 Solution

On a logarithmic strength–density materials selection chart, constant $M_A$ gives ${\sigma }_f=M_A\rho$, a guideline of slope one; constant $M_B$ gives ${\sigma }_f=M_B^2{\rho }^2$, of slope two. Moving each guideline upwards selects better candidates. A suitably qualified longitudinal carbon-fibre-reinforced polymer is a promising candidate for both designs. Composite families occupy favourable high-strength/low-density regions in published course selection charts, but a beam must meet its compressive as well as tensile allowable.^[ref]

For scale, an explicitly illustrative candidate with ${\sigma }_{\lim }=1\mathrm{GPa}$ in both required loading senses and $\rho =1600\mathrm{kg}{\mathrm{m}}^{-3}$ would have $M_A=6.25\times 10^5{\mathrm{m}}^2{\mathrm{s}}^{-2}$ and $M_B=19.76\mathrm{P}{\mathrm{a}}^{1/2}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$, exceeding the best supplied materials. These are assumed screening values, not measured properties of a specified product.