Question:
A plain carbon steel sample of eutectoid composition is heat treated from austenizing temperature to achieve a microstructure consisting of ONLY bainite.
Which ONE of the following is the corresponding heat treatment method?
A plain carbon steel sample of eutectoid composition is heat treated from austenizing temperature to achieve a microstructure consisting of ONLY bainite.
Which ONE of the following is the corresponding heat treatment method?
Which ONE of the following is the corresponding heat treatment method?
Show Hint
Compare the four heat treatments by where the isothermal hold sits on the TTT diagram.
Updated On: Aug 3, 2026
- Normalizing
- Annealing
- Martempering
- Austempering
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We need to pick the heat treatment that gives a fully bainitic structure, with no pearlite, martensite, or ferrite mixed in, for a eutectoid steel.
The key is to remember what each named process does on a TTT (time-temperature-transformation) diagram.
Step 2: Check option (A), Normalizing:
Normalizing means heating to austenite and cooling in air.
Air cooling is fairly fast but not fast enough to miss the pearlite nose, so it gives fine pearlite, not bainite, so this option is wrong.
Step 3: Check option (B), Annealing:
Annealing means very slow cooling inside the furnace.
This gives plenty of time for the pearlite reaction, so the result is coarse pearlite, not bainite, so this option is wrong.
Step 4: Check option (C), Martempering:
Martempering means quenching fast to just above the martensite start temperature, holding briefly to even out the temperature, then air cooling through the martensite range.
The final product is martensite (later tempered), not bainite, so this option is wrong.
Step 5: Check option (D), Austempering:
Austempering means quenching fast enough to miss the pearlite nose, then holding isothermally in the bainite range (above Ms but below the nose) until the austenite fully converts.
Holding in that temperature band is exactly how you get a microstructure that is only bainite, with no martensite formed later, so this option is correct.
\[ \text{Quench past nose} \rightarrow \text{hold in bainite range} \rightarrow \text{100 percent bainite} \]
Final Answer:
Only austempering isothermally transforms all the austenite into bainite. \[ \boxed{\text{Austempering}} \]
We need to pick the heat treatment that gives a fully bainitic structure, with no pearlite, martensite, or ferrite mixed in, for a eutectoid steel.
The key is to remember what each named process does on a TTT (time-temperature-transformation) diagram.
Step 2: Check option (A), Normalizing:
Normalizing means heating to austenite and cooling in air.
Air cooling is fairly fast but not fast enough to miss the pearlite nose, so it gives fine pearlite, not bainite, so this option is wrong.
Step 3: Check option (B), Annealing:
Annealing means very slow cooling inside the furnace.
This gives plenty of time for the pearlite reaction, so the result is coarse pearlite, not bainite, so this option is wrong.
Step 4: Check option (C), Martempering:
Martempering means quenching fast to just above the martensite start temperature, holding briefly to even out the temperature, then air cooling through the martensite range.
The final product is martensite (later tempered), not bainite, so this option is wrong.
Step 5: Check option (D), Austempering:
Austempering means quenching fast enough to miss the pearlite nose, then holding isothermally in the bainite range (above Ms but below the nose) until the austenite fully converts.
Holding in that temperature band is exactly how you get a microstructure that is only bainite, with no martensite formed later, so this option is correct.
\[ \text{Quench past nose} \rightarrow \text{hold in bainite range} \rightarrow \text{100 percent bainite} \]
Final Answer:
Only austempering isothermally transforms all the austenite into bainite. \[ \boxed{\text{Austempering}} \]
Was this answer helpful?
0
0
Top GATE PI Engineering Materials Science Questions
- A thin cylinder has length $L$, diameter $d$, and thickness $t$. It is made of a material with modulus of elasticity $E$ and Poisson’s ratio $\mu$. When the cylinder is subjected to an internal pressure $P$, the change in length is
- An Iron-Carbon alloy at room temperature has pro-eutectoid ferrite as a constituent in the microstructure, that formed above the eutectoid temperature. The alloy also has pearlite as another constituent that has formed below eutectoid temperature.
Which ONE of the following best describes this Iron-Carbon alloy?
Top GATE PI Questions
- In an air-standard Otto cycle, the pressure and temperature of air just before the compression stroke are 200 kPa and 26.85 °C, respectively. The combustion process is assumed to be a constant volume process, where 1.02 MJ/kg heat is added. The cycle efficiency is 50%. The adiabatic index \(γ\) and specific heat at constant volume \(C_v\) can be considered to be constant during the process (corresponding values taken at the mean cycle temperature).
Assuming that the ideal gas law is applicable, \(γ = \frac 43\) and \(c_v= 0.85\ kJ/kg-K\), the maximum pressure (in MPa) reached during the cycle is _____ . (Rounded off to 1 decimal place) - A metallic cylindrical pressure vessel, used to store compressed air in a plant, has 1 m mean radius and 4 mm wall thickness. The maximum allowable normal and shear stresses in the cylindrical portion of the vessel are 100 MPa and 40 MPa, respectively. Considering only these data in the design, the maximum allowable internal gauge pressure (in MPa) of the compressed air is ______ . (Rounded off to 2 decimal places)
- A flat belt drive with pulley of r = 20 cm radius is designed to transmit 6.283 kW power at 600 RPM. In the figure, τ is the corresponding torque. If the coefficient of static friction between the belt and the pulley is 0.3, then the minimum value of the tightening force F (in kN) required to prevent the belt slip is _____ . (Rounded off 2 decimal places)
- Mild steel plates are welded to make butt joints by arc welding with 85% heat transfer efficiency ignoring other losses. The first weld joint is made by selecting arc voltage of 30 V and current of 180 A with a welding speed of 6 mm/s. Using identical plates, a second weld joint is made with the same arc voltage and a welding speed of 8 mm/s. If both the welds have the same heat input, then the welding current (in A) for the second weld joint is _____. (Answer in integer)
- In a single pass cold rolling operation, a flat plate is reduced to a thickness of 3 mm. In this operation, two rolls of diameter 400 mm each are rotating in opposite direction at 300 RPM, and the elastic deflection of these rolls is negligible. The angle of bite is 10°. If the neutral point is present at an angle of 7° from the exit side, then the thickness of the plate (in mm) at the neutral point is _____ . (Rounded off to 1 decimal place)
View More Questions