A tie (tension member) of fixed length L must carry a tensile load F without yielding, and its cross-sectional area may be chosen freely. To make it as light as possible, which performance index should be maximized, with σf the yield strength and ρ the density of the material?

Answer: A. σf/ρ

The condition for not yielding is F/A ≤ σf, so the required area is A = F/σf. The mass of the tie is m = ρAL = FL(ρ/σf), and since F and L are fixed, minimizing mass means choosing the material with the largest σf/ρ (specific strength). E/ρ (specific modulus) is the index to maximize for a tie whose constraint is not strength but the amount of extension, that is, stiffness. ∛(σf²)/ρ is the index for a beam in bending whose cross-section size is varied to avoid yielding; it comes from the bending stress varying inversely with the cube of the section dimension. The key point is that even for “light and strong,” the index changes with how the load is applied.