Fibre-axis

In oriented systems (fibres or stretched films), the scattered image often appears as a two-bar or a four-point pattern with the scattering maximum at or near the meridian (fibre axis). The one-dimensional scattered intensity along the meridian must be calculated by the projection method using the following fonnalism... 

We must therefore hope that CFRP can give us the required anisotropy. The modulus of type-1 carbon fibres is 390 GPa along the fibre axis (although it is only 12 GPa at right angles to this). So... 

The transverse modulus, Ect, may be determined in a matmer similar to that described earlier for the longitudinal modulus. Consider a unidirectional fibre composite subjected to a transverse force, Fct, in the direction perpendicular to the fibre axis. 

The previous analysis is a preparation for the more interesting and practical situation where the applied loading axis does not coincide with the fibre axis. This is illustrated in Fig. 3.10. 

The first step in the analysis of this situation is the transformation of the applied stresses on to the fibre axis. Referring to Fig. 3.10 it may be seen that Ox and Oy may be resolved into the x, y axes as follows (the reader may wish to refer to any standard Strength of Materials text such as Benham, Crawford and Armstrong for more details of this stress transformation) ... 

The behaviour of the lamina when subjected to loading at 6 degrees off the fibre axis is determined using matrix manipulation as follows ... 

For the situation where the loading is applied off the fibre axis, then the above approach involving the Plate Constitutive Equations can be used but it is necessary to use the transformed stiffness matrix terms Q. 

It has been shown that the anisotropy depends on the orientation of the diagonals of indentation relative to the axial direction 14). At least two well defined hardness values for draw ratios A. > 8 emerge. One value (maximum) can be derived from the indentation diagonal parallel to the fibre axis. The second one (minimum) is deduced from the diagonal perpendicular to it. The former value is, in fact, not a physical measure of hardness but responds to an instant elastic recovery of the fibrous network in the draw direction. The latter value defines the plastic component of the oriented material. 

Fig.4 The four normal stresses and the four equal shear stresses acting on the domain in the fibre under a tensile stress a. The chains are parallel to AB and make an angle 0 with the fibre axis...

The function N(0)=p(0)sin0 is proportional to the total number of domains in the fibre at an angle 0 with the fibre axis. The determination of the modulus using sonic frequencies will yield a higher value of g than the method by which the modulus is derived from the initial slope of the tensile curve. For random orientation of the chains in a fibre (sin20),=O.5 and the modulus of an isotropic fibre is given by Eiso 4g. 

In the previous section we discussed the ultimate strength of a polymer fibre o0. This value corresponds to the stress at which all secondary bonds in the fibre are broken. Due to the presence of the chain orientation distribution alone, it follows that even the strength of a polymer fibre without any flaws will never attain this value. Yet, fracture in a real fibre may not always initiate in the most disoriented domains. If there are inhomogeneities that lead to stress concentrations, fracture can also occur in domains at a smaller angle to the fibre axis. 

Fig. 14 The four normal stresses and the four shear stresses acting on a domain resulting from a tensile stress on the fibre. A crack is drawn parallel to the chain direction. The angle in the loaded state between the domain axis and the fibre axis is 6... 

The polymer fibre is considered to be composed of a parallel array of identical fibrils. A single fibril is a series arrangement of domains with a varying angle to the fibre axis described by the orientation function p(Q) in the unloaded state. The strength of a domain is supposed to be given by... 

It is assumed that all chains are oriented parallel to the fibre axis and that equal parts of the fibre contain equal numbers of starting points of chains of arbitrary length u. Thus the starting points are distributed homogeneously along the length of the fibre. The length distribution of the chains is determined by the distribution function /( ) with... 

For, SHG it is desirable to fabricate fibres which allow single mode operation at the fundamental wavelength so that all the launched power is available for conversion. The frequencies for which single mode operation is possible in a step-index uniaxial crystal cored fibre with crystal axis along the fibre axis are given by (30.)... 

Since polymer single crystals prepared these days are too small for x-ray diffraction experiments, the Crystal Structure of a polymer is generally determined from x-ray patterns of a fibre drawn from the polymer. Due to the alignment of the crystalline regions with the long axes of the molecules parallel to the fibre axis, the pattern is essentially identical to a rotation pattern from... 

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