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Aero Eng Y2 > Materials 2 > Flashcards

Materials 2 Flashcards

1
Q

How to translate the design requirements into properties and constraints?

A
  1. Function?
  2. Constraints.
  3. Objectives.
  4. Free variables.
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2
Q

FCC

A

Face- centred cubic.
{111} are close packed.
<110> directions are close packed.

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3
Q

HCP

A

Hexagonal Close Packed.
{0001} planes are close packed.
<1120} directions are close packed.
Is most brittle as it has only 3 or 6 slip systems.

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4
Q

BCC

A

Body centred cubic.
No close packed planes.
<111> directions are close packed.
So, has it has no close packed planes, it is less ductile.

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5
Q

Define a slip system

A

The set of symmetrically identical slip planes (and associated family of slip directions) for which dislocation motion can easily occur and lead to plastic deformation.

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6
Q

Define polymorphism.

A

Any material that can exist in more than one form or structures.

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7
Q

What are the types of solid solutions?

A

Substitutional solid solution.
Interstitial solid solution.

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8
Q

Explain the types of substitutional solid substitution.

A
  1. Random
  2. Ordered (where B atoms prefer A atoms)
  3. Clustered (where B atoms prefer B atoms)
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9
Q

How much solute can we dissolve?

A

If you add too much solute, you may exceed the solubility limit and the solute will precipitate as a new solution.
This could be 2 random solutions (where nothing has an order at all)
or perhaps 1 ordered and 1 random.

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10
Q

Define a phase

A

A region of material with uniform physical and chemical properties. Phases have their own grains.

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11
Q

Define a grain-boundary

A

A grain is a region with the same crystal orientation, but has an eenrgy associated with them.

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12
Q

Define phase boundary

A

For 2-phase materials, these have an energy associated with them to join them together.

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13
Q

Boundaries in metals.

A
  1. Coherent (has the least energy)
  2. Coherency-strain
  3. Semi-coherent (has gaps)
  4. Incoherent (has the most energy)
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14
Q

Interfacial energy

A

E = GB * A
Grain boundary energy * interfacial area.
If the GB is isotropic, then the lowest E is for the shape with the lowest interfacial area.

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15
Q

For isotropic shapes, what is the best interfacial area? in two phase materials?

A

A sphere.
Because the grains want to grow in the direction of the lowest energy - and mathematically, the best shape in this case is a sphere.

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16
Q

For non-isotropic shapes, what is the best interfacial area? in two phase materials?

A

A plate. - it has the lowest gb energy.

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17
Q

What happens if the E(gb) = E(ab)

A

The beta can form their own grains adjacent to the alpha grains. This is the shape that minimises energy.

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18
Q

Define an alloy

A

The result of taking a pure metal and
adding other elements (Or “mixture of
elements”)

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19
Q

Define an Alloy System

A

All the alloys that can be made from the
components

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20
Q

Define a component

A

The elements that make up an alloy.

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21
Q

Define a composition

A

Usually measured in weight % (mass%)

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22
Q

Define constitution

A

A combination of:
The overall composition (e.g. Al-4Cu)
The number of phases present
The composition of each phase
The percentage by weight of each phase

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23
Q

What does the alloy constitution base on?

A

The number of phases present
The volume fraction of phases
The composition of each phase

These depend on:
- overall composition
- temperature.

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24
Q

What happens about the liquidus line

A

All is liquid.

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25
Q

How to say
‘18 wt% Nickel in Copper’

A

Cu-18wt% Ni

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26
Q

How to calculate the proportion by weight of each phase inside the L+α shape?

A

This is a 2 phase area.
1.Draw a horizontal line at the point, until it meets the liquidus and solidus line.
2. Read the weight % at those points.
The one meeting the Liquidus line is Cl, and the other is Cα.

Know that by conservation of mass, Cα + CL = 100%,
and C0 = fL(CL) + fα(Cα)
Can create an equatoin to calculate fα and

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27
Q

Melting point and alloys

A

They don’t have a specific melting point, but have a range instead.

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28
Q

Define a eutectic phase diagram

A

It is the phase diagram for a limited solubility limit.

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29
Q

A Eutectic Phase Diagram: Single phase alloys

A

This is the far left/ right side of the diagram. It results in all alpha or beta grains being formed.

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30
Q

A Eutectic Phase Diagram: Two phase alloys

A

Starts with a liquid, ends with the beta phases forming on the alpha grain boundaries. This is not a good way to store them, because it reduces the strength of the material, allows man more dislocations to form.

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31
Q

A Eutectic Phase Diagram: The eutectic composition

A

The liquid transforms from Liquid into solid alpha+ solid beta at one temperature - called the Eutectic temperature, at the eutectic point.
It forms sort of stripes of alpha and beta alternating, in a laminar structure, called the eutectic phase.

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32
Q

A Eutectic Phase Diagram: Off-eutectic alloys

A

It is all liquid at first, then above the eutectic temperature, solid grains start to form in the liquid. To calculate the fα(T) = (C0 - CL)/(Cα - CL)
At the eutectic temperature, the remaining liquid undergoes eutectic solidification. To calculate the fEUT(T) = (C0 - C(EUT))/(Cα - C(EUT))

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33
Q

Define Eutectoid

A

alpha -> beta + gamma.

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34
Q

Define Peritectic

A

L + alpha -> beta.

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35
Q

Define internal energy (U)

A

The sum of the kinetic potential and electrical energy of the atoms in the material.

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36
Q

Gibbs Energy

A

At equilibrium - the Gibbs Energy will be the lowest. If ∆G is negative, then it means there is a driving force for things to change.

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37
Q

Equation for Gibbs Energy

A

G = H -TS.
G = U +PV -TS

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38
Q

Change in gibbs energy when a solid becomes a liquid.

A

There is no change in gibbs energy, this is a neutral equilibrium.

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39
Q

Change in gibbs energy when a liquid becomes a solid.

A

There is a reduction in Gibbs energy if the liquid becomes a solid.
This is an unstable situation and a driving force for change.
The G(s) < G(L) for pure water at T<Tm. (Temperature lower than the melting point)

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40
Q

Coarsening (basic explanation)

A

If you have fine grains, and you increase the temperature, the size of the grains will increase. Therefore, the gibbs energy decreases, same with the driving force. So, if we have particles of phase Beta within a matric of phase Beta, there will be a change in the GE if the shape of the particles change.

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41
Q

Coarsening (the magnitude of the driving force)

A

If we decrease the size of the phase boundary, then we can minimise the precipitate growth.

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42
Q

Linking the driving force to an undercooling - what happens at a small departure from the melting point?

A

For a small departure from the melting point, the values of H and S can be considered to be independent of T.

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43
Q

Define undercooling and what happens in it? - Equation for it.

A

Cooling the liquid past the point of it melting. The bigger the undercooling, the bigger the driving force.
Undercooling = Tm - T.

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44
Q

Relation between the driving force and the kinetic energy (the change in transformation rate with temperature)

A

The driving force for transformation DeltaG increases as the temperature decreases below the melting point. This increases the undercooling; decreasing the kinetic energy. So the probability that an atom will jump from the liquid to the solid decreases.

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45
Q

Define diffusion

A

Mass transport by atomic motion.

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46
Q

What are structure sensitive properties

A

They are influenced significantly by changes to the microstructure. Density and Young’s Modulus.

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47
Q

What are structure insensitive properties

A

Aren’t influenced significantly by changes to the microstructure.
Yield strength, tensile strength, ductility, fracture toughness.

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48
Q

What happens if the alpha-beta Phase boundary is less than 1/2 the Grain boundary?

A

Then, cos theta = 1, so theta = 0.
This means that the beta phase will spread out along all the alpha grain boundaries.

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49
Q

Define an equilibrium constitution

A

At a given constant T and pressure p, there is no further tendency for the constitution to change with time.

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50
Q

What happens at the eutectic point?

A

L - > alpha + beta.

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51
Q

What happens at the eutectoid point?

A

alpha -> beta + gamma.
Upon cooling, the solid phase transforms isothermally and reversibly into 2 new solid phases that are intimately mixed.

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52
Q

Explain diffusion controlled phase transformation

A

Eutectoid growth is this way.
An element diffuses out of one phase and into another phase.
beta -> alpha + gamma.
alpha has little B, and gamma has a lot of B.

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53
Q

Explain why we need rcrit?

A

Once a nucleus has nucleated, it will not simply just start to grow immediately.
It will only grow if growing means it’ll have less energy.
All atoms that have radius above rcrit will nucleate.

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54
Q

Diffusive Transform.

A

Generally, as T is far from Tm, we expect more nuclei as the temperature decreases below the equilibrium temperature. But not too much, then the atoms don’t have enough thermal energy to form the second phase.

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55
Q

What is displacive transformation?

A

We quench and miss the nose of the TTT curve. Small lenses of BCC-Fe form at the grain boundaries and propagate across the grains at the speed of sound. There is a switch zone in front of the lenses where the FCC_Fe is broken and BCC_Fe is made instead.

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56
Q

How does FCC turn into BCC in displacive transformation?

A

Inside 2 FCC is a BCC, but it’s slightly strained. The growth of the lenses enforces this strain back to a normal BCC. There must be enough Delta_G to be converted to the strain energy and allow the displacive transformation to happen.

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57
Q

Phases in steel:

A

gamma - Austenite. - FCC.
alpha - Ferrite - BCC.
Fe3C - Iron carbide or cementite.

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58
Q

What is pearlite.

A

alpha-Fe and Fe3C eutectoid is called pearlite.
At the eutectoid transformation, the pearlite nodules grow.

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59
Q

Explain the diffusive transformation in hypo-eutectoid steels.

A
  1. There are gamma grains.
  2. Primary alpha nucleates.
  3. Alpha continues to grow.
  4. At the eutectoid composition, the gamma transforms into pearlite.
  5. Then there are grains of primary alpha + nodules of pearlite in the mixture.
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60
Q

Explain the diffusive transformation in hyper-eutectoid steels

A
  1. There are gamma grains.
  2. Then, Fe3C nucleates.
  3. Fe3C grows.
  4. At the eutectoid composition, the gamma becomes pearlite, and grow to consume the remaining gamma. And the alpha-Fe continue to nucleate.
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61
Q

What are the mechanical properties for carbon steels

A

Fraction of Fe3C increases with increasing C.
The yield strength and UTS increases with C content because Fe3C is a hard phase.
Ductility decreases with C content because alpha-Fe-Fe3C interfaces are good at nucleating cracks.

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62
Q

What happens when you add Carbon to gamma-Fe through a dispacive transformation?

A

Well BCC_Fe can only dissolve 0.035 wt% C. But usually, we add more than that. So, upon quenching this, a new shape, called a Body Centred Tetragonal structure forms.
The hardness of Martensite increases with Carbon content, because the carbon distorts the BCT Lattice more. The lattic distortion gives great hardness but is a very brittle material.

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63
Q

Explain tempering steel

A

If we reheat the steel to 300-600 degrees C, the carbon atoms have enough thermal energy to leave the supersaturated BCT-martensite and will react with Fe to form Fe3C.
This causes numerous, small closely packed particles of Fe3C. And the BCT Structure relaxes back to BCC.
This means that the ductility increases while the small Fe3C particles maintain a good strength/ hardness. (Don’t hold it there for too long though, or else the Fe3C will coarsen. which is rlly bad to retarding dislocation motion.)

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64
Q

Stabilisers and solution strengtheners for steel.

A

Mn - stabilises the gamma Austenite phase.
Cr - stabilises the alpha Ferrite phase.

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65
Q

Improving hardenability in steel.

A

Essentially, as martensite is quite hard, we want to form this easily and all the way to the centre of a thick component.
You can add Mo, Mn, Cr and Ni additions- these displace the TTT curves to longer times, enabling the martensite to form even at slow cooling rates.

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66
Q

Carbide forming additions in steel.

A

Certain elements in steel can form carbides - which are more thermodynamically stable than cementite. At high enough concentrations, these will be formed, and not cementite.
Ex: Cr, Mo, V, W and Ti.

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67
Q

What is homogenisation in Al alloys?

A
  1. Solution is solidified first.
  2. The alloy is heated to a temperature below the eutectic temperature and held for a long time.
    This removes the laminar structure that was caused due to the eutectic phase.
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68
Q

What is the need for the precipitation sequence in Al2Cu?

A

When the solution has been quenched, there is a large driving force (because the undercooling is large). But there is very little thermal energy available for diffusion. And all the interfaces between Al2Cu and alpha would be incoherent so the nucleation of Al2Cu is difficult.

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69
Q

Precipitation sequence in Al-Cu

A
  1. Initial supersaturated solution of Cu in alpha.
  2. Cu atoms cluster to form Guinier-Preston Zones. They are disc shaped, the disc faces are fully coherent with alpha and the disc edges have coherency strain.
  3. GP zones act as nucleation sites for unstable beta’’. It is a tetragonal crystal. Disc faces are fully coherent, disc edges have coherency strain.
  4. Another unstable phase beta’ forms, nucleating at dislocations and grows at the expense of beta’’. Disc faces are fully coherent, but disc edges are incoherent.
  5. Eventually, beta-AlCu2 forms. It is a crystal with very different lattice parameters to alpha, and all interfaces are incoherent. Therefore, beta-AlCu2 grows are rounded particles.
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70
Q

When is the peak strength reached during the precipitation sequence?

A

At beta’’. This gives numerous, closely space particles throughout the alpha phase.

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71
Q

What to optimise for the upper wing?

A

Compressive strength
Stiffness.

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72
Q

What to optimise for the lower wing?

A

Tensile strength
Fatigue resistance.

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73
Q

Al alloy selection for wings of subsonic aircraft?

A

7xxx - upper.
2xxx - lower.

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74
Q

What to optimise for the fuselage?

A
  • Axial tensile stress.
  • Circumferential tensile stress.
  • Fatigue critical.
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75
Q

What is the temperature limit for Al alloys?

A

Mach 1.8

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76
Q

Advantages of using Magnesium?

A
  • Light structural metal.
  • Precipitation hardenable
  • Can have a good creep resistance.
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77
Q

Disadvantages of Magnesium?

A
  • Low melting point (compared to Ti or Ni)
  • Low stiffness
  • HCP crystal structure (Dislocation slip only on (0001) planes at Room temp)
  • Corrosion properties are okay.
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78
Q

Specific stiffness for struts?

A

Maximise E/rho.

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79
Q

Specific stiffness for beams?

A

Maximise E^0.5/rho

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80
Q

Specific stiffness for panels?

A

Maximise E^1/3 / rho.

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81
Q

Mg alloys : Mg-Al-Zn alloys

A

Solution heat treatment at 400degrees.
Quench
SSSS -> to Mg17Al12 directly.

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82
Q

Mg alloys : Mg - Y - Nd - Zr

A

Zr addition:
- Zr promotes nucleation of Mg grains of a mcuh smaller grain size. Good for grain boundary strengthening and retarding crack growth.
Creep resistance:
- Numerous, small B/ B’ precipitates are distributed evenly - which obstructs dislocation motion at elevated T.
- Ppts at grain boundaries, relatively stable at 250 - 300.
Inhibits grain boundary sliding at elevated T.

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83
Q

Advantages of WE alloys

A
  • Best properties of all Mg alloys.
  • Service temperature of up to 250-300 degrees.
  • Superior strength to Al alloys at 250 - 300 degrees.
84
Q

Disadvantages of WE alloys

A
  • Expensive.
  • Yttrium is difficult to handle in molten Mg alloys.
85
Q

Advantages of Titanium

A

Excellent corrosion resistance.
High specific strength.
Good compatibility with CFRP.
Excellent properties at elevated temperatures.

86
Q

Disadvantages of Titanium

A

Expensive.
Low wear resistance
Difficult to form.
Pick up of oxygen and nitrogen above 500 degrees.

87
Q

Explain the near alpha alloys in titanium

A

alpha forms between beta upon cooling.
giving it good creep properties and poor fatigure resistance. This is because diffusion is faster in beta-Ti, so a high fraction of alpha-Ti is good for creep resistance.

Or, a basket weave structure of alpha next to beta is formed.
There are OK creep properties, but it has really goo fatigue resistance. The entanglement slows down the crack movement and increases the crack deflection.

In both, you can use solid solution or grain boundary strengthening.
Used as the blades in a gas turbine engine, operating at high temperatures.

88
Q

Explain alpha-beta alloys in Titanium

A

Good balance of creep and fatigue properties.
Used in fan blades, - good strength and superplastic formable.

89
Q

Explain near beta alloys in Titanium

A

Has very good strength at low T.
Used in landing gear, helicopter rotor hub.

90
Q

Explain beta alloys in Titanium

A

Very good strength at low T.
Highest strength Ti allows.
Poor properties at higher T.

91
Q

Explain the low cycle fatigue

A
  1. Plastic deformation roughens the surface in a small number of cycles.
  2. Slip planes cause extrusions and intrusions at the surface.
  3. The cracks then initiate along the slip paths.
  4. The cracks then change their direction to grow perpendicular to the tensile axis.
92
Q

Explain high cycle fatigue

A

There is some local plasticity in regions where a notch or scratch or change in section concentrate the stress. Cracks eventually form here.

93
Q

Explain fatigue fracture surfaces

A
  1. Initiation region: surface.
  2. Propagation: beach marks.
  3. Fast fracture - when the crack length is greater than the af.
94
Q

Predicting fatigue lifetime - Fail safe design

A

Comparing endurance limit or fatigue limit to the delta Kth value (below this value, the crack will not advance any further)

95
Q

Predicting fatigue lifetime - Limited lifetime design

A

With pre-existing cracks, use the Paris law.
It is also useful for determining necessary inspection intervals.

96
Q

How to design for creep - design against dislocation creep

A
  • Choose a materials with a high melting point (decrease T/Tm)
  • Maximise obstacles that are stable at the chosen T.
  • Choose an alloy where the majority phase has a high lattice resistance. (covalent bonding for example)
97
Q

How to design for creep - design against diffusion creep

A
  • Choose a material with a high melting point (decrease T/Tm)
  • Maximise the grain size to give long diffusion distances and little gb diffusion. (aka single crystals)
  • Choose an alloy with precipitates at gb to impede gb sliding.
98
Q

Advantages of single crystals?

A

No grain boundaries.
Superior creep
Superior thermal fatigue resistance.

99
Q

Creep damage and failure

A
  • Voids can form at the grain boundaries lying perpendicular to the stress.
  • Atoms must diffuse to these gb for diffusion to occur, but creep can occur by the growth of voids.
  • Voids decrease the load-bearing area- thus increasing the stress.
100
Q

Alloy requirements

A
  • Creep resistance.
  • Resistance to high temperature oxidation.
  • Toughness.
  • Thermal fatigue resistance
  • Thermal stability
  • Low density.
101
Q

Define Creep

A

Slow, continuous deformation with time at elevated temperature.

102
Q

Explain the 4 types of diffusion mechanisms

A
  1. Interstitial - needs to overcome the energy barrier.
  2. Substitutional - needs to overcome the energy barrier and also a vacancy.
  3. Grain-boundary diffusion - Useful when the grain size is very small.
  4. Dislocation-core diffusion - Can act as fast diffusion wires. If there is a high dislocation density, this can be good.
103
Q

Explain Creep Mechanisms - Dislocation Creep

A

Slide : Slides on slip planes if the stress is high enough for the dislocations to overcome the lattice resistance.
Glide : Whenever the dislocation meets a ppt, it meets it a glide force. There will be a reaction force, with some component acting perpendicular to the glide direction. The dislocation can’t just glide upwards, but it can move up if atoms at the bottom of the half-plane can diffuse away. Bc it requires diffusion - it is temperature dependent. Higher the stress, the higher the climb force - more dislocation glides - higher strain rate.

104
Q

Explain diffusion creep

A

At lower temperature - diffusion occurs across the grain boundaries - Coble Creep.
At higher temperatures - diffusion occurs across the bulk of the material (Nebarro-Herring Creep).
To ensure continuity within the material, the grain arrange themselves by grain boundary sliding.

105
Q

Factors affecting diffusion creep

A

Strain rate is proportional to D and the stress applied. And proportional to 1/d^2 - the larger the diameter, the more the atoms have to diffuse.

106
Q

Which materials will oxidise?

A

If their energy change is negative, they will oxidise and are called oxide stable.

107
Q

When does linear corrosion loss happen

A

If the oxide is volatile.

108
Q

When does linear corrosion growth happen

A

When the oxide breaks up.
If the volume of the oxide is less than the metal, the oxide splits open from the metal.
If the volume of the oxide is greater than the metal, the oxide shrinks from the metal.

109
Q

How to improve the oxidation resistance?

A
  1. Coating with alumnium.
  2. Heating in a furnace. (forms either AlNi and the rest oxidises to form a Al2O3 layer)
110
Q

Disadvantages of oxide films

A
  • Brittle and can crack - especially when T changes.
  • Oxide layers act as initiation sites for thermal fatigue cracks - spreading into the alloy.
111
Q

Define amorphous solids

A

They are made up of randomly orientated atoms, ions or molecules. They don’t have a defined pattern or lattice structure.

112
Q

Define hardness

A

A measure of how easily dislocations can move through a material.

113
Q

Dislocation motion in covalent ceramics

A

Dislocation motion is very difficult because the interatomic bonds must be broke and then reformed.

114
Q

Dislocation motion in ionic ceramics

A

Dislocation motion is easier in some planes but harder on others.

115
Q

Define thermal shock resistance

A

The maximum temperature change a material can withstand upon quenching. (Usually in Kelvin)

116
Q

How does chain length distribution affect number average molar mass, or weight average molar mass or z-average molar mass?

A

Mn - The number of entanglements per chain is proportional to the number of chain ends.
Mw - melt viscosity - higher viscosity due to more entanglements.
Mz - melt elasticity - highest molecular weight fraction determines the melt elasticity.

117
Q

Define isotactic chain regularity?

A

All the side groups are on the same side.

118
Q

Define sindiotactic chain regularity?

A

Alternate in a regular manner

119
Q

Define atactic chain regularity

A

The side groups are in random positions.

120
Q

Why does a rubber become stiffer once heated?

A

When you stretch the polymer, you have reduced its entropy (less disorder because you have correctly oriented the chains). So, when we heat it up, the system will go back and get stiffer.

121
Q

What material processes are used in the rubber region?

A

Vaccuum forming
Bottle blowing.

122
Q

What material process are used in the liquid form?

A

Extrusion
Injection mouldering
Rotor moulding.

123
Q

Polymers - Rubber Compounding

A

Literally to create rubber.
It’s created into thin sheet strips by using a two-roll mill shear mixing.
They are then allowed to cool down.

124
Q

Polymers - Extrusion

A

Plastic pellets are put into the feeder.
These are heat up and a turning screw pushes it out by spinning.
Produces pipes, tubing, window railings, plastic films, sheets, wire insulation.

125
Q

Polymers - Use of twin screw compounding

A

Plasticizing (PVC)
Dispersing of fillers and fibres (composites)
Alloying and blending (polymer blends)
Colouring and additives

126
Q

Polymers - Explain die swell

A

At high shear rates, the non-Newtonian behaviour is good.
Die swell happens because of elastic spring back.
Controlled by Mz.

127
Q

Polymers - Explain melt fracture

A

When the flow rate in increased a lot, the extrudate can break. Usually happens when stretching and cooling occur too fast and cause micro tears. Occurs due to too low temperature of the melt, high molecular weight or poor die design.

128
Q

Polymers - Co-extrusion

A

Extrusion of multiple layers at the same time.
2 or more extruders are used to deliver the output into an extrusion head. The layer thickness is controlled by relative speeds and sizes of the extruders delivering the material.

129
Q

Polymers - Sheet extrusion

A

Die is used and cooled through a set of cooling rolls. They cool and also determine sheet thickness.

130
Q

Polymers - Blow film extrusion

A

Extrusion and then air is blown into the output.
Guide rolls and pinch rolls are used.

131
Q

Polymers - Melt Fibre spinning

A

Hot melt is extruded through a series of small holes in a plate (spinnerets) into air or
water (or another fluid). The fibres are wound up at high speed on a rotating drum (godet)

132
Q

Polymers - Solution Fibre Spinning

A

A diluted polymer solution is spun from a spinnernet. Ultra high molecular weight polyethylene is dissolved in a solvent. Gives a very low enthanglement density, extremely high strength and gives it a nearly 100% crystalline structure and a high orientation. But rather hard to process.

133
Q

Polymers - Aramid fibre spinning

A

They have aromatic chains. after you spin it, they become very orientated.

134
Q

Polymers - Injection moulding

A

The screw here both spins and also pushes back.
Injects material into a mould.

135
Q

Polymers - Rotational moulding

A

involves a heated hollow mould which is filled with a charge or shot weight of
material. It is then slowly rotated (usually around two perpendicular axes) causing the softened material
to disperse and stick to the walls of the mould.

136
Q

Polymers - Vacuum forming (or thermoforming)

A

Use a heater to place the sheet of material over the object. Then use a vacuum to suck all the air out.

137
Q

Pros of composites

A

Weight reduction (high specific stiffness & strength)
Design flexibility & integrated parts – ability to create
novel shapes tailored to integrated part function.
Environmentally friendly? – low energy consumption in manufacture.
Safety – crush structures.
Low maintenance cost (corrosion resistance)

138
Q

Cons of composites

A

Costs of materials and processing still high
Absence of mass production technology for highperformance composites
Recycling of composites is very difficult
Lack of knowledge in designing with anisotropic materials
Uncertainties in predicting failure and life-time

139
Q

Types of reinforcements

A

Particulate
Flake
Fibre reinforced
Laminated

140
Q

Types of composites

A

Uniaxial
Angle-ply
Cross-ply
Random-in-the-plane
Random in 3d.

141
Q

Why does the random in 3d composite have the least mechanical property?

A

It has random directions. And also has discontinuities because the fibres are shorter, which means less strength.

142
Q

Polyethylene

A

Cheap, easy to mould and fabricate.
Plastic bottles, tupperware.

143
Q

PTFE

A

Non stick cooking utensils, Rain gear.
Expensive
Water repellents
Stable.

144
Q

PS - Polystyrene

A

Uses : pens, foams, cups, CD cases.
In its simplest form, it is brittle.

145
Q

PVC - Polyvinylchloride

A

Rigid, brittle, cheap.
CDs, swimming fins, pipes.

146
Q

PMMA

A

Transparent - acrylic.
Most resembles glass in transparency and durability.

147
Q

PC - Polycarbonate

A

Used in:
- safety goggles, fire alarm buttons, helmets.

148
Q

PET

A

Thermoplastic polyester
Food packaging, bottles etc.

149
Q

PEEK

A

High performance thermoplastic.
High stiffness, strength, resistance to heat.
Expensive.
Nuts, bolts etc.

150
Q

Epoxy

A

Resistant to heat and chemical attack.
Adhesives
Surface coatings, filled with other materials.
Expensive and brittle.

151
Q

Unsaturated polyesters

A

Thermosets.
Poor corrosion resistance.
Cheaper than epoxies.

152
Q

Natural rubber

A

Elastomer.
Low hysteresis and high tear resistance.
Tyres, tubes, shoes etc.

153
Q

Composite Architecture

A

Unidirectional prepreg/tapes.
Non crimp fabrics/ uniweave.
Non-woven mat.
Woven - plain weave - good stability, but difficult to drape.
Twill weave - Good drape, good stability too. Smoother and slightly higher mechanical properties.
Satin weave - Very flat and have best degree of drape.
Braided.
Through thickness reinforcement - stitching, tufting and z-pinning.

154
Q

Composites - Hand lay up

A

Simplest way:
Resins are impregnated by hand into fibres which are in the form of woven, knitted, stitched or bonded
fabrics. This is usually accomplished by rollers or brushes, with an increasing use of nip-roller type
impregnators for forcing resin into the fabrics by means of rotating rollers and a bath of resin

155
Q

Composites - Hand lay up - Advantages

A
  • Cheap
  • Used for many years.
  • Wide choice of suppliers and materials.
  • High fibre contents and longer fibres than spray lay-up.
156
Q

Composites - Hand lay up - Disadvantages

A
  • Dependent on the skills of the laminators.
  • Health and safety factors.
157
Q

Composites - Hand lay up - Typical applications

A

Standard wind turbine blade.
Production boats.
Architectural mouldings.

158
Q

Composites - Spray up

A

Fibre (glass roving) is chopped in a hand-held gun and fed into a spray of catalysed polyester resin directed at the mould. The deposited materials are left to cure under standard atmospheric conditions

159
Q

Composites - Spray up - Advantages

A
  • Used for many years.
  • Low cost.
  • Low cost tools.
160
Q

Composites - Spray up - Disadvantages

A
  • Laminates tend to be very resin-rich - so heavy.
  • Only short fibres which limits the properties.
  • Resins need to be low in viscosity to be sprayable.
  • Problems with legislation.
161
Q

Composites - Spray up - Applications

A

Simple enclosures, caravan bodies, bathtubs, shower trays, small dinghies.

162
Q

Composites - Filament winding

A

The rotationing mandrel is a oval sectioned component, such as pipes or tanks. Fibre tows are passed through a resin bath and then wound onto the mandrel in many different orientations.

163
Q

Composites - Filament winding - Advantages

A
  • Fast, and economic.
  • Resin content can be controlled.
  • Fibre cost is minimised bc no secondary processes to convert fibre into fabrics is needed.
  • Good properties - straight fibres can be applied in complex patterns to match the applied loads.
164
Q

Composites - Filament winding - Disadvantages

A
  • Process limited to convex shaped components.
  • Fibre cannot easily be laid along the length of a
    component.
  • Mandrel costs for large components can be
    high.
  • External surface of the component is
    unmoulded, and therefore cosmetically unattractive.
  • Low viscosity resins needed and hence lower
    mechanical properties with health and safety
165
Q

Composites - Filament winding - Applications

A

Chemical storage tanks and
pipelines, gas cylinders, fire-fighters’ breathing tanks.

166
Q

Composites - Pultrusion

A

Fibres into a resin bath, and then a heated die. The die completes the impregnation of the fibre, and cures into the final shape. Cured profile is then cute to length.

167
Q

Composites - Pultrusion - Advantages

A
  • This can be a very fast automated method.
  • Resin content can be controlled.
  • Fibre cost minimised since it is taken from a creel.
  • Structural properties of laminates can be very good since profiles can have very straight fibres and high
    fibre contents.
  • Resin impregnation area can be enclosed thus limiting volatile emissions.
168
Q

Composites - Pultrusion - Disadvantages

A
  • Limited to constant or near constant cross-section components
  • Heated die costs can be high.
169
Q

Composites - Pultrusion - Typical applications

A

Beams and girders used in roof structures,
bridges, ladders, frameworks.

170
Q

Composites - Resin Transfer moulding

A

Fabrics are laid up as a dry stack of materials. These fabrics are sometimes pre-pressed to the mould shape (preforms), and held together by a binder. A second mould tool is then clamped over the first, and resin is injected into the cavity. Vacuum can also be applied to the mould cavity to assist resin in being drawn into the fabrics. This is known as Vacuum Assisted Resin Transfer Moulding (VARTM). Once the fabric is wet out, the resin inlets are closed and the laminate is allowed to cure. Both injection and cure can take place at either ambient or elevated temperature.

171
Q

Composites - Resin Transfer moulding - Advantages

A
  • High fibre volume laminates can be
    obtained with very low void contents.
  • Good health and safety, and
    environmental control due to closed
    mould process.
  • Possible labour reductions.
  • Both sides of the component have
    a (smooth) moulded surface.
172
Q

Composites - Resin Transfer moulding - Disadvantages

A
  • Matched tooling is expensive, and
    heavy in order to withstand
    pressures.
  • Generally limited to smaller
    components.
  • Problems to inject highly viscous
    toughened resins
173
Q

Composites - Resin Transfer moulding - Typical Applications

A

Small complex aircraft and automotive
components, train seats.

174
Q

Composites - Vacuum infusion

A

Same as resin transfer moulding but with a vacuum used as well.
Uses:
Semi-production small yachts, train
and truck body panels.

175
Q

Composites - Bladder Moulding

A
  1. Preform is inserted in.
  2. Draping occurs where air pushes the material to the mould. from the inside.
  3. Component is removed.
  4. Component is injected (by RTM), heated and cured.
176
Q

Composites - Bladder Moulding - typical applications

A

Tennis rackets and bicycles, pipes,

177
Q

Composites - Autoclave Moulding - typical Applications

A

Aircraft structural
components (e.g. wings and tail sections), F1
racing cars, sporting goods such as
professional tennis racquets and skis.

178
Q

What is the heat treatment for Nickel - Aluminium

A
  1. Solutionise in Ni to dissolve the Al into solution.
  2. Quench to create SSSS.
  3. Age at 800 degrees C for 8 hours to control the precipitation of Ni3Al throughout Ni grains.
179
Q

Which is better - to have coherent or incoherent interfaces between the alpha and the precipitates?

A

We want incoherent interfaces. Incoherent interfaces have a greater strain, and they produce a greater resistance on dislocation motion.

180
Q

What is the interface between Ni3Al and Ni?

A

It is a coherent interface, both phases are cubic and have a similar lattice parameter.

181
Q

Explain why Ni-Ni3Al is good despite having coherent interfaces?

A

Dislocations find it hard to move because the Ni3Al is ordered. The passage of a dislocation disrupts this order, the dislocation must pass through in pairs, one which disrupts and the other restores, leading to strengthening.
Also, it is thermally stable, so it minimises coarsening.

182
Q

Explain some Ni superalloy compositions

A
  1. Solid Solution Strengtheners - Co,W, Cr.
  2. Precipitate formers - Al, Ti, Mo, Ta, C.
  3. Oxidation protection - Cr.
183
Q

Designing for creep : Turbines blades .

A
  1. Equiaxed - Having ppts at grain boundaries.
  2. Use columnar ppts - maximise the grain boundaries to give long diffusion distances and little gb diffusion.
  3. Single crystals have no gb, no voids.
184
Q

Nickel based Turbine blade cooling

A

The temperatures in the turbine are often greater than the nickel melting temperature.
Cooling air from the compressor is fed through channels. A film of cool air then envelopes the blade, allowing the blade to cool down.

185
Q

Nickel based Turbine blade - Thermal barrier coating

A

Add cooling.
But also add a coating of ceramic and a bond coat (zirconium oxide), giving:
- protection from oxidation.
- thermal shielding (oxides are poor conductors of heat, so they provide insulation).

186
Q

Composites - Metal Matrix + Ceramic Fibres - Foil diffusion boding (solid state)

A

Put a film down, and then insert the fibres inside. Then add a new film.
Layer them all up and then put inside a vacuum. Heat to fabrication temperature and apply pressure. The matrix flows around the fibres.

187
Q

Composites - Metal Matrix + Ceramic Fibres - Foil diffusion boding (spray deposition)

A

Spraying a liquid metal against the fibres.

188
Q

Composites - Metal Matrix + Ceramic Fibres - Foil diffusion boding (plasma spray deposition)

A

A layer of fibres is laid up on a rotating mandrel, the metal is deposited on the fibres by plasma spraying, a second layer of fibres is put on, and the operations are repeated until the desired thickness and the number of layers is attained.

189
Q

Explain the difference between a thermosetting and thermoplastic polymers

A

Thermoplastic polymers can melt under heat whilst curing, but thermoset polymers tend to hold their shape.

190
Q

For thermoplastic polymers, explain 5 factors that promote brittle fracture

A
  • Reduction in temperature.
  • Increase in strain rate.
  • Presence of a sharp notch.
  • Increase specimen thickness
  • Modifications of the polymer structure.
191
Q

Does deformation by drawing increase or decrease the tensile modulus?

A

Increases it.
Upon drawing, the structure becomes a highly oriented structure, so there is a higher degree of interchain secondary bonding.

192
Q

Common Environmental factors

A

Long-term environmental exposure, moisture.
Cost?
Temperature (-50 - high altitude cruise) to 100 degrees (taxi on runway)
Moisture ranges - dry to hail and lightning.
Impact threats - runway debris, birdstrike, maintenance, collision with ground vehicles.
Impact resistance.
Fatigue resistance.
Fire resistance.
Electrical conductivity - or electrical insulation?
Thermal conductivity?

193
Q

Functions of the matrix phase

A

Bind the fibres together.
Distribute the stress between the fibres.
Protect the surface of the fibres from damage.
Separate the fibres and inhibit crack propagation.
Lateral support to the fibres against buckling.

194
Q

Desired characteristics of the matrix and fibre phases

A

Matrix: Ductile, soft, bonds well to the fibres, low density, and cheap.
Fibre: Stiff and strong, good chemical and thermal compatibility with the matrix.

195
Q

Explain why alloying with aluminium can improve the oxidation performance of a Ni Super-alloy?

A
  • Al has a higher affinity for oxygen so, Al2O3 is formed on the surface.
  • The parabolic growth law is still x^2 = kt, but k is much smaller. Diffusion through the Al oxide is slower, because the motion of electron is very slow.
196
Q

CFRP Advantages

A
  • Low weight
  • Corrosion resistant.
  • Formability/ shaping.
  • Paintability.
  • Aesthetics.
197
Q

CFRP Disadvantages

A
  • Rapid manufacture is difficult.
  • High material cost.
  • Electrical/ thermal conductivity could be an issue.
  • Immature design knowledge.
  • Joining dissimilar materials as issue.
198
Q

Amorphous against crystalline?

A

Most materials - especially thermoset and thermoplastics are amorphous.

199
Q

Define solid solution.

A

A solid state solution of one or more solutes in a solvent. Such a mixture is considered a solution (and not a compound) when the crystal structure of the solvent remains unchanged by the addition of solutions.

200
Q

Loading types

A

Torsion
Aerodynamics Loads - in bending
Fatigue
Impact loads.
Penetration resistance.

201
Q

Under what service condition can fatigue cause failure in steel?

A

They can fail at stresses below the tensile strength of the material if subjected to alternating stress of strain.

202
Q

Explain beach marks

A

They can be observed. Each beach mark represents a period of time over which the crack growth occurred.

203
Q

Describe the 3 stages of fatigue failure shown

A

Stage 1 - No observable fatigue crack growth. The cracks behave as non-propagating cracks.
Stage 2 - Paris law. There is a linear relationship between da/dN and log K.
Stage 3 - Accelerated crack growth.

204
Q

What are the 4 grain refinement mechanisms

A

Precipitation strengthening - adding precipitates to decrease dislocation motion.
Solid solution - Adding atoms of other elements to the crystalline structure. This adds a stress and strain field - impeding dislocation motion.
Grain refinement - Adding more grain boundaries means the dislocations don’t travel in a continuous slip plane - hinder dislocation motion.
Work hardening - Strengthening a metal by plastic deformation, because it increases the hardness.

205
Q

Why do ceramics crack at an extreme temperature change?

A

There is a rapid change in volume due to the thermal expansion/ contraction between different layers of a solid. This can create a stress which is greater than the tensile strength, meaning the ceramic breaks.

206
Q

Explain the modulus of rupture in ceramics

A

Basically, tension tests can’t be done on ceramics - there are many problems, but especially with the clamping of the system.
If you do somehow manage to go through and do a tensile test, then the tensile stress you receive is quite low.
If you test it in bending instead, this gives a modulus of rupture of much greater values.

207
Q

In ceramics - what can you do or change about the ceramic that would allow a greater tensile strength?

A

Increase the value of m - it will reduce the size of defects and distribution.
Choose a materials that is less defect sensitive.
Change the dimensions so it’s smaller, because this has smaller cracks.
Or add a better surface finish.

Aero Eng Y2 (3 decks)

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