readily apparent that volume fractions higher then 90% are impossible and that even 78% fibre loading would be very difficult to achieve. In practice, the maximum volume fraction is around 60% in unidirectional aligned fibre composites. In woven materials, the total volume fraction rarely exceeds 40% in a given layer of cloth and so the effective fibre fraction in either the warp or weft directions is unlikely to exceed 20% for a plain weave, satin or harness weave fabric. For loosely packed fabrics such as chopped strand mat, the total volume fractions of fibres is unlikely to exceed 10% and are normally used to provide filler layers between the outer load bearing layers in a multilayer laminate.
Strength of Fibre Composites
We have already seen that in a simple aligned fibre composite, loaded parallel to the fibres that both the matrix and the fibre experience the same strain (amount of stretch). It would be logical therefore to expect the composite to break at the lower of the matrix fracture strain or the fibre fracture strain. There are two cases to consider, firstly, where the matrix fails first and secondly, where the fibre fails first. The former situation is common in polymer matrix composites with low strength brittle matrices such as polyesters, epoxies or bismelamides, the latter case is observed in metal matrix composites or thermoplastic polymer composites where, because of plastic deformation in the matrix, the failure strain of the fibre is the smaller value.
Matrix Fails First.
At low volume fractions of fibres, the matrix constitutes the major load bearing section and the addition of fibres gradually increases the strength as the applied load is partitioned between the fibres and the matrix. However, when the strain in the composite reaches the fracture strain of the matrix, the matrix will fail. All of the load will then transfer instantly to the fibres, which occupying such a small fraction of the sample area will see a large jump in stress and they too will fail. When the composite is deformed the elastic modulus is linear. At the strain at which the matrix is about to fracture, ε_{m}, the stress in the composite can be determined using Hookes' Law since both the fibre and the matrix are still behaving elastically, i.e.
The stress in the matrix, σ_{m}, is now equal to the matrix fracture stress, but the stress in the fibre is still much less that the fibre fracture stress  we know this because the stress in the fibre is simply calculated using Hookes' Law. What happens next, as the matrix breaks, depends on the mode of loading, either constant deflection (deflection rate) i.e. the end points of the composite are fixed or constant load (loading rate) where there is a dead weight hanging off the end of the composite. Ultimately, the distinction is irrelevant to the overall strength of the composite but affects the shape of the stressstrain curve. We will just consider the case of dead weight loading...
Before the matrix breaks, the load on the composite is
After the matrix breaks only the fibres remain to carry the load and the stress in the fibre jumps by. If this increase takes the stress in the fibre above its fracture strength then the fibres too will snap. This is most likely to happen when f, the volume fraction of fibres is small and when the strength of the matrix is large. This is called MATRIX CONTROLLED FRACTURE. However, if the jump in stress is not sufficient to break the fibres then the load can be increased until the fibres break i.e.
This is known as FIBRE CONTROLLED FRACTURE
The graph above shows how the strength of a fibreglass composite changes as the volume fraction of fibres is increased. At the low fibre fractions, the strength is controlled by the fracture of the matrix; the strength increasing as the fibres are added.
Matrix fracture strength is ~50MPa and the failure strain is 0.010.
Fibre fracture strength is ~1200MPa and the failure strain is 0.016.
Above a fibre content of 10% the fibres begin to dominate the fracture process and while the composite can sustain high stresses, structural integrity would be lost when the matrix fractures because the composite would be full of cracks if loaded to its ultimate tensile strength. The effective strength of the composite is given by the (lower) matrix controlled strength. Even so, for a fibre loading of 40% the strength of the composite would be 330MPa; a very respectable 560% increase over the strength of the matrix alone.
This type of behaviour is typical of the composites used in sailboard components, such as boards, masts, fins and nowadays booms (glassepoxy or graphiteepoxy).
Strength of Aligned Continuous Fibre Composites
Fibres Fail First
We shall now consider the case where the matrix is ductile and the elastic strain to fracture in the fibres is less than the elastic/plastic extension of the matrix as would occur in fibre reinforced metal matrix composites or thermoplastic matrix composites. At low volume fractions of fibres, the chain of events is analogous to the case where the matrix fails first in that the fibres will break and the load will transfer to the matrix which, having a reduced crosssection, will see a sudden jump in stress. Again, what happens next depends on the magnitude of the increase in the stress in the matrix  will it fracture or won't it? The stress on the composite at the point of fibre fracture (σ_{f}) is
The force on the composite is just the product of the stress and the crosssectional area, so the stress on the matrix after the fibres break is
So the stress on the matrix increases by . If the rise in stress is not sufficient to fracture the matrix then it will continue to support the applied load. Thus the fracture strength of the composite will be given by
where σ_{m} is the ultimate tensile strength of the matrix; i.e. the addition of fibres leads to a reduction in the strength of the composite to levels below that of the unreinforced matrix. Fortunately, as the fibre volume fraction increases, the fibres carry more of the applied load. When the fibres break, the load transferred to the matrix is large and the much reduced crosssectional area of the matrix will be unable to support the load and the matrix too will fail. The strength of the composite, like the previous example, is determined by the strength of the fibres i.e.
We can plainly see that the tensile strength of a composite in which the fibres fail at a lower strain that the matrix initially decreases below that of the matrix alone, reaches a minimum and thereafter increase. There is, therefore, a minimum volume fraction, fmin, of fibres that must be added in order for the composite to have a strength at least equal to that of the matrix alone, i.e.
In the example shown above, where glass fibres are used to reinforce a polyAmide matrix, f_{min} is around 9%.

Matrix Modulus (E_{m}) = 5GPa; Tensile Strength (_{m}) = 120 MPa; strain at yield = 0.024; strain at fracture (_{m}) = 0.1.

_{matrix}(_{f}) is the stress in the matrix at the strain at which the fibres break.

Fibre modulus (E_{f}) = 75GPa; Tensile Strength (_{f}) = 800MPa; strain at failure (_{f}) = 0.01
The strength is calculated at the lower strain of

the fibre fractures, or

the (ductile) matrix yields, or

the (brittle) matrix fractures.
Transverse Strength
So far we have only considered the strength of the composite when loaded in a direction parallel to the fibres. However, if the composite is loaded in a direction perpendicular to the fibres then a different set of rules apply  just one of the problems associated with analysing anisotropic materials.

We should recall, that when loaded in the transverse direction, both the fibres and the matrix experience the same stress  so to determine what the strength is we need only look at the weakest link in the composite. Of the two materials that make up the composite, the matrix is invariably the weaker material and so fracture will occur when the stress reaches the matrix fracture stress  or will it? Up to now we have assumed that the join between the matrix and the fibre is perfect and will transmit all the load applied to it. A great deal of effort goes into the engineering of the fibre matrix interface either to make it strong or to deliberately weaken it, depending on the application. We will discuss the fibre matrix interface in a latter class but for now it is safe to assume that the interface is always the weakest link, therefore err on the safe side and set the transverse strength to some fraction of the matrix strength  the exact value can be determined most easily by experiment.
Effect of Orientation on Stiffness
This means that we are now going to look at the effect of loading a composite in a direction that is neither parallel nor perpendicular to the fibres. This section is also going to be a bit heavy on the maths with Tensors and Matrix Algebra. A complete description of this next section can be found in any text on mechanics of composites . If this is somewhat daunting then it's probably best to jump right to the calculator We should be familiar with the tensor representation of the stressstrain relationships which define elastic behaviour. The stiffness matrix, Q, for plane stress is given by the matrix shown below, where is Poisson's ratio representing a strain in the '2' direction resulting from a load applied in the '1' direction, i.e. ; similarly . '1' and '2' are at right angles in the eplane of the composite, '3' is perpendicular to the plane of the sheet  since there won't be any stresses applied perpendicular to the plane of the sheet we are going to ignore the '3' direction completely.
where
When considering fibre reinforced composites we generally deal with thin sheets or plys. In this case plane stress is assumed and therefore there are no through thickness stresses, i.e. . However we must remember that the composite is not isotropic and thus E_{11} and E_{22} are not the same.
Next we shall introduce the compliance matrix S, which is the inverse of the stiffness matrix Q and enables the calculation of strain given a system of applied stresses. Note that the compliance matrix S is much simpler than the stiffness matrix Q. Both matrices are symmetrical about the diagonal.
Below we define the principal axes of the composite and the corresponding elastic constants. We defined the parallel (E_{11})and transverse (E_{22}) elastic moduli in Class 1. G_{21} is the shear modulus and relates the shear stress to the shear strain.
Next we define the rotation from the special '12' coordinate system that is aligned with the fibres to a more general 'xy' coordinate system that is aligned with the direction of loading. is the angle between the two.
If the composite is tested at an angle to the fibre orientation then the elastic properties in the general directions 'xy' (parallel and perpendicular to the testing direction) can be determined in terms of the 'special orthotropic' properties as follows.
1. Translate the strains from the general 'xy' orientation (the loading directions) to the orthotropic '12' orientation. Note that we will need to rewrite the strain tensor in terms of engineering strain not tensor strain  the engineering shear strain is 2 x the tensor shear strain.
The matrices R (R.Tensor Strain = Engineering Strain) and T, the tensor rotation matrix are defined as
2. Since , we can write the orthogonal stresses in the 12 orientation in terms of the special orthogonal elastic properties, Q and the engineering strains in the xy directions (parallel and perpendicular to the applied loads.
3. All that remains is to rotate the special 12 stress tensor into the general xy orientation.
i.e.
4. Now, all the matrix terms T,Q and R can be collected together in a single matrix, which represents the elastic properties of the composite at an arbitrary angle to the fibres.
So much for theory  let’s see how its works in practice...
Effect of Fibre Orientation on the Strength of Aligned Continuous Fibre Composites
When considering the effect of fibre orientation on the strength of a composite material made up of continuous aligned fibres embedded in a matrix, it should be recognised that there are 3 possible modes of failure...

Tensile fracture parallel to the fibres (whether the fibres fail or the matrix fails will depend on the particular combination of fibre and matrix materials as well as the volume fraction of fibres),

Shear failure of the matrix as a result of a large shear stress acting parallel to the fibres ,

Tensile failure of the matrix or fibre/matrix interface when stressed perpendicular to the fibres.
We have already determined suitable expressions for the strength of a composite when tested parallel to the fibres, We'll call this strength X. We also know the tensile strength of the matrix material which we'll call Y. The shear strength of the matrix can be determined using the Tresca criteria and is simply Y/2. In order to examine the effect of orientation on strength we need to make use of Mohr's Circle to establish the state of stress aligned parallel and perpendicular to the fibres and then to equate these stresses with the appropriate failure stress of the composite in each those directions.
For failure to occur, the applied stress must be increase until either
These equations are plotted out below and since failure is a "weakest link" phenomenon, fracture will occur at whichever criterion is reached first and so the mechanism of failure changes from tensile failure of the fibres to shear of the matrix to tensile failure of the matrix as the fibre angle is increased from 0 to 90°.
Failure under Mutliaxial Stress States (Plane Stress)  TsaiHill
When two mutually perpendicular stresses and/or a shear stress is applied to the composite we need to be able to define a failure criterion. Tsai and Hill have established a suitable fracture criteria based on maximum strain energy, rather than considering stress and strain. This maximum strain energy approach allows us to ignore the fact that failure can occur because either a stress has exceeded a critical value (e.g. the stress resolved perpendicular to the fibres has exceeded the tensile strength of the matrix) or the strain has exceeded a particular value (e.g. the strain resolved parallel to the fibres has exceeded the fibre fracture strain). The TsaiHill maximum strain energy formulation is:
Which we can see for the case of a uniaxial stress applied

Parallel to the fibres ,

Perpendicular to the fibres ,

Simple shear
