Course E - Mechanical Behaviours of Materials

Question 1

what is tensile testing, explain the different regions and the graph that is formed

Answer 1

• specimen is stretched
• force and corresponding extension of sample is recorded
• done until failure

1) as material extends, force rises and is linear with extension, in this region all deformation is elastic

2) yield point, from linear/elastic —> non-linear/plastic deformation

3) Ultimate tensile strength, highest point on graph, max force a material can withstand

4) failure

Question 2

what can we say about unloading from a tensile test

Answer 2

• if the only extension that occurred was in the elastic region then it is all recovered
• any plastic deformation is not recovered but the elastic deformation that had occurred is

Question 3

what is the limitation with tensile testing

Answer 3

• it is very dependent on the size/shape of the material

Question 4

define stress, give the equation for (true) stress

Answer 4

stress, σ, is the normal force, F, acting on a surface per unit area, A,

σ = F/A

units = Pa

Question 5

what is an issue that occurs with stress, what alternative expression can we use for stress

Answer 5

• when under stress, the sample can deform and hence the cross-sectional area that the force acts on can change
• hence we tend to just use engineering stress where the initial cross sectional area is taken

σ(eng) = F / Ao

Question 6

define strain/ give the expression for (true) strain

Answer 6

• strain, ε, is defined as the relative change in the linear dimension of the material in the direction of the force

δε = δl/l

integrate between limits (li and lo) gives
ε(true) = ln(li/lo)

Question 7

what is engineering strain, give the expression for it

Answer 7

ε(eng) = li-lo/lo = li/lo -1

Question 8

define shear stress

Answer 8

shear stress, τ, is the force per unit area parallel to a surface

τ = F/Ao

Question 9

define shear strain, γ

Answer 9

• to do with the change in shape that occurs under a shear stress
• generally under a shear stress a sample will ‘squish’/ deform
• focusing on one vertex we can note the distance by which the opposite side has shifted, this is Δyo, the side length is xo

γ = Δyo/xo = tan(φ)

Question 10

define Poisson’s ratio, give the equation for it

Answer 10

• we know that under deformation, the shape of a material changes
• i.e. it elongated parallel to the force
• for a normal stress on z we can say

εx = εy = - v εz

where
v = Poisson’s ratio

• generally (0.2 - 0.5)

Question 11

what is a stress-strain curve, why do we use them, what does the gradient of the linear part give

Answer 11

• similar to force-extension graph but now it is independent of the material geometry
• it follows the exact same shape as a force- ext. graph but has slightly different features
• yield point –> yield stress σy
• ultimate tensile strength –> ultimate tensile stress, σ(UTS)
• strain to failiure εf = strain at failiure
• gradient of linear part = Young’s Modulus, E

Question 12

define elastic strain, give the equations linking stress and strain for normal and shear stresses

Answer 12

“In an elastic deformation, any strain in response to stress will be immediately and completely recoverable on removal of the stress”

σ = Eε
τ = Gγ

E = Young’s modulus
G = shear modulus

Question 13

how can we relate E and G using poisson’s ratio, v

Answer 13

E = 2G(1+v)

Question 14

how can we calculate the elastic strain energy/work done in elastic strain PER UNIT VOLUME for a material

Answer 14

consider work done as
FδL = σAδL = σALδε = VEε δε
W = ∫(0 to εmax) VEε dε = 1/2 Vεmax^2

so
Wv = 1/2 Eεmax^2 (normal stress)
Wv = 1/2 Gγ^2 (shear stress)

Question 15

what is the link between the work done per unit volume in elastic deformation and the stress-strain graph

Answer 15

• the work done PER UNIT VOLUME is the area under a stress-strain graph

Question 16

how can we model bonds/ roughly explain the shape of their potential against bond length graph

Answer 16

• when the atoms are far apart, their binding energy –> 0
• as the atoms approach each other PE decreases, potential well, attractive forces
• Umin = PEmin = equilibrium separation of atoms at 0K
• if atoms are close than the bond length at Umin, ro, then their energy rises/repulsive forces

The shape of the potential can be approximated using the Leonard-Jones potential

U(LJ) = Umin[(ro/r)^12 - 2(ro/r)^6]
DON’T LEARN THIS

Question 17

give a summary of how we can approximate Young’s moduli of a simple cubic crystal using the Leonard Jones potential

Answer 17

for small displacements from ro we assume its roughly linear so
F = dU/dr
(explains F = kx)

we consider stretching a crystal on [100]
If the crystal is simple cubic then:
- each bond has an area ro^2 perp. to normal stress
σ = F/ro^2 = 1/ro^2 dU/dr
E = dσ/dε = dσ/dr dr/dε = (1/ro^2 d^2U/dr^2 |r = ro) (ro)

E = (1/ro) (d^2U/dr^2 | r = ro)

Question 18

why do we use composites

Answer 18

• we can combine ceramics and polymers for better properties
• generally ceramics/polymers are stronger as fibres
• these fibres are embedded in a matrix material
• orientation within a plane can be random or aligned
• plies of aligned fibres can be stacked
• greatest reinforcing effect when fibres are aligned parallel but this makes material V anisotropic

Question 19

how can we do analysis on composites/ what model should we use

Answer 19

we can use a slab model:
- we consider ‘slabs’ of the matrix material and fibre material
- the volumes of the slabs correspond to their true volume fractions

Question 20

how can we determine/estimate the axial modulus of a composite using a slab model

Answer 20

• consider an axial stress on the slabs, this is where the stress acts on the side faces where both slabs are visible
• Voigt Model
• we assume extension by each slab is equal

σc = σfVf + σmVm = σfVf + σm(1-Vf)
σ = Eε
Ecεc = EfεfVf + Emεm(1-Vf)

we assume εc = εf = εm
so

Ec = EfVf + Em(1-Vm)

• this is a reasonable estimate - it is just a weighted average

Question 21

how can we determine/estimate the transverse modulus of a composite using a slab model

Answer 21

We use Reuss model:
- we assume equal stress

• this time we consider stress acting on the two faces ‘top and bottom’ where only one of the slabs is visible

εc = εfVf + εm(1-Vf)

ε = σ/E

σc/Ec = σfVf/Ef + σm(1-Vf)/Em

σc = σf = σm
so
Ec = EfEm / (EmVf + Ef(1-Vf))

• this is a poor model, lower bound estimate as it does not consider shielding of areas around fibres by fibres

Question 22

how is temperature related to kinetic energy

Answer 22

3/2 KT = <1/2mv^2> = avr. Ek

Question 23

microscopically/ on an atomic level, how can we explain thermal expansion

Answer 23

• due to the asymmetry of the Leonard-Jones potential, we can see by taking the mean ro value at different temperatures that ro increases with temp
• this is done by considering higher Y lines on the U(LJ) graph

Question 24

macroscopically give the equation for what happens in thermal expansion

Answer 24

εt = α ΔT
α = coefficient of thermal expansion

Question 25

what can we say about the thermal expansion of materials with stronger/weaker bonds

Answer 25

• materials with strong inter-atomic bonds tend to have a lower α value, and a greater E value, and a deep, sharp U well, so they expand less
• materials with weaker inter-atomic bonds tend to have a higher α value, lower E and a shallower U well, so expand more

Question 26

what can occur to a bimaterial strip when heated/cooled

Answer 26

• if the two materials have different α values then one will expand more than the other
• this can cause bending if the two strips are bonded together

Question 27

which sections of a beam are under which sorts of stress when bent, what is the neutral axis

Answer 27

• when a rod is bent, the top is under tension, the bottom is under compression
• at the halfway point, there is a neutral axis where σ, ε = 0

Question 28

what do we define our radius as when considering the bending of beams

Answer 28

• although θ is constant, arclength changes with height so we define our radius as being to the neutral axis

R = l/θ

l = length of neutral axis

Question 29

how can we calculate σ(axial) and ε(axial) for a bending beam

Answer 29

• we know the neutral axis represents an undeformed length = l = Rθ
• a deformed length is (R+y)θ

Hence
ε(axial) = ((R+y)θ - Rθ) / Rθ = y/R
σ(axial) = Ey/R

Question 30

how can we calculate the total bending moment on a beam

Answer 30

we know the force on a cross-sectional area is F = σA = Ey/R bdy
b = base length
dy = infinitesimal height difference

moment = F x perp. dist. to N.A. = Ey^2/R

Total bending moment = E/R ∫(-h/2 to h/2) y^2 b dy

= EI/R

I = second moment of area

Question 31

what is the equation for beam stiffness

Answer 31

Beam stiffness = Λ = EI
I = second moment of area

Question 32

what is the equation for the second moment of area, what is it for a rectangular section

Answer 32

I = ∫(on the relevant section) y^2 b(y) dy

for a rectangular section
I = bh^3/12

Question 33

what is the equation for the moment on a cantilever beam

Answer 33

M = F(L-x) = d^2y/dx^2 EI

Question 34

what is the equation for the deflection at a point, y for a cantilever beam

Hence what is the equation for max. deflection, δ on a cantilever beam

Answer 34

y = (Fx^2/6EI) (3L-x)

max deflection, δ, occurs where x = L, at end
δ = FL^3 / 3EI

Question 35

in a perfect crystal, what is the only way in which plastic deformation can occur, what is the relevant equation that would give this

Answer 35

in a perfect crystal, plastic deformation can only occur by whole planes of atoms sliding across each other, a ‘Block slip’ model

this would be given by

τ(crit) = Gb / 2πh

b = interatomic spacing
G = shear modulus
h = planar spacing

Question 36

what is wrong with the ‘block slip’ model in perfect crystals, what is the actual way in which plastic deformation occurs, what is a good way to think about it

Answer 36

• the ‘block slip’ model very much overestimates max. strength
• doesn’t explain why only certain planes move
• actual mechanism is through movement of defects/dislocations
• once a defect/dislocation has moved through the entire crystal, it appears as though the whole plane as slipped but it actually only requires movement of a few atoms at a time
• this explains the lower yield stresses
• can be thought of as difference between pulling an entire rug and moving a ‘ruck’ through a rug

Question 37

what are dislocations, how can they be thought of locally

Answer 37

line defects

locally they can be thought of as an ‘extra’ half plane of atoms

Question 38

what are the two vectors that define a dislocation, define them

Answer 38

1) the line vector, l, separates the slipped and unslipped regions of the crystal
2) the Burgers vector, b, describes the magnitude and direction of the displacement caused by the dislocation

Question 39

how can we determine a burgers vector through a burgers circuit

Answer 39

1) draw a circuit/ closed loop around the dislocation
2) redraw circuit/ closed loop in a ‘perfect crystal’ (one without the dislocation)
3) this will form a closure failure, vector from finish F, to start, S is generally defined as the burgers vector

Question 40

what is an edge dislocation, give its features and its formal definition

Answer 40

• the line vector runs along the base of the extra half plane of atoms
• corresponding burgers vector can be made by making a burgers circuit
• the convention for the burgers circuit is clockwise around l and F—>S

“an edge dislocation is defined by b perp. to l”

Question 41

what is a screw dislocation, give its features and its formal definition

Answer 41

• where part of the crystal has sheared
• hard to visualise but imagining a burgers circuit, if a square path is traversed, the end is directly below the start

“Screw dislocations are defined by b parallel to l”

Question 42

what is a mixed dislocation, give some of its features

Answer 42

• rarely are dislocations pure edge or pure screw
• b can be neither parallel nor perpendicular to l
• this is a mixed dislocation
• commonly found in dislocation loops

Question 43

by what process do dislocations move, it is localised?
on what do they move?

Answer 43

• dislocations move in a process called glide
• it is very localised and occurs on a single plane called a slip plane
• it is easiest to visualise in an edge dislocation by considering the extra half plane of atoms ‘moving’ across the crystal as they align with another plane of atoms
• screw dislocations move more through a sort of unzipping process

Question 44

what can we say about the displacement to a section of the crystal that occurs when a dislocation passes

Answer 44

• once the dislocation passes, atoms initially bonded become displaced by a burgers vector
• slip direction // b
• after the dislocation has passed, the structure is perfect again

Question 45

what are the two KEY CONDITIONS for dislocation motion

Answer 45

1) “If an applied shear stress has a sufficient component parallel to the burgers vector then slip occurs parallel to the burgers vector and the dislocation moves perpendicular to its line vector”
SUFFICIENT FORCE // b
MOVES ⟂ to l

2) “A dislocation can only move on a crystallographic plane that contains both b and l, these planes are slip planes”

Question 46

from the definition of a slip plane and screw and edge dislocations, what can we say about the number of slip planes for each type

Answer 46

Edge dislocation:
b ⟂ l, so glide is limited to a single slip plane

Screw dislocation:
b // l, so family of possible slip planes

Question 47

how can we rationalise the movement of a dislocation loop under stress

Answer 47

• we can consider the movement of specific sections under an applied shear stress

By convention:
- a dislocation should be defined by a single line vector, thus the burgers vector is the same around the whole loop

In reality:
- segments on opposite sides of a dislocation loop are of opposite sense
- so their reaction to an applied shear stress // b is opposire

• thus a dislocation loop will expand or contract when a shear stress τ > τcrit is applied

Question 48

what is the shear stress required to move a dislocation in an otherwise perfect crystal

Answer 48

• the Peierls- Nabarro stress

τp = 3G exp(-2πw / b)

G = shear modulus
w = dislocation width, the distance over which atoms are significantly displaced from their ideal positions (generally where displacement > b/4)

b = modulus (burgers vector)

Question 49

what is the work done when a dislocation moves, how can we consider forces acting to do this work

Answer 49

If a dislocation is acted on by τ > τcrit then it moves ⟂ l
- hence we expect work to be done
- we can define a virtual ‘glide’ force (not real but useful to consider)

W = FLd
F = glide force (per unit length)
W = work done
L = length of line of dislocation
d = distance dislocation moves

F = τb (THIS IS IN DATA BOOK)
W = τbLd

Question 50

explain briefly where the expression for the energy of a dislocation comes from

Answer 50

• elastic distortions occur due to dislocations, we can estimate the energy by considering distortion of an Annulus around a screw dislocation

use
V = 2πrldr
γ = b/2πr
shear strain per unit vol = 1/2 Gγ^2
dU = 1/2 Gγ^2 V

then substitute and integrate between the radius at which strain is too great to be considered elastic, ro and some outer limit, then add on the core energy for where there is greater than elastic strain

generally:
ro = b/4
outer limit = half distance between dislocations

DON’T NEED TO FULLY LEARN THIS

Question 51

what are the equations that result from calculating the energy of a dislocation, i,e, overall strain energy and strain energy per unit length of a dislocation

Answer 51

Overall Strain energy:
U = 1/2 Gb^2l

Strain energy per unit length:
Λ = 1/2 Gb^2
(IN DATA BOOK)

Question 52

is the energy of a screw or edge dislocation greater

Answer 52

• the energy of an edge dislocation is a always greater than the energy of a screw dislocation

Question 53

what are 3 conditions we can deduce about a crystal for dislocation slip in real crystals regarding which dislocations form/can slip and their burgers vectors

Answer 53

1) for a crystal structure to be restored after the passage of a dislocation, b = lattice vector

2) we know from the equation for energy and Peierls-Nabarro stress, smaller dislocations (smaller b) are lower in energy and require less stress to slip, this means dislocations will form/move on close-packed planes

3) we also know that the stress required for slip is lower in wider dislocations, this means slip is favoured on low-index planes

Question 54

give the slip systems (and number of slip systems) for bcc, give the burgers vector

Answer 54

{110} = slip planes

<1 bar1 1> = slip directions

12 possible

b = a/2 <1 bar1 1>

Question 55

give the slip systems (and number of slip systems) for fcc, give the burgers vector

Answer 55

{111} = slip planes
<1 bar1 0> = slip directions

12 possible

b = a/2 <1 bar10>

Question 56

give the slip systems (and number of slip systems) for hcp, give the burgers vector

Answer 56

{100} = slip planes
<100> = slip directions

3 possible

b = a<100>

Question 57

when can a dislocation move (regarding slip systems)

Answer 57

• crystals may contain many dislocations
• but ONLY the ones on permitted slip systems will move
• and ONLY if τ> τcrit

Question 58

how can we calculate τcrit using the yield stress

Answer 58

τcrit = σy cos(φ) cos(λ)

φ = angle of slip plane normal to tensile axis (TA)

λ = angle of slip direction to tensile axis (TA)

Question 59

state the two methods we can use to work out which slip system will become active at the lowest macroscopic stress, i.e. for which system does the resolved shear stress reach the critical resolved shear stress first

Answer 59

1) Schmid factor calculations
2) OILS rule

Question 60

explain how to work out which slip system becomes active first using the Schmid factor method

Answer 60

1) calculate cos(φ) and cos(λ) from tensile axis direction/ slip direction / miller indices (represent slip plane normal), using the dot product

2) calculate Schmid factors (S.F.) for all relevant systems = cos(φ) cos(λ)

3) S.F. is between 0 and 0.5, whichever is closest to 0.5 activates first

NOTE: this is V effective but also time consuming

Question 61

explain how to work out which slip system becomes active first using the OILS rule method

Answer 61

OILS = zero intermediate lowest sign
1) write down tensile axis as [UVW]

2) ignoring sign, identify highest, intermediate and lowest numbers in tensile axis, giving [LIH] etc.

3) place a zero in the intermediate place:
- for fcc chose the <1 bar1 0> direction with the 0 in the corresponding place as the slip direction
- for bcc choose the {110} with the in the 0 in the corresponding place as the slip plane
- keep other signs the same (H and L)

4) identify the lowest index, reverse the sign
- for fcc choose the {111} plane such that the sign of L is reversed but the signs of H and I are as original
- for bcc choose the <1 bar1 1> direction following the same rules but for slip direction

NOTE THIS IS ALL IN DATA BOOK

Question 62

what occurs to the geometry of the crystal as slip proceeds

Answer 62

• slip on a single sysem –> lateral displacement and axial extension
• if the grips of a tensile test are aligned to prevent this then the tensile axis rotates towards the slip direction

Question 63

as plastic deformation occurs via dislocation glide, what two crystal/ plane factors remain constant, what effect does this have

Answer 63

1) interplanar spacing, d
2) number of planes, N

this means that as slip proceeds, the slip direction rotates towards the tensile axis

Question 64

how can we consider the rotation of the slip direction towards the tensile axis (under extensive plastic deformation) from the reference frame of the sample, how can we work out which other slip systems will become active

Answer 64

1) the first slip system is active, plastic deformation occurs, the TA rotates towards the slip direction, i.e. λ –> 0

2) this can be modeled by adding multiples of the slip direction to the TA
TA’ = TA(orig) + n[slip direction]

3) this continues until two of the indices (ignoring the sign) are the same

4) at this point 2 slip systems will have the same Schmid factor so slip occurs on both

5) using the new tensile axis, TA’, find both slip systems using the OILS rule

6) this process continues but now adding multiples of both slip directions

Question 65

explain what is different about how slip systems operate in hcp metals

Answer 65

• only 1 slip system operates, this means τcrit is constant in deformation
• although τcrit is constant, the tensile axis still rotates towards the slip direction, this changes the Schmid factor

Question 66

explain the different stages/ regimes of slip in fcc metals

Answer 66

• initial elastic strain first occurs (as always)

Stage 1: ‘Easy glide’
- one slip system operates
- TA rotates towards slip direction, λ –> 0
- extent depends on initial orientation

Stage 2: ‘Duplex slip’
- 2 slip systems operate due to rotation of TA
- dislocations move on both and can interact/interfere with each other
- this causes a rapid increase in τcrit (work hardening)

Stage 3: ‘Cross slip’
- τ(resolved) is high enough to activate other slip systems
- allows some dislocations to bridge obstacles

Question 67

what can we roughly say about the relationship between yield stress and τcrit in polycrystalline materials

Answer 67

• very difficult to predict as grains in many different directions

roughly
σy = 3 τcrit

Question 68

what is the only type of strain field that screw dislocations give

Answer 68

• shear strains, γ

Question 69

what sort of strain fields do edge dislocations give

Answer 69

• complex ones
• there are compressive, tensile and shear strains
• compressive stress in region with ‘extra half plane’
• tensile stress below ‘extra half plane’

Question 70

what are the 4 possible interactions of the stress fields of different edge interactions, why do they interact

Answer 70

• they interact to minimise strain

1) same sign, (i.e. compressive fields, tensile fields on same side) same plane = repel
2) opposite sign, same plane = attract and annihilate
3) same sign, parallel (adjacent or nearby) plane, attract but don’t annihilate
4) opposite sign, parallel (adjacent or nearby) plane, repel

Question 71

what is the effect of repulsive interactions between dislocations

Answer 71

• making passing V difficult
• high energy
• raises τcrit

Question 72

what is the effect of attractive interactions between dislocations

Answer 72

• rearrange to reduce energy
• formation of regular arrays of dislocations
• aligned dislocations are difficult to separate

Question 73

what are the two possible interactions when dislocations meet

Answer 73

• intersection
• combination

Question 74

how can we find the burgers and line vectors of the dislocation that forms when two dislocations combine

how can we then find the slip plane of the new dislocation to find the overall slip system

Answer 74

Burgers vectors:
b3 = b1 + b2

Line vectors:
Use the Weiss Zone law to find the direction common to both of the slip planes, this is the line vector

Slip plane:
use the Weiss zone law again to work out the plane common to both b3 and l3

Question 75

when is it favourable/not favourable for two dislocations to combine

Answer 75

use Frank’s rule:
as the energy of dislocation is proportional to the square of its burgers vector we can say it is favourable for them to combine if:

b3^2 < b1^2 + b2^2

Question 76

how can we find the type of a new dislocation formed by combination

Answer 76

• as usual, use the dot product on b3, l3 to find edge, screw, mixed etc.

Question 77

what is a Lomer lock

Answer 77

• when two dislocations combine, if the new slip plane is not valid in the given materials then the new dislocation will not move
• this is a Lomer lock

Question 78

explain the process of dislocation generation by a frank-read source (5 steps)

Answer 78

1) - a dislocation is pinned at both ends to due to defect
- it lies in a straight line with the lowest possible elastic strain
- it experiences shear stress normal to the dislocation line

2) - it bows outwards under the force as the ends cannot move due to pinning
- there is a balance of line tension and glide force

3) - if the shear stress is sufficient, the bowing expands to a semi-circular shape
- at this point the dislocation loop expands rapidly

4) - the dislocation spirals around the pinning points
- the two ‘arms’ approach each other

5) - segments of the 2 arms annihilate, the original dislocation becomes a free loop
- as the line vectors of the two arms are defined in opposite directions, a new dislocation is formed between the pinning points again

Question 79

state the mechanism by which edge dislocations move out of their plane to avoid obstacles

Answer 79

climb

Question 80

state the mechanism by which screw dislocations move out of their plane to avoid obstacles

Answer 80

cross slip

Question 81

explain the process of climb in edge dislocations

Answer 81

• for an edge dislocation, glide can only occur on one plane but they can move onto parallel planes to avoid obstacles if it is energetically favourable
• vacancies in the crystal migrate to the dislocation
• a segment of the dislocation can absorb vacancies, causing it to rise up a plane, this forms jogs
• the same in reverse can also happen where atoms migrate to the dislocation and the dislocation moves down a plane
• however, many vacancies are needed so lots of diffusion is needed so high temps are needed

Question 82

explain the process of cross-slip in screw dislocations

Answer 82

• the mechanism by which screw dislocations avoid obstacles
• as b // l in screw dislocations, there is a family of planes that it can glide on

1) screw dislocation gliding on plane
2) encounters obstacle, movement impeded, stress for more plastic deformation must increase
3) as the applied stress increases, τ(resultant) increases
4) if τ(resultant) > τcrit for a new slip system then the dislocation can move onto this system

Question 83

define /give the equation for dislocation density

Answer 83

“dislocation density, ρ, is the total dislocation line length per unit volume”

units = m^-2

ρ = 1/l^2

l = avr. spacing of dislocations on an area

Question 84

what are the 3 main strengthening mechanisms in single phase metals

Answer 84

1) forest hardening
2) influence of grain size
3) solid solution strengthening

Question 85

explain how forest hardening works as a mechanism for strengthening in single phase metals

Answer 85

• there are many sessile dislocations on a primary slip plane
• as a dislocation glides along this plane, it will intersect (2nd dislocation interaction mechanism) with these sessile dislocations
• this causes both dislocations to gain ‘jogs’
• this means increasing dislocation length so more energy is required so there’s overall hardening
• some jogs also create additional sessile segments so the process repeats

Question 86

explain how grain size influences yield stress as a mechanism for strengthening in single phase metals, give the equation for this and its name

Answer 86

• grain boundaries are an obstacle to dislocation motion
• dislocations build up at grain boundaries
• they form a ‘pile-up’ until the stress on the neighbouring grain is enough to cause slip in it
• the bigger the pile-up, the less the additional stress required to induce slip
• hence yield stress is lower

this is shown in the Hall-Petch relationship

σy = σ(peierls-nabarro) + k/sqrt(d)

d = grain diameter (avr.)
k = constant

Question 87

what are the two types of solid solution strengthening as a mechanism for strengthening in single phase metals

Answer 87

1) substitutional
2) interstitial

Question 88

how does substitutional solid solution strengthening work as a mechanism for strengthening in single phase metals

Answer 88

• If substitutional solute atom> crystal atoms we get spherically symmetric compressive stress field
• if substitutional solute atom < crystal atom we get spherically symmetric tensile stress field
• stress fields have no shear component so will not interact with screw dislocations

BUT they will interact with edge dislocation stress fields:
- larger substitutional atoms will locate below the slip plane in ‘gap’
- smaller ones will locate above the slip plane in compressive field

• they must be near active slip planes to interact so

OVERALL WEAK EFFECT

Question 89

how does interstitial solid solution strengthening work as a mechanism for strengthening in single phase metals

due to their fast diffusion, what can they form?

Answer 89

• Generally stronger effect than substitutional as larger stress fields
• if asymetric interstice then shear component of stress field too so can interact with both types of dislocation
• also faster diffusion so they can locate at every part of the slip plane by the dislocation
• this gives a Cottrell Atmosphere
• this has a stronger effect because to create yield, all of the favourable configurations must be broken for the dislocations to glide and plastic deformation to occur

Question 90

what are Lüders bands

Answer 90

• usually only occur in low carbon steels
• steel starts to plastically deform at upper yield stress
• yields first at ends where highest stress conc.
• dislocations escape Cottrell atmospheres
• as they are no longer pinned by the Cottrell atmosphere, yield stress decreases so plastic deformation can occur
• during plastic deformation, forest hardening occurs so yield stress increases
• this means the boundary between yielded/unyielded regions moves as it will just start to plastically deform the next bit that isn’t forest hardened
• these are Lüders bands

Question 91

what is the Portevin- Le Chatalier effect

Answer 91

• at low temps, Lüders bands occur in plastic deformation of low carbon steels
• at high temps, C diffusion rates are too quick so they ‘keep up’ with the dislocation motion
• at intermediate temps, they repeatedly escape, re-pin
• this is the Portevin- Le Chatalier effect

Question 92

what are the two main strengthening mechanisms in multi-phase metals

Answer 92

• Precipitate cutting
• Orowan bowing

Question 93

explain how precipitate cutting acts as a strengthening mechanism for a multi-phase metal, also state when cutting occurs and give the suitable equation

Answer 93

• occurs where precipitates are small
• small ppts tend to be coherent with matrix
• dislocations can pass from matrix into ppts and ‘cut’ them i.e. form steps in them as they normally would at the edge of a crystal
• the stress required to move the dislocation through the ppts is greater so there’s a higher τcrit than normal

Δτ proportional to sqrt(r)

Question 94

explain how Orowan bowing acts as a strengthening mechanism for a multi-phase metal, also state when Orowan bowing occurs

Answer 94

• occurs where there are larger precipitates because the interfaces become less coherent so dislocations cannot move through them

1) gliding dislocation encounters ppts
2) dislocation pinned at ppts
3) dislocation bows outwards
4) forms loops around ppts
5) the dislocation is free again as the ‘arms’ of the loops combine to form a ring dislocation around the ppts

Question 95

give the equations of the stress required for Orowan Bowing

Answer 95

• peak energy is when the bowing is at a semi-circle configuration

Λ = line tension = 1/2 G b^2
F = glide force = τb

resolving vertically and integrating over the distance between ppts gives

τ = Gb/L

i.e. τ decreases with 1/r

IN DATA BOOK

NOTE: L is SURFACE to SURFACE distance NOT centre-centre

Question 96

when is the maximum strengthening acheived (regarding ppt size) in multi-phase metal strengthening

Answer 96

• at the crossover point between cutting and bowing
• small ppts = cutting = little hardening
• larger cutting ppts = better
• even larger = bowing = decreases with 1/R
• greater volume fraction of ppts helps both cutting and bowing, higher peak stress
• stronger ppts increases cutting resistance and decreases radius for max stress

Question 97

what are partial dislocations, give examples for an fcc structure, why (energetically) does this occur

Answer 97

• normally the passage of a dislocation must move an atom back to its initial ‘place’ in an atom
• in fcc this is b = a/2 [bar1 01]
• but instead the crystal structure could be reformed by the passage of two smaller dislocations, partial dislocations

this, in fcc moves a b atom first to a c site then to the next b site, they are
b2 = a/6 [bar2 11]
b3 = a/6 [bar1 bar1 2]

this occurs as it is energetically favourable due to Frank’s rule

Question 98

what do partial dislocations lead to in terms of the crystal structure what determines the extent of this

Answer 98

• when a perfect dislocation dissociates, the elastic stress fields of the partials repel
• this creates a stacking fault region between the two partials
• the width of this depends on the stacking fault energy

Question 99

explain the background to order hardening, superpartials, APBs

Answer 99

• some compounds have an abrupt change between a disordered structure with the same average composition on each site and an ordered structure where certain atoms are on certain sites
• what is a permitted lattice vector for a burgers vector in the disordered structure may not be permitted in the ordered structure as it may return an atom to the ‘wrong position’ when passing
• this creates energetically unfavourable like-like bonds, an antiphase boundary (APB)
• it can be fixed by another dislocation passing through, but this creates an APB between them
• The passage of the two dislocations (superpartials) is called a superdislocation
• Superpartial separation is determined by the APB energy

Question 100

explain how order hardening gives a strengthening effect

Answer 100

• the first superpartial experiences a large resistive force as it forms an APB
• this large resistive force strengthens the material
• the second superpartial does not experience a resistive force and corrects the structure

Question 101

explain plastic deformation by cooperative shear and when it occurs

Answer 101

• this is an alternative deformation mechanism, there is no dislocation motion just the shear of successive atomic planes
• it only really occurs where there are high strain rates, very low temperatures, or very limited slip systems (e.g. hcp)
• we only discuss twinning where the structure formed from the cooperative shear is the same as the original but in a different orientation
• there are different twinning planes and directions for different metals

Question 102

briefly explain twinning in fcc metals

Answer 102

• shear occurs in the (111) planes
• each atom is displaced by a/6 [1 1 bar2]
• gives strain on 1/sqrt(2): found by considering the magnitude of this vector over the interplanar spacing

Question 103

define fracture

Answer 103

“The separation or fragmentation of a solid body, under the influence of an applied load - involves creation/propagation of cracks”

Question 104

assuming no defects, how would fracture have to occur

Answer 104

all atomic bonds would have to be broken

Question 105

how do defects influence fracture, how does fracture actually begin from this, what is the maximum stress due to defects

Answer 105

• if a defect-free material is loaded elastically, the force is evenly transmitted
• if a material contains a crack, the material above/below the crack cannot support an applied load
• this concentrates stress at the crack tip

σmax = σo (1+2sqrt(c/r))
c = crack length for surface crack or 1/2 crack length for non-surface crack
r = radius of curvature at crack tip

Question 106

state the basic principle behind the Griffith Criterion

Answer 106

“when a pre-existing crack grows, its extension creates 2 new surfaces, this requires energy. This could be supplied by a combination of elastic energy stored in the material the crack is advancing into and the work done by the applied stress”

Question 107

how can we define the strain energy release rate, what does this expression imply

Answer 107

strain energy release rate, G = crack driving force = rate that stored elastic strain energy is released per unit area of new crack

G = dW / new crack area = π (σo)^2 (c) / E

this expression of
G = (πc(σo)^2)/E
(LEARN)
shows that more energy is released per unit extension for a longer crack

Question 108

how can we use the expression for strain energy release rate, G, to define the Griffith criterion

Answer 108

“In order for a crack to propagate, the strain energy release rate, G, must be greater than the rate of energy absorption”

• in fracture, the energy absorption is equal to the surface energy as the crack propagates = 2γ

so the Griffith criterion is that if
G > Gc = 2γ
G =(πc(σo)^2)/E
then the crack will propagate

Question 109

How can we use the Griffith Criterion to find the critical stress at which a crack will extend

Answer 109

we know a crack will propagate when
G = Gc
so
Gc = 2γ = (πc(σo)^2)/E

rearrange for stress
σ* = sqrt(2γE / πc)

Question 110

How can we use the Griffith Criterion to find the critical flaw size for a brittle material

Answer 110

we know that fracture will occur when
G = Gc
so
Gc = (πc(σo)^2)/E = 2γ
rearrange for c

c* = 2γE / π(σo)^2

Question 111

how can a material be processed to reduce the impact of cracks and reduce the risk of fracture

Answer 111

• polishing
• putting the surface into compression
• any method to close cracks

Question 112

how is fracture different in ductile materials, how can the Griffith criterion be edited to accommodate this

Answer 112

• we know the stress at the crack tip is greatest
• if this stress is greater than the yield stress then a zone of plasticity forms ahead of the crack tip
• this blunts the crack tip and increases its radius of curvature
• work is done in this process and is dissipated as heat
• the Griffith Criterion now becomes:
  G > Gc = 2(γ+γp)
  γp&raquo_space; γ

Question 113

what is the stress intensity factor, what is fracture toughness

Answer 113

• Gc is const. for a material, as is E
  so rearranging Griffith Criterion gives

sqrt(GcE) = σo sqrt(πc)

this is a constant for a material and is known as the stress intensity parameter, K
K = σo sqrt(πc)
Kc = sqrt(GcE)
Kc is a critical value at which fracture occurs, this is the fracture toughness

Question 114

what gives composites a high toughness

Answer 114

• they have energy absorbing mechanisms
• predominantly this is that as a crack advances, fibres are pulled out of their matrix sockets
• this is a process called ‘Fibre pull-out’
• this requires a lot of work to be done
• so toughness increases

Question 115

explain/state the stacking change when twinning occurs in an fcc metal

Answer 115

consider
A
B
C
A
B
C
A
B
C
- after the first set consider twinning occurring
- remember that EACH SUCCESSIVE PLANE shears so we end up with

A
B
C
A –> B
B –> C –> A
C –> A –> B –> C
A –> B –> C –> A –> B
B –> C –> A –> B –> C –> A
etc.

THIS IS IMPORTANT

Question 116

how can we calculate the shear involved in twinning in fcc metals

Answer 116

• we know each atom moves to the next position
  e.g.
  A –> B
  B –> C
  etc.
  hence we can work out the vector required for this
• considering a close packed plane on an fcc structure and sketching it as hexagonal then considering the stacking positions as almost interstitial in hcp we can deduce the vector is
  a/6 <112>
• a/6 because <112> is for two unit cells, but it moves a third of a unit cell

then consider the magnitude of this over the (111) interplanar spacing to get strain as
sqrt(2)/2

Question 117

when calculating Schmid factor for non-cubic systems what should we be careful of

Answer 117

we need to watch out for lattice parameters when doing the dot products

• when doing the dot product for tensile axis/slip direction/slip plane normal in a cubic system the lattice parameters all cancel
• in non-cubic e.g. hexagonal they don’t

Question 118

consider annealing a super-saturated solid solution (SSSS) to form precipitates, what are the 4 factors which will affect strength and change with annealing time, briefly explain them

Answer 118

1) Solute hardening:
- at short aging times, solid solution strengthening is dominant
- this decreases with ageing time as ppts form

2) Coherency strains:
- the coherent and semi-coherent ppts that form in the early stages of ageing introduce coherency strains
- these inhibit dislocation motion, and provide some strengthening
- this strengthening decreases again once they become incoherent

3) Precipitate cutting
- as explained before

4) Orowan Bowing:
- as explained before

Question 119

what does the overall graph for yield strength against ageing time look like for a SSSS forming ppts, explain how each of the 4 factors contribute

Answer 119

1) solid solution strengthening:
- drops off with a sort of exp(-x) shape

2) Cohrerency strengthening
- rises to a peak
- falls off again

3) ppt Cutting:
- as before

4) Orowan bowing:
- as before

Overall this gives a graph which drops initially, rises, plateaus, rises again, then drops off as 1/r
(SEE EH72)

Question 120

How could we calculate the maximum elastic energy stored in a cantilever beam (rectangular)

Answer 120

• the elastic energy stored will be dependent on the maximum strain, this will occur at the failure stress

ε = y/R for a beam
so
εmax = h/2R

we know
M = EI/R = FL

where E = σf / εmax
I = wh^3/12

substitute for I,E to get F
substitute F into
δ = FL^3 / 6EI

then use
Stored energy = 1/2 Fδ

Question 121

derive the expression for hoop stress in a pressurised pipe

Answer 121

consider a pipe of internal pressure P, thickness t, length L, radius r

1) Consider the pressure acting on any given plane longitudinally through the cylinder:

2) on the part of the plane acting within the cylinder we know the force will be the area of the rectangle, 2rL multiplied by the pressure
F = 2rLp

3) Consider the force acting on the part of the same plane intersecting the walls of the cylinder, the pressure in this section is not the same as the internal pressure, it is the hoop stress, σ(θ):
- the area of interest is 2tL
- so the force is
F = 2tLσ(θ)

to prevent breaking these forces must be equal so

2tLσ(θ) = 2rLp
σ(θ) = rP/t

Question 122

derive the expression for axial stress in a pressurised pipe

Answer 122

consider a pipe of internal pressure P, thickness t, length L, radius r

1) consider a plane now intersecting the cylinder across its radius / a cross-section

2) on the part of the cross-section within the pipe, calculate the force (acting axially), this is:
F = PA = P(πr^2)

3) on the part of the cross-section intersecting with the walls of the pipe, consider the force (acting axially), the pressure in this section is not the same as the internal pressure, it is the axial stress σ(z):
F = (π(r+t)^2 - πr^2) σ(z)
so in the limit of small t:
F = 2πrt σ(z)

for the pipe not to break, these forces must be equal
2πrt σ(z) = P(πr^2)
σ(z) = rP/2t

Question 123

Give the expression for hoop stress and axial stress in a pressurised pipe with thin walls, explain using them how a pipe will burst

Answer 123

Axial Stress = σ(z) = rP/2t
Hoop Stress = σ(θ) = rP/t

• the hoop stress is twice the size of the axial stress
• hence the pipe will burst longitudinally as the hoop stress will ‘pull’ a crack open in that direction