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Year 2 > Materials 2 - Section A > Flashcards

Materials 2 - Section A Flashcards

1
Q

Which one of the following statements is correct in relation to single crystal alloys?

a. A single crystal alloy will not undergo yielding prior to fracture because it will not contain dislocations.
b. A single crystal alloy can be assumed to be isotropic because it will not have any grain boundaries.
c. A single crystal alloy will be anisotropic.
d. A single crystal alloy cannot contain precipitates of a second phase.
e. A single crystal alloy will have poor resistance to creep because it does not contain any grain boundaries.

A

c. A single crystal alloy will be anisotropic.

a. Ofc it contains dislocations.
b. Has no grain boundaries but isn’t isotropic.
c. Yes anisotropic
d. Yes ofc it can
e. Has no grain boundaries which means more resistant to creep, not less.

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2
Q

Which one of the following statements is correct in relation to phase diagrams for binary alloys?

a. A phase diagram can reveal the rate at which a new phase will form provided that it forms through a diffusion-based process.
b. Phase diagrams can reveal whether an alloy has a ductile-to-brittle transition temperature.
c. Phase diagrams can only be used with confidence if the homologous temperature (T/Tm) is above 0.4.
d. A phase diagram can reveal both the number of phases in an alloy, and the chemical composition of each phase, when thermodynamic equilibrium is achieved.
e. The Lever Rule can be used to establish whether a material is thermodynamically stable.

A

d. A phase diagram can reveal both the number of phases in an alloy, and the chemical composition of each phase, when thermodynamic equilibrium is achieved.

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3
Q

Which one of the following statements is correct in relation to atomic diffusion?

a. Diffusion in metals only occurs when there is a thermodynamic driving force for a phase transformation.
b. Interstitial diffusion in metals and alloys is contingent on the presence of vacancies in the crystal lattice.
c. The movement of dislocations in a metal can only occur through a process that involves substitutional diffusion.
d. At a given temperature, diffusion in face-centred cubic (FCC) iron is faster than diffusion in body-centred cubic iron because FCC crystals have more close packed directions.
e. At a given temperature, diffusion in BCC iron is faster than diffusion in FCC iron because BCC iron has a lower atomic packing factor.

A

e. At a given temperature, diffusion in BCC iron is faster than diffusion in FCC iron because BCC iron has a lower atomic packing factor.

a. No, diffusion can occur without a driving force.
b. No, interstitial atoms neither replace nor substitute. They fit in the small spaces in-between atoms in the interstitial sites.
c. Can also happen via interstitial.
d. No, diffusion is due to concentration gradients only.
e. Yes, lower APF means less packed, so diffusion is easier.

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4
Q

BCC vs FCC

How many close packed planes and close packed directions?

Atomic packing factor, and its effects?

What are the transitions and properties?

A

BCC: No close packed planes, 4 close packed directions. Lower APF, meaning diffusion is easier as more space. A ductile to brittle transition.

FCC: 4 close packed planes and 12 close packed directions. Higher APF and is ductile.

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5
Q

What are the main differences between theoretical and real strength?

A
  1. Theoretical is calculated by estimating the stress required to separate planes of atoms.
  2. Theoretical assumes crystals are perfect.
  3. Real crystals contain defects (dislocations).
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6
Q

List and describe the 5 strengthening mechanisms in metals?

A
  1. Work hardening: Causes dislocations to move through the lattice, which means higher concentration of defects, hence hardening.
  2. Solid-solution strengthening: Alloyed atoms which are a different size are introduced into the lattice, causing dislocations.
  3. Precipitation hardening.
  4. Grain boundary strengthening: More boundaries means more grains meeting at different orientations. Hence harder to dislocate.
  5. Transformation strengthening: Heating then cooling to introduce phase transformation.
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7
Q

What is meant by “stable”?

A

Stability is defined by the thermodynamic stability in the metal. Thermodynamic stability is achieved when Gibbs Free Energy is at a minimum.

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8
Q

How is Thermodynamic Equilibrium achieved?

A

It can be achieved by either very slow cooling or heating. This gives diffusion more time to take place.

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9
Q

How does diffusion occur?

A

Diffusion occurs due to concentration gradients. I.e. high to low concentration etc.

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10
Q

List and describe the 2 types of diffusion?

A

Diffusion occurs by either vacancy or interstitial diffusion.

Vacancy means alloying elements that are a similar size to the solvent atoms in the structure, move into a vacancy nearby or replace existing, similar sized atoms.

Interstitial means alloying elements that are small enough to fit in between atoms in the interstitial sites. Carbon is a common interstitial alloying element.

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11
Q

What structure is Ferrite?

A

BCC

So a ferritic structure is a BCC structure,

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12
Q

What structure is Austenite?

A

FCC

So an austenitic structure is an FCC structure,

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13
Q

What effects the rate of diffusion?

A
  1. Number of vacancies
  2. Crystal structure
  3. Diffusing atoms
  4. Temperature
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14
Q

What does a phase diagram show?

A

It shows the number of phases present in an alloy at a given weight and temperature. And also the chemical composition of each phase when thermodynamic equilibrium is achieved.

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15
Q

Label the phase diagram?

A
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16
Q

What phases exist at points A, B, and C? And are they BCC or FCC?

A
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17
Q

Label the Stress-Strain diagram

A
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18
Q

Eutectic = x -> Solid A + Solid B

Eutectoid = y -> Solid A + Solid B

A

Eutectic = Liquid -> Solid A + Solid B

Eutectoid = Solid -> Solid A + Solid B

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19
Q

If Carbon % is: <0.8%, =0.8%, >0.8%?

A

< 0.8%, Hypoeutectoid
= 0.8%, Eutectoid
> 0.8%, Hypereutectoid

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20
Q

What is the relationship with carbon concentration and strength, ductility and pearlite?

A

Increased carbon concentration =

Increased strength, decreased ductility, more pearlite.

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21
Q

For Ferrite/Cementite and Pearlite:

Near Equilibrium?
Cooling Rate?
Diffusion?
Crystal Structure?

A

Near Equilibrium - Yes
Cooling Rate - Slow
Diffusion - Both substitutional and interstitial
Crystal Structure - BCC

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22
Q

For Bainite:

Near Equilibrium?
Cooling Rate?
Diffusion?
Crystal Structure?

A

Near Equilibrium - No
Cooling Rate - Medium
Diffusion - Interstitial
Crystal Structure - BCC

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23
Q

For Martensite:

Near Equilibrium?
Cooling Rate?
Diffusion?
Crystal Structure?

A

Near Equilibrium - No
Cooling Rate - Fast
Diffusion - No diffusion at all
Crystal Structure - BCT

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24
Q

What is Bainite and how is it formed?

A

Bainite is formed by shear transformation. The carbon is not trapped and can escape.

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25
Q

What is Martensite?

And explain why tempered Martensite increases toughness?

A

Martensite has a high amount of carbon trapped, and hence is hard and brittle. It has a small grain size due to high nucleation rate.

Tempered Martensite causes trapped carbon to escape the Martensite and form carbides. This decreases hardness and brittleness, but increases toughness. It dramatically changed the distribution of Cementite.

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26
Q

What do Continuous Cooling Transformation Diagrams show?

A

They show the rate of austenite transformation if the temp is dropped at an instant. They indicate which phase will form when equilibrium is not maintained.

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27
Q

How is Pearlite formed?

What are the differences between coarse and fine Pearlite?

A

Pearlite is formed due to the diffusion of carbon. It is alternating layers of Ferrite and Cementite.

Coarse Pearlite is soft and formed at high temps.
Fine Pearlite is hard and formed at low temps.

Fine Pearlite has more interfaces and as a result means it’s harder for dislocation to move, hence stronger.

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28
Q

What influence do alloying elements have on diffusion?

A

They can slow transformations down that depend on diffusion. Imagine large alloying elements that block atoms that want to diffuse.

They make it easier for Martensite to form as by slowing diffusion you can trap the carbon.

They can be added to steel to increase hardenability and strength.

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29
Q

Reheating Steel after Quenching causes what?

A

It causes a decrease in strength but toughness to be restored.

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30
Q

What are Stainless Steels?

A

Steels with >12% Chromium alloyed to prevent oxidation.

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31
Q

What are the effects of adding Chromium and Nickel as an alloying element?

A

We know the structure of the alloying elements promotes the same structure in the steel they’ve been added to. Therefore:

Chromium, which is BCC will promote Ferrite as BCC, but reduce Austenite phase as FCC.

Conversely, Nickel which is FCC will promote Austenite but reduce Ferrite.

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32
Q

What is an effect of alloying regarding solidification?

What influence does the structure (BCC or FCC) of Stainless Steels have on it’s properties?

A

Alloying can lead to a complex solidification process where more than one phase forms.

An FCC structured Stainless Steels promotes ductility and toughness.

A BCC promotes corrosion resistance.

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33
Q

What are the features of Austenitic, Ferritic and Duplex Stainless Steels?

A

Austenitic (FCC) account for 70% of stainless steels, have good corrosion resistance and excellent ductility and toughness.

Ferritic (BCC) are less common, have excellent corrosion resistance, however can be brittle at high temperatures.

Duplex (FCC and BCC mixture) compromise between both. So they provide very good corrosion resistance alongside very good ductility and toughness. They are far cheaper than Austenitic, so can provide a good compromise.

34
Q

What is Sensitisation and how does it occur?

What does the diagram look like?

A

Sensitisation is where chromium carbides form on the grain boundaries.

This occurs due to the stainless steel being heated to 700 degrees and then either cooled slowly or held there for long periods.

At high temps, diffusion of carbon allows chromium carbides to form on grain boundaries. Grain boundaries have poor atomic packing, making diffusion is easy.

35
Q

Why is Sensitisation undesirable and how can it be prevented?

A

It’s undesirable since we need the Chromium to be free to form Chromium Oxide on the grain boundaries. Instead of forming Chromium Carbide.

There are 3 logical solutions:

  1. Don’t heat up to or as high as 700 degrees.
  2. Reduce carbon concentration, reducing the likelihood of Chromium Carbides forming.
  3. Add another alloying element that forms Carbides easier and more quickly than Chromium.
36
Q

What is Sigma Phase Embrittlement, how is it formed, and why is it undersirable?

A

A phase made up of FeCr, Iron and Chromium called Ferrochrome.

It is formed when Ferrite stainless steels are exposed to temps of 500-900 degrees.

Similarly to Sensitisation, it takes away available Chromium for corrosion protection. I.e. no free Chromium to form Chromium Oxide protective layer.

37
Q

What is 475 degree Embrittlement?

A

Embrittlement that occurs mostly at 475 degree, but can be in the range of 300-550. Creates a Chromium rich phase and a Chromium depleted phase.

38
Q

What is creep, when does it occur, and what are the names of the 2 mechanisms?

A

Creep is the permanent deformation of a material over time.

It occurs when materials are at elevated temperatures.

There are 2 mechanisms of creep: diffusion and dislocation.

39
Q

Diffusion Creep:

What temperatures does it occur at?

Describe the process, correlations, and most likely size of grain?

A

It is the most likely mechanism to occur at very high temps close to the melting point, and homologous temps of 0.8-0.9.

Applied stresses cause grains to elongate in the direction of loading.

The distance over which the diffusion takes place is correlated to the size of the grains.

Fine grain materials are more likely to undergo diffusion creep.

40
Q

Dislocation Creep:

What temperatures does it occur at?

Describe the process, and correlations?

A

It is the most likely mechanism to occur at homologous temps of 0.4-0.7.

Involves the movement of dislocations through the lattice.

The rate of dislocation is correlated to the applied stress.

When the applied stress < yield stress, dislocations are pinned. They can be unpinned via a process called dislocation climb.
The dislocations can climb as atoms at the core of a dislocation move into vacancies. Hence, it is reliant on substitutional diffusion.

41
Q

Describe dislocation climb, and how it occurs?

A

When the applied stress < yield stress, dislocations are pinned.

They can be unpinned via a process called dislocation climb.

The dislocations can climb as atoms at the core of a dislocation move into vacancies. Hence, it is reliant on substitutional diffusion.

42
Q

What does the stages of creep graph look like, and describe briefly each phase?

A
43
Q

Describe primary creep?

A
  1. High strain rate that rapidly reduces.
  2. Rapid movement of dislocations.
  3. Work hardening occurs, and the density of dislocations increases, and the strain rate reduces.
  4. Strain rate stabilises
44
Q

Describe secondary creep?

A
  1. Constant strain rate, hence “steady-state” creep.
  2. Balance achieved between work hardening and dislocations unpinning through diffusion.
  3. Spends the majority of time in this stage.
45
Q

Describe tertiary creep?

A
  1. Strain rate increases rapidly.
  2. Large cracks form out of smaller ones combining.
  3. Creep deformation accelerates.
46
Q

Label the following diagram?

A
47
Q

What is the ductile to brittle transition temperature?

A

The DBTT is the temperature that corresponds to an absorbed energy half way between brittle and ductile, i.e. lower and upper shelves.

48
Q

Fill in the gaps of what increases, and decreases the DBTT?

A
49
Q

Which structure is best suited to very low temperatures, and what are the effects?

A

FCC metals are best suited for low temperatures down to -196 degrees.

At temperatures this low, Austenite will transform to Martensite.

50
Q

An increased carbon content does what to the following?

Yield strength
Elongation
Hardness
Charpy absorbed energy

A

Yield strength - Increases (more carbon = stronger).
Elongation - Decreases (stronger = less likely to elongate).
Hardness - Increases
Charpy absorbed energy - Decreases

51
Q

When austenitising and tempering, why is it essential for the steel to fully transform to austenite?

A
  1. When tempering we want both a uniform distribution of carbon, and for carbides to form.
  2. If not fully Austenite, there will be a mixture of different phases, including Austenite.
  3. Therefore the carbon won’t be uniformly distributed, which is undesirable.
52
Q

Why are austenitisation and tempering heat treatments less effective in steels that have low carbon contents?

A
  1. When tempering we want both a uniform distribution of carbon, and for carbides to form.
  2. Less carbon means less carbides.
53
Q

When would we use a:
Phase diagram?
CCT diagram?

A

Phase diagram: When the cooling is slow so diffusion can take place. And when we have thermodynamic equilibrium.

CCT diagram: When the cooling is rapid and instant and when we’re not near equilibrium.

54
Q

What phases would form for Cases 1, 2, and 3?

A

Case 1: Crosses Martensite line only.
Case 2: Crosses both Bainite lines, hence Bainite only.
Case 3: Crosses three lines, Ferrite, Pearlite and Bainite.

55
Q

A structural steel is being considered for an application in which the service temperature can vary between -40 to 50 degrees. The tensile properties and toughness have been measured at room temperature (20 degrees) and they are clearly acceptable. Are any further checks necessary?

A

Yes (obvious by the question). We know steels undergo Ductile to Brittle transformations. So we need to check that at the low temperatures of the service range, so -40 degrees, that the steel doesn’t become too brittle that it loses its toughness.

56
Q

Stainless steels tend to have low carbon contents. Explain why
this is the case?

A

Sensitisation exists. This is where if a Stainless Steel is heated to 700 degrees, and then cooled slowly or held at that temperature, diffusion occurs. This causes Chromium Carbides to form on grain boundaries. This is undesirable since we want Chromium to act as a corrosion resistance layer and form Chromium Oxide. Once it forms Chromium Carbides, it can’t form Oxides and hence the steel becomes prone to corrosion.

To prevent this we can:

  1. Reduce carbon content.
  2. Don’t heat to 700 degrees.
  3. Use another alloying element to form Carbides instead of Cr.
57
Q

Materials with a face-centred cubic crystal structure tend
to be less susceptible to creep deformation than materials
with a body-centred cubic crystal structure. Explain why
this is so?

A
  1. FCC structures have a higher APF.
  2. A higher APF make substitutional diffusion more difficult.
  3. Creep deformation requires substitutional diffusion.
  4. Therefore creep deformation will occur at a slower rate for an FCC compared to a BCC.
58
Q

Single crystal nickel-based alloys are used in gas turbine engines where they are exposed to temperatures that can exceed 1000°C. Giving consideration to the service temperature, explain why it might be desirable to use materials that consist of a single grain?

A
  1. At such high temperatures, creep deformation occurs.
  2. Creep deformation relies upon substitutional diffusion, which takes place at grain boundaries.
  3. Single crystal alloys don’t have any grain boundaries.
  4. Hence, making it difficult for creep to occur.
59
Q

What is pearlite?

A

Pearlite is the specific microstructure of two phases, Ferrite and Cementite in alternating layers.

The two phases have to be cooled slowly to room temperature. Note: They have to be in alternating layers, otherwise Pearlite doesn’t exist.

60
Q

Are phase diagrams reliable at predicting the microstructure of a steel when cooled very quickly?

A

No, because phase diagrams only provide information on an alloy that is under thermodynamic equilibrium.

Thermodynamic equilibrium occurs when all diffusion has taken place.

This takes time, so we know thermodynamic equilibrium has not been achieved.

61
Q

How can the portions of ferrite and austenite differ so much in stainless steels?

A

They differ due to different alloying elements.

Nickel which is FCC, promotes austenite (FCC).

Whilst chromium which is BCC, promotes ferrite (BCC)

62
Q

How would you find the fractions of C1 and C2 using the Lever Rule?

A

Fraction of phase 1 = (C2 - C) / (C2 - C1)

Fraction of phase 2 = (C - C1) / (C2 - C1)

% = | ΔC/Cnot | / | ΔC1/C2 |

63
Q

How would an increase in the carbon content of a plain carbon steel influence its mechanical properties?

A

Increasing carbon content further distorts the lattice. This leads to an increase in strength, but a decrease in toughness.

64
Q

How does the cooling rate influence the coarseness of the resulting microstructure if a phase transformation takes place?

A

Diffusion is dependent on the cooling rate. A slow cooling rate means more diffusion, and hence grains grow in size. Hence, coarse grain size.

Conversely, a fast cooling rate means little diffusion, so small grains. Hence fine grain size.

65
Q

What are the steps involved in quenching and tempering heat treatments?

A
  1. We heat the material above its critical temperature. This ensures that the material is fully transformed to Austenite so we know that the carbon is uniformly distributed.
  2. The material is quenched and is cooled quickly to prevent the diffusion of carbon. This forms martensite which is strong, but brittle.
  3. The material is then tempered to a temperature lower than the critical temperature. This allows diffusion to take place which restores the toughness.
66
Q

What is martensite?

How would the mechanical properties of martensite vary with the carbon concentration?

A

Martensite is a super-saturated solid, with trapped carbon.

Increasing carbon content further distorts the lattice. This leads to an increase in strength, but a decrease in toughness.

67
Q

How do alloying elements enhance the response of a steel to heat treatment?

A

Alloying elements reduce the rate and inhibit diffusion. We know in quenching that we want the steel cooled asap to prevent the diffusion of carbon. Hence, alloying elements prevent the diffusion of carbon making martensite easier to form.

68
Q

Describe the differences between each of the three stages of creep?

A

Primary Creep:
Strain rate is initial very high but reduces quickly.
Rapid movement of dislocations.
Work hardening occurs, and the density of dislocations increases, and the strain rate reduces.

Secondary Creep:
Material spends majority of its time here.
Rate of work hardening and the rate of dislocation movement become balanced.
Strain rate stays constant.

Tertiary Creep:
Strain rate increases exponentially.
Large cracks form out of smaller ones combining.
Creep deformation accelerates.

69
Q

Describe the microstructure you would see when the green line hits room temperature?

A

The line crosses through the bainite start line and martensite finish line.

Hence, we will see a mixture of bainite and martensite.

70
Q

Stainless steels can be prone to intergranular corrosion after prolonged exposures to temperatures in the vicinity of 700°C. Explain why this is the case?

A

If the stainless steels is either held at 700 degrees or slowly cooled from there, we get sensitisation. This is where chromium carbides form on the grain boundaries due to the diffusion of carbon and poor APF on the grain boundaries.

This means that less chromium is available to form chromium oxides, as formed chromium carbides instead. Hence, less corrosion protection.

71
Q

How do you find the homologous temperature?

A

T/Tm in K.

Where T is current temp, and Tm is the melting temp.

72
Q

Above what homologous temperature does creep deformation become an issues?

A

T/Tm > 0.4

73
Q

If a steel has a ductile-to-brittle transition temperature of -10C, could it be suitable for use in applications where the temperature can be as low as -40C?

A

No. As below -10 degrees the steel will be brittle and may fail suddenly.

74
Q

What is the Charpy Impact Test?

A

A standardised test that determines the energy absorbed by a material under impact loading.

  1. A notch is made in the sample.
  2. Axe strikes the other side of that notch.
  3. Energy absorbed = GPE delta. So released height - impact height.
  4. Tests are done at a range of temps. Hence, we can plot a DBTT diagram.
75
Q

Label the following diagram and explain both?

Include:
The job of the filler material?
What affects cracking in…?

A

Fusion Zone:
Undergoes melting and solidification during welding. Becomes a mixture of parent and filler “weld” materials. The filler material is designed to produce microstructures in the fusion zone that have good mechanical properties for cooling rates.

Heat Affected Zone:
Does not melt but undergoes microstructural changes due to the heat. Prone to cracking. This depends on the amount of austenite formed.

76
Q

What are the significant values for the carbon equivalent?

What precautions can we take?

A

CE < 0.35, no problems with cooling rates.

CE > 0.5, problems with cooling rates. So precautions such as must be taken to avoid brittle phases forming or cracks. These precautions include: pre-heating and post-weld treatment.

77
Q

What is hardness testing?

What must we do?

A

How well a material can cope with an indentation made by another harder material, usually diamond.

Diamond + Force = Indentation.
Small indentation = High hardness.
Large indentation = Low hardness.

Since the indentations cause plastic deformation. More indentations should be made away from the original. Ideally more than 3x the size of the indentation away.

78
Q

Some stainless steels do not exhibit a strong tendency to suffer from sensitisation. How can this be the case?

A

Presence of alloying elements that form carbides, instead of chromium.

Lower concentration of carbon, hence less likely for CrC to form.

79
Q

Why would it be bad to accelerate creep tests by either increasing the temperature of the test coupon, or by applying a much greater stress?

A

A large increase in temperature could significantly change the microstructure of the test material. Hence, n the real world, at regular service temperature. making any results completely void as in the real world, at regular service temperate. The test material would be a completely different microstructure.

80
Q

Explain why PWHT is often necessary for low-alloy steels?

A

Welding causes rapid heating and cooling on the material. This means diffusion has very little time to occur. Overtime this significantly weakens the material making it extremely brittle.

To restore toughness, we can do post-weld heat treatment.

81
Q

Explain why PWHT may not be advisable for austenitic stainless steels?

A

Austenitic stainless steels are unlikely to be brittle.

Also there is a risk sensitisation could occur with temperatures close to 700 degrees.

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