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6.2: Stiffness of Long Fiber Composites

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    Before we go into the details it is worth noting that the most suitable material for a given application may well not be the stiffest material. For example, for a given force applied to the free end of a cantilevered composite beam, the minimum deflection per unit mass is achieved by maximizing the merit index E / ρ2.

    Click here for Merit index derivation. (See Optimisation of Materials Properties in Living Systems for more on merit indices)

    An Ashby property map (Young's Modulus against Density) for composites:

    An Ashby property map for Youngs modulus vs density

    The axial and transverse Young's Moduli can be predicted using a simple slab model, in which the fibre and matrix are represented by parallel slabs of material, with thicknesses in proportion to their volume fractions, E and (1- E).

    Axial Loading: Voigt model

    The fibre strain is equal to the matrix strain: EQUAL STRAIN.

    Diagram of axial loading of composite


    \[\varepsilon_{1}=\varepsilon_{1 f}=\frac{\sigma_{1 f}}{E_{f}}=\varepsilon_{1 m}=\frac{\sigma_{1 m}}{E_{m}}=\frac{\sigma_{1}}{E_{1}}\]

    See definitions of terms

    For a composite in which the fibres are much stiffer than the matrix ( Ef >> Em ), the reinforcement fibre is subject to much higher stresses ( σ1f >> σ1m ) than the matrix and there is a redistribution of the load. The overall stress σ1 can be expressed in terms of the two contributions:

    \[\sigma_{1}=(1-f) \sigma_{1 \mathrm{m}}+f \sigma_{1 f}\]

    The Young's modulus of the composite can now be written as

    \[E_{1}=\frac{\sigma_{1}}{\varepsilon_{1}}=\frac{(1-f) \sigma_{1 \mathrm{m}}+f \sigma_{1 \mathrm{f}}}{\left(\frac{\sigma_{1 \mathrm{f}}}{E_{\mathrm{f}}}\right)}=(1-f) E_{\mathrm{m}}+f E_{\mathrm{f}}\]

    This is known as the "Rule of Mixtures" and it shows that the axial stiffness is given by a weighted mean of the stiffnesses of the two components, depending only on the volume fraction of fibres.

    Transverse Loading: Reuss Model

    The stress acting on the reinforcement is equal to the stress acting on the matrix: EQUAL STRESS.

    \[\sigma_{2}=\sigma_{2 f}=\varepsilon_{2 f} E_{f}=\sigma_{2 m}=\varepsilon_{2 m} E_{m}\]

    The net strain is the sum of the contributions from the matrix and the fibre:

    \[\varepsilon_{2}=f \varepsilon_{2 \mathrm{f}}+(1-f) \varepsilon_{2 \mathrm{m}}\]

    from which the composite modulus is given by:

    \[E_{2}=\frac{\sigma_{2}}{\varepsilon_{2}}=\frac{\sigma_{2 \mathrm{f}}}{f \varepsilon_{2 \mathrm{f}}+(1-f) \varepsilon_{2 \mathrm{m}}}=\left[\frac{f}{E_{\mathrm{f}}}+\frac{(1-f)}{E_{\mathrm{m}}}\right]^{-1}\]

    This "Inverse Rule of Mixtures" is actually a poor approximate for E2 since in reality regions of the matrix 'in series' with the fibres, close to them and in line along the loading direction, are subjected to a high stress similar to that carried by the reinforcement fibres; whereas the regions of the matrix 'in parallel' with the fibres (adjacent laterally) are constrained to have the same strain as the fibres and carry a low stress. This leads to non-uniform distributions of stress and strain during transverse loading, which means that the model is inappropriate. The slab model provides the lower bound for the transverse stiffness.

    Diagram of transverse loading of composite

    A more successful estimate is the semi-empirical Halpin-Tsai expression:

    \[E_{2}=\frac{E_{m}(1+\xi \eta f)}{(1-\eta f)}\]


    \[\eta=\frac{\left(\frac{E_{f}}{E_{m}}-1\right)}{\left(\frac{E_{f}}{E_{m}}+\xi\right)} \text { and } \xi \approx 1\]

    An even more powerful, but complex, analytical tool is the Eshelby method (see Hull and Clyne, 1996).

    This page titled 6.2: Stiffness of Long Fiber Composites is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Dissemination of IT for the Promotion of Materials Science (DoITPoMS) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.