Nonferrous metals and alloys
In designing α and α+β titanium alloys, the effect of α-stabilizing elements is summarized by the aluminum equivalent [Al]eq = [Al] + [Sn]/3 + [Zr]/6 + 10[O] (mass%), with a guideline that it should not exceed about 9. For a Ti-6Al-2Sn-4Zr-2Mo alloy containing 0.10 mass% oxygen, which is closest to [Al]eq?
- AAbout 8.3
- BAbout 7.4
- CAbout 13
Show answer and explanation
Answer: A. About 8.3
[Al]eq = 6 + 2/3 + 4/6 + 10 × 0.10 = 6 + 0.67 + 0.67 + 1.0 ≈ 8.3, below the guideline of 9. About 7.4 forgets to multiply oxygen by its coefficient of 10, and about 13 adds Sn and Zr without dividing by their coefficients. If the aluminum equivalent is too high, the ordered phase α₂ (Ti₃Al) precipitates in the α phase, lowering ductility and resistance to stress corrosion cracking. As the fact that a mere 0.1% oxygen counts as much as 1% aluminum shows, interstitial oxygen strongly stabilizes the α phase and raises strength, so the oxygen content is an important control item that distinguishes grades of an alloy. Mo is a β-stabilizing element and does not enter this expression.