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Engineering - Year Three > Mats 301 > Flashcards

Mats 301 Flashcards

1
Q

What is a composite?

A

A material that consists of two or more constituent parts. The material is different to its individual components, but they remain separate and distinct.

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2
Q

Dispersed phase (composites)

A

Particles or fibres.

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3
Q

Matrix phase (composites)

A

Continuous phase that surrounds the dispersed phase.

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4
Q

Glass fibre production process

A
  1. Liquid glass is formed by blending quarry products and heating the mixture in a furnace at very high temp.
  2. Liquid glass is passed through a platinum bushing with very fine holes.
  3. The resulting fibres are cooled using water spray, then drawn together using a ‘size’ to provide filament cohesion, and to protect the glass from abrasion.
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5
Q

Three categories of glass fibre

A
  1. E-glass (electrical) - lower alkali content, reasonably good tensile and compressive strength.
  2. C-glass (chemical) - resistant to chemical attack, used in pipes and tanks.
  3. S-glass - higher tensile strength and modulus than E-glass, achieved by using smaller fibre diameter.
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6
Q

Five advantages of using glass fibres

A
  1. Low cost
  2. Relatively high strength
  3. Heat resistant
  4. Insensitive to moisture
  5. Electrical insulator
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7
Q

Three disadvantages of using glass fibres

A
  1. Low stiffness
  2. Attacked by acids
  3. Relatively poor fatigue resistance
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8
Q

Aramid fibres (kevlar) production process

A
  1. Solid fibres are extruded from a liquid chemical blend using a spinneret.
  2. Fibres are washed in a neutralising bath, then dried and stretched at 500°C to improve their molecular alignment.
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9
Q

Three advantages of aramid fibres

A
  1. High specific tensile strength
  2. High impact and abrasion resistance
  3. High fatigue resistance
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10
Q

Three disadvantages of aramid fibres

A
  1. Poor in compression
  2. Attacked by acids and UV light
  3. Low temperature resistance
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11
Q

Carbon fibre production process

A
  1. Produced by the controlled oxidation, carbonisation and graphitisation of carbon-rich organic precursors which are already in fibre form, the most common being PAN.
  2. PAN fibres are greatly stretched to improve molecular alignment, then oxidised in air at 300°C.
  3. They’re then carbonised at 1500°C to improve crystallinity (nitrogen released) then finally graphitised by heating and stretching at 3000°C
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12
Q

Three advantages of carbon fibres

A
  1. High strength and modulus
  2. High creep and fatigue resistance
  3. Good energy absorption
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13
Q

Three disadvantages of carbon fibres

A
  1. High cost
  2. Poor impact resistance
  3. Electrical conductor
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14
Q

Polyester fibres

A

Low density fibre with good impact resistance, but low modulus.

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15
Q

Polyethylene fibres

A

Drawing procedure orientates the molecules, giving a very high tensile strength.

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16
Q

Quartz fibres

A

Very high silica version of glass fibres, with much higher mechanical properties and excellent resistance to high temperatures.

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17
Q

Boron fibres

A

Carbon or metal fibres coated with a layer of boron. Strong, stiff and light.

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18
Q

Ceramic fibres

A

Very high temperature resistance, but low impact resistance.

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19
Q

Natural fibres

A

Fibrous plant material, e.g. coconut.

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20
Q

Five functions of the matrix phase

A
  1. Bind fibres together
  2. Transmit applied load to fibres
  3. Add ductility/toughness to the composite
  4. Prevent propagation of cracks
  5. Protect the fibres from damage
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21
Q

Resin

A

Refers to a man-made polymer

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22
Q

Resin criteria

A

The resin must be able to deform to at least the same extent as the fibre, otherwise the full mechanical properties of the fibre component won’t be achieved. Must also have good environmental and stress cycling resistance.

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23
Q

Toughness

A

The area under a stress-strain curve. It’s a combination of strength and ductility.

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24
Q

Why is high adhesion between resin and fibres important?

A

To ensure that the loads are transferred efficiently, and to prevent cracking or fibre/resin de-bonding when stressed.

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25
Q

Two types of resin

A

Thermoplastic and thermoset

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26
Q

Thermoplastic resin

A

Behave more like metals. Soften when heated and eventually melt, hardening again when cooled. This can be done as often as desired.

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27
Q

Thermoset resin

A

Formed from a chemical reaction in-situ. Don’t become liquid again when heated, but above a certain temperature the mechanical properties change significantly. This is known as the glass transition temperature.

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28
Q

What happens at the glass transition temperature?

A

The thermoset resin becomes more flexible, meaning a lower stiffness and reduced strength of the composite. This is reversible on cooling.

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29
Q

Two advantages of thermoset resins

A
  1. Liquid at room temp, so easy to work with and get fibres into.
  2. Can exhibit excellent properties at low raw material cost.
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30
Q

One disadvantage of thermoset resins

A

Once a thermoset composite has been formed, it cannot be remelted or reshaped, meaning recycling is very difficult.

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31
Q

Two advantages of thermoplastic resins

A
  1. Increased impact resistance to comparable thermosets.
  2. Can be remelted or reshaped, so recycling is possible.
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32
Q

Two disadvantages of thermoplastic resins

A
  1. Solid state at room temperature, so much more difficult to introduce reinforcing fibres.
  2. Resin must be melted, injected around the fibres under pressure, then cooled under pressure to form the composite. This is complex and expensive.
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33
Q

Four common thermosetting resins

A

Polyester, vinylester, epoxy resin and phenolic resin.

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34
Q

Polyester

A

A thermoset consisting of a polyester solution in a styrene monomer. The addition of styrene helps to make the resin easier to handle by reducing its viscosity.

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35
Q

Vinylester

A

Tougher than polyester. Fewer ester groups means better water resistance.

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36
Q

Epoxy resin

A

Superior mechanical and adhesive properties. Resistant to environmental degradation.

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37
Q

Phenolic resin

A

Excellent fire resistance - retains properties well at high temperatures. Poor properties overall. Very brittle.

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38
Q

Two types of thermoplastic matrix composites

A

Glass mat thermoplastics (GMT) and Advanced thermoplastic composites (ATC).

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39
Q

Glass mat thermoplastics

A

Can use nearly any thermoplastic for the matrix. Short chopped fibres are used.

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40
Q

Advanced thermoplastic composites

A

Low density and high strength. Good toughness and environmental resistance.

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41
Q

Elastic/ Young’s modulus

A

Measure of a material’s resistance to elastic deformation when a force is applied to it. It’s the gradient of a stress-strain curve in the elastic region. A stiffer material will have a higher elastic modulus.

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42
Q

Predictions of modulus, strength and Poisson’s ratio can be made from composites based on six assumptions

A
  1. Perfect bonding between fibre and matrix
  2. High adhesion between resin and fibres
  3. No voids or imperfections
  4. Consistent fibre size
  5. Average values of elastic modulus/strength of the components are used
  6. Uniform strain and stress in the individual components.
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43
Q

Fibre volume fraction + Matrix volume fraction

A

Vf + Vm = 1

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44
Q

What is the UTS of a composite limited by?

A

The UTS of the fibres.

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45
Q

Longitudinal force carried by a composite

A

Fc = Ff + Fm

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46
Q

Longitudinal composite stress (sigma)

A

sigmac = sigmamVm + sigmafVf

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47
Q

State of isostrain (longitudinal loading)

A

For a unidirectional composite loaded longitudinally, the strain in the fibres = strain in matrix = strain in composite

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48
Q

Rule of mixtures (longitudinal loading)

A

Ec = EmVm + EfVf, where E is the modulus.

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49
Q

Major Poisson’s ratio

A

nu12 = numVm + nufVf

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50
Q

State of isostress (transverse loading)

A

sigmac = sigmam = sigmaf

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51
Q

Overall strain in composite in transverse direction, epsilonc

A

epsilonc = epsilonmVm + epsilonfVf

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52
Q

Rule of mixtures (transverse loading)

A

1/Ec = Vm/Em + Vf/Ef

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53
Q

Ratio of forces carried by matrix and fibres in longitudinal direction

A

Ff/Fm = EfVf/EmVm

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54
Q

Isotropic material

A

Material properties are the same in every direction at a point in the body, so the properties are not a function of the orientation.

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55
Q

Orthotropic material

A

Material properties are different in three mutually perpendicular planes at a point in the body, and have three mutually perpendicular planes of symmetry.

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56
Q

Anisotropic material

A

Material properties that are different in all directions at a point in the body.

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57
Q

Poisson’s ratio

A

The negative ratio of lateral strain to axial strain, resulting from a uni-axial stress in the fibre direction.

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58
Q

Poisson’s ratio for an isotropic material

A

nu = -epsilony/epsilonx

59
Q

Major Poisson’s ratio for an orthotropic material

A

nu12 = -epsilon2/epsilon1, where 1 is the fibre direction.

60
Q

Minor Poisson’s ratio for an orthotropic material

A

nu21 = -epsilon1/epsilon2

61
Q

Relationship between the orthotropic Poisson’s ratios

A

nu12/E1 = nu21/E2

62
Q

S11 (reduced compliance matrix)

A

1/E1

63
Q

S12 (reduced compliance matrix)

A

-nu12/E1 = -nu21/E2

64
Q

S22 (reduced compliance matrix)

A

1/E2

65
Q

S66 (reduced compliance matrix)

A

1/G12, where G is the shear modulus.

66
Q

Q11 (reduced stiffness matrix)

A

E1/(1-nu12nu21)

67
Q

Q12 (reduced stiffness matrix)

A

nu12E2/(1-nu12nu21) = nu21E1/(1-nu12nu21)

68
Q

Q22 (reduced stiffness matrix)

A

E2/(1-nu12nu21)

69
Q

Q66 (reduced stiffness matrix)

A

G12

70
Q

Reason why plies aren’t always aligned

A

Laminates are made up of made up of many plies at different orientations, which gives the laminate good material properties in the longitudinal and transverse direction.

71
Q

Two reasons it’s important to use symmetrical laminates

A
  1. Two plies with different Poisson’s ratios or thermal expansion coefficients will expand by different amounts when heated.
  2. When attached together, they can cool at different rates, which can cause warping, but when aligned symmetrically, the forces on either side balance out, resulting in no warping.
72
Q

Four properties of aluminium

A

Low density, high specific strength, good ductility due to FCC and high electrical conductivity.

73
Q

Three methods of increasing aluminium’s yield strength

A
  1. Solid solution strengthening (alloying)
  2. Precipitation strengthening (age hardening)
  3. Work hardening
74
Q

How does solid solution strengthening work?

A

An element is dissolved (solute) into the aluminium (solvent) so the atoms adopt random sites within the FCC lattice. Solute atoms make slip of dislocations more difficult, so material is strengthened.

75
Q

What is alloy yield stress, sigmay, proportional to?

A

∂C^n, where ∂ is the atomic misfit (difference in size between Al and alloy atoms), C is the solute concentration, and n = 0.3-0.5.

76
Q

A high solute concentration, C, is desirable in solid solution strengthening, but what is it limited by?

A

Temperature. The solubility of Al increases with temperature.

77
Q

How can you achieve a supersaturated solution?

A

Cool the alloy sufficiently fast enough (quench). This freezes the alloying atoms in place.

78
Q

Work hardening

A

Working a metal by rolling, drawing or forging to produce a large amount of plastic deformation, and therefore a large number of dislocations. This increases strength, but reduces ductility.

79
Q

Three stages of precipitation strengthening (age hardening)

A
  1. Solution treatment at high temperature to dissolve alloying elements.
  2. Quenching at room temperature to obtain supersaturated solid solution.
  3. Ageing at controlled temperature and time to decompose the supersaturated solid solution, to form finely dispersed precipitates.
80
Q

Goldilocks zone for precipitation strengthening (age hardening)

A

If precipitates are too small, then it’s easy for the dislocations to cut through them. Too big, and the dislocations will just move around them

81
Q

Precipitation strengthening (age hardening) precipitate size

A

The smallest precipitates have the shortest spacing, and are coherent with the Al matrix. There are strain fields around the precipitates, so the dislocations have to overcome this and cut through them. If overaged, the precipitates coarsen and become incoherent.

82
Q

Wrought aluminium stages

A

Melting, casting, homogenisation, fabrication (working) and thermal treatment.

83
Q

Two stages of Al melting

A
  1. Melt is degassed - Hydrogen gas is formed from water vapour in the furnace, which can cause porosity, so this is removed by blowing an inert gas through the melt.
  2. Melt is poured through a glass cloth on porous ceramic foam to remove oxide skin and particulates.
84
Q

Two methods of Al casting

A
  1. Direct-chill - a semi-continuous process where the molten metal solidifies due to cooling from water spray.
  2. Continuous casting - used for lower strength alloys.
85
Q

Characteristic of cast metal

A

It’s polycrystalline, meaning it consists of many grains. A small grain size is desirable.

86
Q

In terms of grain size, what is the yield stress proportional to?

A

1/√d, where d is the grain size

87
Q

Homogenisation (DC ingots)

A

Diffusion of alloying elements from grain boundaries and other solute rich regions.

88
Q

Three types of fabrication for DC ingots

A
  1. Hot working - break down cast structure for uniform grains.
  2. Annealing - removes residual stress, and softens work-hardened metal.
  3. Cold working - increases the yield stress.
89
Q

Five properties of titanium

A

Low density, high yield stress, high modulus, high Tmelt and excellent corrosion resistance.

90
Q

Titanium metallurgy for T<882˚C

A

Alpha-phase hexagonal closed-packed (HCP)

91
Q

Titanium metallurgy for T>882˚C

A

ß-phase body centred cubic (BCC)

92
Q

Key difference between alpha and beta phase Ti

A

Alpha has lower ductility, but higher creep resistance.

93
Q

Ductility

A

The ability of a material to undergo a large plastic deformation before it fractures, measured by change in elongation or reduction in area.

94
Q

Creep

A

Slow, continuous deformation of materials with time under loads, caused by dislocation movement and diffusion. Metals start to creep at 0.3-0.4 Tmelt.

95
Q

What is the relative stability of alpha and beta affected by?

A

The alloying elements.

96
Q

Alpha stabilisers

A

Dissolve in alpha , and raise the alpha/beta transus.

97
Q

Beta stabilisers

A

Lower the alpha/beta transus.

98
Q

Strengthening mechanism for alpha Ti alloys

A

Solution hardening - they cannot be strengthened by heat treatment.

99
Q

Three different heat treatments of commercially pure Ti

A
  1. Hot worked and annealed at 700˚C then slow cooled, resulting in small alpha grains and a high yield strength.
  2. Quenched from beta phase, which gives martensitic alpha, resulting in lower strength and ductility, but improved fracture toughness and creep resistance.
  3. Air-cooled from beta phase, which gives Widmannstatten alpha plates, again resulting in lower strength and ductility, but improved fracture toughness and creep resistance.
100
Q

Four properties of fully-alpha Ti alloys

A
  1. Relatively low tensile strength
  2. Reasonable creep resistance
  3. Good weldability
  4. Limited formability
101
Q

Near alpha Ti alloys

A

Up to 2% beta phase. Better strength due to formation of hexagonal martensite, and excellent creep resistance.

102
Q

Three properties of alpha-beta alloys

A
  1. High yield strength
  2. Better formability
  3. Reduced creep strength above 400˚C
103
Q

Main stabilisers of beta Ti alloys

A

Mo, V, Cr and Fe

104
Q

Structure of beta Ti alloys

A

Contain enough beta stabilisers to retain full beta structure on quenching from above beta transus. Quenched beta is metastable, and alpha precipitates form on ageing.

105
Q

Five properties of beta Ti alloys

A
  1. High yield stress
  2. Hardenability of thick sections
  3. Formability
  4. Higher density
  5. Low creep resistance
106
Q

What’s a monel?

A

Alloys of nickel and copper. They have good corrosion resistance.

107
Q

Two properties of superalloys

A
  1. High creep and temperature resistance
  2. Metallurgy not as complex as Ti - FCC at all temperatures.
108
Q

What is steady state creep, ∂epsilon/∂t, proportional to?

A

Sigma^n, where n = 3-8

109
Q

Five requirements of alloys

A
  1. Resistance to dislocation slip
  2. Resistance to vacancy diffusion
  3. Stability of microstructure at high temperatures.
  4. Elimination of grain boundaries perpendicular to stress
  5. Corrosion/oxidation resistant
110
Q

What is most strength in Ni superalloys due to?

A

Ni3Al and Ti3Al precipitates in the Ni matrix, that make dislocation slip very difficult.

111
Q

What’s similar between the precipitates and the matrix in Ni superalloys?

A

They’re both FCC, but with a different lattice parameter, meaning coherency strain.

112
Q

Superalloy casting

A

Components (aero blades) are cast in their final shape (investment casting) since superalloys are very hard and therefore difficult to machine.

113
Q

Why do we want to minimise the number of grain boundaries in Ni superalloys?

A

Whilst at low temperatures small grains are good, for high temperature applications, creep resistance is more important, so we want to reduce or eliminate grain boundaries.

114
Q

Two methods of reducing grain boundaries in Ni superalloys

A
  1. Directional solidification - gives only columnar grains, which is worth an extra 50-100˚C operating temperature.
  2. Single crystal blades - no grain boundaries. Worth an extra 50-100˚C operating temperature.
115
Q

Investment casting (lost wax) process (5 steps)

A
  1. Produce master mould using a machinable alloy.
  2. Make wax or a low Tmelt alloy patterns.
  3. Coat patterns with 8-10mm thick slurry.
  4. Melt wax, then sinter the ceramic coating. This gives the mould.
  5. Pour molten metal in vacuum.
116
Q

How is blade cooling achieved?

A

Having hollow blades with air holes.

117
Q

What is applied to superalloy blades?

A

A coating to resist high temperature corrosion/oxidation. It also acts as a thermal barrier.

118
Q

What does tensile testing determine?

A

The tensile modulus and Poisson’s ratio.

119
Q

How is a tensile test conducted? (3 steps)

A
  1. Straight-sided specimen are used with end tabs to protect the fibres from the machine grips.
  2. Strain gauges measure the transverse and longitudinal strains - the resulting stress strain curve yields the modulus.
  3. The Poisson’s ratio can also be determined from the strain gauges
120
Q

How are E1 and E2 measured from a tensile test?

A

E1 is measured in specimen whose fibres are aligned along the specimen axis; E2 is measured in samples with fibres aligned at 90˚.

121
Q

How is the shear modulus, G12, determined from a tensile test?

A

A sample with fibres orientated at 45˚ is used. The stress/strain curve is plotted and used to find Ex. G12 is then calculated from the S11 term in the reduced compliance matrix.

122
Q

G12

A

1/((4/Ex)-(1/E1)-(1/E2)+(2nu12/E1))

123
Q

What other method can be used to find G12?

A

V-notched beam method.

124
Q

Why’s it difficult to measure compressive properties of composites?

A

Because they fail easily in buckling. Large tabs are required on the specimen, and it must be supported by an anti-buckling guide.

125
Q

Delamination

A

The separation of the plies that make up the composite under shear and normal loads.

126
Q

Mode 1 loading - interlaminar fracture toughness.

A

Opening up a cracked body, with the crack propagating along the mid-plane.

127
Q

Mode 1 fracture energy, G1c

A

3Pd/2Ba, where P is the force at that point, d is the corresponding displacement, B is the specimen width and a is the total crack length, including a0. The values of G1c are then plotted against the crack length.

128
Q

Mode 2 loading - end notch flexure specimen

A

Shearing a cracked body specimen that’s loaded in 3-point bending. The crack propagates in an unstable manner to the centre of the specimen, directly under the load. The max force, Pmax, is used to determine G2c.

129
Q

Why is G2c usually higher than G1c?

A

Energy is dissipated in friction between the crack surfaces.

130
Q

What is G2c directly related to?

A

The impact resistance of the composite.

131
Q

Four possible manufacturing problems for composites

A
  1. Porosity - gas getting trapped within the material.
  2. Inclusions - inclusion of a foreign substance.
  3. Fibre-matrix de-bonding.
  4. Fibre waviness due to improper handling.
132
Q

Why is non-destructive testing (NDT) useful?

A

Because it allows the inspection of a component without impairing serviceability.

133
Q

Four NDT methods

A
  1. Ultrasonic
  2. X - ray radiography
  3. Acoustic emission
  4. Thermography
134
Q

Two types of ultrasonic NDT method

A

Pulse echo and through transmission.

135
Q

Pulse echo

A

Like a pregnancy scan. Glass panel placed behind composite. The measured difference in the emitted/received energy gives information on the presence of defects. Transmitter/receiver needs to be in water or have gel applied for good transmission.

136
Q

Three types of pulse echo

A
  1. A-scan - signal displayed on oscilloscope. Depth, size and nature of defect can be estimated.
  2. B-scan - data from a series of linear A-scans can be combined to obtain a representation of the cross-section of the material.
  3. C-scan - data combined over the entire surface. Light areas show damage down to 2mm.
137
Q

Through transmission

A

A separate receiver is placed behind the structure. The signal only passes through the material once.

138
Q

What does the level of absorption of X-rays passing through a structure depend on?

A

The density of the material.

139
Q

How does X-ray radiography work in NDT?

A

The presence of different densities (de-bonding or inclusions) will show up as contrast on the film. Often, the specimen is immersed into an X-ray absorbing liquid prior to inspection, which seeps into the cracks and better shows up defects.

140
Q

Acoustic emission NDT process (3 steps)

A
  1. When a crack develops under loading, stress waves are released that propagate to the surface and produce small displacements.
  2. Surface-mounted transducers detect these displacements and convert them into electrical signals.
  3. No information is provided about the size of damage or failure mode, but the location can be obtained using several transducers and measuring the time of flight of the signal to each sensor.
141
Q

Why isn’t acoustic emission truly non-destructive?

A

Because it depends on damage initiating from pre-existing damage under the application of an external load.

142
Q

Thermography NDT

A

Subjecting the composite to a rapid pulse of heat and measuring the resulting temperature rise. Defects such as delamination change the heat diffusion rate through the composite, which can be detected using an infra-red camera.

143
Q

Two types of thermography

A
  1. Single sided method - source and camera are on the same side of the composite. The defects appear as hot spots.
  2. Double sided method - source and camera are on different sides of the composite. The defects appear as cooler regions.

Engineering - Year Three (9 decks)

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