Heat treatment and metallic materials
Learn TTT and CCT diagrams, martensite, quenching and tempering, precipitation hardening, solid-solution strengthening, the Hall–Petch relation, and the basics of steels and aluminum alloys with multiple-choice questions.
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Heat treatment and metallic materials: key points
Heat treatment of steel is the art of producing different microstructures by controlling how fast austenite is cooled. Cooling faster than the nose of the TTT or CCT diagram produces martensite by a diffusionless, shear-type transformation. Martensite is hard and brittle, so it is tempered to restore toughness before use. Alloying elements shift the nose to longer times and raise hardenability.
Metals are strengthened by solid-solution strengthening, work hardening, grain refinement (the Hall–Petch relation) and precipitation hardening. In alloys such as Al-Cu, fine precipitates formed by solution treatment, quenching and aging give large strengthening, but aging too long softens the alloy by overaging. Make sure you know which strengthening mechanisms are used in which alloys.
All Heat treatment and metallic materials questions (12)
- Which statement about the martensitic transformation in steel is correct?
- Which correctly describes the difference between a TTT (isothermal transformation) diagram and a CCT (continuous cooling transformation) diagram?
- Adding alloying elements such as Mn, Cr and Mo to steel improves its hardenability. Which is the correct reason?
- What is the main purpose of tempering a quenched steel?
- The yield stress of a steel follows the Hall–Petch relation σy = σ₀ + k/√d with σ₀ = 70 MPa and k = 0.74 MPa·√m. Which value is closest to the yield stress for an average grain size d = 25 µm?
- The yield stress of a metal was 150 MPa at an average grain size of 100 µm and 250 MPa at 25 µm. Assuming the Hall–Petch relation σy = σ₀ + k/√d, which value is closest to the yield stress at a grain size of 4 µm?
- Which is the correct sequence for precipitation hardening an Al-Cu alloy (duralumin type)?
- Which of the following aluminum alloys cannot be strengthened by precipitation-hardening heat treatment (that is, is non-heat-treatable)?
- Which statement about substitutional solid-solution strengthening is correct?
- A copper wire 10.0 mm in diameter is drawn down to 8.0 mm. What is the percent cold work (reduction in area)?
- Which statement about recrystallization when a cold-worked metal is heated is correct?
- An aluminum alloy reaches peak hardness after 10 hours of aging at 190 °C. If the aging rate follows the Arrhenius form with an activation energy of 120 kJ/mol, which value is closest to the time needed to reach peak hardness at 160 °C? Take R = 8.314 J/(mol·K).