Course E - Mechanical Behaviours of Materials Flashcards
what is tensile testing, explain the different regions and the graph that is formed
- 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
what can we say about unloading from a tensile test
- 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
what is the limitation with tensile testing
- it is very dependent on the size/shape of the material
define stress, give the equation for (true) stress
stress, σ, is the normal force, F, acting on a surface per unit area, A,
σ = F/A
units = Pa
what is an issue that occurs with stress, what alternative expression can we use for stress
- 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
define strain/ give the expression for (true) strain
- 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)
what is engineering strain, give the expression for it
ε(eng) = li-lo/lo = li/lo -1
define shear stress
shear stress, τ, is the force per unit area parallel to a surface
τ = F/Ao
define shear strain, γ
- 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)
define Poisson’s ratio, give the equation for it
- 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)
what is a stress-strain curve, why do we use them, what does the gradient of the linear part give
- 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
define elastic strain, give the equations linking stress and strain for normal and shear stresses
“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
how can we relate E and G using poisson’s ratio, v
E = 2G(1+v)
how can we calculate the elastic strain energy/work done in elastic strain PER UNIT VOLUME for a material
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)
what is the link between the work done per unit volume in elastic deformation and the stress-strain graph
- the work done PER UNIT VOLUME is the area under a stress-strain graph
how can we model bonds/ roughly explain the shape of their potential against bond length graph
- 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
give a summary of how we can approximate Young’s moduli of a simple cubic crystal using the Leonard Jones potential
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)
why do we use composites
- 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
how can we do analysis on composites/ what model should we use
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
how can we determine/estimate the axial modulus of a composite using a slab model
- 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
how can we determine/estimate the transverse modulus of a composite using a slab model
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
how is temperature related to kinetic energy
3/2 KT = <1/2mv^2> = avr. Ek
microscopically/ on an atomic level, how can we explain thermal expansion
- 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
macroscopically give the equation for what happens in thermal expansion
εt = α ΔT
α = coefficient of thermal expansion
what can we say about the thermal expansion of materials with stronger/weaker bonds
- 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
what can occur to a bimaterial strip when heated/cooled
- 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
which sections of a beam are under which sorts of stress when bent, what is the neutral axis
- 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
what do we define our radius as when considering the bending of beams
- 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
how can we calculate σ(axial) and ε(axial) for a bending beam
- 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
how can we calculate the total bending moment on a beam
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
what is the equation for beam stiffness
Beam stiffness = Λ = EI
I = second moment of area
what is the equation for the second moment of area, what is it for a rectangular section
I = ∫(on the relevant section) y^2 b(y) dy
for a rectangular section
I = bh^3/12
what is the equation for the moment on a cantilever beam
M = F(L-x) = d^2y/dx^2 EI
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
y = (Fx^2/6EI) (3L-x)
max deflection, δ, occurs where x = L, at end
δ = FL^3 / 3EI
in a perfect crystal, what is the only way in which plastic deformation can occur, what is the relevant equation that would give this
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
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
- 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
what are dislocations, how can they be thought of locally
line defects
locally they can be thought of as an ‘extra’ half plane of atoms
what are the two vectors that define a dislocation, define them
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
how can we determine a burgers vector through a burgers circuit
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
what is an edge dislocation, give its features and its formal definition
- 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”
what is a screw dislocation, give its features and its formal definition
- 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”
what is a mixed dislocation, give some of its features
- 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
by what process do dislocations move, it is localised?
on what do they move?
- 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
what can we say about the displacement to a section of the crystal that occurs when a dislocation passes
- 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
what are the two KEY CONDITIONS for dislocation motion
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”
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
Edge dislocation:
b ⟂ l, so glide is limited to a single slip plane
Screw dislocation:
b // l, so family of possible slip planes
how can we rationalise the movement of a dislocation loop under stress
- 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
what is the shear stress required to move a dislocation in an otherwise perfect crystal
- 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)
what is the work done when a dislocation moves, how can we consider forces acting to do this work
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
explain briefly where the expression for the energy of a dislocation comes from
- 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
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
Overall Strain energy:
U = 1/2 Gb^2l
Strain energy per unit length:
Λ = 1/2 Gb^2
(IN DATA BOOK)
is the energy of a screw or edge dislocation greater
- the energy of an edge dislocation is a always greater than the energy of a screw dislocation
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
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
give the slip systems (and number of slip systems) for bcc, give the burgers vector
{110} = slip planes
<1 bar1 1> = slip directions
12 possible
b = a/2 <1 bar1 1>
give the slip systems (and number of slip systems) for fcc, give the burgers vector
{111} = slip planes
<1 bar1 0> = slip directions
12 possible
b = a/2 <1 bar10>
give the slip systems (and number of slip systems) for hcp, give the burgers vector
{100} = slip planes
<100> = slip directions
3 possible
b = a<100>
when can a dislocation move (regarding slip systems)
- crystals may contain many dislocations
- but ONLY the ones on permitted slip systems will move
- and ONLY if τ> τcrit
how can we calculate τcrit using the yield stress
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)
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
1) Schmid factor calculations
2) OILS rule
explain how to work out which slip system becomes active first using the Schmid factor method
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
explain how to work out which slip system becomes active first using the OILS rule method
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
what occurs to the geometry of the crystal as slip proceeds
- 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
as plastic deformation occurs via dislocation glide, what two crystal/ plane factors remain constant, what effect does this have
1) interplanar spacing, d
2) number of planes, N
this means that as slip proceeds, the slip direction rotates towards the tensile axis
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
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
explain what is different about how slip systems operate in hcp metals
- 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
explain the different stages/ regimes of slip in fcc metals
- 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
what can we roughly say about the relationship between yield stress and τcrit in polycrystalline materials
- very difficult to predict as grains in many different directions
roughly
σy = 3 τcrit
what is the only type of strain field that screw dislocations give
- shear strains, γ
what sort of strain fields do edge dislocations give
- complex ones
- there are compressive, tensile and shear strains
- compressive stress in region with ‘extra half plane’
- tensile stress below ‘extra half plane’
what are the 4 possible interactions of the stress fields of different edge interactions, why do they interact
- 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
what is the effect of repulsive interactions between dislocations
- making passing V difficult
- high energy
- raises τcrit
what is the effect of attractive interactions between dislocations
- rearrange to reduce energy
- formation of regular arrays of dislocations
- aligned dislocations are difficult to separate
what are the two possible interactions when dislocations meet
- intersection
- combination
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
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
when is it favourable/not favourable for two dislocations to combine
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
how can we find the type of a new dislocation formed by combination
- as usual, use the dot product on b3, l3 to find edge, screw, mixed etc.
what is a Lomer lock
- 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
explain the process of dislocation generation by a frank-read source (5 steps)
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
state the mechanism by which edge dislocations move out of their plane to avoid obstacles
climb
state the mechanism by which screw dislocations move out of their plane to avoid obstacles
cross slip
explain the process of climb in edge dislocations
- 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
explain the process of cross-slip in screw dislocations
- 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
define /give the equation for dislocation density
“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
what are the 3 main strengthening mechanisms in single phase metals
1) forest hardening
2) influence of grain size
3) solid solution strengthening
explain how forest hardening works as a mechanism for strengthening in single phase metals
- 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
explain how grain size influences yield stress as a mechanism for strengthening in single phase metals, give the equation for this and its name
- 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
what are the two types of solid solution strengthening as a mechanism for strengthening in single phase metals
1) substitutional
2) interstitial
how does substitutional solid solution strengthening work as a mechanism for strengthening in single phase metals
- 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
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?
- 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
what are Lüders bands
- 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
what is the Portevin- Le Chatalier effect
- 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
what are the two main strengthening mechanisms in multi-phase metals
- Precipitate cutting
- Orowan bowing
explain how precipitate cutting acts as a strengthening mechanism for a multi-phase metal, also state when cutting occurs and give the suitable equation
- 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)
explain how Orowan bowing acts as a strengthening mechanism for a multi-phase metal, also state when Orowan bowing occurs
- 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
give the equations of the stress required for Orowan Bowing
- 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
when is the maximum strengthening acheived (regarding ppt size) in multi-phase metal strengthening
- 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
what are partial dislocations, give examples for an fcc structure, why (energetically) does this occur
- 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
what do partial dislocations lead to in terms of the crystal structure what determines the extent of this
- 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
explain the background to order hardening, superpartials, APBs
- 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
explain how order hardening gives a strengthening effect
- 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
explain plastic deformation by cooperative shear and when it occurs
- 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
briefly explain twinning in fcc metals
- 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
define fracture
“The separation or fragmentation of a solid body, under the influence of an applied load - involves creation/propagation of cracks”
assuming no defects, how would fracture have to occur
all atomic bonds would have to be broken
how do defects influence fracture, how does fracture actually begin from this, what is the maximum stress due to defects
- 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
state the basic principle behind the Griffith Criterion
“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”
how can we define the strain energy release rate, what does this expression imply
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
how can we use the expression for strain energy release rate, G, to define the Griffith criterion
“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
How can we use the Griffith Criterion to find the critical stress at which a crack will extend
we know a crack will propagate when
G = Gc
so
Gc = 2γ = (πc(σo)^2)/E
rearrange for stress
σ* = sqrt(2γE / πc)
How can we use the Griffith Criterion to find the critical flaw size for a brittle material
we know that fracture will occur when
G = Gc
so
Gc = (πc(σo)^2)/E = 2γ
rearrange for c
c* = 2γE / π(σo)^2
how can a material be processed to reduce the impact of cracks and reduce the risk of fracture
- polishing
- putting the surface into compression
- any method to close cracks
how is fracture different in ductile materials, how can the Griffith criterion be edited to accommodate this
- 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»_space; γ
what is the stress intensity factor, what is fracture toughness
- 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
what gives composites a high toughness
- 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
explain/state the stacking change when twinning occurs in an fcc metal
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
how can we calculate the shear involved in twinning in fcc metals
- 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
when calculating Schmid factor for non-cubic systems what should we be careful of
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
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
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
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
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)
How could we calculate the maximum elastic energy stored in a cantilever beam (rectangular)
- 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
derive the expression for hoop stress in a pressurised pipe
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
derive the expression for axial stress in a pressurised pipe
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
Give the expression for hoop stress and axial stress in a pressurised pipe with thin walls, explain using them how a pipe will burst
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
