Elastic Properties of Aligned Continuous Fibre CompositesParallel to Aligned Continuous FibresIf 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, Ff and the matrix Fm. 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, sf, multiplied by the cross-sectional 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 cross-sectional 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 cross-sectional area of the matrix in the composite, i.e. (1-f).A . Since the cross-sectional 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 cross-sectional areas.  The stress in the fibre and the stress in the matrix are generally 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 this 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 with the volume fraction of fibres. Now what about the stiffness perpendicular to the fibres? Perpendicular to Aligned Continuous FibresIf 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 (df) and the matrix (dm).  Since deflection is just strain multiplied by the initial length we can substitute for the deflections and, as L appears in every term of the resulting equation we can eliminate it. 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. In both instances the limiting values of the stiffness of the composite, either parallel or perpendicular to the fibres are given by the elastic moduli of the fibre and matrix materials. There is however, one important caveat that we must be aware of! In the derivations above we have assumed that both the matrix and fibre are elastically isotropic and thus the fibres have the same elastic modulus when stretched parallel to the fibre axis and when loaded in a radial direction. This is certainly the case for our example composite, fibreglass, but it is not the case when composites containing organic fibres such as graphite, kevlar, pbo etc., are manufactured. Due to the highly aligned and linear nature of the organuic molecules making up the fibre, the elastic modulus of the fibre in the radial direction is often 10 to 20 times lower than in the axial direction - the reason? - in the axial direction stiffness arises primarily from the resistance to rotation and stretching of the C-C bonds that make up the molecular structure of the fibre - in the transverse or radial direction it is the relatively weak intermolecular Van der Waals bonds that are being stretched. SO REMEMBER THIS! i.e.  where Eft is the transverse elastic modulus of the fibre. So far, we have considered the fraction of fibres, f, in terms of volume fraction but it is sometimes necessary in manufacturing to look at weight fractions and convert back and forth between the two. Also, we have to recognize that in a real composite, there is actually a limit to the amount of the reinforcement that can be added to the matrix in order that the matrix can transfer the load to the reinforcing material.
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