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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# Merit index derivation

Say the beam is of length L and has a square cross section with sides H, and the applied force has magnitude F

The deflection, δ, of the end of the beam is: \[\delta = \frac{{FL}}{{3EI}}\]

I is the second moment of area of the beam and is given by:\[I = \frac{{{H^4}}}{{12}}\]

\[∴\;\;\delta = \frac{{4F{L^3}}}{{E{H^4}}}\]

Now the mass of the beam is given by m = L H^{2} ρ

\[∴\;\;\delta = \frac{{4F{L^5}{\rho ^2}}}{{{m^2}E}}\]

Therefore, in order to minimise δ, we must maximise E/ρ^{2}