Question

TTT diagram of a eutectoid steel is shown below. Match the heat treatment cycle (in Column I) with its microstructure (in Column II).

• A. P-1, Q-2, R-4
• B. P-2, Q-3, R-2
• C. P-2, Q-4, R-1
• D. P-2, Q-3, R-1

Hint:

- Slow cooling above TTT nose → Pearlite only.
- Quenching to $M_s$ region → Martensite (with some pearlite).
- Cooling into intermediate region → Bainite only.

Answer 1

The Correct Option is C

Step 1: Understand the TTT diagram.
- The TTT diagram of eutectoid steel shows transformations of austenite under different cooling paths. - Key phases: Pearlite (high-temp transformation), Bainite (intermediate), and Martensite (fast quench).
Step 2: Analyze path P.
- Path P involves slow cooling above the “nose” of the TTT curve. - This gives **100% pearlite**. \[ P = 2 (\text{Pearlite only}) \]
Step 3: Analyze path Q.
- Path Q involves quenching directly below $M_s$ (martensite start). - Here, austenite transforms mostly to **martensite**. - But since cooling rate is not infinite, a small amount of pearlite also forms. \[ Q = 4 (\text{Pearlite + Martensite}) \]
Step 4: Analyze path R.
- Path R cools into the bainite region (intermediate temperatures). - Transformation results in **bainite only**. \[ R = 1 (\text{Bainite only}) \]

Step 5: Final Matching.
\[ P-2,\; Q-4,\; R-1 \] Final Answer: \[ \boxed{\text{(C) P-2, Q-4, R-1}} \]