Mechanics of Fibre-reinforced Composites
AimsBefore you startIntroductionStiffness of long fibre compositesStrength of long fibre compositesComposite vaulting poles – why don´t they break?Toughness of composites and fibre pull–outOff–axis loading of a laminaStiffness of laminatesTensile–shear interactions and balanced laminatesOut–of–plane stresses and symmetric laminatesFailure of laminates and the Tsai–Hill criterionSummaryQuestionsGoing furtherTLP creditsTLP contentsShow all contentViewing and downloading resourcesAbout the TLPsTerms of useFeedbackCredits Print this page
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Transverse strength derivation
Quick proof
End-on view |
Side-on view |
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| Fibre volume fraction = f = π;r2L / 4R2L = πr2 / 4R2 | Maximum fibre cross-sectional area fraction = 2r L / 2R L = \(2\)\( \sqrt{\frac{f}{\pi }} \) = maximum fractional reduction in cross-sectional area |
Minimum matrix cross-sectional area fraction = X = 1 – \(2\)\( \sqrt{\frac{f}{\pi }} \)
The composite ultimate transverse tensile stress is therefore a fraction X of the matrix ultimate tensile stress:
\[{\sigma _{2u}} = {\sigma _{mu}}\left[ {1 - 2{{\left( {\frac{f}{\pi }} \right)}^{1/2}}} \right]\]