Elastic Properties Parallel to 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, 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?