Microstructure Property Relationships in Simple Steels 6-49

Question 1

6. Briefly explain the current status of the steel industry, its approximate net worth and name the top three end-use industries.

Answer 1

Steel industry is the second largest after oil and gas
It is worth US$900 billion a year
0.9% decrease in world steel demand due to COVID-19

Question 2

7. Briefly explain three reasons why steels are the no. 1 structural materials.

Answer 2

Adjustable mechanical properties (strength and ductility) - phase transformation
Austenite/ferrite/pearlite/martensite/bainite using temperature
Corrosion resistance, temperatures, wear resistance - Alloying
Availability - Plentiful abundance
Price - economical extraction from ores
Plastic formability - cubic unit cell - lots of slip systems for deformation
Mechanical properties - young’s modulus (stiffness)
ferro magnetism - incomplete 3-d shell
Alloying capabilities to improve properties -> corrosion resistance, increased tensile strength, wear resistance

Question 3

8. What is the main difference between the metastable and the stable Iron-Carbon phase diagram? What influence does a stable transformation behaviour have on the eutectic and eutectoid temperatures? What alloying element provokes stable phase transformations in steels?

Answer 3

Metastable - Fe and Fe3C (cementite)
Stable - Fe and C, in an equilibrium system with infinite cooling time
The stable transformation behaviour increases both the eutectic (1147 -> 1153) and eutectoid (738 -> 723) temperatures.
Silicon suppresses Fe3C and provokes stable phase transformation. Theorised that because Si doesn’t dissolve in Fe3C, it suppresses the development of this phase.

Question 4

9. What are the temperatures, compositions, and overall reactions of (a) the peritectic, (b) eutectic, and (c) eutectoid phase transformations in the metastable Fe-Fe3C system?

Answer 4

Different temperatures from different techniques used by researchers
Undercooling is required to monitor phase transformation (723/727)
Peritectic - L + \delta (ferrite) → \gamma (austenite) - 1493
Eutectic - L → \gamma (austenite) + Fe3C - 1147
Eutectoid - \gamma (austenite)→ \alpha (ferrite) + Fe3C - 727

Question 5

10. Explain the phase transformations during cooling of pure Iron from 1600ºC to room temperature (temperatures and crystal structures).

Answer 5

1600 liquid Fe -> 1538 (Mt) Solidification of delta-ferrite (delta) -> 1394 phase transformation into austenite (gamma) -> 912 phase transformation into ferrite (alpha)

Question 6

11. Why do textbook Fe-Fe3C phase diagrams never go beyond 6.67 wt.% C content? What is the different between the steel and cast iron regions?

Answer 6

Fe3C phase field to 6.70wt% C- past 6.67% C it would be a solid solution with Fe3C and C
Steel has carbon limit of 2.06%
Cast iron has carbon limits from 1.7% to 6.67%C
These regions exist due to the max solubility of carbon in austenite and that properties of cast iron prevents workability (hence, only casting)

Question 7

12. List the four important solid phases in the metastable Fe-Fe3C diagram, their crystal structures and maximum C-solubilities.

Answer 7

Ferrite (a-solid solution)
• bcc, lattice parameter at room temperature: 2.86 Å
• Low hardness (Vickers hardness ~100 HV)
• maximum C-solubility of 0.02 % at 723 °C
• below 769°C ferromagnetic (a-Fe), above this temperature (Curie-temperature) paramagnetic
o extra centre atom
δ-Ferrite (δ-solid solution)
• bcc, lattice parameter at 1400°C: 2.92 Å
• maximum C-solubility of 0.1 % at 1493°C
Austenite (g-solid solution)
• fcc, lattice parameter at 911°C: 3.64 Å
• in Fe-C steels only above 723°C @ 0.8 wt% C
• higher packing density than ferrite, but larger interstitial sites higher solubility for C, C atoms occupy octahedral sites
• paramagnetic
o extra Fe on each face
Cementite (Fe3C)
• rhombohedral unit cell
• no suitable glide systems available, high hardness (Vickers hardness above 800 HV)
• below 215°C magnetic
• transforms into Fe and C at high temperatures
• primary cementite: precipitation of Fe3Cprim between below liquidus line CD
• Primary as the Carbon comes out of the liquid steel
• secondary cementite: precipitation of Fe3Csec along SE line, between eutectic (and eutectoid temperature
• tertiary cementite: precipitation of Fe3Ctert along PQ line. Only visible at low C contents (< 0.2 %)
Pearlite (microstructural constituent)
• eutectoid mixture of ferrite and cementite at 0.8 % C, lamellar arrangement (12 % Fe3C and 88 % ferrite)
• hardness: ~250 HV

Question 8

13. Explain the microstructural evolution of a eutectoid steel during slow cooling from the gamma-phase field to room temperature. At what temperature will phase transformations occur? Describe the final microstructure at room temperature (microstructural components and phases)?

Answer 8

0.8 wt%C
Austenite (gamma) has a higher carbon solubility than ferrite (alpha) and as a result, the carbon diffuses out during the phase transformation that occurs after the temperature passes below 727 degrees. The temperature will vary case by case due to the variation in the undercooling that is required for different metal samples (e.g. the coarsity of the austenitic grains will affect the undercooling).
Ferritic grains will nucleate which will locally expel carbon and increase the surrounding carbon concentrations. These carbon-rick regions will form cementite which creates a diffusion force which pulls carbon from surrounding regions. This push and pull of the carbon in the matrix results in regions that are locally carbon-rich and carbon-deficient which results in the laminar structure of the pearlite.
A pure eutectoid steel will have a 100% pearlite structure when cooled slowly past the eutectoid temperature.

Question 9

14. Explain the microstructural evolution of a hypereutectoid steel during slow cooling from the gamma-phase field to room temperature. At what temperature will phase transformations occur? Describe the final microstructure at room temperature (microstructural components and phases)?

Answer 9

> 0.8 wt%C
Phase transformation will start to occur at a temperature higher than the eutectoid temperature (727). This is dependent on the composition of the hypereutectoid steel (higher C concentrations leads to higher temperatures, probably draw the phase diagram) and at this point secondary cementite will start to nucleate and grow in the austenitic grain boundaries.

Once cooled below the eutectoid temperature, pearlite grains will start to form as described above within the austenitic matrix.

The final microstructure is composed of pearlite grains decorated with cementite grains which are located where the previous austenitic grain boundaries were.

Question 10

15. Explain the microstructural evolution of a hypoeutectoid steel during slow cooling from the gamma-phase field to room temperature. At what temperature will phase transformations occur? Describe the final microstructure at room temperature (microstructural components and phases)?

Answer 10

<0.8 wt%
E.g 0.3 wt%C
The first phase transformations occur at a temperature higher than the eutectoid temperature depending on the composition of the hypoeutectoid steel (lower concentrations of carbon lead to higher temperatures, draw diagram to demonstrate). At this point, ferritic grains will begin to nucleate and grow within the austenitic grain boundaries. Below the eutectoid temperature (727), the remaining austenitic matrix will be converted to pearlite grains.

The final microstructure will have pearlite grains surrounded by ferritic grains which are located where the austenitic grain boundaries previously were. The more hypoeutectoid the composition, the greater the proportion of ferritic grains in the final structure.

Question 11

16. What is the Carbon content of a hypoeutectoid steel with 75% ferrite and 25% pearlite? Why can we easily estimate this without doing any further experiments?

Answer 11

0.2%wt C
The bulk density of pearlite and ferrite is comparable and therefore, the carbon content can be estimated based on the volume composition of the steel. The range is from approximately 0.8% at pure pearlite and below. Therefore, the composition is 25%*0.8%=0.2 wt% C.

Question 12

17. List and briefly explain three methods to discern between grain boundary ferrite and grain boundary cementite.

Answer 12

Image analysis- grain boundary ferrite is thicker and has more rounded grains. Grain boundary cementite is a lot thinner
Hardness- hypoeutectoid steel with grain boundary ferrite is a lot softer than hypereutectoid steel with grain boundary cementite
XRD?- compositional analysis where the ferrite would have a much smaller amount of carbon then cementite.
Secondary electron analysis on scanning electron microscopes can be used where ferritic grains will show up whiter than carbon electrons due to the higher density of electrons in the ferrite (higher iron concentration)

Question 13

18. What is the difference between primary, secondary and tertiary cementite? What are the reasons for precipitation of secondary and tertiary cementite during cooling?

Answer 13

Cool down cast iron
Primary cementite- solidifies between the liquidus and the eutectic temperature, 1538 C and 1147 C. This cementite is precipitating from the liquid. Usually pretty coarse. /

Secondary cementite -form from gamma matrix, precipitates in austenite between the eutectic and the eutectoid temperature, 1147 C and 727 C. This precipitates due to carbon being displaced from the austenitic phase as the carbon solubility in austenite decreases below 1147 C (2.14 wt% C).

Tertiary cementite-precipitate out of alpha, precipitates below the eutectoid temperature, 727 C. This is because there is a major decrease in the carbon solubility in ferrite from austenite below the eutectoid temperature, 727 C from 0.76 to 0.022. This mainly produces the laminar cementite present in the pearlite microstructure. Additionally, as the steel is cooled, the carbon solubility in ferrite decreases further, resulting in more cementite precipitation but this is minimal compared to the initial cementite precipitation.

Different carbon solubilities at different temperatures(different matrix, i.e alpha, gamma)→ formation of secondary and tertiary cementite
Decrease in carbon solubility→ excess carbon goes into the precipitation of cementite

Question 14

19. How can we discern ferrite and austenite via light optical microscopy (name three differences in the morphology of these phases

Answer 14

Austenite:
Typically exhibits twin boundaries. Appears like to parallel grain boundaries on the micrograph
It is also typical to exhibit a junction where three grain boundaries meet at approximately 120 degrees
Ferrite
There are no twin boundaries in ferrite
Cured grain boundaries?
Curvy?

Question 15

20. Explain why pearlite appears as dark-brown and white stripes in the light optical microscope after standard Nital etching. How does this relate to cabbage in a bucket of water?

Answer 15

The cabbage in a bucket of water is meant to be a real-world example of a 3D pearlite structure.

Question 16

21. Explain the consequences of different C-solubility in the austenite versus the ferrite during a diffusive phase transformation. What happens at the transformation front?

Answer 16

The difference between the carbon solubility in ferrite and austenite causes local displacement of the carbon as ferrite grains nucleate. The ferrite has maximal solubility of 0.022 wt% carbon compared to the 2.14 wt% for austenite

This creates locally carbon-rich regions around the ferrite grains which creates a driving force to produce cementite which draws in carbon from other surrounding regions,. The effect of this is the laminar structure of pearlite. (Also refer to question 13)

Question 17

22. What is the influence of grain size on yield strength, toughness and elongation after fracture on a ferritic steel?

Answer 17

Decreasing grain size, increases both strength and toughness and decreases elongation.

Hall-Petch relationship- the yield stress of the steel is inversely proportional to square root of the grain diameter.

A small grain size increases the density of grain boundaries in the material which resists the motion of dislocations in the steel. The dislocations are often stopped by grain boundaries and are unable to continue through to the next grain. This resists plastic deformation and elongation as this requires the flow of dislocations to occur.

Smaller grains also result in increased deflection points for growing cracks as these cracks tend to follow the grain boundaries. These deflections increase the energy absorption of the steel and increases the toughness.

Question 18

23. What is the influence of cold deformation on the yield strength of a ferritic steel?

Answer 18

Increases the yield strength due to strain hardening. Dislocation interlocking etc.
Called strain hardening

Strain induced precipitation? Probably more relevant for hot-rolling

Question 19

24. What is the influence of C-content on the tensile strength of a hypoeutectoid 1 wt.% Mn steel?

Answer 19

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Question 20

25. Rank in terms of increasing cooling rate: martensite, pearlite, ferrite, lower bainite, upper bainite. Sorbite and troostite are different morphologies of which microstructural component, what is the difference and why do they have their own names although they belong to the same parent microstructural component?

Answer 20

Ferrite/pearlite
Upper bainite
Lower bainite
Martensite

Sorbite and troostite are different morphologies of pearlite. These two morphologies are differentiated by the size of the lamellae in the pearlite with troostite having thinner lamellae. The reason that these exist is because before the use of SEM and TEM microscopy techniques, it was not easy to determine the presence of the lamellae and therefore, troostite and sorbite were named due to their different colours.

Question 21

26. Explain the consequences of different C-solubility in the austenite versus the ferrite during a displacive phase transformation. Explain the position of C atoms in the crystal lattice during the phase transformation?

Answer 21

In a FCC, austenitic unit cell, the carbon atoms are located at the octahedral interstitial sites.
Displacive reactions are martensitic reactions.

Question 22

27. How would a micrograph of martensite and retained austenite look like in the light optical microscope? Why do we sometimes get retained austenite and what could we do to get rid of this phase (give one example)?

Answer 22

A micrograph of martensite and retained austenite from a light optical microscope would depict dark needles of martensite amongst lighter grains of retained austenite. The austenite grains are the parent matrix for the martensitic grains and therefore, the shape of the austenitic grains are not definitive.

Retained austenite is present due their not being enough driving force to produce more martensite from this phase. To achieve further martensitic growth, the temperature will need to be colder.

Question 23

28. How would you discern upper versus lower bainite apart in the light optical microscope versus in the transmission electron microscope?

Answer 23

Upper bainite appears to be more similar to pearlite with sheaves of needles and carbides. There is a small amount of time for carbon diffusion to occur and therefore, bainitic ferrite is produced and as well as discontinuous decoration of cementite
Lower bainite is more similar to martnesite with chaotic fine needles. There are very fine particles of Fe3C that precipitate in the needles of the bainite which differentiates it from martensite. However, these precipitates are unlikely to be seen using an optical light microscope. It would be clearer on a transmission electron microscope

Question 24

29. Upper bainite, lower bainite, martensite and tempered martensite may look similar in the transmission electron microscope. What phase can you use to tell them apart and what would be the typical morphology of this phase in each case?

Answer 24

Cementite would allow you to tell the difference
Upper bainite: more similar to ~equilibrium pearlite with discontinuous decorations of cementite
Lower bainite: more similar to martensite except for the fine cementite particles that have nucleated in the matrix of the needles.
Martnesite: there is a minute amount of fine cementite precipitates in the needle grains due to minimal time for diffusion. Most of the carbon however is locked in the martensitic matrix.
Tempered martensite: diffusion allowed to occur resulting in …

Question 25

30. Briefly explain at least three characteristics of martensitic phase transformations.

Answer 25

High undercooling (DT) required
Martensitic transformations occur at low temperatures in an ‚athermal‘ manner, i.e. they start immediately during cooling once a critical temperature (MS) has been reached and stop immediately if further undercooling does not occur. These transformations will only continue if further driving forces are applied. →  cascadal transformations within split seconds 
If there is no further cooling, transformation will cease, resulting in retained austenite, (no DT ® no transformation) 
Transformed areas appear as needle-shaped structures on previously polished surfaces

Question 26

31. What do we mean by the Kurdjumov-Sachs orientation relationship? What is the consequence of C atoms in octahedral lattice sites in the austenite on the reaction product of a displacive phase transformation?

Answer 26

Kurdjumov-Sachs orientation relationship
To transform austenite to martensite, there is a 45 degree rotation of the lattice structure.

Questions seems similar to 26

Question 27

32. Summarize the individual strengthening contributions of fresh martensite with 0.4 wt.% C.

Answer 27

pictures

Question 28

33. You have a small piece of steel with 0.4 wt.% C that was annealed at 900ºC for 10 mins and then quenched in a bucket of ice-water. What is its final microstructure and its mechanical properties? What would you do next to use this piece of steel to make a tool?

Answer 28

Martensite
Mechanical properties - high strength, high hardness, very brittle
Temper to achieve better mechanical properties like ductility
You would temper at approx. 400 degrees (not to hot or cold)

Question 29

34. Why is hardening and tempering of steels done? What is a typical time-temperature-cooling schedule?

Answer 29

Hardened martensite is too brittle for structural applications. Therefore it must be tempered (to reduce brittleness and increase toughness/reduction in area)

picture….

Question 30

35. Explain the tempering stages up to 500ºC of a plain Carbon martensitic steel.

Answer 30

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Question 31

36. Explain the influence of tempering temperature on tensile strength, yield strength and elongation at fracture in a martensitic steel. Give rough estimates of typical strength values in [MPa] and reduction of area in [%] at 200 versus 600ºC.

Answer 31

Temperature (C)
200
600
Tensile Strength (MPa)
275
150
Yield Strength (MPa)
245
140
Reduction in area (%)
42
58

Question 32

37. Explain the influence of C content on the Brinell hardness of fresh martensite, tempered martensite, fine pearlite, and coarse pearlite.

Answer 32

Martensite has highest Brinell hardness regardless of the carbon composition. However, it also increases much more dramatically at lower carbon concentrations and then looks to plateau with increasing concentrations.

Next, hardest is tempered martensite which initially appears to have a linear increase with increasing carbon and then plateaus at around 1 wt%.

Lowest hardness is fine pearlite which has a linear relation with increasing carbon up to 1 wt%.

Question 33

38. Briefly explain two common thermal analysis techniques to study the phase transformation behaviour of steels. What are the signals recorded?

Answer 33

Dilatometry- induction heating process used. The apparatus determines changes in volume (dilation which is change of length over the original length) as a result of phase transformations.
Differential Scanning Calorimetry- 2 samples used with 1 being the one tested and the other being a reference sample with known properties. The machine acts to maintain the same temperature for both the samples. A phase transformation is typically an endothermic or exothermic reaction and therefore, there is an energy input adjustment that is made and recorded to maintain the same temperature for both samples.

Question 34

39. Explain three different scenarios and relate to phase transformations in steels: (a) group of cadets boarding a bus, (b) group of UNSW Sydney first year students boarding the 891 bus, and (c) cadets and their twitchy kindergarten siblings boarding a bus.

Answer 34

Austenite pearlite represents the group of 1st years that board the bus in whatever order they please. Some people may desire to sit closer to others and they are able to do so.
Austenite martensite represents the group of ordered cadets that board the bus.
Bainite austenite relates to c. The cadets remain ordered with the kids representing the atoms that are diffusing through the lattice.

Question 35

40. Compare (upper/lower) bainite to martensite and pearlite. Name two similarities each.

Answer 35

Bainite is intermediate

picture…

Carbon diffusion
C- redistribution film like cementite

Upper Bainite
Linear laths called sheaves

Lower Bainite
Fine needles, messy
Carbides inside the individual laths, no time to diffuse to grain boundaries

Martensite - Military;
No carbon diffusion and distortion of the crystal structure

Question 36

41. What is a carbide-free bainitic steel, what does its final microstructure look like and why is this great for making next generation premium cars?

Answer 36

Carbide-free bainitic steels exhibit excellent mechanical properties due to the fine microstructure containing small plates of bainitic ferrite separated by films of retained austenite (through alloying with Si which suppresses Fe3C formation).
Nano-Composite of strong bainitic ferrite laths and ductile retained austenite.
Overcome strength-toughness trade-off

Question 37

42. What do we mean by TTT diagrams, what are the two main types and for what kinds of heat treatment are they suitable?

Answer 37

Time temperature transformation
Isothermal heat treatment- this demonstrates the different transformations that occur when a material is brought up to a certain temperature and held there. This is ideal for a tempered product which is held at a given temperature for a certain amount of time.
Continuous cooling - designed to show the transformations that occur for a material with a constant cooling rate. This is ideal for an air-cooled or quenched product to determine the transformations.

Question 38

43. Why do isothermal TTT diagrams always have such a characteristic C/nose-shape?

Answer 38

Diffusion
Driving force dT
These two phenomena trade off between each other during the isothermal TTT diagrams. At higher temperatures, diffusion dominates but there is little driving force to stimulate the transformation. At lower, temperatures there is a large driving force but atoms are not able to diffuse as effectively to complete the transformation. Therefore, there typically exists a middle ground where diffusion and the thermodynamic driving force are not maximised individually, but together provide the fastest transformation.

Question 39

44. What would be the most characteristic difference between isothermal TTT diagrams of a hypo-/hyper- versus eutectoid steel.

Answer 39

Hypo - addition of ferrite above pearlite (ferrite + pearlite)
Hyper - grain boundary cementite

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Question 40

45. Why is the martensitic transformation not C-shaped in continuous TTT diagrams, what do the numbers along cooling curves mean, and what is the encircled number at the bottom of each cooling curve?

Answer 40

Martensitic transformation is not a diffusional process
Martensitic transformations are athermal transformations
The numbers along the cooling curves indicate the proportion of the respective matrix component that will form
The encircled number at the bottom of the curves is indicative of the hardness of the final product

picture,…

Question 41

46. Why do we need to give the steel composition and austenitizing temperature and time and/or grain size in TTT diagrams? What influence would a larger grain size have on the phase transformation behaviour?

Answer 41

?

Question 42

47. What do we mean by gamma- and alpha-stabilizers for steels and name at least three each.

Answer 42

They extend the respective phase fields (some have high solubility and others has restricted solubility). Austinitic steel requires the stabilisers to get the best quality for lower price

gamma - NiCCoCuMnN
alpha - CrAlTiTaSuNiVW

Question 43

48. What is intergranular corrosion in stainless steels? What is the so-called sigma-phase?

Answer 43

Cr-Ni steels → austenite, 18/8 or 18/10
Twin boundaries, triple points
Ni is austenite stabiliser
Austenite (FCC) - nickel is a substitutional alloy
Sigma phase (fe cr) ~45 wt% not desirable as it leaches Cr from steel into grain boundary - under certain circumstances, the heat will cause Cr to diffuse out and deplete Cr from steel. This exposes the steel matrix to corrosion.

Question 44

49. What is general the influence on the phase transformation behaviour of steels? Give three examples of alloying elements and their impact on selected phase transformations.

Answer 44

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