Problem 1 · Materials selection
A beam of fixed length must meet a specified bending stiffness at minimum mass. Its square cross section may be sized freely, so mass is minimized by the material with the largest value of E^(1/2)/ρ. Candidate properties are: carbon steel E = 200 GPa, ρ = 7.85 g/cm³; aluminum alloy E = 70 GPa, ρ = 2.70 g/cm³; titanium alloy E = 114 GPa, ρ = 4.43 g/cm³; magnesium alloy E = 45 GPa, ρ = 1.77 g/cm³. The best choice is:
- A.Carbon steel
- B.Aluminum alloy
- C.Titanium alloy
- D.Magnesium alloy
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Answer: D: Magnesium alloy
E^(1/2)/ρ in (GPa)^(1/2)/(g/cm³): steel √200/7.85 = 1.80; aluminum √70/2.70 = 3.10; titanium √114/4.43 = 2.41; magnesium √45/1.77 = 3.79. Magnesium has the largest index and gives the lightest beam. Choosing by E alone (steel) or by E/ρ (which is nearly equal for all four metals) misses that a freely sized bending section rewards low density more than high modulus.
Materials selection · Reference Handbook: Materials Science/Structure of Matter
