Plastic Deformation of Materials Flashcards

1
Q

What convention is used for finding a Burgers vector of a dislocation?

A

FSRH

Finish, start, right hand.

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2
Q

What direction is the Burgers vector for a screw dislocation?

A

Parallel or antiparallel to the line direction.

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3
Q

What are screw dislocations defined as either?

A

Left or right handed screws.

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4
Q

When is a screw dislocation left or right handed?

A

It is right handed when the line direction and Burgers vector are parallel and left handed when antiparallel.

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5
Q

How can we treat mixed dislocations for stress analysis?

A

As equivalent to the sum of edge and screw components independently.

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6
Q

Why can we treat mixed dislocations as a sum of edge and screw components?

A

The stress fields around each are orthogonal (have no components in common).

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7
Q

Sketch a dislocation loop with the Burgers vector in the slip plane (shear loop) and include views down different directions.

A

Check slide 8

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8
Q

Can dislocations ever terminate in perfect crystals?

A

No. They will always form closed loops, junctions or terminate at free surfaces or grain boundaries.

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9
Q

Sketch a pure edge dislocation loop.

A

Check slide 8

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10
Q

How can dislocations be imaged using TEM?

A

Around the dislocation, the strain field bends the lattice planes which causes them to no longer satisfy Bragg condition, thus they appear dark in a bright fields image

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11
Q

Describe the long range stress field around a dislocation.

A

Can be modelled using linear elasticity.
Diffuse strain energy stored in a large volume.
No variation with core position relative to atomic structure.

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12
Q

Describe the core structure of a dislocation.

A

Strains too great to be treated by linear elasticity.
Intense strain energy stored in small volume.
May have large fluctuations of energy with core position.

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13
Q

How does the elastic field around a dislocation affect its interactions within a material?

A
The elastic field controls how dislocations react to distant microstructure features with their own elastic stress fields such as:
Other dislocations
Mis-fitting precipitates
Mis-fitting solute atoms
Twins
Applied stresses
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14
Q

Describe how the core structure of a dislocation affects its interactions within a material.

A

The core structure controls how the dislocation interacts with the crystal lattice and atomic structure:
Dislocation dissociation
Core spreading

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15
Q

Roughly how big is the core of a dislocation considered to be?

A

The radius is approximately 4 Burgers vectors which is roughly 1nm

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16
Q

States the stresses for the stress tensor in each direction.

A

Check slide 18

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17
Q

Derive an expression for the stain field of a straight screw dislocation.

A

Check slide 19

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18
Q

Derive an expression for the stress field of a straight screw dislocation.

A

Check slide 20

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19
Q

Describe the stress and strain fields around a straight screw dislocation.

A

Both fields are pure shear.
The fields have radial symmetry
The stresses and strains α 1/r
(Since infinite stresses cannot exist in a real material, the assumption of linear elasticity breakdown at r(0), ≈1nm)

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20
Q

State expressions for the stress field around a straight edge dislocation.

A

Check slide 21

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21
Q

State how you would derive a set of expressions for the stress field around a straight edge dislocation.

A

Take a hollow cylinder along the z axis of the dislocation.
Cut on a plane parallel to the z-axis
Displace the free surface LMNO by b in the x-direction
This situation is plane strain so no displacements in the z-direction,

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22
Q

Describe the stress and strain fields around a straight edge dislocation

A

The stress and strain fields are not pure shear.
The stresses and strains α 1/r
(Since infinite stresses cannot exist in a real material, the assumption of linear elasticity breakdown at r(0), ≈1nm)

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23
Q

Derive an expression for the strain energy around a screw dislocation must know.

A

Check slide 23

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24
Q

Derive an expression for the strain energy around an edge dislocation.

A

Check slide 24

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25
Q

Link the expressions for the strain energy around an edge dislocation with that of a screw.

A

E(edge) = E(screw)/(1-nu) ≈1.4E(screw)

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26
Q

How to find the core energy of a dislocation in its relaxed state.

A

The vacancy formation energy over the coordination number.

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27
Q

What is Perierls-Nabarro stress?

A

The minimum stress required to move a dislocation line.

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28
Q

Derive an expression for the approximate elastic energy of a dislocation.

A

Check slide 25

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29
Q

What energy of a dislocation dominates the total energy of the dislocation motion?

A

The diffuse elastic energy (energy fluctuations depend on core energy term).
E(el) ≈ 10E(core) so elastic term dominates

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30
Q

Sketch a plot of energy against displacement for a dislocation.

A

Check slide 25

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31
Q

Under what conditions will an edge dislocation glide?

A

If there isa shear component of stress in the direction of b.

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32
Q

Under what conditions will an edge dislocation climb?

A

If there is a dilation stress parallel to b, and if the temperature is high enough for vacancies to diffuse.

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33
Q

What is meant by the “configurational” force on a dislocation?

A

The rate of change of energy of the system as the dislocation moves.

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34
Q

Why is glide of an edge dislocation conservative motion?

A

The shape of the material changes but the volume doesn’t. Edge dislocation climb is non-conservative in that the shape and volume both change.

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35
Q

Derive an expression for the force on a dislocation

A

Check slide 27

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36
Q

State the Peach-Kohler equation

A

Check slide 27

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37
Q

What do systems want to minimise?

A

Elastic E.

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38
Q

Show that like dislocations repel and opposite dislocations attract.

A

Check slide 29

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39
Q

Derive an expression for the force between two screw dislocations.

A

Check slide 30

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40
Q

Derive an expressions for the forces between two edge dislocations.

A

Check slide 31

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41
Q

Sketch a plot of F(glide) against ∆x/∆y for like and oppositely signed dislocations stating where each pair have a stable equilibrium.

A

Check slide 32

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42
Q

What stable arrangement do like dislocations tend to take?

A

Planar stacks such as a low angle tilt boundary.

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43
Q

What stable arrangement do like dislocations tend to take?

A

They can form a Taylor lattice or dipole dispersion, both of which have minimal long range stress fields.

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44
Q

What is the traction of a dislocation?

A

The force per unit area that an dislocation exerts on a surface.

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45
Q

What are image forces?

A

Image forces are the result of a dislocation near a free surface causing a traction on it. At a free surface there should be no overall forces on the surface a virtual dislocation is introduced as a mirror to the real one.

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46
Q

Derive an expression for dislocation line tension.

A

Check slide 34

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47
Q

Derive an expression for the critical shear stress required to operate a Frank-Read source.

A

Check slide 36

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48
Q

Sketch a plot of curvature against line length for the operation of a Frank-Read source.

A

Check slide 36

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49
Q

Describe how slip bands can widen by multiple cross-slip.

A

Screw dislocations have a high enough resolved shear stress for glide on more than one slip plane.
Cross slip can occur which leaves segments of the dislocation on the primary slip plane.
The dislocation can then cross slip back onto a parallel slip plane to the primary slip plane where it forms a new dislocation source as it has pinned ends.
This repeats when new dislocations are produced, resulting in the widening of the slip band.

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50
Q

Derive an expression for the strain rate from the motion of dislocations.

A

Check slide 47

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51
Q

What affect does dislocation core width have on the mobility of dislocations?

A

Wider cores need lower force to move dislocation.

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52
Q

Derive an expression for the Peierls stress of a dislocation.

A

Check slide 54

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53
Q

Why do kinks in dislocations form and what limits their formation?

A

Dislocation kinks form when parts of the dislocation line fall into different Peierls valleys. This is opposed by the line tension and mutual attraction of opposite kinks.

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54
Q

How can a kink form when two dislocations intersect?

A

When an edge and a screw intersect.
For the screw the kink is effectively an atomic length section of edge.
On same glide plane as rest of dislocation
Acts as a “secondary dislocation”
Sideways motion of kink can act as glide mechanism for main screw.
Diagram on slide 58

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55
Q

When can jogs form?

A

When 2 screws intersect or an edge and a screw intersect.

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56
Q

What is the difference between a kink and a jog forming?

A

A kink forms on the same slip plane as the main dislocation whereas a jog forms on a new slip plane.

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57
Q

Describe how a jog forms when a screw and edge intersect.

A

The jog forms on a new slip plane than the edge dislocation and can move with the dislocation it is attached to for glide.
Diagram slide 59

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58
Q

Describe how a jog forms when 2 screws intersect.

A

The 2 screws intersect as in the diagram on slide 60
The jog acts as a piece of edge in each dislocation.
The edge dislocation is likely to be on a non-glide plane.
This means can only move this part of the dislocation by climb.

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59
Q

Which dislocations can climb?

A

Only edge

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60
Q

Which dislocations can cross slip?

A

Only screw

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61
Q

Will an edge dislocation climb if vacancy diffusion is great enough?

A

Yes but without a driving force there will be no net climb.

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62
Q

What driving forces can cause climb of an edge dislocation?

A

Mechanical driving force - stress applied.

Chemical driving force - excess vacancy conc in the crystal (dislocation core can act as a vacancy sink).

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63
Q

Does climb occur uniformly along the length of an edge dislocation?

A

No, each vacancy that “lands” creates a pair of unit height jogs (diagram on slide 62).

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64
Q

What is the primary slip system in ccp?

A

{111}

<110>

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65
Q

Describe what happens in dislocation dissociation in cap metals.

A

The perfect a/2<110) Burgers vector forms 2 partials of a/6<211> Burgers vector. This is not a whole lattice vector and creates an intrinsic stacking fault.

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66
Q

Why do dislocations dissociate?

A

The two partial dislocations that form have a combined energy lower than that of the perfect dislocation. This is due to their shorter Burgers vector.

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67
Q

Describe the energetics of a stacking fault that results from dislocation dissociation.

A

As the stacking fault width increases (and it’s energy as it is has an excess energy) the dislocation energy decreases from Ga^2/4 to Ga^2/6 (perfect to partials).

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68
Q

Why do ccp metals have dislocations that dissociate but bcc not?

A

ccp metals meet the following criteria:
A stacking fault structure of low energy
Partial dislocations with a lower combined energy than the perfect.
bcc metals have a high stacking fault energy.

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69
Q

Sketch a plot of dislocation energy, fault energy and total energy for the dissociation of perfect dislocations into partials.

A

Check slide 69

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70
Q

Sketch the geometry of a perfect cap dislocation and its partials.

A

Check slide 69

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71
Q

Derive an expression for the equilibrium spacing of the stacking fault when a perfect dislocation dissociates in a cup metal.

A

Check slide 70

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72
Q

What is Thompson Tetrahedron used for?

A

To visualise the order in which a perfect dislocation will dissociate to creak an intrinsic stacking fault.

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73
Q

What phrase to remember for Thompson Tetrahedron?

A

Look down line of dislocation. Greek-Roman on left, Roman-Greek on right. Greek-Roman always first partial, Roman-Greek always second.

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74
Q

What is a Lomer-Cottrell lock?

A

When the first dislocation in two sets of partial dislocations on different slip systems meet and form a new full dislocation in a non-primary slip system which causes it to be locked along with the second partials behind them.

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75
Q

In order for a screw dislocation to cross slip in a ccp metal, what must first happen?

A

The partials must recombine in order for the whole dislocation to cross slip.

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76
Q

What is constriction in the sense of cross slip in ccp metals?

A

It is the force to recombine the dissociated dislocations. The constriction can be assisted by the stacking fault energy and materials with higher stacking fault energies recombine more easily. This means materials with lower stacking fault energies require more energy to recombine the dislocations and cross slip.

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77
Q

How does cross slip affect monotonic loading?

A

In early work hardening it causes glide band spreading, formational of new dislocation sources and can by pass precipitates.
In later work hardening it also bypasses locks and dislocation tangles.

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78
Q

How is cross slip important in fatigue?

A

Irreversible slip in cyclic loading.

Stability of persistent slip band structures.

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79
Q

When vacancies collect to form an intrinsic stacking fault, what else forms?

A

A loop bound by pure edge dislocation.
The loop must have Burgers vector of a/3<111> in ccp
This is normal to the {111} plane.
This is a Frank loop

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80
Q

How can a Frank loop be removed from a material?

A

The loop is sessile so can only grow or be removed by addition of vacancies (by climb).
It could also be removed by a dislocation reaction. If it combines with the right Schockley partial a prefect dislocation forms.

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81
Q

Describe how a Frank loop be converted to a prismatic loop.

A

A Schockley partial nucleates within the Frank loop.
The partial traverse the loop and converts it to a perfect dislocation loop.
This loop can glide, but not in its own loop plane.
This is a prismatic dislocation loop.

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82
Q

In hip systems, what are the favoured slip systems?

A

If c/a≥1.633, basal slip is favoured

If c/a≤1.633 prism slip is favoured

83
Q

In hcp systems, are stacking faults possible with partial dislocations?

A

Yes but only when basal slip is favoured as these are the close packed planes.

84
Q

What must happen for glide of dislocations in hip systems on non-basal planes?

A

The dislocation core must rearrange as they have non-planar cores in non-basal planes. This is why non-basal glide has a higher CRSS.

85
Q

What is the primary slip system in bcc metals?

A

{110}<111>

86
Q

What are the secondary slip systems in bcc?

A

{211}<111>

{321}<111>

87
Q

Can dislocations dissociate in bcc?

A

No as the stacking fault energy is high.

88
Q

Why is cross slip usually seen in bcc metals?

A

There are no stacking faults so the screw dislocations are not dissociated
There are very many slip planes for a given direction.

89
Q

Describe the core structure of screw dislocations in bcc.

A

Screw dislocations have non-planar cores.
The delocalisation causes relative displacement of atoms around the core.
The shifts are concentrated along potential slip directions.
There are no stacking faults and this is not dissociation.
This means glide of a screw is difficult and very temperature dependent as thermal activation assists core rearrangements.

90
Q

Is Schmid’s law obeyed when screw dislocations delocalise in bcc?

A

No as it assumes stress applied normal to the slip plane has no affect, but it does on delocalised screws.

91
Q

When dislocation loops are activated in bcc, what happens?

A

They formal oval shapes as the screw components don’t glide easily but the edge components do.

92
Q

Why does dislocation mobility in bcc vary strongly with impurity content?

A

Cottrell atmospheres can form, increasing the activation stress.
Stress fields around interstitial solutes have strong shear and dilation components that interact with edge and screw dislocations
Thermal activation is needed to pull the dislocation cores away from these solutes.

93
Q

What is the big difference between dislocation motion in ionic crystals compared to metals?

A

There are strong fluctuations in energy that occur when ions of like or opposite charge move close to one another.

94
Q

Why isn’t slip on {111} observed in rock salt?

A

{111} planes contain atoms of the same charge.

Glid on these plans would have very strong interactions of like charges moving past one another.

95
Q

What system does primary slip occur on in rock salt?

A

{110}<110> as these planes are charge neutral.

96
Q

What can make slip easier in ionic structures?

A

Thermal activation can make slip easier. If more slip systems are activated, Von Mises criteria can be met and the material can plastically deform. This often does not happen in ionic systems.

97
Q

How can plastic deformation occur in glasses?

A

Densification (compaction of glass structure) or shear flow (isochoric).
Plasticity can be induced below T(g).

98
Q

When thinking about strengthening metals, what mechanisms are there (toolbox)?

A

Effect of bonding type on ease of dislocation motion.
Dissolved atoms interact with dislocation stress fields.
Interaction between stress fields of dislocations and precipitates.
Precipitates act as local barriers for dislocations.
Grain boundaries act as barriers for dislocations.
Immobile dislocations block mobile dislocations.
Phase changes can produce fine, strained microstructures with precipitates.

99
Q

What intrinsically causes strength in a material?

A

Peierls-Nabarro stress.
High elastic modulus (resistance to bond distortion)
High melting point (resistance to bond breakage)
Localised bonding (core width)

100
Q

Give an expression for the Peierls energy of a dislocation.

A

Check slide 8

101
Q

True or false, materials with similar structure, tend to have similar hardness/YM temperature relationships?

A

False, however if the temperatures are normalised by dividing by melting temperature, they follow similar patterns.

102
Q

How are dislocations affected by lattice distortions?

A

Interaction energies between the stress fields of a moving dislocation and other static distortions give rise to widely varying dislocation energy with position.
This leads to an increases stress to move dislocations through an array of obstacles as F = -dE/dx

103
Q

Why won’t dislocation lines be straight through an array of obstacles?

A

The dislocation moves between minimum energy configurations and it is a compromise between the line tension and attraction/repulsion from solute atoms.

104
Q

How can temperature affect dislocation propagation?

A

Thermal energy allows dislocations to overcome barriers.

105
Q

Derive an expression for how dislocation velocity varies with temperature.

A

Check slide 17

106
Q

Derive an expression for how macroscopic strain rate varies with temperature due to the thermal activation of dislocations.

A

Check slide 17

107
Q

Derive an expression for the critical temperature of dislocation propagation.

A

Check slide 18

108
Q

What is critical temperature in dislocation propagation?

A

The temperature at which there is sufficient thermal energy to overcome barriers without stress.

109
Q

Sketch a plot of shear strength against temperature labelling the thermal and athermal regions.

A

Check slide 19

110
Q

True or false, any interaction with a dislocation in a lattice causes strengthening.

A

True

111
Q

What is solid solution strengthening proportional to?

A

Amount of solute
Solute misfit (increases barrier strength)
Elastic modulus of matrix (increases barrier strength)

112
Q

Give an expression for the interaction energy of a misfitting substitutional solute.

A

E(I) = p∆V

Where p is hydrostatic pressure and ∆V is the volume change.

113
Q

Derive an expression for the change in hole volume of a misfitting substitutional solute.

A

Check slide 25

114
Q

Give an expression for the hydrostatic pressure around an edge dislocation.

A

Check slide 27

115
Q

Give a full expression in terms of the shear modulus, misfit etc of the interaction energy of a misfitting substitutional solute with an edge dislocation.

A

Check slide 27

116
Q

Give a full expression in terms of the shear modulus, misfit etc of the interaction energy of a misfitting substitutional solute with an screw dislocation.

A

Zero, as the hydrostatic pressure around a screw dislocation is zero.

117
Q

Give an expression for the maximum interaction energy between a substitutional solute and an edge dislocation on different planes.

A

Check slide 29

118
Q

Find an expression for the maximum glide resistance (force) to an edge dislocation caused by a substitutional solute on different planes. and state the angle at which it happens

A

Check slide 30

And 60° between edge dislocation and solute measured from the glide plane of the edge dislocation.

119
Q

What is can be said of the shape of the stress and strain fields around a substitutional solute?

A

They are spherically symmetric.

120
Q

What is the result of interstitial solutes such as carbon in iron being bigger than the radius ratio of an atom that will fit in the interstitial site?

A

Each interstitial atom has a large and non-spherically symmetric stress field around it with shears as well as dilation.

121
Q

How can interstitial atoms pin a dislocation?

A

A Cottrell atmosphere can form.
The interstitials have high stress fields that can have lower energy by being at dislocation cores.
The stress fields have both shear and dilation components so pin both edge and screw dislocations.
They can diffuse rapidly at RT.

122
Q

What do Cottrell atmosphere’s do to the yield of a material?

A

Increases until dislocations become unpinned then a lower stress is required to propagate dislocations.
The interstitials will diffuse back over time causing the yield stress to increases again. This is UYS and LYS.

123
Q

Describe the formation of Lüder’s bands

A

Can form in any material with a yield drop.
Slip begins in a grain near surface where there is a higher stress conc (likely surface flaw)
Has maximum Schmid factor.
Slip occurs in grain and dislocations pile up at gb causing stress conc in the second grain on slip plane with angle nearest to 45°
Band of deformed grains crosses specimen at ≈45°
The band then widens when slip is triggered in neighbouring grains
Slip due to Lüder’s bands occurs at the lower yield stress.

124
Q

Give the relationship between misfit of an interstitial solute and the temperature that causes lower core concentration. State an equation too.

A

Bigger misfit implies higher required T to lower the core concentration of interstitials at a dislocation core.
Equation on slide 37

125
Q

How can point defects interact with dislocations in the lattice?

A

The defect-matrix bond may be harder or softer than the pure matrix and it cause cause a large change on modulus close to the point defect.

126
Q

What happens between the dislocation and the defect in the case of a hard or a soft defect?

A

Soft defect causes modulus locally to decrease (eg, vacancy), so there is an attractive interaction with a dislocation.
Hard defect causes modulus locally to increase, so there is a repulsive interaction with a dislocation.

127
Q

What happens if G is changed locally to a point defect in a material to dislocation motion?

A

If G increases, so too does E(I) so the maximum glide force also increases.

128
Q

What electrical interactions can happen in a material with dislocations?

A

Conduction electron density tends to increase in tensile and decrease in compressive regions of an edge dislocation core which causes an electric dipole that may interact with solutes of different valency.
In ionic crystals there is a significant amount of electronic interaction as lack of free electrons which means jogs can become charged

129
Q

How can stacking faults affect the strength of a metal in terms of their relationship with solutes?

A

Stacking fault formed between two partial dislocations.
Solute atoms have different solubility in stacking fault than matrix.
Concentration in stacking fault either increases or decreases by diffusion to lower the stacking fault energy.
This means extra work is needed to move the stacking fault.

130
Q

What 4 types of forces can obstacles exert on dislocations?

A

Strong, weak, local or diffuse.

131
Q

Describe how a strong force acts from obstacles on dislocations.

A

Causes large bowing of dislocation. To get more bowing, the obstacles must have similar spacing along the dislocation as in the material.

132
Q

Describe how a weak force acts from obstacles on dislocations.

A

Obstacles do not cause much bowing of dislocation.

133
Q

Describe how a diffuse force acts from obstacles on dislocations.

A

Dislocation feels force before getting to the obstacle, eg, coherent ppt or misfitting solute. Force exerted over a long length of the dislocation line

134
Q

Describe how a local force acts from obstacles on dislocations.

A

Dislocation does feel force until at the obstacle, eg, incoherent ppt. Force exerted over a small part of the dislocation line.

135
Q

State an expression for the maximum force on a dislocation.

A

F(max) = tau(i)b

136
Q

Give an approximate expression for the radius of curvature when a dislocation bows in terms of G, b, and tau(i).

A

R≈Gb/2tau(i)

137
Q

Hoe can strong and weak forces on a dislocation be defined when the obstacles have spacing Λ.

A

For strong forces: R ≤ Λ

For weak forces: R&raquo_space; Λ

138
Q

Derive an expression for the shear stress required to move a dislocation with strong force and diffuse interactions.

A

Check slide 51

139
Q

Derive an expression for the shear stress required to move a dislocation with weak force and diffuse interactions.

A

Check slide 55

140
Q

Derive an expression for L (length if bow) for weak diffuse interaction.

A

Check slide 53

141
Q

Derive an expression for the maximum (strong) localised force on a dislocation from a regular array of obstacles.

A

Check slide 57

142
Q

Derive an expression for the weak localised force on a dislocation from a regular array of obstacles.

A

Check slide 60

143
Q

State Archimedes Quadrature of the the Parabola

A

Check slide 59

144
Q

What is the relationship between strength and concentration of solute.

A

Check slide 66

145
Q

If there is more than one solute in an alloy, state the approximation for the combined effect.

A

τ = √(τ(1)^2+ τ(2)^2)

146
Q

What long range interactions with precipitates can harden materials?

A

Coherency strains

Semi-coherent precipitates

147
Q

What “cutting” interactions with precipitates can harden

materials?

A
Interfacial energy and morphology
Lattice friction stress
Stacking fault energy
Ordered precipitates
Modulus effects
148
Q

What “by pass” interactions with precipitates can harden

materials?

A

Bowing around precipitates (Orowan mechanism)

149
Q

Describe how coherent precipitate strengthening works

A

Strength increases with concentration misfit

150
Q

When do coherent precipitates produce diffuse strengthening by weak obstacles?

A

When N is small (early stage of precipitation) so Λ is small

151
Q

When do coherent precipitates produce diffuse strengthening by strong obstacles?

A

When precipitates grow, N and Λ increase and dislocations curve around precipitates.

152
Q

What kind of strengthening and obstacles causes cutting of coherent and semi-coherent precipitates.

A

Localised strengthening by weak obstacles

153
Q

Derive an expression for the diffuse interaction around a strong obstacle.

A

See slide 79

154
Q

Derive an expression for the K(max) when cutting through a precipitate.

A

Check slide 81

155
Q

Derive an expression for the strengthening of a material by dislocation bowing.

A

Check slide 82

156
Q

Describe how Orowan strength changes with concentration and Λ.

A

Orowan strength will increase with concentration and decreasing precipitate spacing

157
Q

How can Orowan strength increase even if the precipitate spacing is not large compared to precipitate size?

A

The formation of dislocation loops (Orowan loops) around precipitates.

158
Q

How do stacking faults interact with coherent and semi-coherent precipitates?

A

Either:
Stacking faults energy per unit area is lowered in ppt.
Stacking fault widens
Dislocation stacking fault energy per unit length increases
Dislocation elastic energy per unit length decreases
Overall dislocation energy in ppt decreases
Or:
Stacking fault energy per unit area increases in ppt.
Stacking fault contracts
Dislocation stacking fault energy per unit length decreases
Dislocation elastic energy per unit length increases
Overall dislocation energy in ppt increases
Either way, increased resistance

159
Q

Describe how ordered ppts cause an increase in strength,

A

In cases where a dislocation enters a dislocation enters an ordered ppt and the Burgers vector is not favourable because of other atoms (eg Ni3Al in bcc structure) dislocations “reverse” the ordering forming an anti phase boundary which means dislocations must pair up to cut through. (removes the APB)

160
Q

How do ordered ppts work in superalloys.

A

Ni3Al is the classic example.
γ’ phase is held together by Ni “glue”
APBs can be more stable on non-glide planes so dislocations can flip into a non-glide configuration in the ordered ppt.
APB on (100) but glide on (111)
Causes a lock as stacking faults still on (111)
This is also thermally activated and has a greater effect with increasing γ’

161
Q

What phases are present in the ageing of Al-Cu?

A

GP zones
θ’’
θ’
θ

162
Q

State the coherency of the phases present in the ageing of Al-Cu.

A

GP zones: coherent
θ’’: coherent
θ’: semi-coherent
θ: incoherent

163
Q

What happens in the solid solution stage of ageing Al-Cu that strengthens the alloy?

A

Strengthening only due to weak obstacles (local or diffuse)

164
Q

What happens in the GP zones or θ’’ stages of ageing Al-Cu that strengthens the alloy?

A

Coherent ppts
Strengthening due to coherency strains due to diffuse obstacles
Weak, then strong diffuse obstacles with growth
ppt cutting also occurs (weak localised obstacles)
Strengthening increases with ppt size

165
Q

What happens in the θ’ stage of ageing Al-Cu that strengthens the alloy?

A

semi-coherent ppts
Strengthening due to coherency strains due to diffuse obstacles
Weak, then strong diffuse obstacles with growth
ppt cutting also occurs (weak localised obstacles)
Strengthening increases with ppt size

166
Q

What happens in the θ stage of ageing Al-Cu that strengthens the alloy?

A

Strengthening by localised strong obstacles
Dislocation bowing as ppts coarsening and spacing increases.
Strength decreases with increased ppt size

167
Q

What ways can grain and phase boundaries be joined?

A

They can have a common line
Or
They can have no matching orientations

168
Q

What happens with dislocation propagation when two grain or phase boundaries have a common line?

A

Slip planes in both grains have a common line in g.b. (this may happen at twin boundaries)
Burgers vector lies in the g.b. plane
Still maybe be problems in dislocation propagation if dissociated.

169
Q

What happens with dislocation propagation if there is no common line at grain and phase boundaries?

A

Direct transmission of dislocations is almost impossible

170
Q

How can slip propagate in a neighbouring grain when there is pile up at a boundary?

A

At the yield stress
Every dislocation in the pile up exerts a stress back on the source - tending to stop it operating
Slip must also be transmitted from one grain to the next
This is assisted by a stress concentration at the head of the pile up.
Sufficiently large stresses are concentrated in the next grain to operate a dislocation source.

171
Q

Derive the Hall-Petch relation

A

Check slide 102

172
Q

Pile up stress on a gb depends on when?

A

Applied stress

Number of dislocations

173
Q

Dislocation pile up length depends on what?

A

Depends on grain size

174
Q

Applied stress required to transmit strain across grain boundary depends on what?

A

Grain size

175
Q

Are most dislocations mobile?

A

No, they are just obstacles to the few that are.

176
Q

Sketch a stress strain curve for a metal with stress against true and engineering strain.

A

Check slide 113

177
Q

State an expression for the relationship between strengthening and dislocation density

A

strengthening α √(dislocation density)

178
Q

State an expression for the critical point of necking in a tensile test and explain it.

A

dσ(T)/dε = σ(T)

179
Q

When do Lüder’s bands form in work hardening?

A

When the work hardening rate is zero

180
Q

Derive the Considere Criterion (plastic instability)

A

Check slide 115

181
Q

Explain what happens at plastic instability.

A

The hardening rate (of true stress) is equal to the true stress at the point at which the material can no longer support an increasing load.
Beyond this point on the stress-strain curve, the deformation will tend to localise in a (diffuse) neck.

182
Q

Derive an expression for the Cnsidere’s Construction for true stress against engineering strain.

A

Check slide 116

183
Q

State the Hollomon equation

A

σ(T) = Kε^n

where true strain at the instability ε=n

184
Q

Describe what happens in stage I work hardening.

A

Low work hardening rate

Dislocations move large distances on primary slip systems

185
Q

Describe what happens in stage II work hardening.

A

Higher work hardening rate than stage I

Increases dislocation multiplication and dislocation movement on secondary slip systems.

186
Q

Describe what happens in stage III work hardening.

A

Reduced work hardening rate compared to stage II

Cross slip occurs to overcome or by pass the barriers to dislocation motion.

187
Q

What factors strongly affect the transitions between the stages of work hardening and strain hardening rate?

A

Crystal structure

Temperature

188
Q

State an expression for the shear stress required to move on dislocation past another in terms of G and dislocation density.

A

Check slide 120

189
Q

Describe the process of recovery.

A

Dislocation multiplication leads to tangles.
Most of the work during plastic straining is dissipated as heat but some stored elastically as strain in the increased density of dislocations and point defects.
Cold work stable below ≈0.3 T(m)
Above 0.3 T(m) dislocations become sufficiently mobile to reconfigure by climb and cross-slip to reduced stored energy forming low angle gbs.

190
Q

Is stage I work hardening observed in polycrystalline materials?

A

No, due to Von Mises criterion for grain shape compatibility which requires 5 independent slip systems to be active.

191
Q

Why is the yield stress for polycrystalline materials higher than that for a single crystal.

A

Polycrystalline materials have a Taylor factor that assumes all grains in a polycrystalline aggregate experience the same deformation and that the compatibility between grains is satisfied. The factor is found by averaging the stress over all grains and considering the most favoured slip systems. It assume all grains experience the same deformation and VM criterion is upheld,

192
Q

State expressions for the yield stress and work hardening rate of a polycrystalline material (Taylor factor).

A

Check slide 125

193
Q

What is the Orientation Distribution Function (ODF)?

A

It is a measure of the fraction of grains in a volume with a certain orientation. This can help to define texture.

194
Q

Define macrotexture

A

Description of texture of all grains in a sample without regard to their spatial location, orientation relationship of grains to neighbouring grains.

195
Q

Define microtexture

A

Description of variation of orientation of grains with location or neighbouring grains.

196
Q

Define fibre texture

A

One crystal orientation is aligned along a direction; grains otherwise orientated randomly about that direction.
Result of wire drawing, extrusion etc.

197
Q

Define sheet texture.

A

A plane (hkl) is aligned to the sheet plane and direction [uvw] is parallel to the rolling orientation.

198
Q

How can macro texture be measured?

A

X-ray diffraction

Neutron diffraction.

199
Q

How can microtexture be measured?

A

TEM (direct measurement by HR TEM, SEAD or CBED)

SEM (electron backscatter diffraction)

200
Q

Explain how pole configurations can show texture.

A

Similar to heat maps, can show that grains are orientated along a direction.

201
Q

Explain the properties of a textured material.

A

Anisotropic crystal structure leading to anisotropic mechanical properties.

202
Q

Explain the defects of a textured material.

A

Can cause planar anisotropy and formation of east in drawing
Further rolling requires higher pressures which can lead to cracking due to internal stresses
Pitting of stringers can cause stress corrosion cracking

203
Q

How is the Taylor factor related to texture?

A

Taylor factor determines the relationship between crystal properties and the polycrystalline.