Dissemination of IT for the Promotion of Materials Science (DoITPoMS)

# 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 H2 ρ

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

Therefore, in order to minimise δ, we must maximise E/ρ2