Polymer Fracture and Failure Flashcards
(121 cards)
1
Q
Why do Polymers have a short service life?
A
Even though they are a ductile material, the failures are invariable brittle in nature resulting in a short service life
2
Q
What are the causes of polymer failure?
A
Incorrect material selection
Chemical and environmental interactions
Response to long term loads
Processing errors
Inappropriate design
Effect of additives
3
Q
What is incorrect material selection?
A
The use of an incorrect material for an application is a
common cause of polymer failure, lack of understanding of
the interaction of polymers with chemicals and environments
and
long term response commonly leads to premature failure, e.g.
brittle & ductile plastics
4
Q
What are chemical and environmental interactions?
A
For many materials specific interactions with common
household products can lead to rapid crazing, cracking,
fracture and product failure,
e.g. HDPE used for underground water pipes (environmental
stress cracking (ESC))
5
Q
What is environmental stress cracking(ESC)?
A
a brittle fracture failure
mode that results from exposure to mechanical stress in the presence of a
chemical that initiates stress relief
6
Q
What is the response to long term loads?
A
Plastic and rubber components are sensitive to long term loads.
These can be either static loads through creep or fatigue mechanisms.
These processes mean that the strength and stiffness of polymer
components in service conditions are frequently significantly
lower than quoted on a standard data sheet.
7
Q
what are processing errors?
A
incorrect processing of polymer materials causes failure through degradation and embrittlement process,
high residual stresses,
material inhomogeneity,
introduction of product faults, defects or contaminants
8
Q
What is an inappropriate design?
A
Inappropriate design
product design is fundamental to ensuring polymer
component long term durability.
incorrect design will highlight polymeric material
weaknesses to
long term loads,
chemical environments,
high speed loading,
fatigue loads
resulting in failure of the product within a short service life.
9
Q
What is the effect of additives?
A
Even though every additives is intended to enhance or
assure satisfactory performance, they can contribute to or
cause failure for any number possible reasons, such as:
Incorrect amount – too much or too little
Incorrect additive or combination thereof – low
compatibility in plastics (plasticisers, colourants), too high
volatility in a certain plastic
10
Q
What are the intentional additives composition ? - 2 examples
A
Additives and modifiers - antioxidants, brighteners, adhesion promoters, colouring aid, emulsifiers, flame retardant
Fillers and reinforcements - 36 polymerics(cellulose, reclaimed rubber) & inorganics( calcium carbonate, asbestos)
Reinforcements - glass fiber, carbon fiber, aramid fiber and synthetic fibers (fabrics, filaments)
11
Q
What are intentional additives intended to do?
A
Migration to the surface – dependent on compatibility,
required for some applications (antistatic agents); too much
may interfere with printing and adhesion; undesirable for
other applications; environmental stress-cracking
Processing requirements that may adversely affect the
product
Incomplete or non-uniform dispersion in the product
Unanticipated secondary effects (enhanced crystallinity due
to an additive)
12
Q
What are types of unintentional additives?
A
Extraneous lint, dirt and other contaminant materials
Residual monomer or solvent
Water
Compounding process aids
Additives to formulation ingredients to improve their performance Ionic impurities from water in service
Ionic impurities in carbon black
Trace metal from extruder barrel and screw coatings; and
Impurities in intentional additives
13
Q
What is the composition of unintentional additives?
A
Residual monomer or solvent
Food packaging
Adhesive tapes in skin contact for medical purpose
Water
Hydrolysis of condensation type polymers in melt processing
– reduction in MW
Appearance problem due to water in melt processing –
cloudy appearance
Voids formed by water in melting process – water boils at
100°C
Shrinkage and expansion of moulding – in close tight fit,
water could affect the performance
Impurities in intentional additives
A low quality grade of mineral oil – colour change in product
14
Q
How to understand the failure of polymers?
A
Need to acquire knowledge of the properties of polymer materials
The correct selection of a polymer material for a given application.
Mechanical properties data were used to predict the response of materials under mechanical loads.
Expressed in terms of forces which may deform materials or even cause them to fail completely.
15
Q
How to analyse failure? -3
A
- What is the nature of the load?
Continuous and uniform or rising steadily:
IMPACT (e.g. hammering action, accidental drop)- Alternating (periodic application of an force):
FATIGUE (e.g. vibration, rotation in loaded components)
2.The geometry of the loaded component can be designed to deal with these conditions. - The physical nature of the material has to ensure that the component can survive in service.
16
Q
What is mechanical failure in polymers caused by ? -4
A
Excessive deformation
Ductile failure
Brittle failure
Crazing
17
Q
what is excessive deformation and how does it occur?
A
Very large deformations are possible in low modulus polymers are able to accommodate large strains before failure.
Such deformations could occur without fracture design features and other considerations might only tolerate deformations to a prescribed ceiling
value.
The case in rubbery thermoplastics, such as flexible PVC or EVA, for pressurised tubing.
18
Q
Why can polymer failure not be wholly brittle or ductile caused?
A
Because of the viscoelastic character of polymers
19
Q
What does the proportion of ductile to brittle depend on in polymer failure?
A
The speed (and time of loading) and the temperature of the sample?
20
Q
What is the most common rupture in polymers?
A
Creep. Rupture, Fatigue failure and Impact failure.
21
Q
what is ductile failure? and what is yielding?
A
Encountered in materials that are able to undergo large-scale irreversible plastic deformation under loading, known as yielding, before fracturing.
Yielding marks the onset of failure setting the upper limit to stress in service to be below the yield
point is common practice.
Estimate loading conditions likely to cause yielding (yield criteria), in order to design components with a view to avoid it in service.
22
Q
What is brittle failure?
A
This is a type of failure involving low strains accompanied by negligible permanent deformation and is frequently characterised by
“clean” fracture surfaces.
It occurs in components that contain geometrical discontinuities that act as stress concentrations.
23
Q
What are the effective stress concentrating discontinuities in brittle failure? -3
A
cracks,
badly distributed or oversized additive
particulates,
impurities etc.
24
Q
What is crazing?
A
It occurs at a strain level which is below the level required for brittle fracture and although undesirable, this type of “failure” is not catastrophic. Crazing is often observed in highly strained regions during bending.
Crazes are made up of micro-cavities whose surfaces are joined by highly oriented, or fibrillar, material.
They are initiated near structural discontinuities, such as impurities, and are collectively visible at the strained surface because they become large enough to reflect light.
Crazes are not cracks and can continue to sustain loads after they are formed. However, they can transform into cracks via the breakage of the fibrils.
25
What are the 2 fracture/failure mechanics of polymers?
Linear elastic fracture mechanics - brittle
Elastic plastic fracture mechanics - ductile
26
What does linear elastic fracture mechanics observe behaviour in?
Observes linear elastic material behaviour in
brittle polymer materials,
filled or fibre-reinforced material systems, or
specimens of great thickness or below the glass
transition temperature Tg
LEFM with small-scale yielding, in the case where
plastic zone at the crack tip is taken into consideration.
27
What do linear elastic fracture mechanics do ?
As a parameter, the stress intensity factor (SIF) Kdescribing the linear elastic area in front of the crack tip is
used.
The critical value of the SIF under conditions of plane strain is called the fracture or crack toughness KIc (static loading), KId (dynamic loading),
KII or KIII (I, II, III refer to crack-tip-open modes). In practice,
mode I has the highest importance.
28
What are the three types of crack-opening modes?
Mode I (opening)
Mode II (in-plane shear)
Mode III (out-of-plan shear)
29
How can crack instability be characterised?
The tensile stress increases linearly with the strain, followed by an abrupt drop of the load at the instant of crack
growth initiation. When characterising the crack
instability by the stress, a solely
stress-dependent term, the stress
intensity factor K, is used.
K = σi√a
where σi is the external stress, a is the
diameter of the flaw. This is called the
critical stress intensity factor Kc in the
case of failure.
30
What is elastic plastic fracture mechanics for?
For evaluating the toughness in the case of non-negligible
elastic-plastic material behaviour and extensive plastic
area in front of the crack tip, the concepts of elastic-plastic
fracture mechanics have to be used.
31
What are the two most important concepts of elastic plastic fracture mechanics?
the crack-tip opening displacement (CTOD) and the J-integral.
32
What is the crack tip opening (CTOD) based on? What is the measure of this?
The fracture process is not
controlled by stress intensity any more, but by plastic deformation in front of the crack tip.
A measure of this is the widening at the crack tip, called the crack-tip-opening displacement or crack-opening
displacement δ.
evaluated against the crack resistance concept.
33
What is the J integral?
After the load or the deformation has been determined J (the energy) can be determine to ascertain whether there is unstable crack growth, stable/unstable crack growth or stable crack growth.
34
What are the three basic categories of molecular variables that affects the polymer properties?
Composition(polymer or additives), Molecular weight(weight distribution or cross linking) and intermolecular order(crystallinity, orientation, degree of fusion or thermal transitions)
35
How can polymers be classified?
Natural polymers or synthetic polymers
36
What are natural polymers?
Found in plants and animals
proteins,
cellulose,
starch,
natural rubber
37
What are synthetic polymers?
Synthesises by chemical methods
plastic (polyethylene, polypropylene, PVC, Polystyrene…),
synthetic fibres (nylon 6,6, Kevlar… ) and
synthetic rubbers (polybutadiene)
38
What are copolymers?
two or more monomers polymerised together
39
What are the 4 categories for copolymers?
random – A and B randomly
positioned along chain
alternating – A and B
alternate in polymer chain
block – large blocks of A
units alternate with large
blocks of B units
graft – chains of B units
grafted onto A backbone
40
What is the architecture of polymers? - 3
1.Linear Polymers: These polymers consist of long and straight
chains.
high density polythene,
PVC,
2.Branched Polymers: These polymers contain linear chains having
some branches,
low density polythene.
3. Cross-linked Polymers: These are usually formed from the
monomers with more than two functional groups and contain
strong covalent bonds between various linear polymer chains,
vulcanized rubber,
urea-formaldehyde resins,
Epoxy.
41
What does high molecular weight allow polymers to do?
High molecular weight, or long chain length, is the single most
important property of polymers that distinguished them from
other materials, such as metals and ceramics.
Molecular weight in the range of about 10,000 – 100,000
accounts for the ability of polymers to perform in many
applications.
42
What properties does molecular weight allow polymers to have?
Resistance to stress in tensile, flexural and shear modes, toughness, creep resistance, and environmental stress cracking are some of the product performance criteria that are strongly affected by molecular weight.
43
How does tensile strength and molecular weight interact?
When tensile strength increases the molecular weight increases
44
How is the molecular weight distribution done? what is the equation?
A polymer normally does not
have a single MW. The
number of monomer units
comprising a polymer chain is
not exactly the same for each
and every molecule in a
particular polymer.
There is actually a distribution
of chain length, which is called
molecular weight distribution
(MWD).
MWD = Mw / Mn The broader the distribution, the
higher polydispersity index.
45
What is the relaxation modulus?
Characteristic of material viscoelasticity used to describe stress relaxation of materials with time
45
What is the effect of temperature on the mechanical properties of polymers?
As the temperature increases and the relaxation modulus decreases logarithmicly, the polymer moves from a glassy state to a leathery state and then it continues to be rubbers -> rubbery flow -> viscous flow at the lowest modulus and highest temperatures.
45
What is the relationship of relaxation modulus versus temperature for crystalline isostatic
(curve A), lightly crosslinked atactic (curve B) and amorphous (curve C) polystyrene ?
Crystalline isostatic - High modulus at low temperature, sharp drop near T_m (melting temperature) as the material softens.
Lightly crosslinked atactic- Gradual decrease in modulus with temperature, less sharp than crystalline due to crosslinking, but still softens at high temperatures then plateaus in the rubber region.
Amorphous polystyrene -Sharp drop at Tg, showing a glass-to-rubber transition, then a slight plateau in the rubbery state before decreasing further with increasing temperature.
46
How does molecular weight effect the rubbery plateau region?
The higher the molecular weight with a increasing temperature, the longer it shall plateau in the rubbery region.
47
What are the two thermal transition types?
Glass transition - amorphous materials
Melt transition- crystalline materials
the transition is a change of state relating to changing temperature or pressure
semi crystalline can show both.
48
What is the glass transition temperature Tg?
Both transition types originate from (inter) molecular forces - thus typically Tg~0.5=0.65Tm
The glass transition is not a phase transition and arises from the decoupling of the vibrational and
translational motions ie at and above the glass transition temperature the flow of entire chains is possible (for polymers).
It is the temperature at which the polymer transforms from rubber to glass in a LONG RANGE SEGMENTAL MOTION. Lowering temperature reduces free volume.
49
What happens to the motion of the individual chain segments around the glass transition temperature?
At temperatures above Tg, 10 to 50 repeat units of the polymer backbone are relatively free to move in cooperative thermal motion to provide conformational rearrangement of the backbone.
Below Tg, the motion of these individual chains segments becomes frozen with only small scale molecular motion remaining, involving individual or small groups of atoms.
Below Tg polymers are hard and glassy
Above Tg polymers are soft and leathery
50
How does free volume vary with glass transition?
The actual volume of the molecules stays the same
through Tg, but the free volume (the volume through
which they can move) increases.
Free volume is the space in a solid or liquid sample
which is not occupied by polymer molecules, i.e. the
“empty-space” between molecules.
51
What factors affect the melting and glass transition temperature?
Both Tm and Tg increase with increasing chain stiffness
Chain stiffness increased by presence of
Bulky side-groups
Polar groups or sidegroups
Chain double bonds and aromatic chain groups
Regularity of repeat unit arrangements – affects Tm
only
52
What is the viscoelasticity of polymer deformation a result of and force relationship between molecules?
Polymers exhibit rate-dependent viscoelastic deformation, which is a direct result of their molecular
structure.
Two neighbouring molecules, or different segments of a single molecule that is folded back upon itself, experience weak attractive force called Van der Waals bonds.
These secondary bonds resist any external force that attempts to pull the molecules apart.
53
How does the strain rate affect the materials ability to deform ?
Deforming a polymer requires cooperative motion among molecules.
The material is relatively compliant if the imposed strain rate is sufficiently low to provide molecules sufficient time to move.
At faster strain rate, however, the forced molecular motion produces friction, and a higher stress is required to deform the material.
If the load is removed, the material attempts to return to its original shape, but molecular entanglements prevent instantaneous elastic recovery.
54
What are the 3 main morphologies of polymers and what do the chain arrangements look like?
1. Crystalline materials have their molecules arranged in repeating patterns. As such, they all tend to have highly ordered and regular structures.
2. Amorphous materials have their molecules arranged randomly and in long chains which twist and curve around one-another, making large regions of highly structured morphology unlikely.
3. The morphology of most polymers is semi-crystalline: a combination with the tangled and disordered regions surrounding the crystalline areas.
55
How does crystallinity effect the polymer properties ? +ves and -ves
For the most part, crystallinity is intentional and can improve the
strength, toughness and thermal stability of the plastics.
However, crystallinity can produce shrinkage, unexpected internal
stress, and environmental stress-cracking as well.
Polymers have variab
56
What is the equation for degree of crystallinity?
% crystallinity = [ρc(ρs – ρa) / ρs (ρc – ρa
)] x 100
where ρs is the density of a specimen for which the percentage crystallinity is
to be determined, ρa is the density of the totally amorphous polymer, and ρc is the density of perfectly crystalline polymer.
57
How does polymer processing vary between crystalline and amorphous polymers ? In regard to melting and glass temperatures.
Intermolecular order - Crystallinity Processing requires heating above the melting point. For high melting polymers, like nylon, processing at high
temperature possesses the danger of degradation, especially if traces of water are present.
For an amorphous polymer, the polymer is in the rubbery or “soft” state above Tg. This limits its service temperature range.
For a crystalline polymer the crystalline form retains dimensional stability almost up to the melting point. Thus, crystalline polymers can be used at higher service temperatures.
58
What does crosslinking do for polymers?
The polymer chains are chemically bonded to each other via crosslinks.
Crosslinking raises the Tg
and increase melt viscosity above Tg. The higher of the degree of crosslinks, the higher of the modulus (the
more brittle) of the polymers.
59
How is orientation caused? and is it wanted?
Many moulded products have at least some orientation resulting
from processing.
In some cases, orientation is intentional and is required for the
product to perform such as fibres and heat shrink tubing.
Much of the time, however, orientation is not wanted and
processing is carried out so as to minimise it. Frozen-in stress in
a plastic product is the result of such orientation.
In their normal state most polymer molecules are random coils with no particular shape, physically intertwined with each other.
60
How does different processing option affect the amount of orientation in the polymer?
Processing at high speed and high shear rate extends these
random coils into an orientated configuration.
Rapid cooling during processing does not allow enough time for
complete relaxation, leaving a substantial amount of orientation
and resultant frozen-in stress in the part.
The higher the molecular weight, the longer it takes for orientated
molecules to relax at any given temperature.
With time or temperature, this stress can be relieved, allowing the
molecules to become more like their original relaxed, coiled configuration.
61
How can frozen in stress be minimised for orientation due to processing?
The frozen-in stress can be removed by heating above the material’s softening temperature.
The greater the shrinkage and distortion on heating, the greater the frozen-in stress. Generally, the smaller the change on heating, the
less likely is the part to fail.
62
What is the difference between the glass transition temperature for both amorphous and crystalline polymers?
Amorphous materials undergo a “glass transition”, Tg
Crystalline materials have a “melt transition”, Tm
Below Tg polymers are hard and glassy
Above Tg polymers are soft and leathery
63
What are the main factors that affect the deformation of polymeric materials?
Tension
Compression
Shear
Torsion
Flexure (bending stress) containing tensile, compressive and shear components
Hydrostatic pressure
64
What about the polymer governs the mechanical response it will have?
the covalent bonds between carbon atoms and
the secondary Van der Waals forces between molecular segments
Ultimate fracture normally requires breaking the former, but the secondary bonds often play a major role in the deformation mechanisms that lead to fracture.
65
What are the 3 main factors that govern the toughness and ductility of polymers?
strain rate,
temperature and
molecular structure.
66
What does the area under a stress strain curve represent?
the absorbed energy
67
What are the two main factors to aid in the degree of fracture in polymers?
1. high rates or low temperatures (relative to Tg) polymers tend to be brittle, because there is insufficient time for the polymer chains to respond to stress with large-scale viscoelastic deformation or yielding.
2. Highly cross-linked polymers are also incapable of large-scale viscoelastic deformation. Primary bonds between chain segments must be broken for these materials to deform.
68
What is chain scission?
Fracture at atomic level (severing bonds). The crack-like flaws can produce significant local stress
concentrations to overcome the actual bond strength than what is measured.
69
What aids chain scission to take place?
Molecules are not stressed uniformly. When a stress is applied certain chain segments carry a different proportion of the load (sufficient to exceed the bond strength).
70
How does the degree of nonuniformity vary for amorphous and crystalline polymers in chain scission?
The degree of nonuniformity in stress is more pronounced in amorphous polymers, while the degree of symmetry in crystalline polymers tends to distribute stress more evenly.
71
How can chain scission be detected?
Free radicals form when covalent bonds in
polymers are severed.
Consequently, chain scission can be detected
experimentally by infrared spectroscopy (IR).
72
What is chain disentanglement in polymers?
fracture occurs where molecules separate from one another intact. The likelihood of chain disentanglement depends on the length of the molecules and the degree to which they are interwoven.
73
What is the likelihood of chain scission occurring?
Chain scission can occur at relatively low strains in cross-linked or highly aligned polymers, but the mechanical response of isotropic polymers with low cross link density is governed by secondary bonds at low strains.
At low strains, many polymers yield before fracture.
74
What 2 types can the failure of polymers be categorised into?
Type I - tend to fail by crazing if the craze initiation stress (σci) < yield stress (σy)
Type II - tend to fail by shear yielding if (σci) > (σy)
If (σci) ≈ (σy), the both mechanisms can occur
simultaneously
75
What are Type I and type II polymers?
1. Type I polymers
brittle at 10-20°C below their Tg
low initiation energies
low crack propagation energies, leading to low un-notched and notched impact strength such as PS
2. Type II polymers
High crack initiation energies
High crack propagation energies, leading to high unnotched impact strength, but low notched value
Show a brittle to ductile (tough) transition (Tbt), such as PC, Nylons and PET
76
How does shear yielding exhibit itself in polymers?
Similar to the plastic flow in metals.
Molecules slide with respect to one another when
subjected to a critical shear stress.
The shear yielding criteria can be calculated based on the maximum shear stress (Tresca and von Mises yield criteria)
77
how and why is it important to establish accurate yield criteria?
To accurately describe the yielding behaviour of ductile solids,
Experiement can verify the yield criteria have shown a marked dependence of yield stress on hydrostatic pressure, and on the stress system associated with various mechanisms of loading
Specification of the most accurate and appropriate yield criterion for polymers is still subject to some debate
78
How is evidence of shear yielding seen?
Yielding criteria are often quoted in terms of critical
shear stress parameters.
Evidence of shear yielding is often seen in the form of
shear bands, which occur on a plane of maximum
resolved shear stress.
79
What is the difference between crazes and cracks?
Crazes have a continuity of material across the craze plane, whereas cracks do not possess
any continuity.
Consequently, crazed zones are capable of bearing loads as opposed to cracked ones.
80
what does dilatational in craze yielding mean?
voids are formed between the oriented
material which bridges the adjacent surfaces of the uncrazed bulk.
crack-like defects are bridged by oriented microfibils – strands of yielded and drawn material which are oriented parallel to the stress direction –
which contribute significantly to further deformation resistance.
81
What does crazing look like on a macroscopic level?
On the macroscopic level, crazing appears as a
stress-whitened region, due to a low refractive index.
The craze zone usually forms perpendicular to the
maximum principal normal stress.
82
How and What do fractures in a craze zone usually originate from?
Fracture in a craze zone usually initiates from
inorganic dust particles that are entrapped in the polymer.
Fracture occurs in a craze
zone when individual fibrils
rupture.
This process can be unstable
if, when a fibril fails, the
redistributed stress is
sufficient to rupture one or
more neighbouring fibrils.
83
What kinds of polymers does crazing most likely occur in?
Crazes occurs most commonly in amorphous, glassy polymers such as PS, SAN, PMMA, PVC, and PC but also in semicrystalline ones (PE, PP, PETP and POM). Crazes have also been observed in epoxy and phenxy resins.
Crazes are sharply bounded regions filled with 'craze matter' consisting of oriented strands interspersed with voids
84
Why can the width of a fibril increases?
The increase in the width of
craze with sample extension is mostly due
to fibrillation of further matrix material rather
than to an increase in fibril extension.
85
What are the 3 main stages involves in polymer fracture by crazing?
Craze initiation/nucleation,
Craze propagation
Craze breakdown
86
What demonstrates that crazing is a time dependent phenonmenon?
Crazes normally initiate on surface grooves or small imperfections in the bulk of the polymer, especially at low stresses.
There can be a considerable time lag between load application and the occurrence of the first visible craze, which indicates that craze initiation is a time-dependent phenomenon.
87
How can crazing and shear yielding be characterised (machine)?
SEM / Optical microscopy
Microindentation
88
What is the micro hardness of the craze zones determined by ?
Indentations are performed on a straight
line along the direction of crack propagation.
The hardness values measured along the crack
zone are lower than the micro-hardness of the
bulk.
The micro-hardness of the craze zones is determined
and the elastic moduli of the crazed material are
calculated.
89
Why does the micro hardness of the craze zone increase rapidly than compared with the hardness of the bulk?
Intense plastic deformation → strain hardening
Formation of oriented fibrils → local strengthening
Triaxial stress and constraint → densification
Energy absorption mechanisms → increased local resistance
90
Why are rubbers used for plastics?
to improve the toughness of plastics. both the crazing and shear yielding mechanisms are involved in
rubber toughening plastics
91
How is the homopolymer of styrene(PS) used to toughen plastics? 2 mechanisms
Increase the breaking stress by the introduction of polar groups. For example, styrene and acrylonitrile are copolymerized (SAN).
Increase the elongation without an accompanying lowering of strength. This can be performed by the addition of rubber in particle form to PS.
92
How is ABS and dispersed rubber particles disperse into polystyrene chains?
It is important that the dispersed rubber particles are grafted onto polystyrene chains (not a simple milling-in of rubber with PS matrix).
Poly(acrylonitrile/butadiene/styrene) (ABS) is a two-phase material consisting of elastomer particles in a glassy polymer matrix of SAN.
93
What are the 2 main phases of rubber toughening?
Upon loading, the presence of the elastomeric phase
concentrates and makes the stress distribution more
complex in the surrounding matrix;
Yield stress is reduced, and toughening occurs by the
resultant plastic deformation, either by
crazing (crazing mechanism), or by
the formation of shear bands (shear yield mechanism).
94
How is rubber toughening achieved for crazing?
by the dispersed rubber particles acting as craze initiators. The rubber particles should be
Small enough to prevent catastrophic crack
propagation
Large enough so as not to be engulfed by a craze
In the order of microns
95
How is rubber toughening achieved for shear yielding?
By the dispersed rubber particles acting as the initiators of shear bands
96
How does shear yielding protect against crazes?
Shear yielding acts as energy absorbing process
Shear bands are barriers to the propagation of crazes, and therefore, also of crack growth, i.e. delay failure
Particles act as producers of large number of triaxial tension in the matrix to initiate extensive shear yielding
Rubber particle cavitation may also be involved, which
explains stress whitening
97
How is the addition of a rubber second phase to a polymer good for toughness?
-significantly increases toughness by making
craze/shear yielding initiation easier.
The low modulus particles provide sites for void
nucleation, thereby lowering the stress required for
craze/shear yielding formation.
- Energy absorbed by the rubber particles
Only a small portion of the enhanced impact energies is attributed to the deformation of the rubber phase during the impact.
Matrix crazing
Crazes initiate and terminate at rubber particles. Interactions between particles can cause craze path deviation.
98
How can crazing be visually characterised?
Method of toughness enhancement in commercial HIPS compounds, and is characterised by severe stress whitening in the vicinity of maximum stress; this
is caused by light scattering due to refractive index difference in crazed and un-yielded material.
The effects of a shear-yielding toughening mechanism are less easily identified, since no such small-scale changes in specific volume occur. Moreover, separation of the two mechanisms is not always easy, since there is evidence to suggest that synergistic effects exist between them.
99
What do the relative contributions of both crazing and shear yielding at the same time be factors of?
rubber particle size, dispersion and concentration,
the matrix and
the temperature.
100
What is the main difference of the mechanisms between crazing and shear yielding?
voids and crazing introduce an increase in
volume,
while shear yielding leads to essentially no
change in volume.
101
Why are crazes load bearing but cracks are not?
crazes are highly oriented polymer segments (fibrils)
102
Why is ABS/HIPS tougher than PS?
ABS are composed of a two-phase rubber-toughened plastic with a continuous poly(styrene-co-acrylonitrile) (SAN) copolymer glassy phase containing droplets of a dispersed polybutadiene latex (spheric salami
morphology).
The reasons for the improvement in toughness (compared to PS) are many simultaneously occurring dissipative processes, including Multi crazing
and/or Shear deformation processes initiated and localised at the particles.
Crazing - the dispersed rubber particles acting as craze initiators. Crazes initiate and terminate at rubber particles. Interactions between particles can
cause craze path deviation.
Shear yielding - the dispersed rubber particles acting as the initiators of shear bands, which act as energy absorbing process and also as barriers to
the propagation of cracks.
Energy absorbed by the rubber particles by the deformation of the rubber phase during the impact.
Debonding at the interface of the rubber and plastic can also be important
103
What factors will affect the rubber toughening in plastics?
Concentration of the rubber addition
Particle size of the rubber added
The temperature
104
What criteria must the rubber used for rubber toughened plastics meet?
the rubber should exist as a discrete second phase, usually as spheroids;
compounding processes or the copolymerisation must disperse the rubber phase effectively;
adhesion between the rubber and the matrix polymer must be optimised.
the optimum particle size and distribution of the rubber may vary according to the yield characteristics of the matrix;
the elastomer Tg should be well below the minimum
expected service temperature of the toughened plastics
105
How can high impact polystyrene (HIPS) morphology be improved with increasing rubber content?
Within each micron size domain, many spherical PS sub-domains are distributed, which are separated by
rubbery polybutadiene (PB) phase, resulting in a hierarchical morphology similar to ‘‘salami’’.
With increasing rubber content, the impact strength increases, but the tensile and bending strength dropped dramatically, which is always a primary conflict in toughening plastic with rubbers.
The size of rubber domains can be controlled to produce bimodal size distribution, the surface of a larger particle is more likely to initiate crazing and small particles will reduce the avg ligament thickness between particles = toughness increase.
106
Why is (High Impact Polystyrene )HIPS better than just PS with rubber particles?
The PS component of HIPS provides the resin’s rigidity, and the PB phase brings five times higher toughness than pure PS, which is the decisive factor for the wide application of HIPS resin.
The hierarchical salami morphology can effectively accommodate displacements to avoid the formation of premature flaws in the interphase.
107
What is the base morphology of ABS (Acrylonitrile–butadiene–styrene )?
Composes of 2 phase rubber toughened plastic with a continuous poly(styrene-co-acrylonitrile) (SAN) co polymer glassy phase. Rubber particles are larger and contain SAN occlusions.
108
What is the darker colour in ABS a result of?
a sample pre-treatment process, called staining,
and this staining process works specifically on carbon-carbon double covalent bonds that are present in the rubber.
109
How is ABS made? And what properties can this give ABS?
ABS is a terpolymer made by polymerising styrene and acrylonitrile in the presence of polybutadiene.
The result is a long chain of polybutadiene crisscrossed with shorter chains of poly(styrene-co-acrylonitrile). The nitrile groups from neighbouring chains, being
polar, attract each other and bind the chains together, making ABS stronger than pure polystyrene.
contributes chemical resistance, fatigue resistance,
hardness, and rigidity, while increasing the heat deflection temperature = shiny impervious surface.
Polybutadiene, a rubbery substance, provides toughness and ductility at low temperatures, at the cost of heat resistance and rigidity
110
How and why is the SAN matric be improved in ABS?
To improve the phase interaction of the elastomeric particles with the SAN matrix, they are grafted with styrene and acrylonitrile. Acrylonitrile polar groups provide a highly cohesive and very tough SAN matrix.
111
Why does introducing a soft rubbery phase into a brittle ABS material matrix improve impact toughness?
The reasons for this are many simultaneously occurring
dissipative processes, such as Multi crazing and/or Shear deformation processes initiated and localised at the particles.
These deformation processes are caused by stress
concentrations in the equatorial regions of the elastomeric particles.
With increasing deformation, stress are restricted and more mechanical energy can be dissipated before fracture. Only if T>70 deg C does shear deformation dominate the deformation behaviour.
For ABS the matrix modulus is typically 100 to 1000 times larger than the modulus of the dispersed elastomeric particles.
Therefore, a heterogeneous stress distribution occurs in the phases if the material is mechanically loaded.
112
What deformation mechanism dominates near the crack in rubber-toughened polymers?
Shear deformation mechanisms like shear yield dominate due to higher temperatures and lower stress levels near the crack.
113
What microscopic process precedes shear yielding near the crack?
Cavitation of rubber particles occurs first, enabling the shear yielding process.
114
What visual effect is caused by cavitation and multiple crazing?
Stress whitening, a whitening of the material due to localized deformation
115
How does cavitation affect stress levels and shear yielding?
Cavitation reduces local stress, which thermodynamically stabilizes the shear yield process
116
When is shear yield most dominant in these materials?
At high temperatures and high rubber content, where rubber particles are closely spaced
117
What happens in areas far from the crack tip?
The stress must be high enough to initiate plastic deformation, but only multiple crazing (not shear yield) typically occurs.
118
How do HIPS vs ABS compare as rubber toughening plastics?
They are both thermoplastics.
ABS is a more robust material that has better impact strength, temperature resistance, and dimensional stability than HIPS.
However, HIPS is more affordable and easier to process than ABS.
119
How do the proportion of the domains vary for ABS and HIPS ? Which has best impact resistance?
The dispersed phases are small spherical and uniform in ABS but ABS only becomes high impact when the domains rich in polybutadiene present small dimensions.
The domains of the dispersed phase for HIPS are larger than that for ABS and basically spherical with high variation of size from 0.16 μm to 3.03 μm, resulting in the lower impact resistance of the HIPS compared to the ABS.
